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| score
float64 -1
3.99
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
908
|
C
|
New Year and Curling
|
PROGRAMMING
| 1,500
|
[
"brute force",
"geometry",
"implementation",
"math"
] | null | null |
Carol is currently curling.
She has *n* disks each with radius *r* on the 2D plane.
Initially she has all these disks above the line *y*<==<=10100.
She then will slide the disks towards the line *y*<==<=0 one by one in order from 1 to *n*.
When she slides the *i*-th disk, she will place its center at the point (*x**i*,<=10100). She will then push it so the disk’s *y* coordinate continuously decreases, and *x* coordinate stays constant. The disk stops once it touches the line *y*<==<=0 or it touches any previous disk. Note that once a disk stops moving, it will not move again, even if hit by another disk.
Compute the *y*-coordinates of centers of all the disks after all disks have been pushed.
|
The first line will contain two integers *n* and *r* (1<=≤<=*n*,<=*r*<=≤<=1<=000), the number of disks, and the radius of the disks, respectively.
The next line will contain *n* integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=1<=000) — the *x*-coordinates of the disks.
|
Print a single line with *n* numbers. The *i*-th number denotes the *y*-coordinate of the center of the *i*-th disk. The output will be accepted if it has absolute or relative error at most 10<=-<=6.
Namely, let's assume that your answer for a particular value of a coordinate is *a* and the answer of the jury is *b*. The checker program will consider your answer correct if for all coordinates.
|
[
"6 2\n5 5 6 8 3 12\n"
] |
[
"2 6.0 9.87298334621 13.3370849613 12.5187346573 13.3370849613\n"
] |
The final positions of the disks will look as follows:
In particular, note the position of the last disk.
| 1,000
|
[
{
"input": "6 2\n5 5 6 8 3 12",
"output": "2 6.0 9.87298334621 13.3370849613 12.5187346573 13.3370849613"
},
{
"input": "1 1\n5",
"output": "1"
},
{
"input": "5 300\n939 465 129 611 532",
"output": "300 667.864105343 1164.9596696 1522.27745533 2117.05388391"
},
{
"input": "5 1\n416 387 336 116 81",
"output": "1 1 1 1 1"
},
{
"input": "3 10\n1 100 1000",
"output": "10 10 10"
},
{
"input": "2 1\n2 20",
"output": "1 1"
},
{
"input": "3 2\n10 10 100",
"output": "2 6.0 2"
}
] | 1,573,272,575
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 15
| 655
| 307,200
|
import math
# input is a circle with centre (x1,y1)
# and the 2nd centre has x-coordinate x2
# find the y-coordinate of the 2nd circle, larger than y1, when it touches the 1st circle
def touch_y(x1, y1, x2):
if abs(x1-x2) > 2*r: # no intersection
return r
x_touch_offset = abs(x1 - x2)/2.0
y_touch = y1 + 2*math.sqrt(r*r - x_touch_offset*x_touch_offset)
return y_touch
# n^2 complexity, loop over the input xi and find the maximal y coordinate of a touching circle with the already computed solutions
# finally, print out the list of y coordinates
def solve(xi, r):
solution = []
for x in xi:
if len(solution) == 0:
solution = [r]
else:
max_y = 0
for i in range(len(solution)):
max_y = max(max_y, touch_y(xi[i], solution[i], x))
solution.append(max_y)
print(' '.join([str(s) for s in solution]))
n_and_r = input()
r = float(n_and_r.split(' ')[1])
xi = [float(n) for n in input().split(' ')]
solve(xi, r)
|
Title: New Year and Curling
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Carol is currently curling.
She has *n* disks each with radius *r* on the 2D plane.
Initially she has all these disks above the line *y*<==<=10100.
She then will slide the disks towards the line *y*<==<=0 one by one in order from 1 to *n*.
When she slides the *i*-th disk, she will place its center at the point (*x**i*,<=10100). She will then push it so the disk’s *y* coordinate continuously decreases, and *x* coordinate stays constant. The disk stops once it touches the line *y*<==<=0 or it touches any previous disk. Note that once a disk stops moving, it will not move again, even if hit by another disk.
Compute the *y*-coordinates of centers of all the disks after all disks have been pushed.
Input Specification:
The first line will contain two integers *n* and *r* (1<=≤<=*n*,<=*r*<=≤<=1<=000), the number of disks, and the radius of the disks, respectively.
The next line will contain *n* integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=1<=000) — the *x*-coordinates of the disks.
Output Specification:
Print a single line with *n* numbers. The *i*-th number denotes the *y*-coordinate of the center of the *i*-th disk. The output will be accepted if it has absolute or relative error at most 10<=-<=6.
Namely, let's assume that your answer for a particular value of a coordinate is *a* and the answer of the jury is *b*. The checker program will consider your answer correct if for all coordinates.
Demo Input:
['6 2\n5 5 6 8 3 12\n']
Demo Output:
['2 6.0 9.87298334621 13.3370849613 12.5187346573 13.3370849613\n']
Note:
The final positions of the disks will look as follows:
In particular, note the position of the last disk.
|
```python
import math
# input is a circle with centre (x1,y1)
# and the 2nd centre has x-coordinate x2
# find the y-coordinate of the 2nd circle, larger than y1, when it touches the 1st circle
def touch_y(x1, y1, x2):
if abs(x1-x2) > 2*r: # no intersection
return r
x_touch_offset = abs(x1 - x2)/2.0
y_touch = y1 + 2*math.sqrt(r*r - x_touch_offset*x_touch_offset)
return y_touch
# n^2 complexity, loop over the input xi and find the maximal y coordinate of a touching circle with the already computed solutions
# finally, print out the list of y coordinates
def solve(xi, r):
solution = []
for x in xi:
if len(solution) == 0:
solution = [r]
else:
max_y = 0
for i in range(len(solution)):
max_y = max(max_y, touch_y(xi[i], solution[i], x))
solution.append(max_y)
print(' '.join([str(s) for s in solution]))
n_and_r = input()
r = float(n_and_r.split(' ')[1])
xi = [float(n) for n in input().split(' ')]
solve(xi, r)
```
| 3
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
You have *n* devices that you want to use simultaneously.
The *i*-th device uses *a**i* units of power per second. This usage is continuous. That is, in λ seconds, the device will use λ·*a**i* units of power. The *i*-th device currently has *b**i* units of power stored. All devices can store an arbitrary amount of power.
You have a single charger that can plug to any single device. The charger will add *p* units of power per second to a device. This charging is continuous. That is, if you plug in a device for λ seconds, it will gain λ·*p* units of power. You can switch which device is charging at any arbitrary unit of time (including real numbers), and the time it takes to switch is negligible.
You are wondering, what is the maximum amount of time you can use the devices until one of them hits 0 units of power.
If you can use the devices indefinitely, print -1. Otherwise, print the maximum amount of time before any one device hits 0 power.
|
The first line contains two integers, *n* and *p* (1<=≤<=*n*<=≤<=100<=000, 1<=≤<=*p*<=≤<=109) — the number of devices and the power of the charger.
This is followed by *n* lines which contain two integers each. Line *i* contains the integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=100<=000) — the power of the device and the amount of power stored in the device in the beginning.
|
If you can use the devices indefinitely, print -1. Otherwise, print the maximum amount of time before any one device hits 0 power.
Your answer will be considered correct if its absolute or relative error does not exceed 10<=-<=4.
Namely, let's assume that your answer is *a* and the answer of the jury is *b*. The checker program will consider your answer correct if .
|
[
"2 1\n2 2\n2 1000\n",
"1 100\n1 1\n",
"3 5\n4 3\n5 2\n6 1\n"
] |
[
"2.0000000000",
"-1\n",
"0.5000000000"
] |
In sample test 1, you can charge the first device for the entire time until it hits zero power. The second device has enough power to last this time without being charged.
In sample test 2, you can use the device indefinitely.
In sample test 3, we can charge the third device for 2 / 5 of a second, then switch to charge the second device for a 1 / 10 of a second.
| 0
|
[
{
"input": "2 1\n2 2\n2 1000",
"output": "2.0000000000"
},
{
"input": "1 100\n1 1",
"output": "-1"
},
{
"input": "3 5\n4 3\n5 2\n6 1",
"output": "0.5000000000"
},
{
"input": "1 1\n1 87",
"output": "-1"
},
{
"input": "1 1\n100 77",
"output": "0.7777777778"
},
{
"input": "5 10\n3 81\n3 49\n1 20\n1 12\n1 30",
"output": "-1"
},
{
"input": "5 10\n4 3\n1 54\n2 57\n2 31\n1 99",
"output": "-1"
},
{
"input": "5 10\n2 81\n3 31\n4 49\n1 35\n1 67",
"output": "263.0000000000"
},
{
"input": "10 1\n1 92\n1 92\n1 92\n1 92\n1 92\n1 92\n1 92\n1 92\n1 92\n1 92",
"output": "102.2222222222"
},
{
"input": "10 1\n1 16\n1 16\n1 16\n1 16\n1 16\n1 16\n1 16\n1 16\n1 16\n1 16",
"output": "17.7777777778"
},
{
"input": "10 1\n1 40\n1 40\n1 40\n1 40\n1 40\n1 40\n1 40\n1 40\n1 40\n1 40",
"output": "44.4444444444"
},
{
"input": "2 1\n1 10\n1 10",
"output": "20.0000000000"
},
{
"input": "20 16807\n75250 50074\n43659 8931\n11273 27545\n50879 77924\n37710 64441\n38166 84493\n43043 7988\n22504 82328\n31730 78841\n42613 44304\n33170 17710\n97158 29561\n70934 93100\n80279 51817\n95336 99098\n7827 13513\n29268 23811\n77634 80980\n79150 36580\n58822 11968",
"output": "0.2244225704"
}
] | 1,492,360,660
| 3,760
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
PRETESTS
| 12
| 2,000
| 7,372,800
|
(N, P) = map(int, input().split())
As = []
Bs = []
for _ in range(N):
(A, B) = map(int, input().split())
As.append(A)
Bs.append(B)
EPS = 1E-18
def check(t):
tmp = 0.0
for i in range(N):
tmp += max(As[i] * t - Bs[i], 0) / P
return tmp < t or abs(tmp - t) < EPS
if sum(As) <= P:
print(-1)
else:
l = 0.0
r = 1E18
cnt = 0
while EPS < (r - l) and cnt < 1000:
cnt += 1
c = (l + r) / 2
if check(c):
l = c
else:
r = c
print(l)
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have *n* devices that you want to use simultaneously.
The *i*-th device uses *a**i* units of power per second. This usage is continuous. That is, in λ seconds, the device will use λ·*a**i* units of power. The *i*-th device currently has *b**i* units of power stored. All devices can store an arbitrary amount of power.
You have a single charger that can plug to any single device. The charger will add *p* units of power per second to a device. This charging is continuous. That is, if you plug in a device for λ seconds, it will gain λ·*p* units of power. You can switch which device is charging at any arbitrary unit of time (including real numbers), and the time it takes to switch is negligible.
You are wondering, what is the maximum amount of time you can use the devices until one of them hits 0 units of power.
If you can use the devices indefinitely, print -1. Otherwise, print the maximum amount of time before any one device hits 0 power.
Input Specification:
The first line contains two integers, *n* and *p* (1<=≤<=*n*<=≤<=100<=000, 1<=≤<=*p*<=≤<=109) — the number of devices and the power of the charger.
This is followed by *n* lines which contain two integers each. Line *i* contains the integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=100<=000) — the power of the device and the amount of power stored in the device in the beginning.
Output Specification:
If you can use the devices indefinitely, print -1. Otherwise, print the maximum amount of time before any one device hits 0 power.
Your answer will be considered correct if its absolute or relative error does not exceed 10<=-<=4.
Namely, let's assume that your answer is *a* and the answer of the jury is *b*. The checker program will consider your answer correct if .
Demo Input:
['2 1\n2 2\n2 1000\n', '1 100\n1 1\n', '3 5\n4 3\n5 2\n6 1\n']
Demo Output:
['2.0000000000', '-1\n', '0.5000000000']
Note:
In sample test 1, you can charge the first device for the entire time until it hits zero power. The second device has enough power to last this time without being charged.
In sample test 2, you can use the device indefinitely.
In sample test 3, we can charge the third device for 2 / 5 of a second, then switch to charge the second device for a 1 / 10 of a second.
|
```python
(N, P) = map(int, input().split())
As = []
Bs = []
for _ in range(N):
(A, B) = map(int, input().split())
As.append(A)
Bs.append(B)
EPS = 1E-18
def check(t):
tmp = 0.0
for i in range(N):
tmp += max(As[i] * t - Bs[i], 0) / P
return tmp < t or abs(tmp - t) < EPS
if sum(As) <= P:
print(-1)
else:
l = 0.0
r = 1E18
cnt = 0
while EPS < (r - l) and cnt < 1000:
cnt += 1
c = (l + r) / 2
if check(c):
l = c
else:
r = c
print(l)
```
| 0
|
|
71
|
A
|
Way Too Long Words
|
PROGRAMMING
| 800
|
[
"strings"
] |
A. Way Too Long Words
|
1
|
256
|
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
|
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
|
[
"4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n"
] |
[
"word\nl10n\ni18n\np43s\n"
] |
none
| 500
|
[
{
"input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis",
"output": "word\nl10n\ni18n\np43s"
},
{
"input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm",
"output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m"
},
{
"input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd",
"output": "n20n\nj27b\nk15d"
},
{
"input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt",
"output": "t41t"
},
{
"input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings",
"output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz",
"output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz"
},
{
"input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij",
"output": "a98j"
},
{
"input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco",
"output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco"
},
{
"input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt",
"output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t"
},
{
"input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr",
"output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr"
}
] | 1,668,753,858
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
for i in range(int(input()):
s=int(input())
if(len(s)>10):
s1=len(s)-2
print(s[0],str(s1),s[len(s)-1],sep='')
else:
print(s)
|
Title: Way Too Long Words
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output Specification:
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
Demo Input:
['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n']
Demo Output:
['word\nl10n\ni18n\np43s\n']
Note:
none
|
```python
for i in range(int(input()):
s=int(input())
if(len(s)>10):
s1=len(s)-2
print(s[0],str(s1),s[len(s)-1],sep='')
else:
print(s)
```
| -1
|
445
|
B
|
DZY Loves Chemistry
|
PROGRAMMING
| 1,400
|
[
"dfs and similar",
"dsu",
"greedy"
] | null | null |
DZY loves chemistry, and he enjoys mixing chemicals.
DZY has *n* chemicals, and *m* pairs of them will react. He wants to pour these chemicals into a test tube, and he needs to pour them in one by one, in any order.
Let's consider the danger of a test tube. Danger of an empty test tube is 1. And every time when DZY pours a chemical, if there are already one or more chemicals in the test tube that can react with it, the danger of the test tube will be multiplied by 2. Otherwise the danger remains as it is.
Find the maximum possible danger after pouring all the chemicals one by one in optimal order.
|
The first line contains two space-separated integers *n* and *m* .
Each of the next *m* lines contains two space-separated integers *x**i* and *y**i* (1<=≤<=*x**i*<=<<=*y**i*<=≤<=*n*). These integers mean that the chemical *x**i* will react with the chemical *y**i*. Each pair of chemicals will appear at most once in the input.
Consider all the chemicals numbered from 1 to *n* in some order.
|
Print a single integer — the maximum possible danger.
|
[
"1 0\n",
"2 1\n1 2\n",
"3 2\n1 2\n2 3\n"
] |
[
"1\n",
"2\n",
"4\n"
] |
In the first sample, there's only one way to pour, and the danger won't increase.
In the second sample, no matter we pour the 1st chemical first, or pour the 2nd chemical first, the answer is always 2.
In the third sample, there are four ways to achieve the maximum possible danger: 2-1-3, 2-3-1, 1-2-3 and 3-2-1 (that is the numbers of the chemicals in order of pouring).
| 1,000
|
[
{
"input": "1 0",
"output": "1"
},
{
"input": "2 1\n1 2",
"output": "2"
},
{
"input": "3 2\n1 2\n2 3",
"output": "4"
},
{
"input": "10 10\n1 8\n4 10\n4 6\n5 10\n2 3\n1 7\n3 4\n3 6\n6 9\n3 7",
"output": "512"
},
{
"input": "20 20\n6 8\n13 20\n7 13\n6 17\n5 15\n1 12\n2 15\n5 17\n5 14\n6 14\n12 20\n7 20\n1 6\n1 7\n2 19\n14 17\n1 10\n11 15\n9 18\n2 12",
"output": "32768"
},
{
"input": "30 30\n7 28\n16 26\n14 24\n16 18\n20 29\n4 28\n19 21\n8 26\n1 25\n14 22\n13 23\n4 15\n15 16\n2 19\n29 30\n12 20\n3 4\n3 26\n3 11\n22 27\n5 16\n2 24\n2 18\n7 16\n17 21\n17 25\n8 15\n23 27\n12 21\n5 30",
"output": "67108864"
},
{
"input": "40 40\n28 33\n15 21\n12 29\n14 31\n2 26\n3 12\n25 34\n6 30\n6 25\n5 28\n9 17\n23 29\n30 36\n3 21\n35 37\n7 25\n29 39\n15 19\n12 35\n24 34\n15 25\n19 33\n26 31\n7 29\n1 40\n11 27\n6 9\n6 27\n36 39\n10 14\n6 16\n23 25\n2 38\n3 24\n30 31\n29 30\n4 12\n11 13\n14 40\n22 39",
"output": "34359738368"
},
{
"input": "50 50\n16 21\n23 47\n23 30\n2 12\n23 41\n3 16\n14 20\n4 49\n2 47\n19 29\n13 42\n5 8\n24 38\n13 32\n34 37\n38 46\n3 20\n27 50\n7 42\n33 45\n2 48\n41 47\n9 48\n15 26\n27 37\n32 34\n17 24\n1 39\n27 30\n10 33\n38 47\n32 33\n14 39\n35 50\n2 19\n3 12\n27 34\n18 25\n12 23\n31 44\n5 35\n28 45\n38 39\n13 44\n34 38\n16 46\n5 15\n26 30\n47 49\n2 10",
"output": "4398046511104"
},
{
"input": "50 0",
"output": "1"
},
{
"input": "50 7\n16 32\n31 34\n4 16\n4 39\n1 50\n43 49\n1 33",
"output": "128"
},
{
"input": "7 20\n2 3\n3 6\n1 6\n1 2\n3 5\n1 7\n4 5\n4 7\n1 3\n2 6\n2 7\n4 6\n3 4\n1 4\n3 7\n1 5\n2 5\n5 6\n5 7\n2 4",
"output": "64"
},
{
"input": "5 4\n1 2\n2 3\n3 4\n4 5",
"output": "16"
},
{
"input": "10 7\n1 2\n2 3\n1 5\n2 7\n7 8\n1 9\n9 10",
"output": "128"
},
{
"input": "20 15\n1 3\n3 4\n3 5\n4 6\n1 7\n1 8\n1 9\n7 11\n8 12\n5 13\n3 16\n1 17\n3 18\n1 19\n17 20",
"output": "32768"
},
{
"input": "30 24\n2 3\n3 4\n1 5\n4 6\n6 7\n1 8\n1 9\n4 10\n9 11\n5 12\n6 13\n10 14\n14 15\n12 16\n14 17\n2 18\n8 19\n3 20\n10 21\n11 24\n3 25\n1 26\n7 27\n4 29",
"output": "16777216"
},
{
"input": "40 28\n1 2\n2 4\n3 5\n1 7\n1 8\n7 9\n6 10\n7 11\n2 12\n9 13\n11 15\n12 16\n1 18\n10 19\n7 21\n7 23\n20 25\n24 27\n14 28\n9 29\n23 30\n27 31\n11 34\n21 35\n32 36\n23 38\n7 39\n20 40",
"output": "268435456"
},
{
"input": "50 41\n1 2\n1 3\n2 4\n1 5\n2 7\n4 8\n7 9\n2 11\n10 13\n11 14\n12 15\n14 16\n4 19\n7 20\n14 21\n8 23\n16 24\n16 25\n16 26\n19 27\n2 28\n3 29\n21 30\n12 31\n20 32\n23 33\n30 34\n6 35\n34 36\n34 37\n33 38\n34 40\n30 41\n3 42\n39 43\n5 44\n8 45\n40 46\n20 47\n31 49\n34 50",
"output": "2199023255552"
},
{
"input": "50 39\n1 2\n1 4\n5 6\n4 7\n5 8\n7 9\n9 10\n10 11\n2 12\n8 14\n11 15\n11 17\n3 18\n13 19\n17 20\n7 21\n6 22\n22 23\n14 24\n22 25\n23 26\n26 27\n27 28\n15 29\n8 30\n26 31\n32 33\n21 35\n14 36\n30 37\n17 38\n12 40\n11 42\n14 43\n12 44\n1 45\n29 46\n22 47\n47 50",
"output": "549755813888"
},
{
"input": "50 38\n1 2\n2 3\n3 4\n3 5\n4 7\n5 10\n9 11\n9 12\n11 13\n12 14\n6 15\n8 16\n2 18\n15 19\n3 20\n10 21\n4 22\n9 24\n2 25\n23 26\n3 28\n20 29\n14 30\n4 32\n24 33\n20 36\n1 38\n19 39\n39 40\n22 41\n18 42\n19 43\n40 45\n45 46\n9 47\n6 48\n9 49\n25 50",
"output": "274877906944"
},
{
"input": "50 41\n1 3\n1 4\n2 5\n2 7\n1 8\n2 10\n4 11\n5 12\n12 13\n4 14\n10 17\n1 18\n1 21\n5 22\n14 23\n19 24\n13 25\n3 26\n11 27\n6 28\n26 29\n21 30\n17 31\n15 32\n1 33\n12 34\n23 36\n6 37\n15 38\n37 39\n31 40\n15 41\n25 42\n19 43\n20 44\n32 45\n44 46\n31 47\n2 48\n32 49\n27 50",
"output": "2199023255552"
},
{
"input": "50 47\n1 2\n1 3\n1 4\n1 5\n5 6\n2 7\n2 8\n2 9\n2 10\n8 11\n5 12\n11 13\n10 14\n6 15\n9 16\n1 17\n1 18\n8 19\n5 20\n5 21\n11 22\n2 23\n22 24\n24 25\n5 26\n21 27\n27 28\n8 29\n2 30\n4 31\n11 32\n17 33\n22 34\n25 35\n28 36\n28 37\n11 38\n17 39\n19 42\n6 43\n11 44\n29 45\n2 46\n24 47\n7 48\n3 49\n44 50",
"output": "140737488355328"
},
{
"input": "11 20\n3 6\n2 6\n2 9\n4 5\n9 11\n6 8\n5 6\n1 6\n4 11\n9 10\n5 10\n4 6\n3 8\n2 3\n1 7\n1 11\n2 7\n1 3\n3 7\n1 8",
"output": "1024"
},
{
"input": "26 17\n1 2\n2 3\n1 6\n6 7\n7 8\n2 9\n4 10\n3 11\n11 12\n9 13\n6 14\n2 16\n5 18\n6 19\n11 22\n15 24\n6 26",
"output": "131072"
},
{
"input": "48 43\n1 2\n1 3\n3 4\n4 5\n2 6\n5 7\n7 9\n4 10\n6 11\n3 12\n6 13\n3 14\n6 15\n13 16\n4 17\n12 18\n18 19\n1 20\n1 21\n16 22\n9 23\n3 24\n22 25\n2 26\n10 27\n18 28\n13 30\n3 31\n24 33\n29 34\n15 35\n16 36\n23 37\n21 38\n34 39\n37 40\n39 41\n19 42\n15 43\n23 44\n22 45\n14 47\n10 48",
"output": "8796093022208"
},
{
"input": "8 5\n1 2\n1 3\n1 4\n5 6\n7 8",
"output": "32"
},
{
"input": "8 7\n1 2\n2 3\n3 4\n1 4\n5 6\n6 7\n7 8",
"output": "64"
}
] | 1,620,732,438
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 62
| 6,963,200
|
n,m=map(int,input().split())
l=[]
for i in range(m):
a,b=map(int,input().split())
f=0
for i in l:
if a in i:
f=1
i.append(b)
elif b in i:
f=1
i.append(a)
if f!=1:
l.append([a,b])
#print(l)
tot=0
#print(l)
for i in range(len(l)):
for j in range(i+1,len(l)):
if set(l[i]).intersection(set(l[j]))!=0:
for k in l[j]:
l[i].append(k)
l[i]=list(set(l[i]))
l[j]=[]
#print(l)
for i in l:
if i!=[]:
tot+=(2**(len(i)-1))
if tot==0:
print(1)
else:
print(tot)
|
Title: DZY Loves Chemistry
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
DZY loves chemistry, and he enjoys mixing chemicals.
DZY has *n* chemicals, and *m* pairs of them will react. He wants to pour these chemicals into a test tube, and he needs to pour them in one by one, in any order.
Let's consider the danger of a test tube. Danger of an empty test tube is 1. And every time when DZY pours a chemical, if there are already one or more chemicals in the test tube that can react with it, the danger of the test tube will be multiplied by 2. Otherwise the danger remains as it is.
Find the maximum possible danger after pouring all the chemicals one by one in optimal order.
Input Specification:
The first line contains two space-separated integers *n* and *m* .
Each of the next *m* lines contains two space-separated integers *x**i* and *y**i* (1<=≤<=*x**i*<=<<=*y**i*<=≤<=*n*). These integers mean that the chemical *x**i* will react with the chemical *y**i*. Each pair of chemicals will appear at most once in the input.
Consider all the chemicals numbered from 1 to *n* in some order.
Output Specification:
Print a single integer — the maximum possible danger.
Demo Input:
['1 0\n', '2 1\n1 2\n', '3 2\n1 2\n2 3\n']
Demo Output:
['1\n', '2\n', '4\n']
Note:
In the first sample, there's only one way to pour, and the danger won't increase.
In the second sample, no matter we pour the 1st chemical first, or pour the 2nd chemical first, the answer is always 2.
In the third sample, there are four ways to achieve the maximum possible danger: 2-1-3, 2-3-1, 1-2-3 and 3-2-1 (that is the numbers of the chemicals in order of pouring).
|
```python
n,m=map(int,input().split())
l=[]
for i in range(m):
a,b=map(int,input().split())
f=0
for i in l:
if a in i:
f=1
i.append(b)
elif b in i:
f=1
i.append(a)
if f!=1:
l.append([a,b])
#print(l)
tot=0
#print(l)
for i in range(len(l)):
for j in range(i+1,len(l)):
if set(l[i]).intersection(set(l[j]))!=0:
for k in l[j]:
l[i].append(k)
l[i]=list(set(l[i]))
l[j]=[]
#print(l)
for i in l:
if i!=[]:
tot+=(2**(len(i)-1))
if tot==0:
print(1)
else:
print(tot)
```
| 0
|
|
195
|
B
|
After Training
|
PROGRAMMING
| 1,300
|
[
"data structures",
"implementation",
"math"
] | null | null |
After a team finished their training session on Euro football championship, Valeric was commissioned to gather the balls and sort them into baskets. Overall the stadium has *n* balls and *m* baskets. The baskets are positioned in a row from left to right and they are numbered with numbers from 1 to *m*, correspondingly. The balls are numbered with numbers from 1 to *n*.
Valeric decided to sort the balls in the order of increasing of their numbers by the following scheme. He will put each new ball in the basket with the least number of balls. And if he's got several variants, he chooses the basket which stands closer to the middle. That means that he chooses the basket for which is minimum, where *i* is the number of the basket. If in this case Valeric still has multiple variants, he chooses the basket with the minimum number.
For every ball print the number of the basket where it will go according to Valeric's scheme.
Note that the balls are sorted into baskets in the order of increasing numbers, that is, the first ball goes first, then goes the second ball and so on.
|
The first line contains two space-separated integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=105) — the number of balls and baskets, correspondingly.
|
Print *n* numbers, one per line. The *i*-th line must contain the number of the basket for the *i*-th ball.
|
[
"4 3\n",
"3 1\n"
] |
[
"2\n1\n3\n2\n",
"1\n1\n1\n"
] |
none
| 1,000
|
[
{
"input": "4 3",
"output": "2\n1\n3\n2"
},
{
"input": "3 1",
"output": "1\n1\n1"
},
{
"input": "10 3",
"output": "2\n1\n3\n2\n1\n3\n2\n1\n3\n2"
},
{
"input": "6 5",
"output": "3\n2\n4\n1\n5\n3"
},
{
"input": "2 6",
"output": "3\n4"
},
{
"input": "5 2",
"output": "1\n2\n1\n2\n1"
},
{
"input": "85702 100000",
"output": "50000\n50001\n49999\n50002\n49998\n50003\n49997\n50004\n49996\n50005\n49995\n50006\n49994\n50007\n49993\n50008\n49992\n50009\n49991\n50010\n49990\n50011\n49989\n50012\n49988\n50013\n49987\n50014\n49986\n50015\n49985\n50016\n49984\n50017\n49983\n50018\n49982\n50019\n49981\n50020\n49980\n50021\n49979\n50022\n49978\n50023\n49977\n50024\n49976\n50025\n49975\n50026\n49974\n50027\n49973\n50028\n49972\n50029\n49971\n50030\n49970\n50031\n49969\n50032\n49968\n50033\n49967\n50034\n49966\n50035\n49965\n50036\n49964\n..."
},
{
"input": "9 2",
"output": "1\n2\n1\n2\n1\n2\n1\n2\n1"
},
{
"input": "45 88",
"output": "44\n45\n43\n46\n42\n47\n41\n48\n40\n49\n39\n50\n38\n51\n37\n52\n36\n53\n35\n54\n34\n55\n33\n56\n32\n57\n31\n58\n30\n59\n29\n60\n28\n61\n27\n62\n26\n63\n25\n64\n24\n65\n23\n66\n22"
},
{
"input": "61 51",
"output": "26\n25\n27\n24\n28\n23\n29\n22\n30\n21\n31\n20\n32\n19\n33\n18\n34\n17\n35\n16\n36\n15\n37\n14\n38\n13\n39\n12\n40\n11\n41\n10\n42\n9\n43\n8\n44\n7\n45\n6\n46\n5\n47\n4\n48\n3\n49\n2\n50\n1\n51\n26\n25\n27\n24\n28\n23\n29\n22\n30\n21"
},
{
"input": "21 57",
"output": "29\n28\n30\n27\n31\n26\n32\n25\n33\n24\n34\n23\n35\n22\n36\n21\n37\n20\n38\n19\n39"
},
{
"input": "677 787",
"output": "394\n393\n395\n392\n396\n391\n397\n390\n398\n389\n399\n388\n400\n387\n401\n386\n402\n385\n403\n384\n404\n383\n405\n382\n406\n381\n407\n380\n408\n379\n409\n378\n410\n377\n411\n376\n412\n375\n413\n374\n414\n373\n415\n372\n416\n371\n417\n370\n418\n369\n419\n368\n420\n367\n421\n366\n422\n365\n423\n364\n424\n363\n425\n362\n426\n361\n427\n360\n428\n359\n429\n358\n430\n357\n431\n356\n432\n355\n433\n354\n434\n353\n435\n352\n436\n351\n437\n350\n438\n349\n439\n348\n440\n347\n441\n346\n442\n345\n443\n344\n444\n343\n4..."
},
{
"input": "37 849",
"output": "425\n424\n426\n423\n427\n422\n428\n421\n429\n420\n430\n419\n431\n418\n432\n417\n433\n416\n434\n415\n435\n414\n436\n413\n437\n412\n438\n411\n439\n410\n440\n409\n441\n408\n442\n407\n443"
},
{
"input": "453 855",
"output": "428\n427\n429\n426\n430\n425\n431\n424\n432\n423\n433\n422\n434\n421\n435\n420\n436\n419\n437\n418\n438\n417\n439\n416\n440\n415\n441\n414\n442\n413\n443\n412\n444\n411\n445\n410\n446\n409\n447\n408\n448\n407\n449\n406\n450\n405\n451\n404\n452\n403\n453\n402\n454\n401\n455\n400\n456\n399\n457\n398\n458\n397\n459\n396\n460\n395\n461\n394\n462\n393\n463\n392\n464\n391\n465\n390\n466\n389\n467\n388\n468\n387\n469\n386\n470\n385\n471\n384\n472\n383\n473\n382\n474\n381\n475\n380\n476\n379\n477\n378\n478\n377\n4..."
},
{
"input": "165 374",
"output": "187\n188\n186\n189\n185\n190\n184\n191\n183\n192\n182\n193\n181\n194\n180\n195\n179\n196\n178\n197\n177\n198\n176\n199\n175\n200\n174\n201\n173\n202\n172\n203\n171\n204\n170\n205\n169\n206\n168\n207\n167\n208\n166\n209\n165\n210\n164\n211\n163\n212\n162\n213\n161\n214\n160\n215\n159\n216\n158\n217\n157\n218\n156\n219\n155\n220\n154\n221\n153\n222\n152\n223\n151\n224\n150\n225\n149\n226\n148\n227\n147\n228\n146\n229\n145\n230\n144\n231\n143\n232\n142\n233\n141\n234\n140\n235\n139\n236\n138\n237\n137\n238\n1..."
},
{
"input": "328 3",
"output": "2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3\n2\n1\n3..."
},
{
"input": "8 80",
"output": "40\n41\n39\n42\n38\n43\n37\n44"
},
{
"input": "90 544",
"output": "272\n273\n271\n274\n270\n275\n269\n276\n268\n277\n267\n278\n266\n279\n265\n280\n264\n281\n263\n282\n262\n283\n261\n284\n260\n285\n259\n286\n258\n287\n257\n288\n256\n289\n255\n290\n254\n291\n253\n292\n252\n293\n251\n294\n250\n295\n249\n296\n248\n297\n247\n298\n246\n299\n245\n300\n244\n301\n243\n302\n242\n303\n241\n304\n240\n305\n239\n306\n238\n307\n237\n308\n236\n309\n235\n310\n234\n311\n233\n312\n232\n313\n231\n314\n230\n315\n229\n316\n228\n317"
},
{
"input": "85 60",
"output": "30\n31\n29\n32\n28\n33\n27\n34\n26\n35\n25\n36\n24\n37\n23\n38\n22\n39\n21\n40\n20\n41\n19\n42\n18\n43\n17\n44\n16\n45\n15\n46\n14\n47\n13\n48\n12\n49\n11\n50\n10\n51\n9\n52\n8\n53\n7\n54\n6\n55\n5\n56\n4\n57\n3\n58\n2\n59\n1\n60\n30\n31\n29\n32\n28\n33\n27\n34\n26\n35\n25\n36\n24\n37\n23\n38\n22\n39\n21\n40\n20\n41\n19\n42\n18"
},
{
"input": "392 5",
"output": "3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3..."
},
{
"input": "8 87",
"output": "44\n43\n45\n42\n46\n41\n47\n40"
},
{
"input": "6 358",
"output": "179\n180\n178\n181\n177\n182"
},
{
"input": "501 70",
"output": "35\n36\n34\n37\n33\n38\n32\n39\n31\n40\n30\n41\n29\n42\n28\n43\n27\n44\n26\n45\n25\n46\n24\n47\n23\n48\n22\n49\n21\n50\n20\n51\n19\n52\n18\n53\n17\n54\n16\n55\n15\n56\n14\n57\n13\n58\n12\n59\n11\n60\n10\n61\n9\n62\n8\n63\n7\n64\n6\n65\n5\n66\n4\n67\n3\n68\n2\n69\n1\n70\n35\n36\n34\n37\n33\n38\n32\n39\n31\n40\n30\n41\n29\n42\n28\n43\n27\n44\n26\n45\n25\n46\n24\n47\n23\n48\n22\n49\n21\n50\n20\n51\n19\n52\n18\n53\n17\n54\n16\n55\n15\n56\n14\n57\n13\n58\n12\n59\n11\n60\n10\n61\n9\n62\n8\n63\n7\n64\n6\n65\n5\n6..."
},
{
"input": "3834 1",
"output": "1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1..."
},
{
"input": "1 8828",
"output": "4414"
},
{
"input": "69230 89906",
"output": "44953\n44954\n44952\n44955\n44951\n44956\n44950\n44957\n44949\n44958\n44948\n44959\n44947\n44960\n44946\n44961\n44945\n44962\n44944\n44963\n44943\n44964\n44942\n44965\n44941\n44966\n44940\n44967\n44939\n44968\n44938\n44969\n44937\n44970\n44936\n44971\n44935\n44972\n44934\n44973\n44933\n44974\n44932\n44975\n44931\n44976\n44930\n44977\n44929\n44978\n44928\n44979\n44927\n44980\n44926\n44981\n44925\n44982\n44924\n44983\n44923\n44984\n44922\n44985\n44921\n44986\n44920\n44987\n44919\n44988\n44918\n44989\n44917\n..."
},
{
"input": "27646 59913",
"output": "29957\n29956\n29958\n29955\n29959\n29954\n29960\n29953\n29961\n29952\n29962\n29951\n29963\n29950\n29964\n29949\n29965\n29948\n29966\n29947\n29967\n29946\n29968\n29945\n29969\n29944\n29970\n29943\n29971\n29942\n29972\n29941\n29973\n29940\n29974\n29939\n29975\n29938\n29976\n29937\n29977\n29936\n29978\n29935\n29979\n29934\n29980\n29933\n29981\n29932\n29982\n29931\n29983\n29930\n29984\n29929\n29985\n29928\n29986\n29927\n29987\n29926\n29988\n29925\n29989\n29924\n29990\n29923\n29991\n29922\n29992\n29921\n29993\n..."
},
{
"input": "37006 54783",
"output": "27392\n27391\n27393\n27390\n27394\n27389\n27395\n27388\n27396\n27387\n27397\n27386\n27398\n27385\n27399\n27384\n27400\n27383\n27401\n27382\n27402\n27381\n27403\n27380\n27404\n27379\n27405\n27378\n27406\n27377\n27407\n27376\n27408\n27375\n27409\n27374\n27410\n27373\n27411\n27372\n27412\n27371\n27413\n27370\n27414\n27369\n27415\n27368\n27416\n27367\n27417\n27366\n27418\n27365\n27419\n27364\n27420\n27363\n27421\n27362\n27422\n27361\n27423\n27360\n27424\n27359\n27425\n27358\n27426\n27357\n27427\n27356\n27428\n..."
},
{
"input": "1 100000",
"output": "50000"
},
{
"input": "100000 1",
"output": "1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1..."
},
{
"input": "100000 100000",
"output": "50000\n50001\n49999\n50002\n49998\n50003\n49997\n50004\n49996\n50005\n49995\n50006\n49994\n50007\n49993\n50008\n49992\n50009\n49991\n50010\n49990\n50011\n49989\n50012\n49988\n50013\n49987\n50014\n49986\n50015\n49985\n50016\n49984\n50017\n49983\n50018\n49982\n50019\n49981\n50020\n49980\n50021\n49979\n50022\n49978\n50023\n49977\n50024\n49976\n50025\n49975\n50026\n49974\n50027\n49973\n50028\n49972\n50029\n49971\n50030\n49970\n50031\n49969\n50032\n49968\n50033\n49967\n50034\n49966\n50035\n49965\n50036\n49964\n..."
},
{
"input": "100000 13",
"output": "7\n6\n8\n5\n9\n4\n10\n3\n11\n2\n12\n1\n13\n7\n6\n8\n5\n9\n4\n10\n3\n11\n2\n12\n1\n13\n7\n6\n8\n5\n9\n4\n10\n3\n11\n2\n12\n1\n13\n7\n6\n8\n5\n9\n4\n10\n3\n11\n2\n12\n1\n13\n7\n6\n8\n5\n9\n4\n10\n3\n11\n2\n12\n1\n13\n7\n6\n8\n5\n9\n4\n10\n3\n11\n2\n12\n1\n13\n7\n6\n8\n5\n9\n4\n10\n3\n11\n2\n12\n1\n13\n7\n6\n8\n5\n9\n4\n10\n3\n11\n2\n12\n1\n13\n7\n6\n8\n5\n9\n4\n10\n3\n11\n2\n12\n1\n13\n7\n6\n8\n5\n9\n4\n10\n3\n11\n2\n12\n1\n13\n7\n6\n8\n5\n9\n4\n10\n3\n11\n2\n12\n1\n13\n7\n6\n8\n5\n9\n4\n10\n3\n11\n2\n12\n1\n..."
},
{
"input": "100000 44",
"output": "22\n23\n21\n24\n20\n25\n19\n26\n18\n27\n17\n28\n16\n29\n15\n30\n14\n31\n13\n32\n12\n33\n11\n34\n10\n35\n9\n36\n8\n37\n7\n38\n6\n39\n5\n40\n4\n41\n3\n42\n2\n43\n1\n44\n22\n23\n21\n24\n20\n25\n19\n26\n18\n27\n17\n28\n16\n29\n15\n30\n14\n31\n13\n32\n12\n33\n11\n34\n10\n35\n9\n36\n8\n37\n7\n38\n6\n39\n5\n40\n4\n41\n3\n42\n2\n43\n1\n44\n22\n23\n21\n24\n20\n25\n19\n26\n18\n27\n17\n28\n16\n29\n15\n30\n14\n31\n13\n32\n12\n33\n11\n34\n10\n35\n9\n36\n8\n37\n7\n38\n6\n39\n5\n40\n4\n41\n3\n42\n2\n43\n1\n44\n22\n23\n21..."
},
{
"input": "100000 37820",
"output": "18910\n18911\n18909\n18912\n18908\n18913\n18907\n18914\n18906\n18915\n18905\n18916\n18904\n18917\n18903\n18918\n18902\n18919\n18901\n18920\n18900\n18921\n18899\n18922\n18898\n18923\n18897\n18924\n18896\n18925\n18895\n18926\n18894\n18927\n18893\n18928\n18892\n18929\n18891\n18930\n18890\n18931\n18889\n18932\n18888\n18933\n18887\n18934\n18886\n18935\n18885\n18936\n18884\n18937\n18883\n18938\n18882\n18939\n18881\n18940\n18880\n18941\n18879\n18942\n18878\n18943\n18877\n18944\n18876\n18945\n18875\n18946\n18874\n..."
},
{
"input": "99999 77777",
"output": "38889\n38888\n38890\n38887\n38891\n38886\n38892\n38885\n38893\n38884\n38894\n38883\n38895\n38882\n38896\n38881\n38897\n38880\n38898\n38879\n38899\n38878\n38900\n38877\n38901\n38876\n38902\n38875\n38903\n38874\n38904\n38873\n38905\n38872\n38906\n38871\n38907\n38870\n38908\n38869\n38909\n38868\n38910\n38867\n38911\n38866\n38912\n38865\n38913\n38864\n38914\n38863\n38915\n38862\n38916\n38861\n38917\n38860\n38918\n38859\n38919\n38858\n38920\n38857\n38921\n38856\n38922\n38855\n38923\n38854\n38924\n38853\n38925\n..."
},
{
"input": "1991 1935",
"output": "968\n967\n969\n966\n970\n965\n971\n964\n972\n963\n973\n962\n974\n961\n975\n960\n976\n959\n977\n958\n978\n957\n979\n956\n980\n955\n981\n954\n982\n953\n983\n952\n984\n951\n985\n950\n986\n949\n987\n948\n988\n947\n989\n946\n990\n945\n991\n944\n992\n943\n993\n942\n994\n941\n995\n940\n996\n939\n997\n938\n998\n937\n999\n936\n1000\n935\n1001\n934\n1002\n933\n1003\n932\n1004\n931\n1005\n930\n1006\n929\n1007\n928\n1008\n927\n1009\n926\n1010\n925\n1011\n924\n1012\n923\n1013\n922\n1014\n921\n1015\n920\n1016\n919\n1017..."
},
{
"input": "17 812",
"output": "406\n407\n405\n408\n404\n409\n403\n410\n402\n411\n401\n412\n400\n413\n399\n414\n398"
},
{
"input": "30078 300",
"output": "150\n151\n149\n152\n148\n153\n147\n154\n146\n155\n145\n156\n144\n157\n143\n158\n142\n159\n141\n160\n140\n161\n139\n162\n138\n163\n137\n164\n136\n165\n135\n166\n134\n167\n133\n168\n132\n169\n131\n170\n130\n171\n129\n172\n128\n173\n127\n174\n126\n175\n125\n176\n124\n177\n123\n178\n122\n179\n121\n180\n120\n181\n119\n182\n118\n183\n117\n184\n116\n185\n115\n186\n114\n187\n113\n188\n112\n189\n111\n190\n110\n191\n109\n192\n108\n193\n107\n194\n106\n195\n105\n196\n104\n197\n103\n198\n102\n199\n101\n200\n100\n201\n9..."
},
{
"input": "10500 5",
"output": "3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3\n2\n4\n1\n5\n3..."
},
{
"input": "90091 322",
"output": "161\n162\n160\n163\n159\n164\n158\n165\n157\n166\n156\n167\n155\n168\n154\n169\n153\n170\n152\n171\n151\n172\n150\n173\n149\n174\n148\n175\n147\n176\n146\n177\n145\n178\n144\n179\n143\n180\n142\n181\n141\n182\n140\n183\n139\n184\n138\n185\n137\n186\n136\n187\n135\n188\n134\n189\n133\n190\n132\n191\n131\n192\n130\n193\n129\n194\n128\n195\n127\n196\n126\n197\n125\n198\n124\n199\n123\n200\n122\n201\n121\n202\n120\n203\n119\n204\n118\n205\n117\n206\n116\n207\n115\n208\n114\n209\n113\n210\n112\n211\n111\n212\n1..."
},
{
"input": "8471 92356",
"output": "46178\n46179\n46177\n46180\n46176\n46181\n46175\n46182\n46174\n46183\n46173\n46184\n46172\n46185\n46171\n46186\n46170\n46187\n46169\n46188\n46168\n46189\n46167\n46190\n46166\n46191\n46165\n46192\n46164\n46193\n46163\n46194\n46162\n46195\n46161\n46196\n46160\n46197\n46159\n46198\n46158\n46199\n46157\n46200\n46156\n46201\n46155\n46202\n46154\n46203\n46153\n46204\n46152\n46205\n46151\n46206\n46150\n46207\n46149\n46208\n46148\n46209\n46147\n46210\n46146\n46211\n46145\n46212\n46144\n46213\n46143\n46214\n46142\n..."
},
{
"input": "1 2",
"output": "1"
},
{
"input": "2 1",
"output": "1\n1"
},
{
"input": "52097 88310",
"output": "44155\n44156\n44154\n44157\n44153\n44158\n44152\n44159\n44151\n44160\n44150\n44161\n44149\n44162\n44148\n44163\n44147\n44164\n44146\n44165\n44145\n44166\n44144\n44167\n44143\n44168\n44142\n44169\n44141\n44170\n44140\n44171\n44139\n44172\n44138\n44173\n44137\n44174\n44136\n44175\n44135\n44176\n44134\n44177\n44133\n44178\n44132\n44179\n44131\n44180\n44130\n44181\n44129\n44182\n44128\n44183\n44127\n44184\n44126\n44185\n44125\n44186\n44124\n44187\n44123\n44188\n44122\n44189\n44121\n44190\n44120\n44191\n44119\n..."
}
] | 1,649,105,913
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 44
| 528
| 0
|
n, m = map(int, input().split())
# i%m vai de 0 ate m-1
for i in range(n):
if (m+i%m)%2 > 0:
# pega o valor medio, valores entre (m+1)/2 até m
print(str(int((m+i%m+1)/2)))
else:
# Pega o valor médio que seja menor ou igual a metade de m.
# Portanto valores entre 1 até m/2
print(str(int((m-(i%m))/2)))
|
Title: After Training
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After a team finished their training session on Euro football championship, Valeric was commissioned to gather the balls and sort them into baskets. Overall the stadium has *n* balls and *m* baskets. The baskets are positioned in a row from left to right and they are numbered with numbers from 1 to *m*, correspondingly. The balls are numbered with numbers from 1 to *n*.
Valeric decided to sort the balls in the order of increasing of their numbers by the following scheme. He will put each new ball in the basket with the least number of balls. And if he's got several variants, he chooses the basket which stands closer to the middle. That means that he chooses the basket for which is minimum, where *i* is the number of the basket. If in this case Valeric still has multiple variants, he chooses the basket with the minimum number.
For every ball print the number of the basket where it will go according to Valeric's scheme.
Note that the balls are sorted into baskets in the order of increasing numbers, that is, the first ball goes first, then goes the second ball and so on.
Input Specification:
The first line contains two space-separated integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=105) — the number of balls and baskets, correspondingly.
Output Specification:
Print *n* numbers, one per line. The *i*-th line must contain the number of the basket for the *i*-th ball.
Demo Input:
['4 3\n', '3 1\n']
Demo Output:
['2\n1\n3\n2\n', '1\n1\n1\n']
Note:
none
|
```python
n, m = map(int, input().split())
# i%m vai de 0 ate m-1
for i in range(n):
if (m+i%m)%2 > 0:
# pega o valor medio, valores entre (m+1)/2 até m
print(str(int((m+i%m+1)/2)))
else:
# Pega o valor médio que seja menor ou igual a metade de m.
# Portanto valores entre 1 até m/2
print(str(int((m-(i%m))/2)))
```
| 3
|
|
555
|
B
|
Case of Fugitive
|
PROGRAMMING
| 2,000
|
[
"data structures",
"greedy",
"sortings"
] | null | null |
Andrewid the Android is a galaxy-famous detective. He is now chasing a criminal hiding on the planet Oxa-5, the planet almost fully covered with water.
The only dry land there is an archipelago of *n* narrow islands located in a row. For more comfort let's represent them as non-intersecting segments on a straight line: island *i* has coordinates [*l**i*,<=*r**i*], besides, *r**i*<=<<=*l**i*<=+<=1 for 1<=≤<=*i*<=≤<=*n*<=-<=1.
To reach the goal, Andrewid needs to place a bridge between each pair of adjacent islands. A bridge of length *a* can be placed between the *i*-th and the (*i*<=+<=1)-th islads, if there are such coordinates of *x* and *y*, that *l**i*<=≤<=*x*<=≤<=*r**i*, *l**i*<=+<=1<=≤<=*y*<=≤<=*r**i*<=+<=1 and *y*<=-<=*x*<==<=*a*.
The detective was supplied with *m* bridges, each bridge can be used at most once. Help him determine whether the bridges he got are enough to connect each pair of adjacent islands.
|
The first line contains integers *n* (2<=≤<=*n*<=≤<=2·105) and *m* (1<=≤<=*m*<=≤<=2·105) — the number of islands and bridges.
Next *n* lines each contain two integers *l**i* and *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=1018) — the coordinates of the island endpoints.
The last line contains *m* integer numbers *a*1,<=*a*2,<=...,<=*a**m* (1<=≤<=*a**i*<=≤<=1018) — the lengths of the bridges that Andrewid got.
|
If it is impossible to place a bridge between each pair of adjacent islands in the required manner, print on a single line "No" (without the quotes), otherwise print in the first line "Yes" (without the quotes), and in the second line print *n*<=-<=1 numbers *b*1,<=*b*2,<=...,<=*b**n*<=-<=1, which mean that between islands *i* and *i*<=+<=1 there must be used a bridge number *b**i*.
If there are multiple correct answers, print any of them. Note that in this problem it is necessary to print "Yes" and "No" in correct case.
|
[
"4 4\n1 4\n7 8\n9 10\n12 14\n4 5 3 8\n",
"2 2\n11 14\n17 18\n2 9\n",
"2 1\n1 1\n1000000000000000000 1000000000000000000\n999999999999999999\n"
] |
[
"Yes\n2 3 1 \n",
"No\n",
"Yes\n1 \n"
] |
In the first sample test you can, for example, place the second bridge between points 3 and 8, place the third bridge between points 7 and 10 and place the first bridge between points 10 and 14.
In the second sample test the first bridge is too short and the second bridge is too long, so the solution doesn't exist.
| 750
|
[
{
"input": "4 4\n1 4\n7 8\n9 10\n12 14\n4 5 3 8",
"output": "Yes\n2 3 1 "
},
{
"input": "2 2\n11 14\n17 18\n2 9",
"output": "No"
},
{
"input": "2 1\n1 1\n1000000000000000000 1000000000000000000\n999999999999999999",
"output": "Yes\n1 "
},
{
"input": "5 10\n1 2\n3 3\n5 7\n11 13\n14 20\n9 10 2 9 10 4 9 9 9 10",
"output": "No"
},
{
"input": "5 9\n1 2\n3 3\n5 7\n11 13\n14 20\n2 3 4 10 6 2 6 9 5",
"output": "Yes\n1 6 3 2 "
},
{
"input": "6 9\n1 4\n10 18\n23 29\n33 43\n46 57\n59 77\n11 32 32 19 20 17 32 24 32",
"output": "Yes\n1 6 4 5 8 "
},
{
"input": "6 9\n1 2\n8 16\n21 27\n31 46\n49 57\n59 78\n26 27 28 13 2 4 2 2 24",
"output": "No"
},
{
"input": "20 10\n4 9\n10 15\n17 18\n20 21\n25 27\n29 32\n35 36\n46 48\n49 51\n53 56\n59 60\n63 64\n65 68\n69 70\n74 75\n79 80\n81 82\n84 87\n88 91\n98 100\n4 7 6 1 5 4 3 1 5 2",
"output": "No"
},
{
"input": "2 1\n1 2\n5 6\n1",
"output": "No"
},
{
"input": "2 1\n1 1\n100 100\n5",
"output": "No"
},
{
"input": "3 2\n1000000000000000 1000000000000000\n3000000000000000 4000000000000000\n6000000000000000 7000000000000000\n2000000000000000 4000000000000000",
"output": "Yes\n1 2 "
},
{
"input": "3 2\n1 5\n6 12\n14 100000000000\n10000000000 4",
"output": "Yes\n2 1 "
}
] | 1,435,420,248
| 6,048
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
PRETESTS
| 11
| 3,000
| 58,777,600
|
import heapq
def read_data():
n, m = map(int, input().split())
LR = []
for i in range(n):
l, r = map(int, input().split())
LR.append((l, r))
As = list(map(int, input().split()))
return n, m, LR, As
def solve(n, m, LR, As):
if m < n - 1:
return False
mins, maxs = convert2minmax(LR)
mins.sort()
As = [(a, i) for i, a in enumerate(As, 1)]
As.sort()
allocation = [0] * (n - 1)
gaps = []
allocated = 0
mins_cursor = 0
for a, bridge_id in As:
mins_cursor = update_gaps(gaps, a, mins, maxs, mins_cursor)
if gaps:
maxgap, gap_id = heapq.heappop(gaps)
if maxgap < a:
return False
allocation[gap_id] = bridge_id
allocated += 1
if allocated != n - 1:
return False
return allocation
def convert2minmax(LR):
prevL, prevR = LR[0]
mins = []
maxs = []
for i, (L, R) in enumerate(LR[1:]):
mins.append((L - prevR, i))
maxs.append(R - prevL)
prevL, prevR = L, R
return mins, maxs
def update_gaps(gaps, a, mins, maxs, mins_cursor):
cursor = mins_cursor
for mingap, gap_id in mins[mins_cursor:]:
if mingap <= a:
heapq.heappush(gaps, (maxs[gap_id], gap_id))
cursor += 1
else:
break
return cursor
if __name__ == '__main__':
n, m, LR, As = read_data()
result = solve(n, m, LR, As)
if result:
print('Yes')
print(*result)
else:
print('No')
|
Title: Case of Fugitive
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Andrewid the Android is a galaxy-famous detective. He is now chasing a criminal hiding on the planet Oxa-5, the planet almost fully covered with water.
The only dry land there is an archipelago of *n* narrow islands located in a row. For more comfort let's represent them as non-intersecting segments on a straight line: island *i* has coordinates [*l**i*,<=*r**i*], besides, *r**i*<=<<=*l**i*<=+<=1 for 1<=≤<=*i*<=≤<=*n*<=-<=1.
To reach the goal, Andrewid needs to place a bridge between each pair of adjacent islands. A bridge of length *a* can be placed between the *i*-th and the (*i*<=+<=1)-th islads, if there are such coordinates of *x* and *y*, that *l**i*<=≤<=*x*<=≤<=*r**i*, *l**i*<=+<=1<=≤<=*y*<=≤<=*r**i*<=+<=1 and *y*<=-<=*x*<==<=*a*.
The detective was supplied with *m* bridges, each bridge can be used at most once. Help him determine whether the bridges he got are enough to connect each pair of adjacent islands.
Input Specification:
The first line contains integers *n* (2<=≤<=*n*<=≤<=2·105) and *m* (1<=≤<=*m*<=≤<=2·105) — the number of islands and bridges.
Next *n* lines each contain two integers *l**i* and *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=1018) — the coordinates of the island endpoints.
The last line contains *m* integer numbers *a*1,<=*a*2,<=...,<=*a**m* (1<=≤<=*a**i*<=≤<=1018) — the lengths of the bridges that Andrewid got.
Output Specification:
If it is impossible to place a bridge between each pair of adjacent islands in the required manner, print on a single line "No" (without the quotes), otherwise print in the first line "Yes" (without the quotes), and in the second line print *n*<=-<=1 numbers *b*1,<=*b*2,<=...,<=*b**n*<=-<=1, which mean that between islands *i* and *i*<=+<=1 there must be used a bridge number *b**i*.
If there are multiple correct answers, print any of them. Note that in this problem it is necessary to print "Yes" and "No" in correct case.
Demo Input:
['4 4\n1 4\n7 8\n9 10\n12 14\n4 5 3 8\n', '2 2\n11 14\n17 18\n2 9\n', '2 1\n1 1\n1000000000000000000 1000000000000000000\n999999999999999999\n']
Demo Output:
['Yes\n2 3 1 \n', 'No\n', 'Yes\n1 \n']
Note:
In the first sample test you can, for example, place the second bridge between points 3 and 8, place the third bridge between points 7 and 10 and place the first bridge between points 10 and 14.
In the second sample test the first bridge is too short and the second bridge is too long, so the solution doesn't exist.
|
```python
import heapq
def read_data():
n, m = map(int, input().split())
LR = []
for i in range(n):
l, r = map(int, input().split())
LR.append((l, r))
As = list(map(int, input().split()))
return n, m, LR, As
def solve(n, m, LR, As):
if m < n - 1:
return False
mins, maxs = convert2minmax(LR)
mins.sort()
As = [(a, i) for i, a in enumerate(As, 1)]
As.sort()
allocation = [0] * (n - 1)
gaps = []
allocated = 0
mins_cursor = 0
for a, bridge_id in As:
mins_cursor = update_gaps(gaps, a, mins, maxs, mins_cursor)
if gaps:
maxgap, gap_id = heapq.heappop(gaps)
if maxgap < a:
return False
allocation[gap_id] = bridge_id
allocated += 1
if allocated != n - 1:
return False
return allocation
def convert2minmax(LR):
prevL, prevR = LR[0]
mins = []
maxs = []
for i, (L, R) in enumerate(LR[1:]):
mins.append((L - prevR, i))
maxs.append(R - prevL)
prevL, prevR = L, R
return mins, maxs
def update_gaps(gaps, a, mins, maxs, mins_cursor):
cursor = mins_cursor
for mingap, gap_id in mins[mins_cursor:]:
if mingap <= a:
heapq.heappush(gaps, (maxs[gap_id], gap_id))
cursor += 1
else:
break
return cursor
if __name__ == '__main__':
n, m, LR, As = read_data()
result = solve(n, m, LR, As)
if result:
print('Yes')
print(*result)
else:
print('No')
```
| 0
|
|
892
|
B
|
Wrath
|
PROGRAMMING
| 1,200
|
[
"greedy",
"implementation",
"two pointers"
] | null | null |
Hands that shed innocent blood!
There are *n* guilty people in a line, the *i*-th of them holds a claw with length *L**i*. The bell rings and every person kills some of people in front of him. All people kill others at the same time. Namely, the *i*-th person kills the *j*-th person if and only if *j*<=<<=*i* and *j*<=≥<=*i*<=-<=*L**i*.
You are given lengths of the claws. You need to find the total number of alive people after the bell rings.
|
The first line contains one integer *n* (1<=≤<=*n*<=≤<=106) — the number of guilty people.
Second line contains *n* space-separated integers *L*1,<=*L*2,<=...,<=*L**n* (0<=≤<=*L**i*<=≤<=109), where *L**i* is the length of the *i*-th person's claw.
|
Print one integer — the total number of alive people after the bell rings.
|
[
"4\n0 1 0 10\n",
"2\n0 0\n",
"10\n1 1 3 0 0 0 2 1 0 3\n"
] |
[
"1\n",
"2\n",
"3\n"
] |
In first sample the last person kills everyone in front of him.
| 1,000
|
[
{
"input": "4\n0 1 0 10",
"output": "1"
},
{
"input": "2\n0 0",
"output": "2"
},
{
"input": "10\n1 1 3 0 0 0 2 1 0 3",
"output": "3"
},
{
"input": "10\n0 0 2 0 0 3 3 2 2 0",
"output": "2"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "5\n0 0 0 1 0",
"output": "4"
},
{
"input": "6\n3 1 1 0 3 3",
"output": "1"
},
{
"input": "8\n0 0 0 1 0 0 1 2",
"output": "5"
},
{
"input": "1\n1000000000",
"output": "1"
},
{
"input": "2\n1 3",
"output": "1"
},
{
"input": "2\n1000000000 1000000000",
"output": "1"
},
{
"input": "11\n1 0 0 1 1 3 2 0 0 2 3",
"output": "4"
},
{
"input": "1\n1",
"output": "1"
}
] | 1,522,846,396
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 43
| 1,232
| 83,763,200
|
n = int(input())
a = list(map(int, input().split()))[::-1]
l, r = -1, -1
ans = 0
while l < n - 1:
l += 1
if l > r:
ans += 1
r = max(r, l + a[l])
if r > n - 1:
break
print(ans)
|
Title: Wrath
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Hands that shed innocent blood!
There are *n* guilty people in a line, the *i*-th of them holds a claw with length *L**i*. The bell rings and every person kills some of people in front of him. All people kill others at the same time. Namely, the *i*-th person kills the *j*-th person if and only if *j*<=<<=*i* and *j*<=≥<=*i*<=-<=*L**i*.
You are given lengths of the claws. You need to find the total number of alive people after the bell rings.
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=106) — the number of guilty people.
Second line contains *n* space-separated integers *L*1,<=*L*2,<=...,<=*L**n* (0<=≤<=*L**i*<=≤<=109), where *L**i* is the length of the *i*-th person's claw.
Output Specification:
Print one integer — the total number of alive people after the bell rings.
Demo Input:
['4\n0 1 0 10\n', '2\n0 0\n', '10\n1 1 3 0 0 0 2 1 0 3\n']
Demo Output:
['1\n', '2\n', '3\n']
Note:
In first sample the last person kills everyone in front of him.
|
```python
n = int(input())
a = list(map(int, input().split()))[::-1]
l, r = -1, -1
ans = 0
while l < n - 1:
l += 1
if l > r:
ans += 1
r = max(r, l + a[l])
if r > n - 1:
break
print(ans)
```
| 3
|
|
450
|
A
|
Jzzhu and Children
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
There are *n* children in Jzzhu's school. Jzzhu is going to give some candies to them. Let's number all the children from 1 to *n*. The *i*-th child wants to get at least *a**i* candies.
Jzzhu asks children to line up. Initially, the *i*-th child stands at the *i*-th place of the line. Then Jzzhu start distribution of the candies. He follows the algorithm:
1. Give *m* candies to the first child of the line. 1. If this child still haven't got enough candies, then the child goes to the end of the line, else the child go home. 1. Repeat the first two steps while the line is not empty.
Consider all the children in the order they go home. Jzzhu wants to know, which child will be the last in this order?
|
The first line contains two integers *n*,<=*m* (1<=≤<=*n*<=≤<=100; 1<=≤<=*m*<=≤<=100). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100).
|
Output a single integer, representing the number of the last child.
|
[
"5 2\n1 3 1 4 2\n",
"6 4\n1 1 2 2 3 3\n"
] |
[
"4\n",
"6\n"
] |
Let's consider the first sample.
Firstly child 1 gets 2 candies and go home. Then child 2 gets 2 candies and go to the end of the line. Currently the line looks like [3, 4, 5, 2] (indices of the children in order of the line). Then child 3 gets 2 candies and go home, and then child 4 gets 2 candies and goes to the end of the line. Currently the line looks like [5, 2, 4]. Then child 5 gets 2 candies and goes home. Then child 2 gets two candies and goes home, and finally child 4 gets 2 candies and goes home.
Child 4 is the last one who goes home.
| 500
|
[
{
"input": "5 2\n1 3 1 4 2",
"output": "4"
},
{
"input": "6 4\n1 1 2 2 3 3",
"output": "6"
},
{
"input": "7 3\n6 1 5 4 2 3 1",
"output": "4"
},
{
"input": "10 5\n2 7 3 6 2 5 1 3 4 5",
"output": "4"
},
{
"input": "100 1\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100",
"output": "100"
},
{
"input": "9 3\n9 5 2 3 7 1 8 4 6",
"output": "7"
},
{
"input": "20 10\n58 4 32 10 73 7 30 39 47 6 59 21 24 66 79 79 46 13 29 58",
"output": "16"
},
{
"input": "50 5\n89 56 3 2 40 37 56 52 83 59 43 83 43 59 29 74 22 58 53 41 53 67 78 30 57 32 58 29 95 46 45 85 60 49 41 82 8 71 52 40 45 26 6 71 84 91 4 93 40 54",
"output": "48"
},
{
"input": "50 1\n4 3 9 7 6 8 3 7 10 9 8 8 10 2 9 3 2 4 4 10 4 6 8 10 9 9 4 2 8 9 4 4 9 5 1 5 2 4 4 9 10 2 5 10 7 2 8 6 8 1",
"output": "44"
},
{
"input": "50 5\n3 9 10 8 3 3 4 6 8 2 9 9 3 1 2 10 6 8 7 2 7 4 2 7 5 10 2 2 2 5 10 5 6 6 8 7 10 4 3 2 10 8 6 6 8 6 4 4 1 3",
"output": "46"
},
{
"input": "50 2\n56 69 72 15 95 92 51 1 74 87 100 29 46 54 18 81 84 72 84 83 20 63 71 27 45 74 50 89 48 8 21 15 47 3 39 73 80 84 6 99 17 25 56 3 74 64 71 39 89 78",
"output": "40"
},
{
"input": "50 3\n31 39 64 16 86 3 1 9 25 54 98 42 20 3 49 41 73 37 55 62 33 77 64 22 33 82 26 13 10 13 7 40 48 18 46 79 94 72 19 12 11 61 16 37 10 49 14 94 48 69",
"output": "11"
},
{
"input": "50 100\n67 67 61 68 42 29 70 77 12 61 71 27 4 73 87 52 59 38 93 90 31 27 87 47 26 57 76 6 28 72 81 68 50 84 69 79 39 93 52 6 88 12 46 13 90 68 71 38 90 95",
"output": "50"
},
{
"input": "100 3\n4 14 20 11 19 11 14 20 5 7 6 12 11 17 5 11 7 6 2 10 13 5 12 8 5 17 20 18 7 19 11 7 7 20 20 8 10 17 17 19 20 5 15 16 19 7 11 16 4 17 2 10 1 20 20 16 19 9 9 11 5 7 12 9 9 6 20 18 13 19 8 4 8 1 2 4 10 11 15 14 1 7 17 12 13 19 12 2 3 14 15 15 5 17 14 12 17 14 16 9",
"output": "86"
},
{
"input": "100 5\n16 8 14 16 12 11 17 19 19 2 8 9 5 6 19 9 11 18 6 9 14 16 14 18 17 17 17 5 15 20 19 7 7 10 10 5 14 20 5 19 11 16 16 19 17 9 7 12 14 10 2 11 14 5 20 8 10 11 19 2 14 14 19 17 5 10 8 8 4 2 1 10 20 12 14 11 7 6 6 15 1 5 9 15 3 17 16 17 5 14 11 9 16 15 1 11 10 6 15 7",
"output": "93"
},
{
"input": "100 1\n58 94 18 50 17 14 96 62 83 80 75 5 9 22 25 41 3 96 74 45 66 37 2 37 13 85 68 54 77 11 85 19 25 21 52 59 90 61 72 89 82 22 10 16 3 68 61 29 55 76 28 85 65 76 27 3 14 10 56 37 86 18 35 38 56 68 23 88 33 38 52 87 55 83 94 34 100 41 83 56 91 77 32 74 97 13 67 31 57 81 53 39 5 88 46 1 79 4 49 42",
"output": "77"
},
{
"input": "100 2\n1 51 76 62 34 93 90 43 57 59 52 78 3 48 11 60 57 48 5 54 28 81 87 23 44 77 67 61 14 73 29 53 21 89 67 41 47 9 63 37 1 71 40 85 4 14 77 40 78 75 89 74 4 70 32 65 81 95 49 90 72 41 76 55 69 83 73 84 85 93 46 6 74 90 62 37 97 7 7 37 83 30 37 88 34 16 11 59 85 19 57 63 85 20 63 97 97 65 61 48",
"output": "97"
},
{
"input": "100 3\n30 83 14 55 61 66 34 98 90 62 89 74 45 93 33 31 75 35 82 100 63 69 48 18 99 2 36 71 14 30 70 76 96 85 97 90 49 36 6 76 37 94 70 3 63 73 75 48 39 29 13 2 46 26 9 56 1 18 54 53 85 34 2 12 1 93 75 67 77 77 14 26 33 25 55 9 57 70 75 6 87 66 18 3 41 69 73 24 49 2 20 72 39 58 91 54 74 56 66 78",
"output": "20"
},
{
"input": "100 4\n69 92 76 3 32 50 15 38 21 22 14 3 67 41 95 12 10 62 83 52 78 1 18 58 94 35 62 71 58 75 13 73 60 34 50 97 50 70 19 96 53 10 100 26 20 39 62 59 88 26 24 83 70 68 66 8 6 38 16 93 2 91 81 89 78 74 21 8 31 56 28 53 77 5 81 5 94 42 77 75 92 15 59 36 61 18 55 45 69 68 81 51 12 42 85 74 98 31 17 41",
"output": "97"
},
{
"input": "100 5\n2 72 10 60 6 50 72 34 97 77 35 43 80 64 40 53 46 6 90 22 29 70 26 68 52 19 72 88 83 18 55 32 99 81 11 21 39 42 41 63 60 97 30 23 55 78 89 35 24 50 99 52 27 76 24 8 20 27 51 37 17 82 69 18 46 19 26 77 52 83 76 65 43 66 84 84 13 30 66 88 84 23 37 1 17 26 11 50 73 56 54 37 40 29 35 8 1 39 50 82",
"output": "51"
},
{
"input": "100 7\n6 73 7 54 92 33 66 65 80 47 2 53 28 59 61 16 54 89 37 48 77 40 49 59 27 52 17 22 78 80 81 80 8 93 50 7 87 57 29 16 89 55 20 7 51 54 30 98 44 96 27 70 1 1 32 61 22 92 84 98 31 89 91 90 28 56 49 25 86 49 55 16 19 1 18 8 88 47 16 18 73 86 2 96 16 91 74 49 38 98 94 25 34 85 29 27 99 31 31 58",
"output": "97"
},
{
"input": "100 9\n36 4 45 16 19 6 10 87 44 82 71 49 70 35 83 19 40 76 45 94 44 96 10 54 82 77 86 63 11 37 21 3 15 89 80 88 89 16 72 23 25 9 51 25 10 45 96 5 6 18 51 31 42 57 41 51 42 15 89 61 45 82 16 48 61 67 19 40 9 33 90 36 78 36 79 79 16 10 83 87 9 22 84 12 23 76 36 14 2 81 56 33 56 23 57 84 76 55 35 88",
"output": "47"
},
{
"input": "100 10\n75 81 39 64 90 58 92 28 75 9 96 78 92 83 77 68 76 71 14 46 58 60 80 25 78 11 13 63 22 82 65 68 47 6 33 63 90 50 85 43 73 94 80 48 67 11 83 17 22 15 94 80 66 99 66 4 46 35 52 1 62 39 96 57 37 47 97 49 64 12 36 63 90 16 4 75 85 82 85 56 13 4 92 45 44 93 17 35 22 46 18 44 29 7 52 4 100 98 87 51",
"output": "98"
},
{
"input": "100 20\n21 19 61 70 54 97 98 14 61 72 25 94 24 56 55 25 12 80 76 11 35 17 80 26 11 94 52 47 84 61 10 2 74 25 10 21 2 79 55 50 30 75 10 64 44 5 60 96 52 16 74 41 20 77 20 44 8 86 74 36 49 61 99 13 54 64 19 99 50 43 12 73 48 48 83 55 72 73 63 81 30 27 95 9 97 82 24 3 89 90 33 14 47 88 22 78 12 75 58 67",
"output": "94"
},
{
"input": "100 30\n56 79 59 23 11 23 67 82 81 80 99 79 8 58 93 36 98 81 46 39 34 67 3 50 4 68 70 71 2 21 52 30 75 23 33 21 16 100 56 43 8 27 40 8 56 24 17 40 94 10 67 49 61 36 95 87 17 41 7 94 33 19 17 50 26 11 94 54 38 46 77 9 53 35 98 42 50 20 43 6 78 6 38 24 100 45 43 16 1 50 16 46 14 91 95 88 10 1 50 19",
"output": "95"
},
{
"input": "100 40\n86 11 97 17 38 95 11 5 13 83 67 75 50 2 46 39 84 68 22 85 70 23 64 46 59 93 39 80 35 78 93 21 83 19 64 1 49 59 99 83 44 81 70 58 15 82 83 47 55 65 91 10 2 92 4 77 37 32 12 57 78 11 42 8 59 21 96 69 61 30 44 29 12 70 91 14 10 83 11 75 14 10 19 39 8 98 5 81 66 66 79 55 36 29 22 45 19 24 55 49",
"output": "88"
},
{
"input": "100 50\n22 39 95 69 94 53 80 73 33 90 40 60 2 4 84 50 70 38 92 12 36 74 87 70 51 36 57 5 54 6 35 81 52 17 55 100 95 81 32 76 21 1 100 1 95 1 40 91 98 59 84 19 11 51 79 19 47 86 45 15 62 2 59 77 31 68 71 92 17 33 10 33 85 57 5 2 88 97 91 99 63 20 63 54 79 93 24 62 46 27 30 87 3 64 95 88 16 50 79 1",
"output": "99"
},
{
"input": "100 70\n61 48 89 17 97 6 93 13 64 50 66 88 24 52 46 99 6 65 93 64 82 37 57 41 47 1 84 5 97 83 79 46 16 35 40 7 64 15 44 96 37 17 30 92 51 67 26 3 14 56 27 68 66 93 36 39 51 6 40 55 79 26 71 54 8 48 18 2 71 12 55 60 29 37 31 97 26 37 25 68 67 70 3 87 100 41 5 82 65 92 24 66 76 48 89 8 40 93 31 95",
"output": "100"
},
{
"input": "100 90\n87 32 30 15 10 52 93 63 84 1 82 41 27 51 75 32 42 94 39 53 70 13 4 22 99 35 44 38 5 23 18 100 61 80 9 12 42 93 9 77 3 7 60 95 66 78 95 42 69 8 1 88 93 66 96 20 76 63 15 36 92 52 2 72 36 57 48 63 29 20 74 88 49 47 81 61 94 74 70 93 47 3 19 52 59 41 5 40 22 3 76 97 91 37 95 88 91 99 76 15",
"output": "98"
},
{
"input": "100 100\n79 75 7 28 6 96 38 35 57 95 41 74 24 96 32 78 81 13 63 84 24 95 3 23 66 1 60 6 96 49 41 5 14 18 31 97 66 19 49 89 49 70 51 28 20 99 18 1 28 77 24 46 69 21 40 32 31 66 28 6 66 97 9 16 70 90 91 30 34 82 93 41 65 11 39 52 1 88 63 43 80 50 60 49 28 56 18 76 24 57 74 1 28 99 36 35 79 54 18 16",
"output": "100"
},
{
"input": "1 3\n5",
"output": "1"
},
{
"input": "1 1\n100",
"output": "1"
},
{
"input": "2 3\n4 2",
"output": "1"
},
{
"input": "2 5\n99 97",
"output": "2"
},
{
"input": "3 4\n7 5 2",
"output": "2"
},
{
"input": "3 50\n47 86 51",
"output": "3"
},
{
"input": "5 100\n82 100 85 1 37",
"output": "5"
},
{
"input": "5 20\n40 39 21 5 20",
"output": "3"
},
{
"input": "1 27\n81",
"output": "1"
},
{
"input": "20 13\n7 8 29 83 74 28 93 85 7 8 3 9 8 70 49 50 39 41 57 1",
"output": "7"
},
{
"input": "2 1\n100 2",
"output": "1"
},
{
"input": "2 2\n6 4",
"output": "1"
},
{
"input": "5 2\n6 4 4 1 1",
"output": "1"
},
{
"input": "1 4\n3",
"output": "1"
},
{
"input": "3 2\n1 5 3",
"output": "2"
},
{
"input": "3 1\n3 2 2",
"output": "1"
},
{
"input": "3 1\n2 3 2",
"output": "2"
},
{
"input": "5 1\n5 1 1 1 1",
"output": "1"
},
{
"input": "2 3\n7 4",
"output": "1"
}
] | 1,690,424,103
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 49
| 46
| 0
|
import math
n, m = list(map(int, input().split()))
candies = list(map(int, input().split()))
a = []
b = []
for i in range(len(candies)):
x = candies[i] / m
x = math.ceil(x)
a.append(x)
highest_demand = max(a)
for j in range(len(a)):
if a[j] == highest_demand:
b.append(j)
ans = max(b)
print(ans + 1)
|
Title: Jzzhu and Children
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* children in Jzzhu's school. Jzzhu is going to give some candies to them. Let's number all the children from 1 to *n*. The *i*-th child wants to get at least *a**i* candies.
Jzzhu asks children to line up. Initially, the *i*-th child stands at the *i*-th place of the line. Then Jzzhu start distribution of the candies. He follows the algorithm:
1. Give *m* candies to the first child of the line. 1. If this child still haven't got enough candies, then the child goes to the end of the line, else the child go home. 1. Repeat the first two steps while the line is not empty.
Consider all the children in the order they go home. Jzzhu wants to know, which child will be the last in this order?
Input Specification:
The first line contains two integers *n*,<=*m* (1<=≤<=*n*<=≤<=100; 1<=≤<=*m*<=≤<=100). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100).
Output Specification:
Output a single integer, representing the number of the last child.
Demo Input:
['5 2\n1 3 1 4 2\n', '6 4\n1 1 2 2 3 3\n']
Demo Output:
['4\n', '6\n']
Note:
Let's consider the first sample.
Firstly child 1 gets 2 candies and go home. Then child 2 gets 2 candies and go to the end of the line. Currently the line looks like [3, 4, 5, 2] (indices of the children in order of the line). Then child 3 gets 2 candies and go home, and then child 4 gets 2 candies and goes to the end of the line. Currently the line looks like [5, 2, 4]. Then child 5 gets 2 candies and goes home. Then child 2 gets two candies and goes home, and finally child 4 gets 2 candies and goes home.
Child 4 is the last one who goes home.
|
```python
import math
n, m = list(map(int, input().split()))
candies = list(map(int, input().split()))
a = []
b = []
for i in range(len(candies)):
x = candies[i] / m
x = math.ceil(x)
a.append(x)
highest_demand = max(a)
for j in range(len(a)):
if a[j] == highest_demand:
b.append(j)
ans = max(b)
print(ans + 1)
```
| 3
|
|
299
|
A
|
Ksusha and Array
|
PROGRAMMING
| 1,000
|
[
"brute force",
"number theory",
"sortings"
] | null | null |
Ksusha is a beginner coder. Today she starts studying arrays. She has array *a*1,<=*a*2,<=...,<=*a**n*, consisting of *n* positive integers.
Her university teacher gave her a task. Find such number in the array, that all array elements are divisible by it. Help her and find the number!
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=105), showing how many numbers the array has. The next line contains integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the array elements.
|
Print a single integer — the number from the array, such that all array elements are divisible by it. If such number doesn't exist, print -1.
If there are multiple answers, you are allowed to print any of them.
|
[
"3\n2 2 4\n",
"5\n2 1 3 1 6\n",
"3\n2 3 5\n"
] |
[
"2\n",
"1\n",
"-1\n"
] |
none
| 500
|
[
{
"input": "3\n2 2 4",
"output": "2"
},
{
"input": "5\n2 1 3 1 6",
"output": "1"
},
{
"input": "3\n2 3 5",
"output": "-1"
},
{
"input": "1\n331358794",
"output": "331358794"
},
{
"input": "5\n506904227 214303304 136194869 838256937 183952885",
"output": "-1"
},
{
"input": "2\n500000000 1000000000",
"output": "500000000"
},
{
"input": "2\n4 6",
"output": "-1"
},
{
"input": "5\n10 8 6 4 2",
"output": "2"
},
{
"input": "2\n6 10",
"output": "-1"
},
{
"input": "1\n1000000000",
"output": "1000000000"
},
{
"input": "2\n6 8",
"output": "-1"
},
{
"input": "5\n2 2 2 2 1000000000",
"output": "2"
},
{
"input": "2\n6 4",
"output": "-1"
}
] | 1,633,659,872
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 3
| 122
| 6,963,200
|
import math
n = int(input())
a = (input().split(' '))
a1 = int(a[0])
a2 = int(a[1])
f = 0
for x in range(n-2):
a1 = math.gcd(int(a1), int(a2))
if x == 1:
f = 1
if n <= 3:
break
a2 = a[x+2]
if (a1 == 1 and f == 1) or a1 > 1:
print(a1)
else:
print(-1)
|
Title: Ksusha and Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ksusha is a beginner coder. Today she starts studying arrays. She has array *a*1,<=*a*2,<=...,<=*a**n*, consisting of *n* positive integers.
Her university teacher gave her a task. Find such number in the array, that all array elements are divisible by it. Help her and find the number!
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105), showing how many numbers the array has. The next line contains integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the array elements.
Output Specification:
Print a single integer — the number from the array, such that all array elements are divisible by it. If such number doesn't exist, print -1.
If there are multiple answers, you are allowed to print any of them.
Demo Input:
['3\n2 2 4\n', '5\n2 1 3 1 6\n', '3\n2 3 5\n']
Demo Output:
['2\n', '1\n', '-1\n']
Note:
none
|
```python
import math
n = int(input())
a = (input().split(' '))
a1 = int(a[0])
a2 = int(a[1])
f = 0
for x in range(n-2):
a1 = math.gcd(int(a1), int(a2))
if x == 1:
f = 1
if n <= 3:
break
a2 = a[x+2]
if (a1 == 1 and f == 1) or a1 > 1:
print(a1)
else:
print(-1)
```
| -1
|
|
535
|
B
|
Tavas and SaDDas
|
PROGRAMMING
| 1,100
|
[
"bitmasks",
"brute force",
"combinatorics",
"implementation"
] | null | null |
Once again Tavas started eating coffee mix without water! Keione told him that it smells awful, but he didn't stop doing that. That's why Keione told his smart friend, SaDDas to punish him! SaDDas took Tavas' headphones and told him: "If you solve the following problem, I'll return it to you."
The problem is:
You are given a lucky number *n*. Lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
If we sort all lucky numbers in increasing order, what's the 1-based index of *n*?
Tavas is not as smart as SaDDas, so he asked you to do him a favor and solve this problem so he can have his headphones back.
|
The first and only line of input contains a lucky number *n* (1<=≤<=*n*<=≤<=109).
|
Print the index of *n* among all lucky numbers.
|
[
"4\n",
"7\n",
"77\n"
] |
[
"1\n",
"2\n",
"6\n"
] |
none
| 1,000
|
[
{
"input": "4",
"output": "1"
},
{
"input": "7",
"output": "2"
},
{
"input": "77",
"output": "6"
},
{
"input": "4",
"output": "1"
},
{
"input": "474744",
"output": "83"
},
{
"input": "777774",
"output": "125"
},
{
"input": "447",
"output": "8"
},
{
"input": "774",
"output": "13"
},
{
"input": "4",
"output": "1"
},
{
"input": "4447747",
"output": "140"
},
{
"input": "7747474",
"output": "233"
},
{
"input": "4444",
"output": "15"
},
{
"input": "4447",
"output": "16"
},
{
"input": "7",
"output": "2"
},
{
"input": "4",
"output": "1"
},
{
"input": "4447744",
"output": "139"
},
{
"input": "77474",
"output": "57"
},
{
"input": "7747",
"output": "28"
},
{
"input": "444",
"output": "7"
},
{
"input": "7",
"output": "2"
},
{
"input": "7774477",
"output": "242"
},
{
"input": "4477774",
"output": "157"
},
{
"input": "7444",
"output": "23"
},
{
"input": "7474747",
"output": "212"
},
{
"input": "77",
"output": "6"
},
{
"input": "774477",
"output": "114"
},
{
"input": "7",
"output": "2"
},
{
"input": "47",
"output": "4"
},
{
"input": "747777",
"output": "110"
},
{
"input": "444444444",
"output": "511"
},
{
"input": "777777777",
"output": "1022"
},
{
"input": "477477447",
"output": "728"
},
{
"input": "777744747",
"output": "996"
}
] | 1,580,403,378
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 33
| 109
| 0
|
n=int(input())
i=0
s=0
while n>0:
if n%10==4:
s=s+(2**i)
if n%10==7:
s=s+(2**(i+1))
n=n//10
i=i+1
print(s)
|
Title: Tavas and SaDDas
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Once again Tavas started eating coffee mix without water! Keione told him that it smells awful, but he didn't stop doing that. That's why Keione told his smart friend, SaDDas to punish him! SaDDas took Tavas' headphones and told him: "If you solve the following problem, I'll return it to you."
The problem is:
You are given a lucky number *n*. Lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
If we sort all lucky numbers in increasing order, what's the 1-based index of *n*?
Tavas is not as smart as SaDDas, so he asked you to do him a favor and solve this problem so he can have his headphones back.
Input Specification:
The first and only line of input contains a lucky number *n* (1<=≤<=*n*<=≤<=109).
Output Specification:
Print the index of *n* among all lucky numbers.
Demo Input:
['4\n', '7\n', '77\n']
Demo Output:
['1\n', '2\n', '6\n']
Note:
none
|
```python
n=int(input())
i=0
s=0
while n>0:
if n%10==4:
s=s+(2**i)
if n%10==7:
s=s+(2**(i+1))
n=n//10
i=i+1
print(s)
```
| 3
|
|
707
|
A
|
Brain's Photos
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Small, but very brave, mouse Brain was not accepted to summer school of young villains. He was upset and decided to postpone his plans of taking over the world, but to become a photographer instead.
As you may know, the coolest photos are on the film (because you can specify the hashtag #film for such).
Brain took a lot of colourful pictures on colored and black-and-white film. Then he developed and translated it into a digital form. But now, color and black-and-white photos are in one folder, and to sort them, one needs to spend more than one hour!
As soon as Brain is a photographer not programmer now, he asks you to help him determine for a single photo whether it is colored or black-and-white.
Photo can be represented as a matrix sized *n*<=×<=*m*, and each element of the matrix stores a symbol indicating corresponding pixel color. There are only 6 colors:
- 'C' (cyan)- 'M' (magenta)- 'Y' (yellow)- 'W' (white)- 'G' (grey)- 'B' (black)
The photo is considered black-and-white if it has only white, black and grey pixels in it. If there are any of cyan, magenta or yellow pixels in the photo then it is considered colored.
|
The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of photo pixel matrix rows and columns respectively.
Then *n* lines describing matrix rows follow. Each of them contains *m* space-separated characters describing colors of pixels in a row. Each character in the line is one of the 'C', 'M', 'Y', 'W', 'G' or 'B'.
|
Print the "#Black&White" (without quotes), if the photo is black-and-white and "#Color" (without quotes), if it is colored, in the only line.
|
[
"2 2\nC M\nY Y\n",
"3 2\nW W\nW W\nB B\n",
"1 1\nW\n"
] |
[
"#Color",
"#Black&White",
"#Black&White"
] |
none
| 500
|
[
{
"input": "2 2\nC M\nY Y",
"output": "#Color"
},
{
"input": "3 2\nW W\nW W\nB B",
"output": "#Black&White"
},
{
"input": "1 1\nW",
"output": "#Black&White"
},
{
"input": "2 3\nW W W\nB G Y",
"output": "#Color"
},
{
"input": "1 1\nW",
"output": "#Black&White"
},
{
"input": "5 5\nW G B Y M\nG B Y M C\nB Y M C W\nY M C W G\nM C W G B",
"output": "#Color"
},
{
"input": "1 6\nC M Y W G B",
"output": "#Color"
},
{
"input": "1 3\nW G B",
"output": "#Black&White"
},
{
"input": "1 1\nW",
"output": "#Black&White"
},
{
"input": "5 5\nW G B W G\nG B W G B\nB W G B W\nW G B W G\nG B W G B",
"output": "#Black&White"
},
{
"input": "2 3\nW W W\nB G C",
"output": "#Color"
},
{
"input": "2 3\nW W W\nB G M",
"output": "#Color"
},
{
"input": "3 3\nC B W\nB Y M\nB B W",
"output": "#Color"
},
{
"input": "1 3\nW C W",
"output": "#Color"
},
{
"input": "3 3\nB W B\nB C W\nB W W",
"output": "#Color"
},
{
"input": "1 2\nW Y",
"output": "#Color"
},
{
"input": "1 1\nG",
"output": "#Black&White"
},
{
"input": "1 3\nB G W",
"output": "#Black&White"
},
{
"input": "2 2\nW W\nB C",
"output": "#Color"
},
{
"input": "1 1\nM",
"output": "#Color"
},
{
"input": "1 2\nW C",
"output": "#Color"
},
{
"input": "2 3\nW W M\nW W M",
"output": "#Color"
},
{
"input": "3 2\nW W\nW W\nB C",
"output": "#Color"
},
{
"input": "2 3\nW W C\nW W W",
"output": "#Color"
},
{
"input": "1 3\nG G G",
"output": "#Black&White"
},
{
"input": "1 1\nC",
"output": "#Color"
},
{
"input": "1 2\nC W",
"output": "#Color"
},
{
"input": "1 3\nW W C",
"output": "#Color"
},
{
"input": "2 2\nW B\nB G",
"output": "#Black&White"
},
{
"input": "2 2\nB B\nY Y",
"output": "#Color"
},
{
"input": "2 2\nG G\nC C",
"output": "#Color"
},
{
"input": "1 2\nB C",
"output": "#Color"
},
{
"input": "1 4\nG G G C",
"output": "#Color"
},
{
"input": "1 1\nB",
"output": "#Black&White"
},
{
"input": "2 1\nY\nB",
"output": "#Color"
}
] | 1,687,019,845
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 50
| 46
| 0
|
# Link: https://codeforces.com/contest/707/problem/A
color_photo = False
rows, columns = map(int, input().split())
for _ in range(rows):
row_pixels = input().split()
if not color_photo:
if row_pixels.count('C') > 0 or row_pixels.count('M') > 0 or row_pixels.count('Y') > 0:
color_photo = True
print("#Color" if color_photo else "#Black&White")
|
Title: Brain's Photos
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Small, but very brave, mouse Brain was not accepted to summer school of young villains. He was upset and decided to postpone his plans of taking over the world, but to become a photographer instead.
As you may know, the coolest photos are on the film (because you can specify the hashtag #film for such).
Brain took a lot of colourful pictures on colored and black-and-white film. Then he developed and translated it into a digital form. But now, color and black-and-white photos are in one folder, and to sort them, one needs to spend more than one hour!
As soon as Brain is a photographer not programmer now, he asks you to help him determine for a single photo whether it is colored or black-and-white.
Photo can be represented as a matrix sized *n*<=×<=*m*, and each element of the matrix stores a symbol indicating corresponding pixel color. There are only 6 colors:
- 'C' (cyan)- 'M' (magenta)- 'Y' (yellow)- 'W' (white)- 'G' (grey)- 'B' (black)
The photo is considered black-and-white if it has only white, black and grey pixels in it. If there are any of cyan, magenta or yellow pixels in the photo then it is considered colored.
Input Specification:
The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of photo pixel matrix rows and columns respectively.
Then *n* lines describing matrix rows follow. Each of them contains *m* space-separated characters describing colors of pixels in a row. Each character in the line is one of the 'C', 'M', 'Y', 'W', 'G' or 'B'.
Output Specification:
Print the "#Black&White" (without quotes), if the photo is black-and-white and "#Color" (without quotes), if it is colored, in the only line.
Demo Input:
['2 2\nC M\nY Y\n', '3 2\nW W\nW W\nB B\n', '1 1\nW\n']
Demo Output:
['#Color', '#Black&White', '#Black&White']
Note:
none
|
```python
# Link: https://codeforces.com/contest/707/problem/A
color_photo = False
rows, columns = map(int, input().split())
for _ in range(rows):
row_pixels = input().split()
if not color_photo:
if row_pixels.count('C') > 0 or row_pixels.count('M') > 0 or row_pixels.count('Y') > 0:
color_photo = True
print("#Color" if color_photo else "#Black&White")
```
| 3
|
|
610
|
A
|
Pasha and Stick
|
PROGRAMMING
| 1,000
|
[
"combinatorics",
"math"
] | null | null |
Pasha has a wooden stick of some positive integer length *n*. He wants to perform exactly three cuts to get four parts of the stick. Each part must have some positive integer length and the sum of these lengths will obviously be *n*.
Pasha likes rectangles but hates squares, so he wonders, how many ways are there to split a stick into four parts so that it's possible to form a rectangle using these parts, but is impossible to form a square.
Your task is to help Pasha and count the number of such ways. Two ways to cut the stick are considered distinct if there exists some integer *x*, such that the number of parts of length *x* in the first way differ from the number of parts of length *x* in the second way.
|
The first line of the input contains a positive integer *n* (1<=≤<=*n*<=≤<=2·109) — the length of Pasha's stick.
|
The output should contain a single integer — the number of ways to split Pasha's stick into four parts of positive integer length so that it's possible to make a rectangle by connecting the ends of these parts, but is impossible to form a square.
|
[
"6\n",
"20\n"
] |
[
"1\n",
"4\n"
] |
There is only one way to divide the stick in the first sample {1, 1, 2, 2}.
Four ways to divide the stick in the second sample are {1, 1, 9, 9}, {2, 2, 8, 8}, {3, 3, 7, 7} and {4, 4, 6, 6}. Note that {5, 5, 5, 5} doesn't work.
| 500
|
[
{
"input": "6",
"output": "1"
},
{
"input": "20",
"output": "4"
},
{
"input": "1",
"output": "0"
},
{
"input": "2",
"output": "0"
},
{
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{
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{
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{
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{
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{
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{
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{
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{
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{
"input": "1980649112",
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{
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{
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"output": "0"
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{
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{
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{
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{
"input": "1934896224",
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{
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{
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{
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{
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{
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{
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{
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{
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{
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{
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{
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{
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"output": "498971656"
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{
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"output": "0"
},
{
"input": "1996193634",
"output": "499048408"
},
{
"input": "9605180",
"output": "2401294"
},
{
"input": "1996459740",
"output": "499114934"
},
{
"input": "32691948",
"output": "8172986"
},
{
"input": "1975903308",
"output": "493975826"
},
{
"input": "1976637136",
"output": "494159283"
},
{
"input": "29803038",
"output": "7450759"
},
{
"input": "1977979692",
"output": "494494922"
},
{
"input": "1978595336",
"output": "494648833"
},
{
"input": "27379344",
"output": "6844835"
},
{
"input": "1979729912",
"output": "494932477"
},
{
"input": "1980253780",
"output": "495063444"
},
{
"input": "1980751584",
"output": "495187895"
},
{
"input": "53224878",
"output": "13306219"
},
{
"input": "5",
"output": "0"
},
{
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"output": "0"
},
{
"input": "8",
"output": "1"
},
{
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"output": "0"
},
{
"input": "10",
"output": "2"
},
{
"input": "11",
"output": "0"
},
{
"input": "12",
"output": "2"
},
{
"input": "13",
"output": "0"
},
{
"input": "14",
"output": "3"
},
{
"input": "15",
"output": "0"
},
{
"input": "16",
"output": "3"
},
{
"input": "17",
"output": "0"
},
{
"input": "18",
"output": "4"
},
{
"input": "19",
"output": "0"
},
{
"input": "21",
"output": "0"
},
{
"input": "22",
"output": "5"
},
{
"input": "23",
"output": "0"
},
{
"input": "24",
"output": "5"
},
{
"input": "25",
"output": "0"
},
{
"input": "26",
"output": "6"
},
{
"input": "27",
"output": "0"
},
{
"input": "28",
"output": "6"
},
{
"input": "29",
"output": "0"
},
{
"input": "30",
"output": "7"
},
{
"input": "111",
"output": "0"
},
{
"input": "55",
"output": "0"
},
{
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{
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"output": "0"
},
{
"input": "151",
"output": "0"
}
] | 1,664,954,471
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 76
| 62
| 0
|
k = int(input())
if k %2 == 1:
k = 1
n = int(k)//2
print(max(0,(n+1)//2-1))
|
Title: Pasha and Stick
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pasha has a wooden stick of some positive integer length *n*. He wants to perform exactly three cuts to get four parts of the stick. Each part must have some positive integer length and the sum of these lengths will obviously be *n*.
Pasha likes rectangles but hates squares, so he wonders, how many ways are there to split a stick into four parts so that it's possible to form a rectangle using these parts, but is impossible to form a square.
Your task is to help Pasha and count the number of such ways. Two ways to cut the stick are considered distinct if there exists some integer *x*, such that the number of parts of length *x* in the first way differ from the number of parts of length *x* in the second way.
Input Specification:
The first line of the input contains a positive integer *n* (1<=≤<=*n*<=≤<=2·109) — the length of Pasha's stick.
Output Specification:
The output should contain a single integer — the number of ways to split Pasha's stick into four parts of positive integer length so that it's possible to make a rectangle by connecting the ends of these parts, but is impossible to form a square.
Demo Input:
['6\n', '20\n']
Demo Output:
['1\n', '4\n']
Note:
There is only one way to divide the stick in the first sample {1, 1, 2, 2}.
Four ways to divide the stick in the second sample are {1, 1, 9, 9}, {2, 2, 8, 8}, {3, 3, 7, 7} and {4, 4, 6, 6}. Note that {5, 5, 5, 5} doesn't work.
|
```python
k = int(input())
if k %2 == 1:
k = 1
n = int(k)//2
print(max(0,(n+1)//2-1))
```
| 3
|
|
980
|
D
|
Perfect Groups
|
PROGRAMMING
| 2,100
|
[
"dp",
"math",
"number theory"
] | null | null |
SaMer has written the greatest test case of all time for one of his problems. For a given array of integers, the problem asks to find the minimum number of groups the array can be divided into, such that the product of any pair of integers in the same group is a perfect square.
Each integer must be in exactly one group. However, integers in a group do not necessarily have to be contiguous in the array.
SaMer wishes to create more cases from the test case he already has. His test case has an array $A$ of $n$ integers, and he needs to find the number of contiguous subarrays of $A$ that have an answer to the problem equal to $k$ for each integer $k$ between $1$ and $n$ (inclusive).
|
The first line of input contains a single integer $n$ ($1 \leq n \leq 5000$), the size of the array.
The second line contains $n$ integers $a_1$,$a_2$,$\dots$,$a_n$ ($-10^8 \leq a_i \leq 10^8$), the values of the array.
|
Output $n$ space-separated integers, the $k$-th integer should be the number of contiguous subarrays of $A$ that have an answer to the problem equal to $k$.
|
[
"2\n5 5\n",
"5\n5 -4 2 1 8\n",
"1\n0\n"
] |
[
"3 0\n",
"5 5 3 2 0\n",
"1\n"
] |
none
| 2,000
|
[
{
"input": "2\n5 5",
"output": "3 0"
},
{
"input": "5\n5 -4 2 1 8",
"output": "5 5 3 2 0"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "3\n-10 -5 7",
"output": "3 2 1"
},
{
"input": "5\n-6 -7 -2 -3 -10",
"output": "5 4 3 2 1"
},
{
"input": "8\n-5 5 10 -5 9 -2 5 7",
"output": "8 7 7 5 6 3 0 0"
},
{
"input": "19\n-51 -47 -57 57 -62 41 42 53 47 -7 -41 -37 -41 -46 -47 41 -39 -59 -50",
"output": "19 19 18 17 16 15 14 13 12 11 10 11 7 4 3 1 0 0 0"
},
{
"input": "25\n94 -94 61 -98 89 -91 83 61 -3 53 91 61 -3 -3 95 -67 -73 -59 -73 37 -47 -3 94 -79 -73",
"output": "26 26 25 29 27 26 27 27 23 20 14 16 10 10 7 8 4 0 0 0 0 0 0 0 0"
},
{
"input": "80\n8861 -8846 -3257 8263 -8045 4549 9626 -8599 5755 -3559 5813 -7411 9151 -1847 2441 4201 2381 4651 -6571 199 -6737 -6333 -9433 -4967 9041 -9319 6801 5813 -2731 -8599 7681 -1303 -7333 6067 -1847 -6333 6047 -7907 -7333 8677 -937 829 2441 -5702 -8045 -3037 -4882 -7703 9001 -5702 -2039 -7493 9173 -6047 6046 5802 -8609 9173 8083 8291 -1453 2381 7321 -859 8539 -2011 2374 8263 8111 -5897 -2017 -2311 -571 4993 -3389 -7789 1847 -4799 4651 -5702",
"output": "80 79 78 77 77 78 77 76 75 74 73 72 72 71 70 71 69 68 67 66 68 66 69 64 66 65 63 60 58 57 55 56 52 53 48 48 46 45 44 41 42 42 39 39 36 35 33 33 31 29 28 27 25 26 23 22 25 22 15 16 12 13 9 12 5 4 3 0 0 0 0 0 0 0 0 0 0 0 0 0"
},
{
"input": "1\n-3622679",
"output": "1"
},
{
"input": "2\n6649609 -8006683",
"output": "2 1"
},
{
"input": "4\n0 -7955897 3123413 -7955897",
"output": "5 5 0 0"
},
{
"input": "3\n1 2 0",
"output": "4 2 0"
},
{
"input": "2\n-7 0",
"output": "3 0"
},
{
"input": "5\n5 0 0 0 0",
"output": "15 0 0 0 0"
},
{
"input": "3\n0 1 2",
"output": "4 2 0"
},
{
"input": "30\n1 2 0 2 1 0 0 1 2 0 1 2 3 0 4 1 0 0 0 3 2 0 1 0 1 0 5 0 6 0",
"output": "77 90 216 42 40 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0"
},
{
"input": "2\n2 0",
"output": "3 0"
},
{
"input": "10\n0 0 5 0 0 5 0 0 5 0",
"output": "55 0 0 0 0 0 0 0 0 0"
},
{
"input": "2\n0 2",
"output": "3 0"
},
{
"input": "4\n1 0 1 0",
"output": "10 0 0 0"
},
{
"input": "2\n0 1",
"output": "3 0"
},
{
"input": "6\n5 -4 0 2 1 8",
"output": "8 8 3 2 0 0"
},
{
"input": "3\n1 0 1",
"output": "6 0 0"
},
{
"input": "3\n1 0 -1",
"output": "5 1 0"
},
{
"input": "5\n0 1 2 3 4",
"output": "6 4 5 0 0"
}
] | 1,690,487,390
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 46
| 0
|
print("_RANDOM_GUESS_1690487390.8621504")# 1690487390.8621697
|
Title: Perfect Groups
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
SaMer has written the greatest test case of all time for one of his problems. For a given array of integers, the problem asks to find the minimum number of groups the array can be divided into, such that the product of any pair of integers in the same group is a perfect square.
Each integer must be in exactly one group. However, integers in a group do not necessarily have to be contiguous in the array.
SaMer wishes to create more cases from the test case he already has. His test case has an array $A$ of $n$ integers, and he needs to find the number of contiguous subarrays of $A$ that have an answer to the problem equal to $k$ for each integer $k$ between $1$ and $n$ (inclusive).
Input Specification:
The first line of input contains a single integer $n$ ($1 \leq n \leq 5000$), the size of the array.
The second line contains $n$ integers $a_1$,$a_2$,$\dots$,$a_n$ ($-10^8 \leq a_i \leq 10^8$), the values of the array.
Output Specification:
Output $n$ space-separated integers, the $k$-th integer should be the number of contiguous subarrays of $A$ that have an answer to the problem equal to $k$.
Demo Input:
['2\n5 5\n', '5\n5 -4 2 1 8\n', '1\n0\n']
Demo Output:
['3 0\n', '5 5 3 2 0\n', '1\n']
Note:
none
|
```python
print("_RANDOM_GUESS_1690487390.8621504")# 1690487390.8621697
```
| 0
|
|
510
|
A
|
Fox And Snake
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Fox Ciel starts to learn programming. The first task is drawing a fox! However, that turns out to be too hard for a beginner, so she decides to draw a snake instead.
A snake is a pattern on a *n* by *m* table. Denote *c*-th cell of *r*-th row as (*r*,<=*c*). The tail of the snake is located at (1,<=1), then it's body extends to (1,<=*m*), then goes down 2 rows to (3,<=*m*), then goes left to (3,<=1) and so on.
Your task is to draw this snake for Fox Ciel: the empty cells should be represented as dot characters ('.') and the snake cells should be filled with number signs ('#').
Consider sample tests in order to understand the snake pattern.
|
The only line contains two integers: *n* and *m* (3<=≤<=*n*,<=*m*<=≤<=50).
*n* is an odd number.
|
Output *n* lines. Each line should contain a string consisting of *m* characters. Do not output spaces.
|
[
"3 3\n",
"3 4\n",
"5 3\n",
"9 9\n"
] |
[
"###\n..#\n###\n",
"####\n...#\n####\n",
"###\n..#\n###\n#..\n###\n",
"#########\n........#\n#########\n#........\n#########\n........#\n#########\n#........\n#########\n"
] |
none
| 500
|
[
{
"input": "3 3",
"output": "###\n..#\n###"
},
{
"input": "3 4",
"output": "####\n...#\n####"
},
{
"input": "5 3",
"output": "###\n..#\n###\n#..\n###"
},
{
"input": "9 9",
"output": "#########\n........#\n#########\n#........\n#########\n........#\n#########\n#........\n#########"
},
{
"input": "3 5",
"output": "#####\n....#\n#####"
},
{
"input": "3 6",
"output": "######\n.....#\n######"
},
{
"input": "7 3",
"output": "###\n..#\n###\n#..\n###\n..#\n###"
},
{
"input": "7 4",
"output": "####\n...#\n####\n#...\n####\n...#\n####"
},
{
"input": "49 50",
"output": "##################################################\n.................................................#\n##################################################\n#.................................................\n##################################################\n.................................................#\n##################################################\n#.................................................\n##################################################\n.............................................."
},
{
"input": "43 50",
"output": "##################################################\n.................................................#\n##################################################\n#.................................................\n##################################################\n.................................................#\n##################################################\n#.................................................\n##################################################\n.............................................."
},
{
"input": "43 27",
"output": "###########################\n..........................#\n###########################\n#..........................\n###########################\n..........................#\n###########################\n#..........................\n###########################\n..........................#\n###########################\n#..........................\n###########################\n..........................#\n###########################\n#..........................\n###########################\n....................."
},
{
"input": "11 15",
"output": "###############\n..............#\n###############\n#..............\n###############\n..............#\n###############\n#..............\n###############\n..............#\n###############"
},
{
"input": "11 3",
"output": "###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###"
},
{
"input": "19 3",
"output": "###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###"
},
{
"input": "23 50",
"output": "##################################################\n.................................................#\n##################################################\n#.................................................\n##################################################\n.................................................#\n##################################################\n#.................................................\n##################################################\n.............................................."
},
{
"input": "49 49",
"output": "#################################################\n................................................#\n#################################################\n#................................................\n#################################################\n................................................#\n#################################################\n#................................................\n#################################################\n................................................#\n#..."
},
{
"input": "33 43",
"output": "###########################################\n..........................................#\n###########################################\n#..........................................\n###########################################\n..........................................#\n###########################################\n#..........................................\n###########################################\n..........................................#\n###########################################\n#.................."
},
{
"input": "33 44",
"output": "############################################\n...........................................#\n############################################\n#...........................................\n############################################\n...........................................#\n############################################\n#...........................................\n############################################\n...........................................#\n############################################\n#......."
},
{
"input": "45 45",
"output": "#############################################\n............................................#\n#############################################\n#............................................\n#############################################\n............................................#\n#############################################\n#............................................\n#############################################\n............................................#\n#########################################..."
},
{
"input": "45 49",
"output": "#################################################\n................................................#\n#################################################\n#................................................\n#################################################\n................................................#\n#################################################\n#................................................\n#################################################\n................................................#\n#..."
},
{
"input": "45 3",
"output": "###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###"
},
{
"input": "45 4",
"output": "####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####"
},
{
"input": "5 4",
"output": "####\n...#\n####\n#...\n####"
},
{
"input": "5 5",
"output": "#####\n....#\n#####\n#....\n#####"
},
{
"input": "5 6",
"output": "######\n.....#\n######\n#.....\n######"
},
{
"input": "7 5",
"output": "#####\n....#\n#####\n#....\n#####\n....#\n#####"
},
{
"input": "7 6",
"output": "######\n.....#\n######\n#.....\n######\n.....#\n######"
},
{
"input": "7 7",
"output": "#######\n......#\n#######\n#......\n#######\n......#\n#######"
}
] | 1,691,519,675
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 28
| 46
| 0
|
def fox(n, m):
for item in range (1, n+1):
if item%2 == 1:
print('#'*m)
elif item%4 != 0:
print('.'*(m-1), end = '')
print('#')
else:
print('#', end = '')
print('.'*(m-1))
if __name__ == "__main__":
n, m = map(int, input().split())
fox(n, m)
|
Title: Fox And Snake
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Fox Ciel starts to learn programming. The first task is drawing a fox! However, that turns out to be too hard for a beginner, so she decides to draw a snake instead.
A snake is a pattern on a *n* by *m* table. Denote *c*-th cell of *r*-th row as (*r*,<=*c*). The tail of the snake is located at (1,<=1), then it's body extends to (1,<=*m*), then goes down 2 rows to (3,<=*m*), then goes left to (3,<=1) and so on.
Your task is to draw this snake for Fox Ciel: the empty cells should be represented as dot characters ('.') and the snake cells should be filled with number signs ('#').
Consider sample tests in order to understand the snake pattern.
Input Specification:
The only line contains two integers: *n* and *m* (3<=≤<=*n*,<=*m*<=≤<=50).
*n* is an odd number.
Output Specification:
Output *n* lines. Each line should contain a string consisting of *m* characters. Do not output spaces.
Demo Input:
['3 3\n', '3 4\n', '5 3\n', '9 9\n']
Demo Output:
['###\n..#\n###\n', '####\n...#\n####\n', '###\n..#\n###\n#..\n###\n', '#########\n........#\n#########\n#........\n#########\n........#\n#########\n#........\n#########\n']
Note:
none
|
```python
def fox(n, m):
for item in range (1, n+1):
if item%2 == 1:
print('#'*m)
elif item%4 != 0:
print('.'*(m-1), end = '')
print('#')
else:
print('#', end = '')
print('.'*(m-1))
if __name__ == "__main__":
n, m = map(int, input().split())
fox(n, m)
```
| 3
|
|
445
|
A
|
DZY Loves Chessboard
|
PROGRAMMING
| 1,200
|
[
"dfs and similar",
"implementation"
] | null | null |
DZY loves chessboard, and he enjoys playing with it.
He has a chessboard of *n* rows and *m* columns. Some cells of the chessboard are bad, others are good. For every good cell, DZY wants to put a chessman on it. Each chessman is either white or black. After putting all chessmen, DZY wants that no two chessmen with the same color are on two adjacent cells. Two cells are adjacent if and only if they share a common edge.
You task is to find any suitable placement of chessmen on the given chessboard.
|
The first line contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
Each of the next *n* lines contains a string of *m* characters: the *j*-th character of the *i*-th string is either "." or "-". A "." means that the corresponding cell (in the *i*-th row and the *j*-th column) is good, while a "-" means it is bad.
|
Output must contain *n* lines, each line must contain a string of *m* characters. The *j*-th character of the *i*-th string should be either "W", "B" or "-". Character "W" means the chessman on the cell is white, "B" means it is black, "-" means the cell is a bad cell.
If multiple answers exist, print any of them. It is guaranteed that at least one answer exists.
|
[
"1 1\n.\n",
"2 2\n..\n..\n",
"3 3\n.-.\n---\n--."
] |
[
"B\n",
"BW\nWB\n",
"B-B\n---\n--B"
] |
In the first sample, DZY puts a single black chessman. Of course putting a white one is also OK.
In the second sample, all 4 cells are good. No two same chessmen share an edge in the sample output.
In the third sample, no good cells are adjacent. So you can just put 3 chessmen, no matter what their colors are.
| 500
|
[
{
"input": "1 1\n.",
"output": "B"
},
{
"input": "2 2\n..\n..",
"output": "BW\nWB"
},
{
"input": "3 3\n.-.\n---\n--.",
"output": "B-B\n---\n--B"
},
{
"input": "3 1\n-\n.\n.",
"output": "-\nW\nB"
},
{
"input": "11 11\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------",
"output": "-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------"
},
{
"input": "1 1\n-",
"output": "-"
},
{
"input": "4 2\n..\n..\n..\n..",
"output": "BW\nWB\nBW\nWB"
},
{
"input": "5 1\n.\n.\n.\n.\n.",
"output": "B\nW\nB\nW\nB"
}
] | 1,671,616,511
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 1
| 31
| 0
|
n, m = map(int, input().split())
x = []
for i in range(n):
l = list(input().split())
x.append(l)
for i in range(n):
for j in range(m):
if x[i][j] == '.':
if i > 0 and j > 0:
if x[i][j - 1] == 'B':
x[i][j] = 'W'
elif x[i][j - 1] == 'W':
x[i][j] = 'B'
elif x[i][j + 1] == 'B':
x[i][j] = 'W'
elif x[i][j + 1] == 'W':
x[i][j] = 'B'
elif x[i + 1][j] == 'B':
x[i][j] = 'W'
elif x[i + 1][j] == 'W':
x[i][j] = 'B'
elif x[i - 1][j] == 'W':
x[i][j] = 'B'
elif x[i - 1][j] == 'B':
x[i - 1][j] = 'W'
else:
x[i][j] = 'W'
elif i > 0 and j == 0:
if x[i - 1][j] == 'B':
x[i][j] = 'W'
elif x[i - 1][j] == 'W':
x[i][j] = 'B'
elif x[i + 1][j] == 'B':
x[i][j] = 'W'
elif x[i + 1][j] == 'W':
x[i][j] = 'B'
elif x[i][j + 1] == 'B':
x[i][j] = 'W'
elif x[i][j + 1] == 'W':
x[i][j] = 'B'
else:
x[i][j] = 'W'
elif i == 0 and j > 0:
if x[i][j - 1] == 'B':
x[i][j] = 'W'
elif x[i][j - 1] == 'W':
x[i][j] = 'B'
elif x[i][j + 1] == 'B':
x[i][j] = 'W'
elif x[i][j + 1] == 'W':
x[i][j] = 'B'
elif x[i + 1][j] == 'B':
x[i][j] = 'W'
elif x[i + 1][j] == 'W':
x[i][j] = 'B'
else:
x[i][j] = 'W'
else:
x[i][j] = 'B'
for i in range(n):
for j in range(m):
print(x[i][j], end="")
print('')
|
Title: DZY Loves Chessboard
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
DZY loves chessboard, and he enjoys playing with it.
He has a chessboard of *n* rows and *m* columns. Some cells of the chessboard are bad, others are good. For every good cell, DZY wants to put a chessman on it. Each chessman is either white or black. After putting all chessmen, DZY wants that no two chessmen with the same color are on two adjacent cells. Two cells are adjacent if and only if they share a common edge.
You task is to find any suitable placement of chessmen on the given chessboard.
Input Specification:
The first line contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
Each of the next *n* lines contains a string of *m* characters: the *j*-th character of the *i*-th string is either "." or "-". A "." means that the corresponding cell (in the *i*-th row and the *j*-th column) is good, while a "-" means it is bad.
Output Specification:
Output must contain *n* lines, each line must contain a string of *m* characters. The *j*-th character of the *i*-th string should be either "W", "B" or "-". Character "W" means the chessman on the cell is white, "B" means it is black, "-" means the cell is a bad cell.
If multiple answers exist, print any of them. It is guaranteed that at least one answer exists.
Demo Input:
['1 1\n.\n', '2 2\n..\n..\n', '3 3\n.-.\n---\n--.']
Demo Output:
['B\n', 'BW\nWB\n', 'B-B\n---\n--B']
Note:
In the first sample, DZY puts a single black chessman. Of course putting a white one is also OK.
In the second sample, all 4 cells are good. No two same chessmen share an edge in the sample output.
In the third sample, no good cells are adjacent. So you can just put 3 chessmen, no matter what their colors are.
|
```python
n, m = map(int, input().split())
x = []
for i in range(n):
l = list(input().split())
x.append(l)
for i in range(n):
for j in range(m):
if x[i][j] == '.':
if i > 0 and j > 0:
if x[i][j - 1] == 'B':
x[i][j] = 'W'
elif x[i][j - 1] == 'W':
x[i][j] = 'B'
elif x[i][j + 1] == 'B':
x[i][j] = 'W'
elif x[i][j + 1] == 'W':
x[i][j] = 'B'
elif x[i + 1][j] == 'B':
x[i][j] = 'W'
elif x[i + 1][j] == 'W':
x[i][j] = 'B'
elif x[i - 1][j] == 'W':
x[i][j] = 'B'
elif x[i - 1][j] == 'B':
x[i - 1][j] = 'W'
else:
x[i][j] = 'W'
elif i > 0 and j == 0:
if x[i - 1][j] == 'B':
x[i][j] = 'W'
elif x[i - 1][j] == 'W':
x[i][j] = 'B'
elif x[i + 1][j] == 'B':
x[i][j] = 'W'
elif x[i + 1][j] == 'W':
x[i][j] = 'B'
elif x[i][j + 1] == 'B':
x[i][j] = 'W'
elif x[i][j + 1] == 'W':
x[i][j] = 'B'
else:
x[i][j] = 'W'
elif i == 0 and j > 0:
if x[i][j - 1] == 'B':
x[i][j] = 'W'
elif x[i][j - 1] == 'W':
x[i][j] = 'B'
elif x[i][j + 1] == 'B':
x[i][j] = 'W'
elif x[i][j + 1] == 'W':
x[i][j] = 'B'
elif x[i + 1][j] == 'B':
x[i][j] = 'W'
elif x[i + 1][j] == 'W':
x[i][j] = 'B'
else:
x[i][j] = 'W'
else:
x[i][j] = 'B'
for i in range(n):
for j in range(m):
print(x[i][j], end="")
print('')
```
| -1
|
|
755
|
B
|
PolandBall and Game
|
PROGRAMMING
| 1,100
|
[
"binary search",
"data structures",
"games",
"greedy",
"sortings",
"strings"
] | null | null |
PolandBall is playing a game with EnemyBall. The rules are simple. Players have to say words in turns. You cannot say a word which was already said. PolandBall starts. The Ball which can't say a new word loses.
You're given two lists of words familiar to PolandBall and EnemyBall. Can you determine who wins the game, if both play optimally?
|
The first input line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=103) — number of words PolandBall and EnemyBall know, respectively.
Then *n* strings follow, one per line — words familiar to PolandBall.
Then *m* strings follow, one per line — words familiar to EnemyBall.
Note that one Ball cannot know a word more than once (strings are unique), but some words can be known by both players.
Each word is non-empty and consists of no more than 500 lowercase English alphabet letters.
|
In a single line of print the answer — "YES" if PolandBall wins and "NO" otherwise. Both Balls play optimally.
|
[
"5 1\npolandball\nis\na\ncool\ncharacter\nnope\n",
"2 2\nkremowka\nwadowicka\nkremowka\nwiedenska\n",
"1 2\na\na\nb\n"
] |
[
"YES",
"YES",
"NO"
] |
In the first example PolandBall knows much more words and wins effortlessly.
In the second example if PolandBall says kremowka first, then EnemyBall cannot use that word anymore. EnemyBall can only say wiedenska. PolandBall says wadowicka and wins.
| 1,000
|
[
{
"input": "5 1\npolandball\nis\na\ncool\ncharacter\nnope",
"output": "YES"
},
{
"input": "2 2\nkremowka\nwadowicka\nkremowka\nwiedenska",
"output": "YES"
},
{
"input": "1 2\na\na\nb",
"output": "NO"
},
{
"input": "2 2\na\nb\nb\nc",
"output": "YES"
},
{
"input": "2 1\nc\na\na",
"output": "YES"
},
{
"input": "3 3\nab\nbc\ncd\ncd\ndf\nfg",
"output": "YES"
},
{
"input": "3 3\nc\na\nb\na\nd\ng",
"output": "YES"
},
{
"input": "1 1\naa\naa",
"output": "YES"
},
{
"input": "2 1\na\nb\na",
"output": "YES"
},
{
"input": "6 5\na\nb\nc\nd\ne\nf\nf\ne\nd\nz\ny",
"output": "YES"
},
{
"input": "3 2\na\nb\nc\nd\ne",
"output": "YES"
}
] | 1,614,283,083
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 62
| 307,200
|
n, m = map(int, input().split())
poland_words = []
enemy_words = []
shared_words = []
only_poland_words = []
only_enemy_words = []
for i in range(n):
word = input()
poland_words.append(word)
for i in range(m):
word = input()
enemy_words.append(word)
shared_words = [word for word in poland_words if word in enemy_words]
only_poland_words = [word for word in poland_words if word not in shared_words]
only_enemy_words = [word for word in enemy_words if word not in shared_words]
if len(only_poland_words) + len(shared_words) // 2 + 1 > len(only_enemy_words) + len(shared_words) // 2:
print("YES")
else:
print("NO")
|
Title: PolandBall and Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
PolandBall is playing a game with EnemyBall. The rules are simple. Players have to say words in turns. You cannot say a word which was already said. PolandBall starts. The Ball which can't say a new word loses.
You're given two lists of words familiar to PolandBall and EnemyBall. Can you determine who wins the game, if both play optimally?
Input Specification:
The first input line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=103) — number of words PolandBall and EnemyBall know, respectively.
Then *n* strings follow, one per line — words familiar to PolandBall.
Then *m* strings follow, one per line — words familiar to EnemyBall.
Note that one Ball cannot know a word more than once (strings are unique), but some words can be known by both players.
Each word is non-empty and consists of no more than 500 lowercase English alphabet letters.
Output Specification:
In a single line of print the answer — "YES" if PolandBall wins and "NO" otherwise. Both Balls play optimally.
Demo Input:
['5 1\npolandball\nis\na\ncool\ncharacter\nnope\n', '2 2\nkremowka\nwadowicka\nkremowka\nwiedenska\n', '1 2\na\na\nb\n']
Demo Output:
['YES', 'YES', 'NO']
Note:
In the first example PolandBall knows much more words and wins effortlessly.
In the second example if PolandBall says kremowka first, then EnemyBall cannot use that word anymore. EnemyBall can only say wiedenska. PolandBall says wadowicka and wins.
|
```python
n, m = map(int, input().split())
poland_words = []
enemy_words = []
shared_words = []
only_poland_words = []
only_enemy_words = []
for i in range(n):
word = input()
poland_words.append(word)
for i in range(m):
word = input()
enemy_words.append(word)
shared_words = [word for word in poland_words if word in enemy_words]
only_poland_words = [word for word in poland_words if word not in shared_words]
only_enemy_words = [word for word in enemy_words if word not in shared_words]
if len(only_poland_words) + len(shared_words) // 2 + 1 > len(only_enemy_words) + len(shared_words) // 2:
print("YES")
else:
print("NO")
```
| 0
|
|
443
|
A
|
Anton and Letters
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"implementation"
] | null | null |
Recently, Anton has found a set. The set consists of small English letters. Anton carefully wrote out all the letters from the set in one line, separated by a comma. He also added an opening curved bracket at the beginning of the line and a closing curved bracket at the end of the line.
Unfortunately, from time to time Anton would forget writing some letter and write it again. He asks you to count the total number of distinct letters in his set.
|
The first and the single line contains the set of letters. The length of the line doesn't exceed 1000. It is guaranteed that the line starts from an opening curved bracket and ends with a closing curved bracket. Between them, small English letters are listed, separated by a comma. Each comma is followed by a space.
|
Print a single number — the number of distinct letters in Anton's set.
|
[
"{a, b, c}\n",
"{b, a, b, a}\n",
"{}\n"
] |
[
"3\n",
"2\n",
"0\n"
] |
none
| 500
|
[
{
"input": "{a, b, c}",
"output": "3"
},
{
"input": "{b, a, b, a}",
"output": "2"
},
{
"input": "{}",
"output": "0"
},
{
"input": "{a, a, c, b, b, b, c, c, c, c}",
"output": "3"
},
{
"input": "{a, c, b, b}",
"output": "3"
},
{
"input": "{a, b}",
"output": "2"
},
{
"input": "{a}",
"output": "1"
},
{
"input": "{b, a, b, a, b, c, c, b, c, b}",
"output": "3"
},
{
"input": "{e, g, c, e}",
"output": "3"
},
{
"input": "{a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a}",
"output": "1"
},
{
"input": "{a, a, b}",
"output": "2"
},
{
"input": "{a, b, b, b, a, b, a, a, a, a, a, a, b, a, b, a, a, a, a, a, b, a, b, a}",
"output": "2"
},
{
"input": "{j, u, a, c, f, w, e, w, x, t, h, p, v, n, i, l, x, n, i, b, u, c, a, a}",
"output": "16"
},
{
"input": "{x, i, w, c, p, e, h, z, k, i}",
"output": "9"
},
{
"input": "{t, k, o, x, r, d, q, j, k, e, z, w, y, r, z, s, s, e, s, b, k, i}",
"output": "15"
},
{
"input": "{y}",
"output": "1"
},
{
"input": "{x}",
"output": "1"
},
{
"input": "{b, z, a, z}",
"output": "3"
},
{
"input": "{z}",
"output": "1"
},
{
"input": "{a, z}",
"output": "2"
},
{
"input": "{a, b, z}",
"output": "3"
},
{
"input": "{s, q, z, r, t, a, b, h, j, i, o, z, r, q}",
"output": "11"
}
] | 1,691,762,944
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 62
| 0
|
input_line = input() # Input line containing the set of letters
# Remove opening and closing curly brackets
input_line = input_line.strip('{}')
# Split the string into individual letters using comma as delimiter
letters = input_line.split(', ')
# Convert the list of letters into a set to remove duplicates
distinct_letters = set(letters)
# Count the number of distinct letters
num_distinct_letters = len(distinct_letters)
print(num_distinct_letters)
|
Title: Anton and Letters
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently, Anton has found a set. The set consists of small English letters. Anton carefully wrote out all the letters from the set in one line, separated by a comma. He also added an opening curved bracket at the beginning of the line and a closing curved bracket at the end of the line.
Unfortunately, from time to time Anton would forget writing some letter and write it again. He asks you to count the total number of distinct letters in his set.
Input Specification:
The first and the single line contains the set of letters. The length of the line doesn't exceed 1000. It is guaranteed that the line starts from an opening curved bracket and ends with a closing curved bracket. Between them, small English letters are listed, separated by a comma. Each comma is followed by a space.
Output Specification:
Print a single number — the number of distinct letters in Anton's set.
Demo Input:
['{a, b, c}\n', '{b, a, b, a}\n', '{}\n']
Demo Output:
['3\n', '2\n', '0\n']
Note:
none
|
```python
input_line = input() # Input line containing the set of letters
# Remove opening and closing curly brackets
input_line = input_line.strip('{}')
# Split the string into individual letters using comma as delimiter
letters = input_line.split(', ')
# Convert the list of letters into a set to remove duplicates
distinct_letters = set(letters)
# Count the number of distinct letters
num_distinct_letters = len(distinct_letters)
print(num_distinct_letters)
```
| 0
|
|
41
|
A
|
Translation
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] |
A. Translation
|
2
|
256
|
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
|
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
|
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
|
[
"code\nedoc\n",
"abb\naba\n",
"code\ncode\n"
] |
[
"YES\n",
"NO\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "code\nedoc",
"output": "YES"
},
{
"input": "abb\naba",
"output": "NO"
},
{
"input": "code\ncode",
"output": "NO"
},
{
"input": "abacaba\nabacaba",
"output": "YES"
},
{
"input": "q\nq",
"output": "YES"
},
{
"input": "asrgdfngfnmfgnhweratgjkk\nasrgdfngfnmfgnhweratgjkk",
"output": "NO"
},
{
"input": "z\na",
"output": "NO"
},
{
"input": "asd\ndsa",
"output": "YES"
},
{
"input": "abcdef\nfecdba",
"output": "NO"
},
{
"input": "ywjjbirapvskozubvxoemscfwl\ngnduubaogtfaiowjizlvjcu",
"output": "NO"
},
{
"input": "mfrmqxtzvgaeuleubcmcxcfqyruwzenguhgrmkuhdgnhgtgkdszwqyd\nmfxufheiperjnhyczclkmzyhcxntdfskzkzdwzzujdinf",
"output": "NO"
},
{
"input": "bnbnemvybqizywlnghlykniaxxxlkhftppbdeqpesrtgkcpoeqowjwhrylpsziiwcldodcoonpimudvrxejjo\ntiynnekmlalogyvrgptbinkoqdwzuiyjlrldxhzjmmp",
"output": "NO"
},
{
"input": "pwlpubwyhzqvcitemnhvvwkmwcaawjvdiwtoxyhbhbxerlypelevasmelpfqwjk\nstruuzebbcenziscuoecywugxncdwzyfozhljjyizpqcgkyonyetarcpwkqhuugsqjuixsxptmbnlfupdcfigacdhhrzb",
"output": "NO"
},
{
"input": "gdvqjoyxnkypfvdxssgrihnwxkeojmnpdeobpecytkbdwujqfjtxsqspxvxpqioyfagzjxupqqzpgnpnpxcuipweunqch\nkkqkiwwasbhezqcfeceyngcyuogrkhqecwsyerdniqiocjehrpkljiljophqhyaiefjpavoom",
"output": "NO"
},
{
"input": "umeszdawsvgkjhlqwzents\nhxqhdungbylhnikwviuh",
"output": "NO"
},
{
"input": "juotpscvyfmgntshcealgbsrwwksgrwnrrbyaqqsxdlzhkbugdyx\nibqvffmfktyipgiopznsqtrtxiijntdbgyy",
"output": "NO"
},
{
"input": "zbwueheveouatecaglziqmudxemhrsozmaujrwlqmppzoumxhamwugedikvkblvmxwuofmpafdprbcftew\nulczwrqhctbtbxrhhodwbcxwimncnexosksujlisgclllxokrsbnozthajnnlilyffmsyko",
"output": "NO"
},
{
"input": "nkgwuugukzcv\nqktnpxedwxpxkrxdvgmfgoxkdfpbzvwsduyiybynbkouonhvmzakeiruhfmvrktghadbfkmwxduoqv",
"output": "NO"
},
{
"input": "incenvizhqpcenhjhehvjvgbsnfixbatrrjstxjzhlmdmxijztphxbrldlqwdfimweepkggzcxsrwelodpnryntepioqpvk\ndhjbjjftlvnxibkklxquwmzhjfvnmwpapdrslioxisbyhhfymyiaqhlgecpxamqnocizwxniubrmpyubvpenoukhcobkdojlybxd",
"output": "NO"
},
{
"input": "w\nw",
"output": "YES"
},
{
"input": "vz\nzv",
"output": "YES"
},
{
"input": "ry\nyr",
"output": "YES"
},
{
"input": "xou\nuox",
"output": "YES"
},
{
"input": "axg\ngax",
"output": "NO"
},
{
"input": "zdsl\nlsdz",
"output": "YES"
},
{
"input": "kudl\nldku",
"output": "NO"
},
{
"input": "zzlzwnqlcl\nlclqnwzlzz",
"output": "YES"
},
{
"input": "vzzgicnzqooejpjzads\nsdazjpjeooqzncigzzv",
"output": "YES"
},
{
"input": "raqhmvmzuwaykjpyxsykr\nxkysrypjkyawuzmvmhqar",
"output": "NO"
},
{
"input": "ngedczubzdcqbxksnxuavdjaqtmdwncjnoaicvmodcqvhfezew\nwezefhvqcdomvciaonjcnwdmtqajdvauxnskxbqcdzbuzcdegn",
"output": "YES"
},
{
"input": "muooqttvrrljcxbroizkymuidvfmhhsjtumksdkcbwwpfqdyvxtrlymofendqvznzlmim\nmimlznzvqdnefomylrtxvydqfpwwbckdskmutjshhmfvdiumykziorbxcjlrrvttqooum",
"output": "YES"
},
{
"input": "vxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaivg\ngviayyikkitmuomcpiakhbxszgbnhvwyzkftwoagzixaearxpjacrnvpvbuzenvovehkmmxvblqyxvctroddksdsgebcmlluqpxv",
"output": "YES"
},
{
"input": "mnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfdc\ncdfmkdgrdptkpewbsqvszipgxvgvuiuzbkkwuowbafkikgvnqdkxnayzdjygvezmtsgywnupocdntipiyiorblqkrzjpzatxahnm",
"output": "NO"
},
{
"input": "dgxmzbqofstzcdgthbaewbwocowvhqpinehpjatnnbrijcolvsatbblsrxabzrpszoiecpwhfjmwuhqrapvtcgvikuxtzbftydkw\nwkdytfbztxukivgctvparqhuwmjfhwpceiozsprzbaxrslbbqasvlocjirbnntajphenipthvwocowbweabhtgdcztsfoqbzmxgd",
"output": "NO"
},
{
"input": "gxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwgeh\nhegwxvocotmzstqfbmpjvijgkcyodlxyjawrpkczpmdspsuhoiruavnnnuwvtwohglkdxjetshkboalvzqbgjgthoteceixioxg",
"output": "YES"
},
{
"input": "sihxuwvmaambplxvjfoskinghzicyfqebjtkysotattkahssumfcgrkheotdxwjckpvapbkaepqrxseyfrwtyaycmrzsrsngkh\nhkgnsrszrmcyaytwrfyesxrqpeakbpavpkcjwxdtoehkrgcfmusshakttatosyktjbeqfycizhgniksofjvxlpbmaamvwuxhis",
"output": "YES"
},
{
"input": "ycnahksbughnonldzrhkysujmylcgcfuludjvjiahtkyzqvkopzqcnwhltbzfugzojqkjjlggmvnultascmygelkiktmfieok\nkoeifmtkiklegkmcsatlunvmggkjjlqjozgufzbtlhwncqzpokvqzykthaijvjdulufcgclymjusyyhrzdlnonhgubskhancy",
"output": "NO"
},
{
"input": "wbqasaehtkfojruzyhrlgwmtyiovmzyfifslvlemhqheyaelzwnthrenjsbmntwaoryzwfbxmscmypvxlfmzpnkkjlvwvmtz\nztmvwvljkknpzmflxvpymcsmxbfwzyroawtnmbsjnerhtnwzleayehqhmelvlsfifyzmvoiytmwglrhyzurjofktheasaqbw",
"output": "YES"
},
{
"input": "imippqurprbhfugngtgifelytadegwrgaefnfhbjjnmzikvjaccotqzemufqieqldgnbmviisgkynzeldlhqxuqphjfmyij\njiymfjhpquxqhldleznykgsiivmbngdlqeiqfumezqtoccajvkizmnjjbhfnfeagrwgedatylefigtgngufhbrpruqppimi",
"output": "YES"
},
{
"input": "bikydffiuisckpvzqlteqfhegsagimodb\nbdomigasgehfqetlqzvpkcsiuiffdykib",
"output": "YES"
}
] | 1,653,391,466
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 40
| 92
| 0
|
for i in range(1):
n=input()
c=input()
x=n[::-1]
if c==x:
y='YES'
if c!=x:
y='NO'
print(y)
|
Title: Translation
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
Input Specification:
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
Output Specification:
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
Demo Input:
['code\nedoc\n', 'abb\naba\n', 'code\ncode\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none
|
```python
for i in range(1):
n=input()
c=input()
x=n[::-1]
if c==x:
y='YES'
if c!=x:
y='NO'
print(y)
```
| 3.977
|
246
|
A
|
Buggy Sorting
|
PROGRAMMING
| 900
|
[
"constructive algorithms",
"greedy",
"sortings"
] | null | null |
Little boy Valera studies an algorithm of sorting an integer array. After studying the theory, he went on to the practical tasks. As a result, he wrote a program that sorts an array of *n* integers *a*1,<=*a*2,<=...,<=*a**n* in the non-decreasing order. The pseudocode of the program, written by Valera, is given below. The input of the program gets number *n* and array *a*.
But Valera could have made a mistake, because he hasn't yet fully learned the sorting algorithm. If Valera made a mistake in his program, you need to give a counter-example that makes his program work improperly (that is, the example that makes the program sort the array not in the non-decreasing order). If such example for the given value of *n* doesn't exist, print -1.
|
You've got a single integer *n* (1<=≤<=*n*<=≤<=50) — the size of the sorted array.
|
Print *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100) — the counter-example, for which Valera's algorithm won't work correctly. If the counter-example that meets the described conditions is impossible to give, print -1.
If there are several counter-examples, consisting of *n* numbers, you are allowed to print any of them.
|
[
"1\n"
] |
[
"-1\n"
] |
none
| 500
|
[
{
"input": "1",
"output": "-1"
},
{
"input": "2",
"output": "-1"
},
{
"input": "3",
"output": "3 2 1 "
},
{
"input": "4",
"output": "4 3 2 1 "
},
{
"input": "5",
"output": "5 4 3 2 1 "
},
{
"input": "6",
"output": "6 5 4 3 2 1 "
},
{
"input": "7",
"output": "7 6 5 4 3 2 1 "
},
{
"input": "8",
"output": "8 7 6 5 4 3 2 1 "
},
{
"input": "9",
"output": "9 8 7 6 5 4 3 2 1 "
},
{
"input": "50",
"output": "50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 "
},
{
"input": "22",
"output": "22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 "
},
{
"input": "34",
"output": "34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 "
},
{
"input": "50",
"output": "50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 "
},
{
"input": "12",
"output": "12 11 10 9 8 7 6 5 4 3 2 1 "
},
{
"input": "26",
"output": "26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 "
},
{
"input": "38",
"output": "38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 "
},
{
"input": "4",
"output": "4 3 2 1 "
},
{
"input": "18",
"output": "18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 "
},
{
"input": "30",
"output": "30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 "
},
{
"input": "46",
"output": "46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 "
},
{
"input": "32",
"output": "32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 "
}
] | 1,547,280,907
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 21
| 109
| 0
|
m = []
n = int(input())
for i in range(1, n + 1):
m.append(n + 1 - i)
a = list(m)
a.sort()
for i in range(n - 1):
for j in range(i, n - 1):
if m[j] > m[j + 1]:
m[j], m[j + 1] = m[j + 1], m[j]
if m == a:
print(-1)
else:
a.reverse()
print(* a, sep = " ")
|
Title: Buggy Sorting
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little boy Valera studies an algorithm of sorting an integer array. After studying the theory, he went on to the practical tasks. As a result, he wrote a program that sorts an array of *n* integers *a*1,<=*a*2,<=...,<=*a**n* in the non-decreasing order. The pseudocode of the program, written by Valera, is given below. The input of the program gets number *n* and array *a*.
But Valera could have made a mistake, because he hasn't yet fully learned the sorting algorithm. If Valera made a mistake in his program, you need to give a counter-example that makes his program work improperly (that is, the example that makes the program sort the array not in the non-decreasing order). If such example for the given value of *n* doesn't exist, print -1.
Input Specification:
You've got a single integer *n* (1<=≤<=*n*<=≤<=50) — the size of the sorted array.
Output Specification:
Print *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100) — the counter-example, for which Valera's algorithm won't work correctly. If the counter-example that meets the described conditions is impossible to give, print -1.
If there are several counter-examples, consisting of *n* numbers, you are allowed to print any of them.
Demo Input:
['1\n']
Demo Output:
['-1\n']
Note:
none
|
```python
m = []
n = int(input())
for i in range(1, n + 1):
m.append(n + 1 - i)
a = list(m)
a.sort()
for i in range(n - 1):
for j in range(i, n - 1):
if m[j] > m[j + 1]:
m[j], m[j + 1] = m[j + 1], m[j]
if m == a:
print(-1)
else:
a.reverse()
print(* a, sep = " ")
```
| 3
|
|
732
|
A
|
Buy a Shovel
|
PROGRAMMING
| 800
|
[
"brute force",
"constructive algorithms",
"implementation",
"math"
] | null | null |
Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=≤<=*r*<=≤<=9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.
|
The single line of input contains two integers *k* and *r* (1<=≤<=*k*<=≤<=1000, 1<=≤<=*r*<=≤<=9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.
|
Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.
|
[
"117 3\n",
"237 7\n",
"15 2\n"
] |
[
"9\n",
"1\n",
"2\n"
] |
In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change.
| 500
|
[
{
"input": "117 3",
"output": "9"
},
{
"input": "237 7",
"output": "1"
},
{
"input": "15 2",
"output": "2"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 9",
"output": "9"
},
{
"input": "1000 3",
"output": "1"
},
{
"input": "1000 1",
"output": "1"
},
{
"input": "1000 9",
"output": "1"
},
{
"input": "1 2",
"output": "2"
},
{
"input": "999 9",
"output": "1"
},
{
"input": "999 8",
"output": "2"
},
{
"input": "105 6",
"output": "2"
},
{
"input": "403 9",
"output": "3"
},
{
"input": "546 4",
"output": "4"
},
{
"input": "228 9",
"output": "5"
},
{
"input": "57 2",
"output": "6"
},
{
"input": "437 9",
"output": "7"
},
{
"input": "997 6",
"output": "8"
},
{
"input": "109 1",
"output": "9"
},
{
"input": "998 9",
"output": "5"
},
{
"input": "4 2",
"output": "3"
},
{
"input": "9 3",
"output": "7"
},
{
"input": "8 2",
"output": "4"
},
{
"input": "1 3",
"output": "3"
},
{
"input": "1 4",
"output": "4"
},
{
"input": "1 5",
"output": "5"
},
{
"input": "1 6",
"output": "6"
},
{
"input": "1 7",
"output": "7"
},
{
"input": "1 8",
"output": "8"
},
{
"input": "100 3",
"output": "1"
},
{
"input": "1000 2",
"output": "1"
},
{
"input": "1000 4",
"output": "1"
},
{
"input": "1000 5",
"output": "1"
},
{
"input": "1000 6",
"output": "1"
},
{
"input": "1000 7",
"output": "1"
},
{
"input": "1000 8",
"output": "1"
},
{
"input": "23 4",
"output": "8"
},
{
"input": "33 1",
"output": "7"
},
{
"input": "33 2",
"output": "4"
},
{
"input": "666 5",
"output": "5"
},
{
"input": "2 3",
"output": "5"
},
{
"input": "5 5",
"output": "1"
},
{
"input": "3 6",
"output": "2"
},
{
"input": "12 4",
"output": "2"
},
{
"input": "15 5",
"output": "1"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "25 5",
"output": "1"
},
{
"input": "2 9",
"output": "5"
},
{
"input": "6 7",
"output": "5"
},
{
"input": "8 9",
"output": "5"
},
{
"input": "2 7",
"output": "5"
},
{
"input": "4 7",
"output": "5"
},
{
"input": "2 1",
"output": "5"
},
{
"input": "261 1",
"output": "1"
}
] | 1,689,869,014
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 54
| 62
| 0
|
k,y=map(int,input().split())
for i in range(1,11):
if (i*k) %10 == 0 or (k*i) % 10 == y:
print(i)
break
|
Title: Buy a Shovel
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=≤<=*r*<=≤<=9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.
Input Specification:
The single line of input contains two integers *k* and *r* (1<=≤<=*k*<=≤<=1000, 1<=≤<=*r*<=≤<=9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.
Output Specification:
Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.
Demo Input:
['117 3\n', '237 7\n', '15 2\n']
Demo Output:
['9\n', '1\n', '2\n']
Note:
In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change.
|
```python
k,y=map(int,input().split())
for i in range(1,11):
if (i*k) %10 == 0 or (k*i) % 10 == y:
print(i)
break
```
| 3
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
For a given positive integer *n* denote its *k*-rounding as the minimum positive integer *x*, such that *x* ends with *k* or more zeros in base 10 and is divisible by *n*.
For example, 4-rounding of 375 is 375·80<==<=30000. 30000 is the minimum integer such that it ends with 4 or more zeros and is divisible by 375.
Write a program that will perform the *k*-rounding of *n*.
|
The only line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=109, 0<=≤<=*k*<=≤<=8).
|
Print the *k*-rounding of *n*.
|
[
"375 4\n",
"10000 1\n",
"38101 0\n",
"123456789 8\n"
] |
[
"30000\n",
"10000\n",
"38101\n",
"12345678900000000\n"
] |
none
| 0
|
[
{
"input": "375 4",
"output": "30000"
},
{
"input": "10000 1",
"output": "10000"
},
{
"input": "38101 0",
"output": "38101"
},
{
"input": "123456789 8",
"output": "12345678900000000"
},
{
"input": "1 0",
"output": "1"
},
{
"input": "2 0",
"output": "2"
},
{
"input": "100 0",
"output": "100"
},
{
"input": "1000000000 0",
"output": "1000000000"
},
{
"input": "160 2",
"output": "800"
},
{
"input": "3 0",
"output": "3"
},
{
"input": "10 0",
"output": "10"
},
{
"input": "1 1",
"output": "10"
},
{
"input": "2 1",
"output": "10"
},
{
"input": "3 1",
"output": "30"
},
{
"input": "4 1",
"output": "20"
},
{
"input": "5 1",
"output": "10"
},
{
"input": "6 1",
"output": "30"
},
{
"input": "7 1",
"output": "70"
},
{
"input": "8 1",
"output": "40"
},
{
"input": "9 1",
"output": "90"
},
{
"input": "10 1",
"output": "10"
},
{
"input": "11 1",
"output": "110"
},
{
"input": "12 1",
"output": "60"
},
{
"input": "16 2",
"output": "400"
},
{
"input": "2 2",
"output": "100"
},
{
"input": "1 2",
"output": "100"
},
{
"input": "5 2",
"output": "100"
},
{
"input": "15 2",
"output": "300"
},
{
"input": "36 2",
"output": "900"
},
{
"input": "1 8",
"output": "100000000"
},
{
"input": "8 8",
"output": "100000000"
},
{
"input": "96 8",
"output": "300000000"
},
{
"input": "175 8",
"output": "700000000"
},
{
"input": "9999995 8",
"output": "199999900000000"
},
{
"input": "999999999 8",
"output": "99999999900000000"
},
{
"input": "12345678 8",
"output": "617283900000000"
},
{
"input": "78125 8",
"output": "100000000"
},
{
"input": "390625 8",
"output": "100000000"
},
{
"input": "1953125 8",
"output": "500000000"
},
{
"input": "9765625 8",
"output": "2500000000"
},
{
"input": "68359375 8",
"output": "17500000000"
},
{
"input": "268435456 8",
"output": "104857600000000"
},
{
"input": "125829120 8",
"output": "9830400000000"
},
{
"input": "128000 8",
"output": "400000000"
},
{
"input": "300000 8",
"output": "300000000"
},
{
"input": "3711871 8",
"output": "371187100000000"
},
{
"input": "55555 8",
"output": "1111100000000"
},
{
"input": "222222222 8",
"output": "11111111100000000"
},
{
"input": "479001600 8",
"output": "7484400000000"
},
{
"input": "655360001 7",
"output": "6553600010000000"
},
{
"input": "655360001 8",
"output": "65536000100000000"
},
{
"input": "1000000000 1",
"output": "1000000000"
},
{
"input": "1000000000 7",
"output": "1000000000"
},
{
"input": "1000000000 8",
"output": "1000000000"
},
{
"input": "100000000 8",
"output": "100000000"
},
{
"input": "10000000 8",
"output": "100000000"
},
{
"input": "1000000 8",
"output": "100000000"
},
{
"input": "10000009 8",
"output": "1000000900000000"
},
{
"input": "10000005 8",
"output": "200000100000000"
},
{
"input": "10000002 8",
"output": "500000100000000"
},
{
"input": "999999997 8",
"output": "99999999700000000"
},
{
"input": "999999997 7",
"output": "9999999970000000"
},
{
"input": "999999995 8",
"output": "19999999900000000"
},
{
"input": "123 8",
"output": "12300000000"
},
{
"input": "24 2",
"output": "600"
},
{
"input": "16 4",
"output": "10000"
},
{
"input": "123456787 8",
"output": "12345678700000000"
},
{
"input": "100000000 8",
"output": "100000000"
},
{
"input": "7 1",
"output": "70"
},
{
"input": "101 1",
"output": "1010"
},
{
"input": "50 2",
"output": "100"
},
{
"input": "999999818 1",
"output": "4999999090"
},
{
"input": "2 1",
"output": "10"
},
{
"input": "123 1",
"output": "1230"
},
{
"input": "16 1",
"output": "80"
},
{
"input": "1 1",
"output": "10"
},
{
"input": "1000000000 8",
"output": "1000000000"
},
{
"input": "15304 6",
"output": "1913000000"
},
{
"input": "3 8",
"output": "300000000"
},
{
"input": "4 2",
"output": "100"
},
{
"input": "100000 7",
"output": "10000000"
},
{
"input": "5 8",
"output": "100000000"
},
{
"input": "16724 6",
"output": "4181000000"
},
{
"input": "999999999 1",
"output": "9999999990"
},
{
"input": "999999990 8",
"output": "9999999900000000"
},
{
"input": "999999999 8",
"output": "99999999900000000"
},
{
"input": "100000 1",
"output": "100000"
},
{
"input": "8 3",
"output": "1000"
},
{
"input": "16768 6",
"output": "262000000"
},
{
"input": "123456789 1",
"output": "1234567890"
},
{
"input": "2 8",
"output": "100000000"
},
{
"input": "999999937 8",
"output": "99999993700000000"
},
{
"input": "5 1",
"output": "10"
},
{
"input": "2000000 7",
"output": "10000000"
},
{
"input": "1999998 2",
"output": "99999900"
},
{
"input": "125 3",
"output": "1000"
}
] | 1,505,654,072
| 572
|
Python 3
|
OK
|
TESTS
| 96
| 62
| 5,529,600
|
n,k = map(int,input().split())
n2 = n
i = 0
while n2 % 5 == 0:
n2 /= 5
i += 1
j = 0
while n2 % 2 == 0:
n2 /= 2
j += 1
print(n*(5**max(0,k - i))*(2**max(0,k - j)))
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
For a given positive integer *n* denote its *k*-rounding as the minimum positive integer *x*, such that *x* ends with *k* or more zeros in base 10 and is divisible by *n*.
For example, 4-rounding of 375 is 375·80<==<=30000. 30000 is the minimum integer such that it ends with 4 or more zeros and is divisible by 375.
Write a program that will perform the *k*-rounding of *n*.
Input Specification:
The only line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=109, 0<=≤<=*k*<=≤<=8).
Output Specification:
Print the *k*-rounding of *n*.
Demo Input:
['375 4\n', '10000 1\n', '38101 0\n', '123456789 8\n']
Demo Output:
['30000\n', '10000\n', '38101\n', '12345678900000000\n']
Note:
none
|
```python
n,k = map(int,input().split())
n2 = n
i = 0
while n2 % 5 == 0:
n2 /= 5
i += 1
j = 0
while n2 % 2 == 0:
n2 /= 2
j += 1
print(n*(5**max(0,k - i))*(2**max(0,k - j)))
```
| 3
|
|
255
|
A
|
Greg's Workout
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Greg is a beginner bodybuilder. Today the gym coach gave him the training plan. All it had was *n* integers *a*1,<=*a*2,<=...,<=*a**n*. These numbers mean that Greg needs to do exactly *n* exercises today. Besides, Greg should repeat the *i*-th in order exercise *a**i* times.
Greg now only does three types of exercises: "chest" exercises, "biceps" exercises and "back" exercises. Besides, his training is cyclic, that is, the first exercise he does is a "chest" one, the second one is "biceps", the third one is "back", the fourth one is "chest", the fifth one is "biceps", and so on to the *n*-th exercise.
Now Greg wonders, which muscle will get the most exercise during his training. We know that the exercise Greg repeats the maximum number of times, trains the corresponding muscle the most. Help Greg, determine which muscle will get the most training.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=20). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=25) — the number of times Greg repeats the exercises.
|
Print word "chest" (without the quotes), if the chest gets the most exercise, "biceps" (without the quotes), if the biceps gets the most exercise and print "back" (without the quotes) if the back gets the most exercise.
It is guaranteed that the input is such that the answer to the problem is unambiguous.
|
[
"2\n2 8\n",
"3\n5 1 10\n",
"7\n3 3 2 7 9 6 8\n"
] |
[
"biceps\n",
"back\n",
"chest\n"
] |
In the first sample Greg does 2 chest, 8 biceps and zero back exercises, so the biceps gets the most exercises.
In the second sample Greg does 5 chest, 1 biceps and 10 back exercises, so the back gets the most exercises.
In the third sample Greg does 18 chest, 12 biceps and 8 back exercises, so the chest gets the most exercise.
| 500
|
[
{
"input": "2\n2 8",
"output": "biceps"
},
{
"input": "3\n5 1 10",
"output": "back"
},
{
"input": "7\n3 3 2 7 9 6 8",
"output": "chest"
},
{
"input": "4\n5 6 6 2",
"output": "chest"
},
{
"input": "5\n8 2 2 6 3",
"output": "chest"
},
{
"input": "6\n8 7 2 5 3 4",
"output": "chest"
},
{
"input": "8\n7 2 9 10 3 8 10 6",
"output": "chest"
},
{
"input": "9\n5 4 2 3 4 4 5 2 2",
"output": "chest"
},
{
"input": "10\n4 9 8 5 3 8 8 10 4 2",
"output": "biceps"
},
{
"input": "11\n10 9 7 6 1 3 9 7 1 3 5",
"output": "chest"
},
{
"input": "12\n24 22 6 16 5 21 1 7 2 19 24 5",
"output": "chest"
},
{
"input": "13\n24 10 5 7 16 17 2 7 9 20 15 2 24",
"output": "chest"
},
{
"input": "14\n13 14 19 8 5 17 9 16 15 9 5 6 3 7",
"output": "back"
},
{
"input": "15\n24 12 22 21 25 23 21 5 3 24 23 13 12 16 12",
"output": "chest"
},
{
"input": "16\n12 6 18 6 25 7 3 1 1 17 25 17 6 8 17 8",
"output": "biceps"
},
{
"input": "17\n13 8 13 4 9 21 10 10 9 22 14 23 22 7 6 14 19",
"output": "chest"
},
{
"input": "18\n1 17 13 6 11 10 25 13 24 9 21 17 3 1 17 12 25 21",
"output": "back"
},
{
"input": "19\n22 22 24 25 19 10 7 10 4 25 19 14 1 14 3 18 4 19 24",
"output": "chest"
},
{
"input": "20\n9 8 22 11 18 14 15 10 17 11 2 1 25 20 7 24 4 25 9 20",
"output": "chest"
},
{
"input": "1\n10",
"output": "chest"
},
{
"input": "2\n15 3",
"output": "chest"
},
{
"input": "3\n21 11 19",
"output": "chest"
},
{
"input": "4\n19 24 13 15",
"output": "chest"
},
{
"input": "5\n4 24 1 9 19",
"output": "biceps"
},
{
"input": "6\n6 22 24 7 15 24",
"output": "back"
},
{
"input": "7\n10 8 23 23 14 18 14",
"output": "chest"
},
{
"input": "8\n5 16 8 9 17 16 14 7",
"output": "biceps"
},
{
"input": "9\n12 3 10 23 6 4 22 13 12",
"output": "chest"
},
{
"input": "10\n1 9 20 18 20 17 7 24 23 2",
"output": "back"
},
{
"input": "11\n22 25 8 2 18 15 1 13 1 11 4",
"output": "biceps"
},
{
"input": "12\n20 12 14 2 15 6 24 3 11 8 11 14",
"output": "chest"
},
{
"input": "13\n2 18 8 8 8 20 5 22 15 2 5 19 18",
"output": "back"
},
{
"input": "14\n1 6 10 25 17 13 21 11 19 4 15 24 5 22",
"output": "biceps"
},
{
"input": "15\n13 5 25 13 17 25 19 21 23 17 12 6 14 8 6",
"output": "back"
},
{
"input": "16\n10 15 2 17 22 12 14 14 6 11 4 13 9 8 21 14",
"output": "chest"
},
{
"input": "17\n7 22 9 22 8 7 20 22 23 5 12 11 1 24 17 20 10",
"output": "biceps"
},
{
"input": "18\n18 15 4 25 5 11 21 25 12 14 25 23 19 19 13 6 9 17",
"output": "chest"
},
{
"input": "19\n3 1 3 15 15 25 10 25 23 10 9 21 13 23 19 3 24 21 14",
"output": "back"
},
{
"input": "20\n19 18 11 3 6 14 3 3 25 3 1 19 25 24 23 12 7 4 8 6",
"output": "back"
},
{
"input": "1\n19",
"output": "chest"
},
{
"input": "2\n1 7",
"output": "biceps"
},
{
"input": "3\n18 18 23",
"output": "back"
},
{
"input": "4\n12 15 1 13",
"output": "chest"
},
{
"input": "5\n11 14 25 21 21",
"output": "biceps"
},
{
"input": "6\n11 9 12 11 22 18",
"output": "biceps"
},
{
"input": "7\n11 1 16 20 21 25 20",
"output": "chest"
},
{
"input": "8\n1 2 20 9 3 22 17 4",
"output": "back"
},
{
"input": "9\n19 2 10 19 15 20 3 1 13",
"output": "back"
},
{
"input": "10\n11 2 11 8 21 16 2 3 19 9",
"output": "back"
},
{
"input": "20\n25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 24",
"output": "chest"
},
{
"input": "12\n4 24 21 3 13 24 22 13 12 21 1 15",
"output": "back"
},
{
"input": "13\n14 14 16 2 13 5 1 14 9 4 16 8 3",
"output": "biceps"
},
{
"input": "14\n1 9 15 4 11 8 25 3 9 14 13 2 1 11",
"output": "biceps"
},
{
"input": "15\n4 19 10 6 16 12 5 11 7 23 1 24 11 7 17",
"output": "back"
},
{
"input": "16\n2 8 2 8 13 22 20 12 22 23 18 13 18 22 11 17",
"output": "chest"
},
{
"input": "17\n24 5 5 16 10 8 22 6 4 13 10 10 5 23 8 20 8",
"output": "chest"
},
{
"input": "18\n14 8 9 12 11 18 24 1 14 24 18 5 12 17 1 10 1 22",
"output": "chest"
},
{
"input": "19\n21 2 10 6 9 1 24 5 2 19 10 13 10 7 19 2 6 13 24",
"output": "chest"
},
{
"input": "20\n7 1 14 17 6 6 18 13 12 3 25 4 3 19 22 24 16 14 1 23",
"output": "biceps"
},
{
"input": "1\n19",
"output": "chest"
},
{
"input": "20\n2 1 2 2 1 2 2 1 2 1 1 1 1 1 1 1 1 1 1 22",
"output": "biceps"
}
] | 1,646,290,915
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 61
| 92
| 0
|
n=int(input())
a=[int(x)for x in input().split()]
exercise=[0,0,0]
i=0
j=0
while i<n:
exercise[j]+=a[i]
i+=1
j+=1
if j==3:
j=0
m=max(exercise)
if m==exercise[0]:
print("chest")
elif m==exercise[1]:
print("biceps")
else:
print("back")
|
Title: Greg's Workout
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Greg is a beginner bodybuilder. Today the gym coach gave him the training plan. All it had was *n* integers *a*1,<=*a*2,<=...,<=*a**n*. These numbers mean that Greg needs to do exactly *n* exercises today. Besides, Greg should repeat the *i*-th in order exercise *a**i* times.
Greg now only does three types of exercises: "chest" exercises, "biceps" exercises and "back" exercises. Besides, his training is cyclic, that is, the first exercise he does is a "chest" one, the second one is "biceps", the third one is "back", the fourth one is "chest", the fifth one is "biceps", and so on to the *n*-th exercise.
Now Greg wonders, which muscle will get the most exercise during his training. We know that the exercise Greg repeats the maximum number of times, trains the corresponding muscle the most. Help Greg, determine which muscle will get the most training.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=20). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=25) — the number of times Greg repeats the exercises.
Output Specification:
Print word "chest" (without the quotes), if the chest gets the most exercise, "biceps" (without the quotes), if the biceps gets the most exercise and print "back" (without the quotes) if the back gets the most exercise.
It is guaranteed that the input is such that the answer to the problem is unambiguous.
Demo Input:
['2\n2 8\n', '3\n5 1 10\n', '7\n3 3 2 7 9 6 8\n']
Demo Output:
['biceps\n', 'back\n', 'chest\n']
Note:
In the first sample Greg does 2 chest, 8 biceps and zero back exercises, so the biceps gets the most exercises.
In the second sample Greg does 5 chest, 1 biceps and 10 back exercises, so the back gets the most exercises.
In the third sample Greg does 18 chest, 12 biceps and 8 back exercises, so the chest gets the most exercise.
|
```python
n=int(input())
a=[int(x)for x in input().split()]
exercise=[0,0,0]
i=0
j=0
while i<n:
exercise[j]+=a[i]
i+=1
j+=1
if j==3:
j=0
m=max(exercise)
if m==exercise[0]:
print("chest")
elif m==exercise[1]:
print("biceps")
else:
print("back")
```
| 3
|
|
496
|
A
|
Minimum Difficulty
|
PROGRAMMING
| 900
|
[
"brute force",
"implementation",
"math"
] | null | null |
Mike is trying rock climbing but he is awful at it.
There are *n* holds on the wall, *i*-th hold is at height *a**i* off the ground. Besides, let the sequence *a**i* increase, that is, *a**i*<=<<=*a**i*<=+<=1 for all *i* from 1 to *n*<=-<=1; we will call such sequence a track. Mike thinks that the track *a*1, ..., *a**n* has difficulty . In other words, difficulty equals the maximum distance between two holds that are adjacent in height.
Today Mike decided to cover the track with holds hanging on heights *a*1, ..., *a**n*. To make the problem harder, Mike decided to remove one hold, that is, remove one element of the sequence (for example, if we take the sequence (1,<=2,<=3,<=4,<=5) and remove the third element from it, we obtain the sequence (1,<=2,<=4,<=5)). However, as Mike is awful at climbing, he wants the final difficulty (i.e. the maximum difference of heights between adjacent holds after removing the hold) to be as small as possible among all possible options of removing a hold. The first and last holds must stay at their positions.
Help Mike determine the minimum difficulty of the track after removing one hold.
|
The first line contains a single integer *n* (3<=≤<=*n*<=≤<=100) — the number of holds.
The next line contains *n* space-separated integers *a**i* (1<=≤<=*a**i*<=≤<=1000), where *a**i* is the height where the hold number *i* hangs. The sequence *a**i* is increasing (i.e. each element except for the first one is strictly larger than the previous one).
|
Print a single number — the minimum difficulty of the track after removing a single hold.
|
[
"3\n1 4 6\n",
"5\n1 2 3 4 5\n",
"5\n1 2 3 7 8\n"
] |
[
"5\n",
"2\n",
"4\n"
] |
In the first sample you can remove only the second hold, then the sequence looks like (1, 6), the maximum difference of the neighboring elements equals 5.
In the second test after removing every hold the difficulty equals 2.
In the third test you can obtain sequences (1, 3, 7, 8), (1, 2, 7, 8), (1, 2, 3, 8), for which the difficulty is 4, 5 and 5, respectively. Thus, after removing the second element we obtain the optimal answer — 4.
| 500
|
[
{
"input": "3\n1 4 6",
"output": "5"
},
{
"input": "5\n1 2 3 4 5",
"output": "2"
},
{
"input": "5\n1 2 3 7 8",
"output": "4"
},
{
"input": "3\n1 500 1000",
"output": "999"
},
{
"input": "10\n1 2 3 4 5 6 7 8 9 10",
"output": "2"
},
{
"input": "10\n1 4 9 16 25 36 49 64 81 100",
"output": "19"
},
{
"input": "10\n300 315 325 338 350 365 379 391 404 416",
"output": "23"
},
{
"input": "15\n87 89 91 92 93 95 97 99 101 103 105 107 109 111 112",
"output": "2"
},
{
"input": "60\n3 5 7 8 15 16 18 21 24 26 40 41 43 47 48 49 50 51 52 54 55 60 62 71 74 84 85 89 91 96 406 407 409 412 417 420 423 424 428 431 432 433 436 441 445 446 447 455 458 467 469 471 472 475 480 485 492 493 497 500",
"output": "310"
},
{
"input": "3\n159 282 405",
"output": "246"
},
{
"input": "81\n6 7 22 23 27 38 40 56 59 71 72 78 80 83 86 92 95 96 101 122 125 127 130 134 154 169 170 171 172 174 177 182 184 187 195 197 210 211 217 223 241 249 252 253 256 261 265 269 274 277 291 292 297 298 299 300 302 318 338 348 351 353 381 386 387 397 409 410 419 420 428 430 453 460 461 473 478 493 494 500 741",
"output": "241"
},
{
"input": "10\n218 300 388 448 535 629 680 740 836 925",
"output": "111"
},
{
"input": "100\n6 16 26 36 46 56 66 76 86 96 106 116 126 136 146 156 166 176 186 196 206 216 226 236 246 256 266 276 286 296 306 316 326 336 346 356 366 376 386 396 406 416 426 436 446 456 466 476 486 496 506 516 526 536 546 556 566 576 586 596 606 616 626 636 646 656 666 676 686 696 706 716 726 736 746 756 766 776 786 796 806 816 826 836 846 856 866 876 886 896 906 916 926 936 946 956 966 976 986 996",
"output": "20"
},
{
"input": "100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 951 952 953 954 955 956 957 958 959 960 961 962 963 964 965 966 967 968 969 970 971 972 973 974 975 976 977 978 979 980 981 982 983 984 985 986 987 988 989 990 991 992 993 994 995 996 997 998 999 1000",
"output": "901"
},
{
"input": "100\n1 9 15 17 28 29 30 31 32 46 48 49 52 56 62 77 82 85 90 91 94 101 102 109 111 113 116 118 124 125 131 132 136 138 139 143 145 158 161 162 165 167 171 173 175 177 179 183 189 196 801 802 804 806 817 819 827 830 837 840 842 846 850 855 858 862 863 866 869 870 878 881 883 884 896 898 899 901 904 906 908 909 910 911 912 917 923 924 925 935 939 943 945 956 963 964 965 972 976 978",
"output": "605"
},
{
"input": "100\n2 43 47 49 50 57 59 67 74 98 901 903 904 906 907 908 909 910 911 912 913 914 915 916 917 918 919 920 921 922 923 924 925 926 927 928 929 930 931 932 933 934 935 936 938 939 940 942 943 944 945 946 947 948 949 950 952 953 954 956 957 958 959 960 961 962 963 965 966 967 968 969 970 971 972 973 974 975 976 978 979 980 981 982 983 984 985 986 987 988 989 990 991 992 993 994 995 996 998 999",
"output": "803"
},
{
"input": "72\n178 186 196 209 217 226 236 248 260 273 281 291 300 309 322 331 343 357 366 377 389 399 409 419 429 442 450 459 469 477 491 501 512 524 534 548 557 568 582 593 602 616 630 643 652 660 670 679 693 707 715 728 737 750 759 768 776 789 797 807 815 827 837 849 863 873 881 890 901 910 920 932",
"output": "17"
},
{
"input": "38\n1 28 55 82 109 136 163 190 217 244 271 298 325 352 379 406 433 460 487 514 541 568 595 622 649 676 703 730 757 784 811 838 865 892 919 946 973 1000",
"output": "54"
},
{
"input": "28\n1 38 75 112 149 186 223 260 297 334 371 408 445 482 519 556 593 630 667 704 741 778 815 852 889 926 963 1000",
"output": "74"
}
] | 1,538,998,555
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 19
| 124
| 0
|
n = int(input())
l = list(map(int,input().split()))
m = 0
for i in range(0,n-1):
m = max(m,l[i+1]-l[i])
a = 10000
for i in range(1,n-1):
a = min(a,l[i+1]-l[i-1])
print([m,a][m<a])
|
Title: Minimum Difficulty
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mike is trying rock climbing but he is awful at it.
There are *n* holds on the wall, *i*-th hold is at height *a**i* off the ground. Besides, let the sequence *a**i* increase, that is, *a**i*<=<<=*a**i*<=+<=1 for all *i* from 1 to *n*<=-<=1; we will call such sequence a track. Mike thinks that the track *a*1, ..., *a**n* has difficulty . In other words, difficulty equals the maximum distance between two holds that are adjacent in height.
Today Mike decided to cover the track with holds hanging on heights *a*1, ..., *a**n*. To make the problem harder, Mike decided to remove one hold, that is, remove one element of the sequence (for example, if we take the sequence (1,<=2,<=3,<=4,<=5) and remove the third element from it, we obtain the sequence (1,<=2,<=4,<=5)). However, as Mike is awful at climbing, he wants the final difficulty (i.e. the maximum difference of heights between adjacent holds after removing the hold) to be as small as possible among all possible options of removing a hold. The first and last holds must stay at their positions.
Help Mike determine the minimum difficulty of the track after removing one hold.
Input Specification:
The first line contains a single integer *n* (3<=≤<=*n*<=≤<=100) — the number of holds.
The next line contains *n* space-separated integers *a**i* (1<=≤<=*a**i*<=≤<=1000), where *a**i* is the height where the hold number *i* hangs. The sequence *a**i* is increasing (i.e. each element except for the first one is strictly larger than the previous one).
Output Specification:
Print a single number — the minimum difficulty of the track after removing a single hold.
Demo Input:
['3\n1 4 6\n', '5\n1 2 3 4 5\n', '5\n1 2 3 7 8\n']
Demo Output:
['5\n', '2\n', '4\n']
Note:
In the first sample you can remove only the second hold, then the sequence looks like (1, 6), the maximum difference of the neighboring elements equals 5.
In the second test after removing every hold the difficulty equals 2.
In the third test you can obtain sequences (1, 3, 7, 8), (1, 2, 7, 8), (1, 2, 3, 8), for which the difficulty is 4, 5 and 5, respectively. Thus, after removing the second element we obtain the optimal answer — 4.
|
```python
n = int(input())
l = list(map(int,input().split()))
m = 0
for i in range(0,n-1):
m = max(m,l[i+1]-l[i])
a = 10000
for i in range(1,n-1):
a = min(a,l[i+1]-l[i-1])
print([m,a][m<a])
```
| 3
|
|
731
|
A
|
Night at the Museum
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] | null | null |
Grigoriy, like the hero of one famous comedy film, found a job as a night security guard at the museum. At first night he received embosser and was to take stock of the whole exposition.
Embosser is a special devise that allows to "print" the text of a plastic tape. Text is printed sequentially, character by character. The device consists of a wheel with a lowercase English letters written in a circle, static pointer to the current letter and a button that print the chosen letter. At one move it's allowed to rotate the alphabetic wheel one step clockwise or counterclockwise. Initially, static pointer points to letter 'a'. Other letters are located as shown on the picture:
After Grigoriy add new item to the base he has to print its name on the plastic tape and attach it to the corresponding exhibit. It's not required to return the wheel to its initial position with pointer on the letter 'a'.
Our hero is afraid that some exhibits may become alive and start to attack him, so he wants to print the names as fast as possible. Help him, for the given string find the minimum number of rotations of the wheel required to print it.
|
The only line of input contains the name of some exhibit — the non-empty string consisting of no more than 100 characters. It's guaranteed that the string consists of only lowercase English letters.
|
Print one integer — the minimum number of rotations of the wheel, required to print the name given in the input.
|
[
"zeus\n",
"map\n",
"ares\n"
] |
[
"18\n",
"35\n",
"34\n"
] |
To print the string from the first sample it would be optimal to perform the following sequence of rotations:
1. from 'a' to 'z' (1 rotation counterclockwise), 1. from 'z' to 'e' (5 clockwise rotations), 1. from 'e' to 'u' (10 rotations counterclockwise), 1. from 'u' to 's' (2 counterclockwise rotations).
| 500
|
[
{
"input": "zeus",
"output": "18"
},
{
"input": "map",
"output": "35"
},
{
"input": "ares",
"output": "34"
},
{
"input": "l",
"output": "11"
},
{
"input": "abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuv",
"output": "99"
},
{
"input": "gngvi",
"output": "44"
},
{
"input": "aaaaa",
"output": "0"
},
{
"input": "a",
"output": "0"
},
{
"input": "z",
"output": "1"
},
{
"input": "vyadeehhikklnoqrs",
"output": "28"
},
{
"input": "jjiihhhhgggfedcccbazyxx",
"output": "21"
},
{
"input": "fyyptqqxuciqvwdewyppjdzur",
"output": "117"
},
{
"input": "fqcnzmzmbobmancqcoalzmanaobpdse",
"output": "368"
},
{
"input": "zzzzzaaaaaaazzzzzzaaaaaaazzzzzzaaaazzzza",
"output": "8"
},
{
"input": "aucnwhfixuruefkypvrvnvznwtjgwlghoqtisbkhuwxmgzuljvqhmnwzisnsgjhivnjmbknptxatdkelhzkhsuxzrmlcpeoyukiy",
"output": "644"
},
{
"input": "sssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssss",
"output": "8"
},
{
"input": "nypjygrdtpzpigzyrisqeqfriwgwlengnezppgttgtndbrryjdl",
"output": "421"
},
{
"input": "pnllnnmmmmoqqqqqrrtssssuuvtsrpopqoonllmonnnpppopnonoopooqpnopppqppqstuuuwwwwvxzxzzaa",
"output": "84"
},
{
"input": "btaoahqgxnfsdmzsjxgvdwjukcvereqeskrdufqfqgzqfsftdqcthtkcnaipftcnco",
"output": "666"
},
{
"input": "eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeerrrrrrrrrrrrrrrrwwwwwwwwww",
"output": "22"
},
{
"input": "uyknzcrwjyzmscqucclvacmorepdgmnyhmakmmnygqwglrxkxhkpansbmruwxdeoprxzmpsvwackopujxbbkpwyeggsvjykpxh",
"output": "643"
},
{
"input": "gzwpooohffcxwtpjgfzwtooiccxsrrokezutoojdzwsrmmhecaxwrojcbyrqlfdwwrliiib",
"output": "245"
},
{
"input": "dbvnkktasjdwqsrzfwwtmjgbcxggdxsoeilecihduypktkkbwfbruxzzhlttrssicgdwqruddwrlbtxgmhdbatzvdxbbro",
"output": "468"
},
{
"input": "mdtvowlktxzzbuaeiuebfeorgbdczauxsovbucactkvyvemsknsjfhifqgycqredzchipmkvzbxdjkcbyukomjlzvxzoswumned",
"output": "523"
},
{
"input": "kkkkkkkaaaaxxaaaaaaaxxxxxxxxaaaaaaxaaaaaaaaaakkkkkkkkkaaaaaaannnnnxxxxkkkkkkkkaannnnnnna",
"output": "130"
},
{
"input": "dffiknqqrsvwzcdgjkmpqtuwxadfhkkkmpqrtwxyadfggjmpppsuuwyyzcdgghhknnpsvvvwwwyabccffiloqruwwyyzabeeehh",
"output": "163"
},
{
"input": "qpppmmkjihgecbyvvsppnnnkjiffeebaaywutrrqpmkjhgddbzzzywtssssqnmmljheddbbaxvusrqonmlifedbbzyywwtqnkheb",
"output": "155"
},
{
"input": "wvvwwwvvwxxxyyyxxwwvwwvuttttttuvvwxxwxxyxxwwwwwvvuttssrssstsssssrqpqqppqrssrsrrssrssssrrsrqqrrqpppqp",
"output": "57"
},
{
"input": "dqcpcobpcobnznamznamzlykxkxlxlylzmaobnaobpbnanbpcoaobnboaoboanzlymzmykylymylzlylymanboanaocqdqesfrfs",
"output": "1236"
},
{
"input": "nnnnnnnnnnnnnnnnnnnnaaaaaaaaaaaaaaaaaaaakkkkkkkkkkkkkkkkkkkkkkaaaaaaaaaaaaaaaaaaaaxxxxxxxxxxxxxxxxxx",
"output": "49"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "0"
},
{
"input": "cgilqsuwzaffilptwwbgmnttyyejkorxzflqvzbddhmnrvxchijpuwaeiimosxyycejlpquuwbfkpvbgijkqvxybdjjjptxcfkqt",
"output": "331"
},
{
"input": "ufsepwgtzgtgjssxaitgpailuvgqweoppszjwhoxdhhhpwwdorwfrdjwcdekxiktwziqwbkvbknrtvajpyeqbjvhiikxxaejjpte",
"output": "692"
},
{
"input": "uhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuh",
"output": "1293"
},
{
"input": "vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvgggggggggggggggggggggggggggggggggggggggggggggggggg",
"output": "16"
},
{
"input": "lyidmjyzbszgiwkxhhpnnthfwcvvstueionspfrvqgkvngmwyhezlosrpdnbvtcjjxxsykixwnepbumaacdzadlqhnjlcejovple",
"output": "616"
},
{
"input": "etzqqbaveffalkdguunfmyyrzkccnxmlluxeasqmopxzfvlkbhipqdwjgrttoemruohgwukfisdhznqyvhswbbypoxgtxyappcrl",
"output": "605"
},
{
"input": "lizussgedcbdjhrbeskhgatyozvwwekanlggcstijrniivupmcoofbaxfqrxddyzzptwxcftlhajsmmkkriarrqtkoauhcqefyud",
"output": "549"
},
{
"input": "dvjuvgfdogpknmbowlsfjzcimnygbtjiucyeeroqwhmzwpjqxlbjkqawrdtmvxbiqufllfuqibxvmtdrwaqkjblxqjpwzmhwqore",
"output": "688"
},
{
"input": "eeycuijtbgynmiczjfslwobmnkpgodfgvujvduyfeqchuaoktqrrairkkmmsjahltfcxwtpzzyddxrqfxabfoocmpuviinrjitsc",
"output": "604"
},
{
"input": "cgglnakewwvzoytaghksebrhjdbcdegssuzilrcppayxtgxopybbwshvyqnzhdsifkuwghourmeottrgjwdqpihbklvfzxpomqsa",
"output": "572"
},
{
"input": "aexullmxncckzryymfnuugdklaffevabqqztelpvojecljnhqldazdcaamubpenwxikysxxjjctvbndprsolzehywmgnvkgqvrfp",
"output": "609"
},
{
"input": "psnoieutsvvcwfhtnnphhxkwigzsbzyjmdiyl",
"output": "223"
},
{
"input": "aa",
"output": "0"
}
] | 1,641,902,061
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 44
| 46
| 0
|
inp = input()
start = ord('a')
steps = 0
for i in inp:
inner_dis = abs(ord(i)-start)
outer_dis = 26-inner_dis
steps += min(outer_dis, inner_dis)
start = ord(i)
print(steps)
|
Title: Night at the Museum
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Grigoriy, like the hero of one famous comedy film, found a job as a night security guard at the museum. At first night he received embosser and was to take stock of the whole exposition.
Embosser is a special devise that allows to "print" the text of a plastic tape. Text is printed sequentially, character by character. The device consists of a wheel with a lowercase English letters written in a circle, static pointer to the current letter and a button that print the chosen letter. At one move it's allowed to rotate the alphabetic wheel one step clockwise or counterclockwise. Initially, static pointer points to letter 'a'. Other letters are located as shown on the picture:
After Grigoriy add new item to the base he has to print its name on the plastic tape and attach it to the corresponding exhibit. It's not required to return the wheel to its initial position with pointer on the letter 'a'.
Our hero is afraid that some exhibits may become alive and start to attack him, so he wants to print the names as fast as possible. Help him, for the given string find the minimum number of rotations of the wheel required to print it.
Input Specification:
The only line of input contains the name of some exhibit — the non-empty string consisting of no more than 100 characters. It's guaranteed that the string consists of only lowercase English letters.
Output Specification:
Print one integer — the minimum number of rotations of the wheel, required to print the name given in the input.
Demo Input:
['zeus\n', 'map\n', 'ares\n']
Demo Output:
['18\n', '35\n', '34\n']
Note:
To print the string from the first sample it would be optimal to perform the following sequence of rotations:
1. from 'a' to 'z' (1 rotation counterclockwise), 1. from 'z' to 'e' (5 clockwise rotations), 1. from 'e' to 'u' (10 rotations counterclockwise), 1. from 'u' to 's' (2 counterclockwise rotations).
|
```python
inp = input()
start = ord('a')
steps = 0
for i in inp:
inner_dis = abs(ord(i)-start)
outer_dis = 26-inner_dis
steps += min(outer_dis, inner_dis)
start = ord(i)
print(steps)
```
| 3
|
|
12
|
A
|
Super Agent
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Super Agent
|
2
|
256
|
There is a very secret base in Potatoland where potato mash is made according to a special recipe. The neighbours from Porridgia decided to seize this recipe and to sell it to Pilauland. For this mission they have been preparing special agent Pearlo for many years. When, finally, Pearlo learned all secrets of espionage, he penetrated into the Potatoland territory and reached the secret base.
Now he is standing at the entrance, but to get inside he need to pass combination lock. Minute ago one of the workers entered the password on the terminal and opened the door. The terminal is a square digital keyboard 3<=×<=3 with digits from 1 to 9.
Pearlo knows that the password consists from distinct digits and is probably symmetric with respect to the central button of the terminal. He has heat sensor which allowed him to detect the digits which the worker pressed. Now he wants to check whether the password entered by the worker is symmetric with respect to the central button of the terminal. This fact can Help Pearlo to reduce the number of different possible password combinations.
|
Input contains the matrix of three rows of three symbols each. Symbol «X» means that the corresponding button was pressed, and «.» means that is was not pressed. The matrix may contain no «X», also it may contain no «.».
|
Print YES if the password is symmetric with respect to the central button of the terminal and NO otherwise.
|
[
"XX.\n...\n.XX\n",
"X.X\nX..\n...\n"
] |
[
"YES\n",
"NO\n"
] |
If you are not familiar with the term «central symmetry», you may look into http://en.wikipedia.org/wiki/Central_symmetry
| 0
|
[
{
"input": "XX.\n...\n.XX",
"output": "YES"
},
{
"input": ".X.\n.X.\n.X.",
"output": "YES"
},
{
"input": "XXX\nXXX\nXXX",
"output": "YES"
},
{
"input": "XXX\nX.X\nXXX",
"output": "YES"
},
{
"input": "X..\n.X.\n..X",
"output": "YES"
},
{
"input": "...\nX.X\nX..",
"output": "NO"
},
{
"input": ".X.\nX.X\n.X.",
"output": "YES"
},
{
"input": "X.X\n.X.\nX.X",
"output": "YES"
},
{
"input": "...\n...\n..X",
"output": "NO"
},
{
"input": "XXX\n...\nXXX",
"output": "YES"
},
{
"input": "..X\nX..\n..X",
"output": "NO"
},
{
"input": ".X.\n...\nX.X",
"output": "NO"
},
{
"input": "X.X\nX.X\nX.X",
"output": "YES"
},
{
"input": ".X.\nX.X\nXX.",
"output": "NO"
},
{
"input": "...\nXXX\nXXX",
"output": "NO"
},
{
"input": "XXX\n..X\nXXX",
"output": "NO"
},
{
"input": "X..\nX.X\n.X.",
"output": "NO"
},
{
"input": "...\n..X\nXXX",
"output": "NO"
},
{
"input": "..X\nX.X\nX..",
"output": "YES"
},
{
"input": "..X\n..X\nXXX",
"output": "NO"
},
{
"input": "X..\nX..\nX..",
"output": "NO"
},
{
"input": "XXX\n.X.\nXXX",
"output": "YES"
},
{
"input": "..X\n...\nX..",
"output": "YES"
},
{
"input": "...\n...\nX..",
"output": "NO"
},
{
"input": "...\n...\n.X.",
"output": "NO"
},
{
"input": "...\n...\n..X",
"output": "NO"
},
{
"input": "...\n.X.\nX..",
"output": "NO"
}
] | 1,639,372,520
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 40
| 154
| 0
|
arr = []
for _ in range(3):
s = list(input())
arr.extend(s)
if arr[0] == arr[8] and arr[1] == arr[7] and arr[2] == arr[6] and arr[3] == arr[5]:
print("YES")
else:
print("NO")
|
Title: Super Agent
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
There is a very secret base in Potatoland where potato mash is made according to a special recipe. The neighbours from Porridgia decided to seize this recipe and to sell it to Pilauland. For this mission they have been preparing special agent Pearlo for many years. When, finally, Pearlo learned all secrets of espionage, he penetrated into the Potatoland territory and reached the secret base.
Now he is standing at the entrance, but to get inside he need to pass combination lock. Minute ago one of the workers entered the password on the terminal and opened the door. The terminal is a square digital keyboard 3<=×<=3 with digits from 1 to 9.
Pearlo knows that the password consists from distinct digits and is probably symmetric with respect to the central button of the terminal. He has heat sensor which allowed him to detect the digits which the worker pressed. Now he wants to check whether the password entered by the worker is symmetric with respect to the central button of the terminal. This fact can Help Pearlo to reduce the number of different possible password combinations.
Input Specification:
Input contains the matrix of three rows of three symbols each. Symbol «X» means that the corresponding button was pressed, and «.» means that is was not pressed. The matrix may contain no «X», also it may contain no «.».
Output Specification:
Print YES if the password is symmetric with respect to the central button of the terminal and NO otherwise.
Demo Input:
['XX.\n...\n.XX\n', 'X.X\nX..\n...\n']
Demo Output:
['YES\n', 'NO\n']
Note:
If you are not familiar with the term «central symmetry», you may look into http://en.wikipedia.org/wiki/Central_symmetry
|
```python
arr = []
for _ in range(3):
s = list(input())
arr.extend(s)
if arr[0] == arr[8] and arr[1] == arr[7] and arr[2] == arr[6] and arr[3] == arr[5]:
print("YES")
else:
print("NO")
```
| 3.9615
|
50
|
A
|
Domino piling
|
PROGRAMMING
| 800
|
[
"greedy",
"math"
] |
A. Domino piling
|
2
|
256
|
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
|
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
|
Output one number — the maximal number of dominoes, which can be placed.
|
[
"2 4\n",
"3 3\n"
] |
[
"4\n",
"4\n"
] |
none
| 500
|
[
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "2 6",
"output": "6"
},
{
"input": "2 7",
"output": "7"
},
{
"input": "2 14",
"output": "14"
},
{
"input": "2 15",
"output": "15"
},
{
"input": "1 4",
"output": "2"
},
{
"input": "2 16",
"output": "16"
},
{
"input": "3 5",
"output": "7"
},
{
"input": "3 6",
"output": "9"
},
{
"input": "3 10",
"output": "15"
},
{
"input": "3 14",
"output": "21"
},
{
"input": "3 15",
"output": "22"
},
{
"input": "3 16",
"output": "24"
},
{
"input": "5 7",
"output": "17"
},
{
"input": "16 16",
"output": "128"
},
{
"input": "15 16",
"output": "120"
},
{
"input": "2 3",
"output": "3"
},
{
"input": "15 15",
"output": "112"
},
{
"input": "14 16",
"output": "112"
},
{
"input": "11 13",
"output": "71"
},
{
"input": "5 16",
"output": "40"
},
{
"input": "8 15",
"output": "60"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "3 4",
"output": "6"
},
{
"input": "4 4",
"output": "8"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "1 3",
"output": "1"
},
{
"input": "14 15",
"output": "105"
}
] | 1,683,955,671
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 35
| 92
| 0
|
import math
test_on = input()
test_num = test_on.split(" ")
loerd = int(test_num[0])
boubd = int(test_num[1])
operatiomn = math.floor((loerd * boubd) / 2)
print(operatiomn)
|
Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
Input Specification:
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output Specification:
Output one number — the maximal number of dominoes, which can be placed.
Demo Input:
['2 4\n', '3 3\n']
Demo Output:
['4\n', '4\n']
Note:
none
|
```python
import math
test_on = input()
test_num = test_on.split(" ")
loerd = int(test_num[0])
boubd = int(test_num[1])
operatiomn = math.floor((loerd * boubd) / 2)
print(operatiomn)
```
| 3.977
|
248
|
A
|
Cupboards
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
One foggy Stockholm morning, Karlsson decided to snack on some jam in his friend Lillebror Svantenson's house. Fortunately for Karlsson, there wasn't anybody in his friend's house. Karlsson was not going to be hungry any longer, so he decided to get some food in the house.
Karlsson's gaze immediately fell on *n* wooden cupboards, standing in the kitchen. He immediately realized that these cupboards have hidden jam stocks. Karlsson began to fly greedily around the kitchen, opening and closing the cupboards' doors, grab and empty all the jars of jam that he could find.
And now all jars of jam are empty, Karlsson has had enough and does not want to leave traces of his stay, so as not to let down his friend. Each of the cupboards has two doors: the left one and the right one. Karlsson remembers that when he rushed to the kitchen, all the cupboards' left doors were in the same position (open or closed), similarly, all the cupboards' right doors were in the same position (open or closed). Karlsson wants the doors to meet this condition as well by the time the family returns. Karlsson does not remember the position of all the left doors, also, he cannot remember the position of all the right doors. Therefore, it does not matter to him in what position will be all left or right doors. It is important to leave all the left doors in the same position, and all the right doors in the same position. For example, all the left doors may be closed, and all the right ones may be open.
Karlsson needs one second to open or close a door of a cupboard. He understands that he has very little time before the family returns, so he wants to know the minimum number of seconds *t*, in which he is able to bring all the cupboard doors in the required position.
Your task is to write a program that will determine the required number of seconds *t*.
|
The first input line contains a single integer *n* — the number of cupboards in the kitchen (2<=≤<=*n*<=≤<=104). Then follow *n* lines, each containing two integers *l**i* and *r**i* (0<=≤<=*l**i*,<=*r**i*<=≤<=1). Number *l**i* equals one, if the left door of the *i*-th cupboard is opened, otherwise number *l**i* equals zero. Similarly, number *r**i* equals one, if the right door of the *i*-th cupboard is opened, otherwise number *r**i* equals zero.
The numbers in the lines are separated by single spaces.
|
In the only output line print a single integer *t* — the minimum number of seconds Karlsson needs to change the doors of all cupboards to the position he needs.
|
[
"5\n0 1\n1 0\n0 1\n1 1\n0 1\n"
] |
[
"3\n"
] |
none
| 500
|
[
{
"input": "5\n0 1\n1 0\n0 1\n1 1\n0 1",
"output": "3"
},
{
"input": "2\n0 0\n0 0",
"output": "0"
},
{
"input": "3\n0 1\n1 1\n1 1",
"output": "1"
},
{
"input": "8\n0 1\n1 0\n0 1\n1 1\n0 1\n1 0\n0 1\n1 0",
"output": "7"
},
{
"input": "8\n1 0\n1 0\n1 0\n0 1\n0 1\n1 1\n1 1\n0 1",
"output": "6"
},
{
"input": "15\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "5\n1 0\n1 0\n1 0\n0 1\n0 1",
"output": "4"
}
] | 1,617,453,072
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 92
| 0
|
n = int(input())
l1 = []
l2 = []
d1 = {}
d2 = {}
r = 0
for i in range(n):
x, y = input().split()
l1.append(x)
l2.append(y)
l1c1 = 0
l1c2 = 0
for j in l1:
if j == "0":
l1c1 = l1c1 + 1
else:
l1c2 = l1c2 + 1
l2c1 = 0
l2c2 = 0
for j in l2:
if j == "0":
l2c1 = l2c1 + 1
else:
l2c2 = l2c2 + 1
if l1c1 > l2c1:
r = r + l2c1
else:
r = r + l1c1
if l1c2 > l2c2:
r = r + l2c2
else:
r = r + l1c2
print(r)
|
Title: Cupboards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One foggy Stockholm morning, Karlsson decided to snack on some jam in his friend Lillebror Svantenson's house. Fortunately for Karlsson, there wasn't anybody in his friend's house. Karlsson was not going to be hungry any longer, so he decided to get some food in the house.
Karlsson's gaze immediately fell on *n* wooden cupboards, standing in the kitchen. He immediately realized that these cupboards have hidden jam stocks. Karlsson began to fly greedily around the kitchen, opening and closing the cupboards' doors, grab and empty all the jars of jam that he could find.
And now all jars of jam are empty, Karlsson has had enough and does not want to leave traces of his stay, so as not to let down his friend. Each of the cupboards has two doors: the left one and the right one. Karlsson remembers that when he rushed to the kitchen, all the cupboards' left doors were in the same position (open or closed), similarly, all the cupboards' right doors were in the same position (open or closed). Karlsson wants the doors to meet this condition as well by the time the family returns. Karlsson does not remember the position of all the left doors, also, he cannot remember the position of all the right doors. Therefore, it does not matter to him in what position will be all left or right doors. It is important to leave all the left doors in the same position, and all the right doors in the same position. For example, all the left doors may be closed, and all the right ones may be open.
Karlsson needs one second to open or close a door of a cupboard. He understands that he has very little time before the family returns, so he wants to know the minimum number of seconds *t*, in which he is able to bring all the cupboard doors in the required position.
Your task is to write a program that will determine the required number of seconds *t*.
Input Specification:
The first input line contains a single integer *n* — the number of cupboards in the kitchen (2<=≤<=*n*<=≤<=104). Then follow *n* lines, each containing two integers *l**i* and *r**i* (0<=≤<=*l**i*,<=*r**i*<=≤<=1). Number *l**i* equals one, if the left door of the *i*-th cupboard is opened, otherwise number *l**i* equals zero. Similarly, number *r**i* equals one, if the right door of the *i*-th cupboard is opened, otherwise number *r**i* equals zero.
The numbers in the lines are separated by single spaces.
Output Specification:
In the only output line print a single integer *t* — the minimum number of seconds Karlsson needs to change the doors of all cupboards to the position he needs.
Demo Input:
['5\n0 1\n1 0\n0 1\n1 1\n0 1\n']
Demo Output:
['3\n']
Note:
none
|
```python
n = int(input())
l1 = []
l2 = []
d1 = {}
d2 = {}
r = 0
for i in range(n):
x, y = input().split()
l1.append(x)
l2.append(y)
l1c1 = 0
l1c2 = 0
for j in l1:
if j == "0":
l1c1 = l1c1 + 1
else:
l1c2 = l1c2 + 1
l2c1 = 0
l2c2 = 0
for j in l2:
if j == "0":
l2c1 = l2c1 + 1
else:
l2c2 = l2c2 + 1
if l1c1 > l2c1:
r = r + l2c1
else:
r = r + l1c1
if l1c2 > l2c2:
r = r + l2c2
else:
r = r + l1c2
print(r)
```
| 0
|
|
228
|
A
|
Is your horseshoe on the other hoof?
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Valera the Horse is going to the party with friends. He has been following the fashion trends for a while, and he knows that it is very popular to wear all horseshoes of different color. Valera has got four horseshoes left from the last year, but maybe some of them have the same color. In this case he needs to go to the store and buy some few more horseshoes, not to lose face in front of his stylish comrades.
Fortunately, the store sells horseshoes of all colors under the sun and Valera has enough money to buy any four of them. However, in order to save the money, he would like to spend as little money as possible, so you need to help Valera and determine what is the minimum number of horseshoes he needs to buy to wear four horseshoes of different colors to a party.
|
The first line contains four space-separated integers *s*1,<=*s*2,<=*s*3,<=*s*4 (1<=≤<=*s*1,<=*s*2,<=*s*3,<=*s*4<=≤<=109) — the colors of horseshoes Valera has.
Consider all possible colors indexed with integers.
|
Print a single integer — the minimum number of horseshoes Valera needs to buy.
|
[
"1 7 3 3\n",
"7 7 7 7\n"
] |
[
"1\n",
"3\n"
] |
none
| 500
|
[
{
"input": "1 7 3 3",
"output": "1"
},
{
"input": "7 7 7 7",
"output": "3"
},
{
"input": "81170865 673572653 756938629 995577259",
"output": "0"
},
{
"input": "3491663 217797045 522540872 715355328",
"output": "0"
},
{
"input": "251590420 586975278 916631563 586975278",
"output": "1"
},
{
"input": "259504825 377489979 588153796 377489979",
"output": "1"
},
{
"input": "652588203 931100304 931100304 652588203",
"output": "2"
},
{
"input": "391958720 651507265 391958720 651507265",
"output": "2"
},
{
"input": "90793237 90793237 90793237 90793237",
"output": "3"
},
{
"input": "551651653 551651653 551651653 551651653",
"output": "3"
},
{
"input": "156630260 609654355 668943582 973622757",
"output": "0"
},
{
"input": "17061017 110313588 434481173 796661222",
"output": "0"
},
{
"input": "24975422 256716298 337790533 690960249",
"output": "0"
},
{
"input": "255635360 732742923 798648949 883146723",
"output": "0"
},
{
"input": "133315691 265159773 734556507 265159773",
"output": "1"
},
{
"input": "28442865 741657755 978106882 978106882",
"output": "1"
},
{
"input": "131245479 174845575 497483467 131245479",
"output": "1"
},
{
"input": "139159884 616215581 958341883 616215581",
"output": "1"
},
{
"input": "147784432 947653080 947653080 947653080",
"output": "2"
},
{
"input": "94055790 756126496 756126496 94055790",
"output": "2"
},
{
"input": "240458500 511952208 240458500 511952208",
"output": "2"
},
{
"input": "681828506 972810624 972810624 681828506",
"output": "2"
},
{
"input": "454961014 454961014 454961014 454961014",
"output": "3"
},
{
"input": "915819430 915819430 915819430 915819430",
"output": "3"
},
{
"input": "671645142 671645142 671645142 671645142",
"output": "3"
},
{
"input": "132503558 132503558 132503558 132503558",
"output": "3"
},
{
"input": "5 5 999999 6",
"output": "1"
},
{
"input": "1 1 2 5",
"output": "1"
},
{
"input": "2 1 2 3",
"output": "1"
},
{
"input": "1 1 3 5",
"output": "1"
},
{
"input": "1 1 3 3",
"output": "2"
},
{
"input": "2 2 2 1",
"output": "2"
},
{
"input": "3 1 1 1",
"output": "2"
},
{
"input": "1 2 2 2",
"output": "2"
}
] | 1,686,431,353
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 34
| 92
| 0
|
l=list(map(int,input().split()))
distinct=set(l)
ans=len(l) - len(distinct)
print(ans)
|
Title: Is your horseshoe on the other hoof?
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera the Horse is going to the party with friends. He has been following the fashion trends for a while, and he knows that it is very popular to wear all horseshoes of different color. Valera has got four horseshoes left from the last year, but maybe some of them have the same color. In this case he needs to go to the store and buy some few more horseshoes, not to lose face in front of his stylish comrades.
Fortunately, the store sells horseshoes of all colors under the sun and Valera has enough money to buy any four of them. However, in order to save the money, he would like to spend as little money as possible, so you need to help Valera and determine what is the minimum number of horseshoes he needs to buy to wear four horseshoes of different colors to a party.
Input Specification:
The first line contains four space-separated integers *s*1,<=*s*2,<=*s*3,<=*s*4 (1<=≤<=*s*1,<=*s*2,<=*s*3,<=*s*4<=≤<=109) — the colors of horseshoes Valera has.
Consider all possible colors indexed with integers.
Output Specification:
Print a single integer — the minimum number of horseshoes Valera needs to buy.
Demo Input:
['1 7 3 3\n', '7 7 7 7\n']
Demo Output:
['1\n', '3\n']
Note:
none
|
```python
l=list(map(int,input().split()))
distinct=set(l)
ans=len(l) - len(distinct)
print(ans)
```
| 3
|
|
433
|
B
|
Kuriyama Mirai's Stones
|
PROGRAMMING
| 1,200
|
[
"dp",
"implementation",
"sortings"
] | null | null |
Kuriyama Mirai has killed many monsters and got many (namely *n*) stones. She numbers the stones from 1 to *n*. The cost of the *i*-th stone is *v**i*. Kuriyama Mirai wants to know something about these stones so she will ask you two kinds of questions:
1. She will tell you two numbers, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), and you should tell her . 1. Let *u**i* be the cost of the *i*-th cheapest stone (the cost that will be on the *i*-th place if we arrange all the stone costs in non-decreasing order). This time she will tell you two numbers, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), and you should tell her .
For every question you should give the correct answer, or Kuriyama Mirai will say "fuyukai desu" and then become unhappy.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=105). The second line contains *n* integers: *v*1,<=*v*2,<=...,<=*v**n* (1<=≤<=*v**i*<=≤<=109) — costs of the stones.
The third line contains an integer *m* (1<=≤<=*m*<=≤<=105) — the number of Kuriyama Mirai's questions. Then follow *m* lines, each line contains three integers *type*, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*; 1<=≤<=*type*<=≤<=2), describing a question. If *type* equal to 1, then you should output the answer for the first question, else you should output the answer for the second one.
|
Print *m* lines. Each line must contain an integer — the answer to Kuriyama Mirai's question. Print the answers to the questions in the order of input.
|
[
"6\n6 4 2 7 2 7\n3\n2 3 6\n1 3 4\n1 1 6\n",
"4\n5 5 2 3\n10\n1 2 4\n2 1 4\n1 1 1\n2 1 4\n2 1 2\n1 1 1\n1 3 3\n1 1 3\n1 4 4\n1 2 2\n"
] |
[
"24\n9\n28\n",
"10\n15\n5\n15\n5\n5\n2\n12\n3\n5\n"
] |
Please note that the answers to the questions may overflow 32-bit integer type.
| 1,500
|
[
{
"input": "6\n6 4 2 7 2 7\n3\n2 3 6\n1 3 4\n1 1 6",
"output": "24\n9\n28"
},
{
"input": "4\n5 5 2 3\n10\n1 2 4\n2 1 4\n1 1 1\n2 1 4\n2 1 2\n1 1 1\n1 3 3\n1 1 3\n1 4 4\n1 2 2",
"output": "10\n15\n5\n15\n5\n5\n2\n12\n3\n5"
},
{
"input": "4\n2 2 3 6\n9\n2 2 3\n1 1 3\n2 2 3\n2 2 3\n2 2 2\n1 1 3\n1 1 3\n2 1 4\n1 1 2",
"output": "5\n7\n5\n5\n2\n7\n7\n13\n4"
},
{
"input": "18\n26 46 56 18 78 88 86 93 13 77 21 84 59 61 5 74 72 52\n25\n1 10 10\n1 9 13\n2 13 17\n1 8 14\n2 2 6\n1 12 16\n2 15 17\n2 3 6\n1 3 13\n2 8 9\n2 17 17\n1 17 17\n2 5 10\n2 1 18\n1 4 16\n1 1 13\n1 1 8\n2 7 11\n2 6 12\n1 5 9\n1 4 5\n2 7 15\n1 8 8\n1 8 14\n1 3 7",
"output": "77\n254\n413\n408\n124\n283\n258\n111\n673\n115\n88\n72\n300\n1009\n757\n745\n491\n300\n420\n358\n96\n613\n93\n408\n326"
},
{
"input": "56\n43 100 44 66 65 11 26 75 96 77 5 15 75 96 11 44 11 97 75 53 33 26 32 33 90 26 68 72 5 44 53 26 33 88 68 25 84 21 25 92 1 84 21 66 94 35 76 51 11 95 67 4 61 3 34 18\n27\n1 20 38\n1 11 46\n2 42 53\n1 8 11\n2 11 42\n2 35 39\n2 37 41\n1 48 51\n1 32 51\n1 36 40\n1 31 56\n1 18 38\n2 9 51\n1 7 48\n1 15 52\n1 27 31\n2 5 19\n2 35 50\n1 31 34\n1 2 7\n2 15 33\n2 46 47\n1 26 28\n2 3 29\n1 23 45\n2 29 55\n1 14 29",
"output": "880\n1727\n1026\n253\n1429\n335\n350\n224\n1063\n247\n1236\n1052\n2215\n2128\n1840\n242\n278\n1223\n200\n312\n722\n168\n166\n662\n1151\n2028\n772"
},
{
"input": "18\n38 93 48 14 69 85 26 47 71 11 57 9 38 65 72 78 52 47\n38\n2 10 12\n1 6 18\n2 2 2\n1 3 15\n2 1 16\n2 5 13\n1 9 17\n1 2 15\n2 5 17\n1 15 15\n2 4 11\n2 3 4\n2 2 5\n2 1 17\n2 6 16\n2 8 16\n2 8 14\n1 9 12\n2 8 13\n2 1 14\n2 5 13\n1 2 3\n1 9 14\n2 12 15\n2 3 3\n2 9 13\n2 4 12\n2 11 14\n2 6 16\n1 8 14\n1 12 15\n2 3 4\n1 3 5\n2 4 14\n1 6 6\n2 7 14\n2 7 18\n1 8 12",
"output": "174\n658\n11\n612\n742\n461\n453\n705\n767\n72\n353\n40\n89\n827\n644\n559\n409\n148\n338\n592\n461\n141\n251\n277\n14\n291\n418\n262\n644\n298\n184\n40\n131\n558\n85\n456\n784\n195"
},
{
"input": "1\n2\n10\n1 1 1\n1 1 1\n2 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n2 1 1\n1 1 1\n1 1 1",
"output": "2\n2\n2\n2\n2\n2\n2\n2\n2\n2"
},
{
"input": "2\n1 5\n8\n2 1 2\n1 1 1\n1 1 2\n1 1 1\n2 2 2\n2 1 2\n1 1 1\n1 2 2",
"output": "6\n1\n6\n1\n5\n6\n1\n5"
},
{
"input": "8\n2 6 4 6 8 4 7 7\n9\n2 6 8\n1 2 3\n2 3 4\n2 1 2\n2 4 5\n1 3 8\n2 4 4\n2 2 4\n1 2 4",
"output": "22\n10\n10\n6\n12\n36\n6\n14\n16"
},
{
"input": "4\n1 1 2 8\n9\n1 2 4\n1 2 4\n1 1 2\n2 3 4\n2 1 1\n1 2 3\n2 3 4\n1 1 4\n1 1 4",
"output": "11\n11\n2\n10\n1\n3\n10\n12\n12"
},
{
"input": "5\n8 8 1 10 7\n5\n1 2 2\n1 1 4\n1 3 3\n1 2 2\n1 2 5",
"output": "8\n27\n1\n8\n26"
},
{
"input": "10\n10 3 3 2 1 2 7 1 1 1\n4\n2 10 10\n2 6 6\n1 5 10\n1 6 8",
"output": "10\n2\n13\n10"
},
{
"input": "6\n6 6 3 8 5 4\n3\n2 2 3\n2 1 5\n1 1 5",
"output": "9\n24\n28"
},
{
"input": "1\n1\n2\n1 1 1\n2 1 1",
"output": "1\n1"
}
] | 1,679,954,247
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 46
| 1,060
| 17,817,600
|
n = int(input())
l = list(map(int, input().split()))
l1 = sorted(l)
lsum = []
l1sum = []
sum = 0
sum1 = 0
for i in range(n):
sum += l[i]
lsum.append(sum)
for i in range(n):
sum1 += l1[i]
l1sum.append(sum1)
for i in range(int(input())):
t, l, r = map(int, input().split())
if t == 1:
if l<=1:
print(lsum[r-1])
else:
print(lsum[r-1] - lsum[l-2])
else:
if l<=1:
print(l1sum[r-1])
else:
print(l1sum[r-1] - l1sum[l-2])
|
Title: Kuriyama Mirai's Stones
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kuriyama Mirai has killed many monsters and got many (namely *n*) stones. She numbers the stones from 1 to *n*. The cost of the *i*-th stone is *v**i*. Kuriyama Mirai wants to know something about these stones so she will ask you two kinds of questions:
1. She will tell you two numbers, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), and you should tell her . 1. Let *u**i* be the cost of the *i*-th cheapest stone (the cost that will be on the *i*-th place if we arrange all the stone costs in non-decreasing order). This time she will tell you two numbers, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), and you should tell her .
For every question you should give the correct answer, or Kuriyama Mirai will say "fuyukai desu" and then become unhappy.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=105). The second line contains *n* integers: *v*1,<=*v*2,<=...,<=*v**n* (1<=≤<=*v**i*<=≤<=109) — costs of the stones.
The third line contains an integer *m* (1<=≤<=*m*<=≤<=105) — the number of Kuriyama Mirai's questions. Then follow *m* lines, each line contains three integers *type*, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*; 1<=≤<=*type*<=≤<=2), describing a question. If *type* equal to 1, then you should output the answer for the first question, else you should output the answer for the second one.
Output Specification:
Print *m* lines. Each line must contain an integer — the answer to Kuriyama Mirai's question. Print the answers to the questions in the order of input.
Demo Input:
['6\n6 4 2 7 2 7\n3\n2 3 6\n1 3 4\n1 1 6\n', '4\n5 5 2 3\n10\n1 2 4\n2 1 4\n1 1 1\n2 1 4\n2 1 2\n1 1 1\n1 3 3\n1 1 3\n1 4 4\n1 2 2\n']
Demo Output:
['24\n9\n28\n', '10\n15\n5\n15\n5\n5\n2\n12\n3\n5\n']
Note:
Please note that the answers to the questions may overflow 32-bit integer type.
|
```python
n = int(input())
l = list(map(int, input().split()))
l1 = sorted(l)
lsum = []
l1sum = []
sum = 0
sum1 = 0
for i in range(n):
sum += l[i]
lsum.append(sum)
for i in range(n):
sum1 += l1[i]
l1sum.append(sum1)
for i in range(int(input())):
t, l, r = map(int, input().split())
if t == 1:
if l<=1:
print(lsum[r-1])
else:
print(lsum[r-1] - lsum[l-2])
else:
if l<=1:
print(l1sum[r-1])
else:
print(l1sum[r-1] - l1sum[l-2])
```
| 3
|
|
918
|
A
|
Eleven
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation"
] | null | null |
Eleven wants to choose a new name for herself. As a bunch of geeks, her friends suggested an algorithm to choose a name for her. Eleven wants her name to have exactly *n* characters.
Her friend suggested that her name should only consist of uppercase and lowercase letters 'O'. More precisely, they suggested that the *i*-th letter of her name should be 'O' (uppercase) if *i* is a member of Fibonacci sequence, and 'o' (lowercase) otherwise. The letters in the name are numbered from 1 to *n*. Fibonacci sequence is the sequence *f* where
- *f*1<==<=1, - *f*2<==<=1, - *f**n*<==<=*f**n*<=-<=2<=+<=*f**n*<=-<=1 (*n*<=><=2).
As her friends are too young to know what Fibonacci sequence is, they asked you to help Eleven determine her new name.
|
The first and only line of input contains an integer *n* (1<=≤<=*n*<=≤<=1000).
|
Print Eleven's new name on the first and only line of output.
|
[
"8\n",
"15\n"
] |
[
"OOOoOooO\n",
"OOOoOooOooooOoo\n"
] |
none
| 500
|
[
{
"input": "8",
"output": "OOOoOooO"
},
{
"input": "15",
"output": "OOOoOooOooooOoo"
},
{
"input": "85",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooo"
},
{
"input": "381",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooo"
},
{
"input": "805",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "1000",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "1",
"output": "O"
},
{
"input": "2",
"output": "OO"
},
{
"input": "3",
"output": "OOO"
},
{
"input": "5",
"output": "OOOoO"
},
{
"input": "17",
"output": "OOOoOooOooooOoooo"
},
{
"input": "49",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooo"
},
{
"input": "256",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooo"
},
{
"input": "512",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "933",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "61",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooo"
},
{
"input": "781",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "999",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
}
] | 1,570,111,684
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 18
| 124
| 307,200
|
a=int(input())
i=0
s=[0,1]
while i in range(1000):
q=s[i]+s[i+1]
s.append(q)
i+=1
s1=''
for k in range(1,a+1):
if s.count(k)==0:
s1+='o'
else:
s1+='O'
print(s1)
|
Title: Eleven
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Eleven wants to choose a new name for herself. As a bunch of geeks, her friends suggested an algorithm to choose a name for her. Eleven wants her name to have exactly *n* characters.
Her friend suggested that her name should only consist of uppercase and lowercase letters 'O'. More precisely, they suggested that the *i*-th letter of her name should be 'O' (uppercase) if *i* is a member of Fibonacci sequence, and 'o' (lowercase) otherwise. The letters in the name are numbered from 1 to *n*. Fibonacci sequence is the sequence *f* where
- *f*1<==<=1, - *f*2<==<=1, - *f**n*<==<=*f**n*<=-<=2<=+<=*f**n*<=-<=1 (*n*<=><=2).
As her friends are too young to know what Fibonacci sequence is, they asked you to help Eleven determine her new name.
Input Specification:
The first and only line of input contains an integer *n* (1<=≤<=*n*<=≤<=1000).
Output Specification:
Print Eleven's new name on the first and only line of output.
Demo Input:
['8\n', '15\n']
Demo Output:
['OOOoOooO\n', 'OOOoOooOooooOoo\n']
Note:
none
|
```python
a=int(input())
i=0
s=[0,1]
while i in range(1000):
q=s[i]+s[i+1]
s.append(q)
i+=1
s1=''
for k in range(1,a+1):
if s.count(k)==0:
s1+='o'
else:
s1+='O'
print(s1)
```
| 3
|
|
143
|
A
|
Help Vasilisa the Wise 2
|
PROGRAMMING
| 1,000
|
[
"brute force",
"math"
] | null | null |
Vasilisa the Wise from the Kingdom of Far Far Away got a magic box with a secret as a present from her friend Hellawisa the Wise from the Kingdom of A Little Closer. However, Vasilisa the Wise does not know what the box's secret is, since she cannot open it again. She hopes that you will help her one more time with that.
The box's lock looks as follows: it contains 4 identical deepenings for gems as a 2<=×<=2 square, and some integer numbers are written at the lock's edge near the deepenings. The example of a lock is given on the picture below.
The box is accompanied with 9 gems. Their shapes match the deepenings' shapes and each gem contains one number from 1 to 9 (each number is written on exactly one gem). The box will only open after it is decorated with gems correctly: that is, each deepening in the lock should be filled with exactly one gem. Also, the sums of numbers in the square's rows, columns and two diagonals of the square should match the numbers written at the lock's edge. For example, the above lock will open if we fill the deepenings with gems with numbers as is shown on the picture below.
Now Vasilisa the Wise wants to define, given the numbers on the box's lock, which gems she should put in the deepenings to open the box. Help Vasilisa to solve this challenging task.
|
The input contains numbers written on the edges of the lock of the box. The first line contains space-separated integers *r*1 and *r*2 that define the required sums of numbers in the rows of the square. The second line contains space-separated integers *c*1 and *c*2 that define the required sums of numbers in the columns of the square. The third line contains space-separated integers *d*1 and *d*2 that define the required sums of numbers on the main and on the side diagonals of the square (1<=≤<=*r*1,<=*r*2,<=*c*1,<=*c*2,<=*d*1,<=*d*2<=≤<=20). Correspondence between the above 6 variables and places where they are written is shown on the picture below. For more clarifications please look at the second sample test that demonstrates the example given in the problem statement.
|
Print the scheme of decorating the box with stones: two lines containing two space-separated integers from 1 to 9. The numbers should be pairwise different. If there is no solution for the given lock, then print the single number "-1" (without the quotes).
If there are several solutions, output any.
|
[
"3 7\n4 6\n5 5\n",
"11 10\n13 8\n5 16\n",
"1 2\n3 4\n5 6\n",
"10 10\n10 10\n10 10\n"
] |
[
"1 2\n3 4\n",
"4 7\n9 1\n",
"-1\n",
"-1\n"
] |
Pay attention to the last test from the statement: it is impossible to open the box because for that Vasilisa the Wise would need 4 identical gems containing number "5". However, Vasilisa only has one gem with each number from 1 to 9.
| 500
|
[
{
"input": "3 7\n4 6\n5 5",
"output": "1 2\n3 4"
},
{
"input": "11 10\n13 8\n5 16",
"output": "4 7\n9 1"
},
{
"input": "1 2\n3 4\n5 6",
"output": "-1"
},
{
"input": "10 10\n10 10\n10 10",
"output": "-1"
},
{
"input": "5 13\n8 10\n11 7",
"output": "3 2\n5 8"
},
{
"input": "12 17\n10 19\n13 16",
"output": "-1"
},
{
"input": "11 11\n17 5\n12 10",
"output": "9 2\n8 3"
},
{
"input": "12 11\n11 12\n16 7",
"output": "-1"
},
{
"input": "5 9\n7 7\n8 6",
"output": "3 2\n4 5"
},
{
"input": "10 7\n4 13\n11 6",
"output": "-1"
},
{
"input": "18 10\n16 12\n12 16",
"output": "-1"
},
{
"input": "13 6\n10 9\n6 13",
"output": "-1"
},
{
"input": "14 16\n16 14\n18 12",
"output": "-1"
},
{
"input": "16 10\n16 10\n12 14",
"output": "-1"
},
{
"input": "11 9\n12 8\n11 9",
"output": "-1"
},
{
"input": "5 14\n10 9\n10 9",
"output": "-1"
},
{
"input": "2 4\n1 5\n3 3",
"output": "-1"
},
{
"input": "17 16\n14 19\n18 15",
"output": "-1"
},
{
"input": "12 12\n14 10\n16 8",
"output": "9 3\n5 7"
},
{
"input": "15 11\n16 10\n9 17",
"output": "7 8\n9 2"
},
{
"input": "8 10\n9 9\n13 5",
"output": "6 2\n3 7"
},
{
"input": "13 7\n10 10\n5 15",
"output": "4 9\n6 1"
},
{
"input": "14 11\n9 16\n16 9",
"output": "-1"
},
{
"input": "12 8\n14 6\n8 12",
"output": "-1"
},
{
"input": "10 6\n6 10\n4 12",
"output": "-1"
},
{
"input": "10 8\n10 8\n4 14",
"output": "-1"
},
{
"input": "14 13\n9 18\n14 13",
"output": "-1"
},
{
"input": "9 14\n8 15\n8 15",
"output": "-1"
},
{
"input": "3 8\n2 9\n6 5",
"output": "-1"
},
{
"input": "14 17\n18 13\n15 16",
"output": "-1"
},
{
"input": "16 14\n15 15\n17 13",
"output": "9 7\n6 8"
},
{
"input": "14 11\n16 9\n13 12",
"output": "9 5\n7 4"
},
{
"input": "13 10\n11 12\n7 16",
"output": "4 9\n7 3"
},
{
"input": "14 8\n11 11\n13 9",
"output": "8 6\n3 5"
},
{
"input": "12 11\n13 10\n10 13",
"output": "-1"
},
{
"input": "6 5\n2 9\n5 6",
"output": "-1"
},
{
"input": "7 8\n8 7\n12 3",
"output": "-1"
},
{
"input": "7 11\n7 11\n6 12",
"output": "-1"
},
{
"input": "8 5\n11 2\n8 5",
"output": "-1"
},
{
"input": "10 16\n14 12\n14 12",
"output": "-1"
},
{
"input": "7 9\n4 12\n5 11",
"output": "-1"
},
{
"input": "11 13\n19 5\n12 12",
"output": "-1"
},
{
"input": "8 12\n5 15\n11 9",
"output": "2 6\n3 9"
},
{
"input": "16 5\n13 8\n10 11",
"output": "9 7\n4 1"
},
{
"input": "3 14\n8 9\n10 7",
"output": "2 1\n6 8"
},
{
"input": "12 14\n11 15\n9 17",
"output": "3 9\n8 6"
},
{
"input": "13 7\n9 11\n14 6",
"output": "8 5\n1 6"
},
{
"input": "13 8\n15 6\n11 10",
"output": "9 4\n6 2"
},
{
"input": "9 6\n5 10\n3 12",
"output": "1 8\n4 2"
},
{
"input": "12 7\n5 14\n8 11",
"output": "3 9\n2 5"
},
{
"input": "9 12\n3 17\n10 10",
"output": "-1"
},
{
"input": "3 8\n4 6\n5 5",
"output": "-1"
}
] | 1,656,384,331
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 3
| 404
| 2,662,400
|
from math import *
import string
import sys
from bisect import bisect, bisect_left, bisect_right
from collections import Counter, defaultdict, deque
from functools import cache, lru_cache
input = sys.stdin.readline
M = 10**9+7
def mod(x):
return x % M
def inp():
return(int(input()))
def inlt():
return(list(map(int, input().split())))
def floatl():
return(list(map(float, input().split())))
def insr():
s = input()
return(list(s[:len(s) - 1]))
def ins():
s = input()
return s[:len(s) - 1]
def invr():
return(map(int, input().split()))
r1, r2 = inlt()
c1, c2 = inlt()
d1, d2 = inlt()
def solve():
for a00 in range(1, min(r1, c1, d1)):
a01 = r1-a00
a10 = c1-a00
a11 = d1-a00
if a10 + a11 == r2 and a01 + a11 == c2 and a01 + a10 == d2:
print(a00, a01)
print(a10, a11)
return
print(-1)
solve()
|
Title: Help Vasilisa the Wise 2
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasilisa the Wise from the Kingdom of Far Far Away got a magic box with a secret as a present from her friend Hellawisa the Wise from the Kingdom of A Little Closer. However, Vasilisa the Wise does not know what the box's secret is, since she cannot open it again. She hopes that you will help her one more time with that.
The box's lock looks as follows: it contains 4 identical deepenings for gems as a 2<=×<=2 square, and some integer numbers are written at the lock's edge near the deepenings. The example of a lock is given on the picture below.
The box is accompanied with 9 gems. Their shapes match the deepenings' shapes and each gem contains one number from 1 to 9 (each number is written on exactly one gem). The box will only open after it is decorated with gems correctly: that is, each deepening in the lock should be filled with exactly one gem. Also, the sums of numbers in the square's rows, columns and two diagonals of the square should match the numbers written at the lock's edge. For example, the above lock will open if we fill the deepenings with gems with numbers as is shown on the picture below.
Now Vasilisa the Wise wants to define, given the numbers on the box's lock, which gems she should put in the deepenings to open the box. Help Vasilisa to solve this challenging task.
Input Specification:
The input contains numbers written on the edges of the lock of the box. The first line contains space-separated integers *r*1 and *r*2 that define the required sums of numbers in the rows of the square. The second line contains space-separated integers *c*1 and *c*2 that define the required sums of numbers in the columns of the square. The third line contains space-separated integers *d*1 and *d*2 that define the required sums of numbers on the main and on the side diagonals of the square (1<=≤<=*r*1,<=*r*2,<=*c*1,<=*c*2,<=*d*1,<=*d*2<=≤<=20). Correspondence between the above 6 variables and places where they are written is shown on the picture below. For more clarifications please look at the second sample test that demonstrates the example given in the problem statement.
Output Specification:
Print the scheme of decorating the box with stones: two lines containing two space-separated integers from 1 to 9. The numbers should be pairwise different. If there is no solution for the given lock, then print the single number "-1" (without the quotes).
If there are several solutions, output any.
Demo Input:
['3 7\n4 6\n5 5\n', '11 10\n13 8\n5 16\n', '1 2\n3 4\n5 6\n', '10 10\n10 10\n10 10\n']
Demo Output:
['1 2\n3 4\n', '4 7\n9 1\n', '-1\n', '-1\n']
Note:
Pay attention to the last test from the statement: it is impossible to open the box because for that Vasilisa the Wise would need 4 identical gems containing number "5". However, Vasilisa only has one gem with each number from 1 to 9.
|
```python
from math import *
import string
import sys
from bisect import bisect, bisect_left, bisect_right
from collections import Counter, defaultdict, deque
from functools import cache, lru_cache
input = sys.stdin.readline
M = 10**9+7
def mod(x):
return x % M
def inp():
return(int(input()))
def inlt():
return(list(map(int, input().split())))
def floatl():
return(list(map(float, input().split())))
def insr():
s = input()
return(list(s[:len(s) - 1]))
def ins():
s = input()
return s[:len(s) - 1]
def invr():
return(map(int, input().split()))
r1, r2 = inlt()
c1, c2 = inlt()
d1, d2 = inlt()
def solve():
for a00 in range(1, min(r1, c1, d1)):
a01 = r1-a00
a10 = c1-a00
a11 = d1-a00
if a10 + a11 == r2 and a01 + a11 == c2 and a01 + a10 == d2:
print(a00, a01)
print(a10, a11)
return
print(-1)
solve()
```
| 0
|
|
525
|
A
|
Vitaliy and Pie
|
PROGRAMMING
| 1,100
|
[
"greedy",
"hashing",
"strings"
] | null | null |
After a hard day Vitaly got very hungry and he wants to eat his favorite potato pie. But it's not that simple. Vitaly is in the first room of the house with *n* room located in a line and numbered starting from one from left to right. You can go from the first room to the second room, from the second room to the third room and so on — you can go from the (*n*<=-<=1)-th room to the *n*-th room. Thus, you can go to room *x* only from room *x*<=-<=1.
The potato pie is located in the *n*-th room and Vitaly needs to go there.
Each pair of consecutive rooms has a door between them. In order to go to room *x* from room *x*<=-<=1, you need to open the door between the rooms with the corresponding key.
In total the house has several types of doors (represented by uppercase Latin letters) and several types of keys (represented by lowercase Latin letters). The key of type *t* can open the door of type *T* if and only if *t* and *T* are the same letter, written in different cases. For example, key f can open door F.
Each of the first *n*<=-<=1 rooms contains exactly one key of some type that Vitaly can use to get to next rooms. Once the door is open with some key, Vitaly won't get the key from the keyhole but he will immediately run into the next room. In other words, each key can open no more than one door.
Vitaly realizes that he may end up in some room without the key that opens the door to the next room. Before the start his run for the potato pie Vitaly can buy any number of keys of any type that is guaranteed to get to room *n*.
Given the plan of the house, Vitaly wants to know what is the minimum number of keys he needs to buy to surely get to the room *n*, which has a delicious potato pie. Write a program that will help Vitaly find out this number.
|
The first line of the input contains a positive integer *n* (2<=≤<=*n*<=≤<=105) — the number of rooms in the house.
The second line of the input contains string *s* of length 2·*n*<=-<=2. Let's number the elements of the string from left to right, starting from one.
The odd positions in the given string *s* contain lowercase Latin letters — the types of the keys that lie in the corresponding rooms. Thus, each odd position *i* of the given string *s* contains a lowercase Latin letter — the type of the key that lies in room number (*i*<=+<=1)<=/<=2.
The even positions in the given string contain uppercase Latin letters — the types of doors between the rooms. Thus, each even position *i* of the given string *s* contains an uppercase letter — the type of the door that leads from room *i*<=/<=2 to room *i*<=/<=2<=+<=1.
|
Print the only integer — the minimum number of keys that Vitaly needs to buy to surely get from room one to room *n*.
|
[
"3\naAbB\n",
"4\naBaCaB\n",
"5\nxYyXzZaZ\n"
] |
[
"0\n",
"3\n",
"2\n"
] |
none
| 250
|
[
{
"input": "3\naAbB",
"output": "0"
},
{
"input": "4\naBaCaB",
"output": "3"
},
{
"input": "5\nxYyXzZaZ",
"output": "2"
},
{
"input": "26\naAbBcCdDeEfFgGhHiIjJkKlLmMnNoOpPqQrRsStTuUvVwWxXyY",
"output": "0"
},
{
"input": "26\nzAyBxCwDvEuFtGsHrIqJpKoLnMmNlOkPjQiRhSgTfUeVdWcXbY",
"output": "13"
},
{
"input": "5\naArRaRaR",
"output": "2"
},
{
"input": "2\ndA",
"output": "1"
},
{
"input": "2\ncB",
"output": "1"
},
{
"input": "10\nhNcMeXsSlHsUwYeMcA",
"output": "7"
},
{
"input": "100\nqDpInBmCrFwXpDbFgOzVvOcEmJrUcToAdEwEgTvBvBfWwRpGyEaXgDdRwVlQnYgWmWhMrHaIzPyXvGaFlRsVzHhZrOuVpXrKxFzAmWwPlFtNfPtJxVmLuHjKfYyArHrEnSwSzOvDpQhCgCqLlAcNpGhXrEeFuCmAqIkXyYtSsQwIxJzNiIuTgEbVuWrMwPrAlLyKaZ",
"output": "42"
},
{
"input": "2\ndD",
"output": "0"
},
{
"input": "2\ndE",
"output": "1"
},
{
"input": "3\ndDdD",
"output": "0"
},
{
"input": "3\ndEdD",
"output": "1"
},
{
"input": "3\ndEeD",
"output": "1"
},
{
"input": "3\ndEeF",
"output": "2"
}
] | 1,636,376,861
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
from collections import defaultdict
import sys
input = sys.stdin.readline
############ ---- Input Functions ---- ############
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def insr():
s = input()
return(list(s[:len(s) - 1]))
def invr():
return(map(int,input().split()))
hashdict = defaultdict(int)
n = inp()
key_rooms = insr()
ans = 0
for counter, ch in enumerate(key_rooms[0]):
if counter % 2 == 1:
key = ch.lower()
if hashdict[key] > 0:
hashdict[ch] -= 1
else:
ans += 1
else:
hashdict[ch] += 1
return ans
|
Title: Vitaliy and Pie
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After a hard day Vitaly got very hungry and he wants to eat his favorite potato pie. But it's not that simple. Vitaly is in the first room of the house with *n* room located in a line and numbered starting from one from left to right. You can go from the first room to the second room, from the second room to the third room and so on — you can go from the (*n*<=-<=1)-th room to the *n*-th room. Thus, you can go to room *x* only from room *x*<=-<=1.
The potato pie is located in the *n*-th room and Vitaly needs to go there.
Each pair of consecutive rooms has a door between them. In order to go to room *x* from room *x*<=-<=1, you need to open the door between the rooms with the corresponding key.
In total the house has several types of doors (represented by uppercase Latin letters) and several types of keys (represented by lowercase Latin letters). The key of type *t* can open the door of type *T* if and only if *t* and *T* are the same letter, written in different cases. For example, key f can open door F.
Each of the first *n*<=-<=1 rooms contains exactly one key of some type that Vitaly can use to get to next rooms. Once the door is open with some key, Vitaly won't get the key from the keyhole but he will immediately run into the next room. In other words, each key can open no more than one door.
Vitaly realizes that he may end up in some room without the key that opens the door to the next room. Before the start his run for the potato pie Vitaly can buy any number of keys of any type that is guaranteed to get to room *n*.
Given the plan of the house, Vitaly wants to know what is the minimum number of keys he needs to buy to surely get to the room *n*, which has a delicious potato pie. Write a program that will help Vitaly find out this number.
Input Specification:
The first line of the input contains a positive integer *n* (2<=≤<=*n*<=≤<=105) — the number of rooms in the house.
The second line of the input contains string *s* of length 2·*n*<=-<=2. Let's number the elements of the string from left to right, starting from one.
The odd positions in the given string *s* contain lowercase Latin letters — the types of the keys that lie in the corresponding rooms. Thus, each odd position *i* of the given string *s* contains a lowercase Latin letter — the type of the key that lies in room number (*i*<=+<=1)<=/<=2.
The even positions in the given string contain uppercase Latin letters — the types of doors between the rooms. Thus, each even position *i* of the given string *s* contains an uppercase letter — the type of the door that leads from room *i*<=/<=2 to room *i*<=/<=2<=+<=1.
Output Specification:
Print the only integer — the minimum number of keys that Vitaly needs to buy to surely get from room one to room *n*.
Demo Input:
['3\naAbB\n', '4\naBaCaB\n', '5\nxYyXzZaZ\n']
Demo Output:
['0\n', '3\n', '2\n']
Note:
none
|
```python
from collections import defaultdict
import sys
input = sys.stdin.readline
############ ---- Input Functions ---- ############
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def insr():
s = input()
return(list(s[:len(s) - 1]))
def invr():
return(map(int,input().split()))
hashdict = defaultdict(int)
n = inp()
key_rooms = insr()
ans = 0
for counter, ch in enumerate(key_rooms[0]):
if counter % 2 == 1:
key = ch.lower()
if hashdict[key] > 0:
hashdict[ch] -= 1
else:
ans += 1
else:
hashdict[ch] += 1
return ans
```
| -1
|
|
291
|
A
|
Spyke Talks
|
PROGRAMMING
| 800
|
[
"*special",
"implementation",
"sortings"
] | null | null |
Polycarpus is the director of a large corporation. There are *n* secretaries working for the corporation, each of them corresponds via the famous Spyke VoIP system during the day. We know that when two people call each other via Spyke, the Spyke network assigns a unique ID to this call, a positive integer session number.
One day Polycarpus wondered which secretaries are talking via the Spyke and which are not. For each secretary, he wrote out either the session number of his call or a 0 if this secretary wasn't talking via Spyke at that moment.
Help Polycarpus analyze these data and find out the number of pairs of secretaries that are talking. If Polycarpus has made a mistake in the data and the described situation could not have taken place, say so.
Note that the secretaries can correspond via Spyke not only with each other, but also with the people from other places. Also, Spyke conferences aren't permitted — that is, one call connects exactly two people.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=103) — the number of secretaries in Polycarpus's corporation. The next line contains *n* space-separated integers: *id*1,<=*id*2,<=...,<=*id**n* (0<=≤<=*id**i*<=≤<=109). Number *id**i* equals the number of the call session of the *i*-th secretary, if the secretary is talking via Spyke, or zero otherwise.
Consider the secretaries indexed from 1 to *n* in some way.
|
Print a single integer — the number of pairs of chatting secretaries, or -1 if Polycarpus's got a mistake in his records and the described situation could not have taken place.
|
[
"6\n0 1 7 1 7 10\n",
"3\n1 1 1\n",
"1\n0\n"
] |
[
"2\n",
"-1\n",
"0\n"
] |
In the first test sample there are two Spyke calls between secretaries: secretary 2 and secretary 4, secretary 3 and secretary 5.
In the second test sample the described situation is impossible as conferences aren't allowed.
| 500
|
[
{
"input": "6\n0 1 7 1 7 10",
"output": "2"
},
{
"input": "3\n1 1 1",
"output": "-1"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "5\n2 2 1 1 3",
"output": "2"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "10\n4 21 3 21 21 1 1 2 2 3",
"output": "-1"
},
{
"input": "2\n1 2",
"output": "0"
},
{
"input": "5\n0 0 0 0 0",
"output": "0"
},
{
"input": "6\n6 6 0 8 0 0",
"output": "1"
},
{
"input": "10\n0 0 0 0 0 1 0 1 0 1",
"output": "-1"
},
{
"input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 0 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 3 0 3 0 0 3 0 0 0 0 0 0 3 0 0 3 0 0 0 0 0 0 0 3 0 0 0 0 0",
"output": "-1"
},
{
"input": "1\n1000000000",
"output": "0"
},
{
"input": "2\n1 0",
"output": "0"
},
{
"input": "2\n1000000000 1000000000",
"output": "1"
},
{
"input": "5\n1 0 0 0 1",
"output": "1"
},
{
"input": "15\n380515742 842209759 945171461 664384656 945171461 474872104 0 0 131648973 131648973 474872104 842209759 664384656 0 380515742",
"output": "6"
},
{
"input": "123\n0 6361 8903 10428 0 258 0 10422 0 0 2642 1958 0 0 0 0 0 8249 1958 0 0 2642 0 0 0 11566 4709 1847 3998 0 1331 0 0 10289 2739 6135 3450 0 0 10994 6069 4337 5854 1331 5854 0 630 630 11244 5928 2706 0 683 214 0 9080 0 0 0 10422 683 11566 10994 0 0 3450 11244 11542 3998 1847 2708 9871 2739 2001 0 12216 6069 0 5928 0 10289 1307 0 1307 8903 0 6361 6135 6632 10428 0 0 632 258 9080 12216 4709 4967 2706 0 11542 2001 6632 0 8249 214 0 10301 4967 10301 7296 7296 10914 2708 4337 0 0 632 0 10914 0 9871 0",
"output": "40"
},
{
"input": "10\n0 3 2 3 2 0 1 3 3 0",
"output": "-1"
},
{
"input": "20\n0 1 2 0 0 0 0 5 3 4 0 0 1 1 3 0 4 0 1 0",
"output": "-1"
},
{
"input": "47\n1 6 0 6 1 1 6 4 3 6 5 3 6 3 2 2 5 1 4 7 3 5 6 1 6 7 4 5 6 3 3 3 7 4 1 6 1 1 7 1 3 1 5 5 1 3 6",
"output": "-1"
},
{
"input": "74\n0 0 0 0 0 37 0 0 0 0 0 0 0 8 0 0 9 0 0 0 0 0 0 0 0 0 8 0 0 0 0 0 9 0 7 0 0 0 0 0 0 19 19 0 0 0 0 0 0 0 0 0 0 0 0 17 0 30 0 0 0 0 0 0 30 0 0 0 0 0 0 0 37 0",
"output": "5"
},
{
"input": "3\n1 1 1",
"output": "-1"
},
{
"input": "2\n2 3",
"output": "0"
},
{
"input": "2\n2 2",
"output": "1"
},
{
"input": "5\n10000 10000 1 1 10000",
"output": "-1"
}
] | 1,599,544,767
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 7
| 186
| 307,200
|
n=int(input())
l=list(map(int,input().split()))
l.sort()
list1=[]
sum1=1
if n>1:
for i in range(n-1):
if l[i]==l[i+1]:
sum1+=1
if i==n-2:
list1.append(sum1)
else:
list1.append(sum1)
sum1=1
if max(list1)==1:
print(0)
elif max(list1)>2:
print(-1)
elif max(list1)==2:
count=0
for i in list1:
if i==2:
count+=1
print(count)
else:
print(0)
|
Title: Spyke Talks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus is the director of a large corporation. There are *n* secretaries working for the corporation, each of them corresponds via the famous Spyke VoIP system during the day. We know that when two people call each other via Spyke, the Spyke network assigns a unique ID to this call, a positive integer session number.
One day Polycarpus wondered which secretaries are talking via the Spyke and which are not. For each secretary, he wrote out either the session number of his call or a 0 if this secretary wasn't talking via Spyke at that moment.
Help Polycarpus analyze these data and find out the number of pairs of secretaries that are talking. If Polycarpus has made a mistake in the data and the described situation could not have taken place, say so.
Note that the secretaries can correspond via Spyke not only with each other, but also with the people from other places. Also, Spyke conferences aren't permitted — that is, one call connects exactly two people.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=103) — the number of secretaries in Polycarpus's corporation. The next line contains *n* space-separated integers: *id*1,<=*id*2,<=...,<=*id**n* (0<=≤<=*id**i*<=≤<=109). Number *id**i* equals the number of the call session of the *i*-th secretary, if the secretary is talking via Spyke, or zero otherwise.
Consider the secretaries indexed from 1 to *n* in some way.
Output Specification:
Print a single integer — the number of pairs of chatting secretaries, or -1 if Polycarpus's got a mistake in his records and the described situation could not have taken place.
Demo Input:
['6\n0 1 7 1 7 10\n', '3\n1 1 1\n', '1\n0\n']
Demo Output:
['2\n', '-1\n', '0\n']
Note:
In the first test sample there are two Spyke calls between secretaries: secretary 2 and secretary 4, secretary 3 and secretary 5.
In the second test sample the described situation is impossible as conferences aren't allowed.
|
```python
n=int(input())
l=list(map(int,input().split()))
l.sort()
list1=[]
sum1=1
if n>1:
for i in range(n-1):
if l[i]==l[i+1]:
sum1+=1
if i==n-2:
list1.append(sum1)
else:
list1.append(sum1)
sum1=1
if max(list1)==1:
print(0)
elif max(list1)>2:
print(-1)
elif max(list1)==2:
count=0
for i in list1:
if i==2:
count+=1
print(count)
else:
print(0)
```
| 0
|
|
115
|
A
|
Party
|
PROGRAMMING
| 900
|
[
"dfs and similar",
"graphs",
"trees"
] | null | null |
A company has *n* employees numbered from 1 to *n*. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee *A* is said to be the superior of another employee *B* if at least one of the following is true:
- Employee *A* is the immediate manager of employee *B* - Employee *B* has an immediate manager employee *C* such that employee *A* is the superior of employee *C*.
The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all *n* employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees *A* and *B* such that *A* is the superior of *B*.
What is the minimum number of groups that must be formed?
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=2000) — the number of employees.
The next *n* lines contain the integers *p**i* (1<=≤<=*p**i*<=≤<=*n* or *p**i*<==<=-1). Every *p**i* denotes the immediate manager for the *i*-th employee. If *p**i* is -1, that means that the *i*-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (*p**i*<=≠<=*i*). Also, there will be no managerial cycles.
|
Print a single integer denoting the minimum number of groups that will be formed in the party.
|
[
"5\n-1\n1\n2\n1\n-1\n"
] |
[
"3\n"
] |
For the first example, three groups are sufficient, for example:
- Employee 1 - Employees 2 and 4 - Employees 3 and 5
| 500
|
[
{
"input": "5\n-1\n1\n2\n1\n-1",
"output": "3"
},
{
"input": "4\n-1\n1\n2\n3",
"output": "4"
},
{
"input": "12\n-1\n1\n2\n3\n-1\n5\n6\n7\n-1\n9\n10\n11",
"output": "4"
},
{
"input": "6\n-1\n-1\n2\n3\n1\n1",
"output": "3"
},
{
"input": "3\n-1\n1\n1",
"output": "2"
},
{
"input": "1\n-1",
"output": "1"
},
{
"input": "2\n2\n-1",
"output": "2"
},
{
"input": "2\n-1\n-1",
"output": "1"
},
{
"input": "3\n2\n-1\n1",
"output": "3"
},
{
"input": "3\n-1\n-1\n-1",
"output": "1"
},
{
"input": "5\n4\n5\n1\n-1\n4",
"output": "3"
},
{
"input": "12\n-1\n1\n1\n1\n1\n1\n3\n4\n3\n3\n4\n7",
"output": "4"
},
{
"input": "12\n-1\n-1\n1\n-1\n1\n1\n5\n11\n8\n6\n6\n4",
"output": "5"
},
{
"input": "12\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n2\n-1\n-1\n-1",
"output": "2"
},
{
"input": "12\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1",
"output": "1"
},
{
"input": "12\n3\n4\n2\n8\n7\n1\n10\n12\n5\n-1\n9\n11",
"output": "12"
},
{
"input": "12\n5\n6\n7\n1\n-1\n9\n12\n4\n8\n-1\n3\n2",
"output": "11"
},
{
"input": "12\n-1\n9\n11\n6\n6\n-1\n6\n3\n8\n6\n1\n6",
"output": "6"
},
{
"input": "12\n7\n8\n4\n12\n7\n9\n-1\n-1\n-1\n8\n6\n-1",
"output": "3"
},
{
"input": "12\n-1\n10\n-1\n1\n-1\n5\n9\n12\n-1\n-1\n3\n-1",
"output": "2"
},
{
"input": "12\n-1\n7\n9\n12\n1\n7\n-1\n-1\n8\n5\n4\n-1",
"output": "3"
},
{
"input": "12\n11\n11\n8\n9\n1\n1\n2\n-1\n10\n3\n-1\n8",
"output": "5"
},
{
"input": "12\n-1\n8\n9\n-1\n4\n2\n11\n1\n-1\n6\n-1\n10",
"output": "6"
},
{
"input": "12\n7\n4\n4\n-1\n6\n7\n9\n7\n-1\n7\n4\n1",
"output": "4"
},
{
"input": "12\n-1\n6\n10\n6\n1\n11\n5\n9\n11\n5\n1\n9",
"output": "4"
},
{
"input": "12\n8\n6\n6\n8\n6\n-1\n6\n6\n8\n6\n8\n8",
"output": "3"
},
{
"input": "12\n3\n8\n9\n12\n2\n8\n11\n12\n2\n9\n-1\n11",
"output": "7"
},
{
"input": "7\n7\n6\n-1\n-1\n4\n4\n-1",
"output": "3"
},
{
"input": "77\n53\n-1\n48\n-1\n51\n-1\n54\n43\n70\n19\n64\n31\n45\n58\n6\n61\n41\n56\n34\n-1\n54\n-1\n66\n65\n74\n16\n38\n1\n2\n61\n-1\n34\n-1\n-1\n7\n19\n57\n13\n61\n42\n-1\n34\n65\n16\n31\n61\n6\n5\n34\n15\n-1\n13\n4\n22\n74\n7\n-1\n70\n38\n61\n45\n53\n46\n22\n-1\n65\n34\n19\n43\n48\n33\n58\n36\n-1\n47\n24\n34",
"output": "6"
},
{
"input": "5\n2\n3\n4\n5\n-1",
"output": "5"
}
] | 1,698,371,199
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 3
| 92
| 0
|
def bl(n, managers):
adj_list = [[] for _ in range(n)]
for i, manager in enumerate(managers):
if manager != -1:
adj_list[manager - 1].append(i)
return adj_list
def find_minimum_groups(adj_list, current_employee):
if not adj_list[current_employee]:
return 1
group_sizes = [find_minimum_groups(adj_list, employee) for employee in adj_list[current_employee]]
return max(group_sizes) + 1
n = int(input())
managers = [int(input()) for _ in range(n)]
adj_list = bl(n, managers)
result = find_minimum_groups(adj_list, 0)
print(result)
|
Title: Party
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A company has *n* employees numbered from 1 to *n*. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee *A* is said to be the superior of another employee *B* if at least one of the following is true:
- Employee *A* is the immediate manager of employee *B* - Employee *B* has an immediate manager employee *C* such that employee *A* is the superior of employee *C*.
The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all *n* employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees *A* and *B* such that *A* is the superior of *B*.
What is the minimum number of groups that must be formed?
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=2000) — the number of employees.
The next *n* lines contain the integers *p**i* (1<=≤<=*p**i*<=≤<=*n* or *p**i*<==<=-1). Every *p**i* denotes the immediate manager for the *i*-th employee. If *p**i* is -1, that means that the *i*-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (*p**i*<=≠<=*i*). Also, there will be no managerial cycles.
Output Specification:
Print a single integer denoting the minimum number of groups that will be formed in the party.
Demo Input:
['5\n-1\n1\n2\n1\n-1\n']
Demo Output:
['3\n']
Note:
For the first example, three groups are sufficient, for example:
- Employee 1 - Employees 2 and 4 - Employees 3 and 5
|
```python
def bl(n, managers):
adj_list = [[] for _ in range(n)]
for i, manager in enumerate(managers):
if manager != -1:
adj_list[manager - 1].append(i)
return adj_list
def find_minimum_groups(adj_list, current_employee):
if not adj_list[current_employee]:
return 1
group_sizes = [find_minimum_groups(adj_list, employee) for employee in adj_list[current_employee]]
return max(group_sizes) + 1
n = int(input())
managers = [int(input()) for _ in range(n)]
adj_list = bl(n, managers)
result = find_minimum_groups(adj_list, 0)
print(result)
```
| 0
|
|
1,009
|
D
|
Relatively Prime Graph
|
PROGRAMMING
| 1,700
|
[
"brute force",
"constructive algorithms",
"graphs",
"greedy",
"math"
] | null | null |
Let's call an undirected graph $G = (V, E)$ relatively prime if and only if for each edge $(v, u) \in E$ $GCD(v, u) = 1$ (the greatest common divisor of $v$ and $u$ is $1$). If there is no edge between some pair of vertices $v$ and $u$ then the value of $GCD(v, u)$ doesn't matter. The vertices are numbered from $1$ to $|V|$.
Construct a relatively prime graph with $n$ vertices and $m$ edges such that it is connected and it contains neither self-loops nor multiple edges.
If there exists no valid graph with the given number of vertices and edges then output "Impossible".
If there are multiple answers then print any of them.
|
The only line contains two integers $n$ and $m$ ($1 \le n, m \le 10^5$) — the number of vertices and the number of edges.
|
If there exists no valid graph with the given number of vertices and edges then output "Impossible".
Otherwise print the answer in the following format:
The first line should contain the word "Possible".
The $i$-th of the next $m$ lines should contain the $i$-th edge $(v_i, u_i)$ of the resulting graph ($1 \le v_i, u_i \le n, v_i \neq u_i$). For each pair $(v, u)$ there can be no more pairs $(v, u)$ or $(u, v)$. The vertices are numbered from $1$ to $n$.
If there are multiple answers then print any of them.
|
[
"5 6\n",
"6 12\n"
] |
[
"Possible\n2 5\n3 2\n5 1\n3 4\n4 1\n5 4\n",
"Impossible\n"
] |
Here is the representation of the graph from the first example: <img class="tex-graphics" src="https://espresso.codeforces.com/7a1353a992545456c007e3071fa0a06fe46fc64e.png" style="max-width: 100.0%;max-height: 100.0%;"/>
| 0
|
[
{
"input": "5 6",
"output": "Possible\n2 1\n3 1\n4 1\n5 1\n3 2\n5 2"
},
{
"input": "6 12",
"output": "Impossible"
},
{
"input": "572 99643",
"output": "Possible\n2 1\n3 1\n4 1\n5 1\n6 1\n7 1\n8 1\n9 1\n10 1\n11 1\n12 1\n13 1\n14 1\n15 1\n16 1\n17 1\n18 1\n19 1\n20 1\n21 1\n22 1\n23 1\n24 1\n25 1\n26 1\n27 1\n28 1\n29 1\n30 1\n31 1\n32 1\n33 1\n34 1\n35 1\n36 1\n37 1\n38 1\n39 1\n40 1\n41 1\n42 1\n43 1\n44 1\n45 1\n46 1\n47 1\n48 1\n49 1\n50 1\n51 1\n52 1\n53 1\n54 1\n55 1\n56 1\n57 1\n58 1\n59 1\n60 1\n61 1\n62 1\n63 1\n64 1\n65 1\n66 1\n67 1\n68 1\n69 1\n70 1\n71 1\n72 1\n73 1\n74 1\n75 1\n76 1\n77 1\n78 1\n79 1\n80 1\n81 1\n82 1\n83 1\n84 1\n85 1\n86 1\n..."
},
{
"input": "571 99403",
"output": "Possible\n2 1\n3 1\n4 1\n5 1\n6 1\n7 1\n8 1\n9 1\n10 1\n11 1\n12 1\n13 1\n14 1\n15 1\n16 1\n17 1\n18 1\n19 1\n20 1\n21 1\n22 1\n23 1\n24 1\n25 1\n26 1\n27 1\n28 1\n29 1\n30 1\n31 1\n32 1\n33 1\n34 1\n35 1\n36 1\n37 1\n38 1\n39 1\n40 1\n41 1\n42 1\n43 1\n44 1\n45 1\n46 1\n47 1\n48 1\n49 1\n50 1\n51 1\n52 1\n53 1\n54 1\n55 1\n56 1\n57 1\n58 1\n59 1\n60 1\n61 1\n62 1\n63 1\n64 1\n65 1\n66 1\n67 1\n68 1\n69 1\n70 1\n71 1\n72 1\n73 1\n74 1\n75 1\n76 1\n77 1\n78 1\n79 1\n80 1\n81 1\n82 1\n83 1\n84 1\n85 1\n86 1\n..."
},
{
"input": "100 3043",
"output": "Possible\n2 1\n3 1\n4 1\n5 1\n6 1\n7 1\n8 1\n9 1\n10 1\n11 1\n12 1\n13 1\n14 1\n15 1\n16 1\n17 1\n18 1\n19 1\n20 1\n21 1\n22 1\n23 1\n24 1\n25 1\n26 1\n27 1\n28 1\n29 1\n30 1\n31 1\n32 1\n33 1\n34 1\n35 1\n36 1\n37 1\n38 1\n39 1\n40 1\n41 1\n42 1\n43 1\n44 1\n45 1\n46 1\n47 1\n48 1\n49 1\n50 1\n51 1\n52 1\n53 1\n54 1\n55 1\n56 1\n57 1\n58 1\n59 1\n60 1\n61 1\n62 1\n63 1\n64 1\n65 1\n66 1\n67 1\n68 1\n69 1\n70 1\n71 1\n72 1\n73 1\n74 1\n75 1\n76 1\n77 1\n78 1\n79 1\n80 1\n81 1\n82 1\n83 1\n84 1\n85 1\n86 1\n..."
},
{
"input": "10 31",
"output": "Possible\n2 1\n3 1\n4 1\n5 1\n6 1\n7 1\n8 1\n9 1\n10 1\n3 2\n5 2\n7 2\n9 2\n4 3\n5 3\n7 3\n8 3\n10 3\n5 4\n7 4\n9 4\n6 5\n7 5\n8 5\n9 5\n7 6\n8 7\n9 7\n10 7\n9 8\n10 9"
},
{
"input": "1 1",
"output": "Impossible"
},
{
"input": "2 1",
"output": "Possible\n2 1"
},
{
"input": "100000 1",
"output": "Impossible"
},
{
"input": "100000 99999",
"output": "Possible\n2 1\n3 1\n4 1\n5 1\n6 1\n7 1\n8 1\n9 1\n10 1\n11 1\n12 1\n13 1\n14 1\n15 1\n16 1\n17 1\n18 1\n19 1\n20 1\n21 1\n22 1\n23 1\n24 1\n25 1\n26 1\n27 1\n28 1\n29 1\n30 1\n31 1\n32 1\n33 1\n34 1\n35 1\n36 1\n37 1\n38 1\n39 1\n40 1\n41 1\n42 1\n43 1\n44 1\n45 1\n46 1\n47 1\n48 1\n49 1\n50 1\n51 1\n52 1\n53 1\n54 1\n55 1\n56 1\n57 1\n58 1\n59 1\n60 1\n61 1\n62 1\n63 1\n64 1\n65 1\n66 1\n67 1\n68 1\n69 1\n70 1\n71 1\n72 1\n73 1\n74 1\n75 1\n76 1\n77 1\n78 1\n79 1\n80 1\n81 1\n82 1\n83 1\n84 1\n85 1\n86 1\n..."
},
{
"input": "100000 100000",
"output": "Possible\n2 1\n3 1\n4 1\n5 1\n6 1\n7 1\n8 1\n9 1\n10 1\n11 1\n12 1\n13 1\n14 1\n15 1\n16 1\n17 1\n18 1\n19 1\n20 1\n21 1\n22 1\n23 1\n24 1\n25 1\n26 1\n27 1\n28 1\n29 1\n30 1\n31 1\n32 1\n33 1\n34 1\n35 1\n36 1\n37 1\n38 1\n39 1\n40 1\n41 1\n42 1\n43 1\n44 1\n45 1\n46 1\n47 1\n48 1\n49 1\n50 1\n51 1\n52 1\n53 1\n54 1\n55 1\n56 1\n57 1\n58 1\n59 1\n60 1\n61 1\n62 1\n63 1\n64 1\n65 1\n66 1\n67 1\n68 1\n69 1\n70 1\n71 1\n72 1\n73 1\n74 1\n75 1\n76 1\n77 1\n78 1\n79 1\n80 1\n81 1\n82 1\n83 1\n84 1\n85 1\n86 1\n..."
},
{
"input": "1000 100000",
"output": "Possible\n2 1\n3 1\n4 1\n5 1\n6 1\n7 1\n8 1\n9 1\n10 1\n11 1\n12 1\n13 1\n14 1\n15 1\n16 1\n17 1\n18 1\n19 1\n20 1\n21 1\n22 1\n23 1\n24 1\n25 1\n26 1\n27 1\n28 1\n29 1\n30 1\n31 1\n32 1\n33 1\n34 1\n35 1\n36 1\n37 1\n38 1\n39 1\n40 1\n41 1\n42 1\n43 1\n44 1\n45 1\n46 1\n47 1\n48 1\n49 1\n50 1\n51 1\n52 1\n53 1\n54 1\n55 1\n56 1\n57 1\n58 1\n59 1\n60 1\n61 1\n62 1\n63 1\n64 1\n65 1\n66 1\n67 1\n68 1\n69 1\n70 1\n71 1\n72 1\n73 1\n74 1\n75 1\n76 1\n77 1\n78 1\n79 1\n80 1\n81 1\n82 1\n83 1\n84 1\n85 1\n86 1\n..."
},
{
"input": "572 99644",
"output": "Impossible"
},
{
"input": "571 99404",
"output": "Impossible"
},
{
"input": "100 3044",
"output": "Impossible"
},
{
"input": "10 32",
"output": "Impossible"
},
{
"input": "100000 99998",
"output": "Impossible"
},
{
"input": "55910 88645",
"output": "Possible\n2 1\n3 1\n4 1\n5 1\n6 1\n7 1\n8 1\n9 1\n10 1\n11 1\n12 1\n13 1\n14 1\n15 1\n16 1\n17 1\n18 1\n19 1\n20 1\n21 1\n22 1\n23 1\n24 1\n25 1\n26 1\n27 1\n28 1\n29 1\n30 1\n31 1\n32 1\n33 1\n34 1\n35 1\n36 1\n37 1\n38 1\n39 1\n40 1\n41 1\n42 1\n43 1\n44 1\n45 1\n46 1\n47 1\n48 1\n49 1\n50 1\n51 1\n52 1\n53 1\n54 1\n55 1\n56 1\n57 1\n58 1\n59 1\n60 1\n61 1\n62 1\n63 1\n64 1\n65 1\n66 1\n67 1\n68 1\n69 1\n70 1\n71 1\n72 1\n73 1\n74 1\n75 1\n76 1\n77 1\n78 1\n79 1\n80 1\n81 1\n82 1\n83 1\n84 1\n85 1\n86 1\n..."
},
{
"input": "72050 72069",
"output": "Possible\n2 1\n3 1\n4 1\n5 1\n6 1\n7 1\n8 1\n9 1\n10 1\n11 1\n12 1\n13 1\n14 1\n15 1\n16 1\n17 1\n18 1\n19 1\n20 1\n21 1\n22 1\n23 1\n24 1\n25 1\n26 1\n27 1\n28 1\n29 1\n30 1\n31 1\n32 1\n33 1\n34 1\n35 1\n36 1\n37 1\n38 1\n39 1\n40 1\n41 1\n42 1\n43 1\n44 1\n45 1\n46 1\n47 1\n48 1\n49 1\n50 1\n51 1\n52 1\n53 1\n54 1\n55 1\n56 1\n57 1\n58 1\n59 1\n60 1\n61 1\n62 1\n63 1\n64 1\n65 1\n66 1\n67 1\n68 1\n69 1\n70 1\n71 1\n72 1\n73 1\n74 1\n75 1\n76 1\n77 1\n78 1\n79 1\n80 1\n81 1\n82 1\n83 1\n84 1\n85 1\n86 1\n..."
},
{
"input": "53599 55493",
"output": "Possible\n2 1\n3 1\n4 1\n5 1\n6 1\n7 1\n8 1\n9 1\n10 1\n11 1\n12 1\n13 1\n14 1\n15 1\n16 1\n17 1\n18 1\n19 1\n20 1\n21 1\n22 1\n23 1\n24 1\n25 1\n26 1\n27 1\n28 1\n29 1\n30 1\n31 1\n32 1\n33 1\n34 1\n35 1\n36 1\n37 1\n38 1\n39 1\n40 1\n41 1\n42 1\n43 1\n44 1\n45 1\n46 1\n47 1\n48 1\n49 1\n50 1\n51 1\n52 1\n53 1\n54 1\n55 1\n56 1\n57 1\n58 1\n59 1\n60 1\n61 1\n62 1\n63 1\n64 1\n65 1\n66 1\n67 1\n68 1\n69 1\n70 1\n71 1\n72 1\n73 1\n74 1\n75 1\n76 1\n77 1\n78 1\n79 1\n80 1\n81 1\n82 1\n83 1\n84 1\n85 1\n86 1\n..."
},
{
"input": "56557 100000",
"output": "Possible\n2 1\n3 1\n4 1\n5 1\n6 1\n7 1\n8 1\n9 1\n10 1\n11 1\n12 1\n13 1\n14 1\n15 1\n16 1\n17 1\n18 1\n19 1\n20 1\n21 1\n22 1\n23 1\n24 1\n25 1\n26 1\n27 1\n28 1\n29 1\n30 1\n31 1\n32 1\n33 1\n34 1\n35 1\n36 1\n37 1\n38 1\n39 1\n40 1\n41 1\n42 1\n43 1\n44 1\n45 1\n46 1\n47 1\n48 1\n49 1\n50 1\n51 1\n52 1\n53 1\n54 1\n55 1\n56 1\n57 1\n58 1\n59 1\n60 1\n61 1\n62 1\n63 1\n64 1\n65 1\n66 1\n67 1\n68 1\n69 1\n70 1\n71 1\n72 1\n73 1\n74 1\n75 1\n76 1\n77 1\n78 1\n79 1\n80 1\n81 1\n82 1\n83 1\n84 1\n85 1\n86 1\n..."
},
{
"input": "1001 100000",
"output": "Possible\n2 1\n3 1\n4 1\n5 1\n6 1\n7 1\n8 1\n9 1\n10 1\n11 1\n12 1\n13 1\n14 1\n15 1\n16 1\n17 1\n18 1\n19 1\n20 1\n21 1\n22 1\n23 1\n24 1\n25 1\n26 1\n27 1\n28 1\n29 1\n30 1\n31 1\n32 1\n33 1\n34 1\n35 1\n36 1\n37 1\n38 1\n39 1\n40 1\n41 1\n42 1\n43 1\n44 1\n45 1\n46 1\n47 1\n48 1\n49 1\n50 1\n51 1\n52 1\n53 1\n54 1\n55 1\n56 1\n57 1\n58 1\n59 1\n60 1\n61 1\n62 1\n63 1\n64 1\n65 1\n66 1\n67 1\n68 1\n69 1\n70 1\n71 1\n72 1\n73 1\n74 1\n75 1\n76 1\n77 1\n78 1\n79 1\n80 1\n81 1\n82 1\n83 1\n84 1\n85 1\n86 1\n..."
},
{
"input": "50000 100000",
"output": "Possible\n2 1\n3 1\n4 1\n5 1\n6 1\n7 1\n8 1\n9 1\n10 1\n11 1\n12 1\n13 1\n14 1\n15 1\n16 1\n17 1\n18 1\n19 1\n20 1\n21 1\n22 1\n23 1\n24 1\n25 1\n26 1\n27 1\n28 1\n29 1\n30 1\n31 1\n32 1\n33 1\n34 1\n35 1\n36 1\n37 1\n38 1\n39 1\n40 1\n41 1\n42 1\n43 1\n44 1\n45 1\n46 1\n47 1\n48 1\n49 1\n50 1\n51 1\n52 1\n53 1\n54 1\n55 1\n56 1\n57 1\n58 1\n59 1\n60 1\n61 1\n62 1\n63 1\n64 1\n65 1\n66 1\n67 1\n68 1\n69 1\n70 1\n71 1\n72 1\n73 1\n74 1\n75 1\n76 1\n77 1\n78 1\n79 1\n80 1\n81 1\n82 1\n83 1\n84 1\n85 1\n86 1\n..."
},
{
"input": "530 100000",
"output": "Impossible"
},
{
"input": "46133 100000",
"output": "Possible\n2 1\n3 1\n4 1\n5 1\n6 1\n7 1\n8 1\n9 1\n10 1\n11 1\n12 1\n13 1\n14 1\n15 1\n16 1\n17 1\n18 1\n19 1\n20 1\n21 1\n22 1\n23 1\n24 1\n25 1\n26 1\n27 1\n28 1\n29 1\n30 1\n31 1\n32 1\n33 1\n34 1\n35 1\n36 1\n37 1\n38 1\n39 1\n40 1\n41 1\n42 1\n43 1\n44 1\n45 1\n46 1\n47 1\n48 1\n49 1\n50 1\n51 1\n52 1\n53 1\n54 1\n55 1\n56 1\n57 1\n58 1\n59 1\n60 1\n61 1\n62 1\n63 1\n64 1\n65 1\n66 1\n67 1\n68 1\n69 1\n70 1\n71 1\n72 1\n73 1\n74 1\n75 1\n76 1\n77 1\n78 1\n79 1\n80 1\n81 1\n82 1\n83 1\n84 1\n85 1\n86 1\n..."
},
{
"input": "2 2",
"output": "Impossible"
},
{
"input": "3 2",
"output": "Possible\n2 1\n3 1"
},
{
"input": "1 1",
"output": "Impossible"
},
{
"input": "1 2",
"output": "Impossible"
},
{
"input": "1 3",
"output": "Impossible"
},
{
"input": "1 4",
"output": "Impossible"
},
{
"input": "1 5",
"output": "Impossible"
},
{
"input": "2 1",
"output": "Possible\n2 1"
},
{
"input": "2 2",
"output": "Impossible"
},
{
"input": "2 3",
"output": "Impossible"
},
{
"input": "2 4",
"output": "Impossible"
},
{
"input": "2 5",
"output": "Impossible"
},
{
"input": "3 1",
"output": "Impossible"
},
{
"input": "3 2",
"output": "Possible\n2 1\n3 1"
},
{
"input": "3 3",
"output": "Possible\n2 1\n3 1\n3 2"
},
{
"input": "3 4",
"output": "Impossible"
},
{
"input": "3 5",
"output": "Impossible"
},
{
"input": "4 1",
"output": "Impossible"
},
{
"input": "4 2",
"output": "Impossible"
},
{
"input": "4 3",
"output": "Possible\n2 1\n3 1\n4 1"
},
{
"input": "4 4",
"output": "Possible\n2 1\n3 1\n4 1\n3 2"
},
{
"input": "4 5",
"output": "Possible\n2 1\n3 1\n4 1\n3 2\n4 3"
},
{
"input": "5 1",
"output": "Impossible"
},
{
"input": "5 2",
"output": "Impossible"
},
{
"input": "5 3",
"output": "Impossible"
},
{
"input": "5 4",
"output": "Possible\n2 1\n3 1\n4 1\n5 1"
},
{
"input": "5 5",
"output": "Possible\n2 1\n3 1\n4 1\n5 1\n3 2"
}
] | 1,656,659,575
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
import math
def draw(n, m):
if m < n - 1:
print('Impossible')
return
for p in range(1, n + 1):
for q in range(p + 1, n + 1):
if math.gcd(p, q) == 1:
m -= 1
if m == 0:
print("Possible")
return
print('Impossible')
if __name__ == '__main__':
n, m = (int(i) for i in input().split())
draw(n, m)
|
Title: Relatively Prime Graph
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let's call an undirected graph $G = (V, E)$ relatively prime if and only if for each edge $(v, u) \in E$ $GCD(v, u) = 1$ (the greatest common divisor of $v$ and $u$ is $1$). If there is no edge between some pair of vertices $v$ and $u$ then the value of $GCD(v, u)$ doesn't matter. The vertices are numbered from $1$ to $|V|$.
Construct a relatively prime graph with $n$ vertices and $m$ edges such that it is connected and it contains neither self-loops nor multiple edges.
If there exists no valid graph with the given number of vertices and edges then output "Impossible".
If there are multiple answers then print any of them.
Input Specification:
The only line contains two integers $n$ and $m$ ($1 \le n, m \le 10^5$) — the number of vertices and the number of edges.
Output Specification:
If there exists no valid graph with the given number of vertices and edges then output "Impossible".
Otherwise print the answer in the following format:
The first line should contain the word "Possible".
The $i$-th of the next $m$ lines should contain the $i$-th edge $(v_i, u_i)$ of the resulting graph ($1 \le v_i, u_i \le n, v_i \neq u_i$). For each pair $(v, u)$ there can be no more pairs $(v, u)$ or $(u, v)$. The vertices are numbered from $1$ to $n$.
If there are multiple answers then print any of them.
Demo Input:
['5 6\n', '6 12\n']
Demo Output:
['Possible\n2 5\n3 2\n5 1\n3 4\n4 1\n5 4\n', 'Impossible\n']
Note:
Here is the representation of the graph from the first example: <img class="tex-graphics" src="https://espresso.codeforces.com/7a1353a992545456c007e3071fa0a06fe46fc64e.png" style="max-width: 100.0%;max-height: 100.0%;"/>
|
```python
import math
def draw(n, m):
if m < n - 1:
print('Impossible')
return
for p in range(1, n + 1):
for q in range(p + 1, n + 1):
if math.gcd(p, q) == 1:
m -= 1
if m == 0:
print("Possible")
return
print('Impossible')
if __name__ == '__main__':
n, m = (int(i) for i in input().split())
draw(n, m)
```
| 0
|
|
61
|
A
|
Ultra-Fast Mathematician
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Ultra-Fast Mathematician
|
2
|
256
|
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
|
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
|
Write one line — the corresponding answer. Do not omit the leading 0s.
|
[
"1010100\n0100101\n",
"000\n111\n",
"1110\n1010\n",
"01110\n01100\n"
] |
[
"1110001\n",
"111\n",
"0100\n",
"00010\n"
] |
none
| 500
|
[
{
"input": "1010100\n0100101",
"output": "1110001"
},
{
"input": "000\n111",
"output": "111"
},
{
"input": "1110\n1010",
"output": "0100"
},
{
"input": "01110\n01100",
"output": "00010"
},
{
"input": "011101\n000001",
"output": "011100"
},
{
"input": "10\n01",
"output": "11"
},
{
"input": "00111111\n11011101",
"output": "11100010"
},
{
"input": "011001100\n101001010",
"output": "110000110"
},
{
"input": "1100100001\n0110101100",
"output": "1010001101"
},
{
"input": "00011101010\n10010100101",
"output": "10001001111"
},
{
"input": "100000101101\n111010100011",
"output": "011010001110"
},
{
"input": "1000001111010\n1101100110001",
"output": "0101101001011"
},
{
"input": "01011111010111\n10001110111010",
"output": "11010001101101"
},
{
"input": "110010000111100\n001100101011010",
"output": "111110101100110"
},
{
"input": "0010010111110000\n0000000011010110",
"output": "0010010100100110"
},
{
"input": "00111110111110000\n01111100001100000",
"output": "01000010110010000"
},
{
"input": "101010101111010001\n001001111101111101",
"output": "100011010010101100"
},
{
"input": "0110010101111100000\n0011000101000000110",
"output": "0101010000111100110"
},
{
"input": "11110100011101010111\n00001000011011000000",
"output": "11111100000110010111"
},
{
"input": "101010101111101101001\n111010010010000011111",
"output": "010000111101101110110"
},
{
"input": "0000111111100011000010\n1110110110110000001010",
"output": "1110001001010011001000"
},
{
"input": "10010010101000110111000\n00101110100110111000111",
"output": "10111100001110001111111"
},
{
"input": "010010010010111100000111\n100100111111100011001110",
"output": "110110101101011111001001"
},
{
"input": "0101110100100111011010010\n0101100011010111001010001",
"output": "0000010111110000010000011"
},
{
"input": "10010010100011110111111011\n10000110101100000001000100",
"output": "00010100001111110110111111"
},
{
"input": "000001111000000100001000000\n011100111101111001110110001",
"output": "011101000101111101111110001"
},
{
"input": "0011110010001001011001011100\n0000101101000011101011001010",
"output": "0011011111001010110010010110"
},
{
"input": "11111000000000010011001101111\n11101110011001010100010000000",
"output": "00010110011001000111011101111"
},
{
"input": "011001110000110100001100101100\n001010000011110000001000101001",
"output": "010011110011000100000100000101"
},
{
"input": "1011111010001100011010110101111\n1011001110010000000101100010101",
"output": "0000110100011100011111010111010"
},
{
"input": "10111000100001000001010110000001\n10111000001100101011011001011000",
"output": "00000000101101101010001111011001"
},
{
"input": "000001010000100001000000011011100\n111111111001010100100001100000111",
"output": "111110101001110101100001111011011"
},
{
"input": "1101000000000010011011101100000110\n1110000001100010011010000011011110",
"output": "0011000001100000000001101111011000"
},
{
"input": "01011011000010100001100100011110001\n01011010111000001010010100001110000",
"output": "00000001111010101011110000010000001"
},
{
"input": "000011111000011001000110111100000100\n011011000110000111101011100111000111",
"output": "011000111110011110101101011011000011"
},
{
"input": "1001000010101110001000000011111110010\n0010001011010111000011101001010110000",
"output": "1011001001111001001011101010101000010"
},
{
"input": "00011101011001100101111111000000010101\n10010011011011001011111000000011101011",
"output": "10001110000010101110000111000011111110"
},
{
"input": "111011100110001001101111110010111001010\n111111101101111001110010000101101000100",
"output": "000100001011110000011101110111010001110"
},
{
"input": "1111001001101000001000000010010101001010\n0010111100111110001011000010111110111001",
"output": "1101110101010110000011000000101011110011"
},
{
"input": "00100101111000000101011111110010100011010\n11101110001010010101001000111110101010100",
"output": "11001011110010010000010111001100001001110"
},
{
"input": "101011001110110100101001000111010101101111\n100111100110101011010100111100111111010110",
"output": "001100101000011111111101111011101010111001"
},
{
"input": "1111100001100101000111101001001010011100001\n1000110011000011110010001011001110001000001",
"output": "0111010010100110110101100010000100010100000"
},
{
"input": "01100111011111010101000001101110000001110101\n10011001011111110000000101011001001101101100",
"output": "11111110000000100101000100110111001100011001"
},
{
"input": "110010100111000100100101100000011100000011001\n011001111011100110000110111001110110100111011",
"output": "101011011100100010100011011001101010100100010"
},
{
"input": "0001100111111011010110100100111000000111000110\n1100101011000000000001010010010111001100110001",
"output": "1101001100111011010111110110101111001011110111"
},
{
"input": "00000101110110110001110010100001110100000100000\n10010000110011110001101000111111101010011010001",
"output": "10010101000101000000011010011110011110011110001"
},
{
"input": "110000100101011100100011001111110011111110010001\n101011111001011100110110111101110011010110101100",
"output": "011011011100000000010101110010000000101000111101"
},
{
"input": "0101111101011111010101011101000011101100000000111\n0000101010110110001110101011011110111001010100100",
"output": "0101010111101001011011110110011101010101010100011"
},
{
"input": "11000100010101110011101000011111001010110111111100\n00001111000111001011111110000010101110111001000011",
"output": "11001011010010111000010110011101100100001110111111"
},
{
"input": "101000001101111101101111111000001110110010101101010\n010011100111100001100000010001100101000000111011011",
"output": "111011101010011100001111101001101011110010010110001"
},
{
"input": "0011111110010001010100010110111000110011001101010100\n0111000000100010101010000100101000000100101000111001",
"output": "0100111110110011111110010010010000110111100101101101"
},
{
"input": "11101010000110000011011010000001111101000111011111100\n10110011110001010100010110010010101001010111100100100",
"output": "01011001110111010111001100010011010100010000111011000"
},
{
"input": "011000100001000001101000010110100110011110100111111011\n111011001000001001110011001111011110111110110011011111",
"output": "100011101001001000011011011001111000100000010100100100"
},
{
"input": "0111010110010100000110111011010110100000000111110110000\n1011100100010001101100000100111111101001110010000100110",
"output": "1100110010000101101010111111101001001001110101110010110"
},
{
"input": "10101000100111000111010001011011011011110100110101100011\n11101111000000001100100011111000100100000110011001101110",
"output": "01000111100111001011110010100011111111110010101100001101"
},
{
"input": "000000111001010001000000110001001011100010011101010011011\n110001101000010010000101000100001111101001100100001010010",
"output": "110001010001000011000101110101000100001011111001011001001"
},
{
"input": "0101011100111010000111110010101101111111000000111100011100\n1011111110000010101110111001000011100000100111111111000111",
"output": "1110100010111000101001001011101110011111100111000011011011"
},
{
"input": "11001000001100100111100111100100101011000101001111001001101\n10111110100010000011010100110100100011101001100000001110110",
"output": "01110110101110100100110011010000001000101100101111000111011"
},
{
"input": "010111011011101000000110000110100110001110100001110110111011\n101011110011101011101101011111010100100001100111100100111011",
"output": "111100101000000011101011011001110010101111000110010010000000"
},
{
"input": "1001011110110110000100011001010110000100011010010111010101110\n1101111100001000010111110011010101111010010100000001000010111",
"output": "0100100010111110010011101010000011111110001110010110010111001"
},
{
"input": "10000010101111100111110101111000010100110111101101111111111010\n10110110101100101010011001011010100110111011101100011001100111",
"output": "00110100000011001101101100100010110010001100000001100110011101"
},
{
"input": "011111010011111000001010101001101001000010100010111110010100001\n011111001011000011111001000001111001010110001010111101000010011",
"output": "000000011000111011110011101000010000010100101000000011010110010"
},
{
"input": "1111000000110001011101000100100100001111011100001111001100011111\n1101100110000101100001100000001001011011111011010101000101001010",
"output": "0010100110110100111100100100101101010100100111011010001001010101"
},
{
"input": "01100000101010010011001110100110110010000110010011011001100100011\n10110110010110111100100111000111000110010000000101101110000010111",
"output": "11010110111100101111101001100001110100010110010110110111100110100"
},
{
"input": "001111111010000100001100001010011001111110011110010111110001100111\n110000101001011000100010101100100110000111100000001101001110010111",
"output": "111111010011011100101110100110111111111001111110011010111111110000"
},
{
"input": "1011101011101101011110101101011101011000010011100101010101000100110\n0001000001001111010111100100111101100000000001110001000110000000110",
"output": "1010101010100010001001001001100000111000010010010100010011000100000"
},
{
"input": "01000001011001010011011100010000100100110101111011011011110000001110\n01011110000110011011000000000011000111100001010000000011111001110000",
"output": "00011111011111001000011100010011100011010100101011011000001001111110"
},
{
"input": "110101010100110101000001111110110100010010000100111110010100110011100\n111010010111111011100110101011001011001110110111110100000110110100111",
"output": "001111000011001110100111010101111111011100110011001010010010000111011"
},
{
"input": "1001101011000001011111100110010010000011010001001111011100010100110001\n1111100111110101001111010001010000011001001001010110001111000000100101",
"output": "0110001100110100010000110111000010011010011000011001010011010100010100"
},
{
"input": "00000111110010110001110110001010010101000111011001111111100110011110010\n00010111110100000100110101000010010001100001100011100000001100010100010",
"output": "00010000000110110101000011001000000100100110111010011111101010001010000"
},
{
"input": "100101011100101101000011010001011001101110101110001100010001010111001110\n100001111100101011011111110000001111000111001011111110000010101110111001",
"output": "000100100000000110011100100001010110101001100101110010010011111001110111"
},
{
"input": "1101100001000111001101001011101000111000011110000001001101101001111011010\n0101011101010100011011010110101000010010110010011110101100000110110001000",
"output": "1000111100010011010110011101000000101010101100011111100001101111001010010"
},
{
"input": "01101101010011110101100001110101111011100010000010001101111000011110111111\n00101111001101001100111010000101110000100101101111100111101110010100011011",
"output": "01000010011110111001011011110000001011000111101101101010010110001010100100"
},
{
"input": "101100101100011001101111110110110010100110110010100001110010110011001101011\n000001011010101011110011111101001110000111000010001101000010010000010001101",
"output": "101101110110110010011100001011111100100001110000101100110000100011011100110"
},
{
"input": "0010001011001010001100000010010011110110011000100000000100110000101111001110\n1100110100111000110100001110111001011101001100001010100001010011100110110001",
"output": "1110111111110010111000001100101010101011010100101010100101100011001001111111"
},
{
"input": "00101101010000000101011001101011001100010001100000101011101110000001111001000\n10010110010111000000101101000011101011001010000011011101101011010000000011111",
"output": "10111011000111000101110100101000100111011011100011110110000101010001111010111"
},
{
"input": "111100000100100000101001100001001111001010001000001000000111010000010101101011\n001000100010100101111011111011010110101100001111011000010011011011100010010110",
"output": "110100100110000101010010011010011001100110000111010000010100001011110111111101"
},
{
"input": "0110001101100100001111110101101000100101010010101010011001101001001101110000000\n0111011000000010010111011110010000000001000110001000011001101000000001110100111",
"output": "0001010101100110011000101011111000100100010100100010000000000001001100000100111"
},
{
"input": "10001111111001000101001011110101111010100001011010101100111001010001010010001000\n10000111010010011110111000111010101100000011110001101111001000111010100000000001",
"output": "00001000101011011011110011001111010110100010101011000011110001101011110010001001"
},
{
"input": "100110001110110000100101001110000011110110000110000000100011110100110110011001101\n110001110101110000000100101001101011111100100100001001000110000001111100011110110",
"output": "010111111011000000100001100111101000001010100010001001100101110101001010000111011"
},
{
"input": "0000010100100000010110111100011111111010011101000000100000011001001101101100111010\n0100111110011101010110101011110110010111001111000110101100101110111100101000111111",
"output": "0100101010111101000000010111101001101101010010000110001100110111110001000100000101"
},
{
"input": "11000111001010100001110000001001011010010010110000001110100101000001010101100110111\n11001100100100100001101010110100000111100011101110011010110100001001000011011011010",
"output": "00001011101110000000011010111101011101110001011110010100010001001000010110111101101"
},
{
"input": "010110100010001000100010101001101010011010111110100001000100101000111011100010100001\n110000011111101101010011111000101010111010100001001100001001100101000000111000000000",
"output": "100110111101100101110001010001000000100000011111101101001101001101111011011010100001"
},
{
"input": "0000011110101110010101110110110101100001011001101010101001000010000010000000101001101\n1100111111011100000110000111101110011111100111110001011001000010011111100001001100011",
"output": "1100100001110010010011110001011011111110111110011011110000000000011101100001100101110"
},
{
"input": "10100000101101110001100010010010100101100011010010101000110011100000101010110010000000\n10001110011011010010111011011101101111000111110000111000011010010101001100000001010011",
"output": "00101110110110100011011001001111001010100100100010010000101001110101100110110011010011"
},
{
"input": "001110000011111101101010011111000101010111010100001001100001001100101000000111000000000\n111010000000000000101001110011001000111011001100101010011001000011101001001011110000011",
"output": "110100000011111101000011101100001101101100011000100011111000001111000001001100110000011"
},
{
"input": "1110111100111011010101011011001110001010010010110011110010011111000010011111010101100001\n1001010101011001001010100010101100000110111101011000100010101111111010111100001110010010",
"output": "0111101001100010011111111001100010001100101111101011010000110000111000100011011011110011"
},
{
"input": "11100010001100010011001100001100010011010001101110011110100101110010101101011101000111111\n01110000000110111010110100001010000101011110100101010011000110101110101101110111011110001",
"output": "10010010001010101001111000000110010110001111001011001101100011011100000000101010011001110"
},
{
"input": "001101011001100101101100110000111000101011001001100100000100101000100000110100010111111101\n101001111110000010111101111110001001111001111101111010000110111000100100110010010001011111",
"output": "100100100111100111010001001110110001010010110100011110000010010000000100000110000110100010"
},
{
"input": "1010110110010101000110010010110101011101010100011001101011000110000000100011100100011000000\n0011011111100010001111101101000111001011101110100000110111100100101111010110101111011100011",
"output": "1001101001110111001001111111110010010110111010111001011100100010101111110101001011000100011"
},
{
"input": "10010010000111010111011111110010100101100000001100011100111011100010000010010001011100001100\n00111010100010110010000100010111010001111110100100100011101000101111111111001101101100100100",
"output": "10101000100101100101011011100101110100011110101000111111010011001101111101011100110000101000"
},
{
"input": "010101110001010101100000010111010000000111110011001101100011001000000011001111110000000010100\n010010111011100101010101111110110000000111000100001101101001001000001100101110001010000100001",
"output": "000111001010110000110101101001100000000000110111000000001010000000001111100001111010000110101"
},
{
"input": "1100111110011001000111101001001011000110011010111111100010111111001100111111011101100111101011\n1100000011001000110100110111000001011001010111101000010010100011000001100100111101101000010110",
"output": "0000111101010001110011011110001010011111001101010111110000011100001101011011100000001111111101"
},
{
"input": "00011000100100110111100101100100000000010011110111110010101110110011100001010111010011110100101\n00011011111011111011100101100111100101001110010111000010000111000100100100000001110101111011011",
"output": "00000011011111001100000000000011100101011101100000110000101001110111000101010110100110001111110"
},
{
"input": "000101011001001100000111100010110101111011110101111101000110001101011010111110110011100100000001\n011000101010011111011000111000100000000011011000000001111110001000001111101010110000011100001111",
"output": "011101110011010011011111011010010101111000101101111100111000000101010101010100000011111000001110"
},
{
"input": "1000101001011010000100100100010010011101011001110101111011101111111110010101001101010001010101001\n0110110010011100011111011111110111000000010001110100001010111110101011010011111011111110000110000",
"output": "1110011011000110011011111011100101011101001000000001110001010001010101000110110110101111010011001"
},
{
"input": "01111010010101100111110111111011011010100001011101010000111100101101101110111011001100101011100111\n00001100110110100001111011000010001001001100000010111101000001111011100000010111010010000011000010",
"output": "01110110100011000110001100111001010011101101011111101101111101010110001110101100011110101000100101"
},
{
"input": "111101011101111000001011001101111010110111001101110100100011111011011101001101010101011010111000110\n101111100101101101001100110011000001111010011101110111110110101110011011110011111100001001110101101",
"output": "010010111000010101000111111110111011001101010000000011010101010101000110111110101001010011001101011"
},
{
"input": "1010000011010110011101001101110001110010000111011101110010110110111111001001110100101100010101010001\n0111101000111100101100000101111010100100001000011101010100110011100011010011010101000100101011100011",
"output": "1101101011101010110001001000001011010110001111000000100110000101011100011010100001101000111110110010"
},
{
"input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001\n1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111010",
"output": "1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111011"
},
{
"input": "0\n0",
"output": "0"
},
{
"input": "0\n1",
"output": "1"
}
] | 1,550,338,422
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 109
| 0
|
a=int(input())
b=int(input())
c=str(a+b)
k=str("")
for i in range(len(c)):
if c[i]=="0" or c[i]=="2":
k=k+"0"
else:
k=k+"1"
print(k)
|
Title: Ultra-Fast Mathematician
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
Input Specification:
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
Output Specification:
Write one line — the corresponding answer. Do not omit the leading 0s.
Demo Input:
['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n']
Demo Output:
['1110001\n', '111\n', '0100\n', '00010\n']
Note:
none
|
```python
a=int(input())
b=int(input())
c=str(a+b)
k=str("")
for i in range(len(c)):
if c[i]=="0" or c[i]=="2":
k=k+"0"
else:
k=k+"1"
print(k)
```
| 0
|
415
|
B
|
Mashmokh and Tokens
|
PROGRAMMING
| 1,500
|
[
"binary search",
"greedy",
"implementation",
"math"
] | null | null |
Bimokh is Mashmokh's boss. For the following *n* days he decided to pay to his workers in a new way. At the beginning of each day he will give each worker a certain amount of tokens. Then at the end of each day each worker can give some of his tokens back to get a certain amount of money. The worker can save the rest of tokens but he can't use it in any other day to get more money. If a worker gives back *w* tokens then he'll get dollars.
Mashmokh likes the tokens however he likes money more. That's why he wants to save as many tokens as possible so that the amount of money he gets is maximal possible each day. He has *n* numbers *x*1,<=*x*2,<=...,<=*x**n*. Number *x**i* is the number of tokens given to each worker on the *i*-th day. Help him calculate for each of *n* days the number of tokens he can save.
|
The first line of input contains three space-separated integers *n*,<=*a*,<=*b* (1<=≤<=*n*<=≤<=105; 1<=≤<=*a*,<=*b*<=≤<=109). The second line of input contains *n* space-separated integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=109).
|
Output *n* space-separated integers. The *i*-th of them is the number of tokens Mashmokh can save on the *i*-th day.
|
[
"5 1 4\n12 6 11 9 1\n",
"3 1 2\n1 2 3\n",
"1 1 1\n1\n"
] |
[
"0 2 3 1 1 ",
"1 0 1 ",
"0 "
] |
none
| 1,000
|
[
{
"input": "5 1 4\n12 6 11 9 1",
"output": "0 2 3 1 1 "
},
{
"input": "3 1 2\n1 2 3",
"output": "1 0 1 "
},
{
"input": "1 1 1\n1",
"output": "0 "
},
{
"input": "1 1 1000000000\n1000000000",
"output": "0 "
},
{
"input": "1 1 1000000000\n999999999",
"output": "999999999 "
},
{
"input": "10 1 100000000\n999999999 999999999 999999999 999999999 999999999 999999999 999999999 999999999 999999999 999999999",
"output": "99999999 99999999 99999999 99999999 99999999 99999999 99999999 99999999 99999999 99999999 "
}
] | 1,677,190,751
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 47
| 218
| 8,806,400
|
a,b,c=list(map(int,input().split()))
aa=list(map(int,input().split()))
from math import gcd,floor
for i in aa:
m=(i*b)%c
print(m//b,end=" ")
print()
|
Title: Mashmokh and Tokens
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bimokh is Mashmokh's boss. For the following *n* days he decided to pay to his workers in a new way. At the beginning of each day he will give each worker a certain amount of tokens. Then at the end of each day each worker can give some of his tokens back to get a certain amount of money. The worker can save the rest of tokens but he can't use it in any other day to get more money. If a worker gives back *w* tokens then he'll get dollars.
Mashmokh likes the tokens however he likes money more. That's why he wants to save as many tokens as possible so that the amount of money he gets is maximal possible each day. He has *n* numbers *x*1,<=*x*2,<=...,<=*x**n*. Number *x**i* is the number of tokens given to each worker on the *i*-th day. Help him calculate for each of *n* days the number of tokens he can save.
Input Specification:
The first line of input contains three space-separated integers *n*,<=*a*,<=*b* (1<=≤<=*n*<=≤<=105; 1<=≤<=*a*,<=*b*<=≤<=109). The second line of input contains *n* space-separated integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=109).
Output Specification:
Output *n* space-separated integers. The *i*-th of them is the number of tokens Mashmokh can save on the *i*-th day.
Demo Input:
['5 1 4\n12 6 11 9 1\n', '3 1 2\n1 2 3\n', '1 1 1\n1\n']
Demo Output:
['0 2 3 1 1 ', '1 0 1 ', '0 ']
Note:
none
|
```python
a,b,c=list(map(int,input().split()))
aa=list(map(int,input().split()))
from math import gcd,floor
for i in aa:
m=(i*b)%c
print(m//b,end=" ")
print()
```
| 3
|
|
592
|
A
|
PawnChess
|
PROGRAMMING
| 1,200
|
[
"implementation"
] | null | null |
Galois is one of the strongest chess players of Byteforces. He has even invented a new variant of chess, which he named «PawnChess».
This new game is played on a board consisting of 8 rows and 8 columns. At the beginning of every game some black and white pawns are placed on the board. The number of black pawns placed is not necessarily equal to the number of white pawns placed.
Lets enumerate rows and columns with integers from 1 to 8. Rows are numbered from top to bottom, while columns are numbered from left to right. Now we denote as (*r*,<=*c*) the cell located at the row *r* and at the column *c*.
There are always two players A and B playing the game. Player A plays with white pawns, while player B plays with black ones. The goal of player A is to put any of his pawns to the row 1, while player B tries to put any of his pawns to the row 8. As soon as any of the players completes his goal the game finishes immediately and the succeeded player is declared a winner.
Player A moves first and then they alternate turns. On his move player A must choose exactly one white pawn and move it one step upward and player B (at his turn) must choose exactly one black pawn and move it one step down. Any move is possible only if the targeted cell is empty. It's guaranteed that for any scenario of the game there will always be at least one move available for any of the players.
Moving upward means that the pawn located in (*r*,<=*c*) will go to the cell (*r*<=-<=1,<=*c*), while moving down means the pawn located in (*r*,<=*c*) will go to the cell (*r*<=+<=1,<=*c*). Again, the corresponding cell must be empty, i.e. not occupied by any other pawn of any color.
Given the initial disposition of the board, determine who wins the game if both players play optimally. Note that there will always be a winner due to the restriction that for any game scenario both players will have some moves available.
|
The input consists of the board description given in eight lines, each line contains eight characters. Character 'B' is used to denote a black pawn, and character 'W' represents a white pawn. Empty cell is marked with '.'.
It's guaranteed that there will not be white pawns on the first row neither black pawns on the last row.
|
Print 'A' if player A wins the game on the given board, and 'B' if player B will claim the victory. Again, it's guaranteed that there will always be a winner on the given board.
|
[
"........\n........\n.B....B.\n....W...\n........\n..W.....\n........\n........\n",
"..B.....\n..W.....\n......B.\n........\n.....W..\n......B.\n........\n........\n"
] |
[
"A\n",
"B\n"
] |
In the first sample player A is able to complete his goal in 3 steps by always moving a pawn initially located at (4, 5). Player B needs at least 5 steps for any of his pawns to reach the row 8. Hence, player A will be the winner.
| 500
|
[
{
"input": ".BB.B.B.\nB..B..B.\n.B.BB...\nBB.....B\nBBB....B\nB..BB...\nBB.B...B\n....WWW.",
"output": "B"
},
{
"input": "B.B.BB.B\nW.WWW.WW\n.WWWWW.W\nW.BB.WBW\n.W..BBWB\nBB.WWBBB\n.W.W.WWB\nWWW..WW.",
"output": "A"
},
{
"input": "BB..BB..\nBW.W.W.B\n..B.....\n.....BB.\n.B..B..B\n........\n...BB.B.\nW.WWWW.W",
"output": "A"
},
{
"input": "BB......\nW....BBW\n........\n.B.B.BBB\n....BB..\nB....BB.\n...WWWW.\n....WW..",
"output": "A"
},
{
"input": ".B.B..B.\nB.B....B\n...B.B.B\n..B.W..B\n.BBB.B.B\nB.BB.B.B\nBB..BBBB\nW.W.W.WW",
"output": "B"
},
{
"input": "..BB....\n.B.B.B.B\n..B.B...\n..B..B.B\nWWWBWWB.\n.BB...B.\n..BBB...\n......W.",
"output": "B"
},
{
"input": "..BB....\n.WBWBWBB\n.....BBB\n..WW....\n.W.W...W\nWWW...W.\n.W....W.\nW...W.W.",
"output": "A"
},
{
"input": "....BB..\nBB......\n.B.....B\nWW..WWW.\n...BB.B.\nB...BB..\n..W..WWW\n...W...W",
"output": "B"
},
{
"input": "B...BBBB\n...BBB..\nBBWBWW.W\n.B..BB.B\nW..W..WW\nW.WW....\n........\nWW.....W",
"output": "A"
},
{
"input": ".B......\n.B....B.\n...W....\n......W.\nW.WWWW.W\nW.WW....\n..WWW...\n..W...WW",
"output": "A"
},
{
"input": "B.......\nBBB.....\n.B....B.\n.W.BWB.W\n......B.\nW..WW...\n...W....\nW...W..W",
"output": "A"
},
{
"input": ".....B..\n........\n........\n.BB..B..\n..BB....\n........\n....WWW.\n......W.",
"output": "B"
},
{
"input": "B.B...B.\n...BBBBB\n....B...\n...B...B\nB.B.B..B\n........\n........\nWWW..WW.",
"output": "B"
},
{
"input": "B.B...B.\n........\n.......B\n.BB....B\n.....W..\n.W.WW.W.\n...W.WW.\nW..WW..W",
"output": "A"
},
{
"input": "......B.\nB....B..\n...B.BB.\n...B....\n........\n..W....W\nWW......\n.W....W.",
"output": "B"
},
{
"input": ".BBB....\nB.B.B...\nB.BB.B..\nB.BB.B.B\n........\n........\nW.....W.\n..WW..W.",
"output": "B"
},
{
"input": "..B..BBB\n........\n........\n........\n...W.W..\n...W..W.\nW.......\n..W...W.",
"output": "A"
},
{
"input": "........\n.B.B....\n...B..BB\n........\n........\nW...W...\nW...W...\nW.WW.W..",
"output": "A"
},
{
"input": "B....BB.\n...B...B\n.B......\n........\n........\n........\n........\n....W..W",
"output": "B"
},
{
"input": "...BB.BB\nBB...B..\n........\n........\n........\n........\n..W..W..\n......W.",
"output": "A"
},
{
"input": "...BB...\n........\n........\n........\n........\n........\n......W.\nWW...WW.",
"output": "A"
},
{
"input": "...B.B..\n........\n........\n........\n........\n........\n........\nWWW...WW",
"output": "A"
},
{
"input": "BBBBBBB.\n........\n........\n........\n........\n........\n........\n.WWWWWWW",
"output": "A"
},
{
"input": ".BBBBBB.\nB.......\n........\n........\n........\n........\n........\n.WWWWWWW",
"output": "B"
},
{
"input": ".BBBBBBB\n........\n........\n........\n........\n........\n........\nWWWWWWW.",
"output": "A"
},
{
"input": ".BBBBBB.\n.......B\n........\n........\n........\n........\n........\nWWWWWWW.",
"output": "B"
},
{
"input": "B..BB...\n..B...B.\n.WBB...B\nBW......\nW.B...W.\n..BBW.B.\nBW..BB..\n......W.",
"output": "B"
},
{
"input": "B.BBBBBB\nB..BBB.B\nW.BB.W.B\nB.BWBB.B\nBWBWBBBB\n...BBBBB\nB.B...BB\nWW..WW.W",
"output": "B"
},
{
"input": "BBBB.BBB\nBBBB.B.B\nB.B..BBB\nB.BB.BWW\nB.BB.BBB\nB.BB.BBB\n..BW.BB.\nW.WWWWWW",
"output": "B"
},
{
"input": "BBBB.BBB\n.B....WB\nBB.B...B\nWWWW.WWB\nBB...BWW\nWWW..BBB\nW.BW.BB.\nWWWWWWW.",
"output": "B"
},
{
"input": "B.BBBBBB\nW.WWBBBW\nW.BB.WBB\nW.W.BBBW\nW.BWW.WB\nB..B..BB\nB.B.W.BB\nWWWWW.WW",
"output": "B"
},
{
"input": "BBBBBB.B\n.BBWBB.B\nWWW..B.W\n..WW.W.W\nBWB..W.W\n..BW.B.W\nB..B....\nWWWW.WWW",
"output": "B"
},
{
"input": ".B...BB.\nWBB.BWBB\n.BWBW...\n..W...B.\nWB.BWW..\nWBW.....\n.W..W.B.\n.W.W.WW.",
"output": "A"
},
{
"input": ".B..BBBB\nBB...WWB\nB..B.W.B\nWB.W...B\n...W.WW.\nW.....W.\nWB.W.W.W\n.WW...WW",
"output": "A"
},
{
"input": "B.BBBBBB\nW.BB.W.B\nW.BBW...\n..WWWW.B\n....W..B\n.WW.W..W\n.W..WW.W\nW.W....W",
"output": "A"
},
{
"input": "........\n.B......\n.W......\n........\n....B...\n........\n........\n.......W",
"output": "B"
}
] | 1,602,690,465
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 43
| 140
| 0
|
def on_same_col(point, arr, flag):
for p in arr:
if (flag and p[0] < point[0] and p[1] == point[1]
or not flag and p[0] > point[0] and p[1] == point[1]):
return True
return False
whites = []
blacks = []
min_white = 10
min_black = 10
for row in range(8):
row_str = input()
for col in range(8):
if row_str[col] == "W":
whites.append((row, col))
# на пути нет черных
if not on_same_col((row, col), blacks, True) \
and row < min_white:
min_white = row
if row_str[col] == "B":
blacks.append((row, col))
# на пути нет черных
if not on_same_col((row, col), whites, False) \
and 7 - row < min_black:
min_black = 7 - row
if min_white <= min_black:
print("A")
else:
print("B")
|
Title: PawnChess
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Galois is one of the strongest chess players of Byteforces. He has even invented a new variant of chess, which he named «PawnChess».
This new game is played on a board consisting of 8 rows and 8 columns. At the beginning of every game some black and white pawns are placed on the board. The number of black pawns placed is not necessarily equal to the number of white pawns placed.
Lets enumerate rows and columns with integers from 1 to 8. Rows are numbered from top to bottom, while columns are numbered from left to right. Now we denote as (*r*,<=*c*) the cell located at the row *r* and at the column *c*.
There are always two players A and B playing the game. Player A plays with white pawns, while player B plays with black ones. The goal of player A is to put any of his pawns to the row 1, while player B tries to put any of his pawns to the row 8. As soon as any of the players completes his goal the game finishes immediately and the succeeded player is declared a winner.
Player A moves first and then they alternate turns. On his move player A must choose exactly one white pawn and move it one step upward and player B (at his turn) must choose exactly one black pawn and move it one step down. Any move is possible only if the targeted cell is empty. It's guaranteed that for any scenario of the game there will always be at least one move available for any of the players.
Moving upward means that the pawn located in (*r*,<=*c*) will go to the cell (*r*<=-<=1,<=*c*), while moving down means the pawn located in (*r*,<=*c*) will go to the cell (*r*<=+<=1,<=*c*). Again, the corresponding cell must be empty, i.e. not occupied by any other pawn of any color.
Given the initial disposition of the board, determine who wins the game if both players play optimally. Note that there will always be a winner due to the restriction that for any game scenario both players will have some moves available.
Input Specification:
The input consists of the board description given in eight lines, each line contains eight characters. Character 'B' is used to denote a black pawn, and character 'W' represents a white pawn. Empty cell is marked with '.'.
It's guaranteed that there will not be white pawns on the first row neither black pawns on the last row.
Output Specification:
Print 'A' if player A wins the game on the given board, and 'B' if player B will claim the victory. Again, it's guaranteed that there will always be a winner on the given board.
Demo Input:
['........\n........\n.B....B.\n....W...\n........\n..W.....\n........\n........\n', '..B.....\n..W.....\n......B.\n........\n.....W..\n......B.\n........\n........\n']
Demo Output:
['A\n', 'B\n']
Note:
In the first sample player A is able to complete his goal in 3 steps by always moving a pawn initially located at (4, 5). Player B needs at least 5 steps for any of his pawns to reach the row 8. Hence, player A will be the winner.
|
```python
def on_same_col(point, arr, flag):
for p in arr:
if (flag and p[0] < point[0] and p[1] == point[1]
or not flag and p[0] > point[0] and p[1] == point[1]):
return True
return False
whites = []
blacks = []
min_white = 10
min_black = 10
for row in range(8):
row_str = input()
for col in range(8):
if row_str[col] == "W":
whites.append((row, col))
# на пути нет черных
if not on_same_col((row, col), blacks, True) \
and row < min_white:
min_white = row
if row_str[col] == "B":
blacks.append((row, col))
# на пути нет черных
if not on_same_col((row, col), whites, False) \
and 7 - row < min_black:
min_black = 7 - row
if min_white <= min_black:
print("A")
else:
print("B")
```
| 0
|
|
816
|
B
|
Karen and Coffee
|
PROGRAMMING
| 1,400
|
[
"binary search",
"data structures",
"implementation"
] | null | null |
To stay woke and attentive during classes, Karen needs some coffee!
Karen, a coffee aficionado, wants to know the optimal temperature for brewing the perfect cup of coffee. Indeed, she has spent some time reading several recipe books, including the universally acclaimed "The Art of the Covfefe".
She knows *n* coffee recipes. The *i*-th recipe suggests that coffee should be brewed between *l**i* and *r**i* degrees, inclusive, to achieve the optimal taste.
Karen thinks that a temperature is admissible if at least *k* recipes recommend it.
Karen has a rather fickle mind, and so she asks *q* questions. In each question, given that she only wants to prepare coffee with a temperature between *a* and *b*, inclusive, can you tell her how many admissible integer temperatures fall within the range?
|
The first line of input contains three integers, *n*, *k* (1<=≤<=*k*<=≤<=*n*<=≤<=200000), and *q* (1<=≤<=*q*<=≤<=200000), the number of recipes, the minimum number of recipes a certain temperature must be recommended by to be admissible, and the number of questions Karen has, respectively.
The next *n* lines describe the recipes. Specifically, the *i*-th line among these contains two integers *l**i* and *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=200000), describing that the *i*-th recipe suggests that the coffee be brewed between *l**i* and *r**i* degrees, inclusive.
The next *q* lines describe the questions. Each of these lines contains *a* and *b*, (1<=≤<=*a*<=≤<=*b*<=≤<=200000), describing that she wants to know the number of admissible integer temperatures between *a* and *b* degrees, inclusive.
|
For each question, output a single integer on a line by itself, the number of admissible integer temperatures between *a* and *b* degrees, inclusive.
|
[
"3 2 4\n91 94\n92 97\n97 99\n92 94\n93 97\n95 96\n90 100\n",
"2 1 1\n1 1\n200000 200000\n90 100\n"
] |
[
"3\n3\n0\n4\n",
"0\n"
] |
In the first test case, Karen knows 3 recipes.
1. The first one recommends brewing the coffee between 91 and 94 degrees, inclusive. 1. The second one recommends brewing the coffee between 92 and 97 degrees, inclusive. 1. The third one recommends brewing the coffee between 97 and 99 degrees, inclusive.
A temperature is admissible if at least 2 recipes recommend it.
She asks 4 questions.
In her first question, she wants to know the number of admissible integer temperatures between 92 and 94 degrees, inclusive. There are 3: 92, 93 and 94 degrees are all admissible.
In her second question, she wants to know the number of admissible integer temperatures between 93 and 97 degrees, inclusive. There are 3: 93, 94 and 97 degrees are all admissible.
In her third question, she wants to know the number of admissible integer temperatures between 95 and 96 degrees, inclusive. There are none.
In her final question, she wants to know the number of admissible integer temperatures between 90 and 100 degrees, inclusive. There are 4: 92, 93, 94 and 97 degrees are all admissible.
In the second test case, Karen knows 2 recipes.
1. The first one, "wikiHow to make Cold Brew Coffee", recommends brewing the coffee at exactly 1 degree. 1. The second one, "What good is coffee that isn't brewed at at least 36.3306 times the temperature of the surface of the sun?", recommends brewing the coffee at exactly 200000 degrees.
A temperature is admissible if at least 1 recipe recommends it.
In her first and only question, she wants to know the number of admissible integer temperatures that are actually reasonable. There are none.
| 1,000
|
[
{
"input": "3 2 4\n91 94\n92 97\n97 99\n92 94\n93 97\n95 96\n90 100",
"output": "3\n3\n0\n4"
},
{
"input": "2 1 1\n1 1\n200000 200000\n90 100",
"output": "0"
},
{
"input": "1 1 1\n1 1\n1 1",
"output": "1"
},
{
"input": "1 1 1\n200000 200000\n200000 200000",
"output": "1"
}
] | 1,643,176,031
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 8
| 717
| 24,371,200
|
from typing import List, Tuple
def karen_and_coffee(arr: List[Tuple[int, int]], n: int, k: int, q: int, minn,
maxx):
diff = [0 for _ in range(maxx - minn + 1)]
for v in arr:
diff[v[0] - minn] -= 1
if v[1] - minn < maxx - minn:
diff[v[1] - minn + 1] += 1
diff[0] = -diff[0]
for i in range(1, maxx - minn + 1):
diff[i] = diff[i - 1] - diff[i]
res = [0 for _ in range(maxx-minn+2)]
total = 0
for i in range(1, maxx-minn+2):
if diff[i-1] >= k:
total += 1
res[i] = total
for _ in range(q):
left, right = map(int, input().split())
left = max(left - minn + 1, 1)
right = min(maxx - minn + 1, right - minn + 1)
print(res[right] - res[left-1])
def main():
arr = []
n, k, q = map(int, input().split())
arr = []
minn = 200001
maxx = 0
for _ in range(n):
t1, t2 = map(int, input().split())
minn = min(t1, t2, minn)
maxx = max(t1, t2, maxx)
arr.append((t1, t2))
karen_and_coffee(arr, n, k, q, minn, maxx)
if __name__ == '__main__':
main()
|
Title: Karen and Coffee
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
To stay woke and attentive during classes, Karen needs some coffee!
Karen, a coffee aficionado, wants to know the optimal temperature for brewing the perfect cup of coffee. Indeed, she has spent some time reading several recipe books, including the universally acclaimed "The Art of the Covfefe".
She knows *n* coffee recipes. The *i*-th recipe suggests that coffee should be brewed between *l**i* and *r**i* degrees, inclusive, to achieve the optimal taste.
Karen thinks that a temperature is admissible if at least *k* recipes recommend it.
Karen has a rather fickle mind, and so she asks *q* questions. In each question, given that she only wants to prepare coffee with a temperature between *a* and *b*, inclusive, can you tell her how many admissible integer temperatures fall within the range?
Input Specification:
The first line of input contains three integers, *n*, *k* (1<=≤<=*k*<=≤<=*n*<=≤<=200000), and *q* (1<=≤<=*q*<=≤<=200000), the number of recipes, the minimum number of recipes a certain temperature must be recommended by to be admissible, and the number of questions Karen has, respectively.
The next *n* lines describe the recipes. Specifically, the *i*-th line among these contains two integers *l**i* and *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=200000), describing that the *i*-th recipe suggests that the coffee be brewed between *l**i* and *r**i* degrees, inclusive.
The next *q* lines describe the questions. Each of these lines contains *a* and *b*, (1<=≤<=*a*<=≤<=*b*<=≤<=200000), describing that she wants to know the number of admissible integer temperatures between *a* and *b* degrees, inclusive.
Output Specification:
For each question, output a single integer on a line by itself, the number of admissible integer temperatures between *a* and *b* degrees, inclusive.
Demo Input:
['3 2 4\n91 94\n92 97\n97 99\n92 94\n93 97\n95 96\n90 100\n', '2 1 1\n1 1\n200000 200000\n90 100\n']
Demo Output:
['3\n3\n0\n4\n', '0\n']
Note:
In the first test case, Karen knows 3 recipes.
1. The first one recommends brewing the coffee between 91 and 94 degrees, inclusive. 1. The second one recommends brewing the coffee between 92 and 97 degrees, inclusive. 1. The third one recommends brewing the coffee between 97 and 99 degrees, inclusive.
A temperature is admissible if at least 2 recipes recommend it.
She asks 4 questions.
In her first question, she wants to know the number of admissible integer temperatures between 92 and 94 degrees, inclusive. There are 3: 92, 93 and 94 degrees are all admissible.
In her second question, she wants to know the number of admissible integer temperatures between 93 and 97 degrees, inclusive. There are 3: 93, 94 and 97 degrees are all admissible.
In her third question, she wants to know the number of admissible integer temperatures between 95 and 96 degrees, inclusive. There are none.
In her final question, she wants to know the number of admissible integer temperatures between 90 and 100 degrees, inclusive. There are 4: 92, 93, 94 and 97 degrees are all admissible.
In the second test case, Karen knows 2 recipes.
1. The first one, "wikiHow to make Cold Brew Coffee", recommends brewing the coffee at exactly 1 degree. 1. The second one, "What good is coffee that isn't brewed at at least 36.3306 times the temperature of the surface of the sun?", recommends brewing the coffee at exactly 200000 degrees.
A temperature is admissible if at least 1 recipe recommends it.
In her first and only question, she wants to know the number of admissible integer temperatures that are actually reasonable. There are none.
|
```python
from typing import List, Tuple
def karen_and_coffee(arr: List[Tuple[int, int]], n: int, k: int, q: int, minn,
maxx):
diff = [0 for _ in range(maxx - minn + 1)]
for v in arr:
diff[v[0] - minn] -= 1
if v[1] - minn < maxx - minn:
diff[v[1] - minn + 1] += 1
diff[0] = -diff[0]
for i in range(1, maxx - minn + 1):
diff[i] = diff[i - 1] - diff[i]
res = [0 for _ in range(maxx-minn+2)]
total = 0
for i in range(1, maxx-minn+2):
if diff[i-1] >= k:
total += 1
res[i] = total
for _ in range(q):
left, right = map(int, input().split())
left = max(left - minn + 1, 1)
right = min(maxx - minn + 1, right - minn + 1)
print(res[right] - res[left-1])
def main():
arr = []
n, k, q = map(int, input().split())
arr = []
minn = 200001
maxx = 0
for _ in range(n):
t1, t2 = map(int, input().split())
minn = min(t1, t2, minn)
maxx = max(t1, t2, maxx)
arr.append((t1, t2))
karen_and_coffee(arr, n, k, q, minn, maxx)
if __name__ == '__main__':
main()
```
| -1
|
|
22
|
A
|
Second Order Statistics
|
PROGRAMMING
| 800
|
[
"brute force"
] |
A. Second Order Statistics
|
2
|
256
|
Once Bob needed to find the second order statistics of a sequence of integer numbers. Lets choose each number from the sequence exactly once and sort them. The value on the second position is the second order statistics of the given sequence. In other words it is the smallest element strictly greater than the minimum. Help Bob solve this problem.
|
The first input line contains integer *n* (1<=≤<=*n*<=≤<=100) — amount of numbers in the sequence. The second line contains *n* space-separated integer numbers — elements of the sequence. These numbers don't exceed 100 in absolute value.
|
If the given sequence has the second order statistics, output this order statistics, otherwise output NO.
|
[
"4\n1 2 2 -4\n",
"5\n1 2 3 1 1\n"
] |
[
"1\n",
"2\n"
] |
none
| 0
|
[
{
"input": "4\n1 2 2 -4",
"output": "1"
},
{
"input": "5\n1 2 3 1 1",
"output": "2"
},
{
"input": "1\n28",
"output": "NO"
},
{
"input": "2\n-28 12",
"output": "12"
},
{
"input": "3\n-83 40 -80",
"output": "-80"
},
{
"input": "8\n93 77 -92 26 21 -48 53 91",
"output": "-48"
},
{
"input": "20\n-72 -9 -86 80 7 -10 40 -27 -94 92 96 56 28 -19 79 36 -3 -73 -63 -49",
"output": "-86"
},
{
"input": "49\n-74 -100 -80 23 -8 -83 -41 -20 48 17 46 -73 -55 67 85 4 40 -60 -69 -75 56 -74 -42 93 74 -95 64 -46 97 -47 55 0 -78 -34 -31 40 -63 -49 -76 48 21 -1 -49 -29 -98 -11 76 26 94",
"output": "-98"
},
{
"input": "88\n63 48 1 -53 -89 -49 64 -70 -49 71 -17 -16 76 81 -26 -50 67 -59 -56 97 2 100 14 18 -91 -80 42 92 -25 -88 59 8 -56 38 48 -71 -78 24 -14 48 -1 69 73 -76 54 16 -92 44 47 33 -34 -17 -81 21 -59 -61 53 26 10 -76 67 35 -29 70 65 -13 -29 81 80 32 74 -6 34 46 57 1 -45 -55 69 79 -58 11 -2 22 -18 -16 -89 -46",
"output": "-91"
},
{
"input": "100\n34 32 88 20 76 53 -71 -39 -98 -10 57 37 63 -3 -54 -64 -78 -82 73 20 -30 -4 22 75 51 -64 -91 29 -52 -48 83 19 18 -47 46 57 -44 95 89 89 -30 84 -83 67 58 -99 -90 -53 92 -60 -5 -56 -61 27 68 -48 52 -95 64 -48 -30 -67 66 89 14 -33 -31 -91 39 7 -94 -54 92 -96 -99 -83 -16 91 -28 -66 81 44 14 -85 -21 18 40 16 -13 -82 -33 47 -10 -40 -19 10 25 60 -34 -89",
"output": "-98"
},
{
"input": "2\n-1 -1",
"output": "NO"
},
{
"input": "3\n-2 -2 -2",
"output": "NO"
},
{
"input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "NO"
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 -100 100 100 100 100 100 100 100 100 100 100 100 -100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 -100 100 100 100 100 100 100 100 100 100 100 -100 100 100 100 100 -100 100 100 100 100 100 100 100 100 100 100 100",
"output": "100"
},
{
"input": "10\n40 71 -85 -85 40 -85 -85 64 -85 47",
"output": "40"
},
{
"input": "23\n-90 -90 -41 -64 -64 -90 -15 10 -43 -90 -64 -64 89 -64 36 47 38 -90 -64 -90 -90 68 -90",
"output": "-64"
},
{
"input": "39\n-97 -93 -42 -93 -97 -93 56 -97 -97 -97 76 -33 -60 91 7 82 17 47 -97 -97 -93 73 -97 12 -97 -97 -97 -97 56 -92 -83 -93 -93 49 -93 -97 -97 -17 -93",
"output": "-93"
},
{
"input": "51\n-21 6 -35 -98 -86 -98 -86 -43 -65 32 -98 -40 96 -98 -98 -98 -98 -86 -86 -98 56 -86 -98 -98 -30 -98 -86 -31 -98 -86 -86 -86 -86 -30 96 -86 -86 -86 -60 25 88 -86 -86 58 31 -47 57 -86 37 44 -83",
"output": "-86"
},
{
"input": "66\n-14 -95 65 -95 -95 -97 -90 -71 -97 -97 70 -95 -95 -97 -95 -27 35 -87 -95 -5 -97 -97 87 34 -49 -95 -97 -95 -97 -95 -30 -95 -97 47 -95 -17 -97 -95 -97 -69 51 -97 -97 -95 -75 87 59 21 63 56 76 -91 98 -97 6 -97 -95 -95 -97 -73 11 -97 -35 -95 -95 -43",
"output": "-95"
},
{
"input": "77\n-67 -93 -93 -92 97 29 93 -93 -93 -5 -93 -7 60 -92 -93 44 -84 68 -92 -93 69 -92 -37 56 43 -93 35 -92 -93 19 -79 18 -92 -93 -93 -37 -93 -47 -93 -92 -92 74 67 19 40 -92 -92 -92 -92 -93 -93 -41 -93 -92 -93 -93 -92 -93 51 -80 6 -42 -92 -92 -66 -12 -92 -92 -3 93 -92 -49 -93 40 62 -92 -92",
"output": "-92"
},
{
"input": "89\n-98 40 16 -87 -98 63 -100 55 -96 -98 -21 -100 -93 26 -98 -98 -100 -89 -98 -5 -65 -28 -100 -6 -66 67 -100 -98 -98 10 -98 -98 -70 7 -98 2 -100 -100 -98 25 -100 -100 -98 23 -68 -100 -98 3 98 -100 -98 -98 -98 -98 -24 -100 -100 -9 -98 35 -100 99 -5 -98 -100 -100 37 -100 -84 57 -98 40 -47 -100 -1 -92 -76 -98 -98 -100 -100 -100 -63 30 21 -100 -100 -100 -12",
"output": "-98"
},
{
"input": "99\n10 -84 -100 -100 73 -64 -100 -94 33 -100 -100 -100 -100 71 64 24 7 -100 -32 -100 -100 77 -100 62 -12 55 45 -100 -100 -80 -100 -100 -100 -100 -100 -100 -100 -100 -100 -39 -48 -100 -34 47 -100 -100 -100 -100 -100 -77 -100 -100 -100 -100 -100 -100 -52 40 -55 -100 -44 -100 72 33 70 -100 -100 -78 -100 -3 100 -77 22 -100 95 -30 -100 10 -69 -100 -100 -100 -100 52 -39 -100 -100 -100 7 -100 -98 -66 95 -17 -100 52 -100 68 -100",
"output": "-98"
},
{
"input": "100\n-99 -98 -64 89 53 57 -99 29 -78 18 -3 -54 76 -98 -99 -98 37 -98 19 -47 89 73 -98 -91 -99 -99 -98 -48 -99 22 -99 -98 -99 -99 -98 -60 84 67 -99 -98 20 -98 88 -98 46 -98 -99 -98 -99 -71 -99 -98 -98 -39 83 95 -98 63 -98 -99 32 -98 -99 -64 57 -30 -53 -83 -4 -99 58 20 -98 -10 -99 -44 -99 -99 -99 -99 75 34 -98 -52 -98 -30 -98 -99 -98 -98 51 -99 -99 -99 -98 -99 -99 -82 -90 92",
"output": "-98"
},
{
"input": "3\n1 2 3",
"output": "2"
},
{
"input": "3\n1 3 2",
"output": "2"
},
{
"input": "3\n2 1 3",
"output": "2"
},
{
"input": "3\n2 3 1",
"output": "2"
},
{
"input": "3\n3 1 2",
"output": "2"
},
{
"input": "3\n3 2 1",
"output": "2"
},
{
"input": "9\n99 99 99 99 100 100 100 100 100",
"output": "100"
},
{
"input": "5\n-100 -100 -100 -100 -100",
"output": "NO"
}
] | 1,497,832,298
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 62
| 0
|
n = int(input().strip())
A = list(map(int, input().strip().split()))
A = list(set(A))
if len(A) <= 2:
print('NO')
else:
print(sorted(A)[1])
|
Title: Second Order Statistics
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Once Bob needed to find the second order statistics of a sequence of integer numbers. Lets choose each number from the sequence exactly once and sort them. The value on the second position is the second order statistics of the given sequence. In other words it is the smallest element strictly greater than the minimum. Help Bob solve this problem.
Input Specification:
The first input line contains integer *n* (1<=≤<=*n*<=≤<=100) — amount of numbers in the sequence. The second line contains *n* space-separated integer numbers — elements of the sequence. These numbers don't exceed 100 in absolute value.
Output Specification:
If the given sequence has the second order statistics, output this order statistics, otherwise output NO.
Demo Input:
['4\n1 2 2 -4\n', '5\n1 2 3 1 1\n']
Demo Output:
['1\n', '2\n']
Note:
none
|
```python
n = int(input().strip())
A = list(map(int, input().strip().split()))
A = list(set(A))
if len(A) <= 2:
print('NO')
else:
print(sorted(A)[1])
```
| 0
|
721
|
B
|
Passwords
|
PROGRAMMING
| 1,100
|
[
"implementation",
"math",
"sortings",
"strings"
] | null | null |
Vanya is managed to enter his favourite site Codehorses. Vanya uses *n* distinct passwords for sites at all, however he can't remember which one exactly he specified during Codehorses registration.
Vanya will enter passwords in order of non-decreasing their lengths, and he will enter passwords of same length in arbitrary order. Just when Vanya will have entered the correct password, he is immediately authorized on the site. Vanya will not enter any password twice.
Entering any passwords takes one second for Vanya. But if Vanya will enter wrong password *k* times, then he is able to make the next try only 5 seconds after that. Vanya makes each try immediately, that is, at each moment when Vanya is able to enter password, he is doing that.
Determine how many seconds will Vanya need to enter Codehorses in the best case for him (if he spends minimum possible number of second) and in the worst case (if he spends maximum possible amount of seconds).
|
The first line of the input contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100) — the number of Vanya's passwords and the number of failed tries, after which the access to the site is blocked for 5 seconds.
The next *n* lines contains passwords, one per line — pairwise distinct non-empty strings consisting of latin letters and digits. Each password length does not exceed 100 characters.
The last line of the input contains the Vanya's Codehorses password. It is guaranteed that the Vanya's Codehorses password is equal to some of his *n* passwords.
|
Print two integers — time (in seconds), Vanya needs to be authorized to Codehorses in the best case for him and in the worst case respectively.
|
[
"5 2\ncba\nabc\nbb1\nabC\nABC\nabc\n",
"4 100\n11\n22\n1\n2\n22\n"
] |
[
"1 15\n",
"3 4\n"
] |
Consider the first sample case. As soon as all passwords have the same length, Vanya can enter the right password at the first try as well as at the last try. If he enters it at the first try, he spends exactly 1 second. Thus in the best case the answer is 1. If, at the other hand, he enters it at the last try, he enters another 4 passwords before. He spends 2 seconds to enter first 2 passwords, then he waits 5 seconds as soon as he made 2 wrong tries. Then he spends 2 more seconds to enter 2 wrong passwords, again waits 5 seconds and, finally, enters the correct password spending 1 more second. In summary in the worst case he is able to be authorized in 15 seconds.
Consider the second sample case. There is no way of entering passwords and get the access to the site blocked. As soon as the required password has length of 2, Vanya enters all passwords of length 1 anyway, spending 2 seconds for that. Then, in the best case, he immediately enters the correct password and the answer for the best case is 3, but in the worst case he enters wrong password of length 2 and only then the right one, spending 4 seconds at all.
| 1,000
|
[
{
"input": "5 2\ncba\nabc\nbb1\nabC\nABC\nabc",
"output": "1 15"
},
{
"input": "4 100\n11\n22\n1\n2\n22",
"output": "3 4"
},
{
"input": "1 1\na1\na1",
"output": "1 1"
},
{
"input": "1 100\na1\na1",
"output": "1 1"
},
{
"input": "2 1\nabc\nAbc\nAbc",
"output": "1 7"
},
{
"input": "2 2\nabc\nAbc\nabc",
"output": "1 2"
},
{
"input": "2 1\nab\nabc\nab",
"output": "1 1"
},
{
"input": "2 2\nab\nabc\nab",
"output": "1 1"
},
{
"input": "2 1\nab\nabc\nabc",
"output": "7 7"
},
{
"input": "2 2\nab\nabc\nabc",
"output": "2 2"
},
{
"input": "10 3\nOIbV1igi\no\nZS\nQM\n9woLzI\nWreboD\nQ7yl\nA5Rb\nS9Lno72TkP\nfT97o\no",
"output": "1 1"
},
{
"input": "10 3\nHJZNMsT\nLaPcH2C\nlrhqIO\n9cxw\noTC1XwjW\nGHL9Ul6\nUyIs\nPuzwgR4ZKa\nyIByoKR5\nd3QA\nPuzwgR4ZKa",
"output": "25 25"
},
{
"input": "20 5\nvSyC787KlIL8kZ2Uv5sw\nWKWOP\n7i8J3E8EByIq\nNW2VyGweL\nmyR2sRNu\nmXusPP0\nf4jgGxra\n4wHRzRhOCpEt\npPz9kybGb\nOtSpePCRoG5nkjZ2VxRy\nwHYsSttWbJkg\nKBOP9\nQfiOiFyHPPsw3GHo8J8\nxB8\nqCpehZEeEhdq\niOLjICK6\nQ91\nHmCsfMGTFKoFFnv238c\nJKjhg\ngkEUh\nKBOP9",
"output": "3 11"
},
{
"input": "15 2\nw6S9WyU\nMVh\nkgUhQHW\nhGQNOF\nUuym\n7rGQA\nBM8vLPRB\n9E\nDs32U\no\nz1aV2C5T\n8\nzSXjrqQ\n1FO\n3kIt\nBM8vLPRB",
"output": "44 50"
},
{
"input": "20 2\ni\n5Rp6\nE4vsr\nSY\nORXx\nh13C\nk6tzC\ne\nN\nKQf4C\nWZcdL\ndiA3v\n0InQT\nuJkAr\nGCamp\nBuIRd\nY\nM\nxZYx7\n0a5A\nWZcdL",
"output": "36 65"
},
{
"input": "20 2\naWLQ6\nSgQ9r\nHcPdj\n2BNaO\n3TjNb\nnvwFM\nqsKt7\nFnb6N\nLoc0p\njxuLq\nBKAjf\nEKgZB\nBfOSa\nsMIvr\nuIWcR\nIura3\nLAqSf\ntXq3G\n8rQ8I\n8otAO\nsMIvr",
"output": "1 65"
},
{
"input": "20 15\n0ZpQugVlN7\nm0SlKGnohN\nRFXTqhNGcn\n1qm2ZbB\nQXtJWdf78P\nbc2vH\nP21dty2Z1P\nm2c71LFhCk\n23EuP1Dvh3\nanwri5RhQN\n55v6HYv288\n1u5uKOjM5r\n6vg0GC1\nDAPYiA3ns1\nUZaaJ3Gmnk\nwB44x7V4Zi\n4hgB2oyU8P\npYFQpy8gGK\ndbz\nBv\n55v6HYv288",
"output": "6 25"
},
{
"input": "3 1\na\nb\naa\naa",
"output": "13 13"
},
{
"input": "6 3\nab\nac\nad\nabc\nabd\nabe\nabc",
"output": "9 11"
},
{
"input": "4 2\n1\n2\n11\n22\n22",
"output": "8 9"
},
{
"input": "2 1\n1\n12\n12",
"output": "7 7"
},
{
"input": "3 1\nab\nabc\nabd\nabc",
"output": "7 13"
},
{
"input": "2 1\na\nab\nab",
"output": "7 7"
},
{
"input": "5 2\na\nb\nc\nab\naa\naa",
"output": "9 15"
},
{
"input": "6 1\n1\n2\n11\n22\n111\n2222\n22",
"output": "13 19"
},
{
"input": "3 1\n1\n2\n11\n11",
"output": "13 13"
},
{
"input": "10 4\na\nb\nc\nd\ne\nf\nab\ncd\nac\nad\nac",
"output": "12 20"
},
{
"input": "4 2\na\nb\nc\nd\na",
"output": "1 9"
},
{
"input": "4 1\n1\n2\n3\n4\n4",
"output": "1 19"
},
{
"input": "5 1\na\nb\nc\nd\nef\nef",
"output": "25 25"
},
{
"input": "6 4\n1\n2\n22\n33\n44\n555\n555",
"output": "11 11"
},
{
"input": "5 2\na\nb\nc\nd\nab\nab",
"output": "15 15"
},
{
"input": "6 2\n1\n2\n3\n4\n5\n23\n23",
"output": "16 16"
},
{
"input": "4 2\na\nb\naa\nbb\naa",
"output": "8 9"
},
{
"input": "5 4\na\nbb\ncc\ndd\nee\nbb",
"output": "2 10"
},
{
"input": "4 1\na\nb\nc\nab\nab",
"output": "19 19"
},
{
"input": "7 100\na\nb\nc\nd\ne\ng\nab\nab",
"output": "7 7"
},
{
"input": "6 1\na\nb\nc\nd\ne\naa\naa",
"output": "31 31"
},
{
"input": "4 1\na\nas\nasd\nasde\nasde",
"output": "19 19"
},
{
"input": "5 2\n1\n2\n3\n11\n22\n22",
"output": "9 15"
},
{
"input": "10 2\na\nb\nc\nd\nee\nff\ngg\nhh\nii\njj\nii",
"output": "15 30"
},
{
"input": "3 1\na\nab\nbc\nab",
"output": "7 13"
},
{
"input": "6 4\na\nb\nc\nbb\nbc\ncc\ncc",
"output": "4 11"
}
] | 1,590,888,748
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 66
| 109
| 307,200
|
def myFunc(e):
return len(e)
n ,k =map(int,input().split())
liste = []
compteur = 1
compteurmax = 0
i = 0
r = 0
for loop in range(n):
texte = input()
liste.append(texte)
liste.sort(reverse = False,key=myFunc)
solution = input()
while len(liste[i]) != len(solution):
compteur +=1
i +=1
r += 1
if r%k==0:
compteur +=5
compteurmax = compteur
i +=1
if i<len(liste):
while len(liste[i]) == len(solution):
i +=1
r +=1
compteurmax +=1
if r%k == 0:
compteurmax+=5
if i ==len(liste):
break
print("{} {}".format(compteur,compteurmax))
|
Title: Passwords
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vanya is managed to enter his favourite site Codehorses. Vanya uses *n* distinct passwords for sites at all, however he can't remember which one exactly he specified during Codehorses registration.
Vanya will enter passwords in order of non-decreasing their lengths, and he will enter passwords of same length in arbitrary order. Just when Vanya will have entered the correct password, he is immediately authorized on the site. Vanya will not enter any password twice.
Entering any passwords takes one second for Vanya. But if Vanya will enter wrong password *k* times, then he is able to make the next try only 5 seconds after that. Vanya makes each try immediately, that is, at each moment when Vanya is able to enter password, he is doing that.
Determine how many seconds will Vanya need to enter Codehorses in the best case for him (if he spends minimum possible number of second) and in the worst case (if he spends maximum possible amount of seconds).
Input Specification:
The first line of the input contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100) — the number of Vanya's passwords and the number of failed tries, after which the access to the site is blocked for 5 seconds.
The next *n* lines contains passwords, one per line — pairwise distinct non-empty strings consisting of latin letters and digits. Each password length does not exceed 100 characters.
The last line of the input contains the Vanya's Codehorses password. It is guaranteed that the Vanya's Codehorses password is equal to some of his *n* passwords.
Output Specification:
Print two integers — time (in seconds), Vanya needs to be authorized to Codehorses in the best case for him and in the worst case respectively.
Demo Input:
['5 2\ncba\nabc\nbb1\nabC\nABC\nabc\n', '4 100\n11\n22\n1\n2\n22\n']
Demo Output:
['1 15\n', '3 4\n']
Note:
Consider the first sample case. As soon as all passwords have the same length, Vanya can enter the right password at the first try as well as at the last try. If he enters it at the first try, he spends exactly 1 second. Thus in the best case the answer is 1. If, at the other hand, he enters it at the last try, he enters another 4 passwords before. He spends 2 seconds to enter first 2 passwords, then he waits 5 seconds as soon as he made 2 wrong tries. Then he spends 2 more seconds to enter 2 wrong passwords, again waits 5 seconds and, finally, enters the correct password spending 1 more second. In summary in the worst case he is able to be authorized in 15 seconds.
Consider the second sample case. There is no way of entering passwords and get the access to the site blocked. As soon as the required password has length of 2, Vanya enters all passwords of length 1 anyway, spending 2 seconds for that. Then, in the best case, he immediately enters the correct password and the answer for the best case is 3, but in the worst case he enters wrong password of length 2 and only then the right one, spending 4 seconds at all.
|
```python
def myFunc(e):
return len(e)
n ,k =map(int,input().split())
liste = []
compteur = 1
compteurmax = 0
i = 0
r = 0
for loop in range(n):
texte = input()
liste.append(texte)
liste.sort(reverse = False,key=myFunc)
solution = input()
while len(liste[i]) != len(solution):
compteur +=1
i +=1
r += 1
if r%k==0:
compteur +=5
compteurmax = compteur
i +=1
if i<len(liste):
while len(liste[i]) == len(solution):
i +=1
r +=1
compteurmax +=1
if r%k == 0:
compteurmax+=5
if i ==len(liste):
break
print("{} {}".format(compteur,compteurmax))
```
| 3
|
|
71
|
A
|
Way Too Long Words
|
PROGRAMMING
| 800
|
[
"strings"
] |
A. Way Too Long Words
|
1
|
256
|
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
|
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
|
[
"4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n"
] |
[
"word\nl10n\ni18n\np43s\n"
] |
none
| 500
|
[
{
"input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis",
"output": "word\nl10n\ni18n\np43s"
},
{
"input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm",
"output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m"
},
{
"input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd",
"output": "n20n\nj27b\nk15d"
},
{
"input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt",
"output": "t41t"
},
{
"input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings",
"output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz",
"output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz"
},
{
"input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij",
"output": "a98j"
},
{
"input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco",
"output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco"
},
{
"input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt",
"output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t"
},
{
"input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr",
"output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr"
}
] | 1,675,731,098
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 20
| 46
| 0
|
def main():
n = int(input().strip())
list = []
for i in range(0,n):
w = str(input().strip())
list.append(w)
for i in list:
if len(i) > 10:
print(f"{i[0]}{len(i)-2}{i[-1]}")
else:
print(i)
if __name__ == '__main__':
main()
|
Title: Way Too Long Words
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output Specification:
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
Demo Input:
['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n']
Demo Output:
['word\nl10n\ni18n\np43s\n']
Note:
none
|
```python
def main():
n = int(input().strip())
list = []
for i in range(0,n):
w = str(input().strip())
list.append(w)
for i in list:
if len(i) > 10:
print(f"{i[0]}{len(i)-2}{i[-1]}")
else:
print(i)
if __name__ == '__main__':
main()
```
| 3.977
|
69
|
A
|
Young Physicist
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] |
A. Young Physicist
|
2
|
256
|
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
|
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
|
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
|
[
"3\n4 1 7\n-2 4 -1\n1 -5 -3\n",
"3\n3 -1 7\n-5 2 -4\n2 -1 -3\n"
] |
[
"NO",
"YES"
] |
none
| 500
|
[
{
"input": "3\n4 1 7\n-2 4 -1\n1 -5 -3",
"output": "NO"
},
{
"input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3",
"output": "YES"
},
{
"input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41",
"output": "NO"
},
{
"input": "10\n25 -33 43\n-27 -42 28\n-35 -20 19\n41 -42 -1\n49 -39 -4\n-49 -22 7\n-19 29 41\n8 -27 -43\n8 34 9\n-11 -3 33",
"output": "NO"
},
{
"input": "10\n-6 21 18\n20 -11 -8\n37 -11 41\n-5 8 33\n29 23 32\n30 -33 -11\n39 -49 -36\n28 34 -49\n22 29 -34\n-18 -6 7",
"output": "NO"
},
{
"input": "10\n47 -2 -27\n0 26 -14\n5 -12 33\n2 18 3\n45 -30 -49\n4 -18 8\n-46 -44 -41\n-22 -10 -40\n-35 -21 26\n33 20 38",
"output": "NO"
},
{
"input": "13\n-3 -36 -46\n-11 -50 37\n42 -11 -15\n9 42 44\n-29 -12 24\n3 9 -40\n-35 13 50\n14 43 18\n-13 8 24\n-48 -15 10\n50 9 -50\n21 0 -50\n0 0 -6",
"output": "YES"
},
{
"input": "14\n43 23 17\n4 17 44\n5 -5 -16\n-43 -7 -6\n47 -48 12\n50 47 -45\n2 14 43\n37 -30 15\n4 -17 -11\n17 9 -45\n-50 -3 -8\n-50 0 0\n-50 0 0\n-16 0 0",
"output": "YES"
},
{
"input": "13\n29 49 -11\n38 -11 -20\n25 1 -40\n-11 28 11\n23 -19 1\n45 -41 -17\n-3 0 -19\n-13 -33 49\n-30 0 28\n34 17 45\n-50 9 -27\n-50 0 0\n-37 0 0",
"output": "YES"
},
{
"input": "12\n3 28 -35\n-32 -44 -17\n9 -25 -6\n-42 -22 20\n-19 15 38\n-21 38 48\n-1 -37 -28\n-10 -13 -50\n-5 21 29\n34 28 50\n50 11 -49\n34 0 0",
"output": "YES"
},
{
"input": "37\n-64 -79 26\n-22 59 93\n-5 39 -12\n77 -9 76\n55 -86 57\n83 100 -97\n-70 94 84\n-14 46 -94\n26 72 35\n14 78 -62\n17 82 92\n-57 11 91\n23 15 92\n-80 -1 1\n12 39 18\n-23 -99 -75\n-34 50 19\n-39 84 -7\n45 -30 -39\n-60 49 37\n45 -16 -72\n33 -51 -56\n-48 28 5\n97 91 88\n45 -82 -11\n-21 -15 -90\n-53 73 -26\n-74 85 -90\n-40 23 38\n100 -13 49\n32 -100 -100\n0 -100 -70\n0 -100 0\n0 -100 0\n0 -100 0\n0 -100 0\n0 -37 0",
"output": "YES"
},
{
"input": "4\n68 3 100\n68 21 -100\n-100 -24 0\n-36 0 0",
"output": "YES"
},
{
"input": "33\n-1 -46 -12\n45 -16 -21\n-11 45 -21\n-60 -42 -93\n-22 -45 93\n37 96 85\n-76 26 83\n-4 9 55\n7 -52 -9\n66 8 -85\n-100 -54 11\n-29 59 74\n-24 12 2\n-56 81 85\n-92 69 -52\n-26 -97 91\n54 59 -51\n58 21 -57\n7 68 56\n-47 -20 -51\n-59 77 -13\n-85 27 91\n79 60 -56\n66 -80 5\n21 -99 42\n-31 -29 98\n66 93 76\n-49 45 61\n100 -100 -100\n100 -100 -100\n66 -75 -100\n0 0 -100\n0 0 -87",
"output": "YES"
},
{
"input": "3\n1 2 3\n3 2 1\n0 0 0",
"output": "NO"
},
{
"input": "2\n5 -23 12\n0 0 0",
"output": "NO"
},
{
"input": "1\n0 0 0",
"output": "YES"
},
{
"input": "1\n1 -2 0",
"output": "NO"
},
{
"input": "2\n-23 77 -86\n23 -77 86",
"output": "YES"
},
{
"input": "26\n86 7 20\n-57 -64 39\n-45 6 -93\n-44 -21 100\n-11 -49 21\n73 -71 -80\n-2 -89 56\n-65 -2 7\n5 14 84\n57 41 13\n-12 69 54\n40 -25 27\n-17 -59 0\n64 -91 -30\n-53 9 42\n-54 -8 14\n-35 82 27\n-48 -59 -80\n88 70 79\n94 57 97\n44 63 25\n84 -90 -40\n-100 100 -100\n-92 100 -100\n0 10 -100\n0 0 -82",
"output": "YES"
},
{
"input": "42\n11 27 92\n-18 -56 -57\n1 71 81\n33 -92 30\n82 83 49\n-87 -61 -1\n-49 45 49\n73 26 15\n-22 22 -77\n29 -93 87\n-68 44 -90\n-4 -84 20\n85 67 -6\n-39 26 77\n-28 -64 20\n65 -97 24\n-72 -39 51\n35 -75 -91\n39 -44 -8\n-25 -27 -57\n91 8 -46\n-98 -94 56\n94 -60 59\n-9 -95 18\n-53 -37 98\n-8 -94 -84\n-52 55 60\n15 -14 37\n65 -43 -25\n94 12 66\n-8 -19 -83\n29 81 -78\n-58 57 33\n24 86 -84\n-53 32 -88\n-14 7 3\n89 97 -53\n-5 -28 -91\n-100 100 -6\n-84 100 0\n0 100 0\n0 70 0",
"output": "YES"
},
{
"input": "3\n96 49 -12\n2 -66 28\n-98 17 -16",
"output": "YES"
},
{
"input": "5\n70 -46 86\n-100 94 24\n-27 63 -63\n57 -100 -47\n0 -11 0",
"output": "YES"
},
{
"input": "18\n-86 -28 70\n-31 -89 42\n31 -48 -55\n95 -17 -43\n24 -95 -85\n-21 -14 31\n68 -18 81\n13 31 60\n-15 28 99\n-42 15 9\n28 -61 -62\n-16 71 29\n-28 75 -48\n-77 -67 36\n-100 83 89\n100 100 -100\n57 34 -100\n0 0 -53",
"output": "YES"
},
{
"input": "44\n52 -54 -29\n-82 -5 -94\n-54 43 43\n91 16 71\n7 80 -91\n3 15 29\n-99 -6 -77\n-3 -77 -64\n73 67 34\n25 -10 -18\n-29 91 63\n-72 86 -16\n-68 85 -81\n-3 36 44\n-74 -14 -80\n34 -96 -97\n-76 -78 -33\n-24 44 -58\n98 12 77\n95 -63 -6\n-51 3 -90\n-92 -10 72\n7 3 -68\n57 -53 71\n29 57 -48\n35 -60 10\n79 -70 -61\n-20 77 55\n-86 -15 -35\n84 -88 -18\n100 -42 77\n-20 46 8\n-41 -43 -65\n38 -98 -23\n-100 65 45\n-7 -91 -63\n46 88 -85\n48 59 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 1",
"output": "YES"
},
{
"input": "18\n-14 -64 -91\n-8 -66 -86\n-23 92 -40\n6 -3 -53\n57 41 78\n-79 42 -22\n-88 -17 45\n4 -45 44\n83 -18 -25\n34 86 -92\n75 -30 12\n44 99 11\n-67 -13 72\n22 83 -56\n-37 71 72\n-9 -100 100\n0 -100 31\n0 -58 0",
"output": "YES"
},
{
"input": "23\n-70 37 78\n42 84 6\n28 -94 -24\n-49 76 95\n-67 18 84\n-53 78 -5\n65 -63 -64\n-66 -64 -69\n81 -93 95\n10 52 -79\n-89 -61 -64\n-64 -47 43\n-81 -35 55\n80 82 73\n-60 -81 -18\n49 -10 -19\n-58 70 12\n-24 -15 -93\n98 -93 -54\n-28 -75 11\n100 100 -63\n100 100 0\n56 34 0",
"output": "YES"
},
{
"input": "3\n0 2 -2\n1 -1 3\n-3 0 0",
"output": "NO"
}
] | 1,655,620,080
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 81
| 92
| 0
|
ls=[]
l=[]
m=[]
count=0
for _ in range(int(input())):
x,y,z=map(int,input().split())
ls.append(x)
l.append(y)
m.append(z)
if(sum(ls)==0 and sum(l)==0 and sum(m)==0):
print("YES")
else:
print("NO")
|
Title: Young Physicist
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
Output Specification:
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Demo Input:
['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n']
Demo Output:
['NO', 'YES']
Note:
none
|
```python
ls=[]
l=[]
m=[]
count=0
for _ in range(int(input())):
x,y,z=map(int,input().split())
ls.append(x)
l.append(y)
m.append(z)
if(sum(ls)==0 and sum(l)==0 and sum(m)==0):
print("YES")
else:
print("NO")
```
| 3.977
|
703
|
A
|
Mishka and Game
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Mishka is a little polar bear. As known, little bears loves spending their free time playing dice for chocolates. Once in a wonderful sunny morning, walking around blocks of ice, Mishka met her friend Chris, and they started playing the game.
Rules of the game are very simple: at first number of rounds *n* is defined. In every round each of the players throws a cubical dice with distinct numbers from 1 to 6 written on its faces. Player, whose value after throwing the dice is greater, wins the round. In case if player dice values are equal, no one of them is a winner.
In average, player, who won most of the rounds, is the winner of the game. In case if two players won the same number of rounds, the result of the game is draw.
Mishka is still very little and can't count wins and losses, so she asked you to watch their game and determine its result. Please help her!
|
The first line of the input contains single integer *n* *n* (1<=≤<=*n*<=≤<=100) — the number of game rounds.
The next *n* lines contains rounds description. *i*-th of them contains pair of integers *m**i* and *c**i* (1<=≤<=*m**i*,<=<=*c**i*<=≤<=6) — values on dice upper face after Mishka's and Chris' throws in *i*-th round respectively.
|
If Mishka is the winner of the game, print "Mishka" (without quotes) in the only line.
If Chris is the winner of the game, print "Chris" (without quotes) in the only line.
If the result of the game is draw, print "Friendship is magic!^^" (without quotes) in the only line.
|
[
"3\n3 5\n2 1\n4 2\n",
"2\n6 1\n1 6\n",
"3\n1 5\n3 3\n2 2\n"
] |
[
"Mishka",
"Friendship is magic!^^",
"Chris"
] |
In the first sample case Mishka loses the first round, but wins second and third rounds and thus she is the winner of the game.
In the second sample case Mishka wins the first round, Chris wins the second round, and the game ends with draw with score 1:1.
In the third sample case Chris wins the first round, but there is no winner of the next two rounds. The winner of the game is Chris.
| 500
|
[
{
"input": "3\n3 5\n2 1\n4 2",
"output": "Mishka"
},
{
"input": "2\n6 1\n1 6",
"output": "Friendship is magic!^^"
},
{
"input": "3\n1 5\n3 3\n2 2",
"output": "Chris"
},
{
"input": "6\n4 1\n4 2\n5 3\n5 1\n5 3\n4 1",
"output": "Mishka"
},
{
"input": "8\n2 4\n1 4\n1 5\n2 6\n2 5\n2 5\n2 4\n2 5",
"output": "Chris"
},
{
"input": "8\n4 1\n2 6\n4 2\n2 5\n5 2\n3 5\n5 2\n1 5",
"output": "Friendship is magic!^^"
},
{
"input": "9\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n1 3",
"output": "Mishka"
},
{
"input": "9\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "9\n1 2\n1 2\n1 2\n1 2\n1 2\n6 1\n6 1\n6 1\n6 1",
"output": "Chris"
},
{
"input": "9\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "10\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n1 4",
"output": "Mishka"
},
{
"input": "10\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "10\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n6 1\n6 1\n6 1\n6 1",
"output": "Chris"
},
{
"input": "10\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "100\n2 4\n6 6\n3 2\n1 5\n5 2\n1 5\n1 5\n3 1\n6 5\n4 3\n1 1\n5 1\n3 3\n2 4\n1 5\n3 4\n5 1\n5 5\n2 5\n2 1\n4 3\n6 5\n1 1\n2 1\n1 3\n1 1\n6 4\n4 6\n6 4\n2 1\n2 5\n6 2\n3 4\n5 5\n1 4\n4 6\n3 4\n1 6\n5 1\n4 3\n3 4\n2 2\n1 2\n2 3\n1 3\n4 4\n5 5\n4 5\n4 4\n3 1\n4 5\n2 3\n2 6\n6 5\n6 1\n6 6\n2 3\n6 4\n3 3\n2 5\n4 4\n3 1\n2 4\n6 1\n3 2\n1 3\n5 4\n6 6\n2 5\n5 1\n1 1\n2 5\n6 5\n3 6\n5 6\n4 3\n3 4\n3 4\n6 5\n5 2\n4 2\n1 1\n3 1\n2 6\n1 6\n1 2\n6 1\n3 4\n1 6\n3 1\n5 3\n1 3\n5 6\n2 1\n6 4\n3 1\n1 6\n6 3\n3 3\n4 3",
"output": "Chris"
},
{
"input": "100\n4 1\n3 4\n4 6\n4 5\n6 5\n5 3\n6 2\n6 3\n5 2\n4 5\n1 5\n5 4\n1 4\n4 5\n4 6\n1 6\n4 4\n5 1\n6 4\n6 4\n4 6\n2 3\n6 2\n4 6\n1 4\n2 3\n4 3\n1 3\n6 2\n3 1\n3 4\n2 6\n4 5\n5 4\n2 2\n2 5\n4 1\n2 2\n3 3\n1 4\n5 6\n6 4\n4 2\n6 1\n5 5\n4 1\n2 1\n6 4\n4 4\n4 3\n5 3\n4 5\n5 3\n3 5\n6 3\n1 1\n3 4\n6 3\n6 1\n5 1\n2 4\n4 3\n2 2\n5 5\n1 5\n5 3\n4 6\n1 4\n6 3\n4 3\n2 4\n3 2\n2 4\n3 4\n6 2\n5 6\n1 2\n1 5\n5 5\n2 6\n5 1\n1 6\n5 3\n3 5\n2 6\n4 6\n6 2\n3 1\n5 5\n6 1\n3 6\n4 4\n1 1\n4 6\n5 3\n4 2\n5 1\n3 3\n2 1\n1 4",
"output": "Mishka"
},
{
"input": "100\n6 3\n4 5\n4 3\n5 4\n5 1\n6 3\n4 2\n4 6\n3 1\n2 4\n2 2\n4 6\n5 3\n5 5\n4 2\n6 2\n2 3\n4 4\n6 4\n3 5\n2 4\n2 2\n5 2\n3 5\n2 4\n4 4\n3 5\n6 5\n1 3\n1 6\n2 2\n2 4\n3 2\n5 4\n1 6\n3 4\n4 1\n1 5\n1 4\n5 3\n2 2\n4 5\n6 3\n4 4\n1 1\n4 1\n2 4\n4 1\n4 5\n5 3\n1 1\n1 6\n5 6\n6 6\n4 2\n4 3\n3 4\n3 6\n3 4\n6 5\n3 4\n5 4\n5 1\n5 3\n5 1\n1 2\n2 6\n3 4\n6 5\n4 3\n1 1\n5 5\n5 1\n3 3\n5 2\n1 3\n6 6\n5 6\n1 4\n4 4\n1 4\n3 6\n6 5\n3 3\n3 6\n1 5\n1 2\n3 6\n3 6\n4 1\n5 2\n1 2\n5 2\n3 3\n4 4\n4 2\n6 2\n5 4\n6 1\n6 3",
"output": "Mishka"
},
{
"input": "8\n4 1\n6 2\n4 1\n5 3\n4 1\n5 3\n6 2\n5 3",
"output": "Mishka"
},
{
"input": "5\n3 6\n3 5\n3 5\n1 6\n3 5",
"output": "Chris"
},
{
"input": "4\n4 1\n2 4\n5 3\n3 6",
"output": "Friendship is magic!^^"
},
{
"input": "6\n6 3\n5 1\n6 3\n4 3\n4 3\n5 2",
"output": "Mishka"
},
{
"input": "7\n3 4\n1 4\n2 5\n1 6\n1 6\n1 5\n3 4",
"output": "Chris"
},
{
"input": "6\n6 2\n2 5\n5 2\n3 6\n4 3\n1 6",
"output": "Friendship is magic!^^"
},
{
"input": "8\n6 1\n5 3\n4 3\n4 1\n5 1\n4 2\n4 2\n4 1",
"output": "Mishka"
},
{
"input": "9\n2 5\n2 5\n1 4\n2 6\n2 4\n2 5\n2 6\n1 5\n2 5",
"output": "Chris"
},
{
"input": "4\n6 2\n2 4\n4 2\n3 6",
"output": "Friendship is magic!^^"
},
{
"input": "9\n5 2\n4 1\n4 1\n5 1\n6 2\n6 1\n5 3\n6 1\n6 2",
"output": "Mishka"
},
{
"input": "8\n2 4\n3 6\n1 6\n1 6\n2 4\n3 4\n3 6\n3 4",
"output": "Chris"
},
{
"input": "6\n5 3\n3 6\n6 2\n1 6\n5 1\n3 5",
"output": "Friendship is magic!^^"
},
{
"input": "6\n5 2\n5 1\n6 1\n5 2\n4 2\n5 1",
"output": "Mishka"
},
{
"input": "5\n1 4\n2 5\n3 4\n2 6\n3 4",
"output": "Chris"
},
{
"input": "4\n6 2\n3 4\n5 1\n1 6",
"output": "Friendship is magic!^^"
},
{
"input": "93\n4 3\n4 1\n4 2\n5 2\n5 3\n6 3\n4 3\n6 2\n6 3\n5 1\n4 2\n4 2\n5 1\n6 2\n6 3\n6 1\n4 1\n6 2\n5 3\n4 3\n4 1\n4 2\n5 2\n6 3\n5 2\n5 2\n6 3\n5 1\n6 2\n5 2\n4 1\n5 2\n5 1\n4 1\n6 1\n5 2\n4 3\n5 3\n5 3\n5 1\n4 3\n4 3\n4 2\n4 1\n6 2\n6 1\n4 1\n5 2\n5 2\n6 2\n5 3\n5 1\n6 2\n5 1\n6 3\n5 2\n6 2\n6 2\n4 2\n5 2\n6 1\n6 3\n6 3\n5 1\n5 1\n4 1\n5 1\n4 3\n5 3\n6 3\n4 1\n4 3\n6 1\n6 1\n4 2\n6 2\n4 2\n5 2\n4 1\n5 2\n4 1\n5 1\n5 2\n5 1\n4 1\n6 3\n6 2\n4 3\n4 1\n5 2\n4 3\n5 2\n5 1",
"output": "Mishka"
},
{
"input": "11\n1 6\n1 6\n2 4\n2 5\n3 4\n1 5\n1 6\n1 5\n1 6\n2 6\n3 4",
"output": "Chris"
},
{
"input": "70\n6 1\n3 6\n4 3\n2 5\n5 2\n1 4\n6 2\n1 6\n4 3\n1 4\n5 3\n2 4\n5 3\n1 6\n5 1\n3 5\n4 2\n2 4\n5 1\n3 5\n6 2\n1 5\n4 2\n2 5\n5 3\n1 5\n4 2\n1 4\n5 2\n2 6\n4 3\n1 5\n6 2\n3 4\n4 2\n3 5\n6 3\n3 4\n5 1\n1 4\n4 2\n1 4\n6 3\n2 6\n5 2\n1 6\n6 1\n2 6\n5 3\n1 5\n5 1\n1 6\n4 1\n1 5\n4 2\n2 4\n5 1\n2 5\n6 3\n1 4\n6 3\n3 6\n5 1\n1 4\n5 3\n3 5\n4 2\n3 4\n6 2\n1 4",
"output": "Friendship is magic!^^"
},
{
"input": "59\n4 1\n5 3\n6 1\n4 2\n5 1\n4 3\n6 1\n5 1\n4 3\n4 3\n5 2\n5 3\n4 1\n6 2\n5 1\n6 3\n6 3\n5 2\n5 2\n6 1\n4 1\n6 1\n4 3\n5 3\n5 3\n4 3\n4 2\n4 2\n6 3\n6 3\n6 1\n4 3\n5 1\n6 2\n6 1\n4 1\n6 1\n5 3\n4 2\n5 1\n6 2\n6 2\n4 3\n5 3\n4 3\n6 3\n5 2\n5 2\n4 3\n5 1\n5 3\n6 1\n6 3\n6 3\n4 3\n5 2\n5 2\n5 2\n4 3",
"output": "Mishka"
},
{
"input": "42\n1 5\n1 6\n1 6\n1 4\n2 5\n3 6\n1 6\n3 4\n2 5\n2 5\n2 4\n1 4\n3 4\n2 4\n2 6\n1 5\n3 6\n2 6\n2 6\n3 5\n1 4\n1 5\n2 6\n3 6\n1 4\n3 4\n2 4\n1 6\n3 4\n2 4\n2 6\n1 6\n1 4\n1 6\n1 6\n2 4\n1 5\n1 6\n2 5\n3 6\n3 5\n3 4",
"output": "Chris"
},
{
"input": "78\n4 3\n3 5\n4 3\n1 5\n5 1\n1 5\n4 3\n1 4\n6 3\n1 5\n4 1\n2 4\n4 3\n2 4\n5 1\n3 6\n4 2\n3 6\n6 3\n3 4\n4 3\n3 6\n5 3\n1 5\n4 1\n2 6\n4 2\n2 4\n4 1\n3 5\n5 2\n3 6\n4 3\n2 4\n6 3\n1 6\n4 3\n3 5\n6 3\n2 6\n4 1\n2 4\n6 2\n1 6\n4 2\n1 4\n4 3\n1 4\n4 3\n2 4\n6 2\n3 5\n6 1\n3 6\n5 3\n1 6\n6 1\n2 6\n4 2\n1 5\n6 2\n2 6\n6 3\n2 4\n4 2\n3 5\n6 1\n2 5\n5 3\n2 6\n5 1\n3 6\n4 3\n3 6\n6 3\n2 5\n6 1\n2 6",
"output": "Friendship is magic!^^"
},
{
"input": "76\n4 1\n5 2\n4 3\n5 2\n5 3\n5 2\n6 1\n4 2\n6 2\n5 3\n4 2\n6 2\n4 1\n4 2\n5 1\n5 1\n6 2\n5 2\n5 3\n6 3\n5 2\n4 3\n6 3\n6 1\n4 3\n6 2\n6 1\n4 1\n6 1\n5 3\n4 1\n5 3\n4 2\n5 2\n4 3\n6 1\n6 2\n5 2\n6 1\n5 3\n4 3\n5 1\n5 3\n4 3\n5 1\n5 1\n4 1\n4 1\n4 1\n4 3\n5 3\n6 3\n6 3\n5 2\n6 2\n6 3\n5 1\n6 3\n5 3\n6 1\n5 3\n4 1\n5 3\n6 1\n4 2\n6 2\n4 3\n4 1\n6 2\n4 3\n5 3\n5 2\n5 3\n5 1\n6 3\n5 2",
"output": "Mishka"
},
{
"input": "84\n3 6\n3 4\n2 5\n2 4\n1 6\n3 4\n1 5\n1 6\n3 5\n1 6\n2 4\n2 6\n2 6\n2 4\n3 5\n1 5\n3 6\n3 6\n3 4\n3 4\n2 6\n1 6\n1 6\n3 5\n3 4\n1 6\n3 4\n3 5\n2 4\n2 5\n2 5\n3 5\n1 6\n3 4\n2 6\n2 6\n3 4\n3 4\n2 5\n2 5\n2 4\n3 4\n2 5\n3 4\n3 4\n2 6\n2 6\n1 6\n2 4\n1 5\n3 4\n2 5\n2 5\n3 4\n2 4\n2 6\n2 6\n1 4\n3 5\n3 5\n2 4\n2 5\n3 4\n1 5\n1 5\n2 6\n1 5\n3 5\n2 4\n2 5\n3 4\n2 6\n1 6\n2 5\n3 5\n3 5\n3 4\n2 5\n2 6\n3 4\n1 6\n2 5\n2 6\n1 4",
"output": "Chris"
},
{
"input": "44\n6 1\n1 6\n5 2\n1 4\n6 2\n2 5\n5 3\n3 6\n5 2\n1 6\n4 1\n2 4\n6 1\n3 4\n6 3\n3 6\n4 3\n2 4\n6 1\n3 4\n6 1\n1 6\n4 1\n3 5\n6 1\n3 6\n4 1\n1 4\n4 2\n2 6\n6 1\n2 4\n6 2\n1 4\n6 2\n2 4\n5 2\n3 6\n6 3\n2 6\n5 3\n3 4\n5 3\n2 4",
"output": "Friendship is magic!^^"
},
{
"input": "42\n5 3\n5 1\n5 2\n4 1\n6 3\n6 1\n6 2\n4 1\n4 3\n4 1\n5 1\n5 3\n5 1\n4 1\n4 2\n6 1\n6 3\n5 1\n4 1\n4 1\n6 3\n4 3\n6 3\n5 2\n6 1\n4 1\n5 3\n4 3\n5 2\n6 3\n6 1\n5 1\n4 2\n4 3\n5 2\n5 3\n6 3\n5 2\n5 1\n5 3\n6 2\n6 1",
"output": "Mishka"
},
{
"input": "50\n3 6\n2 6\n1 4\n1 4\n1 4\n2 5\n3 4\n3 5\n2 6\n1 6\n3 5\n1 5\n2 6\n2 4\n2 4\n3 5\n1 6\n1 5\n1 5\n1 4\n3 5\n1 6\n3 5\n1 4\n1 5\n1 4\n3 6\n1 6\n1 4\n1 4\n1 4\n1 5\n3 6\n1 6\n1 6\n2 4\n1 5\n2 6\n2 5\n3 5\n3 6\n3 4\n2 4\n2 6\n3 4\n2 5\n3 6\n3 5\n2 4\n2 4",
"output": "Chris"
},
{
"input": "86\n6 3\n2 4\n6 3\n3 5\n6 3\n1 5\n5 2\n2 4\n4 3\n2 6\n4 1\n2 6\n5 2\n1 4\n5 1\n2 4\n4 1\n1 4\n6 2\n3 5\n4 2\n2 4\n6 2\n1 5\n5 3\n2 5\n5 1\n1 6\n6 1\n1 4\n4 3\n3 4\n5 2\n2 4\n5 3\n2 5\n4 3\n3 4\n4 1\n1 5\n6 3\n3 4\n4 3\n3 4\n4 1\n3 4\n5 1\n1 6\n4 2\n1 6\n5 1\n2 4\n5 1\n3 6\n4 1\n1 5\n5 2\n1 4\n4 3\n2 5\n5 1\n1 5\n6 2\n2 6\n4 2\n2 4\n4 1\n2 5\n5 3\n3 4\n5 1\n3 4\n6 3\n3 4\n4 3\n2 6\n6 2\n2 5\n5 2\n3 5\n4 2\n3 6\n6 2\n3 4\n4 2\n2 4",
"output": "Friendship is magic!^^"
},
{
"input": "84\n6 1\n6 3\n6 3\n4 1\n4 3\n4 2\n6 3\n5 3\n6 1\n6 3\n4 3\n5 2\n5 3\n5 1\n6 2\n6 2\n6 1\n4 1\n6 3\n5 2\n4 1\n5 3\n6 3\n4 2\n6 2\n6 3\n4 3\n4 1\n4 3\n5 1\n5 1\n5 1\n4 1\n6 1\n4 3\n6 2\n5 1\n5 1\n6 2\n5 2\n4 1\n6 1\n6 1\n6 3\n6 2\n4 3\n6 3\n6 2\n5 2\n5 1\n4 3\n6 2\n4 1\n6 2\n6 1\n5 2\n5 1\n6 2\n6 1\n5 3\n5 2\n6 1\n6 3\n5 2\n6 1\n6 3\n4 3\n5 1\n6 3\n6 1\n5 3\n4 3\n5 2\n5 1\n6 2\n5 3\n6 1\n5 1\n4 1\n5 1\n5 1\n5 2\n5 2\n5 1",
"output": "Mishka"
},
{
"input": "92\n1 5\n2 4\n3 5\n1 6\n2 5\n1 6\n3 6\n1 6\n2 4\n3 4\n3 4\n3 6\n1 5\n2 5\n1 5\n1 5\n2 6\n2 4\n3 6\n1 4\n1 6\n2 6\n3 4\n2 6\n2 6\n1 4\n3 5\n2 5\n2 6\n1 5\n1 4\n1 5\n3 6\n3 5\n2 5\n1 5\n3 5\n3 6\n2 6\n2 6\n1 5\n3 4\n2 4\n3 6\n2 5\n1 5\n2 4\n1 4\n2 6\n2 6\n2 6\n1 5\n3 6\n3 6\n2 5\n1 4\n2 4\n3 4\n1 5\n2 5\n2 4\n2 5\n3 5\n3 4\n3 6\n2 6\n3 5\n1 4\n3 4\n1 6\n3 6\n2 6\n1 4\n3 6\n3 6\n2 5\n2 6\n1 6\n2 6\n3 5\n2 5\n3 6\n2 5\n2 6\n1 5\n2 4\n1 4\n2 4\n1 5\n2 5\n2 5\n2 6",
"output": "Chris"
},
{
"input": "20\n5 1\n1 4\n4 3\n1 5\n4 2\n3 6\n6 2\n1 6\n4 1\n1 4\n5 2\n3 4\n5 1\n1 6\n5 1\n2 6\n6 3\n2 5\n6 2\n2 4",
"output": "Friendship is magic!^^"
},
{
"input": "100\n4 3\n4 3\n4 2\n4 3\n4 1\n4 3\n5 2\n5 2\n6 2\n4 2\n5 1\n4 2\n5 2\n6 1\n4 1\n6 3\n5 3\n5 1\n5 1\n5 1\n5 3\n6 1\n6 1\n4 1\n5 2\n5 2\n6 1\n6 3\n4 2\n4 1\n5 3\n4 1\n5 3\n5 1\n6 3\n6 3\n6 1\n5 2\n5 3\n5 3\n6 1\n4 1\n6 2\n6 1\n6 2\n6 3\n4 3\n4 3\n6 3\n4 2\n4 2\n5 3\n5 2\n5 2\n4 3\n5 3\n5 2\n4 2\n5 1\n4 2\n5 1\n5 3\n6 3\n5 3\n5 3\n4 2\n4 1\n4 2\n4 3\n6 3\n4 3\n6 2\n6 1\n5 3\n5 2\n4 1\n6 1\n5 2\n6 2\n4 2\n6 3\n4 3\n5 1\n6 3\n5 2\n4 3\n5 3\n5 3\n4 3\n6 3\n4 3\n4 1\n5 1\n6 2\n6 3\n5 3\n6 1\n6 3\n5 3\n6 1",
"output": "Mishka"
},
{
"input": "100\n1 5\n1 4\n1 5\n2 4\n2 6\n3 6\n3 5\n1 5\n2 5\n3 6\n3 5\n1 6\n1 4\n1 5\n1 6\n2 6\n1 5\n3 5\n3 4\n2 6\n2 6\n2 5\n3 4\n1 6\n1 4\n2 4\n1 5\n1 6\n3 5\n1 6\n2 6\n3 5\n1 6\n3 4\n3 5\n1 6\n3 6\n2 4\n2 4\n3 5\n2 6\n1 5\n3 5\n3 6\n2 4\n2 4\n2 6\n3 4\n3 4\n1 5\n1 4\n2 5\n3 4\n1 4\n2 6\n2 5\n2 4\n2 4\n2 5\n1 5\n1 6\n1 5\n1 5\n1 5\n1 6\n3 4\n2 4\n3 5\n3 5\n1 6\n3 5\n1 5\n1 6\n3 6\n3 4\n1 5\n3 5\n3 6\n1 4\n3 6\n1 5\n3 5\n3 6\n3 5\n1 4\n3 4\n2 4\n2 4\n2 5\n3 6\n3 5\n1 5\n2 4\n1 4\n3 4\n1 5\n3 4\n3 6\n3 5\n3 4",
"output": "Chris"
},
{
"input": "100\n4 3\n3 4\n5 1\n2 5\n5 3\n1 5\n6 3\n2 4\n5 2\n2 6\n5 2\n1 5\n6 3\n1 5\n6 3\n3 4\n5 2\n1 5\n6 1\n1 5\n4 2\n3 5\n6 3\n2 6\n6 3\n1 4\n6 2\n3 4\n4 1\n3 6\n5 1\n2 4\n5 1\n3 4\n6 2\n3 5\n4 1\n2 6\n4 3\n2 6\n5 2\n3 6\n6 2\n3 5\n4 3\n1 5\n5 3\n3 6\n4 2\n3 4\n6 1\n3 4\n5 2\n2 6\n5 2\n2 4\n6 2\n3 6\n4 3\n2 4\n4 3\n2 6\n4 2\n3 4\n6 3\n2 4\n6 3\n3 5\n5 2\n1 5\n6 3\n3 6\n4 3\n1 4\n5 2\n1 6\n4 1\n2 5\n4 1\n2 4\n4 2\n2 5\n6 1\n2 4\n6 3\n1 5\n4 3\n2 6\n6 3\n2 6\n5 3\n1 5\n4 1\n1 5\n6 2\n2 5\n5 1\n3 6\n4 3\n3 4",
"output": "Friendship is magic!^^"
},
{
"input": "99\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n1 3",
"output": "Mishka"
},
{
"input": "99\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "99\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1",
"output": "Chris"
},
{
"input": "99\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "100\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n1 4",
"output": "Mishka"
},
{
"input": "100\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "100\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1",
"output": "Chris"
},
{
"input": "100\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "84\n6 2\n1 5\n6 2\n2 3\n5 5\n1 2\n3 4\n3 4\n6 5\n6 4\n2 5\n4 1\n1 2\n1 1\n1 4\n2 5\n5 6\n6 3\n2 4\n5 5\n2 6\n3 4\n5 1\n3 3\n5 5\n4 6\n4 6\n2 4\n4 1\n5 2\n2 2\n3 6\n3 3\n4 6\n1 1\n2 4\n6 5\n5 2\n6 5\n5 5\n2 5\n6 4\n1 1\n6 2\n3 6\n6 5\n4 4\n1 5\n5 6\n4 4\n3 5\n6 1\n3 4\n1 5\n4 6\n4 6\n4 1\n3 6\n6 2\n1 1\n4 5\n5 4\n5 3\n3 4\n6 4\n1 1\n5 2\n6 5\n6 1\n2 2\n2 4\n3 3\n4 6\n1 3\n6 6\n5 2\n1 6\n6 2\n6 6\n4 1\n3 6\n6 4\n2 3\n3 4",
"output": "Chris"
},
{
"input": "70\n3 4\n2 3\n2 3\n6 5\n6 6\n4 3\n2 3\n3 1\n3 5\n5 6\n1 6\n2 5\n5 3\n2 5\n4 6\n5 1\n6 1\n3 1\n3 3\n5 3\n2 1\n3 3\n6 4\n6 3\n4 3\n4 5\n3 5\n5 5\n5 2\n1 6\n3 4\n5 2\n2 4\n1 6\n4 3\n4 3\n6 2\n1 3\n1 5\n6 1\n3 1\n1 1\n1 3\n2 2\n3 2\n6 4\n1 1\n4 4\n3 1\n4 5\n4 2\n6 3\n4 4\n3 2\n1 2\n2 6\n3 3\n1 5\n1 1\n6 5\n2 2\n3 1\n5 4\n5 2\n6 4\n6 3\n6 6\n6 3\n3 3\n5 4",
"output": "Mishka"
},
{
"input": "56\n6 4\n3 4\n6 1\n3 3\n1 4\n2 3\n1 5\n2 5\n1 5\n5 5\n2 3\n1 1\n3 2\n3 5\n4 6\n4 4\n5 2\n4 3\n3 1\n3 6\n2 3\n3 4\n5 6\n5 2\n5 6\n1 5\n1 5\n4 1\n6 3\n2 2\n2 1\n5 5\n2 1\n4 1\n5 4\n2 5\n4 1\n6 2\n3 4\n4 2\n6 4\n5 4\n4 2\n4 3\n6 2\n6 2\n3 1\n1 4\n3 6\n5 1\n5 5\n3 6\n6 4\n2 3\n6 5\n3 3",
"output": "Mishka"
},
{
"input": "94\n2 4\n6 4\n1 6\n1 4\n5 1\n3 3\n4 3\n6 1\n6 5\n3 2\n2 3\n5 1\n5 3\n1 2\n4 3\n3 2\n2 3\n4 6\n1 3\n6 3\n1 1\n3 2\n4 3\n1 5\n4 6\n3 2\n6 3\n1 6\n1 1\n1 2\n3 5\n1 3\n3 5\n4 4\n4 2\n1 4\n4 5\n1 3\n1 2\n1 1\n5 4\n5 5\n6 1\n2 1\n2 6\n6 6\n4 2\n3 6\n1 6\n6 6\n1 5\n3 2\n1 2\n4 4\n6 4\n4 1\n1 5\n3 3\n1 3\n3 4\n4 4\n1 1\n2 5\n4 5\n3 1\n3 1\n3 6\n3 2\n1 4\n1 6\n6 3\n2 4\n1 1\n2 2\n2 2\n2 1\n5 4\n1 2\n6 6\n2 2\n3 3\n6 3\n6 3\n1 6\n2 3\n2 4\n2 3\n6 6\n2 6\n6 3\n3 5\n1 4\n1 1\n3 5",
"output": "Chris"
},
{
"input": "81\n4 2\n1 2\n2 3\n4 5\n6 2\n1 6\n3 6\n3 4\n4 6\n4 4\n3 5\n4 6\n3 6\n3 5\n3 1\n1 3\n5 3\n3 4\n1 1\n4 1\n1 2\n6 1\n1 3\n6 5\n4 5\n4 2\n4 5\n6 2\n1 2\n2 6\n5 2\n1 5\n2 4\n4 3\n5 4\n1 2\n5 3\n2 6\n6 4\n1 1\n1 3\n3 1\n3 1\n6 5\n5 5\n6 1\n6 6\n5 2\n1 3\n1 4\n2 3\n5 5\n3 1\n3 1\n4 4\n1 6\n6 4\n2 2\n4 6\n4 4\n2 6\n2 4\n2 4\n4 1\n1 6\n1 4\n1 3\n6 5\n5 1\n1 3\n5 1\n1 4\n3 5\n2 6\n1 3\n5 6\n3 5\n4 4\n5 5\n5 6\n4 3",
"output": "Chris"
},
{
"input": "67\n6 5\n3 6\n1 6\n5 3\n5 4\n5 1\n1 6\n1 1\n3 2\n4 4\n3 1\n4 1\n1 5\n5 3\n3 3\n6 4\n2 4\n2 2\n4 3\n1 4\n1 4\n6 1\n1 2\n2 2\n5 1\n6 2\n3 5\n5 5\n2 2\n6 5\n6 2\n4 4\n3 1\n4 2\n6 6\n6 4\n5 1\n2 2\n4 5\n5 5\n4 6\n1 5\n6 3\n4 4\n1 5\n6 4\n3 6\n3 4\n1 6\n2 4\n2 1\n2 5\n6 5\n6 4\n4 1\n3 2\n1 2\n5 1\n5 6\n1 5\n3 5\n3 1\n5 3\n3 2\n5 1\n4 6\n6 6",
"output": "Mishka"
},
{
"input": "55\n6 6\n6 5\n2 2\n2 2\n6 4\n5 5\n6 5\n5 3\n1 3\n2 2\n5 6\n3 3\n3 3\n6 5\n3 5\n5 5\n1 2\n1 1\n4 6\n1 2\n5 5\n6 2\n6 3\n1 2\n5 1\n1 3\n3 3\n4 4\n2 5\n1 1\n5 3\n4 3\n2 2\n4 5\n5 6\n4 5\n6 3\n1 6\n6 4\n3 6\n1 6\n5 2\n6 3\n2 3\n5 5\n4 3\n3 1\n4 2\n1 1\n2 5\n5 3\n2 2\n6 3\n4 5\n2 2",
"output": "Mishka"
},
{
"input": "92\n2 3\n1 3\n2 6\n5 1\n5 5\n3 2\n5 6\n2 5\n3 1\n3 6\n4 5\n2 5\n1 2\n2 3\n6 5\n3 6\n4 4\n6 2\n4 5\n4 4\n5 1\n6 1\n3 4\n3 5\n6 6\n3 2\n6 4\n2 2\n3 5\n6 4\n6 3\n6 6\n3 4\n3 3\n6 1\n5 4\n6 2\n2 6\n5 6\n1 4\n4 6\n6 3\n3 1\n4 1\n6 6\n3 5\n6 3\n6 1\n1 6\n3 2\n6 6\n4 3\n3 4\n1 3\n3 5\n5 3\n6 5\n4 3\n5 5\n4 1\n1 5\n6 4\n2 3\n2 3\n1 5\n1 2\n5 2\n4 3\n3 6\n5 5\n5 4\n1 4\n3 3\n1 6\n5 6\n5 4\n5 3\n1 1\n6 2\n5 5\n2 5\n4 3\n6 6\n5 1\n1 1\n4 6\n4 6\n3 1\n6 4\n2 4\n2 2\n2 1",
"output": "Chris"
},
{
"input": "79\n5 3\n4 6\n3 6\n2 1\n5 2\n2 3\n4 4\n6 2\n2 5\n1 6\n6 6\n2 6\n3 3\n4 5\n6 2\n2 1\n1 5\n5 1\n2 1\n2 6\n5 3\n6 2\n2 6\n2 3\n1 5\n4 4\n6 3\n5 2\n3 2\n1 3\n1 3\n6 3\n2 6\n3 6\n5 3\n4 5\n6 1\n3 5\n3 5\n6 5\n1 5\n4 2\n6 2\n2 3\n4 6\n3 6\n2 5\n4 4\n1 1\n4 6\n2 6\n6 4\n3 2\n4 1\n1 2\n6 4\n5 6\n1 4\n2 2\n5 4\n3 2\n1 2\n2 4\n2 5\n2 1\n3 6\n3 3\n1 1\n2 2\n4 4\n4 5\n3 3\n5 3\n6 2\n4 5\n6 5\n2 5\n5 6\n2 2",
"output": "Chris"
},
{
"input": "65\n1 1\n5 1\n2 2\n5 4\n4 5\n2 5\n3 2\n5 6\n6 3\n1 1\n6 1\n1 5\n1 1\n5 2\n6 4\n1 6\n1 1\n4 3\n2 3\n5 6\n4 4\n6 2\n1 3\n4 3\n1 3\n6 3\n3 5\n4 2\n4 1\n6 1\n3 2\n2 6\n3 2\n3 5\n6 3\n4 3\n1 5\n2 6\n1 3\n4 1\n4 1\n2 5\n2 5\n6 2\n5 3\n3 1\n3 3\n5 1\n2 4\n5 3\n3 3\n1 1\n6 3\n3 3\n5 1\n1 6\n4 5\n6 6\n5 5\n2 5\n4 1\n2 2\n1 4\n1 6\n6 5",
"output": "Mishka"
},
{
"input": "1\n1 1",
"output": "Friendship is magic!^^"
}
] | 1,666,978,438
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 69
| 46
| 0
|
m = 0
for i in range(int(input())):
x = [int(i) for i in input().split()]
m+=1 if x[0] > x[1] else -1 if x[0] < x[1] else 0
print("Mishka") if m > 0 else print("Chris") if m<0 else print("Friendship is magic!^^")
|
Title: Mishka and Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mishka is a little polar bear. As known, little bears loves spending their free time playing dice for chocolates. Once in a wonderful sunny morning, walking around blocks of ice, Mishka met her friend Chris, and they started playing the game.
Rules of the game are very simple: at first number of rounds *n* is defined. In every round each of the players throws a cubical dice with distinct numbers from 1 to 6 written on its faces. Player, whose value after throwing the dice is greater, wins the round. In case if player dice values are equal, no one of them is a winner.
In average, player, who won most of the rounds, is the winner of the game. In case if two players won the same number of rounds, the result of the game is draw.
Mishka is still very little and can't count wins and losses, so she asked you to watch their game and determine its result. Please help her!
Input Specification:
The first line of the input contains single integer *n* *n* (1<=≤<=*n*<=≤<=100) — the number of game rounds.
The next *n* lines contains rounds description. *i*-th of them contains pair of integers *m**i* and *c**i* (1<=≤<=*m**i*,<=<=*c**i*<=≤<=6) — values on dice upper face after Mishka's and Chris' throws in *i*-th round respectively.
Output Specification:
If Mishka is the winner of the game, print "Mishka" (without quotes) in the only line.
If Chris is the winner of the game, print "Chris" (without quotes) in the only line.
If the result of the game is draw, print "Friendship is magic!^^" (without quotes) in the only line.
Demo Input:
['3\n3 5\n2 1\n4 2\n', '2\n6 1\n1 6\n', '3\n1 5\n3 3\n2 2\n']
Demo Output:
['Mishka', 'Friendship is magic!^^', 'Chris']
Note:
In the first sample case Mishka loses the first round, but wins second and third rounds and thus she is the winner of the game.
In the second sample case Mishka wins the first round, Chris wins the second round, and the game ends with draw with score 1:1.
In the third sample case Chris wins the first round, but there is no winner of the next two rounds. The winner of the game is Chris.
|
```python
m = 0
for i in range(int(input())):
x = [int(i) for i in input().split()]
m+=1 if x[0] > x[1] else -1 if x[0] < x[1] else 0
print("Mishka") if m > 0 else print("Chris") if m<0 else print("Friendship is magic!^^")
```
| 3
|
|
300
|
C
|
Beautiful Numbers
|
PROGRAMMING
| 1,800
|
[
"brute force",
"combinatorics"
] | null | null |
Vitaly is a very weird man. He's got two favorite digits *a* and *b*. Vitaly calls a positive integer good, if the decimal representation of this integer only contains digits *a* and *b*. Vitaly calls a good number excellent, if the sum of its digits is a good number.
For example, let's say that Vitaly's favourite digits are 1 and 3, then number 12 isn't good and numbers 13 or 311 are. Also, number 111 is excellent and number 11 isn't.
Now Vitaly is wondering, how many excellent numbers of length exactly *n* are there. As this number can be rather large, he asks you to count the remainder after dividing it by 1000000007 (109<=+<=7).
A number's length is the number of digits in its decimal representation without leading zeroes.
|
The first line contains three integers: *a*, *b*, *n* (1<=≤<=*a*<=<<=*b*<=≤<=9,<=1<=≤<=*n*<=≤<=106).
|
Print a single integer — the answer to the problem modulo 1000000007 (109<=+<=7).
|
[
"1 3 3\n",
"2 3 10\n"
] |
[
"1\n",
"165\n"
] |
none
| 2,000
|
[
{
"input": "1 3 3",
"output": "1"
},
{
"input": "2 3 10",
"output": "165"
},
{
"input": "6 8 14215",
"output": "651581472"
},
{
"input": "4 9 104671",
"output": "329390901"
},
{
"input": "6 7 78755",
"output": "0"
},
{
"input": "1 8 265",
"output": "461320265"
},
{
"input": "3 9 37413",
"output": "461358757"
},
{
"input": "1 7 49055",
"output": "461364774"
},
{
"input": "3 4 11028",
"output": "461668105"
},
{
"input": "2 6 32377",
"output": "887598327"
},
{
"input": "3 5 80791",
"output": "999993599"
},
{
"input": "4 8 11857",
"output": "999991923"
},
{
"input": "1 3 10785",
"output": "999952603"
},
{
"input": "4 6 11808",
"output": "999925731"
},
{
"input": "1 2 11857",
"output": "999991923"
},
{
"input": "2 4 88193",
"output": "999976846"
},
{
"input": "1 4 37226",
"output": "999970594"
},
{
"input": "2 5 53049",
"output": "259705254"
},
{
"input": "3 6 1000000",
"output": "786609214"
},
{
"input": "7 9 999999",
"output": "53911803"
},
{
"input": "8 9 999999",
"output": "447886447"
},
{
"input": "3 8 1000000",
"output": "0"
},
{
"input": "2 8 999999",
"output": "0"
},
{
"input": "1 6 997695",
"output": "0"
},
{
"input": "1 5 997694",
"output": "0"
},
{
"input": "5 9 997693",
"output": "0"
},
{
"input": "5 8 997690",
"output": "21735480"
},
{
"input": "7 8 2",
"output": "0"
},
{
"input": "6 9 1",
"output": "2"
},
{
"input": "8 9 111111",
"output": "900401372"
},
{
"input": "8 9 1000000",
"output": "573697309"
},
{
"input": "1 2 1000000",
"output": "786609214"
}
] | 1,641,779,766
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 32
| 560
| 33,996,800
|
mod = 10 ** 9 + 7
a, b, n = map(int, input().split())
N1 = n + 1
fac = [1]
for nn in range(1, N1):
fac.append(nn * fac[-1] % mod)
fac_inv = [pow(fac[-1], mod - 2, mod)]
for nn in range(N1 - 1, 0, -1):
fac_inv.append(fac_inv[-1] * nn % mod)
fac_inv.reverse()
cnt = 0
diff = b - a
start = a * n
sab = {str(a), str(b)}
for i in range(N1):
s = str(start)
for j in s:
if j not in sab:
break
else:
cnt = (cnt + fac[n] * fac_inv[i] * fac_inv[n - i]) % mod
start += diff
print(cnt)
|
Title: Beautiful Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vitaly is a very weird man. He's got two favorite digits *a* and *b*. Vitaly calls a positive integer good, if the decimal representation of this integer only contains digits *a* and *b*. Vitaly calls a good number excellent, if the sum of its digits is a good number.
For example, let's say that Vitaly's favourite digits are 1 and 3, then number 12 isn't good and numbers 13 or 311 are. Also, number 111 is excellent and number 11 isn't.
Now Vitaly is wondering, how many excellent numbers of length exactly *n* are there. As this number can be rather large, he asks you to count the remainder after dividing it by 1000000007 (109<=+<=7).
A number's length is the number of digits in its decimal representation without leading zeroes.
Input Specification:
The first line contains three integers: *a*, *b*, *n* (1<=≤<=*a*<=<<=*b*<=≤<=9,<=1<=≤<=*n*<=≤<=106).
Output Specification:
Print a single integer — the answer to the problem modulo 1000000007 (109<=+<=7).
Demo Input:
['1 3 3\n', '2 3 10\n']
Demo Output:
['1\n', '165\n']
Note:
none
|
```python
mod = 10 ** 9 + 7
a, b, n = map(int, input().split())
N1 = n + 1
fac = [1]
for nn in range(1, N1):
fac.append(nn * fac[-1] % mod)
fac_inv = [pow(fac[-1], mod - 2, mod)]
for nn in range(N1 - 1, 0, -1):
fac_inv.append(fac_inv[-1] * nn % mod)
fac_inv.reverse()
cnt = 0
diff = b - a
start = a * n
sab = {str(a), str(b)}
for i in range(N1):
s = str(start)
for j in s:
if j not in sab:
break
else:
cnt = (cnt + fac[n] * fac_inv[i] * fac_inv[n - i]) % mod
start += diff
print(cnt)
```
| 3
|
|
71
|
A
|
Way Too Long Words
|
PROGRAMMING
| 800
|
[
"strings"
] |
A. Way Too Long Words
|
1
|
256
|
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
|
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
|
[
"4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n"
] |
[
"word\nl10n\ni18n\np43s\n"
] |
none
| 500
|
[
{
"input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis",
"output": "word\nl10n\ni18n\np43s"
},
{
"input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm",
"output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m"
},
{
"input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd",
"output": "n20n\nj27b\nk15d"
},
{
"input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt",
"output": "t41t"
},
{
"input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings",
"output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz",
"output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz"
},
{
"input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij",
"output": "a98j"
},
{
"input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco",
"output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco"
},
{
"input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt",
"output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t"
},
{
"input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr",
"output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr"
}
] | 1,671,788,561
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
n=input()
l=len(n)
if l>10:
print(n[0]+""+str(l-2)+""+n[l-1])
else:
print(n)
|
Title: Way Too Long Words
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output Specification:
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
Demo Input:
['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n']
Demo Output:
['word\nl10n\ni18n\np43s\n']
Note:
none
|
```python
n=input()
l=len(n)
if l>10:
print(n[0]+""+str(l-2)+""+n[l-1])
else:
print(n)
```
| 0
|
706
|
B
|
Interesting drink
|
PROGRAMMING
| 1,100
|
[
"binary search",
"dp",
"implementation"
] | null | null |
Vasiliy likes to rest after a hard work, so you may often meet him in some bar nearby. As all programmers do, he loves the famous drink "Beecola", which can be bought in *n* different shops in the city. It's known that the price of one bottle in the shop *i* is equal to *x**i* coins.
Vasiliy plans to buy his favorite drink for *q* consecutive days. He knows, that on the *i*-th day he will be able to spent *m**i* coins. Now, for each of the days he want to know in how many different shops he can buy a bottle of "Beecola".
|
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of shops in the city that sell Vasiliy's favourite drink.
The second line contains *n* integers *x**i* (1<=≤<=*x**i*<=≤<=100<=000) — prices of the bottles of the drink in the *i*-th shop.
The third line contains a single integer *q* (1<=≤<=*q*<=≤<=100<=000) — the number of days Vasiliy plans to buy the drink.
Then follow *q* lines each containing one integer *m**i* (1<=≤<=*m**i*<=≤<=109) — the number of coins Vasiliy can spent on the *i*-th day.
|
Print *q* integers. The *i*-th of them should be equal to the number of shops where Vasiliy will be able to buy a bottle of the drink on the *i*-th day.
|
[
"5\n3 10 8 6 11\n4\n1\n10\n3\n11\n"
] |
[
"0\n4\n1\n5\n"
] |
On the first day, Vasiliy won't be able to buy a drink in any of the shops.
On the second day, Vasiliy can buy a drink in the shops 1, 2, 3 and 4.
On the third day, Vasiliy can buy a drink only in the shop number 1.
Finally, on the last day Vasiliy can buy a drink in any shop.
| 1,000
|
[
{
"input": "5\n3 10 8 6 11\n4\n1\n10\n3\n11",
"output": "0\n4\n1\n5"
},
{
"input": "5\n868 987 714 168 123\n10\n424\n192\n795\n873\n117\n914\n735\n158\n631\n471",
"output": "2\n2\n3\n4\n0\n4\n3\n1\n2\n2"
},
{
"input": "3\n435 482 309\n7\n245\n241\n909\n745\n980\n29\n521",
"output": "0\n0\n3\n3\n3\n0\n3"
},
{
"input": "1\n653\n9\n903\n980\n80\n770\n965\n874\n381\n657\n969",
"output": "1\n1\n0\n1\n1\n1\n0\n1\n1"
},
{
"input": "12\n35345 58181 32223 84621 35905 73863 99537 30666 67771 39229 36847 29038\n23\n55052824\n82504840\n35160556\n78141700\n73401989\n86305919\n39430705\n31939373\n23501765\n4406029\n61436920\n14295390\n34275309\n28028753\n85724689\n70158847\n2396455\n66994588\n84024224\n30175981\n65622319\n76517111\n76605341",
"output": "12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12"
},
{
"input": "4\n698 193 1584 76192\n4\n553640\n310518\n274280\n429192",
"output": "4\n4\n4\n4"
},
{
"input": "1\n1\n1\n1000000000",
"output": "1"
},
{
"input": "1\n1\n1\n100005",
"output": "1"
},
{
"input": "3\n1 2 3\n1\n100005",
"output": "3"
},
{
"input": "2\n1 2\n3\n100004\n100005\n100006",
"output": "2\n2\n2"
},
{
"input": "3\n1 1 1\n1\n1",
"output": "3"
},
{
"input": "1\n1\n1\n1",
"output": "1"
},
{
"input": "3\n1 2 3\n1\n100000",
"output": "3"
},
{
"input": "5\n1 1 1 1 10\n2\n1\n10",
"output": "4\n5"
},
{
"input": "5\n3 10 8 6 11\n1\n10000000",
"output": "5"
}
] | 1,688,371,270
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 103
| 732
| 13,619,200
|
import math
import bisect
n=int(input())
a=list(map(int,input().split()))
a.sort()
q=int(input())
for i in range(q):
m=int(input())
print(bisect.bisect_right(a,m))
|
Title: Interesting drink
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasiliy likes to rest after a hard work, so you may often meet him in some bar nearby. As all programmers do, he loves the famous drink "Beecola", which can be bought in *n* different shops in the city. It's known that the price of one bottle in the shop *i* is equal to *x**i* coins.
Vasiliy plans to buy his favorite drink for *q* consecutive days. He knows, that on the *i*-th day he will be able to spent *m**i* coins. Now, for each of the days he want to know in how many different shops he can buy a bottle of "Beecola".
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of shops in the city that sell Vasiliy's favourite drink.
The second line contains *n* integers *x**i* (1<=≤<=*x**i*<=≤<=100<=000) — prices of the bottles of the drink in the *i*-th shop.
The third line contains a single integer *q* (1<=≤<=*q*<=≤<=100<=000) — the number of days Vasiliy plans to buy the drink.
Then follow *q* lines each containing one integer *m**i* (1<=≤<=*m**i*<=≤<=109) — the number of coins Vasiliy can spent on the *i*-th day.
Output Specification:
Print *q* integers. The *i*-th of them should be equal to the number of shops where Vasiliy will be able to buy a bottle of the drink on the *i*-th day.
Demo Input:
['5\n3 10 8 6 11\n4\n1\n10\n3\n11\n']
Demo Output:
['0\n4\n1\n5\n']
Note:
On the first day, Vasiliy won't be able to buy a drink in any of the shops.
On the second day, Vasiliy can buy a drink in the shops 1, 2, 3 and 4.
On the third day, Vasiliy can buy a drink only in the shop number 1.
Finally, on the last day Vasiliy can buy a drink in any shop.
|
```python
import math
import bisect
n=int(input())
a=list(map(int,input().split()))
a.sort()
q=int(input())
for i in range(q):
m=int(input())
print(bisect.bisect_right(a,m))
```
| 3
|
|
415
|
B
|
Mashmokh and Tokens
|
PROGRAMMING
| 1,500
|
[
"binary search",
"greedy",
"implementation",
"math"
] | null | null |
Bimokh is Mashmokh's boss. For the following *n* days he decided to pay to his workers in a new way. At the beginning of each day he will give each worker a certain amount of tokens. Then at the end of each day each worker can give some of his tokens back to get a certain amount of money. The worker can save the rest of tokens but he can't use it in any other day to get more money. If a worker gives back *w* tokens then he'll get dollars.
Mashmokh likes the tokens however he likes money more. That's why he wants to save as many tokens as possible so that the amount of money he gets is maximal possible each day. He has *n* numbers *x*1,<=*x*2,<=...,<=*x**n*. Number *x**i* is the number of tokens given to each worker on the *i*-th day. Help him calculate for each of *n* days the number of tokens he can save.
|
The first line of input contains three space-separated integers *n*,<=*a*,<=*b* (1<=≤<=*n*<=≤<=105; 1<=≤<=*a*,<=*b*<=≤<=109). The second line of input contains *n* space-separated integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=109).
|
Output *n* space-separated integers. The *i*-th of them is the number of tokens Mashmokh can save on the *i*-th day.
|
[
"5 1 4\n12 6 11 9 1\n",
"3 1 2\n1 2 3\n",
"1 1 1\n1\n"
] |
[
"0 2 3 1 1 ",
"1 0 1 ",
"0 "
] |
none
| 1,000
|
[
{
"input": "5 1 4\n12 6 11 9 1",
"output": "0 2 3 1 1 "
},
{
"input": "3 1 2\n1 2 3",
"output": "1 0 1 "
},
{
"input": "1 1 1\n1",
"output": "0 "
},
{
"input": "1 1 1000000000\n1000000000",
"output": "0 "
},
{
"input": "1 1 1000000000\n999999999",
"output": "999999999 "
},
{
"input": "10 1 100000000\n999999999 999999999 999999999 999999999 999999999 999999999 999999999 999999999 999999999 999999999",
"output": "99999999 99999999 99999999 99999999 99999999 99999999 99999999 99999999 99999999 99999999 "
}
] | 1,585,900,361
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 0
| 124
| 0
|
import math
n,a,b=map(int,input().split())
arr = list(map(int,input().split()))
for i in range(n):
arr[i]-=((math.floor((arr[i]*a)/b)))
print(*arr)
|
Title: Mashmokh and Tokens
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bimokh is Mashmokh's boss. For the following *n* days he decided to pay to his workers in a new way. At the beginning of each day he will give each worker a certain amount of tokens. Then at the end of each day each worker can give some of his tokens back to get a certain amount of money. The worker can save the rest of tokens but he can't use it in any other day to get more money. If a worker gives back *w* tokens then he'll get dollars.
Mashmokh likes the tokens however he likes money more. That's why he wants to save as many tokens as possible so that the amount of money he gets is maximal possible each day. He has *n* numbers *x*1,<=*x*2,<=...,<=*x**n*. Number *x**i* is the number of tokens given to each worker on the *i*-th day. Help him calculate for each of *n* days the number of tokens he can save.
Input Specification:
The first line of input contains three space-separated integers *n*,<=*a*,<=*b* (1<=≤<=*n*<=≤<=105; 1<=≤<=*a*,<=*b*<=≤<=109). The second line of input contains *n* space-separated integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=109).
Output Specification:
Output *n* space-separated integers. The *i*-th of them is the number of tokens Mashmokh can save on the *i*-th day.
Demo Input:
['5 1 4\n12 6 11 9 1\n', '3 1 2\n1 2 3\n', '1 1 1\n1\n']
Demo Output:
['0 2 3 1 1 ', '1 0 1 ', '0 ']
Note:
none
|
```python
import math
n,a,b=map(int,input().split())
arr = list(map(int,input().split()))
for i in range(n):
arr[i]-=((math.floor((arr[i]*a)/b)))
print(*arr)
```
| 0
|
|
527
|
A
|
Playing with Paper
|
PROGRAMMING
| 1,100
|
[
"implementation",
"math"
] | null | null |
One day Vasya was sitting on a not so interesting Maths lesson and making an origami from a rectangular *a* mm <=×<= *b* mm sheet of paper (*a*<=><=*b*). Usually the first step in making an origami is making a square piece of paper from the rectangular sheet by folding the sheet along the bisector of the right angle, and cutting the excess part.
After making a paper ship from the square piece, Vasya looked on the remaining (*a*<=-<=*b*) mm <=×<= *b* mm strip of paper. He got the idea to use this strip of paper in the same way to make an origami, and then use the remainder (if it exists) and so on. At the moment when he is left with a square piece of paper, he will make the last ship from it and stop.
Can you determine how many ships Vasya will make during the lesson?
|
The first line of the input contains two integers *a*, *b* (1<=≤<=*b*<=<<=*a*<=≤<=1012) — the sizes of the original sheet of paper.
|
Print a single integer — the number of ships that Vasya will make.
|
[
"2 1\n",
"10 7\n",
"1000000000000 1\n"
] |
[
"2\n",
"6\n",
"1000000000000\n"
] |
Pictures to the first and second sample test.
| 500
|
[
{
"input": "2 1",
"output": "2"
},
{
"input": "10 7",
"output": "6"
},
{
"input": "1000000000000 1",
"output": "1000000000000"
},
{
"input": "3 1",
"output": "3"
},
{
"input": "4 1",
"output": "4"
},
{
"input": "3 2",
"output": "3"
},
{
"input": "4 2",
"output": "2"
},
{
"input": "1000 700",
"output": "6"
},
{
"input": "959986566087 524054155168",
"output": "90"
},
{
"input": "4 3",
"output": "4"
},
{
"input": "7 6",
"output": "7"
},
{
"input": "1000 999",
"output": "1000"
},
{
"input": "1000 998",
"output": "500"
},
{
"input": "1000 997",
"output": "336"
},
{
"input": "42 1",
"output": "42"
},
{
"input": "1000 1",
"output": "1000"
},
{
"input": "8 5",
"output": "5"
},
{
"input": "13 8",
"output": "6"
},
{
"input": "987 610",
"output": "15"
},
{
"input": "442 42",
"output": "22"
},
{
"input": "754 466",
"output": "13"
},
{
"input": "1000000000000 999999999999",
"output": "1000000000000"
},
{
"input": "1000000000000 999999999998",
"output": "500000000000"
},
{
"input": "941 14",
"output": "74"
},
{
"input": "998 2",
"output": "499"
},
{
"input": "1000 42",
"output": "32"
},
{
"input": "1000 17",
"output": "66"
},
{
"input": "5 1",
"output": "5"
},
{
"input": "5 2",
"output": "4"
},
{
"input": "5 3",
"output": "4"
},
{
"input": "5 4",
"output": "5"
},
{
"input": "293 210",
"output": "17"
},
{
"input": "787878787878 424242424242",
"output": "8"
},
{
"input": "956722026041 591286729879",
"output": "58"
},
{
"input": "956722026041 365435296162",
"output": "58"
},
{
"input": "628625247282 464807889701",
"output": "102"
},
{
"input": "695928431619 424778620208",
"output": "167"
},
{
"input": "1000000000000 42",
"output": "23809523821"
},
{
"input": "987654345678 23",
"output": "42941493300"
},
{
"input": "10000000001 2",
"output": "5000000002"
},
{
"input": "1000000000000 2",
"output": "500000000000"
},
{
"input": "1000000000000 3",
"output": "333333333336"
},
{
"input": "100000000000 3",
"output": "33333333336"
},
{
"input": "100000000000 23",
"output": "4347826109"
},
{
"input": "999999999997 7",
"output": "142857142861"
},
{
"input": "8589934592 4294967296",
"output": "2"
}
] | 1,502,339,140
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 46
| 62
| 4,608,000
|
a, b = [int(x) for x in input().split()]
pieces = int(a/b)
a -= b * pieces
c = a
while c != 0:
a = b
b = c
pieces += int(a/b)
c = a % b
print(pieces)
|
Title: Playing with Paper
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Vasya was sitting on a not so interesting Maths lesson and making an origami from a rectangular *a* mm <=×<= *b* mm sheet of paper (*a*<=><=*b*). Usually the first step in making an origami is making a square piece of paper from the rectangular sheet by folding the sheet along the bisector of the right angle, and cutting the excess part.
After making a paper ship from the square piece, Vasya looked on the remaining (*a*<=-<=*b*) mm <=×<= *b* mm strip of paper. He got the idea to use this strip of paper in the same way to make an origami, and then use the remainder (if it exists) and so on. At the moment when he is left with a square piece of paper, he will make the last ship from it and stop.
Can you determine how many ships Vasya will make during the lesson?
Input Specification:
The first line of the input contains two integers *a*, *b* (1<=≤<=*b*<=<<=*a*<=≤<=1012) — the sizes of the original sheet of paper.
Output Specification:
Print a single integer — the number of ships that Vasya will make.
Demo Input:
['2 1\n', '10 7\n', '1000000000000 1\n']
Demo Output:
['2\n', '6\n', '1000000000000\n']
Note:
Pictures to the first and second sample test.
|
```python
a, b = [int(x) for x in input().split()]
pieces = int(a/b)
a -= b * pieces
c = a
while c != 0:
a = b
b = c
pieces += int(a/b)
c = a % b
print(pieces)
```
| 3
|
|
101
|
A
|
Homework
|
PROGRAMMING
| 1,200
|
[
"greedy"
] |
A. Homework
|
2
|
256
|
Once when Gerald studied in the first year at school, his teacher gave the class the following homework. She offered the students a string consisting of *n* small Latin letters; the task was to learn the way the letters that the string contains are written. However, as Gerald is too lazy, he has no desire whatsoever to learn those letters. That's why he decided to lose some part of the string (not necessarily a connected part). The lost part can consist of any number of segments of any length, at any distance from each other. However, Gerald knows that if he loses more than *k* characters, it will be very suspicious.
Find the least number of distinct characters that can remain in the string after no more than *k* characters are deleted. You also have to find any possible way to delete the characters.
|
The first input data line contains a string whose length is equal to *n* (1<=≤<=*n*<=≤<=105). The string consists of lowercase Latin letters. The second line contains the number *k* (0<=≤<=*k*<=≤<=105).
|
Print on the first line the only number *m* — the least possible number of different characters that could remain in the given string after it loses no more than *k* characters.
Print on the second line the string that Gerald can get after some characters are lost. The string should have exactly *m* distinct characters. The final string should be the subsequence of the initial string. If Gerald can get several different strings with exactly *m* distinct characters, print any of them.
|
[
"aaaaa\n4\n",
"abacaba\n4\n",
"abcdefgh\n10\n"
] |
[
"1\naaaaa\n",
"1\naaaa\n",
"0\n\n"
] |
In the first sample the string consists of five identical letters but you are only allowed to delete 4 of them so that there was at least one letter left. Thus, the right answer is 1 and any string consisting of characters "a" from 1 to 5 in length.
In the second sample you are allowed to delete 4 characters. You cannot delete all the characters, because the string has length equal to 7. However, you can delete all characters apart from "a" (as they are no more than four), which will result in the "aaaa" string.
In the third sample you are given a line whose length is equal to 8, and *k* = 10, so that the whole line can be deleted. The correct answer is 0 and an empty string.
| 500
|
[
{
"input": "aaaaa\n4",
"output": "1\naaaaa"
},
{
"input": "abacaba\n4",
"output": "1\naaaa"
},
{
"input": "abcdefgh\n10",
"output": "0"
},
{
"input": "aaaaaaaaaaaaaaaaaaaa\n19",
"output": "1\naaaaaaaaaaaaaaaaaaaa"
},
{
"input": "abcdefghijjihgedcba\n0",
"output": "10\nabcdefghijjihgedcba"
},
{
"input": "aababcabcdabcde\n9",
"output": "2\naabababab"
},
{
"input": "xyzuvwxyz\n4",
"output": "3\nxyzxyz"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\n99",
"output": "1\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx\n0",
"output": "1\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx"
},
{
"input": "abcdefghijklmnopqrstuvwxyz\n17",
"output": "9\nrstuvwxyz"
},
{
"input": "abcdefghijklmnopqrstuvwxyz\n0",
"output": "26\nabcdefghijklmnopqrstuvwxyz"
},
{
"input": "abcdefghijklmnopqrsttsrqponmlkjihgfedcba\n0",
"output": "20\nabcdefghijklmnopqrsttsrqponmlkjihgfedcba"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaeeeeeeeeeeeeeeeeeeee\n20",
"output": "1\naaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "xyxjvqrbehasypiekxwjhurlrnegtkiplbogkgxwubzhlyvjwj\n24",
"output": "8\nxyxjrhykxwjhrlrklkxwhlyjwj"
},
{
"input": "clpdaxnimfkubdxtpjwtjkqh\n21",
"output": "2\nxxtt"
},
{
"input": "jeliuewohkqtghdneuuhcputwiddnmkbhhnlxxbfjunhcd\n50",
"output": "0"
},
{
"input": "zgwmpjfeiwtfagp\n62",
"output": "0"
},
{
"input": "halasouqgfxfcrwhqgllaqiphaxekljz\n87",
"output": "0"
},
{
"input": "zimxucbrzojfqvizcopkplrpnvihveqpgvzszkubftoozrydxijokjxfhdfjracjonqupmnhadtsotxrxmwgno\n51",
"output": "7\nzxrzojvzopprpvvpvzzoozrxjojxjrjopoxrxo"
},
{
"input": "geovcaxzjyhxbpnbkbsxfpkyofopxquzzxeigdflfumisevzsjdywehxconimkkbvjyxbqlnmaphvnngcjqoefqkfzmiruubbcmv\n24",
"output": "16\neovxzjyxbpnbkbxfpkyofopxquzzxeiffumievzjyexonimkkbvjyxbqnmpvnnjqoefqkfzmiuubbmv"
},
{
"input": "jsreqtehsewsiwzqbpniwuhbgcrrkxlgbhuobphjigfuinffvvatrcxnzbcxjazrrxyiwxncuiifzndpvqahwpdfo\n67",
"output": "4\nrwiwiwrrxiirxxrrxiwxiiw"
},
{
"input": "uwvkcydkhbmbqyfjuryqnxcxhoanwnjubuvpgfipdeserodhh\n65",
"output": "0"
},
{
"input": "xclfxmeqhfjwurwmazpysafoxepb\n26",
"output": "1\nxxx"
},
{
"input": "hlhugwawagrnpojcmzfiqtffrzuqfovcmxnfqukgzxilglfbtsqgtsweblymqrdskcxjtuytodzujgtivkmiktvskvoqpegoiw\n27",
"output": "15\nlugwwgomzfiqtffzuqfovmxfqukgzxilglftsqgtswlmqskxtutozugtivkmiktvskvoqgoiw"
},
{
"input": "cky\n79",
"output": "0"
},
{
"input": "oodcvb\n16",
"output": "0"
},
{
"input": "lfbfwdoeggorzdsxqnpophbcjcatphjsewamrgzjszf\n20",
"output": "8\nffwoggozspopjpjswgzjszf"
},
{
"input": "ksnizygvqy\n42",
"output": "0"
},
{
"input": "myenljgyxkwcfyzjcpffsucstschcevbzh\n44",
"output": "0"
},
{
"input": "yumufcicodkpuhvifnvi\n36",
"output": "0"
},
{
"input": "fntrmjfquczybyjllywsqwllsxdmqynmyfcqhakftitvvfbxtqktbfsvvvanjbkqubyxu\n63",
"output": "1\nyyyyyy"
},
{
"input": "smiclwubkoobnapkkletsnbbsvihqbvikochzteaewjonkzvsqrbjkywsfcvczwretmhscowapcraof\n45",
"output": "6\nscwbkoobkksbbsbkocwoksbkwsccwscowco"
},
{
"input": "lwkjydpagifuvbhifryskegmzuexfksazfurlsnzfrgvuxcazitfchimmvomdnbdirzccstmuvlpghwskinayvucodiwn\n16",
"output": "17\nlwkydagifuvifryskgmzufksazfurlsnzfrgvucazifcimmvmdndirzccsmuvlgwskinayvucdiwn"
},
{
"input": "a\n0",
"output": "1\na"
},
{
"input": "bbb\n100000",
"output": "0"
},
{
"input": "aa\n2",
"output": "0"
},
{
"input": "a\n1",
"output": "0"
},
{
"input": "aaaa\n4",
"output": "0"
}
] | 1,591,289,529
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 216
| 0
|
import sys
from sys import stdin, stdout
n, m = map(int, sys.stdin.readline().split())
arr = list(map(int, sys.stdin.readline().strip().split()))
curPage = 0
for i in range(n):
t = 0
for j in range(arr[i]):
curPage += 1
if curPage == m:
curPage = 0
t += 1
sys.stdout.write(str(t)+" ")
|
Title: Homework
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Once when Gerald studied in the first year at school, his teacher gave the class the following homework. She offered the students a string consisting of *n* small Latin letters; the task was to learn the way the letters that the string contains are written. However, as Gerald is too lazy, he has no desire whatsoever to learn those letters. That's why he decided to lose some part of the string (not necessarily a connected part). The lost part can consist of any number of segments of any length, at any distance from each other. However, Gerald knows that if he loses more than *k* characters, it will be very suspicious.
Find the least number of distinct characters that can remain in the string after no more than *k* characters are deleted. You also have to find any possible way to delete the characters.
Input Specification:
The first input data line contains a string whose length is equal to *n* (1<=≤<=*n*<=≤<=105). The string consists of lowercase Latin letters. The second line contains the number *k* (0<=≤<=*k*<=≤<=105).
Output Specification:
Print on the first line the only number *m* — the least possible number of different characters that could remain in the given string after it loses no more than *k* characters.
Print on the second line the string that Gerald can get after some characters are lost. The string should have exactly *m* distinct characters. The final string should be the subsequence of the initial string. If Gerald can get several different strings with exactly *m* distinct characters, print any of them.
Demo Input:
['aaaaa\n4\n', 'abacaba\n4\n', 'abcdefgh\n10\n']
Demo Output:
['1\naaaaa\n', '1\naaaa\n', '0\n\n']
Note:
In the first sample the string consists of five identical letters but you are only allowed to delete 4 of them so that there was at least one letter left. Thus, the right answer is 1 and any string consisting of characters "a" from 1 to 5 in length.
In the second sample you are allowed to delete 4 characters. You cannot delete all the characters, because the string has length equal to 7. However, you can delete all characters apart from "a" (as they are no more than four), which will result in the "aaaa" string.
In the third sample you are given a line whose length is equal to 8, and *k* = 10, so that the whole line can be deleted. The correct answer is 0 and an empty string.
|
```python
import sys
from sys import stdin, stdout
n, m = map(int, sys.stdin.readline().split())
arr = list(map(int, sys.stdin.readline().strip().split()))
curPage = 0
for i in range(n):
t = 0
for j in range(arr[i]):
curPage += 1
if curPage == m:
curPage = 0
t += 1
sys.stdout.write(str(t)+" ")
```
| -1
|
587
|
A
|
Duff and Weight Lifting
|
PROGRAMMING
| 1,500
|
[
"greedy"
] | null | null |
Recently, Duff has been practicing weight lifting. As a hard practice, Malek gave her a task. He gave her a sequence of weights. Weight of *i*-th of them is 2*w**i* pounds. In each step, Duff can lift some of the remaining weights and throw them away. She does this until there's no more weight left. Malek asked her to minimize the number of steps.
Duff is a competitive programming fan. That's why in each step, she can only lift and throw away a sequence of weights 2*a*1,<=...,<=2*a**k* if and only if there exists a non-negative integer *x* such that 2*a*1<=+<=2*a*2<=+<=...<=+<=2*a**k*<==<=2*x*, i. e. the sum of those numbers is a power of two.
Duff is a competitive programming fan, but not a programmer. That's why she asked for your help. Help her minimize the number of steps.
|
The first line of input contains integer *n* (1<=≤<=*n*<=≤<=106), the number of weights.
The second line contains *n* integers *w*1,<=...,<=*w**n* separated by spaces (0<=≤<=*w**i*<=≤<=106 for each 1<=≤<=*i*<=≤<=*n*), the powers of two forming the weights values.
|
Print the minimum number of steps in a single line.
|
[
"5\n1 1 2 3 3\n",
"4\n0 1 2 3\n"
] |
[
"2\n",
"4\n"
] |
In the first sample case: One optimal way would be to throw away the first three in the first step and the rest in the second step. Also, it's not possible to do it in one step because their sum is not a power of two.
In the second sample case: The only optimal way is to throw away one weight in each step. It's not possible to do it in less than 4 steps because there's no subset of weights with more than one weight and sum equal to a power of two.
| 500
|
[
{
"input": "5\n1 1 2 3 3",
"output": "2"
},
{
"input": "4\n0 1 2 3",
"output": "4"
},
{
"input": "1\n120287",
"output": "1"
},
{
"input": "2\n28288 0",
"output": "2"
},
{
"input": "2\n95745 95745",
"output": "1"
},
{
"input": "13\n92 194 580495 0 10855 41704 13 96429 33 213 0 92 140599",
"output": "11"
},
{
"input": "13\n688743 688743 1975 688743 688743 688743 688743 688743 688743 0 0 688743 688743",
"output": "4"
},
{
"input": "35\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "3"
},
{
"input": "35\n130212 3176 77075 8071 18 1369 7539 1683 80757 1847 0 1374 122 8524 4 2 21333 270264 4 9254 151921 0 1 33596 73002 54382 0 1 29233 75952 15 38892 1877 6167 4",
"output": "31"
},
{
"input": "35\n0 0 298 0 0 0 0 0 689063 65442 0 984598 2054 43668 0 369 0 2054 0 996220 0 16327 369 0 996220 0 0 0 4693 2054 348 0 118 0 0",
"output": "16"
},
{
"input": "100\n196 1681 196 0 61 93 196 196 196 196 196 0 0 96 18 1576 0 93 666463 18 93 1 1278 8939 93 196 196 1278 3 0 67416 869956 10 56489 196 745 39 783 196 8939 196 81 69634 4552 39 3 14 20 25 8 10 4 7302 0 19579 20 1140 15990 7302 0 19579 4142 11 1354 75252 93 311 1278 0 79475 10 75252 93 7302 0 81 408441 19579 10 39 19 37748 4364 31135 47700 105818 47700 10 4142 543356 3 30647 45917 60714 8939 18 22925 7302 93 75252",
"output": "59"
}
] | 1,663,410,227
| 2,147,483,647
|
PyPy 3-64
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 10
| 1,000
| 109,977,600
|
import sys
input = sys.stdin.readline
from heapq import heappop, heappush
n = int(input())
h = sorted(map(int, input().split()))
b = -1
c = 0
while h:
a = heappop(h)
if h:
b = heappop(h)
if a == b:
heappush(h, a+1)
else:
heappush(h, b)
c += 1
else:
c += 1
print(c)
|
Title: Duff and Weight Lifting
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently, Duff has been practicing weight lifting. As a hard practice, Malek gave her a task. He gave her a sequence of weights. Weight of *i*-th of them is 2*w**i* pounds. In each step, Duff can lift some of the remaining weights and throw them away. She does this until there's no more weight left. Malek asked her to minimize the number of steps.
Duff is a competitive programming fan. That's why in each step, she can only lift and throw away a sequence of weights 2*a*1,<=...,<=2*a**k* if and only if there exists a non-negative integer *x* such that 2*a*1<=+<=2*a*2<=+<=...<=+<=2*a**k*<==<=2*x*, i. e. the sum of those numbers is a power of two.
Duff is a competitive programming fan, but not a programmer. That's why she asked for your help. Help her minimize the number of steps.
Input Specification:
The first line of input contains integer *n* (1<=≤<=*n*<=≤<=106), the number of weights.
The second line contains *n* integers *w*1,<=...,<=*w**n* separated by spaces (0<=≤<=*w**i*<=≤<=106 for each 1<=≤<=*i*<=≤<=*n*), the powers of two forming the weights values.
Output Specification:
Print the minimum number of steps in a single line.
Demo Input:
['5\n1 1 2 3 3\n', '4\n0 1 2 3\n']
Demo Output:
['2\n', '4\n']
Note:
In the first sample case: One optimal way would be to throw away the first three in the first step and the rest in the second step. Also, it's not possible to do it in one step because their sum is not a power of two.
In the second sample case: The only optimal way is to throw away one weight in each step. It's not possible to do it in less than 4 steps because there's no subset of weights with more than one weight and sum equal to a power of two.
|
```python
import sys
input = sys.stdin.readline
from heapq import heappop, heappush
n = int(input())
h = sorted(map(int, input().split()))
b = -1
c = 0
while h:
a = heappop(h)
if h:
b = heappop(h)
if a == b:
heappush(h, a+1)
else:
heappush(h, b)
c += 1
else:
c += 1
print(c)
```
| 0
|
|
71
|
A
|
Way Too Long Words
|
PROGRAMMING
| 800
|
[
"strings"
] |
A. Way Too Long Words
|
1
|
256
|
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
|
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
|
[
"4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n"
] |
[
"word\nl10n\ni18n\np43s\n"
] |
none
| 500
|
[
{
"input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis",
"output": "word\nl10n\ni18n\np43s"
},
{
"input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm",
"output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m"
},
{
"input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd",
"output": "n20n\nj27b\nk15d"
},
{
"input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt",
"output": "t41t"
},
{
"input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings",
"output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz",
"output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz"
},
{
"input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij",
"output": "a98j"
},
{
"input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco",
"output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco"
},
{
"input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt",
"output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t"
},
{
"input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr",
"output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr"
}
] | 1,688,812,644
| 2,147,483,647
|
PyPy 3-64
|
RUNTIME_ERROR
|
TESTS
| 0
| 77
| 2,867,200
|
a=int(input()) #4
word=[a]
nword =input()
word.append(nword)
numw=len(nword)
while word < a :
if numw > 10 :
print(f"{nword[0] + numw + nword[-1]}")
elif numw<10:
print(word)
a+=1
|
Title: Way Too Long Words
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output Specification:
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
Demo Input:
['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n']
Demo Output:
['word\nl10n\ni18n\np43s\n']
Note:
none
|
```python
a=int(input()) #4
word=[a]
nword =input()
word.append(nword)
numw=len(nword)
while word < a :
if numw > 10 :
print(f"{nword[0] + numw + nword[-1]}")
elif numw<10:
print(word)
a+=1
```
| -1
|
610
|
A
|
Pasha and Stick
|
PROGRAMMING
| 1,000
|
[
"combinatorics",
"math"
] | null | null |
Pasha has a wooden stick of some positive integer length *n*. He wants to perform exactly three cuts to get four parts of the stick. Each part must have some positive integer length and the sum of these lengths will obviously be *n*.
Pasha likes rectangles but hates squares, so he wonders, how many ways are there to split a stick into four parts so that it's possible to form a rectangle using these parts, but is impossible to form a square.
Your task is to help Pasha and count the number of such ways. Two ways to cut the stick are considered distinct if there exists some integer *x*, such that the number of parts of length *x* in the first way differ from the number of parts of length *x* in the second way.
|
The first line of the input contains a positive integer *n* (1<=≤<=*n*<=≤<=2·109) — the length of Pasha's stick.
|
The output should contain a single integer — the number of ways to split Pasha's stick into four parts of positive integer length so that it's possible to make a rectangle by connecting the ends of these parts, but is impossible to form a square.
|
[
"6\n",
"20\n"
] |
[
"1\n",
"4\n"
] |
There is only one way to divide the stick in the first sample {1, 1, 2, 2}.
Four ways to divide the stick in the second sample are {1, 1, 9, 9}, {2, 2, 8, 8}, {3, 3, 7, 7} and {4, 4, 6, 6}. Note that {5, 5, 5, 5} doesn't work.
| 500
|
[
{
"input": "6",
"output": "1"
},
{
"input": "20",
"output": "4"
},
{
"input": "1",
"output": "0"
},
{
"input": "2",
"output": "0"
},
{
"input": "3",
"output": "0"
},
{
"input": "4",
"output": "0"
},
{
"input": "2000000000",
"output": "499999999"
},
{
"input": "1924704072",
"output": "481176017"
},
{
"input": "73740586",
"output": "18435146"
},
{
"input": "1925088820",
"output": "481272204"
},
{
"input": "593070992",
"output": "148267747"
},
{
"input": "1925473570",
"output": "481368392"
},
{
"input": "629490186",
"output": "157372546"
},
{
"input": "1980649112",
"output": "495162277"
},
{
"input": "36661322",
"output": "9165330"
},
{
"input": "1943590793",
"output": "0"
},
{
"input": "71207034",
"output": "17801758"
},
{
"input": "1757577394",
"output": "439394348"
},
{
"input": "168305294",
"output": "42076323"
},
{
"input": "1934896224",
"output": "483724055"
},
{
"input": "297149088",
"output": "74287271"
},
{
"input": "1898001634",
"output": "474500408"
},
{
"input": "176409698",
"output": "44102424"
},
{
"input": "1873025522",
"output": "468256380"
},
{
"input": "5714762",
"output": "1428690"
},
{
"input": "1829551192",
"output": "457387797"
},
{
"input": "16269438",
"output": "4067359"
},
{
"input": "1663283390",
"output": "415820847"
},
{
"input": "42549941",
"output": "0"
},
{
"input": "1967345604",
"output": "491836400"
},
{
"input": "854000",
"output": "213499"
},
{
"input": "1995886626",
"output": "498971656"
},
{
"input": "10330019",
"output": "0"
},
{
"input": "1996193634",
"output": "499048408"
},
{
"input": "9605180",
"output": "2401294"
},
{
"input": "1996459740",
"output": "499114934"
},
{
"input": "32691948",
"output": "8172986"
},
{
"input": "1975903308",
"output": "493975826"
},
{
"input": "1976637136",
"output": "494159283"
},
{
"input": "29803038",
"output": "7450759"
},
{
"input": "1977979692",
"output": "494494922"
},
{
"input": "1978595336",
"output": "494648833"
},
{
"input": "27379344",
"output": "6844835"
},
{
"input": "1979729912",
"output": "494932477"
},
{
"input": "1980253780",
"output": "495063444"
},
{
"input": "1980751584",
"output": "495187895"
},
{
"input": "53224878",
"output": "13306219"
},
{
"input": "5",
"output": "0"
},
{
"input": "7",
"output": "0"
},
{
"input": "8",
"output": "1"
},
{
"input": "9",
"output": "0"
},
{
"input": "10",
"output": "2"
},
{
"input": "11",
"output": "0"
},
{
"input": "12",
"output": "2"
},
{
"input": "13",
"output": "0"
},
{
"input": "14",
"output": "3"
},
{
"input": "15",
"output": "0"
},
{
"input": "16",
"output": "3"
},
{
"input": "17",
"output": "0"
},
{
"input": "18",
"output": "4"
},
{
"input": "19",
"output": "0"
},
{
"input": "21",
"output": "0"
},
{
"input": "22",
"output": "5"
},
{
"input": "23",
"output": "0"
},
{
"input": "24",
"output": "5"
},
{
"input": "25",
"output": "0"
},
{
"input": "26",
"output": "6"
},
{
"input": "27",
"output": "0"
},
{
"input": "28",
"output": "6"
},
{
"input": "29",
"output": "0"
},
{
"input": "30",
"output": "7"
},
{
"input": "111",
"output": "0"
},
{
"input": "55",
"output": "0"
},
{
"input": "105",
"output": "0"
},
{
"input": "199",
"output": "0"
},
{
"input": "151",
"output": "0"
}
] | 1,640,861,757
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 2
| 46
| 0
|
n=int(input())
if n%2==0:
count=0
for i in range(2,int(n/2),2):
count=count+1
print(count)
|
Title: Pasha and Stick
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pasha has a wooden stick of some positive integer length *n*. He wants to perform exactly three cuts to get four parts of the stick. Each part must have some positive integer length and the sum of these lengths will obviously be *n*.
Pasha likes rectangles but hates squares, so he wonders, how many ways are there to split a stick into four parts so that it's possible to form a rectangle using these parts, but is impossible to form a square.
Your task is to help Pasha and count the number of such ways. Two ways to cut the stick are considered distinct if there exists some integer *x*, such that the number of parts of length *x* in the first way differ from the number of parts of length *x* in the second way.
Input Specification:
The first line of the input contains a positive integer *n* (1<=≤<=*n*<=≤<=2·109) — the length of Pasha's stick.
Output Specification:
The output should contain a single integer — the number of ways to split Pasha's stick into four parts of positive integer length so that it's possible to make a rectangle by connecting the ends of these parts, but is impossible to form a square.
Demo Input:
['6\n', '20\n']
Demo Output:
['1\n', '4\n']
Note:
There is only one way to divide the stick in the first sample {1, 1, 2, 2}.
Four ways to divide the stick in the second sample are {1, 1, 9, 9}, {2, 2, 8, 8}, {3, 3, 7, 7} and {4, 4, 6, 6}. Note that {5, 5, 5, 5} doesn't work.
|
```python
n=int(input())
if n%2==0:
count=0
for i in range(2,int(n/2),2):
count=count+1
print(count)
```
| -1
|
|
910
|
A
|
The Way to Home
|
PROGRAMMING
| 800
|
[
"dfs and similar",
"dp",
"greedy",
"implementation"
] | null | null |
A frog lives on the axis *Ox* and needs to reach home which is in the point *n*. She starts from the point 1. The frog can jump to the right at a distance not more than *d*. So, after she jumped from the point *x* she can reach the point *x*<=+<=*a*, where *a* is an integer from 1 to *d*.
For each point from 1 to *n* is known if there is a lily flower in it. The frog can jump only in points with a lilies. Guaranteed that there are lilies in the points 1 and *n*.
Determine the minimal number of jumps that the frog needs to reach home which is in the point *n* from the point 1. Consider that initially the frog is in the point 1. If the frog can not reach home, print -1.
|
The first line contains two integers *n* and *d* (2<=≤<=*n*<=≤<=100, 1<=≤<=*d*<=≤<=*n*<=-<=1) — the point, which the frog wants to reach, and the maximal length of the frog jump.
The second line contains a string *s* of length *n*, consisting of zeros and ones. If a character of the string *s* equals to zero, then in the corresponding point there is no lily flower. In the other case, in the corresponding point there is a lily flower. Guaranteed that the first and the last characters of the string *s* equal to one.
|
If the frog can not reach the home, print -1.
In the other case, print the minimal number of jumps that the frog needs to reach the home which is in the point *n* from the point 1.
|
[
"8 4\n10010101\n",
"4 2\n1001\n",
"8 4\n11100101\n",
"12 3\n101111100101\n"
] |
[
"2\n",
"-1\n",
"3\n",
"4\n"
] |
In the first example the from can reach home in two jumps: the first jump from the point 1 to the point 4 (the length of the jump is three), and the second jump from the point 4 to the point 8 (the length of the jump is four).
In the second example the frog can not reach home, because to make it she need to jump on a distance three, but the maximum length of her jump equals to two.
| 500
|
[
{
"input": "8 4\n10010101",
"output": "2"
},
{
"input": "4 2\n1001",
"output": "-1"
},
{
"input": "8 4\n11100101",
"output": "3"
},
{
"input": "12 3\n101111100101",
"output": "4"
},
{
"input": "5 4\n11011",
"output": "1"
},
{
"input": "5 4\n10001",
"output": "1"
},
{
"input": "10 7\n1101111011",
"output": "2"
},
{
"input": "10 9\n1110000101",
"output": "1"
},
{
"input": "10 9\n1100000001",
"output": "1"
},
{
"input": "20 5\n11111111110111101001",
"output": "4"
},
{
"input": "20 11\n11100000111000011011",
"output": "2"
},
{
"input": "20 19\n10100000000000000001",
"output": "1"
},
{
"input": "50 13\n10011010100010100111010000010000000000010100000101",
"output": "5"
},
{
"input": "50 8\n11010100000011001100001100010001110000101100110011",
"output": "8"
},
{
"input": "99 4\n111111111111111111111111111111111111111111111111111111111011111111111111111111111111111111111111111",
"output": "25"
},
{
"input": "99 98\n100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001",
"output": "1"
},
{
"input": "100 5\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111",
"output": "20"
},
{
"input": "100 4\n1111111111111111111111111111111111111111111111111111111111111111111111111111110111111111111111111111",
"output": "25"
},
{
"input": "100 4\n1111111111111111111111111111111111111111111111111111111111111101111111011111111111111111111111111111",
"output": "25"
},
{
"input": "100 3\n1111110111111111111111111111111111111111101111111111111111111111111101111111111111111111111111111111",
"output": "34"
},
{
"input": "100 8\n1111111111101110111111111111111111111111111111111111111111111111111111110011111111111111011111111111",
"output": "13"
},
{
"input": "100 7\n1011111111111111111011101111111011111101111111111101111011110111111111111111111111110111111011111111",
"output": "15"
},
{
"input": "100 9\n1101111110111110101111111111111111011001110111011101011111111111010101111111100011011111111010111111",
"output": "12"
},
{
"input": "100 6\n1011111011111111111011010110011001010101111110111111000111011011111110101101110110101111110000100111",
"output": "18"
},
{
"input": "100 7\n1110001111101001110011111111111101111101101001010001101000101100000101101101011111111101101000100001",
"output": "16"
},
{
"input": "100 11\n1000010100011100011011100000010011001111011110100100001011010100011011111001101101110110010110001101",
"output": "10"
},
{
"input": "100 9\n1001001110000011100100000001000110111101101010101001000101001010011001101100110011011110110011011111",
"output": "13"
},
{
"input": "100 7\n1010100001110101111011000111000001110100100110110001110110011010100001100100001110111100110000101001",
"output": "18"
},
{
"input": "100 10\n1110110000000110000000101110100000111000001011100000100110010001110111001010101000011000000001011011",
"output": "12"
},
{
"input": "100 13\n1000000100000000100011000010010000101010011110000000001000011000110100001000010001100000011001011001",
"output": "9"
},
{
"input": "100 11\n1000000000100000010000100001000100000000010000100100000000100100001000000001011000110001000000000101",
"output": "12"
},
{
"input": "100 22\n1000100000001010000000000000000001000000100000000000000000010000000000001000000000000000000100000001",
"output": "7"
},
{
"input": "100 48\n1000000000000000011000000000000000000000000000000001100000000000000000000000000000000000000000000001",
"output": "3"
},
{
"input": "100 48\n1000000000000000000000100000000000000000000000000000000000000000000001000000000000000000100000000001",
"output": "3"
},
{
"input": "100 75\n1000000100000000000000000000000000000000000000000000000000000000000000000000000001000000000000000001",
"output": "3"
},
{
"input": "100 73\n1000000000000000000000000000000100000000000000000000000000000000000000000000000000000000000000000001",
"output": "2"
},
{
"input": "100 99\n1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001",
"output": "1"
},
{
"input": "100 1\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111",
"output": "99"
},
{
"input": "100 2\n1111111111111111111111111111111110111111111111111111111111111111111111111111111111111111111111111111",
"output": "50"
},
{
"input": "100 1\n1111111111111111011111111111111111111111111111111111111111111111111101111111111111111111111111111111",
"output": "-1"
},
{
"input": "100 3\n1111111111111111111111111101111111111111111111111011111111111111111111111111111011111111111111111111",
"output": "33"
},
{
"input": "100 1\n1101111111111111111111101111111111111111111111111111111111111011111111101111101111111111111111111111",
"output": "-1"
},
{
"input": "100 6\n1111111111111111111111101111111101011110001111111111111111110111111111111111111111111110010111111111",
"output": "17"
},
{
"input": "100 2\n1111111101111010110111011011110111101111111011111101010101011111011111111111111011111001101111101111",
"output": "-1"
},
{
"input": "100 8\n1100110101111001101001111000111100110100011110111011001011111110000110101000001110111011100111011011",
"output": "14"
},
{
"input": "100 10\n1000111110100000001001101100000010011100010101001100010011111001001101111110110111101111001010001101",
"output": "11"
},
{
"input": "100 7\n1110000011010001110101011010000011110001000000011101110111010110001000011101111010010001101111110001",
"output": "-1"
},
{
"input": "100 3\n1111010001000001011011000011001111000100101000101101000010111101111000010000011110110011001101010111",
"output": "-1"
},
{
"input": "100 9\n1101010101101100010111011000010100001010000101010011001001100010110110000000010000101000000001101101",
"output": "13"
},
{
"input": "100 14\n1010100000000000010101000010001100000000000011100010000001000001011010001110001010100000100001101101",
"output": "9"
},
{
"input": "100 13\n1000000001101001110000010000011001000000000000001010000000100001001010000000000000000100010000000001",
"output": "-1"
},
{
"input": "100 18\n1000000000000000110000000000000000010000000001000001000001000000000100000000000010000000000000000001",
"output": "-1"
},
{
"input": "100 32\n1000000000000000000000000001000000000000000000000101000000000000000000000000000000000001000000000001",
"output": "-1"
},
{
"input": "100 79\n1000000001000000000101000000000000000000000000000000000000000000000000000000000000000000000000000001",
"output": "2"
},
{
"input": "100 41\n1000000000000000000000000000000000010000000000000000000000000000000000000000100000000000000000000001",
"output": "3"
},
{
"input": "100 82\n1000000000000000000100000000000000000000000000000000000000000000000000000000000000000000000000000001",
"output": "2"
},
{
"input": "100 96\n1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001",
"output": "-1"
},
{
"input": "43 30\n1001000001111111010100100100110101011101101",
"output": "2"
},
{
"input": "7 1\n1111111",
"output": "6"
},
{
"input": "9 3\n101000001",
"output": "-1"
},
{
"input": "10 3\n1100000001",
"output": "-1"
},
{
"input": "8 2\n10000101",
"output": "-1"
},
{
"input": "2 1\n11",
"output": "1"
}
] | 1,514,376,641
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 63
| 62
| 5,632,000
|
n,k=map(int,input().split())
s=input()
t=i=p=0
while i<n:
#print(i)
if s[i]=='1':t+=1;p=i;i+=k
else:
while i!=p and s[i]=='0' or i>=n:i-=1
if i==p:exit(print(-1))
if p!=n-1:t+=1
print(t-1)
|
Title: The Way to Home
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A frog lives on the axis *Ox* and needs to reach home which is in the point *n*. She starts from the point 1. The frog can jump to the right at a distance not more than *d*. So, after she jumped from the point *x* she can reach the point *x*<=+<=*a*, where *a* is an integer from 1 to *d*.
For each point from 1 to *n* is known if there is a lily flower in it. The frog can jump only in points with a lilies. Guaranteed that there are lilies in the points 1 and *n*.
Determine the minimal number of jumps that the frog needs to reach home which is in the point *n* from the point 1. Consider that initially the frog is in the point 1. If the frog can not reach home, print -1.
Input Specification:
The first line contains two integers *n* and *d* (2<=≤<=*n*<=≤<=100, 1<=≤<=*d*<=≤<=*n*<=-<=1) — the point, which the frog wants to reach, and the maximal length of the frog jump.
The second line contains a string *s* of length *n*, consisting of zeros and ones. If a character of the string *s* equals to zero, then in the corresponding point there is no lily flower. In the other case, in the corresponding point there is a lily flower. Guaranteed that the first and the last characters of the string *s* equal to one.
Output Specification:
If the frog can not reach the home, print -1.
In the other case, print the minimal number of jumps that the frog needs to reach the home which is in the point *n* from the point 1.
Demo Input:
['8 4\n10010101\n', '4 2\n1001\n', '8 4\n11100101\n', '12 3\n101111100101\n']
Demo Output:
['2\n', '-1\n', '3\n', '4\n']
Note:
In the first example the from can reach home in two jumps: the first jump from the point 1 to the point 4 (the length of the jump is three), and the second jump from the point 4 to the point 8 (the length of the jump is four).
In the second example the frog can not reach home, because to make it she need to jump on a distance three, but the maximum length of her jump equals to two.
|
```python
n,k=map(int,input().split())
s=input()
t=i=p=0
while i<n:
#print(i)
if s[i]=='1':t+=1;p=i;i+=k
else:
while i!=p and s[i]=='0' or i>=n:i-=1
if i==p:exit(print(-1))
if p!=n-1:t+=1
print(t-1)
```
| 3
|
|
263
|
A
|
Beautiful Matrix
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
You've got a 5<=×<=5 matrix, consisting of 24 zeroes and a single number one. Let's index the matrix rows by numbers from 1 to 5 from top to bottom, let's index the matrix columns by numbers from 1 to 5 from left to right. In one move, you are allowed to apply one of the two following transformations to the matrix:
1. Swap two neighboring matrix rows, that is, rows with indexes *i* and *i*<=+<=1 for some integer *i* (1<=≤<=*i*<=<<=5). 1. Swap two neighboring matrix columns, that is, columns with indexes *j* and *j*<=+<=1 for some integer *j* (1<=≤<=*j*<=<<=5).
You think that a matrix looks beautiful, if the single number one of the matrix is located in its middle (in the cell that is on the intersection of the third row and the third column). Count the minimum number of moves needed to make the matrix beautiful.
|
The input consists of five lines, each line contains five integers: the *j*-th integer in the *i*-th line of the input represents the element of the matrix that is located on the intersection of the *i*-th row and the *j*-th column. It is guaranteed that the matrix consists of 24 zeroes and a single number one.
|
Print a single integer — the minimum number of moves needed to make the matrix beautiful.
|
[
"0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n",
"0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n"
] |
[
"3\n",
"1\n"
] |
none
| 500
|
[
{
"input": "0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "0"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1",
"output": "4"
},
{
"input": "0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "4"
},
{
"input": "0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "4"
},
{
"input": "0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0",
"output": "4"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0",
"output": "3"
}
] | 1,695,727,063
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 25
| 92
| 0
|
import math as m
x, y = 0, 0
for i in range(5):
row = input().split()
if '1' not in row:
continue
x = row.index('1')
y = i
break
ans = m.fabs(2-x) + m.fabs(2-y)
print(int(ans))
|
Title: Beautiful Matrix
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got a 5<=×<=5 matrix, consisting of 24 zeroes and a single number one. Let's index the matrix rows by numbers from 1 to 5 from top to bottom, let's index the matrix columns by numbers from 1 to 5 from left to right. In one move, you are allowed to apply one of the two following transformations to the matrix:
1. Swap two neighboring matrix rows, that is, rows with indexes *i* and *i*<=+<=1 for some integer *i* (1<=≤<=*i*<=<<=5). 1. Swap two neighboring matrix columns, that is, columns with indexes *j* and *j*<=+<=1 for some integer *j* (1<=≤<=*j*<=<<=5).
You think that a matrix looks beautiful, if the single number one of the matrix is located in its middle (in the cell that is on the intersection of the third row and the third column). Count the minimum number of moves needed to make the matrix beautiful.
Input Specification:
The input consists of five lines, each line contains five integers: the *j*-th integer in the *i*-th line of the input represents the element of the matrix that is located on the intersection of the *i*-th row and the *j*-th column. It is guaranteed that the matrix consists of 24 zeroes and a single number one.
Output Specification:
Print a single integer — the minimum number of moves needed to make the matrix beautiful.
Demo Input:
['0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n', '0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n']
Demo Output:
['3\n', '1\n']
Note:
none
|
```python
import math as m
x, y = 0, 0
for i in range(5):
row = input().split()
if '1' not in row:
continue
x = row.index('1')
y = i
break
ans = m.fabs(2-x) + m.fabs(2-y)
print(int(ans))
```
| 3
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
Fox Ciel has a robot on a 2D plane. Initially it is located in (0, 0). Fox Ciel code a command to it. The command was represented by string *s*. Each character of *s* is one move operation. There are four move operations at all:
- 'U': go up, (x, y) <=→<= (x, y+1); - 'D': go down, (x, y) <=→<= (x, y-1); - 'L': go left, (x, y) <=→<= (x-1, y); - 'R': go right, (x, y) <=→<= (x+1, y).
The robot will do the operations in *s* from left to right, and repeat it infinite times. Help Fox Ciel to determine if after some steps the robot will located in (*a*,<=*b*).
|
The first line contains two integers *a* and *b*, (<=-<=109<=≤<=*a*,<=*b*<=≤<=109). The second line contains a string *s* (1<=≤<=|*s*|<=≤<=100, *s* only contains characters 'U', 'D', 'L', 'R') — the command.
|
Print "Yes" if the robot will be located at (*a*,<=*b*), and "No" otherwise.
|
[
"2 2\nRU\n",
"1 2\nRU\n",
"-1 1000000000\nLRRLU\n",
"0 0\nD\n"
] |
[
"Yes\n",
"No\n",
"Yes\n",
"Yes\n"
] |
In the first and second test case, command string is "RU", so the robot will go right, then go up, then right, and then up and so on.
The locations of its moves are (0, 0) → (1, 0) → (1, 1) → (2, 1) → (2, 2) → ...
So it can reach (2, 2) but not (1, 2).
| 0
|
[
{
"input": "2 2\nRU",
"output": "Yes"
},
{
"input": "1 2\nRU",
"output": "No"
},
{
"input": "-1 1000000000\nLRRLU",
"output": "Yes"
},
{
"input": "0 0\nD",
"output": "Yes"
},
{
"input": "0 0\nUURRDL",
"output": "Yes"
},
{
"input": "987654321 987654321\nUURRDL",
"output": "Yes"
},
{
"input": "4 2\nUURRDL",
"output": "No"
},
{
"input": "4 3\nUURRDL",
"output": "Yes"
},
{
"input": "4 4\nUURRDL",
"output": "Yes"
},
{
"input": "4 6\nUURRDL",
"output": "Yes"
},
{
"input": "4 7\nUURRDL",
"output": "No"
},
{
"input": "1000000000 1000000000\nUURRDL",
"output": "Yes"
},
{
"input": "-1 -1\nUR",
"output": "No"
},
{
"input": "1 1\nUURRDDLL",
"output": "No"
},
{
"input": "987654321 2\nUURDD",
"output": "Yes"
},
{
"input": "0 123456789\nRRULL",
"output": "Yes"
},
{
"input": "4 4\nUUUURRRRDDDDLLLL",
"output": "Yes"
},
{
"input": "-491226083 -49122610\nUDRLDURLDLLLDUDURLRDUUDDUUULUDRDRDUULURDRLLDDDLUDUURLUUDLLDULLLLDDLDDUU",
"output": "Yes"
},
{
"input": "-261597957 418556728\nLLLDLUDUULLRDDULLRRUDRDLULRLRLLRRUUDRRLRUDLRRLUDRDLLUUDUULRURLDLULUUULDDUURLRUDURRL",
"output": "Yes"
},
{
"input": "-771928144 -3\nRUDULULDRDLLLULDDUDDDDUDULRULRUULDDDURUDLUURULLLDLLDDRDDRLRURUULRUURRUDLDLDDRLLULRRDRRLLUULUDRUUDRRD",
"output": "Yes"
},
{
"input": "397346346 1\nDDURRUURLDLRRLULD",
"output": "Yes"
},
{
"input": "-528551525 0\nUDRLRRLDLDLURRRRULDLRLRLURUUDDLRLLDRRULLUDLURDLUUULLLRUUUDRRURLDUDULDDRDDDRDL",
"output": "Yes"
},
{
"input": "311692421 -129871846\nLLLDURULDDDDUDDURRLUUDRLDDRDURDDRUDUURLUDUDLDRUDDDUUURDRRUDRDRDURLLDURUUDRLDLDURRRRRRDULURDRU",
"output": "Yes"
},
{
"input": "485940814 728911221\nURURU",
"output": "Yes"
},
{
"input": "-843450986 632588242\nLURLULULRUDUDULRDDLUL",
"output": "Yes"
},
{
"input": "647999516 -809999401\nUDLDDLLULUDDLLDUULRRRDLUDDLDDLRLRRDRURURDRRDRULUDRDULRULLRRLLDDRLRRUDRURDUULUDLRRLRDR",
"output": "Yes"
},
{
"input": "352820537 -764444491\nRDDUDLUDDUDLRRRDRRRDRRDUDUDDURLRRLDRLLRLLLLUULUDRURRDRLDDLLDRDURDUDRUDDLUDRLURUDRURDRDDLDRLDLDLLU",
"output": "Yes"
},
{
"input": "-284973644 -1\nDLULLDLRUUDRR",
"output": "Yes"
},
{
"input": "356922591 -2\nRRLDLDUDRUUUULUUDDULDDUDD",
"output": "No"
},
{
"input": "27033101 54066203\nUDDDRDLLLRUUDDLRDLDRLRUDDULRLLRULR",
"output": "No"
},
{
"input": "-199335150 39867031\nLLURRDUULRUDDRDUUULDLDRDDLURDRLDRLLLRRRRRULRRRUUDD",
"output": "No"
},
{
"input": "609504072 609504074\nULRLUDLDDR",
"output": "No"
},
{
"input": "497684357 829473929\nRRLDUUURULURRLLRRLRLURRLDU",
"output": "Yes"
},
{
"input": "551922835 183974295\nDUDUUULDRLRURRDULRRUDDLRLLUULLRLRDRDRR",
"output": "No"
},
{
"input": "825368095 -825368096\nRD",
"output": "No"
},
{
"input": "-458990423 -229495204\nDLLDDRLUDLRLUL",
"output": "No"
},
{
"input": "285102789 570205594\nRRDULRULULRRDUURRLURUDDULLRDUL",
"output": "No"
},
{
"input": "109928480 219856920\nLRURLRLURDRDLDRDLRDDUUDDLULDRRUUURRUDLLUULUUUR",
"output": "No"
},
{
"input": "-532674020 532674026\nUURLLL",
"output": "No"
},
{
"input": "999999999 0\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "Yes"
},
{
"input": "0 0\nUDLRUDLRUDLRUDLRUDLRUDLRUDLRUDLRUDLRUDLRUDLRUDLRUDLRUDLRUDLRUDLRUDLRUDLRUDLRUDLRUDLRUDLRUDLRUDLRUDLR",
"output": "Yes"
},
{
"input": "1 1\nUDLRUDLRUDLRUDLRUDLRUDLRUDLRUDLRUDLRUDLRUDLRUDLRUDLRUDLRUDLRUDLRUDLRUDLRUDLRUDLRUDLRUDLRUDLRUDLRUDLR",
"output": "No"
},
{
"input": "-1000000000 -1000000000\nDLDLDLDLDLDLDLDLDLDLDLDLDLDLDLDLDLDLDLDLDLDLDLDLDLDLDLDLDLDLDLDLDLDLDLDLDLDLDLDLDLDLDLDLDLDLDLDLDLDL",
"output": "Yes"
},
{
"input": "3 3\nUURR",
"output": "No"
},
{
"input": "-2 -2\nUR",
"output": "No"
},
{
"input": "5 5\nUDLR",
"output": "No"
},
{
"input": "0 -1\nU",
"output": "No"
},
{
"input": "-1 0\nR",
"output": "No"
},
{
"input": "1000000000 1000000000\nURURURUR",
"output": "Yes"
},
{
"input": "-1 -1\nRU",
"output": "No"
},
{
"input": "1 1\nLD",
"output": "No"
},
{
"input": "-2 -2\nUURR",
"output": "No"
},
{
"input": "1000000000 0\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "Yes"
},
{
"input": "2 6\nRUUUURLDDDL",
"output": "Yes"
},
{
"input": "0 1\nLUUR",
"output": "No"
},
{
"input": "1 1\nURDLDL",
"output": "Yes"
},
{
"input": "-10 -10\nRU",
"output": "No"
},
{
"input": "1000000000 1000000000\nRURURURURURURURURURURURURURURURURURURURURURURURURURURURURURURURURURURURURURURURURURURURURURURURURURU",
"output": "Yes"
},
{
"input": "-1000000000 -500000000\nURR",
"output": "No"
},
{
"input": "-2 0\nULLLDDRRRR",
"output": "Yes"
},
{
"input": "999999999 -999999999\nRRRRRRRRRRRRRRRRRRRRRRRRRDDDDDDDDDDDDDDDDDDDDDDDDDLLLLLLLLLLLLLLLLLLLLLLLUUUUUUUUUUUUUUUUUUUUUUU",
"output": "Yes"
},
{
"input": "-100 -100\nRU",
"output": "No"
},
{
"input": "100 100\nRUL",
"output": "No"
},
{
"input": "0 1\nUDLR",
"output": "Yes"
},
{
"input": "0 1\nD",
"output": "No"
},
{
"input": "0 -3\nRDDL",
"output": "No"
}
] | 1,372,490,186
| 1,886
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 13
| 62
| 102,400
|
a, b = map(int, input().split())
s = input().strip()
c2d = dict()
c2d['U'] = (0, 1)
c2d['D'] = (0, -1)
c2d['L'] = (-1, 0)
c2d['R'] = (1, 0)
dx = dy = 0
for c in s:
dx += c2d[c][0]; dy += c2d[c][1]
ans = not a and not b
x = 0; y = 0
for c in s:
x += c2d[c][0]; y += c2d[c][1]
xx = a - x
yy = b - y
if ((xx % dx == 0 and xx // dx >= 0) if dx else xx) and ((yy % dy == 0 and yy // dy >= 0) if dy else yy) and (not dx or not dy or xx // dx == yy // dy):
ans = True
print("Yes" if ans else "No")
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Fox Ciel has a robot on a 2D plane. Initially it is located in (0, 0). Fox Ciel code a command to it. The command was represented by string *s*. Each character of *s* is one move operation. There are four move operations at all:
- 'U': go up, (x, y) <=→<= (x, y+1); - 'D': go down, (x, y) <=→<= (x, y-1); - 'L': go left, (x, y) <=→<= (x-1, y); - 'R': go right, (x, y) <=→<= (x+1, y).
The robot will do the operations in *s* from left to right, and repeat it infinite times. Help Fox Ciel to determine if after some steps the robot will located in (*a*,<=*b*).
Input Specification:
The first line contains two integers *a* and *b*, (<=-<=109<=≤<=*a*,<=*b*<=≤<=109). The second line contains a string *s* (1<=≤<=|*s*|<=≤<=100, *s* only contains characters 'U', 'D', 'L', 'R') — the command.
Output Specification:
Print "Yes" if the robot will be located at (*a*,<=*b*), and "No" otherwise.
Demo Input:
['2 2\nRU\n', '1 2\nRU\n', '-1 1000000000\nLRRLU\n', '0 0\nD\n']
Demo Output:
['Yes\n', 'No\n', 'Yes\n', 'Yes\n']
Note:
In the first and second test case, command string is "RU", so the robot will go right, then go up, then right, and then up and so on.
The locations of its moves are (0, 0) → (1, 0) → (1, 1) → (2, 1) → (2, 2) → ...
So it can reach (2, 2) but not (1, 2).
|
```python
a, b = map(int, input().split())
s = input().strip()
c2d = dict()
c2d['U'] = (0, 1)
c2d['D'] = (0, -1)
c2d['L'] = (-1, 0)
c2d['R'] = (1, 0)
dx = dy = 0
for c in s:
dx += c2d[c][0]; dy += c2d[c][1]
ans = not a and not b
x = 0; y = 0
for c in s:
x += c2d[c][0]; y += c2d[c][1]
xx = a - x
yy = b - y
if ((xx % dx == 0 and xx // dx >= 0) if dx else xx) and ((yy % dy == 0 and yy // dy >= 0) if dy else yy) and (not dx or not dy or xx // dx == yy // dy):
ans = True
print("Yes" if ans else "No")
```
| 0
|
|
301
|
E
|
Yaroslav and Arrangements
|
PROGRAMMING
| 2,800
|
[
"dp"
] | null | null |
Yaroslav calls an array of *r* integers *a*1,<=*a*2,<=...,<=*a**r* good, if it meets the following conditions: |*a*1<=-<=*a*2|<==<=1,<=|*a*2<=-<=*a*3|<==<=1,<=...,<=|*a**r*<=-<=1<=-<=*a**r*|<==<=1,<=|*a**r*<=-<=*a*1|<==<=1, at that .
An array of integers *b*1,<=*b*2,<=...,<=*b**r* is called great, if it meets the following conditions:
1. The elements in it do not decrease (*b**i*<=≤<=*b**i*<=+<=1). 1. If the inequalities 1<=≤<=*r*<=≤<=*n* and 1<=≤<=*b**i*<=≤<=*m* hold. 1. If we can rearrange its elements and get at least one and at most *k* distinct good arrays.
Yaroslav has three integers *n*,<=*m*,<=*k*. He needs to count the number of distinct great arrays. Help Yaroslav! As the answer may be rather large, print the remainder after dividing it by 1000000007 (109<=+<=7).
Two arrays are considered distinct if there is a position in which they have distinct numbers.
|
The single line contains three integers *n*, *m*, *k* (1<=≤<=*n*,<=*m*,<=*k*<=≤<=100).
|
In a single line print the remainder after dividing the answer to the problem by number 1000000007 (109<=+<=7).
|
[
"1 1 1\n",
"3 3 3\n"
] |
[
"0\n",
"2\n"
] |
none
| 2,500
|
[] | 1,520,945,355
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 374
| 19,148,800
|
mod=1000000007
dp=[[[[0 for i in range(110)] for i in range(110) ] for i in range(110)] for i in range(2)]
c=[[0 for i in range(110)] for i in range(110)]
c[0][0]=1
for i in range(1,110):
for j in range(0,110):
if j==0:
c[i][0]=1
else:
c[i][j]=(c[i-1][j]+c[i-1][j-1])%mod
n,m,p=map(int,input().split())
n=n+1
dp[0][0][1][1]=1
ret=0
for i in range(1,m+1):
for j in range(0,n+1):
for k in range(0,n-j+1):
for t in range(1,p+1):
if j>1 and k==0:
ret=(ret+dp[~i&1][j][k][t]*(m-i+2)%mod)%mod
continue
for w in range(k,n-j+1):
if t*c[w-1][k-1]<=p:
dp[i&1][j+w][w-k][t*c[w-1][k-1]]=(dp[i&1][j+w][w-k][t*c[w-1][k-1]]+dp[~i&1][j][k][t])%mod
dp[~i&1][j][k][t]=0
print(ret)
|
Title: Yaroslav and Arrangements
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Yaroslav calls an array of *r* integers *a*1,<=*a*2,<=...,<=*a**r* good, if it meets the following conditions: |*a*1<=-<=*a*2|<==<=1,<=|*a*2<=-<=*a*3|<==<=1,<=...,<=|*a**r*<=-<=1<=-<=*a**r*|<==<=1,<=|*a**r*<=-<=*a*1|<==<=1, at that .
An array of integers *b*1,<=*b*2,<=...,<=*b**r* is called great, if it meets the following conditions:
1. The elements in it do not decrease (*b**i*<=≤<=*b**i*<=+<=1). 1. If the inequalities 1<=≤<=*r*<=≤<=*n* and 1<=≤<=*b**i*<=≤<=*m* hold. 1. If we can rearrange its elements and get at least one and at most *k* distinct good arrays.
Yaroslav has three integers *n*,<=*m*,<=*k*. He needs to count the number of distinct great arrays. Help Yaroslav! As the answer may be rather large, print the remainder after dividing it by 1000000007 (109<=+<=7).
Two arrays are considered distinct if there is a position in which they have distinct numbers.
Input Specification:
The single line contains three integers *n*, *m*, *k* (1<=≤<=*n*,<=*m*,<=*k*<=≤<=100).
Output Specification:
In a single line print the remainder after dividing the answer to the problem by number 1000000007 (109<=+<=7).
Demo Input:
['1 1 1\n', '3 3 3\n']
Demo Output:
['0\n', '2\n']
Note:
none
|
```python
mod=1000000007
dp=[[[[0 for i in range(110)] for i in range(110) ] for i in range(110)] for i in range(2)]
c=[[0 for i in range(110)] for i in range(110)]
c[0][0]=1
for i in range(1,110):
for j in range(0,110):
if j==0:
c[i][0]=1
else:
c[i][j]=(c[i-1][j]+c[i-1][j-1])%mod
n,m,p=map(int,input().split())
n=n+1
dp[0][0][1][1]=1
ret=0
for i in range(1,m+1):
for j in range(0,n+1):
for k in range(0,n-j+1):
for t in range(1,p+1):
if j>1 and k==0:
ret=(ret+dp[~i&1][j][k][t]*(m-i+2)%mod)%mod
continue
for w in range(k,n-j+1):
if t*c[w-1][k-1]<=p:
dp[i&1][j+w][w-k][t*c[w-1][k-1]]=(dp[i&1][j+w][w-k][t*c[w-1][k-1]]+dp[~i&1][j][k][t])%mod
dp[~i&1][j][k][t]=0
print(ret)
```
| 0
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
Student Vladislav came to his programming exam completely unprepared as usual. He got a question about some strange algorithm on a graph — something that will definitely never be useful in real life. He asked a girl sitting next to him to lend him some cheat papers for this questions and found there the following definition:
The minimum spanning tree *T* of graph *G* is such a tree that it contains all the vertices of the original graph *G*, and the sum of the weights of its edges is the minimum possible among all such trees.
Vladislav drew a graph with *n* vertices and *m* edges containing no loops and multiple edges. He found one of its minimum spanning trees and then wrote for each edge its weight and whether it is included in the found tree or not. Unfortunately, the piece of paper where the graph was painted is gone and the teacher is getting very angry and demands to see the original graph. Help Vladislav come up with a graph so that the information about the minimum spanning tree remains correct.
|
The first line of the input contains two integers *n* and *m* () — the number of vertices and the number of edges in the graph.
Each of the next *m* lines describes an edge of the graph and consists of two integers *a**j* and *b**j* (1<=≤<=*a**j*<=≤<=109,<=*b**j*<==<={0,<=1}). The first of these numbers is the weight of the edge and the second number is equal to 1 if this edge was included in the minimum spanning tree found by Vladislav, or 0 if it was not.
It is guaranteed that exactly *n*<=-<=1 number {*b**j*} are equal to one and exactly *m*<=-<=*n*<=+<=1 of them are equal to zero.
|
If Vladislav has made a mistake and such graph doesn't exist, print <=-<=1.
Otherwise print *m* lines. On the *j*-th line print a pair of vertices (*u**j*,<=*v**j*) (1<=≤<=*u**j*,<=*v**j*<=≤<=*n*,<=*u**j*<=≠<=*v**j*), that should be connected by the *j*-th edge. The edges are numbered in the same order as in the input. The graph, determined by these edges, must be connected, contain no loops or multiple edges and its edges with *b**j*<==<=1 must define the minimum spanning tree. In case there are multiple possible solutions, print any of them.
|
[
"4 5\n2 1\n3 1\n4 0\n1 1\n5 0\n",
"3 3\n1 0\n2 1\n3 1\n"
] |
[
"2 4\n1 4\n3 4\n3 1\n3 2\n",
"-1\n"
] |
none
| 0
|
[
{
"input": "4 5\n2 1\n3 1\n4 0\n1 1\n5 0",
"output": "2 4\n1 4\n3 4\n3 1\n3 2"
},
{
"input": "3 3\n1 0\n2 1\n3 1",
"output": "-1"
},
{
"input": "2 1\n7 1",
"output": "1 2"
},
{
"input": "3 2\n8 1\n9 1",
"output": "1 2\n1 3"
},
{
"input": "3 3\n4 1\n5 0\n7 1",
"output": "-1"
},
{
"input": "3 3\n4 1\n5 1\n7 0",
"output": "1 2\n1 3\n2 3"
},
{
"input": "3 3\n4 1\n4 0\n4 1",
"output": "1 2\n2 3\n1 3"
},
{
"input": "3 3\n4 0\n5 1\n4 1",
"output": "-1"
},
{
"input": "3 3\n5 0\n4 1\n5 1",
"output": "2 3\n1 2\n1 3"
},
{
"input": "4 4\n2 1\n3 0\n3 1\n4 1",
"output": "1 2\n2 3\n1 3\n1 4"
},
{
"input": "4 5\n4 1\n4 1\n4 0\n4 0\n6 1",
"output": "-1"
},
{
"input": "4 6\n2 1\n4 0\n3 0\n1 1\n4 1\n5 0",
"output": "1 3\n2 4\n2 3\n1 2\n1 4\n3 4"
},
{
"input": "4 4\n2 1\n6 0\n7 1\n7 1",
"output": "-1"
},
{
"input": "4 4\n2 1\n8 0\n8 1\n8 1",
"output": "1 2\n2 3\n1 3\n1 4"
},
{
"input": "4 4\n2 0\n2 1\n8 1\n2 1",
"output": "2 3\n1 2\n1 4\n1 3"
},
{
"input": "4 4\n2 1\n3 1\n1 1\n4 0",
"output": "1 3\n1 4\n1 2\n2 3"
},
{
"input": "4 5\n3 1\n4 1\n4 0\n6 0\n6 1",
"output": "1 2\n1 3\n2 3\n2 4\n1 4"
},
{
"input": "4 5\n7 0\n3 0\n1 1\n5 1\n7 1",
"output": "-1"
},
{
"input": "4 6\n2 1\n7 1\n3 0\n1 1\n7 0\n6 0",
"output": "-1"
},
{
"input": "4 6\n1 1\n3 1\n2 0\n2 1\n3 0\n3 0",
"output": "1 2\n1 4\n2 3\n1 3\n2 4\n3 4"
},
{
"input": "4 6\n1 1\n4 1\n2 0\n2 1\n4 0\n3 0",
"output": "-1"
},
{
"input": "10 15\n900000012 1\n900000010 1\n900000007 0\n900000005 0\n900000014 1\n900000000 1\n900000004 0\n900000006 1\n900000009 0\n900000002 0\n900000008 0\n900000001 1\n900000011 1\n900000003 1\n900000013 1",
"output": "1 8\n1 6\n2 5\n3 4\n1 10\n1 2\n2 4\n1 5\n4 5\n2 3\n3 5\n1 3\n1 7\n1 4\n1 9"
},
{
"input": "10 15\n900000007 1\n900000002 1\n900000004 0\n900000002 1\n900000006 1\n900000000 1\n900000006 1\n900000008 1\n900000002 0\n900000003 0\n900000002 0\n900000005 0\n900000001 0\n900000000 1\n900000008 1",
"output": "1 8\n1 4\n3 5\n1 5\n1 6\n1 2\n1 7\n1 9\n2 4\n2 5\n3 4\n4 5\n2 3\n1 3\n1 10"
},
{
"input": "10 15\n900000004 0\n900000006 1\n900000001 1\n900000004 1\n900000007 1\n900000007 1\n900000004 1\n900000008 1\n900000004 0\n900000004 0\n900000007 1\n900000005 0\n900000004 0\n900000002 0\n900000000 1",
"output": "2 4\n1 6\n1 3\n1 4\n1 7\n1 8\n1 5\n1 10\n3 4\n2 5\n1 9\n4 5\n3 5\n2 3\n1 2"
},
{
"input": "10 15\n900000006 1\n900000000 1\n900000004 0\n900000000 1\n900000004 0\n900000006 1\n900000000 1\n900000006 1\n900000005 1\n900000001 0\n900000003 1\n900000006 1\n900000000 0\n900000003 0\n900000000 0",
"output": "1 7\n1 2\n3 5\n1 3\n4 5\n1 8\n1 4\n1 9\n1 6\n3 4\n1 5\n1 10\n2 3\n2 5\n2 4"
},
{
"input": "10 15\n900000000 1\n900000003 1\n900000000 1\n900000000 0\n900000003 0\n900000005 1\n900000005 1\n900000005 1\n900000001 0\n900000002 0\n900000002 0\n900000004 1\n900000002 0\n900000000 1\n900000004 1",
"output": "-1"
},
{
"input": "10 15\n900000001 1\n900000001 1\n900000002 1\n900000001 1\n900000001 0\n900000001 1\n900000001 0\n900000001 0\n900000001 0\n900000001 1\n900000001 0\n900000001 0\n900000004 1\n900000000 1\n900000001 1",
"output": "1 3\n1 4\n1 9\n1 5\n2 3\n1 6\n2 4\n3 4\n2 5\n1 7\n3 5\n4 5\n1 10\n1 2\n1 8"
},
{
"input": "10 15\n900000001 1\n900000001 1\n900000001 0\n900000000 1\n900000001 0\n900000002 1\n900000000 1\n900000002 1\n900000001 0\n900000001 0\n900000001 0\n900000002 1\n900000000 0\n900000002 1\n900000000 1",
"output": "1 5\n1 6\n2 4\n1 2\n3 4\n1 7\n1 3\n1 8\n2 5\n3 5\n4 5\n1 9\n2 3\n1 10\n1 4"
},
{
"input": "5 5\n1 1\n2 1\n3 0\n4 1\n5 1",
"output": "1 2\n1 3\n2 3\n1 4\n1 5"
},
{
"input": "5 6\n1 1\n2 1\n3 0\n4 1\n5 0\n6 1",
"output": "1 2\n1 3\n2 3\n1 4\n2 4\n1 5"
},
{
"input": "5 6\n1 1\n2 1\n3 0\n4 0\n5 1\n6 1",
"output": "-1"
},
{
"input": "5 7\n1 1\n1 1\n1 0\n2 0\n1 0\n2 1\n2 1",
"output": "-1"
},
{
"input": "5 8\n1 0\n1 1\n1 1\n2 0\n1 0\n2 1\n1 0\n1 1",
"output": "2 3\n1 2\n1 3\n2 5\n2 4\n1 5\n3 4\n1 4"
},
{
"input": "5 8\n1 0\n1 1\n1 1\n3 0\n1 0\n3 1\n2 0\n1 1",
"output": "2 3\n1 2\n1 3\n2 5\n2 4\n1 5\n3 4\n1 4"
},
{
"input": "5 8\n1 0\n1 1\n1 1\n3 0\n1 0\n4 1\n2 0\n1 1",
"output": "-1"
},
{
"input": "5 9\n1 1\n2 1\n3 0\n4 1\n5 0\n6 0\n7 1\n8 0\n9 0",
"output": "1 2\n1 3\n2 3\n1 4\n2 4\n3 4\n1 5\n2 5\n3 5"
},
{
"input": "5 9\n1 1\n2 1\n3 0\n4 1\n5 0\n6 0\n7 0\n8 1\n9 0",
"output": "-1"
},
{
"input": "5 10\n1 1\n1 1\n1 0\n1 1\n2 0\n2 0\n2 1\n2 0\n2 0\n2 0",
"output": "1 2\n1 3\n2 3\n1 4\n2 4\n3 4\n1 5\n2 5\n3 5\n4 5"
},
{
"input": "5 10\n1 1\n1 1\n1 0\n1 1\n2 0\n2 0\n3 1\n2 0\n3 0\n3 0",
"output": "-1"
},
{
"input": "10 15\n761759620 0\n761759620 1\n787655728 1\n761759620 0\n294001884 1\n465325912 1\n787655728 0\n683571303 1\n683571303 0\n761759620 0\n787655728 0\n391499930 1\n758807870 1\n611782565 1\n132266542 1",
"output": "2 4\n1 9\n1 10\n3 4\n1 3\n1 5\n3 5\n1 7\n2 3\n2 5\n4 5\n1 4\n1 8\n1 6\n1 2"
},
{
"input": "10 15\n752087443 1\n537185872 1\n439895449 1\n494086747 1\n718088132 1\n93444012 0\n670136349 1\n545547453 0\n718088132 0\n853059674 0\n853059674 1\n762928724 1\n762928724 0\n853059674 0\n156495293 1",
"output": "-1"
},
{
"input": "10 15\n417559883 0\n300974070 1\n292808458 1\n469395226 0\n469395226 1\n564721882 1\n125636288 1\n417559883 0\n417559883 1\n469395226 0\n376390930 1\n233782394 1\n780369860 1\n564721882 0\n417559883 0",
"output": "2 3\n1 5\n1 4\n2 5\n1 8\n1 9\n1 2\n2 4\n1 7\n3 5\n1 6\n1 3\n1 10\n4 5\n3 4"
}
] | 1,482,049,451
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 76
| 935
| 23,449,600
|
def graph(e):
e = [(w, i, b) for i, (w, b) in enumerate(e)]
e.sort()
g = [None]*len(e)
mst = [(w, i) for w, i, b in e if b]
for n, (w, i) in enumerate(mst, 2):
g[i] = 1, n
cm = 0
available = []
for w, i, b in e:
if not b:
if not available:
cm += 1
if mst[cm][0] > w:
return None
available = [(u, cm+2) for u in range(2, cm+2)]
g[i] = available.pop()
return g
if __name__ == '__main__':
n, m = map(int, input().split())
e = []
for _ in range(m):
a, b = map(int, input().split())
e.append((a, b))
g = graph(e)
if g:
for u, v in g:
print("{} {}".format(u, v))
else:
print(-1)
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Student Vladislav came to his programming exam completely unprepared as usual. He got a question about some strange algorithm on a graph — something that will definitely never be useful in real life. He asked a girl sitting next to him to lend him some cheat papers for this questions and found there the following definition:
The minimum spanning tree *T* of graph *G* is such a tree that it contains all the vertices of the original graph *G*, and the sum of the weights of its edges is the minimum possible among all such trees.
Vladislav drew a graph with *n* vertices and *m* edges containing no loops and multiple edges. He found one of its minimum spanning trees and then wrote for each edge its weight and whether it is included in the found tree or not. Unfortunately, the piece of paper where the graph was painted is gone and the teacher is getting very angry and demands to see the original graph. Help Vladislav come up with a graph so that the information about the minimum spanning tree remains correct.
Input Specification:
The first line of the input contains two integers *n* and *m* () — the number of vertices and the number of edges in the graph.
Each of the next *m* lines describes an edge of the graph and consists of two integers *a**j* and *b**j* (1<=≤<=*a**j*<=≤<=109,<=*b**j*<==<={0,<=1}). The first of these numbers is the weight of the edge and the second number is equal to 1 if this edge was included in the minimum spanning tree found by Vladislav, or 0 if it was not.
It is guaranteed that exactly *n*<=-<=1 number {*b**j*} are equal to one and exactly *m*<=-<=*n*<=+<=1 of them are equal to zero.
Output Specification:
If Vladislav has made a mistake and such graph doesn't exist, print <=-<=1.
Otherwise print *m* lines. On the *j*-th line print a pair of vertices (*u**j*,<=*v**j*) (1<=≤<=*u**j*,<=*v**j*<=≤<=*n*,<=*u**j*<=≠<=*v**j*), that should be connected by the *j*-th edge. The edges are numbered in the same order as in the input. The graph, determined by these edges, must be connected, contain no loops or multiple edges and its edges with *b**j*<==<=1 must define the minimum spanning tree. In case there are multiple possible solutions, print any of them.
Demo Input:
['4 5\n2 1\n3 1\n4 0\n1 1\n5 0\n', '3 3\n1 0\n2 1\n3 1\n']
Demo Output:
['2 4\n1 4\n3 4\n3 1\n3 2\n', '-1\n']
Note:
none
|
```python
def graph(e):
e = [(w, i, b) for i, (w, b) in enumerate(e)]
e.sort()
g = [None]*len(e)
mst = [(w, i) for w, i, b in e if b]
for n, (w, i) in enumerate(mst, 2):
g[i] = 1, n
cm = 0
available = []
for w, i, b in e:
if not b:
if not available:
cm += 1
if mst[cm][0] > w:
return None
available = [(u, cm+2) for u in range(2, cm+2)]
g[i] = available.pop()
return g
if __name__ == '__main__':
n, m = map(int, input().split())
e = []
for _ in range(m):
a, b = map(int, input().split())
e.append((a, b))
g = graph(e)
if g:
for u, v in g:
print("{} {}".format(u, v))
else:
print(-1)
```
| 3
|
|
408
|
B
|
Garland
|
PROGRAMMING
| 1,200
|
[
"implementation"
] | null | null |
Once little Vasya read an article in a magazine on how to make beautiful handmade garland from colored paper. Vasya immediately went to the store and bought *n* colored sheets of paper, the area of each sheet is 1 square meter.
The garland must consist of exactly *m* pieces of colored paper of arbitrary area, each piece should be of a certain color. To make the garland, Vasya can arbitrarily cut his existing colored sheets into pieces. Vasya is not obliged to use all the sheets to make the garland.
Vasya wants the garland to be as attractive as possible, so he wants to maximize the total area of *m* pieces of paper in the garland. Calculate what the maximum total area of the pieces of paper in the garland Vasya can get.
|
The first line contains a non-empty sequence of *n* (1<=≤<=*n*<=≤<=1000) small English letters ("a"..."z"). Each letter means that Vasya has a sheet of paper of the corresponding color.
The second line contains a non-empty sequence of *m* (1<=≤<=*m*<=≤<=1000) small English letters that correspond to the colors of the pieces of paper in the garland that Vasya wants to make.
|
Print an integer that is the maximum possible total area of the pieces of paper in the garland Vasya wants to get or -1, if it is impossible to make the garland from the sheets he's got. It is guaranteed that the answer is always an integer.
|
[
"aaabbac\naabbccac\n",
"a\nz\n"
] |
[
"6\n",
"-1"
] |
In the first test sample Vasya can make an garland of area 6: he can use both sheets of color *b*, three (but not four) sheets of color *a* and cut a single sheet of color *c* in three, for example, equal pieces. Vasya can use the resulting pieces to make a garland of area 6.
In the second test sample Vasya cannot make a garland at all — he doesn't have a sheet of color *z*.
| 1,000
|
[
{
"input": "aaabbac\naabbccac",
"output": "6"
},
{
"input": "a\nz",
"output": "-1"
},
{
"input": "r\nr",
"output": "1"
},
{
"input": "stnsdn\nndnndsn",
"output": "4"
},
{
"input": "yqfqfp\ntttwtqq",
"output": "-1"
},
{
"input": "zzbbrrtrtzr\ntbbtrrrzr",
"output": "9"
},
{
"input": "ivvfisvsvii\npaihjinno",
"output": "-1"
},
{
"input": "zbvwnlgkshqerxptyod\nz",
"output": "1"
},
{
"input": "xlktwjymocqrahnbesf\nfoo",
"output": "2"
},
{
"input": "bbzmzqazmbambnmzaabznmbabzqnaabmabmnnabbmnzaanzzezebzabqaabzqaemeqqammmbazmmz\naznnbbmeebmanbeemzmemqbaeebnqenqzzbanebmnzqqebqmmnmqqzmmeqqqaaezemmazqqmqaqnnqqzbzeeazammmenbbamzbmnaenemenaaaebnmanebqmqnznqbenmqqnnnaeaebqmamennmqqeaaqqbammnzqmnmqnqbbezmemznqmanzmmqzzzzembqnzqbanamezqaqbazenenqqznqaebzaeezbqqbmeeaqnmmbnqbbnmaqqemaeaezaabmbnbzzaae",
"output": "77"
},
{
"input": "lccfdfnfflncddlksndcwnfcllnnaswcdnldafcalssfcdnkkaklwnnacsncfwanwnwfadawcsdcfwflnnlfsfclkfknlaldna\nuaaldlllhedgugugueahddhedbuddaedhaaeugdubaealbgbagedldbl",
"output": "-1"
},
{
"input": "hvewdvwdwudrwarwmddwnazmwauzatrmwptwwevavpmwrtruwnpwantewrnwmepdwvtmnveanunrvrtwpvhhnuhnmpptdttzmmndtvudmzhhannmmnahdpzamuvhzaavnhtnumwrwvttdetvuewaaennddwuvzvaptdzrzhtetwwzmzedrwuvrwznprhdvnavrruvvhzuwpdtmpwmzrwvermrhdamv\nuvzhwtpuputnahwwarduzddhpnwwvettprwavdmnztdnrddmarmvuevtdezndnezvarhmppwwnmvnrtddzhhnzrwuhvpwmezuurundarwdazwptrpeurrnwautddnhdmhtwhwvvtavdzezumetzezpazndhuentmrhamutrtttpevtuutemdnvwnwnmnvmznatneweuaahdavmaddhnrdenwwztrwh",
"output": "199"
},
{
"input": "aaccddff\nabcdf",
"output": "-1"
}
] | 1,597,908,371
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 93
| 307,200
|
z=[0]*27
z1=[0]*27
for i in input():z[ord(i)-97]+=1
for i in input():z1[ord(i)-97]+=1
s=0
for i in range(27):
if z[i]>=z1[i]:s+=z1[i]
elif z1[i]!=0 and z[i]!=0:s+=z[i]
if s:print(s)
else:print(-1)
|
Title: Garland
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Once little Vasya read an article in a magazine on how to make beautiful handmade garland from colored paper. Vasya immediately went to the store and bought *n* colored sheets of paper, the area of each sheet is 1 square meter.
The garland must consist of exactly *m* pieces of colored paper of arbitrary area, each piece should be of a certain color. To make the garland, Vasya can arbitrarily cut his existing colored sheets into pieces. Vasya is not obliged to use all the sheets to make the garland.
Vasya wants the garland to be as attractive as possible, so he wants to maximize the total area of *m* pieces of paper in the garland. Calculate what the maximum total area of the pieces of paper in the garland Vasya can get.
Input Specification:
The first line contains a non-empty sequence of *n* (1<=≤<=*n*<=≤<=1000) small English letters ("a"..."z"). Each letter means that Vasya has a sheet of paper of the corresponding color.
The second line contains a non-empty sequence of *m* (1<=≤<=*m*<=≤<=1000) small English letters that correspond to the colors of the pieces of paper in the garland that Vasya wants to make.
Output Specification:
Print an integer that is the maximum possible total area of the pieces of paper in the garland Vasya wants to get or -1, if it is impossible to make the garland from the sheets he's got. It is guaranteed that the answer is always an integer.
Demo Input:
['aaabbac\naabbccac\n', 'a\nz\n']
Demo Output:
['6\n', '-1']
Note:
In the first test sample Vasya can make an garland of area 6: he can use both sheets of color *b*, three (but not four) sheets of color *a* and cut a single sheet of color *c* in three, for example, equal pieces. Vasya can use the resulting pieces to make a garland of area 6.
In the second test sample Vasya cannot make a garland at all — he doesn't have a sheet of color *z*.
|
```python
z=[0]*27
z1=[0]*27
for i in input():z[ord(i)-97]+=1
for i in input():z1[ord(i)-97]+=1
s=0
for i in range(27):
if z[i]>=z1[i]:s+=z1[i]
elif z1[i]!=0 and z[i]!=0:s+=z[i]
if s:print(s)
else:print(-1)
```
| 0
|
|
583
|
A
|
Asphalting Roads
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
City X consists of *n* vertical and *n* horizontal infinite roads, forming *n*<=×<=*n* intersections. Roads (both vertical and horizontal) are numbered from 1 to *n*, and the intersections are indicated by the numbers of the roads that form them.
Sand roads have long been recognized out of date, so the decision was made to asphalt them. To do this, a team of workers was hired and a schedule of work was made, according to which the intersections should be asphalted.
Road repairs are planned for *n*2 days. On the *i*-th day of the team arrives at the *i*-th intersection in the list and if none of the two roads that form the intersection were already asphalted they asphalt both roads. Otherwise, the team leaves the intersection, without doing anything with the roads.
According to the schedule of road works tell in which days at least one road will be asphalted.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of vertical and horizontal roads in the city.
Next *n*2 lines contain the order of intersections in the schedule. The *i*-th of them contains two numbers *h**i*,<=*v**i* (1<=≤<=*h**i*,<=*v**i*<=≤<=*n*), separated by a space, and meaning that the intersection that goes *i*-th in the timetable is at the intersection of the *h**i*-th horizontal and *v**i*-th vertical roads. It is guaranteed that all the intersections in the timetable are distinct.
|
In the single line print the numbers of the days when road works will be in progress in ascending order. The days are numbered starting from 1.
|
[
"2\n1 1\n1 2\n2 1\n2 2\n",
"1\n1 1\n"
] |
[
"1 4 \n",
"1 \n"
] |
In the sample the brigade acts like that:
1. On the first day the brigade comes to the intersection of the 1-st horizontal and the 1-st vertical road. As none of them has been asphalted, the workers asphalt the 1-st vertical and the 1-st horizontal road; 1. On the second day the brigade of the workers comes to the intersection of the 1-st horizontal and the 2-nd vertical road. The 2-nd vertical road hasn't been asphalted, but as the 1-st horizontal road has been asphalted on the first day, the workers leave and do not asphalt anything; 1. On the third day the brigade of the workers come to the intersection of the 2-nd horizontal and the 1-st vertical road. The 2-nd horizontal road hasn't been asphalted but as the 1-st vertical road has been asphalted on the first day, the workers leave and do not asphalt anything; 1. On the fourth day the brigade come to the intersection formed by the intersection of the 2-nd horizontal and 2-nd vertical road. As none of them has been asphalted, the workers asphalt the 2-nd vertical and the 2-nd horizontal road.
| 500
|
[
{
"input": "2\n1 1\n1 2\n2 1\n2 2",
"output": "1 4 "
},
{
"input": "1\n1 1",
"output": "1 "
},
{
"input": "2\n1 1\n2 2\n1 2\n2 1",
"output": "1 2 "
},
{
"input": "2\n1 2\n2 2\n2 1\n1 1",
"output": "1 3 "
},
{
"input": "3\n2 2\n1 2\n3 2\n3 3\n1 1\n2 3\n1 3\n3 1\n2 1",
"output": "1 4 5 "
},
{
"input": "3\n1 3\n3 1\n2 1\n1 1\n1 2\n2 2\n3 2\n3 3\n2 3",
"output": "1 2 6 "
},
{
"input": "4\n1 3\n2 3\n2 4\n4 4\n3 1\n1 1\n3 4\n2 1\n1 4\n4 3\n4 1\n3 2\n1 2\n4 2\n2 2\n3 3",
"output": "1 3 5 14 "
},
{
"input": "4\n3 3\n4 2\n2 3\n3 4\n4 4\n1 2\n3 2\n2 2\n1 4\n3 1\n4 1\n2 1\n1 3\n1 1\n4 3\n2 4",
"output": "1 2 9 12 "
},
{
"input": "9\n4 5\n2 3\n8 3\n5 6\n9 3\n4 4\n5 4\n4 7\n1 7\n8 4\n1 4\n1 5\n5 7\n7 8\n7 1\n9 9\n8 7\n7 5\n3 7\n6 6\n7 3\n5 2\n3 6\n7 4\n9 6\n5 8\n9 7\n6 3\n7 9\n1 2\n1 1\n6 2\n5 3\n7 2\n1 6\n4 1\n6 1\n8 9\n2 2\n3 9\n2 9\n7 7\n2 8\n9 4\n2 5\n8 6\n3 4\n2 1\n2 7\n6 5\n9 1\n3 3\n3 8\n5 5\n4 3\n3 1\n1 9\n6 4\n3 2\n6 8\n2 6\n5 9\n8 5\n8 8\n9 5\n6 9\n9 2\n3 5\n4 9\n4 8\n2 4\n5 1\n4 6\n7 6\n9 8\n1 3\n4 2\n8 1\n8 2\n6 7\n1 8",
"output": "1 2 4 9 10 14 16 32 56 "
},
{
"input": "8\n1 1\n1 2\n1 3\n1 4\n1 5\n8 6\n1 7\n1 8\n2 1\n8 5\n2 3\n2 4\n2 5\n2 6\n4 3\n2 2\n3 1\n3 2\n3 3\n3 4\n3 5\n3 6\n5 6\n3 8\n4 1\n4 2\n2 7\n4 4\n8 8\n4 6\n4 7\n4 8\n5 1\n5 2\n5 3\n6 5\n5 5\n3 7\n5 7\n5 8\n6 1\n6 2\n6 3\n6 4\n5 4\n6 6\n6 7\n6 8\n7 1\n7 2\n7 3\n7 4\n7 5\n7 6\n7 7\n7 8\n8 1\n8 2\n8 3\n8 4\n2 8\n1 6\n8 7\n4 5",
"output": "1 6 11 18 28 36 39 56 "
},
{
"input": "9\n9 9\n5 5\n8 8\n3 3\n2 2\n6 6\n4 4\n1 1\n7 7\n8 4\n1 4\n1 5\n5 7\n7 8\n7 1\n1 7\n8 7\n7 5\n3 7\n5 6\n7 3\n5 2\n3 6\n7 4\n9 6\n5 8\n9 7\n6 3\n7 9\n1 2\n4 5\n6 2\n5 3\n7 2\n1 6\n4 1\n6 1\n8 9\n2 3\n3 9\n2 9\n5 4\n2 8\n9 4\n2 5\n8 6\n3 4\n2 1\n2 7\n6 5\n9 1\n8 3\n3 8\n9 3\n4 3\n3 1\n1 9\n6 4\n3 2\n6 8\n2 6\n5 9\n8 5\n4 7\n9 5\n6 9\n9 2\n3 5\n4 9\n4 8\n2 4\n5 1\n4 6\n7 6\n9 8\n1 3\n4 2\n8 1\n8 2\n6 7\n1 8",
"output": "1 2 3 4 5 6 7 8 9 "
}
] | 1,443,891,768
| 1,068
|
Python 3
|
WRONG_ANSWER
|
PRETESTS
| 3
| 46
| 0
|
n = int(input())
A = [list(map(int, input().split())) for i in range(n*n)]
day = []
for i in range(n*n):
for j in range(1):
if A[i][j] == A[i][j+1] :
day.append(i)
for dni in day: print(dni+1, end = ' ')
|
Title: Asphalting Roads
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
City X consists of *n* vertical and *n* horizontal infinite roads, forming *n*<=×<=*n* intersections. Roads (both vertical and horizontal) are numbered from 1 to *n*, and the intersections are indicated by the numbers of the roads that form them.
Sand roads have long been recognized out of date, so the decision was made to asphalt them. To do this, a team of workers was hired and a schedule of work was made, according to which the intersections should be asphalted.
Road repairs are planned for *n*2 days. On the *i*-th day of the team arrives at the *i*-th intersection in the list and if none of the two roads that form the intersection were already asphalted they asphalt both roads. Otherwise, the team leaves the intersection, without doing anything with the roads.
According to the schedule of road works tell in which days at least one road will be asphalted.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of vertical and horizontal roads in the city.
Next *n*2 lines contain the order of intersections in the schedule. The *i*-th of them contains two numbers *h**i*,<=*v**i* (1<=≤<=*h**i*,<=*v**i*<=≤<=*n*), separated by a space, and meaning that the intersection that goes *i*-th in the timetable is at the intersection of the *h**i*-th horizontal and *v**i*-th vertical roads. It is guaranteed that all the intersections in the timetable are distinct.
Output Specification:
In the single line print the numbers of the days when road works will be in progress in ascending order. The days are numbered starting from 1.
Demo Input:
['2\n1 1\n1 2\n2 1\n2 2\n', '1\n1 1\n']
Demo Output:
['1 4 \n', '1 \n']
Note:
In the sample the brigade acts like that:
1. On the first day the brigade comes to the intersection of the 1-st horizontal and the 1-st vertical road. As none of them has been asphalted, the workers asphalt the 1-st vertical and the 1-st horizontal road; 1. On the second day the brigade of the workers comes to the intersection of the 1-st horizontal and the 2-nd vertical road. The 2-nd vertical road hasn't been asphalted, but as the 1-st horizontal road has been asphalted on the first day, the workers leave and do not asphalt anything; 1. On the third day the brigade of the workers come to the intersection of the 2-nd horizontal and the 1-st vertical road. The 2-nd horizontal road hasn't been asphalted but as the 1-st vertical road has been asphalted on the first day, the workers leave and do not asphalt anything; 1. On the fourth day the brigade come to the intersection formed by the intersection of the 2-nd horizontal and 2-nd vertical road. As none of them has been asphalted, the workers asphalt the 2-nd vertical and the 2-nd horizontal road.
|
```python
n = int(input())
A = [list(map(int, input().split())) for i in range(n*n)]
day = []
for i in range(n*n):
for j in range(1):
if A[i][j] == A[i][j+1] :
day.append(i)
for dni in day: print(dni+1, end = ' ')
```
| 0
|
|
92
|
A
|
Chips
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] |
A. Chips
|
2
|
256
|
There are *n* walruses sitting in a circle. All of them are numbered in the clockwise order: the walrus number 2 sits to the left of the walrus number 1, the walrus number 3 sits to the left of the walrus number 2, ..., the walrus number 1 sits to the left of the walrus number *n*.
The presenter has *m* chips. The presenter stands in the middle of the circle and starts giving the chips to the walruses starting from walrus number 1 and moving clockwise. The walrus number *i* gets *i* chips. If the presenter can't give the current walrus the required number of chips, then the presenter takes the remaining chips and the process ends. Determine by the given *n* and *m* how many chips the presenter will get in the end.
|
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=50, 1<=≤<=*m*<=≤<=104) — the number of walruses and the number of chips correspondingly.
|
Print the number of chips the presenter ended up with.
|
[
"4 11\n",
"17 107\n",
"3 8\n"
] |
[
"0\n",
"2\n",
"1\n"
] |
In the first sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, four chips to the walrus number 4, then again one chip to the walrus number 1. After that the presenter runs out of chips. He can't give anything to the walrus number 2 and the process finishes.
In the third sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, then again one chip to the walrus number 1. The presenter has one chip left and he can't give two chips to the walrus number 2, that's why the presenter takes the last chip.
| 500
|
[
{
"input": "4 11",
"output": "0"
},
{
"input": "17 107",
"output": "2"
},
{
"input": "3 8",
"output": "1"
},
{
"input": "46 7262",
"output": "35"
},
{
"input": "32 6864",
"output": "0"
},
{
"input": "36 6218",
"output": "14"
},
{
"input": "25 9712",
"output": "11"
},
{
"input": "9 7601",
"output": "5"
},
{
"input": "1 9058",
"output": "0"
},
{
"input": "29 7772",
"output": "26"
},
{
"input": "45 9465",
"output": "14"
},
{
"input": "46 866",
"output": "5"
},
{
"input": "29 1241",
"output": "20"
},
{
"input": "17 4248",
"output": "12"
},
{
"input": "20 8082",
"output": "11"
},
{
"input": "50 9555",
"output": "0"
},
{
"input": "4 7455",
"output": "2"
},
{
"input": "36 880",
"output": "4"
},
{
"input": "24 7440",
"output": "9"
},
{
"input": "44 7888",
"output": "12"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "50 10000",
"output": "40"
},
{
"input": "1 10000",
"output": "0"
},
{
"input": "50 1",
"output": "0"
},
{
"input": "50 50",
"output": "5"
}
] | 1,564,667,156
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 25
| 218
| 0
|
n,m=map(int,input().split())
i=0
while 2>0:
if m>=i:
m-=i
if i==n:
i=0
i+=1
if m<i:
break
print(m)
|
Title: Chips
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
There are *n* walruses sitting in a circle. All of them are numbered in the clockwise order: the walrus number 2 sits to the left of the walrus number 1, the walrus number 3 sits to the left of the walrus number 2, ..., the walrus number 1 sits to the left of the walrus number *n*.
The presenter has *m* chips. The presenter stands in the middle of the circle and starts giving the chips to the walruses starting from walrus number 1 and moving clockwise. The walrus number *i* gets *i* chips. If the presenter can't give the current walrus the required number of chips, then the presenter takes the remaining chips and the process ends. Determine by the given *n* and *m* how many chips the presenter will get in the end.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=50, 1<=≤<=*m*<=≤<=104) — the number of walruses and the number of chips correspondingly.
Output Specification:
Print the number of chips the presenter ended up with.
Demo Input:
['4 11\n', '17 107\n', '3 8\n']
Demo Output:
['0\n', '2\n', '1\n']
Note:
In the first sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, four chips to the walrus number 4, then again one chip to the walrus number 1. After that the presenter runs out of chips. He can't give anything to the walrus number 2 and the process finishes.
In the third sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, then again one chip to the walrus number 1. The presenter has one chip left and he can't give two chips to the walrus number 2, that's why the presenter takes the last chip.
|
```python
n,m=map(int,input().split())
i=0
while 2>0:
if m>=i:
m-=i
if i==n:
i=0
i+=1
if m<i:
break
print(m)
```
| 3.9455
|
80
|
A
|
Panoramix's Prediction
|
PROGRAMMING
| 800
|
[
"brute force"
] |
A. Panoramix's Prediction
|
2
|
256
|
A prime number is a number which has exactly two distinct divisors: one and itself. For example, numbers 2, 7, 3 are prime, and 1, 6, 4 are not.
The next prime number after *x* is the smallest prime number greater than *x*. For example, the next prime number after 2 is 3, and the next prime number after 3 is 5. Note that there is exactly one next prime number after each number. So 5 is not the next prime number for 2.
One cold April morning Panoramix predicted that soon Kakofonix will break free from his straitjacket, and this will be a black day for the residents of the Gallic countryside.
Panoramix's prophecy tells that if some day Asterix and Obelix beat exactly *x* Roman soldiers, where *x* is a prime number, and next day they beat exactly *y* Roman soldiers, where *y* is the next prime number after *x*, then it's time to wait for Armageddon, for nothing can shut Kakofonix up while he sings his infernal song.
Yesterday the Gauls beat *n* Roman soldiers and it turned out that the number *n* was prime! Today their victims were a troop of *m* Romans (*m*<=><=*n*). Determine whether the Gauls should wait for the black day after today's victory of Asterix and Obelix?
|
The first and only input line contains two positive integers — *n* and *m* (2<=≤<=*n*<=<<=*m*<=≤<=50). It is guaranteed that *n* is prime.
Pretests contain all the cases with restrictions 2<=≤<=*n*<=<<=*m*<=≤<=4.
|
Print YES, if *m* is the next prime number after *n*, or NO otherwise.
|
[
"3 5\n",
"7 11\n",
"7 9\n"
] |
[
"YES",
"YES",
"NO"
] |
none
| 500
|
[
{
"input": "3 5",
"output": "YES"
},
{
"input": "7 11",
"output": "YES"
},
{
"input": "7 9",
"output": "NO"
},
{
"input": "2 3",
"output": "YES"
},
{
"input": "2 4",
"output": "NO"
},
{
"input": "3 4",
"output": "NO"
},
{
"input": "3 5",
"output": "YES"
},
{
"input": "5 7",
"output": "YES"
},
{
"input": "7 11",
"output": "YES"
},
{
"input": "11 13",
"output": "YES"
},
{
"input": "13 17",
"output": "YES"
},
{
"input": "17 19",
"output": "YES"
},
{
"input": "19 23",
"output": "YES"
},
{
"input": "23 29",
"output": "YES"
},
{
"input": "29 31",
"output": "YES"
},
{
"input": "31 37",
"output": "YES"
},
{
"input": "37 41",
"output": "YES"
},
{
"input": "41 43",
"output": "YES"
},
{
"input": "43 47",
"output": "YES"
},
{
"input": "2 5",
"output": "NO"
},
{
"input": "2 7",
"output": "NO"
},
{
"input": "2 6",
"output": "NO"
},
{
"input": "2 11",
"output": "NO"
},
{
"input": "3 6",
"output": "NO"
},
{
"input": "3 7",
"output": "NO"
},
{
"input": "3 9",
"output": "NO"
},
{
"input": "5 6",
"output": "NO"
},
{
"input": "5 9",
"output": "NO"
},
{
"input": "5 11",
"output": "NO"
},
{
"input": "5 13",
"output": "NO"
},
{
"input": "5 15",
"output": "NO"
},
{
"input": "7 8",
"output": "NO"
},
{
"input": "7 13",
"output": "NO"
},
{
"input": "13 15",
"output": "NO"
},
{
"input": "13 17",
"output": "YES"
},
{
"input": "19 21",
"output": "NO"
},
{
"input": "13 20",
"output": "NO"
},
{
"input": "41 49",
"output": "NO"
},
{
"input": "43 49",
"output": "NO"
},
{
"input": "47 50",
"output": "NO"
},
{
"input": "47 49",
"output": "NO"
},
{
"input": "47 48",
"output": "NO"
},
{
"input": "23 25",
"output": "NO"
},
{
"input": "2 50",
"output": "NO"
},
{
"input": "31 33",
"output": "NO"
}
] | 1,697,613,430
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 45
| 124
| 0
|
def easy(a):
for j in range(2, int(a ** 0.5) + 1):
if a % j == 0:
return 0
return 1
n, m = map(int, input().split())
cnt = 0
for i in range(n + 1, m):
cnt += easy(i)
if cnt == 0 and easy(m):
print("YES")
else:
print("NO")
|
Title: Panoramix's Prediction
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A prime number is a number which has exactly two distinct divisors: one and itself. For example, numbers 2, 7, 3 are prime, and 1, 6, 4 are not.
The next prime number after *x* is the smallest prime number greater than *x*. For example, the next prime number after 2 is 3, and the next prime number after 3 is 5. Note that there is exactly one next prime number after each number. So 5 is not the next prime number for 2.
One cold April morning Panoramix predicted that soon Kakofonix will break free from his straitjacket, and this will be a black day for the residents of the Gallic countryside.
Panoramix's prophecy tells that if some day Asterix and Obelix beat exactly *x* Roman soldiers, where *x* is a prime number, and next day they beat exactly *y* Roman soldiers, where *y* is the next prime number after *x*, then it's time to wait for Armageddon, for nothing can shut Kakofonix up while he sings his infernal song.
Yesterday the Gauls beat *n* Roman soldiers and it turned out that the number *n* was prime! Today their victims were a troop of *m* Romans (*m*<=><=*n*). Determine whether the Gauls should wait for the black day after today's victory of Asterix and Obelix?
Input Specification:
The first and only input line contains two positive integers — *n* and *m* (2<=≤<=*n*<=<<=*m*<=≤<=50). It is guaranteed that *n* is prime.
Pretests contain all the cases with restrictions 2<=≤<=*n*<=<<=*m*<=≤<=4.
Output Specification:
Print YES, if *m* is the next prime number after *n*, or NO otherwise.
Demo Input:
['3 5\n', '7 11\n', '7 9\n']
Demo Output:
['YES', 'YES', 'NO']
Note:
none
|
```python
def easy(a):
for j in range(2, int(a ** 0.5) + 1):
if a % j == 0:
return 0
return 1
n, m = map(int, input().split())
cnt = 0
for i in range(n + 1, m):
cnt += easy(i)
if cnt == 0 and easy(m):
print("YES")
else:
print("NO")
```
| 3.969
|
330
|
B
|
Road Construction
|
PROGRAMMING
| 1,300
|
[
"constructive algorithms",
"graphs"
] | null | null |
A country has *n* cities. Initially, there is no road in the country. One day, the king decides to construct some roads connecting pairs of cities. Roads can be traversed either way. He wants those roads to be constructed in such a way that it is possible to go from each city to any other city by traversing at most two roads. You are also given *m* pairs of cities — roads cannot be constructed between these pairs of cities.
Your task is to construct the minimum number of roads that still satisfy the above conditions. The constraints will guarantee that this is always possible.
|
The first line consists of two integers *n* and *m* .
Then *m* lines follow, each consisting of two integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*, *a**i*<=≠<=*b**i*), which means that it is not possible to construct a road connecting cities *a**i* and *b**i*. Consider the cities are numbered from 1 to *n*.
It is guaranteed that every pair of cities will appear at most once in the input.
|
You should print an integer *s*: the minimum number of roads that should be constructed, in the first line. Then *s* lines should follow, each consisting of two integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=*a**i*<=≠<=*b**i*), which means that a road should be constructed between cities *a**i* and *b**i*.
If there are several solutions, you may print any of them.
|
[
"4 1\n1 3\n"
] |
[
"3\n1 2\n4 2\n2 3\n"
] |
This is one possible solution of the example:
These are examples of wrong solutions:
| 1,000
|
[
{
"input": "4 1\n1 3",
"output": "3\n1 2\n4 2\n2 3"
},
{
"input": "1000 0",
"output": "999\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84\n1 85\n1 86\n1 87..."
},
{
"input": "484 11\n414 97\n414 224\n444 414\n414 483\n414 399\n414 484\n414 189\n414 246\n414 115\n89 414\n14 414",
"output": "483\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84\n1 85\n1 86\n1 87..."
},
{
"input": "150 3\n112 30\n61 45\n37 135",
"output": "149\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84\n1 85\n1 86\n1 87..."
},
{
"input": "34 7\n10 28\n10 19\n10 13\n24 10\n10 29\n20 10\n10 26",
"output": "33\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34"
},
{
"input": "1000 48\n816 885\n576 357\n878 659\n610 647\n37 670\n192 184\n393 407\n598 160\n547 995\n177 276\n788 44\n14 184\n604 281\n176 97\n176 293\n10 57\n852 579\n223 669\n313 260\n476 691\n667 22\n851 792\n411 489\n526 66\n233 566\n35 396\n964 815\n672 123\n148 210\n163 339\n379 598\n382 675\n132 955\n221 441\n253 490\n856 532\n135 119\n276 319\n525 835\n996 270\n92 778\n434 369\n351 927\n758 983\n798 267\n272 830\n539 728\n166 26",
"output": "999\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84\n1 85\n1 86\n1 87..."
},
{
"input": "534 0",
"output": "533\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84\n1 85\n1 86\n1 87..."
},
{
"input": "226 54\n80 165\n2 53\n191 141\n107 207\n95 196\n61 82\n42 168\n118 94\n205 182\n172 160\n84 224\n113 143\n122 93\n37 209\n176 32\n56 83\n151 81\n70 190\n99 171\n68 204\n212 48\n4 67\n116 7\n206 199\n105 62\n158 51\n178 147\n17 129\n22 47\n72 162\n188 77\n24 111\n184 26\n175 128\n110 89\n139 120\n127 92\n121 39\n217 75\n145 69\n20 161\n30 220\n222 154\n54 46\n21 87\n144 185\n164 115\n73 202\n173 35\n9 132\n74 180\n137 5\n157 117\n31 177",
"output": "225\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84\n1 85\n1 86\n1 87..."
},
{
"input": "84 3\n39 19\n55 73\n42 43",
"output": "83\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84"
},
{
"input": "207 35\n34 116\n184 5\n90 203\n12 195\n138 101\n40 150\n189 109\n115 91\n93 201\n106 18\n51 187\n139 197\n168 130\n182 64\n31 42\n86 107\n158 111\n159 132\n119 191\n53 127\n81 13\n153 112\n38 2\n87 84\n121 82\n120 22\n21 177\n151 202\n23 58\n68 192\n29 46\n105 70\n8 167\n56 54\n149 15",
"output": "206\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84\n1 85\n1 86\n1 87..."
},
{
"input": "91 37\n50 90\n26 82\n61 1\n50 17\n51 73\n45 9\n39 53\n78 35\n12 45\n43 47\n83 20\n9 59\n18 48\n68 31\n47 33\n10 25\n15 78\n5 3\n73 65\n77 4\n62 31\n73 3\n53 7\n29 58\n52 14\n56 20\n6 87\n71 16\n17 19\n77 86\n1 50\n74 79\n15 54\n55 80\n13 77\n4 69\n24 69",
"output": "90\n2 1\n2 3\n2 4\n2 5\n2 6\n2 7\n2 8\n2 9\n2 10\n2 11\n2 12\n2 13\n2 14\n2 15\n2 16\n2 17\n2 18\n2 19\n2 20\n2 21\n2 22\n2 23\n2 24\n2 25\n2 26\n2 27\n2 28\n2 29\n2 30\n2 31\n2 32\n2 33\n2 34\n2 35\n2 36\n2 37\n2 38\n2 39\n2 40\n2 41\n2 42\n2 43\n2 44\n2 45\n2 46\n2 47\n2 48\n2 49\n2 50\n2 51\n2 52\n2 53\n2 54\n2 55\n2 56\n2 57\n2 58\n2 59\n2 60\n2 61\n2 62\n2 63\n2 64\n2 65\n2 66\n2 67\n2 68\n2 69\n2 70\n2 71\n2 72\n2 73\n2 74\n2 75\n2 76\n2 77\n2 78\n2 79\n2 80\n2 81\n2 82\n2 83\n2 84\n2 85\n2 86\n2 87\n..."
},
{
"input": "226 54\n197 107\n181 146\n218 115\n36 169\n199 196\n116 93\n152 75\n213 164\n156 95\n165 58\n90 42\n141 58\n203 221\n179 204\n186 69\n27 127\n76 189\n40 195\n111 29\n85 189\n45 88\n84 135\n82 186\n185 17\n156 217\n8 123\n179 112\n92 137\n114 89\n10 152\n132 24\n135 36\n61 218\n10 120\n155 102\n222 79\n150 92\n184 34\n102 180\n154 196\n171 9\n217 105\n84 207\n56 189\n152 179\n43 165\n115 209\n208 167\n52 14\n92 47\n197 95\n13 78\n222 138\n75 36",
"output": "225\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84\n1 85\n1 86\n1 87..."
},
{
"input": "207 35\n154 79\n174 101\n189 86\n137 56\n66 23\n199 69\n18 28\n32 53\n13 179\n182 170\n199 12\n24 158\n105 133\n25 10\n40 162\n64 72\n108 9\n172 125\n43 190\n15 39\n128 150\n102 129\n90 97\n64 196\n70 123\n163 41\n12 126\n127 186\n107 23\n182 51\n29 46\n46 123\n89 35\n59 80\n206 171",
"output": "206\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84\n1 85\n1 86\n1 87..."
},
{
"input": "84 0",
"output": "83\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84"
},
{
"input": "226 54\n5 29\n130 29\n55 29\n19 29\n29 92\n29 38\n185 29\n29 150\n29 202\n29 25\n29 66\n184 29\n29 189\n177 29\n50 29\n87 29\n138 29\n29 48\n151 29\n125 29\n16 29\n42 29\n29 157\n90 29\n21 29\n29 45\n29 80\n29 67\n29 26\n29 173\n74 29\n29 193\n29 40\n172 29\n29 85\n29 102\n88 29\n29 182\n116 29\n180 29\n161 29\n10 29\n171 29\n144 29\n29 218\n190 29\n213 29\n29 71\n29 191\n29 160\n29 137\n29 58\n29 135\n127 29",
"output": "225\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84\n1 85\n1 86\n1 87..."
},
{
"input": "207 35\n25 61\n188 61\n170 61\n113 61\n35 61\n61 177\n77 61\n61 39\n61 141\n116 61\n61 163\n30 61\n192 61\n19 61\n61 162\n61 133\n185 61\n8 61\n118 61\n61 115\n7 61\n61 105\n107 61\n61 11\n161 61\n61 149\n136 61\n82 61\n20 61\n151 61\n156 61\n12 61\n87 61\n61 205\n61 108",
"output": "206\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84\n1 85\n1 86\n1 87..."
},
{
"input": "34 7\n11 32\n33 29\n17 16\n15 5\n13 25\n8 19\n20 4",
"output": "33\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34"
},
{
"input": "43 21\n38 19\n43 8\n40 31\n3 14\n24 21\n12 17\n1 9\n5 27\n25 37\n11 6\n13 26\n16 22\n10 32\n36 7\n30 29\n42 35\n20 33\n4 23\n18 15\n41 34\n2 28",
"output": "42\n39 1\n39 2\n39 3\n39 4\n39 5\n39 6\n39 7\n39 8\n39 9\n39 10\n39 11\n39 12\n39 13\n39 14\n39 15\n39 16\n39 17\n39 18\n39 19\n39 20\n39 21\n39 22\n39 23\n39 24\n39 25\n39 26\n39 27\n39 28\n39 29\n39 30\n39 31\n39 32\n39 33\n39 34\n39 35\n39 36\n39 37\n39 38\n39 40\n39 41\n39 42\n39 43"
},
{
"input": "34 7\n22 4\n5 25\n15 7\n5 9\n27 7\n34 21\n3 13",
"output": "33\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34"
},
{
"input": "50 7\n19 37\n30 32\n43 20\n48 14\n30 29\n18 36\n9 46",
"output": "49\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50"
},
{
"input": "41 12\n41 12\n29 13\n3 37\n2 20\n4 24\n27 6\n39 20\n28 41\n30 1\n35 9\n5 39\n12 31",
"output": "40\n7 1\n7 2\n7 3\n7 4\n7 5\n7 6\n7 8\n7 9\n7 10\n7 11\n7 12\n7 13\n7 14\n7 15\n7 16\n7 17\n7 18\n7 19\n7 20\n7 21\n7 22\n7 23\n7 24\n7 25\n7 26\n7 27\n7 28\n7 29\n7 30\n7 31\n7 32\n7 33\n7 34\n7 35\n7 36\n7 37\n7 38\n7 39\n7 40\n7 41"
},
{
"input": "26 2\n22 1\n14 10",
"output": "25\n2 1\n2 3\n2 4\n2 5\n2 6\n2 7\n2 8\n2 9\n2 10\n2 11\n2 12\n2 13\n2 14\n2 15\n2 16\n2 17\n2 18\n2 19\n2 20\n2 21\n2 22\n2 23\n2 24\n2 25\n2 26"
},
{
"input": "7 3\n4 7\n4 6\n6 5",
"output": "6\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7"
},
{
"input": "484 11\n459 166\n12 217\n14 231\n477 111\n365 225\n62 412\n221 223\n258 184\n266 404\n425 236\n83 269",
"output": "483\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84\n1 85\n1 86\n1 87..."
},
{
"input": "50 24\n18 36\n11 18\n22 18\n18 47\n18 10\n45 18\n18 29\n13 18\n8 18\n18 30\n26 18\n18 6\n5 18\n18 7\n46 18\n50 18\n18 3\n18 32\n4 18\n18 24\n18 19\n23 18\n35 18\n31 18",
"output": "49\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50"
},
{
"input": "7 3\n3 4\n7 3\n3 2",
"output": "6\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7"
},
{
"input": "1 0",
"output": "0"
},
{
"input": "2 0",
"output": "1\n1 2"
},
{
"input": "4 1\n1 2",
"output": "3\n3 1\n3 2\n3 4"
},
{
"input": "3 1\n1 2",
"output": "2\n3 1\n3 2"
},
{
"input": "5 1\n1 5",
"output": "4\n2 1\n2 3\n2 4\n2 5"
}
] | 1,587,228,509
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 43
| 278
| 614,400
|
import math
import sys
from collections import defaultdict
from collections import Counter
from collections import deque
import bisect
input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
ilele = lambda: map(int,input().split())
alele = lambda: list(map(int, input().split()))
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
INF = 10 ** 18
MOD = 10 ** 9 + 7
n,m = ilele()
mark ={}
for i in range(m):
a,b = ilele()
mark[a] = 1
mark[b] = 1
x = 0
for i in range(1,n+1):
if mark.get(i,-1) == -1:
x = i
break
print(n-1)
for i in range(1,n+1):
if i!=x:
print(i,x)
|
Title: Road Construction
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A country has *n* cities. Initially, there is no road in the country. One day, the king decides to construct some roads connecting pairs of cities. Roads can be traversed either way. He wants those roads to be constructed in such a way that it is possible to go from each city to any other city by traversing at most two roads. You are also given *m* pairs of cities — roads cannot be constructed between these pairs of cities.
Your task is to construct the minimum number of roads that still satisfy the above conditions. The constraints will guarantee that this is always possible.
Input Specification:
The first line consists of two integers *n* and *m* .
Then *m* lines follow, each consisting of two integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*, *a**i*<=≠<=*b**i*), which means that it is not possible to construct a road connecting cities *a**i* and *b**i*. Consider the cities are numbered from 1 to *n*.
It is guaranteed that every pair of cities will appear at most once in the input.
Output Specification:
You should print an integer *s*: the minimum number of roads that should be constructed, in the first line. Then *s* lines should follow, each consisting of two integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=*a**i*<=≠<=*b**i*), which means that a road should be constructed between cities *a**i* and *b**i*.
If there are several solutions, you may print any of them.
Demo Input:
['4 1\n1 3\n']
Demo Output:
['3\n1 2\n4 2\n2 3\n']
Note:
This is one possible solution of the example:
These are examples of wrong solutions:
|
```python
import math
import sys
from collections import defaultdict
from collections import Counter
from collections import deque
import bisect
input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
ilele = lambda: map(int,input().split())
alele = lambda: list(map(int, input().split()))
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
INF = 10 ** 18
MOD = 10 ** 9 + 7
n,m = ilele()
mark ={}
for i in range(m):
a,b = ilele()
mark[a] = 1
mark[b] = 1
x = 0
for i in range(1,n+1):
if mark.get(i,-1) == -1:
x = i
break
print(n-1)
for i in range(1,n+1):
if i!=x:
print(i,x)
```
| 3
|
|
677
|
A
|
Vanya and Fence
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Vanya and his friends are walking along the fence of height *h* and they do not want the guard to notice them. In order to achieve this the height of each of the friends should not exceed *h*. If the height of some person is greater than *h* he can bend down and then he surely won't be noticed by the guard. The height of the *i*-th person is equal to *a**i*.
Consider the width of the person walking as usual to be equal to 1, while the width of the bent person is equal to 2. Friends want to talk to each other while walking, so they would like to walk in a single row. What is the minimum width of the road, such that friends can walk in a row and remain unattended by the guard?
|
The first line of the input contains two integers *n* and *h* (1<=≤<=*n*<=≤<=1000, 1<=≤<=*h*<=≤<=1000) — the number of friends and the height of the fence, respectively.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=2*h*), the *i*-th of them is equal to the height of the *i*-th person.
|
Print a single integer — the minimum possible valid width of the road.
|
[
"3 7\n4 5 14\n",
"6 1\n1 1 1 1 1 1\n",
"6 5\n7 6 8 9 10 5\n"
] |
[
"4\n",
"6\n",
"11\n"
] |
In the first sample, only person number 3 must bend down, so the required width is equal to 1 + 1 + 2 = 4.
In the second sample, all friends are short enough and no one has to bend, so the width 1 + 1 + 1 + 1 + 1 + 1 = 6 is enough.
In the third sample, all the persons have to bend, except the last one. The required minimum width of the road is equal to 2 + 2 + 2 + 2 + 2 + 1 = 11.
| 500
|
[
{
"input": "3 7\n4 5 14",
"output": "4"
},
{
"input": "6 1\n1 1 1 1 1 1",
"output": "6"
},
{
"input": "6 5\n7 6 8 9 10 5",
"output": "11"
},
{
"input": "10 420\n214 614 297 675 82 740 174 23 255 15",
"output": "13"
},
{
"input": "10 561\n657 23 1096 487 785 66 481 554 1000 821",
"output": "15"
},
{
"input": "100 342\n478 143 359 336 162 333 385 515 117 496 310 538 469 539 258 676 466 677 1 296 150 560 26 213 627 221 255 126 617 174 279 178 24 435 70 145 619 46 669 566 300 67 576 251 58 176 441 564 569 194 24 669 73 262 457 259 619 78 400 579 222 626 269 47 80 315 160 194 455 186 315 424 197 246 683 220 68 682 83 233 290 664 273 598 362 305 674 614 321 575 362 120 14 534 62 436 294 351 485 396",
"output": "144"
},
{
"input": "100 290\n244 49 276 77 449 261 468 458 201 424 9 131 300 88 432 394 104 77 13 289 435 259 111 453 168 394 156 412 351 576 178 530 81 271 228 564 125 328 42 372 205 61 180 471 33 360 567 331 222 318 241 117 529 169 188 484 202 202 299 268 246 343 44 364 333 494 59 236 84 485 50 8 428 8 571 227 205 310 210 9 324 472 368 490 114 84 296 305 411 351 569 393 283 120 510 171 232 151 134 366",
"output": "145"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "1 1\n2",
"output": "2"
},
{
"input": "46 71\n30 26 56 138 123 77 60 122 73 45 79 10 130 3 14 1 38 46 128 50 82 16 32 68 28 98 62 106 2 49 131 11 114 39 139 70 40 50 45 137 33 30 35 136 135 19",
"output": "63"
},
{
"input": "20 723\n212 602 293 591 754 91 1135 640 80 495 845 928 1399 498 926 1431 1226 869 814 1386",
"output": "31"
},
{
"input": "48 864\n843 1020 751 1694 18 1429 1395 1174 272 1158 1628 1233 1710 441 765 561 778 748 1501 1200 563 1263 1398 1687 1518 1640 1591 839 500 466 1603 1587 1201 1209 432 868 1159 639 649 628 9 91 1036 147 896 1557 941 518",
"output": "75"
},
{
"input": "26 708\n549 241 821 734 945 1161 566 1268 216 30 1142 730 529 1014 255 168 796 1148 89 113 1328 286 743 871 1259 1397",
"output": "41"
},
{
"input": "75 940\n1620 1745 1599 441 64 1466 1496 1239 1716 1475 778 106 1136 1212 1261 444 781 257 1071 747 626 232 609 1544 682 1326 469 1361 1460 1450 1207 1319 922 625 1737 1057 1698 592 692 80 1016 541 1254 201 682 1007 847 206 1066 809 259 109 240 1611 219 1455 1326 1377 1827 786 42 1002 1382 1592 543 1866 1198 334 1524 1760 340 1566 955 257 1118",
"output": "116"
}
] | 1,697,880,698
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 29
| 46
| 0
|
nh=[int(i) for i in input().split()]
n=nh[0]
h=nh[1]
hi=[int(i) for i in input().split()]
s=0
for i in range(len(hi)):
if hi[i]>h:
s+=2
else:
s+=1
print(s)
|
Title: Vanya and Fence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vanya and his friends are walking along the fence of height *h* and they do not want the guard to notice them. In order to achieve this the height of each of the friends should not exceed *h*. If the height of some person is greater than *h* he can bend down and then he surely won't be noticed by the guard. The height of the *i*-th person is equal to *a**i*.
Consider the width of the person walking as usual to be equal to 1, while the width of the bent person is equal to 2. Friends want to talk to each other while walking, so they would like to walk in a single row. What is the minimum width of the road, such that friends can walk in a row and remain unattended by the guard?
Input Specification:
The first line of the input contains two integers *n* and *h* (1<=≤<=*n*<=≤<=1000, 1<=≤<=*h*<=≤<=1000) — the number of friends and the height of the fence, respectively.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=2*h*), the *i*-th of them is equal to the height of the *i*-th person.
Output Specification:
Print a single integer — the minimum possible valid width of the road.
Demo Input:
['3 7\n4 5 14\n', '6 1\n1 1 1 1 1 1\n', '6 5\n7 6 8 9 10 5\n']
Demo Output:
['4\n', '6\n', '11\n']
Note:
In the first sample, only person number 3 must bend down, so the required width is equal to 1 + 1 + 2 = 4.
In the second sample, all friends are short enough and no one has to bend, so the width 1 + 1 + 1 + 1 + 1 + 1 = 6 is enough.
In the third sample, all the persons have to bend, except the last one. The required minimum width of the road is equal to 2 + 2 + 2 + 2 + 2 + 1 = 11.
|
```python
nh=[int(i) for i in input().split()]
n=nh[0]
h=nh[1]
hi=[int(i) for i in input().split()]
s=0
for i in range(len(hi)):
if hi[i]>h:
s+=2
else:
s+=1
print(s)
```
| 3
|
|
659
|
A
|
Round House
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] | null | null |
Vasya lives in a round building, whose entrances are numbered sequentially by integers from 1 to *n*. Entrance *n* and entrance 1 are adjacent.
Today Vasya got bored and decided to take a walk in the yard. Vasya lives in entrance *a* and he decided that during his walk he will move around the house *b* entrances in the direction of increasing numbers (in this order entrance *n* should be followed by entrance 1). The negative value of *b* corresponds to moving |*b*| entrances in the order of decreasing numbers (in this order entrance 1 is followed by entrance *n*). If *b*<==<=0, then Vasya prefers to walk beside his entrance.
Help Vasya to determine the number of the entrance, near which he will be at the end of his walk.
|
The single line of the input contains three space-separated integers *n*, *a* and *b* (1<=≤<=*n*<=≤<=100,<=1<=≤<=*a*<=≤<=*n*,<=<=-<=100<=≤<=*b*<=≤<=100) — the number of entrances at Vasya's place, the number of his entrance and the length of his walk, respectively.
|
Print a single integer *k* (1<=≤<=*k*<=≤<=*n*) — the number of the entrance where Vasya will be at the end of his walk.
|
[
"6 2 -5\n",
"5 1 3\n",
"3 2 7\n"
] |
[
"3\n",
"4\n",
"3\n"
] |
The first example is illustrated by the picture in the statements.
| 500
|
[
{
"input": "6 2 -5",
"output": "3"
},
{
"input": "5 1 3",
"output": "4"
},
{
"input": "3 2 7",
"output": "3"
},
{
"input": "1 1 0",
"output": "1"
},
{
"input": "1 1 -1",
"output": "1"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "100 1 -1",
"output": "100"
},
{
"input": "100 54 100",
"output": "54"
},
{
"input": "100 37 -100",
"output": "37"
},
{
"input": "99 41 0",
"output": "41"
},
{
"input": "97 37 -92",
"output": "42"
},
{
"input": "99 38 59",
"output": "97"
},
{
"input": "35 34 1",
"output": "35"
},
{
"input": "48 1 -1",
"output": "48"
},
{
"input": "87 65 -76",
"output": "76"
},
{
"input": "76 26 29",
"output": "55"
},
{
"input": "100 65 0",
"output": "65"
},
{
"input": "2 1 100",
"output": "1"
},
{
"input": "3 2 -100",
"output": "1"
},
{
"input": "1 1 100",
"output": "1"
},
{
"input": "1 1 -100",
"output": "1"
},
{
"input": "3 1 -100",
"output": "3"
},
{
"input": "4 3 -100",
"output": "3"
},
{
"input": "3 2 -12",
"output": "2"
},
{
"input": "2 2 -100",
"output": "2"
},
{
"input": "3 2 -90",
"output": "2"
},
{
"input": "6 2 -10",
"output": "4"
},
{
"input": "3 3 -100",
"output": "2"
},
{
"input": "5 2 4",
"output": "1"
},
{
"input": "6 4 5",
"output": "3"
},
{
"input": "3 2 -6",
"output": "2"
},
{
"input": "5 1 -99",
"output": "2"
},
{
"input": "6 2 5",
"output": "1"
},
{
"input": "10 1 -100",
"output": "1"
},
{
"input": "2 2 1",
"output": "1"
},
{
"input": "3 3 1",
"output": "1"
},
{
"input": "6 4 4",
"output": "2"
},
{
"input": "17 17 2",
"output": "2"
},
{
"input": "6 6 1",
"output": "1"
},
{
"input": "5 3 -2",
"output": "1"
},
{
"input": "6 2 -100",
"output": "4"
},
{
"input": "5 3 -100",
"output": "3"
},
{
"input": "5 4 3",
"output": "2"
},
{
"input": "3 2 2",
"output": "1"
},
{
"input": "5 5 2",
"output": "2"
},
{
"input": "3 2 5",
"output": "1"
},
{
"input": "5 5 -1",
"output": "4"
},
{
"input": "5 3 3",
"output": "1"
},
{
"input": "4 2 3",
"output": "1"
},
{
"input": "88 76 74",
"output": "62"
}
] | 1,637,486,469
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 50
| 61
| 0
|
#!/usr/bin/env python
# coding=utf-8
'''
Author: Deean
Date: 2021-11-21 16:58:20
LastEditTime: 2021-11-21 17:18:45
Description: Round House
FilePath: CF659A.py
'''
def func():
n, a, b = map(int, input().strip().split())
if b < 0:
b = n + b % n
if (a + b) % n == 0:
print(n)
else:
print((a + b) % n)
if __name__ == '__main__':
func()
|
Title: Round House
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya lives in a round building, whose entrances are numbered sequentially by integers from 1 to *n*. Entrance *n* and entrance 1 are adjacent.
Today Vasya got bored and decided to take a walk in the yard. Vasya lives in entrance *a* and he decided that during his walk he will move around the house *b* entrances in the direction of increasing numbers (in this order entrance *n* should be followed by entrance 1). The negative value of *b* corresponds to moving |*b*| entrances in the order of decreasing numbers (in this order entrance 1 is followed by entrance *n*). If *b*<==<=0, then Vasya prefers to walk beside his entrance.
Help Vasya to determine the number of the entrance, near which he will be at the end of his walk.
Input Specification:
The single line of the input contains three space-separated integers *n*, *a* and *b* (1<=≤<=*n*<=≤<=100,<=1<=≤<=*a*<=≤<=*n*,<=<=-<=100<=≤<=*b*<=≤<=100) — the number of entrances at Vasya's place, the number of his entrance and the length of his walk, respectively.
Output Specification:
Print a single integer *k* (1<=≤<=*k*<=≤<=*n*) — the number of the entrance where Vasya will be at the end of his walk.
Demo Input:
['6 2 -5\n', '5 1 3\n', '3 2 7\n']
Demo Output:
['3\n', '4\n', '3\n']
Note:
The first example is illustrated by the picture in the statements.
|
```python
#!/usr/bin/env python
# coding=utf-8
'''
Author: Deean
Date: 2021-11-21 16:58:20
LastEditTime: 2021-11-21 17:18:45
Description: Round House
FilePath: CF659A.py
'''
def func():
n, a, b = map(int, input().strip().split())
if b < 0:
b = n + b % n
if (a + b) % n == 0:
print(n)
else:
print((a + b) % n)
if __name__ == '__main__':
func()
```
| 3
|
|
260
|
A
|
Adding Digits
|
PROGRAMMING
| 1,400
|
[
"implementation",
"math"
] | null | null |
Vasya has got two number: *a* and *b*. However, Vasya finds number *a* too short. So he decided to repeat the operation of lengthening number *a* *n* times.
One operation of lengthening a number means adding exactly one digit to the number (in the decimal notation) to the right provided that the resulting number is divisible by Vasya's number *b*. If it is impossible to obtain the number which is divisible by *b*, then the lengthening operation cannot be performed.
Your task is to help Vasya and print the number he can get after applying the lengthening operation to number *a* *n* times.
|
The first line contains three integers: *a*,<=*b*,<=*n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=105).
|
In a single line print the integer without leading zeros, which Vasya can get when he applies the lengthening operations to number *a* *n* times. If no such number exists, then print number -1. If there are multiple possible answers, print any of them.
|
[
"5 4 5\n",
"12 11 1\n",
"260 150 10\n"
] |
[
"524848\n",
"121\n",
"-1\n"
] |
none
| 500
|
[
{
"input": "5 4 5",
"output": "524848"
},
{
"input": "12 11 1",
"output": "121"
},
{
"input": "260 150 10",
"output": "-1"
},
{
"input": "78843 5684 42717",
"output": "-1"
},
{
"input": "93248 91435 1133",
"output": "-1"
},
{
"input": "100000 10 64479",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99999 21 73839",
"output": "9999990000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99991 623 36438",
"output": "9999150000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99999 334 94854",
"output": "9999960000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99252 9827 84849",
"output": "9925270000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99313 9833 10561",
"output": "9931330000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "94885 55815 11417",
"output": "9488550000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99492 58525 53481",
"output": "9949250000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99858 28531 79193",
"output": "9985850000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99136 47208 42607",
"output": "9913680000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "63270 19953 5555",
"output": "-1"
},
{
"input": "10240 128 100000",
"output": "1024000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "12 11 3",
"output": "12100"
},
{
"input": "14 12 99998",
"output": "1440000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "1 11 3",
"output": "1100"
},
{
"input": "3 40 1",
"output": "-1"
},
{
"input": "150 100 10",
"output": "1500000000000"
},
{
"input": "5 10 1",
"output": "50"
},
{
"input": "1 15 10",
"output": "15000000000"
},
{
"input": "3 13 2",
"output": "390"
}
] | 1,502,699,146
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 25
| 62
| 4,812,800
|
a, b, n = list(map(int, input().split()))
a *= 10
for k in range(10):
if (a + k) % b == 0:
a += k
break
if a % b != 0:
print(-1)
else:
print(str(a) + '0' * (n-1))
|
Title: Adding Digits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has got two number: *a* and *b*. However, Vasya finds number *a* too short. So he decided to repeat the operation of lengthening number *a* *n* times.
One operation of lengthening a number means adding exactly one digit to the number (in the decimal notation) to the right provided that the resulting number is divisible by Vasya's number *b*. If it is impossible to obtain the number which is divisible by *b*, then the lengthening operation cannot be performed.
Your task is to help Vasya and print the number he can get after applying the lengthening operation to number *a* *n* times.
Input Specification:
The first line contains three integers: *a*,<=*b*,<=*n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=105).
Output Specification:
In a single line print the integer without leading zeros, which Vasya can get when he applies the lengthening operations to number *a* *n* times. If no such number exists, then print number -1. If there are multiple possible answers, print any of them.
Demo Input:
['5 4 5\n', '12 11 1\n', '260 150 10\n']
Demo Output:
['524848\n', '121\n', '-1\n']
Note:
none
|
```python
a, b, n = list(map(int, input().split()))
a *= 10
for k in range(10):
if (a + k) % b == 0:
a += k
break
if a % b != 0:
print(-1)
else:
print(str(a) + '0' * (n-1))
```
| 3
|
|
75
|
A
|
Life Without Zeros
|
PROGRAMMING
| 1,000
|
[
"implementation"
] |
A. Life Without Zeros
|
2
|
256
|
Can you imagine our life if we removed all zeros from it? For sure we will have many problems.
In this problem we will have a simple example if we removed all zeros from our life, it's the addition operation. Let's assume you are given this equation *a*<=+<=*b*<==<=*c*, where *a* and *b* are positive integers, and *c* is the sum of *a* and *b*. Now let's remove all zeros from this equation. Will the equation remain correct after removing all zeros?
For example if the equation is 101<=+<=102<==<=203, if we removed all zeros it will be 11<=+<=12<==<=23 which is still a correct equation.
But if the equation is 105<=+<=106<==<=211, if we removed all zeros it will be 15<=+<=16<==<=211 which is not a correct equation.
|
The input will consist of two lines, the first line will contain the integer *a*, and the second line will contain the integer *b* which are in the equation as described above (1<=≤<=*a*,<=*b*<=≤<=109). There won't be any leading zeros in both. The value of *c* should be calculated as *c*<==<=*a*<=+<=*b*.
|
The output will be just one line, you should print "YES" if the equation will remain correct after removing all zeros, and print "NO" otherwise.
|
[
"101\n102\n",
"105\n106\n"
] |
[
"YES\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "101\n102",
"output": "YES"
},
{
"input": "105\n106",
"output": "NO"
},
{
"input": "544\n397",
"output": "YES"
},
{
"input": "822\n280",
"output": "NO"
},
{
"input": "101\n413",
"output": "NO"
},
{
"input": "309\n139",
"output": "NO"
},
{
"input": "693\n970",
"output": "NO"
},
{
"input": "981\n1",
"output": "YES"
},
{
"input": "352\n276",
"output": "YES"
},
{
"input": "164\n691",
"output": "YES"
},
{
"input": "110036\n43",
"output": "YES"
},
{
"input": "100\n1000",
"output": "NO"
},
{
"input": "1000000000\n1000000000",
"output": "YES"
},
{
"input": "999999999\n999999999",
"output": "YES"
},
{
"input": "6\n4",
"output": "NO"
},
{
"input": "123456\n876543",
"output": "YES"
},
{
"input": "1234567\n9876543",
"output": "NO"
},
{
"input": "1111111\n1119111",
"output": "NO"
},
{
"input": "10502405\n40403403",
"output": "YES"
},
{
"input": "1\n1",
"output": "YES"
},
{
"input": "5\n4",
"output": "YES"
},
{
"input": "1\n6",
"output": "YES"
},
{
"input": "40\n17",
"output": "NO"
},
{
"input": "35\n81",
"output": "YES"
},
{
"input": "347\n908",
"output": "NO"
},
{
"input": "2081\n7450",
"output": "NO"
},
{
"input": "84355694\n373629503",
"output": "NO"
},
{
"input": "885992048\n510468672",
"output": "NO"
},
{
"input": "540144752\n647307841",
"output": "NO"
},
{
"input": "341781108\n784147010",
"output": "NO"
},
{
"input": "960440938\n572334653",
"output": "NO"
},
{
"input": "657414510\n768771903",
"output": "NO"
},
{
"input": "501871728\n725074574",
"output": "NO"
},
{
"input": "968\n851",
"output": "YES"
},
{
"input": "943\n939",
"output": "YES"
},
{
"input": "347\n908",
"output": "NO"
},
{
"input": "611\n41",
"output": "YES"
},
{
"input": "737\n289",
"output": "NO"
},
{
"input": "168\n571",
"output": "YES"
},
{
"input": "868\n463",
"output": "YES"
},
{
"input": "815732007\n913882008",
"output": "NO"
},
{
"input": "459039901\n453033850",
"output": "NO"
},
{
"input": "975305404\n484601256",
"output": "NO"
}
] | 1,611,384,281
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 43
| 156
| 409,600
|
import sys, os
from collections import defaultdict
from math import gcd, sqrt
input = sys.stdin.readline
read = lambda: list(map(int, input().strip().split()))
def main():
a, b = [int(input()) for i in range(2)]; c = a+b
a, b, c = [int(str(i).replace("0", "")) for i in [a, b, c]]
if a+b == c:print("YES")
else:print("NO")
if __name__ == "__main__":
main()
"""
l = n-a+1
a*l + (l*(l+1))//2
"""
|
Title: Life Without Zeros
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Can you imagine our life if we removed all zeros from it? For sure we will have many problems.
In this problem we will have a simple example if we removed all zeros from our life, it's the addition operation. Let's assume you are given this equation *a*<=+<=*b*<==<=*c*, where *a* and *b* are positive integers, and *c* is the sum of *a* and *b*. Now let's remove all zeros from this equation. Will the equation remain correct after removing all zeros?
For example if the equation is 101<=+<=102<==<=203, if we removed all zeros it will be 11<=+<=12<==<=23 which is still a correct equation.
But if the equation is 105<=+<=106<==<=211, if we removed all zeros it will be 15<=+<=16<==<=211 which is not a correct equation.
Input Specification:
The input will consist of two lines, the first line will contain the integer *a*, and the second line will contain the integer *b* which are in the equation as described above (1<=≤<=*a*,<=*b*<=≤<=109). There won't be any leading zeros in both. The value of *c* should be calculated as *c*<==<=*a*<=+<=*b*.
Output Specification:
The output will be just one line, you should print "YES" if the equation will remain correct after removing all zeros, and print "NO" otherwise.
Demo Input:
['101\n102\n', '105\n106\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none
|
```python
import sys, os
from collections import defaultdict
from math import gcd, sqrt
input = sys.stdin.readline
read = lambda: list(map(int, input().strip().split()))
def main():
a, b = [int(input()) for i in range(2)]; c = a+b
a, b, c = [int(str(i).replace("0", "")) for i in [a, b, c]]
if a+b == c:print("YES")
else:print("NO")
if __name__ == "__main__":
main()
"""
l = n-a+1
a*l + (l*(l+1))//2
"""
```
| 3.960237
|
34
|
A
|
Reconnaissance 2
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Reconnaissance 2
|
2
|
256
|
*n* soldiers stand in a circle. For each soldier his height *a**i* is known. A reconnaissance unit can be made of such two neighbouring soldiers, whose heights difference is minimal, i.e. |*a**i*<=-<=*a**j*| is minimal. So each of them will be less noticeable with the other. Output any pair of soldiers that can form a reconnaissance unit.
|
The first line contains integer *n* (2<=≤<=*n*<=≤<=100) — amount of soldiers. Then follow the heights of the soldiers in their order in the circle — *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000). The soldier heights are given in clockwise or counterclockwise direction.
|
Output two integers — indexes of neighbouring soldiers, who should form a reconnaissance unit. If there are many optimum solutions, output any of them. Remember, that the soldiers stand in a circle.
|
[
"5\n10 12 13 15 10\n",
"4\n10 20 30 40\n"
] |
[
"5 1\n",
"1 2\n"
] |
none
| 500
|
[
{
"input": "5\n10 12 13 15 10",
"output": "5 1"
},
{
"input": "4\n10 20 30 40",
"output": "1 2"
},
{
"input": "6\n744 359 230 586 944 442",
"output": "2 3"
},
{
"input": "5\n826 747 849 687 437",
"output": "1 2"
},
{
"input": "5\n999 999 993 969 999",
"output": "1 2"
},
{
"input": "5\n4 24 6 1 15",
"output": "3 4"
},
{
"input": "2\n511 32",
"output": "1 2"
},
{
"input": "3\n907 452 355",
"output": "2 3"
},
{
"input": "4\n303 872 764 401",
"output": "4 1"
},
{
"input": "10\n684 698 429 694 956 812 594 170 937 764",
"output": "1 2"
},
{
"input": "20\n646 840 437 946 640 564 936 917 487 752 844 734 468 969 674 646 728 642 514 695",
"output": "7 8"
},
{
"input": "30\n996 999 998 984 989 1000 996 993 1000 983 992 999 999 1000 979 992 987 1000 996 1000 1000 989 981 996 995 999 999 989 999 1000",
"output": "12 13"
},
{
"input": "50\n93 27 28 4 5 78 59 24 19 134 31 128 118 36 90 32 32 1 44 32 33 13 31 10 12 25 38 50 25 12 4 22 28 53 48 83 4 25 57 31 71 24 8 7 28 86 23 80 101 58",
"output": "16 17"
},
{
"input": "88\n1000 1000 1000 1000 1000 998 998 1000 1000 1000 1000 999 999 1000 1000 1000 999 1000 997 999 997 1000 999 998 1000 999 1000 1000 1000 999 1000 999 999 1000 1000 999 1000 999 1000 1000 998 1000 1000 1000 998 998 1000 1000 999 1000 1000 1000 1000 1000 1000 1000 998 1000 1000 1000 999 1000 1000 999 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 998 1000 1000 1000 998 1000 1000 998 1000 999 1000 1000 1000 1000",
"output": "1 2"
},
{
"input": "99\n4 4 21 6 5 3 13 2 6 1 3 4 1 3 1 9 11 1 6 17 4 5 20 4 1 9 5 11 3 4 14 1 3 3 1 4 3 5 27 1 1 2 10 7 11 4 19 7 11 6 11 13 3 1 10 7 2 1 16 1 9 4 29 13 2 12 14 2 21 1 9 8 26 12 12 5 2 14 7 8 8 8 9 4 12 2 6 6 7 16 8 14 2 10 20 15 3 7 4",
"output": "1 2"
},
{
"input": "100\n713 572 318 890 577 657 646 146 373 783 392 229 455 871 20 593 573 336 26 381 280 916 907 732 820 713 111 840 570 446 184 711 481 399 788 647 492 15 40 530 549 506 719 782 126 20 778 996 712 761 9 74 812 418 488 175 103 585 900 3 604 521 109 513 145 708 990 361 682 827 791 22 596 780 596 385 450 643 158 496 876 975 319 783 654 895 891 361 397 81 682 899 347 623 809 557 435 279 513 438",
"output": "86 87"
},
{
"input": "100\n31 75 86 68 111 27 22 22 26 30 54 163 107 75 160 122 14 23 17 26 27 20 43 58 59 71 21 148 9 32 43 91 133 286 132 70 90 156 84 14 77 93 23 18 13 72 18 131 33 28 72 175 30 86 249 20 14 208 28 57 63 199 6 10 24 30 62 267 43 479 60 28 138 1 45 3 19 47 7 166 116 117 50 140 28 14 95 85 93 43 61 15 2 70 10 51 7 95 9 25",
"output": "7 8"
},
{
"input": "100\n896 898 967 979 973 709 961 968 806 967 896 967 826 975 936 903 986 856 851 931 852 971 786 837 949 978 686 936 952 909 965 749 908 916 943 973 983 975 939 886 964 928 960 976 907 788 994 773 949 871 947 980 945 985 726 981 887 943 907 990 931 874 840 867 948 951 961 904 888 901 976 967 994 921 828 970 972 722 755 970 860 855 914 869 714 899 969 978 898 862 642 939 904 936 819 934 884 983 955 964",
"output": "1 2"
},
{
"input": "100\n994 927 872 970 815 986 952 996 965 1000 877 986 978 999 950 990 936 997 993 960 921 860 895 869 943 998 983 968 973 953 999 990 995 871 853 979 973 963 953 938 997 989 993 964 960 973 946 975 1000 962 920 746 989 957 904 965 920 979 966 961 1000 993 975 952 846 971 991 979 985 969 984 973 956 1000 952 778 983 974 956 927 995 997 980 997 1000 970 960 970 988 983 947 904 935 972 1000 863 992 996 932 967",
"output": "81 82"
},
{
"input": "100\n48 108 63 21 27 8 49 21 75 8 24 42 149 18 8 28 21 18 25 35 59 70 59 33 40 1 67 34 120 82 4 115 72 87 3 15 15 63 37 12 40 27 83 14 38 20 14 58 93 10 31 3 39 6 197 77 54 16 31 146 9 49 14 8 77 82 5 11 80 116 8 61 50 24 7 103 29 11 3 3 1 12 46 24 21 131 39 29 36 2 107 40 16 99 31 41 29 48 17 17",
"output": "36 37"
}
] | 1,653,410,706
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 62
| 0
|
n = int(input())
m = list(map(int, input().split()))
mmin=min(m)
s=m.index(mmin)
m.remove(mmin)
mmmin=min(m)
s1=m.index(mmmin)
if m[0]==mmin and m[n]==mmmin:
print(s1+2,s+1)
else:
print(s+1,s1+2)
|
Title: Reconnaissance 2
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
*n* soldiers stand in a circle. For each soldier his height *a**i* is known. A reconnaissance unit can be made of such two neighbouring soldiers, whose heights difference is minimal, i.e. |*a**i*<=-<=*a**j*| is minimal. So each of them will be less noticeable with the other. Output any pair of soldiers that can form a reconnaissance unit.
Input Specification:
The first line contains integer *n* (2<=≤<=*n*<=≤<=100) — amount of soldiers. Then follow the heights of the soldiers in their order in the circle — *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000). The soldier heights are given in clockwise or counterclockwise direction.
Output Specification:
Output two integers — indexes of neighbouring soldiers, who should form a reconnaissance unit. If there are many optimum solutions, output any of them. Remember, that the soldiers stand in a circle.
Demo Input:
['5\n10 12 13 15 10\n', '4\n10 20 30 40\n']
Demo Output:
['5 1\n', '1 2\n']
Note:
none
|
```python
n = int(input())
m = list(map(int, input().split()))
mmin=min(m)
s=m.index(mmin)
m.remove(mmin)
mmmin=min(m)
s1=m.index(mmmin)
if m[0]==mmin and m[n]==mmmin:
print(s1+2,s+1)
else:
print(s+1,s1+2)
```
| 0
|
271
|
D
|
Good Substrings
|
PROGRAMMING
| 1,800
|
[
"data structures",
"strings"
] | null | null |
You've got string *s*, consisting of small English letters. Some of the English letters are good, the rest are bad.
A substring *s*[*l*...*r*] (1<=≤<=*l*<=≤<=*r*<=≤<=|*s*|) of string *s*<=<==<=<=*s*1*s*2...*s*|*s*| (where |*s*| is the length of string *s*) is string <=*s**l**s**l*<=+<=1...*s**r*.
The substring *s*[*l*...*r*] is good, if among the letters <=*s**l*,<=*s**l*<=+<=1,<=...,<=*s**r* there are at most *k* bad ones (look at the sample's explanation to understand it more clear).
Your task is to find the number of distinct good substrings of the given string *s*. Two substrings *s*[*x*...*y*] and *s*[*p*...*q*] are considered distinct if their content is different, i.e. *s*[*x*...*y*]<=≠<=*s*[*p*...*q*].
|
The first line of the input is the non-empty string *s*, consisting of small English letters, the string's length is at most 1500 characters.
The second line of the input is the string of characters "0" and "1", the length is exactly 26 characters. If the *i*-th character of this string equals "1", then the *i*-th English letter is good, otherwise it's bad. That is, the first character of this string corresponds to letter "a", the second one corresponds to letter "b" and so on.
The third line of the input consists a single integer *k* (0<=≤<=*k*<=≤<=|*s*|) — the maximum acceptable number of bad characters in a good substring.
|
Print a single integer — the number of distinct good substrings of string *s*.
|
[
"ababab\n01000000000000000000000000\n1\n",
"acbacbacaa\n00000000000000000000000000\n2\n"
] |
[
"5\n",
"8\n"
] |
In the first example there are following good substrings: "a", "ab", "b", "ba", "bab".
In the second example there are following good substrings: "a", "aa", "ac", "b", "ba", "c", "ca", "cb".
| 2,000
|
[
{
"input": "ababab\n01000000000000000000000000\n1",
"output": "5"
},
{
"input": "acbacbacaa\n00000000000000000000000000\n2",
"output": "8"
},
{
"input": "a\n00000000000000000000000000\n0",
"output": "0"
},
{
"input": "aaaa\n00000000000000000000000000\n0",
"output": "0"
},
{
"input": "aaaaaa\n00000000000000000000000000\n1",
"output": "1"
},
{
"input": "bbbbbbbbba\n01000000000000000000000000\n0",
"output": "9"
},
{
"input": "bbbbbbbbba\n10000000000000000000000000\n0",
"output": "1"
},
{
"input": "kqdwdulmgvugvbl\n00101010100100100101101110\n13",
"output": "114"
},
{
"input": "acehqnrtuwaealwbqufdmizce\n10000110100000010011101101\n16",
"output": "316"
},
{
"input": "yqahbyyoxltryqdmvenemaqnbakglgqolxnaifnqtoclnnqiab\n11000001000110100111100001\n41",
"output": "1243"
},
{
"input": "dykhvzcntljuuoqghptioetqnfllwekzohiuaxelgecabvsbibgqodqxvyfkbyjwtgbyhvssntinkwsinwsmalusiwnjmtcoovfj\n10001111101011111101101001\n25",
"output": "4420"
},
{
"input": "dcijflpluxgeqlroaktzcujfaaidnrdzgogzhobhsmbwmjpschtvjmivfapddsmxfvlhhgguymgtdjxpiezbnlmnlvdnuaohqskdbcjlyfdrzicflveffvpeyzhwqqdaenbsghuvetpxvqcutjxbelbfzuqpwlfvveebnmkoryxgodpccanzwhziiiumgtwskxhrhwdkwmyreefwmoedcvaokvetcgundyiidsqkolpqkarpszrrmagrfvpnwharotashtwcnrcnhapdwrbltgkpkkmlrpapfizonyttrikh\n11011010101101110101010001\n54",
"output": "31658"
},
{
"input": "baababbbaa\n01011100110010100100001111\n7",
"output": "41"
},
{
"input": "aaaabaababbbbbbbbbaaabaaabbbababbbbbbabbaaabaabaabaaabaaaabbababbaaaaaaababbaababbabbaabbbabaabbbbababaaabbabbbabaaabbbabbbbbbbbabababaaaaaabaababbbbbaaaaaaaaaaaaaaababbbbabbbbbbababaabbabbbaababaabaaaabbababaaaabaaaaaaabaabaababbbbbbaababbabbaaaababbabbaaaabaaaaaaababbbbabbbbabaaabababaaaabaaaababb\n11011010101101110101010001\n54",
"output": "42890"
},
{
"input": "dykhvzcntljuuoqghptioetqnfllwekzohiuaxelgecabvsbibgqodqxvyfkbyjwtgbyhvssntinkwsinwsmalusiwnjmtcoovfj\n11111111111111111111111111\n25",
"output": "4967"
},
{
"input": "twnwdluhxf\n00000000000000000000000000\n1",
"output": "9"
},
{
"input": "aaaaaaaaaa\n01011100110010100100001111\n7",
"output": "7"
},
{
"input": "eoyirpkwgpvvwzaaaaaaaaaaaaaa\n11111111111111111111111111\n1",
"output": "312"
},
{
"input": "fwnpwwuzszuryaaaaaaaaaaaaa\n11111111111111111111111111\n1",
"output": "269"
},
{
"input": "xnahfslefuigqaaaaaaaaaaaaa\n11111111111111111111111111\n1",
"output": "271"
}
] | 1,691,993,941
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 52
| 154
| 4,608,000
|
s = input()
alpha = input()
k = int(input())
subs = sorted(s[i:] for i in range(len(s)))
prev = ''
ans = 0
for j in subs:
uni = False
bad = 0
for i in range(len(j)):
if i >= len(prev) or prev[i] != j[i]:
uni = True
if alpha[ord(j[i])-ord('a')] == '0':
bad += 1
if bad > k:
break
if uni:
ans += 1
prev = j
print(ans)
|
Title: Good Substrings
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got string *s*, consisting of small English letters. Some of the English letters are good, the rest are bad.
A substring *s*[*l*...*r*] (1<=≤<=*l*<=≤<=*r*<=≤<=|*s*|) of string *s*<=<==<=<=*s*1*s*2...*s*|*s*| (where |*s*| is the length of string *s*) is string <=*s**l**s**l*<=+<=1...*s**r*.
The substring *s*[*l*...*r*] is good, if among the letters <=*s**l*,<=*s**l*<=+<=1,<=...,<=*s**r* there are at most *k* bad ones (look at the sample's explanation to understand it more clear).
Your task is to find the number of distinct good substrings of the given string *s*. Two substrings *s*[*x*...*y*] and *s*[*p*...*q*] are considered distinct if their content is different, i.e. *s*[*x*...*y*]<=≠<=*s*[*p*...*q*].
Input Specification:
The first line of the input is the non-empty string *s*, consisting of small English letters, the string's length is at most 1500 characters.
The second line of the input is the string of characters "0" and "1", the length is exactly 26 characters. If the *i*-th character of this string equals "1", then the *i*-th English letter is good, otherwise it's bad. That is, the first character of this string corresponds to letter "a", the second one corresponds to letter "b" and so on.
The third line of the input consists a single integer *k* (0<=≤<=*k*<=≤<=|*s*|) — the maximum acceptable number of bad characters in a good substring.
Output Specification:
Print a single integer — the number of distinct good substrings of string *s*.
Demo Input:
['ababab\n01000000000000000000000000\n1\n', 'acbacbacaa\n00000000000000000000000000\n2\n']
Demo Output:
['5\n', '8\n']
Note:
In the first example there are following good substrings: "a", "ab", "b", "ba", "bab".
In the second example there are following good substrings: "a", "aa", "ac", "b", "ba", "c", "ca", "cb".
|
```python
s = input()
alpha = input()
k = int(input())
subs = sorted(s[i:] for i in range(len(s)))
prev = ''
ans = 0
for j in subs:
uni = False
bad = 0
for i in range(len(j)):
if i >= len(prev) or prev[i] != j[i]:
uni = True
if alpha[ord(j[i])-ord('a')] == '0':
bad += 1
if bad > k:
break
if uni:
ans += 1
prev = j
print(ans)
```
| 3
|
|
58
|
A
|
Chat room
|
PROGRAMMING
| 1,000
|
[
"greedy",
"strings"
] |
A. Chat room
|
1
|
256
|
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
|
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
|
If Vasya managed to say hello, print "YES", otherwise print "NO".
|
[
"ahhellllloou\n",
"hlelo\n"
] |
[
"YES\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "ahhellllloou",
"output": "YES"
},
{
"input": "hlelo",
"output": "NO"
},
{
"input": "helhcludoo",
"output": "YES"
},
{
"input": "hehwelloho",
"output": "YES"
},
{
"input": "pnnepelqomhhheollvlo",
"output": "YES"
},
{
"input": "tymbzjyqhymedasloqbq",
"output": "NO"
},
{
"input": "yehluhlkwo",
"output": "NO"
},
{
"input": "hatlevhhalrohairnolsvocafgueelrqmlqlleello",
"output": "YES"
},
{
"input": "hhhtehdbllnhwmbyhvelqqyoulretpbfokflhlhreeflxeftelziclrwllrpflflbdtotvlqgoaoqldlroovbfsq",
"output": "YES"
},
{
"input": "rzlvihhghnelqtwlexmvdjjrliqllolhyewgozkuovaiezgcilelqapuoeglnwmnlftxxiigzczlouooi",
"output": "YES"
},
{
"input": "pfhhwctyqdlkrwhebfqfelhyebwllhemtrmeblgrynmvyhioesqklclocxmlffuormljszllpoo",
"output": "YES"
},
{
"input": "lqllcolohwflhfhlnaow",
"output": "NO"
},
{
"input": "heheeellollvoo",
"output": "YES"
},
{
"input": "hellooo",
"output": "YES"
},
{
"input": "o",
"output": "NO"
},
{
"input": "hhqhzeclohlehljlhtesllylrolmomvuhcxsobtsckogdv",
"output": "YES"
},
{
"input": "yoegfuzhqsihygnhpnukluutocvvwuldiighpogsifealtgkfzqbwtmgghmythcxflebrkctlldlkzlagovwlstsghbouk",
"output": "YES"
},
{
"input": "uatqtgbvrnywfacwursctpagasnhydvmlinrcnqrry",
"output": "NO"
},
{
"input": "tndtbldbllnrwmbyhvqaqqyoudrstpbfokfoclnraefuxtftmgzicorwisrpfnfpbdtatvwqgyalqtdtrjqvbfsq",
"output": "NO"
},
{
"input": "rzlvirhgemelnzdawzpaoqtxmqucnahvqnwldklrmjiiyageraijfivigvozgwngiulttxxgzczptusoi",
"output": "YES"
},
{
"input": "kgyelmchocojsnaqdsyeqgnllytbqietpdlgknwwumqkxrexgdcnwoldicwzwofpmuesjuxzrasscvyuqwspm",
"output": "YES"
},
{
"input": "pnyvrcotjvgynbeldnxieghfltmexttuxzyac",
"output": "NO"
},
{
"input": "dtwhbqoumejligbenxvzhjlhosqojetcqsynlzyhfaevbdpekgbtjrbhlltbceobcok",
"output": "YES"
},
{
"input": "crrfpfftjwhhikwzeedrlwzblckkteseofjuxjrktcjfsylmlsvogvrcxbxtffujqshslemnixoeezivksouefeqlhhokwbqjz",
"output": "YES"
},
{
"input": "jhfbndhyzdvhbvhmhmefqllujdflwdpjbehedlsqfdsqlyelwjtyloxwsvasrbqosblzbowlqjmyeilcvotdlaouxhdpoeloaovb",
"output": "YES"
},
{
"input": "hwlghueoemiqtjhhpashjsouyegdlvoyzeunlroypoprnhlyiwiuxrghekaylndhrhllllwhbebezoglydcvykllotrlaqtvmlla",
"output": "YES"
},
{
"input": "wshiaunnqnqxodholbipwhhjmyeblhgpeleblklpzwhdunmpqkbuzloetmwwxmeltkrcomulxauzlwmlklldjodozxryghsnwgcz",
"output": "YES"
},
{
"input": "shvksednttggehroewuiptvvxtrzgidravtnjwuqrlnnkxbplctzkckinpkgjopjfoxdbojtcvsuvablcbkrzajrlhgobkcxeqti",
"output": "YES"
},
{
"input": "hyyhddqhxhekehkwfhlnlsihzefwchzerevcjtokefplholrbvxlltdlafjxrfhleglrvlolojoqaolagtbeyogxlbgfolllslli",
"output": "YES"
},
{
"input": "iaagrdhhelxpdegueiulflquvlzidoprzkehthkldaytrjwhyhprwjxlltinxvuilxohqgjqcvkkdcuoliekcnlilwgqlnlzouoo",
"output": "YES"
},
{
"input": "wfluaeseldgxyvxpwuhkptdmlflnlhktwxiabsvkolsquymrmhzczzoybvlilhmvsuunggvgxzgyyffk",
"output": "NO"
},
{
"input": "loee",
"output": "NO"
},
{
"input": "izdipwylefqmjbuoznfglgcfvedeouflkeehxbasaqmiooppfsqlhprospqgxvzrcpwlfdddngoqvpwvggsnvvxhmjoslwjudjlo",
"output": "NO"
},
{
"input": "pplynflbjlseflaegzqpdxswgunhnlglqhycrbcwdfhclmcrwekkzsoosmfblljeenwfjyifcvxihhhbyhmhlkvtskeglaovjnsi",
"output": "YES"
},
{
"input": "xqjqmenkodmlhzyzmmvofdngktygbbxbzpluzcohohmalkoeuwfikblltaaigv",
"output": "NO"
},
{
"input": "hello",
"output": "YES"
},
{
"input": "oohell",
"output": "NO"
},
{
"input": "hell",
"output": "NO"
},
{
"input": "eloellohoelo",
"output": "NO"
},
{
"input": "helllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllo",
"output": "YES"
}
] | 1,688,976,041
| 941
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 31
| 0
|
s = input()
i = s.find("h")
t = 0
if(i==-1):
t+=1
i = s.find("e", i)
if(i==-1):
t+=1
i = s.find("e", i)
if(i==-1):
t+=1
i = s.find("l", i)
if(i==-1):
t+=1
i = s.find("o", i)
if(i==-1):
t+=1
if(t>0):
print("NO")
else:
print("YES")
|
Title: Chat room
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
Input Specification:
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
Output Specification:
If Vasya managed to say hello, print "YES", otherwise print "NO".
Demo Input:
['ahhellllloou\n', 'hlelo\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none
|
```python
s = input()
i = s.find("h")
t = 0
if(i==-1):
t+=1
i = s.find("e", i)
if(i==-1):
t+=1
i = s.find("e", i)
if(i==-1):
t+=1
i = s.find("l", i)
if(i==-1):
t+=1
i = s.find("o", i)
if(i==-1):
t+=1
if(t>0):
print("NO")
else:
print("YES")
```
| 0
|
152
|
C
|
Pocket Book
|
PROGRAMMING
| 1,400
|
[
"combinatorics"
] | null | null |
One day little Vasya found mom's pocket book. The book had *n* names of her friends and unusually enough, each name was exactly *m* letters long. Let's number the names from 1 to *n* in the order in which they are written.
As mom wasn't home, Vasya decided to play with names: he chose three integers *i*, *j*, *k* (1<=≤<=*i*<=<<=*j*<=≤<=*n*, 1<=≤<=*k*<=≤<=*m*), then he took names number *i* and *j* and swapped their prefixes of length *k*. For example, if we take names "CBDAD" and "AABRD" and swap their prefixes with the length of 3, the result will be names "AABAD" and "CBDRD".
You wonder how many different names Vasya can write instead of name number 1, if Vasya is allowed to perform any number of the described actions. As Vasya performs each action, he chooses numbers *i*, *j*, *k* independently from the previous moves and his choice is based entirely on his will. The sought number can be very large, so you should only find it modulo 1000000007 (109<=+<=7).
|
The first input line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of names and the length of each name, correspondingly. Then *n* lines contain names, each name consists of exactly *m* uppercase Latin letters.
|
Print the single number — the number of different names that could end up in position number 1 in the pocket book after the applying the procedures described above. Print the number modulo 1000000007 (109<=+<=7).
|
[
"2 3\nAAB\nBAA\n",
"4 5\nABABA\nBCGDG\nAAAAA\nYABSA\n"
] |
[
"4\n",
"216\n"
] |
In the first sample Vasya can get the following names in the position number 1: "AAB", "AAA", "BAA" and "BAB".
| 1,500
|
[
{
"input": "2 3\nAAB\nBAA",
"output": "4"
},
{
"input": "4 5\nABABA\nBCGDG\nAAAAA\nYABSA",
"output": "216"
},
{
"input": "1 1\nE",
"output": "1"
},
{
"input": "2 2\nNS\nPD",
"output": "4"
},
{
"input": "3 4\nPJKD\nNFJX\nFGFK",
"output": "81"
},
{
"input": "4 5\nSXFMY\nATHLM\nKDDQW\nZWGDS",
"output": "1024"
},
{
"input": "20 14\nJNFKBBBJYZHWQE\nLBOKZCPFNKDBJY\nXKNWGHQHIOXUPF\nDDNRUKVUGHWMXW\nMTIZFNAAFEAPHX\nIXBQOOHEULZYHU\nMRCSREUEOOMUUN\nHJTSQWKUFYZDQU\nGMCMUZCOPRVEIQ\nXBKKGGJECOBLTH\nXXHTLXCNJZJUAF\nVLJRKXXXWMTPKZ\nPTYMNPTBBCWKAD\nQYJGOBUBHMEDYE\nGTKUUVVNKAHTUI\nZNKXYZPCYLBZFP\nQCBLJTRMBDWNNE\nTDOKJOBKEOVNLZ\nFKZUITYAFJOQIM\nUWQNSGLXEEIRWF",
"output": "515139391"
},
{
"input": "5 14\nAQRXUQQNSKZPGC\nDTTKSPFGGVCLPT\nVLZQWWESCHDTAZ\nCOKOWDWDRUOMHP\nXDTRBIZTTCIDGS",
"output": "124999979"
},
{
"input": "9 23\nOILBYKHRGMPENVFNHLSIUOW\nLPJFHTUQUINAALRDGLSQUXR\nLYYJJEBNZATAFQWTDZSPUNZ\nHSJPIQKKWWERJZIEMLCZUKI\nOJYIEYDGPFWRHCMISJCCUEM\nLMGKZVFYIVDRTIHBWPCNUTG\nUBGGNCITVHAIPKXCLTSAULQ\nOWSAWUOXQDBSXXBHTLSXUVD\nUGQTIZQPBGMASRQPVPSFUWK",
"output": "454717784"
},
{
"input": "25 4\nLVKG\nMICU\nZHKW\nLFGG\nOWQO\nLCQG\nLVXU\nOUKB\nLNQX\nZJTO\nOOQX\nLVQP\nMFQB\nMRQV\nOIQH\nOPXX\nXFKU\nFCQB\nZPKH\nLVCH\nNFCU\nOVQW\nOZKU\nLFHX\nLPXO",
"output": "5733"
},
{
"input": "30 10\nUTNTGOKZYJ\nQHOUHNYZVW\nLTVGHJRZVW\nMZHYHOLZYJ\nERYEUEPZYE\nUZDBFTURYJ\nRVSMQTIZGW\nWDJQHMIRYY\nKCORHQPZYE\nRRPLFOZZVY\nJTXMFNNNYJ\nMVTGGOZZVV\nEHAFFNUZVF\nLBRNWJZNYE\nJVMOHTPZYJ\nWTARFJLZVV\nLVJCWOURVW\nLCLQFJYRVV\nQVBVGNJRYF\nNTZGHOLRYE\nMGQKHOUPYJ\nRRSSBXPZYJ\nRYCRGTLZYJ\nJRDEGNKRVW\nRZKFGHYRVG\nMDJBFNIZYG\nMPLWHXIZYE\nSRZMHMURVE\nMTEBBMRZYJ\nJPJIFOLZYM",
"output": "919913906"
},
{
"input": "40 7\nPNTVVER\nPAHTQDR\nRXMJVAS\nVIQNLYC\nILPUSVX\nYJOXQDJ\nSEFODTO\nOTJMREL\nLIQRZGD\nLBJJPOR\nRUTYHQO\nRIWEPBD\nKQUMFIB\nISTRRYH\nXBTOTGK\nRFQODEY\nHDSTZTP\nYCXFAGL\nAREGRFU\nLELZUYU\nGVABDKH\nFJAMMME\nACVULXE\nJHVPJAS\nAAQNMBX\nJJGUCXG\nOQATILQ\nNEOSHJM\nHFLWOFM\nICYEQHY\nFACGLYP\nPLLXJEQ\nDCHXYPB\nAGDDZJJ\nLSQRXTN\nHDQZXIY\nNAHDDWW\nQCMXRQN\nFDUDSZO\nHKBEVTW",
"output": "206575993"
},
{
"input": "2 2\nAA\nBB",
"output": "4"
},
{
"input": "1 10\nAAAAAAAAAA",
"output": "1"
},
{
"input": "2 8\nAAAAAAAA\nBBBBBBBB",
"output": "256"
},
{
"input": "10 10\nAAAAAAAAAA\nBBBBBBBBBB\nCCCCCCCCCC\nDDDDDDDDDD\nAAAAAAAAAA\nBBBBBBBBBB\nCCCCCCCCCC\nDDDDDDDDDD\nAAAAAAAAAA\nBBBBBBBBBB",
"output": "1048576"
},
{
"input": "1 20\nAAAAAAAAAAAAAAAAAAAA",
"output": "1"
},
{
"input": "20 1\nA\nB\nC\nD\nE\nF\nG\nA\nB\nC\nD\nE\nF\nG\nA\nB\nC\nD\nE\nF",
"output": "7"
},
{
"input": "5 60\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB\nCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC\nDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD\nEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE",
"output": "449874206"
},
{
"input": "50 4\nAAAA\nBBBB\nCCCC\nDDDD\nEEEE\nFFFF\nGGGG\nHHHH\nIIII\nJJJJ\nAAAA\nBBBB\nCCCC\nDDDD\nEEEE\nFFFF\nGGGG\nHHHH\nIIII\nJJJJ\nAAAA\nBBBB\nCCCC\nDDDD\nEEEE\nFFFF\nGGGG\nHHHH\nIIII\nJJJJ\nAAAA\nBBBB\nCCCC\nDDDD\nEEEE\nFFFF\nGGGG\nHHHH\nIIII\nJJJJ\nAAAA\nBBBB\nCCCC\nDDDD\nEEEE\nFFFF\nGGGG\nHHHH\nIIII\nJJJJ",
"output": "10000"
},
{
"input": "1 100\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA",
"output": "1"
},
{
"input": "100 1\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA\nA",
"output": "1"
},
{
"input": "100 1\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB\nA\nB",
"output": "2"
},
{
"input": "100 1\nA\nB\nC\nD\nE\nF\nG\nH\nI\nJ\nK\nL\nM\nN\nA\nB\nC\nD\nE\nF\nG\nH\nI\nJ\nK\nL\nM\nN\nA\nB\nC\nD\nE\nF\nG\nH\nI\nJ\nK\nL\nM\nN\nA\nB\nC\nD\nE\nF\nG\nH\nI\nJ\nK\nL\nM\nN\nA\nB\nC\nD\nE\nF\nG\nH\nI\nJ\nK\nL\nM\nN\nA\nB\nC\nD\nE\nF\nG\nH\nI\nJ\nK\nL\nM\nN\nA\nB\nC\nD\nE\nF\nG\nH\nI\nJ\nK\nL\nM\nN\nA\nB",
"output": "14"
},
{
"input": "100 1\nA\nB\nC\nD\nE\nF\nG\nH\nI\nJ\nK\nL\nM\nN\nO\nP\nQ\nR\nS\nT\nU\nV\nW\nX\nY\nZ\nA\nB\nC\nD\nE\nF\nG\nH\nI\nJ\nK\nL\nM\nN\nO\nP\nQ\nR\nS\nT\nU\nV\nW\nX\nY\nZ\nA\nB\nC\nD\nE\nF\nG\nH\nI\nJ\nK\nL\nM\nN\nO\nP\nQ\nR\nS\nT\nU\nV\nW\nX\nY\nZ\nA\nB\nC\nD\nE\nF\nG\nH\nI\nJ\nK\nL\nM\nN\nO\nP\nQ\nR\nS\nT\nU\nV",
"output": "26"
}
] | 1,644,891,610
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 60
| 186
| 2,252,800
|
mod = 1000000007
n, m = map(int, input().split())
lista = []
arr = [0] * m
for c in range(n):
name = input()
lista.append(name)
for c in range(m):
con = set()
for v in range(n):
con.add(lista[v][c])
arr[c] += len(con)
resp = 1
for c in range(m):
resp = (resp * arr[c]) % mod
print(resp)
|
Title: Pocket Book
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day little Vasya found mom's pocket book. The book had *n* names of her friends and unusually enough, each name was exactly *m* letters long. Let's number the names from 1 to *n* in the order in which they are written.
As mom wasn't home, Vasya decided to play with names: he chose three integers *i*, *j*, *k* (1<=≤<=*i*<=<<=*j*<=≤<=*n*, 1<=≤<=*k*<=≤<=*m*), then he took names number *i* and *j* and swapped their prefixes of length *k*. For example, if we take names "CBDAD" and "AABRD" and swap their prefixes with the length of 3, the result will be names "AABAD" and "CBDRD".
You wonder how many different names Vasya can write instead of name number 1, if Vasya is allowed to perform any number of the described actions. As Vasya performs each action, he chooses numbers *i*, *j*, *k* independently from the previous moves and his choice is based entirely on his will. The sought number can be very large, so you should only find it modulo 1000000007 (109<=+<=7).
Input Specification:
The first input line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of names and the length of each name, correspondingly. Then *n* lines contain names, each name consists of exactly *m* uppercase Latin letters.
Output Specification:
Print the single number — the number of different names that could end up in position number 1 in the pocket book after the applying the procedures described above. Print the number modulo 1000000007 (109<=+<=7).
Demo Input:
['2 3\nAAB\nBAA\n', '4 5\nABABA\nBCGDG\nAAAAA\nYABSA\n']
Demo Output:
['4\n', '216\n']
Note:
In the first sample Vasya can get the following names in the position number 1: "AAB", "AAA", "BAA" and "BAB".
|
```python
mod = 1000000007
n, m = map(int, input().split())
lista = []
arr = [0] * m
for c in range(n):
name = input()
lista.append(name)
for c in range(m):
con = set()
for v in range(n):
con.add(lista[v][c])
arr[c] += len(con)
resp = 1
for c in range(m):
resp = (resp * arr[c]) % mod
print(resp)
```
| 3
|
|
747
|
A
|
Display Size
|
PROGRAMMING
| 800
|
[
"brute force",
"math"
] | null | null |
A big company decided to launch a new series of rectangular displays, and decided that the display must have exactly *n* pixels.
Your task is to determine the size of the rectangular display — the number of lines (rows) of pixels *a* and the number of columns of pixels *b*, so that:
- there are exactly *n* pixels on the display; - the number of rows does not exceed the number of columns, it means *a*<=≤<=*b*; - the difference *b*<=-<=*a* is as small as possible.
|
The first line contains the positive integer *n* (1<=≤<=*n*<=≤<=106) — the number of pixels display should have.
|
Print two integers — the number of rows and columns on the display.
|
[
"8\n",
"64\n",
"5\n",
"999999\n"
] |
[
"2 4\n",
"8 8\n",
"1 5\n",
"999 1001\n"
] |
In the first example the minimum possible difference equals 2, so on the display should be 2 rows of 4 pixels.
In the second example the minimum possible difference equals 0, so on the display should be 8 rows of 8 pixels.
In the third example the minimum possible difference equals 4, so on the display should be 1 row of 5 pixels.
| 500
|
[
{
"input": "8",
"output": "2 4"
},
{
"input": "64",
"output": "8 8"
},
{
"input": "5",
"output": "1 5"
},
{
"input": "999999",
"output": "999 1001"
},
{
"input": "716539",
"output": "97 7387"
},
{
"input": "1",
"output": "1 1"
},
{
"input": "2",
"output": "1 2"
},
{
"input": "3",
"output": "1 3"
},
{
"input": "4",
"output": "2 2"
},
{
"input": "6",
"output": "2 3"
},
{
"input": "7",
"output": "1 7"
},
{
"input": "9",
"output": "3 3"
},
{
"input": "10",
"output": "2 5"
},
{
"input": "11",
"output": "1 11"
},
{
"input": "12",
"output": "3 4"
},
{
"input": "15",
"output": "3 5"
},
{
"input": "100",
"output": "10 10"
},
{
"input": "101",
"output": "1 101"
},
{
"input": "169",
"output": "13 13"
},
{
"input": "179",
"output": "1 179"
},
{
"input": "190",
"output": "10 19"
},
{
"input": "1000",
"output": "25 40"
},
{
"input": "10000",
"output": "100 100"
},
{
"input": "10001",
"output": "73 137"
},
{
"input": "100000",
"output": "250 400"
},
{
"input": "100001",
"output": "11 9091"
},
{
"input": "1000000",
"output": "1000 1000"
},
{
"input": "999983",
"output": "1 999983"
},
{
"input": "524288",
"output": "512 1024"
},
{
"input": "954493",
"output": "971 983"
},
{
"input": "966289",
"output": "983 983"
},
{
"input": "944663",
"output": "961 983"
},
{
"input": "912673",
"output": "97 9409"
},
{
"input": "732641",
"output": "679 1079"
},
{
"input": "232897",
"output": "343 679"
},
{
"input": "16807",
"output": "49 343"
},
{
"input": "999958",
"output": "2 499979"
},
{
"input": "990151",
"output": "1 990151"
},
{
"input": "997002",
"output": "998 999"
},
{
"input": "20",
"output": "4 5"
},
{
"input": "20261",
"output": "1 20261"
},
{
"input": "999123",
"output": "3 333041"
},
{
"input": "901841",
"output": "1 901841"
}
] | 1,516,992,425
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 43
| 77
| 5,632,000
|
n = int(input())
a,b = 1,n
for i in range(2,int(n**.5)+1):
if n%i == 0 :
a,b = i,n//i
print(a,b)
|
Title: Display Size
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A big company decided to launch a new series of rectangular displays, and decided that the display must have exactly *n* pixels.
Your task is to determine the size of the rectangular display — the number of lines (rows) of pixels *a* and the number of columns of pixels *b*, so that:
- there are exactly *n* pixels on the display; - the number of rows does not exceed the number of columns, it means *a*<=≤<=*b*; - the difference *b*<=-<=*a* is as small as possible.
Input Specification:
The first line contains the positive integer *n* (1<=≤<=*n*<=≤<=106) — the number of pixels display should have.
Output Specification:
Print two integers — the number of rows and columns on the display.
Demo Input:
['8\n', '64\n', '5\n', '999999\n']
Demo Output:
['2 4\n', '8 8\n', '1 5\n', '999 1001\n']
Note:
In the first example the minimum possible difference equals 2, so on the display should be 2 rows of 4 pixels.
In the second example the minimum possible difference equals 0, so on the display should be 8 rows of 8 pixels.
In the third example the minimum possible difference equals 4, so on the display should be 1 row of 5 pixels.
|
```python
n = int(input())
a,b = 1,n
for i in range(2,int(n**.5)+1):
if n%i == 0 :
a,b = i,n//i
print(a,b)
```
| 3
|
|
888
|
B
|
Buggy Robot
|
PROGRAMMING
| 1,000
|
[
"greedy"
] | null | null |
Ivan has a robot which is situated on an infinite grid. Initially the robot is standing in the starting cell (0,<=0). The robot can process commands. There are four types of commands it can perform:
- U — move from the cell (*x*,<=*y*) to (*x*,<=*y*<=+<=1); - D — move from (*x*,<=*y*) to (*x*,<=*y*<=-<=1); - L — move from (*x*,<=*y*) to (*x*<=-<=1,<=*y*); - R — move from (*x*,<=*y*) to (*x*<=+<=1,<=*y*).
Ivan entered a sequence of *n* commands, and the robot processed it. After this sequence the robot ended up in the starting cell (0,<=0), but Ivan doubts that the sequence is such that after performing it correctly the robot ends up in the same cell. He thinks that some commands were ignored by robot. To acknowledge whether the robot is severely bugged, he needs to calculate the maximum possible number of commands that were performed correctly. Help Ivan to do the calculations!
|
The first line contains one number *n* — the length of sequence of commands entered by Ivan (1<=≤<=*n*<=≤<=100).
The second line contains the sequence itself — a string consisting of *n* characters. Each character can be U, D, L or R.
|
Print the maximum possible number of commands from the sequence the robot could perform to end up in the starting cell.
|
[
"4\nLDUR\n",
"5\nRRRUU\n",
"6\nLLRRRR\n"
] |
[
"4\n",
"0\n",
"4\n"
] |
none
| 0
|
[
{
"input": "4\nLDUR",
"output": "4"
},
{
"input": "5\nRRRUU",
"output": "0"
},
{
"input": "6\nLLRRRR",
"output": "4"
},
{
"input": "88\nLLUUULRDRRURDDLURRLRDRLLRULRUUDDLLLLRRDDURDURRLDURRLDRRRUULDDLRRRDDRRLUULLURDURUDDDDDLDR",
"output": "76"
},
{
"input": "89\nLDLLLDRDUDURRRRRUDULDDDLLUDLRLRLRLDLDUULRDUDLRRDLUDLURRDDRRDLDUDUUURUUUDRLUDUDLURDLDLLDDU",
"output": "80"
},
{
"input": "90\nRRRDUULLLRDUUDDRLDLRLUDURDRDUUURUURDDRRRURLDDDUUDRLLLULURDRDRURLDRRRRUULDULDDLLLRRLRDLLLLR",
"output": "84"
},
{
"input": "91\nRLDRLRRLLDLULULLURULLRRULUDUULLUDULDUULURUDRUDUURDULDUDDUUUDRRUUDLLRULRULURLDRDLDRURLLLRDDD",
"output": "76"
},
{
"input": "92\nRLRDDLULRLLUURRDDDLDDDLDDUURRRULLRDULDULLLUUULDUDLRLRRDRDRDDULDRLUDRDULDRURUDUULLRDRRLLDRLRR",
"output": "86"
},
{
"input": "93\nRLLURLULRURDDLUURLUDDRDLUURLRDLRRRDUULLRDRRLRLDURRDLLRDDLLLDDDLDRRURLLDRUDULDDRRULRRULRLDRDLR",
"output": "84"
},
{
"input": "94\nRDULDDDLULRDRUDRUUDUUDRRRULDRRUDURUULRDUUDLULLLUDURRDRDLUDRULRRRULUURUDDDDDUDLLRDLDRLLRUUURLUL",
"output": "86"
},
{
"input": "95\nRDLUUULLUURDDRLDLLRRRULRLRDULULRULRUDURLULDDDRLURLDRULDUDUUULLRDDURUULULLDDLDRDRLLLURLRDLLDDDDU",
"output": "86"
},
{
"input": "96\nRDDRLRLLDDULRLRURUDLRLDUDRURLLUUDLLURDLRRUURDRRUDRURLLDLLRDURDURLRLUDURULLLRDUURULUUULRRURRDLURL",
"output": "84"
},
{
"input": "97\nRURDDLRLLRULUDURDLRLLUUDURRLLUDLLLDUDRUULDRUUURURULRDLDRRLLUUUDLLLDDLLLLRLLDUDRRDLLUDLRURUDULRLUR",
"output": "82"
},
{
"input": "98\nRUDURLULLDDLLRDLLRDDLLLLRLDDDDRRRDDRRURLDRLLRUUUDLUUUDDDUDRUURLURUUDUUDRULRRULLRRLRULLULDLUURLULRD",
"output": "92"
},
{
"input": "99\nRRULLDULRRDRULLDUDRUDDDRLLUUDRDDUDURLDDRUUDRRUUURRRURDDLDUURDLRLURRDDLUDDLUDURDRUDDURLURURLRUDRURLD",
"output": "86"
},
{
"input": "100\nUDRLRRLLRRLRRRDDLLDDDLULLDDLURUURUULUDDDRDDLLRDLLUURLRDRLRRLRLLLULDUDDUURRLRDULDRDURRRRRRULDRRDLDRRL",
"output": "88"
},
{
"input": "1\nU",
"output": "0"
},
{
"input": "5\nUUULD",
"output": "2"
},
{
"input": "1\nD",
"output": "0"
},
{
"input": "5\nURLUL",
"output": "2"
},
{
"input": "5\nDDDRU",
"output": "2"
},
{
"input": "2\nLR",
"output": "2"
},
{
"input": "8\nDDRDLDUR",
"output": "4"
},
{
"input": "6\nLLLLUD",
"output": "2"
},
{
"input": "13\nRRRLLLLLLLLLL",
"output": "6"
}
] | 1,691,549,908
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 6
| 46
| 0
|
int(input())
s = input()
print(2*(min(s.count('L'),s.count('U'))+min(s.count('R'), s.count('L'))))
|
Title: Buggy Robot
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ivan has a robot which is situated on an infinite grid. Initially the robot is standing in the starting cell (0,<=0). The robot can process commands. There are four types of commands it can perform:
- U — move from the cell (*x*,<=*y*) to (*x*,<=*y*<=+<=1); - D — move from (*x*,<=*y*) to (*x*,<=*y*<=-<=1); - L — move from (*x*,<=*y*) to (*x*<=-<=1,<=*y*); - R — move from (*x*,<=*y*) to (*x*<=+<=1,<=*y*).
Ivan entered a sequence of *n* commands, and the robot processed it. After this sequence the robot ended up in the starting cell (0,<=0), but Ivan doubts that the sequence is such that after performing it correctly the robot ends up in the same cell. He thinks that some commands were ignored by robot. To acknowledge whether the robot is severely bugged, he needs to calculate the maximum possible number of commands that were performed correctly. Help Ivan to do the calculations!
Input Specification:
The first line contains one number *n* — the length of sequence of commands entered by Ivan (1<=≤<=*n*<=≤<=100).
The second line contains the sequence itself — a string consisting of *n* characters. Each character can be U, D, L or R.
Output Specification:
Print the maximum possible number of commands from the sequence the robot could perform to end up in the starting cell.
Demo Input:
['4\nLDUR\n', '5\nRRRUU\n', '6\nLLRRRR\n']
Demo Output:
['4\n', '0\n', '4\n']
Note:
none
|
```python
int(input())
s = input()
print(2*(min(s.count('L'),s.count('U'))+min(s.count('R'), s.count('L'))))
```
| 0
|
|
427
|
A
|
Police Recruits
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
The police department of your city has just started its journey. Initially, they don’t have any manpower. So, they started hiring new recruits in groups.
Meanwhile, crimes keeps occurring within the city. One member of the police force can investigate only one crime during his/her lifetime.
If there is no police officer free (isn't busy with crime) during the occurrence of a crime, it will go untreated.
Given the chronological order of crime occurrences and recruit hirings, find the number of crimes which will go untreated.
|
The first line of input will contain an integer *n* (1<=≤<=*n*<=≤<=105), the number of events. The next line will contain *n* space-separated integers.
If the integer is -1 then it means a crime has occurred. Otherwise, the integer will be positive, the number of officers recruited together at that time. No more than 10 officers will be recruited at a time.
|
Print a single integer, the number of crimes which will go untreated.
|
[
"3\n-1 -1 1\n",
"8\n1 -1 1 -1 -1 1 1 1\n",
"11\n-1 -1 2 -1 -1 -1 -1 -1 -1 -1 -1\n"
] |
[
"2\n",
"1\n",
"8\n"
] |
Lets consider the second example:
1. Firstly one person is hired. 1. Then crime appears, the last hired person will investigate this crime. 1. One more person is hired. 1. One more crime appears, the last hired person will investigate this crime. 1. Crime appears. There is no free policeman at the time, so this crime will go untreated. 1. One more person is hired. 1. One more person is hired. 1. One more person is hired.
The answer is one, as one crime (on step 5) will go untreated.
| 500
|
[
{
"input": "3\n-1 -1 1",
"output": "2"
},
{
"input": "8\n1 -1 1 -1 -1 1 1 1",
"output": "1"
},
{
"input": "11\n-1 -1 2 -1 -1 -1 -1 -1 -1 -1 -1",
"output": "8"
},
{
"input": "7\n-1 -1 1 1 -1 -1 1",
"output": "2"
},
{
"input": "21\n-1 -1 -1 -1 -1 3 2 -1 6 -1 -1 2 1 -1 2 2 1 6 5 -1 5",
"output": "5"
},
{
"input": "98\n-1 -1 1 -1 -1 -1 -1 1 -1 -1 1 -1 -1 1 -1 1 1 1 -1 1 1 1 1 1 -1 1 -1 -1 -1 -1 1 -1 -1 1 1 -1 1 1 1 -1 -1 -1 -1 -1 -1 1 -1 -1 -1 1 -1 1 -1 1 -1 1 1 1 1 1 1 1 -1 -1 1 1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 1 -1 1 1 1 -1 1 1 -1 -1 -1 1 1 1 -1 -1 -1 1 -1 1 1",
"output": "13"
},
{
"input": "3\n-1 5 4",
"output": "1"
},
{
"input": "146\n4 -1 -1 -1 -1 -1 -1 -1 -1 -1 4 -1 3 -1 3 -1 -1 1 4 -1 2 -1 -1 3 -1 -1 -1 4 1 -1 -1 -1 4 -1 -1 -1 -1 -1 -1 3 2 -1 3 2 4 5 2 4 1 5 -1 -1 2 -1 -1 1 -1 5 3 -1 1 2 2 3 5 3 -1 -1 3 -1 -1 3 5 5 -1 -1 5 -1 4 4 1 -1 -1 -1 2 1 -1 -1 -1 2 5 3 -1 -1 -1 3 -1 5 4 -1 1 -1 -1 3 -1 -1 3 1 1 2 -1 -1 -1 1 3 1 -1 2 -1 -1 5 5 -1 -1 3 4 5 1 -1 2 -1 -1 -1 3 -1 5 3 2 -1 2 -1 -1 5 -1 3 -1",
"output": "5"
},
{
"input": "1\n2",
"output": "0"
},
{
"input": "1\n-1",
"output": "1"
},
{
"input": "2\n1 -1",
"output": "0"
},
{
"input": "2\n-1 1",
"output": "1"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "2\n-1 -1",
"output": "2"
},
{
"input": "4\n10 -1 -1 -1",
"output": "0"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "4\n-1 -1 1 1",
"output": "2"
}
] | 1,689,868,510
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
n=int(input())
a=list(map9int,input().split()))
pol=0
crime=0
for i in a:
if i<0 and pol==o:
crime+=1
elif i>0:
pol+=i
elif(i<0) and (pol>0):
pol-=1
print(crime)
|
Title: Police Recruits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The police department of your city has just started its journey. Initially, they don’t have any manpower. So, they started hiring new recruits in groups.
Meanwhile, crimes keeps occurring within the city. One member of the police force can investigate only one crime during his/her lifetime.
If there is no police officer free (isn't busy with crime) during the occurrence of a crime, it will go untreated.
Given the chronological order of crime occurrences and recruit hirings, find the number of crimes which will go untreated.
Input Specification:
The first line of input will contain an integer *n* (1<=≤<=*n*<=≤<=105), the number of events. The next line will contain *n* space-separated integers.
If the integer is -1 then it means a crime has occurred. Otherwise, the integer will be positive, the number of officers recruited together at that time. No more than 10 officers will be recruited at a time.
Output Specification:
Print a single integer, the number of crimes which will go untreated.
Demo Input:
['3\n-1 -1 1\n', '8\n1 -1 1 -1 -1 1 1 1\n', '11\n-1 -1 2 -1 -1 -1 -1 -1 -1 -1 -1\n']
Demo Output:
['2\n', '1\n', '8\n']
Note:
Lets consider the second example:
1. Firstly one person is hired. 1. Then crime appears, the last hired person will investigate this crime. 1. One more person is hired. 1. One more crime appears, the last hired person will investigate this crime. 1. Crime appears. There is no free policeman at the time, so this crime will go untreated. 1. One more person is hired. 1. One more person is hired. 1. One more person is hired.
The answer is one, as one crime (on step 5) will go untreated.
|
```python
n=int(input())
a=list(map9int,input().split()))
pol=0
crime=0
for i in a:
if i<0 and pol==o:
crime+=1
elif i>0:
pol+=i
elif(i<0) and (pol>0):
pol-=1
print(crime)
```
| -1
|
|
68
|
A
|
Irrational problem
|
PROGRAMMING
| 1,100
|
[
"implementation",
"number theory"
] |
A. Irrational problem
|
2
|
256
|
Little Petya was given this problem for homework:
You are given function (here represents the operation of taking the remainder). His task is to count the number of integers *x* in range [*a*;*b*] with property *f*(*x*)<==<=*x*.
It is a pity that Petya forgot the order in which the remainders should be taken and wrote down only 4 numbers. Each of 24 possible orders of taking the remainder has equal probability of being chosen. For example, if Petya has numbers 1, 2, 3, 4 then he can take remainders in that order or first take remainder modulo 4, then modulo 2, 3, 1. There also are 22 other permutations of these numbers that represent orders in which remainder can be taken. In this problem 4 numbers wrote down by Petya will be pairwise distinct.
Now it is impossible for Petya to complete the task given by teacher but just for fun he decided to find the number of integers with property that probability that *f*(*x*)<==<=*x* is not less than 31.4159265352718281828459045%. In other words, Petya will pick up the number *x* if there exist at least 7 permutations of numbers *p*1,<=*p*2,<=*p*3,<=*p*4, for which *f*(*x*)<==<=*x*.
|
First line of the input will contain 6 integers, separated by spaces: *p*1,<=*p*2,<=*p*3,<=*p*4,<=*a*,<=*b* (1<=≤<=*p*1,<=*p*2,<=*p*3,<=*p*4<=≤<=1000,<=0<=≤<=*a*<=≤<=*b*<=≤<=31415).
It is guaranteed that numbers *p*1,<=*p*2,<=*p*3,<=*p*4 will be pairwise distinct.
|
Output the number of integers in the given range that have the given property.
|
[
"2 7 1 8 2 8\n",
"20 30 40 50 0 100\n",
"31 41 59 26 17 43\n"
] |
[
"0\n",
"20\n",
"9\n"
] |
none
| 500
|
[
{
"input": "2 7 1 8 2 8",
"output": "0"
},
{
"input": "20 30 40 50 0 100",
"output": "20"
},
{
"input": "31 41 59 26 17 43",
"output": "9"
},
{
"input": "1 2 3 4 0 0",
"output": "1"
},
{
"input": "1 2 3 4 1 1",
"output": "0"
},
{
"input": "1 2 999 1000 30 40",
"output": "0"
},
{
"input": "17 18 19 20 17 20",
"output": "0"
},
{
"input": "17 18 19 20 16 20",
"output": "1"
},
{
"input": "41 449 328 474 150 709",
"output": "0"
},
{
"input": "467 329 936 440 117 700",
"output": "212"
},
{
"input": "258 811 952 491 931 993",
"output": "0"
},
{
"input": "823 431 359 590 153 899",
"output": "206"
},
{
"input": "292 370 404 698 699 876",
"output": "0"
},
{
"input": "442 705 757 527 868 893",
"output": "0"
},
{
"input": "642 273 18 885 675 788",
"output": "0"
},
{
"input": "291 303 656 660 126 704",
"output": "165"
},
{
"input": "225 862 522 617 630 725",
"output": "0"
},
{
"input": "17 847 715 732 502 778",
"output": "0"
},
{
"input": "41 449 328 474 15724 19169",
"output": "0"
},
{
"input": "467 329 936 440 5705 28145",
"output": "0"
},
{
"input": "258 811 952 491 2995 11942",
"output": "0"
},
{
"input": "823 431 359 590 153 3902",
"output": "206"
},
{
"input": "292 370 404 698 19718 19895",
"output": "0"
},
{
"input": "442 705 757 527 1869 19912",
"output": "0"
},
{
"input": "642 273 18 885 23811 28703",
"output": "0"
},
{
"input": "291 303 656 660 7711 15141",
"output": "0"
},
{
"input": "225 862 522 617 1246 1341",
"output": "0"
},
{
"input": "17 847 715 732 778 27529",
"output": "0"
},
{
"input": "997 998 999 1000 0 31415",
"output": "997"
},
{
"input": "1 2 3 4 0 31415",
"output": "1"
},
{
"input": "541 931 822 948 131 193",
"output": "63"
},
{
"input": "956 800 909 916 89 194",
"output": "106"
},
{
"input": "735 794 942 991 419 490",
"output": "72"
},
{
"input": "818 926 827 575 153 395",
"output": "243"
},
{
"input": "792 858 887 679 179 356",
"output": "178"
},
{
"input": "937 683 742 515 366 373",
"output": "8"
},
{
"input": "616 747 501 875 146 264",
"output": "119"
},
{
"input": "760 773 638 655 111 196",
"output": "86"
},
{
"input": "697 855 997 589 97 192",
"output": "96"
},
{
"input": "998 834 706 722 277 475",
"output": "199"
},
{
"input": "100 101 102 103 10 20",
"output": "11"
}
] | 1,686,378,708
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 41
| 92
| 0
|
a , b , c , d , x , y = map(int,input().split())
m1 = min(a,b,c,d)
answer = 0
for i in range(x,y+1):
if i == i%m1:
answer+=1
print(answer)
|
Title: Irrational problem
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Little Petya was given this problem for homework:
You are given function (here represents the operation of taking the remainder). His task is to count the number of integers *x* in range [*a*;*b*] with property *f*(*x*)<==<=*x*.
It is a pity that Petya forgot the order in which the remainders should be taken and wrote down only 4 numbers. Each of 24 possible orders of taking the remainder has equal probability of being chosen. For example, if Petya has numbers 1, 2, 3, 4 then he can take remainders in that order or first take remainder modulo 4, then modulo 2, 3, 1. There also are 22 other permutations of these numbers that represent orders in which remainder can be taken. In this problem 4 numbers wrote down by Petya will be pairwise distinct.
Now it is impossible for Petya to complete the task given by teacher but just for fun he decided to find the number of integers with property that probability that *f*(*x*)<==<=*x* is not less than 31.4159265352718281828459045%. In other words, Petya will pick up the number *x* if there exist at least 7 permutations of numbers *p*1,<=*p*2,<=*p*3,<=*p*4, for which *f*(*x*)<==<=*x*.
Input Specification:
First line of the input will contain 6 integers, separated by spaces: *p*1,<=*p*2,<=*p*3,<=*p*4,<=*a*,<=*b* (1<=≤<=*p*1,<=*p*2,<=*p*3,<=*p*4<=≤<=1000,<=0<=≤<=*a*<=≤<=*b*<=≤<=31415).
It is guaranteed that numbers *p*1,<=*p*2,<=*p*3,<=*p*4 will be pairwise distinct.
Output Specification:
Output the number of integers in the given range that have the given property.
Demo Input:
['2 7 1 8 2 8\n', '20 30 40 50 0 100\n', '31 41 59 26 17 43\n']
Demo Output:
['0\n', '20\n', '9\n']
Note:
none
|
```python
a , b , c , d , x , y = map(int,input().split())
m1 = min(a,b,c,d)
answer = 0
for i in range(x,y+1):
if i == i%m1:
answer+=1
print(answer)
```
| 3.977
|
96
|
A
|
Football
|
PROGRAMMING
| 900
|
[
"implementation",
"strings"
] |
A. Football
|
2
|
256
|
Petya loves football very much. One day, as he was watching a football match, he was writing the players' current positions on a piece of paper. To simplify the situation he depicted it as a string consisting of zeroes and ones. A zero corresponds to players of one team; a one corresponds to players of another team. If there are at least 7 players of some team standing one after another, then the situation is considered dangerous. For example, the situation 00100110111111101 is dangerous and 11110111011101 is not. You are given the current situation. Determine whether it is dangerous or not.
|
The first input line contains a non-empty string consisting of characters "0" and "1", which represents players. The length of the string does not exceed 100 characters. There's at least one player from each team present on the field.
|
Print "YES" if the situation is dangerous. Otherwise, print "NO".
|
[
"001001\n",
"1000000001\n"
] |
[
"NO\n",
"YES\n"
] |
none
| 500
|
[
{
"input": "001001",
"output": "NO"
},
{
"input": "1000000001",
"output": "YES"
},
{
"input": "00100110111111101",
"output": "YES"
},
{
"input": "11110111111111111",
"output": "YES"
},
{
"input": "01",
"output": "NO"
},
{
"input": "10100101",
"output": "NO"
},
{
"input": "1010010100000000010",
"output": "YES"
},
{
"input": "101010101",
"output": "NO"
},
{
"input": "000000000100000000000110101100000",
"output": "YES"
},
{
"input": "100001000000110101100000",
"output": "NO"
},
{
"input": "100001000011010110000",
"output": "NO"
},
{
"input": "010",
"output": "NO"
},
{
"input": "10101011111111111111111111111100",
"output": "YES"
},
{
"input": "1001101100",
"output": "NO"
},
{
"input": "1001101010",
"output": "NO"
},
{
"input": "1111100111",
"output": "NO"
},
{
"input": "00110110001110001111",
"output": "NO"
},
{
"input": "11110001001111110001",
"output": "NO"
},
{
"input": "10001111001011111101",
"output": "NO"
},
{
"input": "10000010100000001000110001010100001001001010011",
"output": "YES"
},
{
"input": "01111011111010111100101100001011001010111110000010",
"output": "NO"
},
{
"input": "00100000100100101110011001011011101110110110010100",
"output": "NO"
},
{
"input": "10110100110001001011110101110010100010000000000100101010111110111110100011",
"output": "YES"
},
{
"input": "00011101010101111001011011001101101011111101000010100000111000011100101011",
"output": "NO"
},
{
"input": "01110000110100110101110100111000101101011101011110110100100111100001110111",
"output": "NO"
},
{
"input": "11110110011000100111100111101101011111110100010101011011111101110110110111",
"output": "YES"
},
{
"input": "100100010101110010001011001110100011100010011110100101100011010001001010001001101111001100",
"output": "NO"
},
{
"input": "111110010001011010010011111100110110001111000010100011011100111101111101110010101111011110000001010",
"output": "NO"
},
{
"input": "111110111100010100000100001010111011101011000111011011011010110010100010000101011111000011010011110",
"output": "NO"
},
{
"input": "1011110110111010110111111010010010100011111011110000011000110010011110111010110100011010100010111000",
"output": "NO"
},
{
"input": "0010100111100010110110000011100111110100111110001010000100111111111010111100101101010101001011010110",
"output": "YES"
},
{
"input": "1111010100010100101011101100101101110011000010100010000001111100010011100101010001101111000001011000",
"output": "NO"
},
{
"input": "0101100011001110001110100111100011010101011000000000110110010010111100101111010111100011101100100101",
"output": "YES"
},
{
"input": "0001101110011101110000000010011111101001101111100001001010110000110001100000010001111011011110001101",
"output": "YES"
},
{
"input": "1000010000100000100010000100001000010000100001000010000100001000010000100001000010000100001000010000",
"output": "NO"
},
{
"input": "1000000000100000000010000000001000000000100000000010000000001000000000100000000010000000001000000000",
"output": "YES"
},
{
"input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001",
"output": "YES"
},
{
"input": "1111111111111111111111111111111111111111011111111111111111111111111111111111111111111111111111111111",
"output": "YES"
},
{
"input": "10100101000",
"output": "NO"
},
{
"input": "11110111011101",
"output": "NO"
},
{
"input": "10000000",
"output": "YES"
},
{
"input": "00000001",
"output": "YES"
},
{
"input": "01111111",
"output": "YES"
},
{
"input": "11111110",
"output": "YES"
}
] | 1,672,087,521
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 62
| 0
|
print("YES" if ("0"*7 or "1"*7) in input() else "NO")
|
Title: Football
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Petya loves football very much. One day, as he was watching a football match, he was writing the players' current positions on a piece of paper. To simplify the situation he depicted it as a string consisting of zeroes and ones. A zero corresponds to players of one team; a one corresponds to players of another team. If there are at least 7 players of some team standing one after another, then the situation is considered dangerous. For example, the situation 00100110111111101 is dangerous and 11110111011101 is not. You are given the current situation. Determine whether it is dangerous or not.
Input Specification:
The first input line contains a non-empty string consisting of characters "0" and "1", which represents players. The length of the string does not exceed 100 characters. There's at least one player from each team present on the field.
Output Specification:
Print "YES" if the situation is dangerous. Otherwise, print "NO".
Demo Input:
['001001\n', '1000000001\n']
Demo Output:
['NO\n', 'YES\n']
Note:
none
|
```python
print("YES" if ("0"*7 or "1"*7) in input() else "NO")
```
| 0
|
637
|
B
|
Chat Order
|
PROGRAMMING
| 1,200
|
[
"*special",
"binary search",
"constructive algorithms",
"data structures",
"sortings"
] | null | null |
Polycarp is a big lover of killing time in social networks. A page with a chatlist in his favourite network is made so that when a message is sent to some friend, his friend's chat rises to the very top of the page. The relative order of the other chats doesn't change. If there was no chat with this friend before, then a new chat is simply inserted to the top of the list.
Assuming that the chat list is initially empty, given the sequence of Polycaprus' messages make a list of chats after all of his messages are processed. Assume that no friend wrote any message to Polycarpus.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of Polycarpus' messages. Next *n* lines enlist the message recipients in the order in which the messages were sent. The name of each participant is a non-empty sequence of lowercase English letters of length at most 10.
|
Print all the recipients to who Polycarp talked to in the order of chats with them, from top to bottom.
|
[
"4\nalex\nivan\nroman\nivan\n",
"8\nalina\nmaria\nekaterina\ndarya\ndarya\nekaterina\nmaria\nalina\n"
] |
[
"ivan\nroman\nalex\n",
"alina\nmaria\nekaterina\ndarya\n"
] |
In the first test case Polycarpus first writes to friend by name "alex", and the list looks as follows:
1. alex
Then Polycarpus writes to friend by name "ivan" and the list looks as follows:
1. ivan 1. alex
Polycarpus writes the third message to friend by name "roman" and the list looks as follows:
1. roman 1. ivan 1. alex
Polycarpus writes the fourth message to friend by name "ivan", to who he has already sent a message, so the list of chats changes as follows:
1. ivan 1. roman 1. alex
| 1,000
|
[
{
"input": "4\nalex\nivan\nroman\nivan",
"output": "ivan\nroman\nalex"
},
{
"input": "8\nalina\nmaria\nekaterina\ndarya\ndarya\nekaterina\nmaria\nalina",
"output": "alina\nmaria\nekaterina\ndarya"
},
{
"input": "1\nwdi",
"output": "wdi"
},
{
"input": "2\nypg\nypg",
"output": "ypg"
},
{
"input": "3\nexhll\nexhll\narruapexj",
"output": "arruapexj\nexhll"
},
{
"input": "3\nfv\nle\nle",
"output": "le\nfv"
},
{
"input": "8\nm\nm\nm\nm\nm\nm\nm\nm",
"output": "m"
},
{
"input": "10\nr\nr\ni\nw\nk\nr\nb\nu\nu\nr",
"output": "r\nu\nb\nk\nw\ni"
},
{
"input": "7\ne\nfau\ncmk\nnzs\nby\nwx\ntjmok",
"output": "tjmok\nwx\nby\nnzs\ncmk\nfau\ne"
},
{
"input": "6\nklrj\nwe\nklrj\nwe\nwe\nwe",
"output": "we\nklrj"
},
{
"input": "8\nzncybqmh\naeebef\nzncybqmh\nn\naeebef\nzncybqmh\nzncybqmh\nzncybqmh",
"output": "zncybqmh\naeebef\nn"
},
{
"input": "30\nkqqcbs\nvap\nkymomn\nj\nkqqcbs\nfuzlzoum\nkymomn\ndbh\nfuzlzoum\nkymomn\nvap\nvlgzs\ndbh\nvlgzs\nbvy\ndbh\nkymomn\nkymomn\neoqql\nkymomn\nkymomn\nkqqcbs\nvlgzs\nkqqcbs\nkqqcbs\nfuzlzoum\nvlgzs\nrylgdoo\nvlgzs\nrylgdoo",
"output": "rylgdoo\nvlgzs\nfuzlzoum\nkqqcbs\nkymomn\neoqql\ndbh\nbvy\nvap\nj"
},
{
"input": "40\nji\nv\nv\nns\nji\nn\nji\nv\nfvy\nvje\nns\nvje\nv\nhas\nv\nusm\nhas\nfvy\nvje\nkdb\nn\nv\nji\nji\nn\nhas\nv\nji\nkdb\nr\nvje\nns\nv\nusm\nn\nvje\nhas\nns\nhas\nn",
"output": "n\nhas\nns\nvje\nusm\nv\nr\nkdb\nji\nfvy"
},
{
"input": "50\njcg\nvle\njopb\nepdb\nnkef\nfv\nxj\nufe\nfuy\noqta\ngbc\nyuz\nec\nyji\nkuux\ncwm\ntq\nnno\nhp\nzry\nxxpp\ntjvo\ngyz\nkwo\nvwqz\nyaqc\njnj\nwoav\nqcv\ndcu\ngc\nhovn\nop\nevy\ndc\ntrpu\nyb\nuzfa\npca\noq\nnhxy\nsiqu\nde\nhphy\nc\nwovu\nf\nbvv\ndsik\nlwyg",
"output": "lwyg\ndsik\nbvv\nf\nwovu\nc\nhphy\nde\nsiqu\nnhxy\noq\npca\nuzfa\nyb\ntrpu\ndc\nevy\nop\nhovn\ngc\ndcu\nqcv\nwoav\njnj\nyaqc\nvwqz\nkwo\ngyz\ntjvo\nxxpp\nzry\nhp\nnno\ntq\ncwm\nkuux\nyji\nec\nyuz\ngbc\noqta\nfuy\nufe\nxj\nfv\nnkef\nepdb\njopb\nvle\njcg"
},
{
"input": "100\nvhh\nvhh\nvhh\nfa\nfa\nvhh\nvhh\nvhh\nfa\nfa\nfa\nvhh\nfa\nvhh\nvhh\nvhh\nfa\nvhh\nvhh\nfa\nfa\nfa\nfa\nfa\nfa\nvhh\nfa\nfa\nvhh\nvhh\nvhh\nfa\nfa\nfa\nvhh\nfa\nvhh\nfa\nvhh\nvhh\nfa\nvhh\nfa\nvhh\nvhh\nvhh\nfa\nvhh\nfa\nfa\nvhh\nfa\nvhh\nvhh\nvhh\nvhh\nfa\nvhh\nvhh\nvhh\nvhh\nfa\nvhh\nvhh\nvhh\nvhh\nvhh\nfa\nvhh\nvhh\nfa\nfa\nfa\nvhh\nfa\nfa\nvhh\nfa\nvhh\nfa\nfa\nfa\nfa\nfa\nfa\nvhh\nvhh\nfa\nvhh\nfa\nfa\nvhh\nfa\nfa\nvhh\nfa\nvhh\nvhh\nfa\nvhh",
"output": "vhh\nfa"
},
{
"input": "2\naa\nbb",
"output": "bb\naa"
},
{
"input": "2\naa\na",
"output": "a\naa"
},
{
"input": "3\naa\naa\naa",
"output": "aa"
},
{
"input": "5\naa\na\naa\na\naa",
"output": "aa\na"
},
{
"input": "7\naaaa\naaaa\naaa\na\naa\naaaaaaa\naaa",
"output": "aaa\naaaaaaa\naa\na\naaaa"
},
{
"input": "5\na\naa\naaa\naaaa\na",
"output": "a\naaaa\naaa\naa"
},
{
"input": "12\naaaaa\naaaaaa\naaaa\naaaaaa\naa\naaaa\naaaa\naaaaaa\na\naaa\naaaaaaaa\naa",
"output": "aa\naaaaaaaa\naaa\na\naaaaaa\naaaa\naaaaa"
},
{
"input": "3\na\naa\naaa",
"output": "aaa\naa\na"
},
{
"input": "9\nzzz\nzzzzz\nzzz\nzzzz\nzz\nzzzz\nzzzzz\nzzzz\nzzzzzzz",
"output": "zzzzzzz\nzzzz\nzzzzz\nzz\nzzz"
}
] | 1,457,881,139
| 10,739
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
PRETESTS
| 11
| 3,000
| 3,993,600
|
n=int(input())
mas_name=[0]*n
for i in range(n):
mas_name[i]=input()
mas_name=mas_name[-1::-1]
mas_name = [e for i, e in enumerate(mas_name) if e not in mas_name[:i]]
print("\n".join(map(str, mas_name)))
|
Title: Chat Order
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp is a big lover of killing time in social networks. A page with a chatlist in his favourite network is made so that when a message is sent to some friend, his friend's chat rises to the very top of the page. The relative order of the other chats doesn't change. If there was no chat with this friend before, then a new chat is simply inserted to the top of the list.
Assuming that the chat list is initially empty, given the sequence of Polycaprus' messages make a list of chats after all of his messages are processed. Assume that no friend wrote any message to Polycarpus.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of Polycarpus' messages. Next *n* lines enlist the message recipients in the order in which the messages were sent. The name of each participant is a non-empty sequence of lowercase English letters of length at most 10.
Output Specification:
Print all the recipients to who Polycarp talked to in the order of chats with them, from top to bottom.
Demo Input:
['4\nalex\nivan\nroman\nivan\n', '8\nalina\nmaria\nekaterina\ndarya\ndarya\nekaterina\nmaria\nalina\n']
Demo Output:
['ivan\nroman\nalex\n', 'alina\nmaria\nekaterina\ndarya\n']
Note:
In the first test case Polycarpus first writes to friend by name "alex", and the list looks as follows:
1. alex
Then Polycarpus writes to friend by name "ivan" and the list looks as follows:
1. ivan 1. alex
Polycarpus writes the third message to friend by name "roman" and the list looks as follows:
1. roman 1. ivan 1. alex
Polycarpus writes the fourth message to friend by name "ivan", to who he has already sent a message, so the list of chats changes as follows:
1. ivan 1. roman 1. alex
|
```python
n=int(input())
mas_name=[0]*n
for i in range(n):
mas_name[i]=input()
mas_name=mas_name[-1::-1]
mas_name = [e for i, e in enumerate(mas_name) if e not in mas_name[:i]]
print("\n".join(map(str, mas_name)))
```
| 0
|
|
4
|
A
|
Watermelon
|
PROGRAMMING
| 800
|
[
"brute force",
"math"
] |
A. Watermelon
|
1
|
64
|
One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest and the ripest one, in their opinion. After that the watermelon was weighed, and the scales showed *w* kilos. They rushed home, dying of thirst, and decided to divide the berry, however they faced a hard problem.
Pete and Billy are great fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight.
|
The first (and the only) input line contains integer number *w* (1<=≤<=*w*<=≤<=100) — the weight of the watermelon bought by the boys.
|
Print YES, if the boys can divide the watermelon into two parts, each of them weighing even number of kilos; and NO in the opposite case.
|
[
"8\n"
] |
[
"YES\n"
] |
For example, the boys can divide the watermelon into two parts of 2 and 6 kilos respectively (another variant — two parts of 4 and 4 kilos).
| 0
|
[
{
"input": "8",
"output": "YES"
},
{
"input": "5",
"output": "NO"
},
{
"input": "4",
"output": "YES"
},
{
"input": "3",
"output": "NO"
},
{
"input": "2",
"output": "NO"
},
{
"input": "1",
"output": "NO"
},
{
"input": "7",
"output": "NO"
},
{
"input": "6",
"output": "YES"
},
{
"input": "10",
"output": "YES"
},
{
"input": "9",
"output": "NO"
},
{
"input": "53",
"output": "NO"
},
{
"input": "77",
"output": "NO"
},
{
"input": "32",
"output": "YES"
},
{
"input": "44",
"output": "YES"
},
{
"input": "98",
"output": "YES"
},
{
"input": "99",
"output": "NO"
},
{
"input": "90",
"output": "YES"
},
{
"input": "67",
"output": "NO"
},
{
"input": "100",
"output": "YES"
},
{
"input": "88",
"output": "YES"
}
] | 1,691,553,698
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 7
| 122
| 0
|
w = int(input(""))
wHalf = w / 2
if (w % 2 == 0):
if wHalf % 2 == 0:
print("YES")
else:
print("NO")
else:
print("NO")
|
Title: Watermelon
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest and the ripest one, in their opinion. After that the watermelon was weighed, and the scales showed *w* kilos. They rushed home, dying of thirst, and decided to divide the berry, however they faced a hard problem.
Pete and Billy are great fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight.
Input Specification:
The first (and the only) input line contains integer number *w* (1<=≤<=*w*<=≤<=100) — the weight of the watermelon bought by the boys.
Output Specification:
Print YES, if the boys can divide the watermelon into two parts, each of them weighing even number of kilos; and NO in the opposite case.
Demo Input:
['8\n']
Demo Output:
['YES\n']
Note:
For example, the boys can divide the watermelon into two parts of 2 and 6 kilos respectively (another variant — two parts of 4 and 4 kilos).
|
```python
w = int(input(""))
wHalf = w / 2
if (w % 2 == 0):
if wHalf % 2 == 0:
print("YES")
else:
print("NO")
else:
print("NO")
```
| 0
|
69
|
A
|
Young Physicist
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] |
A. Young Physicist
|
2
|
256
|
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
|
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
|
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
|
[
"3\n4 1 7\n-2 4 -1\n1 -5 -3\n",
"3\n3 -1 7\n-5 2 -4\n2 -1 -3\n"
] |
[
"NO",
"YES"
] |
none
| 500
|
[
{
"input": "3\n4 1 7\n-2 4 -1\n1 -5 -3",
"output": "NO"
},
{
"input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3",
"output": "YES"
},
{
"input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41",
"output": "NO"
},
{
"input": "10\n25 -33 43\n-27 -42 28\n-35 -20 19\n41 -42 -1\n49 -39 -4\n-49 -22 7\n-19 29 41\n8 -27 -43\n8 34 9\n-11 -3 33",
"output": "NO"
},
{
"input": "10\n-6 21 18\n20 -11 -8\n37 -11 41\n-5 8 33\n29 23 32\n30 -33 -11\n39 -49 -36\n28 34 -49\n22 29 -34\n-18 -6 7",
"output": "NO"
},
{
"input": "10\n47 -2 -27\n0 26 -14\n5 -12 33\n2 18 3\n45 -30 -49\n4 -18 8\n-46 -44 -41\n-22 -10 -40\n-35 -21 26\n33 20 38",
"output": "NO"
},
{
"input": "13\n-3 -36 -46\n-11 -50 37\n42 -11 -15\n9 42 44\n-29 -12 24\n3 9 -40\n-35 13 50\n14 43 18\n-13 8 24\n-48 -15 10\n50 9 -50\n21 0 -50\n0 0 -6",
"output": "YES"
},
{
"input": "14\n43 23 17\n4 17 44\n5 -5 -16\n-43 -7 -6\n47 -48 12\n50 47 -45\n2 14 43\n37 -30 15\n4 -17 -11\n17 9 -45\n-50 -3 -8\n-50 0 0\n-50 0 0\n-16 0 0",
"output": "YES"
},
{
"input": "13\n29 49 -11\n38 -11 -20\n25 1 -40\n-11 28 11\n23 -19 1\n45 -41 -17\n-3 0 -19\n-13 -33 49\n-30 0 28\n34 17 45\n-50 9 -27\n-50 0 0\n-37 0 0",
"output": "YES"
},
{
"input": "12\n3 28 -35\n-32 -44 -17\n9 -25 -6\n-42 -22 20\n-19 15 38\n-21 38 48\n-1 -37 -28\n-10 -13 -50\n-5 21 29\n34 28 50\n50 11 -49\n34 0 0",
"output": "YES"
},
{
"input": "37\n-64 -79 26\n-22 59 93\n-5 39 -12\n77 -9 76\n55 -86 57\n83 100 -97\n-70 94 84\n-14 46 -94\n26 72 35\n14 78 -62\n17 82 92\n-57 11 91\n23 15 92\n-80 -1 1\n12 39 18\n-23 -99 -75\n-34 50 19\n-39 84 -7\n45 -30 -39\n-60 49 37\n45 -16 -72\n33 -51 -56\n-48 28 5\n97 91 88\n45 -82 -11\n-21 -15 -90\n-53 73 -26\n-74 85 -90\n-40 23 38\n100 -13 49\n32 -100 -100\n0 -100 -70\n0 -100 0\n0 -100 0\n0 -100 0\n0 -100 0\n0 -37 0",
"output": "YES"
},
{
"input": "4\n68 3 100\n68 21 -100\n-100 -24 0\n-36 0 0",
"output": "YES"
},
{
"input": "33\n-1 -46 -12\n45 -16 -21\n-11 45 -21\n-60 -42 -93\n-22 -45 93\n37 96 85\n-76 26 83\n-4 9 55\n7 -52 -9\n66 8 -85\n-100 -54 11\n-29 59 74\n-24 12 2\n-56 81 85\n-92 69 -52\n-26 -97 91\n54 59 -51\n58 21 -57\n7 68 56\n-47 -20 -51\n-59 77 -13\n-85 27 91\n79 60 -56\n66 -80 5\n21 -99 42\n-31 -29 98\n66 93 76\n-49 45 61\n100 -100 -100\n100 -100 -100\n66 -75 -100\n0 0 -100\n0 0 -87",
"output": "YES"
},
{
"input": "3\n1 2 3\n3 2 1\n0 0 0",
"output": "NO"
},
{
"input": "2\n5 -23 12\n0 0 0",
"output": "NO"
},
{
"input": "1\n0 0 0",
"output": "YES"
},
{
"input": "1\n1 -2 0",
"output": "NO"
},
{
"input": "2\n-23 77 -86\n23 -77 86",
"output": "YES"
},
{
"input": "26\n86 7 20\n-57 -64 39\n-45 6 -93\n-44 -21 100\n-11 -49 21\n73 -71 -80\n-2 -89 56\n-65 -2 7\n5 14 84\n57 41 13\n-12 69 54\n40 -25 27\n-17 -59 0\n64 -91 -30\n-53 9 42\n-54 -8 14\n-35 82 27\n-48 -59 -80\n88 70 79\n94 57 97\n44 63 25\n84 -90 -40\n-100 100 -100\n-92 100 -100\n0 10 -100\n0 0 -82",
"output": "YES"
},
{
"input": "42\n11 27 92\n-18 -56 -57\n1 71 81\n33 -92 30\n82 83 49\n-87 -61 -1\n-49 45 49\n73 26 15\n-22 22 -77\n29 -93 87\n-68 44 -90\n-4 -84 20\n85 67 -6\n-39 26 77\n-28 -64 20\n65 -97 24\n-72 -39 51\n35 -75 -91\n39 -44 -8\n-25 -27 -57\n91 8 -46\n-98 -94 56\n94 -60 59\n-9 -95 18\n-53 -37 98\n-8 -94 -84\n-52 55 60\n15 -14 37\n65 -43 -25\n94 12 66\n-8 -19 -83\n29 81 -78\n-58 57 33\n24 86 -84\n-53 32 -88\n-14 7 3\n89 97 -53\n-5 -28 -91\n-100 100 -6\n-84 100 0\n0 100 0\n0 70 0",
"output": "YES"
},
{
"input": "3\n96 49 -12\n2 -66 28\n-98 17 -16",
"output": "YES"
},
{
"input": "5\n70 -46 86\n-100 94 24\n-27 63 -63\n57 -100 -47\n0 -11 0",
"output": "YES"
},
{
"input": "18\n-86 -28 70\n-31 -89 42\n31 -48 -55\n95 -17 -43\n24 -95 -85\n-21 -14 31\n68 -18 81\n13 31 60\n-15 28 99\n-42 15 9\n28 -61 -62\n-16 71 29\n-28 75 -48\n-77 -67 36\n-100 83 89\n100 100 -100\n57 34 -100\n0 0 -53",
"output": "YES"
},
{
"input": "44\n52 -54 -29\n-82 -5 -94\n-54 43 43\n91 16 71\n7 80 -91\n3 15 29\n-99 -6 -77\n-3 -77 -64\n73 67 34\n25 -10 -18\n-29 91 63\n-72 86 -16\n-68 85 -81\n-3 36 44\n-74 -14 -80\n34 -96 -97\n-76 -78 -33\n-24 44 -58\n98 12 77\n95 -63 -6\n-51 3 -90\n-92 -10 72\n7 3 -68\n57 -53 71\n29 57 -48\n35 -60 10\n79 -70 -61\n-20 77 55\n-86 -15 -35\n84 -88 -18\n100 -42 77\n-20 46 8\n-41 -43 -65\n38 -98 -23\n-100 65 45\n-7 -91 -63\n46 88 -85\n48 59 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 1",
"output": "YES"
},
{
"input": "18\n-14 -64 -91\n-8 -66 -86\n-23 92 -40\n6 -3 -53\n57 41 78\n-79 42 -22\n-88 -17 45\n4 -45 44\n83 -18 -25\n34 86 -92\n75 -30 12\n44 99 11\n-67 -13 72\n22 83 -56\n-37 71 72\n-9 -100 100\n0 -100 31\n0 -58 0",
"output": "YES"
},
{
"input": "23\n-70 37 78\n42 84 6\n28 -94 -24\n-49 76 95\n-67 18 84\n-53 78 -5\n65 -63 -64\n-66 -64 -69\n81 -93 95\n10 52 -79\n-89 -61 -64\n-64 -47 43\n-81 -35 55\n80 82 73\n-60 -81 -18\n49 -10 -19\n-58 70 12\n-24 -15 -93\n98 -93 -54\n-28 -75 11\n100 100 -63\n100 100 0\n56 34 0",
"output": "YES"
},
{
"input": "3\n0 2 -2\n1 -1 3\n-3 0 0",
"output": "NO"
}
] | 1,656,642,517
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 92
| 0
|
n = input()
x = 0
y = 0
z = 0
for i in range(n):
a=map(int,raw_input().split(" "))
x = x+a[0]
y = y+a[1]
z = z+a[2]
if(x==0 and y==0 and z==0):
print("YES")
else:
print("NO")
|
Title: Young Physicist
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
Output Specification:
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Demo Input:
['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n']
Demo Output:
['NO', 'YES']
Note:
none
|
```python
n = input()
x = 0
y = 0
z = 0
for i in range(n):
a=map(int,raw_input().split(" "))
x = x+a[0]
y = y+a[1]
z = z+a[2]
if(x==0 and y==0 and z==0):
print("YES")
else:
print("NO")
```
| -1
|
350
|
A
|
TL
|
PROGRAMMING
| 1,200
|
[
"brute force",
"greedy",
"implementation"
] | null | null |
Valera wanted to prepare a Codesecrof round. He's already got one problem and he wants to set a time limit (TL) on it.
Valera has written *n* correct solutions. For each correct solution, he knows its running time (in seconds). Valera has also wrote *m* wrong solutions and for each wrong solution he knows its running time (in seconds).
Let's suppose that Valera will set *v* seconds TL in the problem. Then we can say that a solution passes the system testing if its running time is at most *v* seconds. We can also say that a solution passes the system testing with some "extra" time if for its running time, *a* seconds, an inequality 2*a*<=≤<=*v* holds.
As a result, Valera decided to set *v* seconds TL, that the following conditions are met:
1. *v* is a positive integer; 1. all correct solutions pass the system testing; 1. at least one correct solution passes the system testing with some "extra" time; 1. all wrong solutions do not pass the system testing; 1. value *v* is minimum among all TLs, for which points 1, 2, 3, 4 hold.
Help Valera and find the most suitable TL or else state that such TL doesn't exist.
|
The first line contains two integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=100). The second line contains *n* space-separated positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100) — the running time of each of the *n* correct solutions in seconds. The third line contains *m* space-separated positive integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=100) — the running time of each of *m* wrong solutions in seconds.
|
If there is a valid TL value, print it. Otherwise, print -1.
|
[
"3 6\n4 5 2\n8 9 6 10 7 11\n",
"3 1\n3 4 5\n6\n"
] |
[
"5",
"-1\n"
] |
none
| 500
|
[
{
"input": "3 6\n4 5 2\n8 9 6 10 7 11",
"output": "5"
},
{
"input": "3 1\n3 4 5\n6",
"output": "-1"
},
{
"input": "2 5\n45 99\n49 41 77 83 45",
"output": "-1"
},
{
"input": "50 50\n18 13 5 34 10 36 36 12 15 11 16 17 14 36 23 45 32 24 31 18 24 32 7 1 31 3 49 8 16 23 3 39 47 43 42 38 40 22 41 1 49 47 9 8 19 15 29 30 16 18\n91 58 86 51 94 94 73 84 98 69 74 56 52 80 88 61 53 99 88 50 55 95 65 84 87 79 51 52 69 60 74 73 93 61 73 59 64 56 95 78 86 72 79 70 93 78 54 61 71 50",
"output": "49"
},
{
"input": "55 44\n93 17 74 15 34 16 41 80 26 54 94 94 86 93 20 44 63 72 39 43 67 4 37 49 76 94 5 51 64 74 11 47 77 97 57 30 42 72 71 26 8 14 67 64 49 57 30 23 40 4 76 78 87 78 79\n38 55 17 65 26 7 36 65 48 28 49 93 18 98 31 90 26 57 1 26 88 56 48 56 23 13 8 67 80 2 51 3 21 33 20 54 2 45 21 36 3 98 62 2",
"output": "-1"
},
{
"input": "32 100\n30 8 4 35 18 41 18 12 33 39 39 18 39 19 33 46 45 33 34 27 14 39 40 21 38 9 42 35 27 10 14 14\n65 49 89 64 47 78 59 52 73 51 84 82 88 63 91 99 67 87 53 99 75 47 85 82 58 47 80 50 65 91 83 90 77 52 100 88 97 74 98 99 50 93 65 61 65 65 65 96 61 51 84 67 79 90 92 83 100 100 100 95 80 54 77 51 98 64 74 62 60 96 73 74 94 55 89 60 92 65 74 79 66 81 53 47 71 51 54 85 74 97 68 72 88 94 100 85 65 63 65 90",
"output": "46"
},
{
"input": "1 50\n7\n65 52 99 78 71 19 96 72 80 15 50 94 20 35 79 95 44 41 45 53 77 50 74 66 59 96 26 84 27 48 56 84 36 78 89 81 67 34 79 74 99 47 93 92 90 96 72 28 78 66",
"output": "14"
},
{
"input": "1 1\n4\n9",
"output": "8"
},
{
"input": "1 1\n2\n4",
"output": "-1"
},
{
"input": "22 56\n49 20 42 68 15 46 98 78 82 8 7 33 50 30 75 96 36 88 35 99 19 87\n15 18 81 24 35 89 25 32 23 3 48 24 52 69 18 32 23 61 48 98 50 38 5 17 70 20 38 32 49 54 68 11 51 81 46 22 19 59 29 38 45 83 18 13 91 17 84 62 25 60 97 32 23 13 83 58",
"output": "-1"
},
{
"input": "1 1\n50\n100",
"output": "-1"
},
{
"input": "1 1\n49\n100",
"output": "98"
},
{
"input": "1 1\n100\n100",
"output": "-1"
},
{
"input": "1 1\n99\n100",
"output": "-1"
},
{
"input": "8 4\n1 2 49 99 99 95 78 98\n100 100 100 100",
"output": "99"
},
{
"input": "68 85\n43 55 2 4 72 45 19 56 53 81 18 90 11 87 47 8 94 88 24 4 67 9 21 70 25 66 65 27 46 13 8 51 65 99 37 43 71 59 71 79 32 56 49 43 57 85 95 81 40 28 60 36 72 81 60 40 16 78 61 37 29 26 15 95 70 27 50 97\n6 6 48 72 54 31 1 50 29 64 93 9 29 93 66 63 25 90 52 1 66 13 70 30 24 87 32 90 84 72 44 13 25 45 31 16 92 60 87 40 62 7 20 63 86 78 73 88 5 36 74 100 64 34 9 5 62 29 58 48 81 46 84 56 27 1 60 14 54 88 31 93 62 7 9 69 27 48 10 5 33 10 53 66 2",
"output": "-1"
},
{
"input": "5 100\n1 1 1 1 1\n77 53 38 29 97 33 64 17 78 100 27 12 42 44 20 24 44 68 58 57 65 90 8 24 4 6 74 68 61 43 25 69 8 62 36 85 67 48 69 30 35 41 42 12 87 66 50 92 53 76 38 67 85 7 80 78 53 76 94 8 37 50 4 100 4 71 10 48 34 47 83 42 25 81 64 72 25 51 53 75 43 98 53 77 94 38 81 15 89 91 72 76 7 36 27 41 88 18 19 75",
"output": "2"
},
{
"input": "3 3\n2 3 4\n8 9 10",
"output": "4"
},
{
"input": "2 1\n2 3\n15",
"output": "4"
},
{
"input": "2 1\n2 4\n4",
"output": "-1"
},
{
"input": "2 3\n4 5\n10 11 12",
"output": "8"
},
{
"input": "3 1\n2 3 3\n5",
"output": "4"
},
{
"input": "2 1\n9 10\n100",
"output": "18"
},
{
"input": "3 3\n3 12 15\n7 8 9",
"output": "-1"
},
{
"input": "2 2\n3 5\n7 8",
"output": "6"
},
{
"input": "3 3\n4 5 6\n10 11 12",
"output": "8"
},
{
"input": "3 5\n2 3 3\n6 6 6 6 2",
"output": "-1"
},
{
"input": "3 6\n4 5 3\n8 9 7 10 7 11",
"output": "6"
},
{
"input": "3 6\n4 5 2\n8 9 6 10 7 4",
"output": "-1"
},
{
"input": "2 1\n4 6\n10",
"output": "8"
},
{
"input": "1 2\n1\n3 1",
"output": "-1"
},
{
"input": "2 1\n5 6\n20",
"output": "10"
},
{
"input": "2 1\n1 5\n5",
"output": "-1"
},
{
"input": "3 2\n10 20 30\n30 40",
"output": "-1"
},
{
"input": "2 2\n5 6\n7 100",
"output": "-1"
},
{
"input": "2 1\n2 5\n7",
"output": "5"
},
{
"input": "1 1\n5\n20",
"output": "10"
},
{
"input": "2 1\n10 11\n100",
"output": "20"
},
{
"input": "1 1\n1\n10",
"output": "2"
},
{
"input": "1 1\n10\n100",
"output": "20"
}
] | 1,598,537,319
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 7
| 278
| 0
|
import sys
reader = (line.rstrip() for line in sys.stdin)
input = reader.__next__
n, m = map(int, input().split())
arr = list(map(int, input().split()))
arr2 = list(map(int, input().split()))
arr.sort()
arr2.sort()
v_max = arr2[0]
v_min = arr[0]
v_max2 = arr[-1]
print(v_max2) if v_max2 >= v_min * 2 and v_max2 < v_max else print("-1")
|
Title: TL
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera wanted to prepare a Codesecrof round. He's already got one problem and he wants to set a time limit (TL) on it.
Valera has written *n* correct solutions. For each correct solution, he knows its running time (in seconds). Valera has also wrote *m* wrong solutions and for each wrong solution he knows its running time (in seconds).
Let's suppose that Valera will set *v* seconds TL in the problem. Then we can say that a solution passes the system testing if its running time is at most *v* seconds. We can also say that a solution passes the system testing with some "extra" time if for its running time, *a* seconds, an inequality 2*a*<=≤<=*v* holds.
As a result, Valera decided to set *v* seconds TL, that the following conditions are met:
1. *v* is a positive integer; 1. all correct solutions pass the system testing; 1. at least one correct solution passes the system testing with some "extra" time; 1. all wrong solutions do not pass the system testing; 1. value *v* is minimum among all TLs, for which points 1, 2, 3, 4 hold.
Help Valera and find the most suitable TL or else state that such TL doesn't exist.
Input Specification:
The first line contains two integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=100). The second line contains *n* space-separated positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100) — the running time of each of the *n* correct solutions in seconds. The third line contains *m* space-separated positive integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=100) — the running time of each of *m* wrong solutions in seconds.
Output Specification:
If there is a valid TL value, print it. Otherwise, print -1.
Demo Input:
['3 6\n4 5 2\n8 9 6 10 7 11\n', '3 1\n3 4 5\n6\n']
Demo Output:
['5', '-1\n']
Note:
none
|
```python
import sys
reader = (line.rstrip() for line in sys.stdin)
input = reader.__next__
n, m = map(int, input().split())
arr = list(map(int, input().split()))
arr2 = list(map(int, input().split()))
arr.sort()
arr2.sort()
v_max = arr2[0]
v_min = arr[0]
v_max2 = arr[-1]
print(v_max2) if v_max2 >= v_min * 2 and v_max2 < v_max else print("-1")
```
| 0
|
|
82
|
A
|
Double Cola
|
PROGRAMMING
| 1,100
|
[
"implementation",
"math"
] |
A. Double Cola
|
1
|
256
|
Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the *n*-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.
|
The input data consist of a single integer *n* (1<=≤<=*n*<=≤<=109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.
|
Print the single line — the name of the person who drinks the *n*-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.
|
[
"1\n",
"6\n",
"1802\n"
] |
[
"Sheldon\n",
"Sheldon\n",
"Penny\n"
] |
none
| 500
|
[
{
"input": "1",
"output": "Sheldon"
},
{
"input": "6",
"output": "Sheldon"
},
{
"input": "1802",
"output": "Penny"
},
{
"input": "1",
"output": "Sheldon"
},
{
"input": "2",
"output": "Leonard"
},
{
"input": "3",
"output": "Penny"
},
{
"input": "4",
"output": "Rajesh"
},
{
"input": "5",
"output": "Howard"
},
{
"input": "10",
"output": "Penny"
},
{
"input": "534",
"output": "Rajesh"
},
{
"input": "5033",
"output": "Howard"
},
{
"input": "10010",
"output": "Howard"
},
{
"input": "500000000",
"output": "Penny"
},
{
"input": "63",
"output": "Rajesh"
},
{
"input": "841",
"output": "Leonard"
},
{
"input": "3667",
"output": "Penny"
},
{
"input": "38614",
"output": "Howard"
},
{
"input": "282798",
"output": "Rajesh"
},
{
"input": "9266286",
"output": "Rajesh"
},
{
"input": "27385966",
"output": "Leonard"
},
{
"input": "121580142",
"output": "Penny"
},
{
"input": "5",
"output": "Howard"
},
{
"input": "300",
"output": "Howard"
},
{
"input": "1745",
"output": "Leonard"
},
{
"input": "8302",
"output": "Rajesh"
},
{
"input": "184518",
"output": "Sheldon"
},
{
"input": "1154414",
"output": "Rajesh"
},
{
"input": "28643950",
"output": "Leonard"
},
{
"input": "159222638",
"output": "Howard"
},
{
"input": "24",
"output": "Penny"
},
{
"input": "505",
"output": "Penny"
},
{
"input": "4425",
"output": "Rajesh"
},
{
"input": "12079",
"output": "Sheldon"
},
{
"input": "469726",
"output": "Penny"
},
{
"input": "3961838",
"output": "Penny"
},
{
"input": "57710446",
"output": "Leonard"
},
{
"input": "80719470",
"output": "Howard"
},
{
"input": "1000000000",
"output": "Penny"
},
{
"input": "999999999",
"output": "Penny"
},
{
"input": "999999998",
"output": "Penny"
},
{
"input": "5",
"output": "Howard"
}
] | 1,621,658,026
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 41
| 62
| 0
|
n=int(input())
cou=1
s = ["Sheldon", "Leonard", "Penny", "Rajesh", "Howard"]
while(cou*5<n):
n=n-(cou*5)
cou*=2
n=n-1
n=n//cou
print(s[n])
|
Title: Double Cola
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the *n*-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.
Input Specification:
The input data consist of a single integer *n* (1<=≤<=*n*<=≤<=109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.
Output Specification:
Print the single line — the name of the person who drinks the *n*-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.
Demo Input:
['1\n', '6\n', '1802\n']
Demo Output:
['Sheldon\n', 'Sheldon\n', 'Penny\n']
Note:
none
|
```python
n=int(input())
cou=1
s = ["Sheldon", "Leonard", "Penny", "Rajesh", "Howard"]
while(cou*5<n):
n=n-(cou*5)
cou*=2
n=n-1
n=n//cou
print(s[n])
```
| 3.969
|
830
|
A
|
Office Keys
|
PROGRAMMING
| 1,800
|
[
"binary search",
"brute force",
"dp",
"greedy",
"sortings"
] | null | null |
There are *n* people and *k* keys on a straight line. Every person wants to get to the office which is located on the line as well. To do that, he needs to reach some point with a key, take the key and then go to the office. Once a key is taken by somebody, it couldn't be taken by anybody else.
You are to determine the minimum time needed for all *n* people to get to the office with keys. Assume that people move a unit distance per 1 second. If two people reach a key at the same time, only one of them can take the key. A person can pass through a point with a key without taking it.
|
The first line contains three integers *n*, *k* and *p* (1<=≤<=*n*<=≤<=1<=000, *n*<=≤<=*k*<=≤<=2<=000, 1<=≤<=*p*<=≤<=109) — the number of people, the number of keys and the office location.
The second line contains *n* distinct integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — positions in which people are located initially. The positions are given in arbitrary order.
The third line contains *k* distinct integers *b*1,<=*b*2,<=...,<=*b**k* (1<=≤<=*b**j*<=≤<=109) — positions of the keys. The positions are given in arbitrary order.
Note that there can't be more than one person or more than one key in the same point. A person and a key can be located in the same point.
|
Print the minimum time (in seconds) needed for all *n* to reach the office with keys.
|
[
"2 4 50\n20 100\n60 10 40 80\n",
"1 2 10\n11\n15 7\n"
] |
[
"50\n",
"7\n"
] |
In the first example the person located at point 20 should take the key located at point 40 and go with it to the office located at point 50. He spends 30 seconds. The person located at point 100 can take the key located at point 80 and go to the office with it. He spends 50 seconds. Thus, after 50 seconds everybody is in office with keys.
| 500
|
[
{
"input": "2 4 50\n20 100\n60 10 40 80",
"output": "50"
},
{
"input": "1 2 10\n11\n15 7",
"output": "7"
},
{
"input": "2 5 15\n10 4\n29 23 21 22 26",
"output": "23"
},
{
"input": "3 10 1500\n106 160 129\n1333 1532 1181 1091 1656 1698 1291 1741 1242 1163",
"output": "1394"
},
{
"input": "5 20 1\n314 316 328 323 321\n30 61 11 83 19 63 97 87 14 79 43 57 75 48 47 95 41 27 8 88",
"output": "327"
},
{
"input": "20 20 1000000000\n911196469 574676950 884047241 984218701 641693148 352743122 616364857 455260052 702604347 921615943 671695009 544819698 768892858 254148055 379968391 65297129 178692403 575557323 307174510 63022600\n1621 106 6866 6420 9307 6985 2741 9477 9837 5909 6757 3085 6139 1876 3726 9334 4321 1531 8534 560",
"output": "1984199027"
},
{
"input": "40 45 1000\n6 55 34 32 20 76 2 84 47 68 31 60 14 70 99 72 21 61 81 79 26 51 96 86 10 1 43 69 87 78 13 11 80 67 50 52 9 29 94 12\n1974 1232 234 28 1456 626 408 1086 1525 1209 1096 940 795 1867 548 1774 1993 1199 1112 1087 1923 1156 876 1715 1815 1027 1658 955 398 910 620 1164 749 996 113 109 500 328 800 826 766 518 1474 1038 1029",
"output": "2449"
},
{
"input": "50 55 2000\n9518 9743 9338 9956 9827 9772 9094 9644 9242 9292 9148 9205 9907 9860 9530 9814 9662 9482 9725 9227 9105 9424 9268 9427 9470 9578 9808 9976 9143 9070 9079 9896 9367 9235 9925 9009 9619 9012 9669 9077 9870 9766 9479 9598 9055 9988 9792 9197 9377 9610\n828 656 345 412 69 506 274 994 384 766 587 126 720 227 66 839 997 602 646 955 256 262 243 676 459 83 507 88 559 595 71 154 867 276 487 895 857 888 368 179 813 407 973 780 588 112 815 290 554 230 768 804 974 3 745",
"output": "10833"
},
{
"input": "1 1 1\n1\n1000000000",
"output": "1999999998"
},
{
"input": "1 1 1\n1000000000\n1",
"output": "999999999"
},
{
"input": "1 1 1000000000\n1000000000\n1",
"output": "1999999998"
},
{
"input": "1 1 1000000000\n1\n1000000000",
"output": "999999999"
},
{
"input": "2 2 4\n3 4\n5 6",
"output": "4"
},
{
"input": "2 2 5\n1 2\n3 1000000000",
"output": "1999999993"
},
{
"input": "1 1 1000000000\n1000000000\n1",
"output": "1999999998"
},
{
"input": "2 2 1\n2 3\n4 100",
"output": "196"
},
{
"input": "2 2 10\n3 12\n1 9",
"output": "11"
},
{
"input": "3 3 1\n1 2 3\n999 1000000000 1",
"output": "1999999996"
},
{
"input": "1 1 1\n1\n1",
"output": "0"
},
{
"input": "1 1 1\n1\n1000000000",
"output": "1999999998"
},
{
"input": "1 1 1000000000\n1000000000\n10",
"output": "1999999980"
},
{
"input": "2 2 7122\n123 456\n1 4444",
"output": "7243"
},
{
"input": "1 1 10\n5\n15",
"output": "15"
},
{
"input": "2 4 1000\n1000 999\n1 1000 2 999",
"output": "1"
},
{
"input": "2 2 1000\n10 1010\n1 1001",
"output": "1008"
},
{
"input": "1 1 1\n2\n1000000000",
"output": "1999999997"
},
{
"input": "2 2 3\n1 5\n5 1",
"output": "2"
},
{
"input": "2 2 5\n2 3\n4 6",
"output": "4"
},
{
"input": "2 2 10\n5 6\n4 6",
"output": "7"
},
{
"input": "3 4 10\n5 7 9\n6 8 14 4",
"output": "7"
},
{
"input": "1 1 10\n10\n10",
"output": "0"
},
{
"input": "1 1 50\n1\n1000000000",
"output": "1999999949"
},
{
"input": "1 1 42\n666\n1337",
"output": "1966"
},
{
"input": "2 2 10\n9 11\n11 8",
"output": "3"
},
{
"input": "3 10 5\n1 2 3\n10000 9999 9998 9997 9996 9995 9994 7 6 5",
"output": "6"
},
{
"input": "1 1 2\n1\n1000000000",
"output": "1999999997"
},
{
"input": "2 2 100\n99 150\n1 150",
"output": "197"
},
{
"input": "3 3 4\n1 101 102\n2 3 100",
"output": "99"
}
] | 1,675,770,641
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 59
| 155
| 921,600
|
import copy
import random
import heapq
import math
import sys
import bisect
import datetime
from functools import lru_cache
from collections import deque
from collections import Counter
from collections import defaultdict
from itertools import combinations
from itertools import permutations
from types import GeneratorType
from functools import cmp_to_key
from decimal import Decimal
from heapq import nlargest
inf = float("inf")
sys.setrecursionlimit(10000000)
class FastIO:
def __init__(self):
return
@staticmethod
def _read():
return sys.stdin.readline().strip()
def read_int(self):
return int(self._read())
def read_float(self):
return float(self._read())
def read_ints(self):
return map(int, self._read().split())
def read_floats(self):
return map(float, self._read().split())
def read_ints_minus_one(self):
return map(lambda x: int(x) - 1, self._read().split())
def read_list_ints(self):
return list(map(int, self._read().split()))
def read_list_floats(self):
return list(map(float, self._read().split()))
def read_list_ints_minus_one(self):
return list(map(lambda x: int(x) - 1, self._read().split()))
def read_str(self):
return self._read()
def read_list_strs(self):
return self._read().split()
def read_list_str(self):
return list(self._read())
@staticmethod
def st(x):
return sys.stdout.write(str(x) + '\n')
@staticmethod
def lst(x):
return sys.stdout.write(" ".join(str(w) for w in x) + '\n')
@staticmethod
def round_5(f):
res = int(f)
if f - res >= 0.5:
res += 1
return res
@staticmethod
def max(a, b):
return a if a > b else b
@staticmethod
def min(a, b):
return a if a < b else b
@staticmethod
def bootstrap(f, queue=[]):
def wrappedfunc(*args, **kwargs):
if queue:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if isinstance(to, GeneratorType):
queue.append(to)
to = next(to)
else:
queue.pop()
if not queue:
break
to = queue[-1].send(to)
return to
return wrappedfunc
def main(ac=FastIO()):
n, k, p = ac.read_ints()
person = sorted(ac.read_list_ints())
keys = sorted(ac.read_list_ints())
def check(x):
lst = []
for y in keys:
if abs(y - p) <= x:
z = x - abs(y - p)
lst.append([y - z, y + z])
lst.sort(key=lambda it: [it[1], it[0]])
m = len(lst)
i = 0
for j in range(n):
y = person[j]
while i < m and not (lst[i][0] <= y <= lst[i][1]):
i += 1
if i == m:
return False
i += 1
return True
low = 0
high = 10**15
while low < high - 1:
mid = low + (high - low) // 2
if check(mid):
high = mid
else:
low = mid
ans = low if check(low) else high
ac.st(ans)
return
main()
|
Title: Office Keys
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* people and *k* keys on a straight line. Every person wants to get to the office which is located on the line as well. To do that, he needs to reach some point with a key, take the key and then go to the office. Once a key is taken by somebody, it couldn't be taken by anybody else.
You are to determine the minimum time needed for all *n* people to get to the office with keys. Assume that people move a unit distance per 1 second. If two people reach a key at the same time, only one of them can take the key. A person can pass through a point with a key without taking it.
Input Specification:
The first line contains three integers *n*, *k* and *p* (1<=≤<=*n*<=≤<=1<=000, *n*<=≤<=*k*<=≤<=2<=000, 1<=≤<=*p*<=≤<=109) — the number of people, the number of keys and the office location.
The second line contains *n* distinct integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — positions in which people are located initially. The positions are given in arbitrary order.
The third line contains *k* distinct integers *b*1,<=*b*2,<=...,<=*b**k* (1<=≤<=*b**j*<=≤<=109) — positions of the keys. The positions are given in arbitrary order.
Note that there can't be more than one person or more than one key in the same point. A person and a key can be located in the same point.
Output Specification:
Print the minimum time (in seconds) needed for all *n* to reach the office with keys.
Demo Input:
['2 4 50\n20 100\n60 10 40 80\n', '1 2 10\n11\n15 7\n']
Demo Output:
['50\n', '7\n']
Note:
In the first example the person located at point 20 should take the key located at point 40 and go with it to the office located at point 50. He spends 30 seconds. The person located at point 100 can take the key located at point 80 and go to the office with it. He spends 50 seconds. Thus, after 50 seconds everybody is in office with keys.
|
```python
import copy
import random
import heapq
import math
import sys
import bisect
import datetime
from functools import lru_cache
from collections import deque
from collections import Counter
from collections import defaultdict
from itertools import combinations
from itertools import permutations
from types import GeneratorType
from functools import cmp_to_key
from decimal import Decimal
from heapq import nlargest
inf = float("inf")
sys.setrecursionlimit(10000000)
class FastIO:
def __init__(self):
return
@staticmethod
def _read():
return sys.stdin.readline().strip()
def read_int(self):
return int(self._read())
def read_float(self):
return float(self._read())
def read_ints(self):
return map(int, self._read().split())
def read_floats(self):
return map(float, self._read().split())
def read_ints_minus_one(self):
return map(lambda x: int(x) - 1, self._read().split())
def read_list_ints(self):
return list(map(int, self._read().split()))
def read_list_floats(self):
return list(map(float, self._read().split()))
def read_list_ints_minus_one(self):
return list(map(lambda x: int(x) - 1, self._read().split()))
def read_str(self):
return self._read()
def read_list_strs(self):
return self._read().split()
def read_list_str(self):
return list(self._read())
@staticmethod
def st(x):
return sys.stdout.write(str(x) + '\n')
@staticmethod
def lst(x):
return sys.stdout.write(" ".join(str(w) for w in x) + '\n')
@staticmethod
def round_5(f):
res = int(f)
if f - res >= 0.5:
res += 1
return res
@staticmethod
def max(a, b):
return a if a > b else b
@staticmethod
def min(a, b):
return a if a < b else b
@staticmethod
def bootstrap(f, queue=[]):
def wrappedfunc(*args, **kwargs):
if queue:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if isinstance(to, GeneratorType):
queue.append(to)
to = next(to)
else:
queue.pop()
if not queue:
break
to = queue[-1].send(to)
return to
return wrappedfunc
def main(ac=FastIO()):
n, k, p = ac.read_ints()
person = sorted(ac.read_list_ints())
keys = sorted(ac.read_list_ints())
def check(x):
lst = []
for y in keys:
if abs(y - p) <= x:
z = x - abs(y - p)
lst.append([y - z, y + z])
lst.sort(key=lambda it: [it[1], it[0]])
m = len(lst)
i = 0
for j in range(n):
y = person[j]
while i < m and not (lst[i][0] <= y <= lst[i][1]):
i += 1
if i == m:
return False
i += 1
return True
low = 0
high = 10**15
while low < high - 1:
mid = low + (high - low) // 2
if check(mid):
high = mid
else:
low = mid
ans = low if check(low) else high
ac.st(ans)
return
main()
```
| 3
|
|
230
|
B
|
T-primes
|
PROGRAMMING
| 1,300
|
[
"binary search",
"implementation",
"math",
"number theory"
] | null | null |
We know that prime numbers are positive integers that have exactly two distinct positive divisors. Similarly, we'll call a positive integer *t* Т-prime, if *t* has exactly three distinct positive divisors.
You are given an array of *n* positive integers. For each of them determine whether it is Т-prime or not.
|
The first line contains a single positive integer, *n* (1<=≤<=*n*<=≤<=105), showing how many numbers are in the array. The next line contains *n* space-separated integers *x**i* (1<=≤<=*x**i*<=≤<=1012).
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is advised to use the cin, cout streams or the %I64d specifier.
|
Print *n* lines: the *i*-th line should contain "YES" (without the quotes), if number *x**i* is Т-prime, and "NO" (without the quotes), if it isn't.
|
[
"3\n4 5 6\n"
] |
[
"YES\nNO\nNO\n"
] |
The given test has three numbers. The first number 4 has exactly three divisors — 1, 2 and 4, thus the answer for this number is "YES". The second number 5 has two divisors (1 and 5), and the third number 6 has four divisors (1, 2, 3, 6), hence the answer for them is "NO".
| 500
|
[
{
"input": "3\n4 5 6",
"output": "YES\nNO\nNO"
},
{
"input": "2\n48 49",
"output": "NO\nYES"
},
{
"input": "10\n10 9 8 7 6 5 4 3 2 1",
"output": "NO\nYES\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nNO"
},
{
"input": "1\n36",
"output": "NO"
},
{
"input": "1\n999966000289",
"output": "YES"
},
{
"input": "1\n999993399999",
"output": "NO"
},
{
"input": "9\n111 121 131 111 121 131 111 121 131",
"output": "NO\nYES\nNO\nNO\nYES\nNO\nNO\nYES\nNO"
},
{
"input": "1\n1",
"output": "NO"
},
{
"input": "1\n10",
"output": "NO"
},
{
"input": "1\n976197352729",
"output": "NO"
},
{
"input": "1\n1000000000000",
"output": "NO"
},
{
"input": "1\n9",
"output": "YES"
},
{
"input": "6\n549755813888 847288609443 762939453125 678223072849 285311670611 137858491849",
"output": "NO\nNO\nNO\nNO\nNO\nNO"
},
{
"input": "3\n223092870 6469693230 200560490130",
"output": "NO\nNO\nNO"
},
{
"input": "2\n81 25",
"output": "NO\nYES"
},
{
"input": "1\n16",
"output": "NO"
},
{
"input": "22\n1 2 3 4 5 6 7 8 9 10 12752041 64 121 144 27550356289 124 24657 23756 135153365 25235235235 42351351 81",
"output": "NO\nNO\nNO\nYES\nNO\nNO\nNO\nNO\nYES\nNO\nYES\nNO\nYES\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nNO\nNO"
},
{
"input": "1\n225",
"output": "NO"
},
{
"input": "1\n1521",
"output": "NO"
},
{
"input": "1\n81",
"output": "NO"
}
] | 1,697,048,896
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 16
| 2,000
| 33,382,400
|
import math
MAX_SIZE = 1000001
isprime = [True] * MAX_SIZE
prime = []
SPF = [None] * (MAX_SIZE)
def manipulated_seive(N):
isprime[0] = isprime[1] = False
for i in range(2, N):
if isprime[i] == True:
prime.append(i)
SPF[i] = i
j = 0
while j < len(prime) and i * prime[j] < N and prime[j] <= SPF[i]:
isprime[i * prime[j]] = False
SPF[i * prime[j]] = prime[j]
j += 1
manipulated_seive(10**6)
b = int(input())
c = list(map(int, input().split()))
for t in c:
if math.sqrt(t).is_integer() and int(math.sqrt(t)) in prime:
print("YES")
else:
print("NO")
|
Title: T-primes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
We know that prime numbers are positive integers that have exactly two distinct positive divisors. Similarly, we'll call a positive integer *t* Т-prime, if *t* has exactly three distinct positive divisors.
You are given an array of *n* positive integers. For each of them determine whether it is Т-prime or not.
Input Specification:
The first line contains a single positive integer, *n* (1<=≤<=*n*<=≤<=105), showing how many numbers are in the array. The next line contains *n* space-separated integers *x**i* (1<=≤<=*x**i*<=≤<=1012).
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is advised to use the cin, cout streams or the %I64d specifier.
Output Specification:
Print *n* lines: the *i*-th line should contain "YES" (without the quotes), if number *x**i* is Т-prime, and "NO" (without the quotes), if it isn't.
Demo Input:
['3\n4 5 6\n']
Demo Output:
['YES\nNO\nNO\n']
Note:
The given test has three numbers. The first number 4 has exactly three divisors — 1, 2 and 4, thus the answer for this number is "YES". The second number 5 has two divisors (1 and 5), and the third number 6 has four divisors (1, 2, 3, 6), hence the answer for them is "NO".
|
```python
import math
MAX_SIZE = 1000001
isprime = [True] * MAX_SIZE
prime = []
SPF = [None] * (MAX_SIZE)
def manipulated_seive(N):
isprime[0] = isprime[1] = False
for i in range(2, N):
if isprime[i] == True:
prime.append(i)
SPF[i] = i
j = 0
while j < len(prime) and i * prime[j] < N and prime[j] <= SPF[i]:
isprime[i * prime[j]] = False
SPF[i * prime[j]] = prime[j]
j += 1
manipulated_seive(10**6)
b = int(input())
c = list(map(int, input().split()))
for t in c:
if math.sqrt(t).is_integer() and int(math.sqrt(t)) in prime:
print("YES")
else:
print("NO")
```
| 0
|
|
143
|
A
|
Help Vasilisa the Wise 2
|
PROGRAMMING
| 1,000
|
[
"brute force",
"math"
] | null | null |
Vasilisa the Wise from the Kingdom of Far Far Away got a magic box with a secret as a present from her friend Hellawisa the Wise from the Kingdom of A Little Closer. However, Vasilisa the Wise does not know what the box's secret is, since she cannot open it again. She hopes that you will help her one more time with that.
The box's lock looks as follows: it contains 4 identical deepenings for gems as a 2<=×<=2 square, and some integer numbers are written at the lock's edge near the deepenings. The example of a lock is given on the picture below.
The box is accompanied with 9 gems. Their shapes match the deepenings' shapes and each gem contains one number from 1 to 9 (each number is written on exactly one gem). The box will only open after it is decorated with gems correctly: that is, each deepening in the lock should be filled with exactly one gem. Also, the sums of numbers in the square's rows, columns and two diagonals of the square should match the numbers written at the lock's edge. For example, the above lock will open if we fill the deepenings with gems with numbers as is shown on the picture below.
Now Vasilisa the Wise wants to define, given the numbers on the box's lock, which gems she should put in the deepenings to open the box. Help Vasilisa to solve this challenging task.
|
The input contains numbers written on the edges of the lock of the box. The first line contains space-separated integers *r*1 and *r*2 that define the required sums of numbers in the rows of the square. The second line contains space-separated integers *c*1 and *c*2 that define the required sums of numbers in the columns of the square. The third line contains space-separated integers *d*1 and *d*2 that define the required sums of numbers on the main and on the side diagonals of the square (1<=≤<=*r*1,<=*r*2,<=*c*1,<=*c*2,<=*d*1,<=*d*2<=≤<=20). Correspondence between the above 6 variables and places where they are written is shown on the picture below. For more clarifications please look at the second sample test that demonstrates the example given in the problem statement.
|
Print the scheme of decorating the box with stones: two lines containing two space-separated integers from 1 to 9. The numbers should be pairwise different. If there is no solution for the given lock, then print the single number "-1" (without the quotes).
If there are several solutions, output any.
|
[
"3 7\n4 6\n5 5\n",
"11 10\n13 8\n5 16\n",
"1 2\n3 4\n5 6\n",
"10 10\n10 10\n10 10\n"
] |
[
"1 2\n3 4\n",
"4 7\n9 1\n",
"-1\n",
"-1\n"
] |
Pay attention to the last test from the statement: it is impossible to open the box because for that Vasilisa the Wise would need 4 identical gems containing number "5". However, Vasilisa only has one gem with each number from 1 to 9.
| 500
|
[
{
"input": "3 7\n4 6\n5 5",
"output": "1 2\n3 4"
},
{
"input": "11 10\n13 8\n5 16",
"output": "4 7\n9 1"
},
{
"input": "1 2\n3 4\n5 6",
"output": "-1"
},
{
"input": "10 10\n10 10\n10 10",
"output": "-1"
},
{
"input": "5 13\n8 10\n11 7",
"output": "3 2\n5 8"
},
{
"input": "12 17\n10 19\n13 16",
"output": "-1"
},
{
"input": "11 11\n17 5\n12 10",
"output": "9 2\n8 3"
},
{
"input": "12 11\n11 12\n16 7",
"output": "-1"
},
{
"input": "5 9\n7 7\n8 6",
"output": "3 2\n4 5"
},
{
"input": "10 7\n4 13\n11 6",
"output": "-1"
},
{
"input": "18 10\n16 12\n12 16",
"output": "-1"
},
{
"input": "13 6\n10 9\n6 13",
"output": "-1"
},
{
"input": "14 16\n16 14\n18 12",
"output": "-1"
},
{
"input": "16 10\n16 10\n12 14",
"output": "-1"
},
{
"input": "11 9\n12 8\n11 9",
"output": "-1"
},
{
"input": "5 14\n10 9\n10 9",
"output": "-1"
},
{
"input": "2 4\n1 5\n3 3",
"output": "-1"
},
{
"input": "17 16\n14 19\n18 15",
"output": "-1"
},
{
"input": "12 12\n14 10\n16 8",
"output": "9 3\n5 7"
},
{
"input": "15 11\n16 10\n9 17",
"output": "7 8\n9 2"
},
{
"input": "8 10\n9 9\n13 5",
"output": "6 2\n3 7"
},
{
"input": "13 7\n10 10\n5 15",
"output": "4 9\n6 1"
},
{
"input": "14 11\n9 16\n16 9",
"output": "-1"
},
{
"input": "12 8\n14 6\n8 12",
"output": "-1"
},
{
"input": "10 6\n6 10\n4 12",
"output": "-1"
},
{
"input": "10 8\n10 8\n4 14",
"output": "-1"
},
{
"input": "14 13\n9 18\n14 13",
"output": "-1"
},
{
"input": "9 14\n8 15\n8 15",
"output": "-1"
},
{
"input": "3 8\n2 9\n6 5",
"output": "-1"
},
{
"input": "14 17\n18 13\n15 16",
"output": "-1"
},
{
"input": "16 14\n15 15\n17 13",
"output": "9 7\n6 8"
},
{
"input": "14 11\n16 9\n13 12",
"output": "9 5\n7 4"
},
{
"input": "13 10\n11 12\n7 16",
"output": "4 9\n7 3"
},
{
"input": "14 8\n11 11\n13 9",
"output": "8 6\n3 5"
},
{
"input": "12 11\n13 10\n10 13",
"output": "-1"
},
{
"input": "6 5\n2 9\n5 6",
"output": "-1"
},
{
"input": "7 8\n8 7\n12 3",
"output": "-1"
},
{
"input": "7 11\n7 11\n6 12",
"output": "-1"
},
{
"input": "8 5\n11 2\n8 5",
"output": "-1"
},
{
"input": "10 16\n14 12\n14 12",
"output": "-1"
},
{
"input": "7 9\n4 12\n5 11",
"output": "-1"
},
{
"input": "11 13\n19 5\n12 12",
"output": "-1"
},
{
"input": "8 12\n5 15\n11 9",
"output": "2 6\n3 9"
},
{
"input": "16 5\n13 8\n10 11",
"output": "9 7\n4 1"
},
{
"input": "3 14\n8 9\n10 7",
"output": "2 1\n6 8"
},
{
"input": "12 14\n11 15\n9 17",
"output": "3 9\n8 6"
},
{
"input": "13 7\n9 11\n14 6",
"output": "8 5\n1 6"
},
{
"input": "13 8\n15 6\n11 10",
"output": "9 4\n6 2"
},
{
"input": "9 6\n5 10\n3 12",
"output": "1 8\n4 2"
},
{
"input": "12 7\n5 14\n8 11",
"output": "3 9\n2 5"
},
{
"input": "9 12\n3 17\n10 10",
"output": "-1"
},
{
"input": "3 8\n4 6\n5 5",
"output": "-1"
}
] | 1,603,286,219
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 52
| 218
| 307,200
|
inn = list(map(int, input().split(" ")))
r1 = inn[0]
r2 = inn[1]
inn = list(map(int, input().split(" ")))
c1 = inn[0]
c2 = inn[1]
inn = list(map(int, input().split(" ")))
d1 = inn[0]
d2 = inn[1]
x = int((d1+c1-r2)/2)
y = int(((2*r1)-d1-c1+r2)/2)
a = int(((2*c1)-d1-c1+r2)/2)
b = int((r2-(2*c1)+d1+c1)/2)
if x == y or x == a or x == b or y == a or y == b or a == b or x < 1 or x > 9 or y < 1 or y > 9 or a < 1 or a > 9 or b < 1 or b > 9:
print(-1)
else:
print(x, y)
print(a, b)
|
Title: Help Vasilisa the Wise 2
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasilisa the Wise from the Kingdom of Far Far Away got a magic box with a secret as a present from her friend Hellawisa the Wise from the Kingdom of A Little Closer. However, Vasilisa the Wise does not know what the box's secret is, since she cannot open it again. She hopes that you will help her one more time with that.
The box's lock looks as follows: it contains 4 identical deepenings for gems as a 2<=×<=2 square, and some integer numbers are written at the lock's edge near the deepenings. The example of a lock is given on the picture below.
The box is accompanied with 9 gems. Their shapes match the deepenings' shapes and each gem contains one number from 1 to 9 (each number is written on exactly one gem). The box will only open after it is decorated with gems correctly: that is, each deepening in the lock should be filled with exactly one gem. Also, the sums of numbers in the square's rows, columns and two diagonals of the square should match the numbers written at the lock's edge. For example, the above lock will open if we fill the deepenings with gems with numbers as is shown on the picture below.
Now Vasilisa the Wise wants to define, given the numbers on the box's lock, which gems she should put in the deepenings to open the box. Help Vasilisa to solve this challenging task.
Input Specification:
The input contains numbers written on the edges of the lock of the box. The first line contains space-separated integers *r*1 and *r*2 that define the required sums of numbers in the rows of the square. The second line contains space-separated integers *c*1 and *c*2 that define the required sums of numbers in the columns of the square. The third line contains space-separated integers *d*1 and *d*2 that define the required sums of numbers on the main and on the side diagonals of the square (1<=≤<=*r*1,<=*r*2,<=*c*1,<=*c*2,<=*d*1,<=*d*2<=≤<=20). Correspondence between the above 6 variables and places where they are written is shown on the picture below. For more clarifications please look at the second sample test that demonstrates the example given in the problem statement.
Output Specification:
Print the scheme of decorating the box with stones: two lines containing two space-separated integers from 1 to 9. The numbers should be pairwise different. If there is no solution for the given lock, then print the single number "-1" (without the quotes).
If there are several solutions, output any.
Demo Input:
['3 7\n4 6\n5 5\n', '11 10\n13 8\n5 16\n', '1 2\n3 4\n5 6\n', '10 10\n10 10\n10 10\n']
Demo Output:
['1 2\n3 4\n', '4 7\n9 1\n', '-1\n', '-1\n']
Note:
Pay attention to the last test from the statement: it is impossible to open the box because for that Vasilisa the Wise would need 4 identical gems containing number "5". However, Vasilisa only has one gem with each number from 1 to 9.
|
```python
inn = list(map(int, input().split(" ")))
r1 = inn[0]
r2 = inn[1]
inn = list(map(int, input().split(" ")))
c1 = inn[0]
c2 = inn[1]
inn = list(map(int, input().split(" ")))
d1 = inn[0]
d2 = inn[1]
x = int((d1+c1-r2)/2)
y = int(((2*r1)-d1-c1+r2)/2)
a = int(((2*c1)-d1-c1+r2)/2)
b = int((r2-(2*c1)+d1+c1)/2)
if x == y or x == a or x == b or y == a or y == b or a == b or x < 1 or x > 9 or y < 1 or y > 9 or a < 1 or a > 9 or b < 1 or b > 9:
print(-1)
else:
print(x, y)
print(a, b)
```
| 3
|
|
976
|
C
|
Nested Segments
|
PROGRAMMING
| 1,500
|
[
"greedy",
"implementation",
"sortings"
] | null | null |
You are given a sequence *a*1,<=*a*2,<=...,<=*a**n* of one-dimensional segments numbered 1 through *n*. Your task is to find two distinct indices *i* and *j* such that segment *a**i* lies within segment *a**j*.
Segment [*l*1,<=*r*1] lies within segment [*l*2,<=*r*2] iff *l*1<=≥<=*l*2 and *r*1<=≤<=*r*2.
Print indices *i* and *j*. If there are multiple answers, print any of them. If no answer exists, print -1 -1.
|
The first line contains one integer *n* (1<=≤<=*n*<=≤<=3·105) — the number of segments.
Each of the next *n* lines contains two integers *l**i* and *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=109) — the *i*-th segment.
|
Print two distinct indices *i* and *j* such that segment *a**i* lies within segment *a**j*. If there are multiple answers, print any of them. If no answer exists, print -1 -1.
|
[
"5\n1 10\n2 9\n3 9\n2 3\n2 9\n",
"3\n1 5\n2 6\n6 20\n"
] |
[
"2 1\n",
"-1 -1\n"
] |
In the first example the following pairs are considered correct:
- (2, 1), (3, 1), (4, 1), (5, 1) — not even touching borders; - (3, 2), (4, 2), (3, 5), (4, 5) — touch one border; - (5, 2), (2, 5) — match exactly.
| 0
|
[
{
"input": "5\n1 10\n2 9\n3 9\n2 3\n2 9",
"output": "2 1"
},
{
"input": "3\n1 5\n2 6\n6 20",
"output": "-1 -1"
},
{
"input": "1\n1 1000000000",
"output": "-1 -1"
},
{
"input": "2\n1 1000000000\n1 1000000000",
"output": "2 1"
},
{
"input": "2\n1 1000000000\n500000000 500000000",
"output": "2 1"
},
{
"input": "2\n1 10\n2 10",
"output": "2 1"
},
{
"input": "2\n10 20\n10 11",
"output": "2 1"
},
{
"input": "3\n1 10\n10 20\n9 11",
"output": "-1 -1"
},
{
"input": "3\n1 1\n2 3\n2 2",
"output": "3 2"
},
{
"input": "4\n1 10\n2 11\n3 10000000\n3 100000000",
"output": "3 4"
},
{
"input": "2\n3 7\n3 9",
"output": "1 2"
},
{
"input": "3\n1 2\n2 3\n1 2",
"output": "3 1"
},
{
"input": "3\n5 6\n4 7\n3 8",
"output": "2 3"
},
{
"input": "3\n2 9\n1 7\n2 8",
"output": "3 1"
},
{
"input": "2\n1 4\n1 5",
"output": "1 2"
},
{
"input": "3\n1 2\n1 3\n4 4",
"output": "1 2"
},
{
"input": "3\n1 2\n1 3\n67 1234567",
"output": "1 2"
},
{
"input": "2\n1 1\n1 1",
"output": "2 1"
},
{
"input": "3\n1 5\n4 7\n3 9",
"output": "2 3"
},
{
"input": "2\n1 1\n1 10",
"output": "1 2"
},
{
"input": "2\n1 2\n1 3",
"output": "1 2"
},
{
"input": "2\n1 10\n1 11",
"output": "1 2"
},
{
"input": "2\n1 1\n1 2",
"output": "1 2"
},
{
"input": "2\n2 3\n2 4",
"output": "1 2"
},
{
"input": "2\n1 3\n3 3",
"output": "2 1"
},
{
"input": "3\n1 10\n11 13\n12 12",
"output": "3 2"
},
{
"input": "2\n2 10\n1 10",
"output": "1 2"
},
{
"input": "3\n1 3\n4 5\n4 4",
"output": "3 2"
},
{
"input": "5\n1 1\n2 6\n3 5\n10 15\n20 25",
"output": "3 2"
},
{
"input": "3\n1 1000\n1001 1007\n1002 1007",
"output": "3 2"
},
{
"input": "3\n1 3\n2 5\n3 4",
"output": "3 2"
},
{
"input": "3\n1 10\n2 11\n3 11",
"output": "3 2"
},
{
"input": "2\n2000000 999999999\n1000000 1000000000",
"output": "1 2"
},
{
"input": "3\n2 10\n11 12\n4 5",
"output": "3 1"
},
{
"input": "2\n1 10\n1 19",
"output": "1 2"
},
{
"input": "4\n1 3\n100 102\n108 110\n1 3",
"output": "4 1"
},
{
"input": "3\n1 3\n5 9\n5 6",
"output": "3 2"
},
{
"input": "3\n1 3\n3 4\n3 5",
"output": "2 3"
},
{
"input": "3\n1 2\n1 3\n1 4",
"output": "2 3"
},
{
"input": "4\n2 3\n1 4\n100 200\n1000 2000",
"output": "1 2"
},
{
"input": "3\n1 1\n2 100\n3 99",
"output": "3 2"
},
{
"input": "3\n1 2\n1 3\n12 1234",
"output": "1 2"
},
{
"input": "3\n1 4\n2 6\n3 5",
"output": "3 2"
},
{
"input": "3\n1 10\n2 12\n1 9",
"output": "3 1"
},
{
"input": "2\n1 3\n1 5",
"output": "1 2"
},
{
"input": "3\n1 2\n2 5\n2 3",
"output": "3 2"
},
{
"input": "4\n1 3\n1 4\n5 10\n11 13",
"output": "1 2"
},
{
"input": "4\n7 15\n6 9\n9 10\n10 11",
"output": "3 1"
},
{
"input": "4\n2 3\n100 200\n1000 2000\n1 4",
"output": "1 4"
},
{
"input": "3\n10 20\n5 9\n11 19",
"output": "3 1"
},
{
"input": "10\n1 2\n2 3\n3 4\n4 5\n5 6\n6 6\n6 7\n7 8\n8 9\n9 10",
"output": "6 7"
},
{
"input": "2\n1 4\n1 7",
"output": "1 2"
},
{
"input": "3\n1 11\n2 12\n2 13",
"output": "2 3"
},
{
"input": "2\n1 4\n1 8",
"output": "1 2"
},
{
"input": "2\n2 5\n1 5",
"output": "1 2"
},
{
"input": "2\n2 9\n1 10",
"output": "1 2"
},
{
"input": "3\n2 4\n2 4\n1 3",
"output": "2 1"
},
{
"input": "6\n10 11\n12 13\n15 16\n15 17\n18 19\n59 60",
"output": "3 4"
},
{
"input": "2\n1 3\n1 7",
"output": "1 2"
},
{
"input": "5\n4 6\n7 60\n80 90\n4 5\n8 80",
"output": "4 1"
},
{
"input": "2\n1 3\n1 4",
"output": "1 2"
},
{
"input": "3\n2 9\n1 7\n2 9",
"output": "3 1"
},
{
"input": "2\n1 4\n1 6",
"output": "1 2"
},
{
"input": "3\n4 4\n2 3\n4 5",
"output": "1 3"
},
{
"input": "2\n1 5\n1 7",
"output": "1 2"
},
{
"input": "2\n1 2\n1 4",
"output": "1 2"
},
{
"input": "4\n1 1\n2 2\n5 10\n2 4",
"output": "2 4"
},
{
"input": "3\n11 12\n11 15\n43 45",
"output": "1 2"
},
{
"input": "3\n2 3\n2 4\n2 5",
"output": "2 3"
},
{
"input": "2\n2 3\n2 5",
"output": "1 2"
},
{
"input": "3\n1 3\n1 4\n1 5",
"output": "2 3"
},
{
"input": "3\n1 1\n1 2\n1 3",
"output": "2 3"
},
{
"input": "2\n2 3\n1 3",
"output": "1 2"
},
{
"input": "11\n22226 28285\n9095 23314\n19162 25530\n255 13298\n4904 25801\n17914 23501\n8441 28117\n11880 29994\n11123 19874\n21505 27971\n7658 14109",
"output": "11 5"
},
{
"input": "8\n4 11\n5 12\n6 13\n7 14\n8 15\n9 16\n10 17\n1 11",
"output": "1 8"
},
{
"input": "4\n1 10\n12 15\n1 3\n17 18",
"output": "3 1"
},
{
"input": "3\n1 5\n1 10\n1 20",
"output": "2 3"
},
{
"input": "3\n1 1000\n1001 1003\n1 1",
"output": "3 1"
},
{
"input": "3\n1 10\n2 11\n2 11",
"output": "3 2"
},
{
"input": "2\n1 1\n1 3",
"output": "1 2"
},
{
"input": "2\n1 5\n1 6",
"output": "1 2"
},
{
"input": "3\n1 5\n3 6\n1 4",
"output": "3 1"
},
{
"input": "4\n2 8\n1 3\n2 9\n1 2",
"output": "4 2"
},
{
"input": "3\n3 6\n1 3\n3 9",
"output": "1 3"
},
{
"input": "6\n2 40\n5 50\n10 60\n3 45\n1 40\n100 111",
"output": "1 5"
},
{
"input": "4\n1 2\n4 4\n3 3\n2 3",
"output": "3 4"
},
{
"input": "4\n1 1\n4 5\n7 9\n1 1",
"output": "4 1"
},
{
"input": "6\n30 35\n18 29\n28 32\n4 9\n1002 129212\n8 281",
"output": "2 6"
},
{
"input": "2\n10 13\n10 14",
"output": "1 2"
},
{
"input": "5\n2 4\n3 6\n4 5\n222 333\n111 444",
"output": "3 2"
},
{
"input": "3\n1 2\n4 5\n1 1",
"output": "3 1"
},
{
"input": "2\n2 100\n1 100",
"output": "1 2"
},
{
"input": "3\n1 10\n9 20\n3 5",
"output": "3 1"
},
{
"input": "2\n1 9\n1 10",
"output": "1 2"
},
{
"input": "3\n1 2\n1 4\n1 6",
"output": "2 3"
}
] | 1,525,149,692
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
N = int(input())
A = []
for i in range(1, N+1):
l, r = map(int, input().split())
A.append([l, r, i])
A.sort(key=lambda x:x[0])
if N == 1:
print(-1, -1)
quit()
a = A[0][1]
for i in range(1, N):
if i == 1 and A[1][0] == A[0][0] and A[0][1] > A[1][1]
print(A[0][2], A[1][2])
elif A[i][1] <= a:
print(A[i][2], A[i-1][2])
quit()
else:
if i == N-1:
print(-1, -1)
quit()
else:
a = A[i][1]
|
Title: Nested Segments
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a sequence *a*1,<=*a*2,<=...,<=*a**n* of one-dimensional segments numbered 1 through *n*. Your task is to find two distinct indices *i* and *j* such that segment *a**i* lies within segment *a**j*.
Segment [*l*1,<=*r*1] lies within segment [*l*2,<=*r*2] iff *l*1<=≥<=*l*2 and *r*1<=≤<=*r*2.
Print indices *i* and *j*. If there are multiple answers, print any of them. If no answer exists, print -1 -1.
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=3·105) — the number of segments.
Each of the next *n* lines contains two integers *l**i* and *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=109) — the *i*-th segment.
Output Specification:
Print two distinct indices *i* and *j* such that segment *a**i* lies within segment *a**j*. If there are multiple answers, print any of them. If no answer exists, print -1 -1.
Demo Input:
['5\n1 10\n2 9\n3 9\n2 3\n2 9\n', '3\n1 5\n2 6\n6 20\n']
Demo Output:
['2 1\n', '-1 -1\n']
Note:
In the first example the following pairs are considered correct:
- (2, 1), (3, 1), (4, 1), (5, 1) — not even touching borders; - (3, 2), (4, 2), (3, 5), (4, 5) — touch one border; - (5, 2), (2, 5) — match exactly.
|
```python
N = int(input())
A = []
for i in range(1, N+1):
l, r = map(int, input().split())
A.append([l, r, i])
A.sort(key=lambda x:x[0])
if N == 1:
print(-1, -1)
quit()
a = A[0][1]
for i in range(1, N):
if i == 1 and A[1][0] == A[0][0] and A[0][1] > A[1][1]
print(A[0][2], A[1][2])
elif A[i][1] <= a:
print(A[i][2], A[i-1][2])
quit()
else:
if i == N-1:
print(-1, -1)
quit()
else:
a = A[i][1]
```
| -1
|
|
119
|
A
|
Epic Game
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Simon and Antisimon play a game. Initially each player receives one fixed positive integer that doesn't change throughout the game. Simon receives number *a* and Antisimon receives number *b*. They also have a heap of *n* stones. The players take turns to make a move and Simon starts. During a move a player should take from the heap the number of stones equal to the greatest common divisor of the fixed number he has received and the number of stones left in the heap. A player loses when he cannot take the required number of stones (i. e. the heap has strictly less stones left than one needs to take).
Your task is to determine by the given *a*, *b* and *n* who wins the game.
|
The only string contains space-separated integers *a*, *b* and *n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=100) — the fixed numbers Simon and Antisimon have received correspondingly and the initial number of stones in the pile.
|
If Simon wins, print "0" (without the quotes), otherwise print "1" (without the quotes).
|
[
"3 5 9\n",
"1 1 100\n"
] |
[
"0",
"1"
] |
The greatest common divisor of two non-negative integers *a* and *b* is such maximum positive integer *k*, that *a* is divisible by *k* without remainder and similarly, *b* is divisible by *k* without remainder. Let *gcd*(*a*, *b*) represent the operation of calculating the greatest common divisor of numbers *a* and *b*. Specifically, *gcd*(*x*, 0) = *gcd*(0, *x*) = *x*.
In the first sample the game will go like that:
- Simon should take *gcd*(3, 9) = 3 stones from the heap. After his move the heap has 6 stones left.- Antisimon should take *gcd*(5, 6) = 1 stone from the heap. After his move the heap has 5 stones left.- Simon should take *gcd*(3, 5) = 1 stone from the heap. After his move the heap has 4 stones left.- Antisimon should take *gcd*(5, 4) = 1 stone from the heap. After his move the heap has 3 stones left.- Simon should take *gcd*(3, 3) = 3 stones from the heap. After his move the heap has 0 stones left.- Antisimon should take *gcd*(5, 0) = 5 stones from the heap. As 0 < 5, it is impossible and Antisimon loses.
In the second sample each player during each move takes one stone from the heap. As *n* is even, Antisimon takes the last stone and Simon can't make a move after that.
| 500
|
[
{
"input": "3 5 9",
"output": "0"
},
{
"input": "1 1 100",
"output": "1"
},
{
"input": "23 12 16",
"output": "1"
},
{
"input": "95 26 29",
"output": "1"
},
{
"input": "73 32 99",
"output": "1"
},
{
"input": "1 1 1",
"output": "0"
},
{
"input": "41 12 65",
"output": "1"
},
{
"input": "13 61 100",
"output": "1"
},
{
"input": "100 100 10",
"output": "0"
},
{
"input": "12 24 26",
"output": "1"
},
{
"input": "73 21 96",
"output": "1"
},
{
"input": "17 22 81",
"output": "1"
},
{
"input": "14 88 97",
"output": "1"
},
{
"input": "42 81 17",
"output": "0"
},
{
"input": "34 36 19",
"output": "1"
},
{
"input": "21 87 12",
"output": "1"
},
{
"input": "97 92 98",
"output": "1"
},
{
"input": "49 25 62",
"output": "1"
},
{
"input": "29 17 44",
"output": "1"
},
{
"input": "10 1 10",
"output": "0"
},
{
"input": "99 98 100",
"output": "1"
},
{
"input": "100 100 1",
"output": "0"
},
{
"input": "11 99 55",
"output": "0"
},
{
"input": "8 13 21",
"output": "0"
},
{
"input": "34 55 89",
"output": "0"
},
{
"input": "21 34 55",
"output": "1"
},
{
"input": "13 21 34",
"output": "1"
},
{
"input": "34 21 55",
"output": "0"
},
{
"input": "10 20 84",
"output": "1"
},
{
"input": "4 2 5",
"output": "0"
},
{
"input": "3 2 3",
"output": "0"
},
{
"input": "31 23 1",
"output": "0"
},
{
"input": "100 100 100",
"output": "0"
},
{
"input": "100 1 100",
"output": "0"
},
{
"input": "1 100 100",
"output": "1"
},
{
"input": "1 1 100",
"output": "1"
},
{
"input": "17 23 100",
"output": "1"
},
{
"input": "99 98 100",
"output": "1"
},
{
"input": "1 1 2",
"output": "1"
},
{
"input": "2 1 1",
"output": "0"
},
{
"input": "1 2 1",
"output": "0"
},
{
"input": "2 1 2",
"output": "0"
},
{
"input": "10 10 3",
"output": "1"
},
{
"input": "2 2 1",
"output": "0"
},
{
"input": "9 5 1",
"output": "0"
}
] | 1,590,670,613
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 45
| 218
| 0
|
from math import gcd
a, b, n = list(map(int, input().split()))
amount = [a, b]
turn = 0
while True:
toPick = gcd(amount[turn], n)
if toPick > n:
print((turn+1)%2)
break
n -= toPick
if n == 0:
print(turn)
break
turn = (turn+1)%2
|
Title: Epic Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Simon and Antisimon play a game. Initially each player receives one fixed positive integer that doesn't change throughout the game. Simon receives number *a* and Antisimon receives number *b*. They also have a heap of *n* stones. The players take turns to make a move and Simon starts. During a move a player should take from the heap the number of stones equal to the greatest common divisor of the fixed number he has received and the number of stones left in the heap. A player loses when he cannot take the required number of stones (i. e. the heap has strictly less stones left than one needs to take).
Your task is to determine by the given *a*, *b* and *n* who wins the game.
Input Specification:
The only string contains space-separated integers *a*, *b* and *n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=100) — the fixed numbers Simon and Antisimon have received correspondingly and the initial number of stones in the pile.
Output Specification:
If Simon wins, print "0" (without the quotes), otherwise print "1" (without the quotes).
Demo Input:
['3 5 9\n', '1 1 100\n']
Demo Output:
['0', '1']
Note:
The greatest common divisor of two non-negative integers *a* and *b* is such maximum positive integer *k*, that *a* is divisible by *k* without remainder and similarly, *b* is divisible by *k* without remainder. Let *gcd*(*a*, *b*) represent the operation of calculating the greatest common divisor of numbers *a* and *b*. Specifically, *gcd*(*x*, 0) = *gcd*(0, *x*) = *x*.
In the first sample the game will go like that:
- Simon should take *gcd*(3, 9) = 3 stones from the heap. After his move the heap has 6 stones left.- Antisimon should take *gcd*(5, 6) = 1 stone from the heap. After his move the heap has 5 stones left.- Simon should take *gcd*(3, 5) = 1 stone from the heap. After his move the heap has 4 stones left.- Antisimon should take *gcd*(5, 4) = 1 stone from the heap. After his move the heap has 3 stones left.- Simon should take *gcd*(3, 3) = 3 stones from the heap. After his move the heap has 0 stones left.- Antisimon should take *gcd*(5, 0) = 5 stones from the heap. As 0 < 5, it is impossible and Antisimon loses.
In the second sample each player during each move takes one stone from the heap. As *n* is even, Antisimon takes the last stone and Simon can't make a move after that.
|
```python
from math import gcd
a, b, n = list(map(int, input().split()))
amount = [a, b]
turn = 0
while True:
toPick = gcd(amount[turn], n)
if toPick > n:
print((turn+1)%2)
break
n -= toPick
if n == 0:
print(turn)
break
turn = (turn+1)%2
```
| 3
|
|
250
|
A
|
Paper Work
|
PROGRAMMING
| 1,000
|
[
"greedy"
] | null | null |
Polycarpus has been working in the analytic department of the "F.R.A.U.D." company for as much as *n* days. Right now his task is to make a series of reports about the company's performance for the last *n* days. We know that the main information in a day report is value *a**i*, the company's profit on the *i*-th day. If *a**i* is negative, then the company suffered losses on the *i*-th day.
Polycarpus should sort the daily reports into folders. Each folder should include data on the company's performance for several consecutive days. Of course, the information on each of the *n* days should be exactly in one folder. Thus, Polycarpus puts information on the first few days in the first folder. The information on the several following days goes to the second folder, and so on.
It is known that the boss reads one daily report folder per day. If one folder has three or more reports for the days in which the company suffered losses (*a**i*<=<<=0), he loses his temper and his wrath is terrible.
Therefore, Polycarpus wants to prepare the folders so that none of them contains information on three or more days with the loss, and the number of folders is minimal.
Write a program that, given sequence *a**i*, will print the minimum number of folders.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=100), *n* is the number of days. The second line contains a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (|*a**i*|<=≤<=100), where *a**i* means the company profit on the *i*-th day. It is possible that the company has no days with the negative *a**i*.
|
Print an integer *k* — the required minimum number of folders. In the second line print a sequence of integers *b*1, *b*2, ..., *b**k*, where *b**j* is the number of day reports in the *j*-th folder.
If there are multiple ways to sort the reports into *k* days, print any of them.
|
[
"11\n1 2 3 -4 -5 -6 5 -5 -6 -7 6\n",
"5\n0 -1 100 -1 0\n"
] |
[
"3\n5 3 3 ",
"1\n5 "
] |
Here goes a way to sort the reports from the first sample into three folders:
In the second sample you can put all five reports in one folder.
| 500
|
[
{
"input": "11\n1 2 3 -4 -5 -6 5 -5 -6 -7 6",
"output": "3\n5 3 3 "
},
{
"input": "5\n0 -1 100 -1 0",
"output": "1\n5 "
},
{
"input": "1\n0",
"output": "1\n1 "
},
{
"input": "1\n-1",
"output": "1\n1 "
},
{
"input": "2\n0 0",
"output": "1\n2 "
},
{
"input": "2\n-2 2",
"output": "1\n2 "
},
{
"input": "2\n-2 -1",
"output": "1\n2 "
},
{
"input": "12\n1 -12 -5 -8 0 -8 -1 -1 -6 12 -9 12",
"output": "4\n3 3 2 4 "
},
{
"input": "4\n1 2 0 3",
"output": "1\n4 "
},
{
"input": "4\n4 -3 3 3",
"output": "1\n4 "
},
{
"input": "4\n0 -3 4 -3",
"output": "1\n4 "
},
{
"input": "4\n-3 -2 4 -3",
"output": "2\n1 3 "
},
{
"input": "4\n-3 -2 -1 -4",
"output": "2\n2 2 "
},
{
"input": "5\n-2 -2 4 0 -1",
"output": "2\n1 4 "
},
{
"input": "5\n-5 -3 -1 2 -1",
"output": "2\n2 3 "
},
{
"input": "5\n-3 -2 -3 -2 -3",
"output": "3\n1 2 2 "
},
{
"input": "10\n0 5 2 3 10 9 4 9 9 3",
"output": "1\n10 "
},
{
"input": "10\n10 2 1 2 9 10 7 4 -4 5",
"output": "1\n10 "
},
{
"input": "10\n1 -3 1 10 -7 -6 7 0 -5 3",
"output": "2\n5 5 "
},
{
"input": "10\n6 5 -10 -4 -3 -7 5 -2 -6 -10",
"output": "4\n3 2 3 2 "
},
{
"input": "10\n-2 -4 -1 -6 -5 -5 -7 0 -7 -8",
"output": "5\n1 2 2 2 3 "
},
{
"input": "100\n48 36 10 85 15 57 100 70 14 82 15 75 67 44 40 83 12 94 80 77 92 40 39 80 11 10 2 22 71 31 93 51 22 29 98 90 33 91 66 64 87 70 46 86 62 13 85 15 37 3 49 11 21 57 26 14 5 80 33 82 9 75 26 76 50 32 48 100 62 11 97 47 67 81 86 80 51 51 44 97 2 22 18 52 43 54 65 91 94 54 22 80 23 63 44 7 52 98 80 69",
"output": "1\n100 "
},
{
"input": "100\n7 51 31 14 17 0 72 29 77 6 32 94 70 94 1 64 85 29 67 66 56 -90 38 85 51 5 69 36 62 99 99 43 43 40 68 88 62 39 45 75 50 95 51 96 69 60 65 27 63 89 23 43 49 39 92 90 1 49 22 78 13 90 97 87 5 100 60 82 50 49 0 11 87 34 67 7 35 65 20 92 89 29 73 48 41 8 14 76 91 34 13 18 42 75 36 14 78 80 74 9",
"output": "1\n100 "
},
{
"input": "100\n83 71 43 50 61 54 -45 44 36 35 44 21 34 65 23 32 73 36 70 17 46 47 10 30 48 25 84 58 63 96 44 88 24 93 26 24 70 69 90 75 20 42 63 11 0 41 54 23 95 99 17 27 43 20 46 100 65 -79 15 72 78 0 13 94 76 72 69 35 61 3 65 83 28 12 27 48 8 37 30 37 40 87 30 76 81 78 71 44 79 92 10 60 5 7 9 33 79 31 86 51",
"output": "1\n100 "
},
{
"input": "100\n78 96 4 24 -66 42 28 16 42 -48 89 0 74 19 12 86 75 21 42 100 2 43 11 -76 85 24 12 51 26 48 22 74 68 73 22 39 53 42 37 -78 100 5 9 58 10 63 19 89 76 42 10 -96 76 49 67 59 86 37 13 66 75 92 48 80 37 59 49 -4 83 1 82 25 0 31 73 40 52 3 -47 17 68 94 51 84 47 76 73 -65 83 72 56 50 62 -5 40 12 81 75 84 -6",
"output": "5\n10 30 28 20 12 "
},
{
"input": "100\n-63 20 79 73 18 82 23 -93 55 8 -31 37 33 24 30 41 70 77 14 34 84 79 -94 88 54 81 7 90 74 35 29 3 75 71 14 28 -61 63 90 79 71 97 -90 74 -33 10 27 34 46 31 9 90 100 -73 58 2 73 51 5 46 -27 -9 30 65 73 28 15 14 1 59 96 21 100 78 12 97 72 37 -28 52 12 0 -42 84 88 8 88 8 -48 39 13 -78 20 56 38 82 32 -87 45 39",
"output": "8\n1 10 26 8 16 18 10 11 "
},
{
"input": "100\n21 40 60 28 85 10 15 -3 -27 -7 26 26 9 93 -3 -65 70 88 68 -85 24 75 24 -69 53 56 44 -53 -15 -74 12 22 37 22 77 90 9 95 40 15 -76 7 -81 65 83 51 -57 59 19 78 34 40 11 17 99 75 56 67 -81 39 22 86 -78 61 19 25 53 13 -91 91 17 71 45 39 63 32 -57 83 70 26 100 -53 7 95 67 -47 84 84 28 56 94 72 48 58 21 -89 91 73 16 93",
"output": "10\n9 6 5 8 2 13 16 10 13 18 "
},
{
"input": "100\n39 -70 7 7 11 27 88 16 -3 94 94 -2 23 91 41 49 69 61 53 -99 98 54 87 44 48 73 62 80 86 -33 34 -87 56 48 4 18 92 14 -37 84 7 42 9 70 0 -78 17 68 54 -82 65 -21 59 90 72 -19 -81 8 92 88 -68 65 -42 -60 98 -39 -2 2 88 24 9 -95 17 75 12 -32 -9 85 7 88 59 14 90 69 19 -88 -73 1 2 72 15 -83 65 18 26 25 -71 3 -51 95",
"output": "13\n2 10 18 9 11 6 5 3 3 9 10 6 8 "
},
{
"input": "100\n-47 -28 -90 -35 28 32 63 77 88 3 -48 18 48 22 47 47 89 2 88 46 25 60 65 44 100 28 73 71 19 -55 44 47 30 -25 50 15 -98 5 73 -56 61 15 15 77 67 59 -64 22 17 70 67 -12 26 -81 -58 -20 1 22 34 52 -45 56 78 29 47 -11 -10 70 -57 -2 62 85 -84 -54 -67 67 85 23 6 -65 -6 -79 -13 -1 12 68 1 71 73 77 48 -48 90 70 52 100 45 38 -43 -93",
"output": "15\n2 2 26 7 10 7 2 10 3 4 2 6 2 9 8 "
},
{
"input": "100\n-34 -61 96 14 87 33 29 64 -76 7 47 -41 54 60 79 -28 -18 88 95 29 -89 -29 52 39 8 13 68 13 15 46 -34 -49 78 -73 64 -56 83 -16 45 17 40 11 -86 55 56 -35 91 81 38 -77 -41 67 16 -37 -56 -84 -42 99 -83 45 46 -56 -14 -15 79 77 -48 -87 94 46 77 18 -32 16 -18 47 67 35 89 95 36 -32 51 46 40 78 0 58 81 -47 41 5 -48 65 89 6 -79 -56 -25 74",
"output": "18\n1 8 7 5 10 3 4 8 5 4 2 5 2 4 7 15 7 3 "
},
{
"input": "100\n14 36 94 -66 24 -24 14 -87 86 94 44 88 -68 59 4 -27 -74 12 -75 92 -31 29 18 -69 -47 45 -85 67 95 -77 7 -56 -80 -46 -40 73 40 71 41 -86 50 87 94 16 43 -96 96 -63 66 24 3 90 16 42 50 41 15 -45 72 32 -94 -93 91 -31 -30 -73 -88 33 45 9 71 18 37 -26 43 -82 87 67 62 -9 29 -70 -34 99 -30 -25 -86 -91 -70 -48 24 51 53 25 90 69 -17 -53 87 -62",
"output": "20\n6 7 4 4 4 5 3 2 11 12 4 3 2 9 6 3 2 2 8 3 "
},
{
"input": "100\n-40 87 -68 72 -49 48 -62 73 95 27 80 53 76 33 -95 -53 31 18 -61 -75 84 40 35 -82 49 47 -13 22 -81 -65 -17 47 -61 21 9 -12 52 67 31 -86 -63 42 18 -25 70 45 -3 -18 94 -62 -28 16 -100 36 -96 -73 83 -65 9 -51 83 36 65 -24 77 38 81 -84 32 -34 75 -50 -92 11 -73 -17 81 -66 -61 33 -47 -50 -72 -95 -58 54 68 -46 -41 8 76 28 58 87 88 100 61 -61 75 -1",
"output": "23\n1 4 10 4 5 5 2 5 5 6 3 3 3 4 8 4 3 3 3 2 2 4 11 "
},
{
"input": "100\n-61 56 1 -37 61 -77 -6 -5 28 36 27 -32 -10 -44 -89 -26 67 100 -94 80 -18 -5 -92 94 81 -38 -76 4 -77 2 79 55 -93 54 -19 10 -35 -12 -42 -32 -23 -67 -95 -62 -16 23 -25 41 -16 -51 3 -45 -1 53 20 0 0 21 87 28 15 62 64 -21 6 45 -19 95 -23 87 15 -35 21 -88 47 -81 89 68 66 -65 95 54 18 -97 65 -7 75 -58 -54 -3 99 -95 -57 -84 98 -6 33 44 81 -56",
"output": "25\n4 3 5 2 2 5 2 4 6 4 2 2 2 2 4 3 12 5 5 6 6 3 3 2 6 "
},
{
"input": "100\n-21 61 -52 47 -25 -42 -48 -46 58 -13 75 -65 52 88 -59 68 -12 -25 33 14 -2 78 32 -41 -79 17 0 85 -39 -80 61 30 -27 -92 -100 66 -53 -11 -59 65 -5 92 -2 -85 87 -72 19 -50 -24 32 -27 -92 -100 14 72 13 67 -22 -27 -56 -84 -90 -74 -70 44 -92 70 -49 -50 11 57 -73 23 68 65 99 82 -18 -93 -34 85 45 89 -58 -80 5 -57 -98 -11 -96 28 30 29 -71 47 50 -15 30 -96 -53",
"output": "28\n1 4 2 3 5 3 6 5 4 2 3 3 3 4 3 2 6 2 2 3 3 9 2 5 3 2 7 3 "
},
{
"input": "100\n-61 15 -88 52 -75 -71 -36 29 93 99 -73 -97 -69 39 -78 80 -28 -20 -36 -89 88 -82 56 -37 -13 33 2 -6 -88 -9 8 -24 40 5 8 -33 -83 -90 -48 55 69 -12 -49 -41 -4 92 42 57 -17 -68 -41 -68 77 -17 -45 -64 -39 24 -78 -3 -49 77 3 -23 84 -36 -19 -16 -72 74 -19 -81 65 -79 -57 64 89 -29 49 69 88 -18 16 26 -86 -58 -91 69 -43 -28 86 6 -87 47 -71 18 81 -55 -42 -30",
"output": "30\n3 3 5 2 4 2 3 3 4 3 5 2 4 2 5 2 3 2 3 4 3 2 3 3 7 4 3 4 5 2 "
},
{
"input": "100\n-21 -98 -66 26 3 -5 86 99 96 -22 78 -16 20 -3 93 22 -67 -37 -27 12 -97 43 -46 -48 -58 -4 -19 26 -87 -61 67 -76 -42 57 -87 -50 -24 -79 -6 43 -68 -42 13 -1 -82 81 -32 -88 -6 -99 46 42 19 -17 89 14 -98 -24 34 -37 -17 49 76 81 -61 23 -46 -79 -48 -5 87 14 -97 -67 -31 94 -77 15 -44 38 -44 -67 -69 -84 -58 -59 -17 -54 3 -15 79 -28 -10 -26 34 -73 -37 -57 -42 -44",
"output": "33\n1 2 7 4 4 3 3 2 3 3 3 2 2 3 3 3 2 7 3 5 3 2 4 3 4 2 2 2 3 3 3 2 2 "
},
{
"input": "100\n-63 -62 -88 -14 -58 -75 -28 19 -71 60 -38 77 98 95 -49 -64 -87 -97 2 -37 -37 -41 -47 -96 -58 -42 -88 12 -90 -65 0 52 -59 87 -79 -68 -66 -90 -19 -4 86 -65 -49 -94 67 93 -61 100 68 -40 -35 -67 -4 -100 -90 -86 15 -3 -75 57 65 -91 -80 -57 51 -88 -61 -54 -13 -46 -64 53 -87 -54 -69 29 -67 -23 -96 -93 -3 -77 -10 85 55 -44 17 24 -78 -82 -33 14 85 79 84 -91 -81 54 -89 -86",
"output": "35\n2 2 2 3 6 2 3 2 2 2 3 4 3 2 2 3 4 4 2 2 3 4 2 3 2 2 3 3 2 2 2 6 2 6 3 "
},
{
"input": "100\n30 -47 -87 -49 -4 -58 -10 -10 -37 -15 -12 -85 4 24 -3 -2 57 57 -60 94 -21 82 1 -54 -39 -98 -72 57 84 -6 -41 82 93 -81 -61 -30 18 -68 -88 17 87 -6 43 -26 72 -14 -40 -75 -69 60 -91 -70 -26 -62 -13 -19 -97 -14 -59 -17 -44 -15 -65 60 -60 74 26 -6 12 -83 -49 82 -76 -96 -31 -98 -100 49 -50 -42 -43 92 -56 -79 -38 -86 -99 -37 -75 -26 -79 -12 -9 -87 -63 -62 -25 -3 -5 -92",
"output": "38\n2 2 2 2 2 2 4 5 4 2 4 4 3 4 4 2 3 2 2 2 2 2 2 5 3 3 2 3 2 3 2 2 2 2 2 2 2 2 "
},
{
"input": "100\n-58 -18 -94 -96 -18 -2 -35 -49 47 69 96 -46 -88 -91 -9 -95 -12 -46 -12 16 44 -53 -96 71 -11 -98 -62 -27 -89 -88 -28 -11 -14 -47 67 -69 -33 -64 15 -24 67 53 -93 -10 -75 -98 -8 -97 -62 67 -52 -59 -9 -89 -39 -23 -37 -61 -83 -89 23 -47 -67 18 -38 -63 -73 -98 -65 -70 -20 13 -33 -46 -50 -30 -33 85 -93 -42 -37 48 -8 -11 -32 0 -58 -70 -27 -79 -52 82 22 -62 -100 -12 -5 -82 88 -74",
"output": "40\n2 2 2 2 5 2 2 2 4 3 2 2 2 2 3 3 4 2 2 3 2 2 2 2 3 3 2 2 2 3 2 3 2 3 3 2 2 4 2 3 "
},
{
"input": "100\n-60 -62 -19 -42 -50 -22 -90 -82 -56 40 87 -1 -30 -76 -8 -32 -57 38 -14 -39 84 -60 -28 -82 -62 -83 -37 -59 -61 -86 -13 48 18 -8 50 -27 -47 -100 -42 -88 -19 -45 30 -93 -46 3 -26 -80 -61 -13 -20 76 -95 -51 -26 -1 39 -92 -41 -76 -67 26 -23 30 79 -26 -51 -40 -29 -14 -2 -43 -30 -19 -62 -65 -1 -90 -66 -38 -50 89 -17 -53 -6 -13 -41 -54 -1 -23 -31 -88 -59 -44 -67 -11 -83 -16 -23 -71",
"output": "43\n1 2 2 2 2 4 2 2 3 3 2 2 2 2 5 2 2 2 3 3 2 3 2 3 2 3 4 2 2 2 2 2 2 2 3 2 2 2 2 2 2 2 2 "
},
{
"input": "100\n-1 -65 76 -28 -58 -63 -86 -54 -62 -66 -39 -3 -62 -35 -2 -86 -6 -16 -85 -30 -6 -41 -88 38 -8 -78 -6 -73 -83 -12 40 -99 -78 -51 -97 -15 81 -76 -1 -78 -38 -14 -24 -2 -70 -80 -24 -28 -51 -50 61 -64 -81 -32 -59 -60 -58 -10 -24 -81 -42 -7 58 -23 -11 -14 -84 -27 -45 2 -31 -32 -20 -72 -2 -81 -31 -6 -8 -91 55 -76 -93 -65 -94 -8 -57 -20 -75 -20 -27 -37 -82 97 -37 -8 -16 49 -90 -3",
"output": "45\n2 3 2 2 2 2 2 2 2 2 2 3 2 2 3 2 3 2 2 2 2 2 2 3 2 2 2 2 3 2 2 3 2 2 2 2 3 2 2 2 2 2 3 2 3 "
},
{
"input": "100\n-75 -29 -14 -2 99 -94 -75 82 -17 -19 -61 -18 -14 -94 -17 16 -16 -4 -41 -8 -81 -26 -65 24 -7 -87 -85 -22 -74 -21 46 -31 -39 -82 -88 -20 -2 -13 -46 -1 -78 -66 -83 -50 -13 -15 -60 -56 36 -79 -99 -52 -96 -80 -97 -74 80 -90 -52 -33 -1 -78 73 -45 -3 -77 62 -4 -85 -44 -62 -74 -33 -35 -44 -14 -80 -20 -17 -83 -32 -40 -74 -13 -90 -62 -15 -16 -59 -15 -40 -50 -98 -33 -73 -25 -86 -35 -84 -41",
"output": "46\n1 2 3 3 2 2 2 3 2 2 3 2 2 3 2 2 2 2 2 2 2 2 3 2 2 3 2 2 3 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 "
},
{
"input": "100\n-43 -90 -65 -70 -7 -49 -90 -93 -43 -80 -2 -47 -13 -5 -70 -42 -71 -68 -60 -71 -27 -84 82 -74 -75 -65 -32 -32 -50 -74 62 -96 -85 -95 -65 -51 -69 49 3 -19 -92 -61 -33 -7 -70 -51 -3 -1 -48 -48 -64 -7 -4 -46 -11 -36 -80 -69 -67 -1 -39 -40 66 -9 -40 -8 -58 -74 -27 66 -52 -26 -62 -72 -48 -25 -41 -13 -65 -82 -50 -68 -94 -52 -77 -91 -37 -18 -8 -51 -19 -22 -52 -95 35 -32 59 -41 -54 -88",
"output": "46\n2 2 2 2 2 2 2 2 2 2 2 3 2 2 3 2 2 4 2 2 2 2 2 2 2 2 2 2 2 3 2 2 3 2 2 2 2 2 2 2 2 2 2 2 4 2 "
},
{
"input": "100\n-67 -100 -7 -13 -9 -78 -55 -68 -31 -18 -92 -23 -4 -99 -54 -97 -45 -24 -33 -95 -42 -20 -63 -24 -89 -25 -55 -35 -84 -30 -1 57 -88 -94 -67 -27 -91 -14 -13 -20 -7 -8 -33 -95 -1 -75 -80 -49 -15 -64 -73 -49 -87 -19 -44 -50 -19 -10 -90 -51 -74 90 -42 -18 -93 -99 -43 51 -96 95 -97 -36 -21 -13 -73 -37 -33 -22 -83 -33 -44 -84 -20 -78 -34 -70 -83 -83 -85 -17 -36 62 83 -73 -6 51 -77 -82 -83 -68",
"output": "47\n1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 2 2 2 2 2 2 2 2 2 2 2 2 2 3 2 2 4 2 2 2 2 2 2 2 2 2 2 4 3 2 "
},
{
"input": "100\n-30 -40 -64 -50 -13 -69 -87 -54 -7 -32 -38 30 -79 -31 57 -50 -3 -6 -13 -94 -28 -57 -95 -67 -82 -49 -83 -39 -41 -12 -73 -20 -17 -46 -92 -31 -36 -31 -80 -47 -37 -67 -41 -65 -7 -95 -85 -53 -87 -18 -52 -61 -98 -85 -6 -80 -96 -95 -72 -9 -19 -49 74 84 -60 -69 -64 -39 -82 -28 -24 -82 -13 -7 -15 -28 -26 -48 -88 -9 -36 -38 -75 -1 9 -15 -12 -47 -11 -45 -3 -10 -60 -62 -54 -60 45 -8 -43 -89",
"output": "47\n2 2 2 2 2 3 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 4 2 2 2 2 2 2 2 2 2 3 2 2 2 2 3 2 "
},
{
"input": "100\n-78 -77 -84 -29 -99 -15 -60 97 -56 -9 -19 -21 -5 -29 -20 -41 -56 -15 -77 -22 -87 -75 -56 -96 -46 -24 -35 -64 63 -5 -16 -27 34 -77 84 -30 -9 -73 -58 -93 -20 -20 -69 -16 -42 -40 -44 -66 -42 -90 -47 -35 -87 -55 -37 -48 -34 -3 -40 -3 -46 -25 -80 -55 -12 -62 -46 -99 -38 -33 -72 -60 -18 -12 -52 -3 -75 -5 -48 -30 -59 -56 99 -52 -59 -72 -41 -15 -19 -19 -26 -28 -16 -23 -46 -93 -92 -38 -12 -75",
"output": "48\n1 2 2 2 3 2 2 2 2 2 2 2 2 2 3 3 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 2 2 2 2 2 2 2 2 "
},
{
"input": "100\n22 -83 -95 -61 -100 -53 -50 -19 -24 -85 -45 -43 -3 -74 -6 -24 -78 -54 -58 -52 -42 -16 -18 -56 -93 -45 -97 -67 -88 -27 83 -7 -72 -85 -24 -45 -22 -82 -83 -94 -75 -79 -22 -44 -22 -44 -42 -44 -61 85 -11 -16 -91 -12 -15 -3 -15 -82 -1 -2 -28 -24 -68 -22 -25 -46 -40 -21 -67 -90 -31 -33 -54 -83 -91 -74 -56 -67 -87 -36 -8 -100 -76 -88 -90 -45 -64 -25 -55 -15 -84 -67 -57 -73 -78 86 -28 -41 -63 -57",
"output": "48\n3 2 2 2 2 2 2 2 2 2 2 2 2 2 3 2 2 2 2 2 2 2 2 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 2 "
},
{
"input": "100\n-13 -43 -95 -61 -62 -94 -97 -48 -16 -88 -96 -74 -26 -58 -79 -44 -72 -22 -18 -66 -8 85 -98 -3 -36 -17 -80 -82 -77 -41 -24 -86 -62 -1 -22 -29 -30 -18 -25 -90 -66 -58 -86 -81 -34 -76 -67 -72 -77 -29 -66 -67 -34 3 -16 -90 -9 -14 -28 -60 -26 -99 75 -74 -94 -55 -54 -23 -30 -34 -4 -92 -88 -46 -52 -63 -98 -6 -89 -99 -80 -100 -97 -62 -70 -97 -75 -85 -22 -2 -32 -47 -85 -44 -23 -4 -21 -30 -6 -34",
"output": "49\n1 2 2 2 2 2 2 2 2 2 2 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 2 2 2 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 "
},
{
"input": "100\n-5 -37 -22 -85 -63 -46 -44 -43 -23 -77 -75 -64 -84 -46 -78 -94 -67 -19 -5 -59 -32 -92 -10 -92 -58 -73 -72 -16 99 -58 -94 -49 -60 -3 -60 -74 -12 -8 -32 -94 -63 -53 -24 -29 -6 -46 -30 -32 -87 -41 -58 -70 -53 -20 -73 -42 -54 -5 -84 -45 -11 -9 -84 -7 -68 -100 -11 -2 -87 -27 -65 -45 -17 -33 -88 -55 90 -58 -89 -13 -66 -1 -46 -90 -69 -74 -84 -90 -50 -32 -62 -37 -44 -51 -25 -94 -73 -43 -1 -44",
"output": "49\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 2 2 2 2 2 2 2 2 2 2 2 "
},
{
"input": "100\n-76 -48 -63 -62 -94 -37 -54 -67 -9 -52 -83 -1 -87 -36 -94 -10 -19 -55 -93 -23 -2 -87 -15 -59 -60 -87 -63 -18 -62 -92 -10 -61 -12 -89 -85 -38 -37 -3 -71 -22 -94 -96 -100 -47 -20 -93 -28 77 -35 -74 -50 -72 -38 -29 -58 -80 -24 -9 -59 -4 -93 -65 -31 -47 -36 -13 -89 -96 -99 -83 -99 -36 -45 -58 -22 -93 -51 -26 -93 -36 -85 -72 -49 -27 -69 -29 -51 -84 -35 -26 -41 -43 -45 -87 -65 -100 -45 -69 -69 -73",
"output": "50\n1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 "
},
{
"input": "100\n-77 -6 -71 -86 -42 -1 -40 -41 -31 -67 -75 -49 -62 -21 -2 -40 -2 -82 -90 -42 -43 -14 -72 -50 -33 -37 -58 -51 -67 -96 -63 -39 -56 -22 -17 -69 -88 -60 -18 -47 -16 -41 -32 -59 -82 -48 -22 -46 -29 -69 -21 -2 -41 -52 -83 -3 -49 -39 -31 -78 -60 -100 -12 -64 -28 -72 -43 -68 -60 -98 -21 -29 -72 -82 -5 -4 -65 -76 -60 -40 -37 -17 -77 -21 -19 -98 -39 -67 -49 -75 -7 -45 -11 -13 -45 -19 -83 -38 -14 -89",
"output": "50\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 "
},
{
"input": "4\n1 2 3 4",
"output": "1\n4 "
},
{
"input": "4\n1 2 3 -4",
"output": "1\n4 "
},
{
"input": "4\n-4 2 1 2",
"output": "1\n4 "
},
{
"input": "1\n-1",
"output": "1\n1 "
},
{
"input": "2\n2 -1",
"output": "1\n2 "
},
{
"input": "2\n-100 100",
"output": "1\n2 "
},
{
"input": "3\n-100 0 -100",
"output": "1\n3 "
},
{
"input": "5\n1 2 3 -1 -1",
"output": "1\n5 "
},
{
"input": "5\n-1 -1 2 3 4",
"output": "1\n5 "
},
{
"input": "3\n-3 -4 -5",
"output": "2\n1 2 "
},
{
"input": "4\n-3 -4 1 -3",
"output": "2\n1 3 "
},
{
"input": "1\n-1",
"output": "1\n1 "
},
{
"input": "2\n-1 0",
"output": "1\n2 "
},
{
"input": "4\n0 0 0 0",
"output": "1\n4 "
},
{
"input": "3\n-1 -1 -1",
"output": "2\n1 2 "
},
{
"input": "6\n-1 -1 0 -1 -1 -1",
"output": "3\n1 3 2 "
},
{
"input": "2\n0 0",
"output": "1\n2 "
},
{
"input": "6\n0 0 -1 -1 -1 0",
"output": "2\n3 3 "
}
] | 1,476,939,657
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
P=print
I=input
I()
a=b=0
t=[]
for x in map(int,I().split()):
if x<0:
if b==2:t+=[a];a=b=1
else:a+=1;b+=1
else:a+=1;b=0
if a:t+=[a]
P(len(t))
for x in t:P(x,end=' ')
|
Title: Paper Work
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus has been working in the analytic department of the "F.R.A.U.D." company for as much as *n* days. Right now his task is to make a series of reports about the company's performance for the last *n* days. We know that the main information in a day report is value *a**i*, the company's profit on the *i*-th day. If *a**i* is negative, then the company suffered losses on the *i*-th day.
Polycarpus should sort the daily reports into folders. Each folder should include data on the company's performance for several consecutive days. Of course, the information on each of the *n* days should be exactly in one folder. Thus, Polycarpus puts information on the first few days in the first folder. The information on the several following days goes to the second folder, and so on.
It is known that the boss reads one daily report folder per day. If one folder has three or more reports for the days in which the company suffered losses (*a**i*<=<<=0), he loses his temper and his wrath is terrible.
Therefore, Polycarpus wants to prepare the folders so that none of them contains information on three or more days with the loss, and the number of folders is minimal.
Write a program that, given sequence *a**i*, will print the minimum number of folders.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100), *n* is the number of days. The second line contains a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (|*a**i*|<=≤<=100), where *a**i* means the company profit on the *i*-th day. It is possible that the company has no days with the negative *a**i*.
Output Specification:
Print an integer *k* — the required minimum number of folders. In the second line print a sequence of integers *b*1, *b*2, ..., *b**k*, where *b**j* is the number of day reports in the *j*-th folder.
If there are multiple ways to sort the reports into *k* days, print any of them.
Demo Input:
['11\n1 2 3 -4 -5 -6 5 -5 -6 -7 6\n', '5\n0 -1 100 -1 0\n']
Demo Output:
['3\n5 3 3 ', '1\n5 ']
Note:
Here goes a way to sort the reports from the first sample into three folders:
In the second sample you can put all five reports in one folder.
|
```python
P=print
I=input
I()
a=b=0
t=[]
for x in map(int,I().split()):
if x<0:
if b==2:t+=[a];a=b=1
else:a+=1;b+=1
else:a+=1;b=0
if a:t+=[a]
P(len(t))
for x in t:P(x,end=' ')
```
| 0
|
|
426
|
A
|
Sereja and Mugs
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Sereja showed an interesting game to his friends. The game goes like that. Initially, there is a table with an empty cup and *n* water mugs on it. Then all players take turns to move. During a move, a player takes a non-empty mug of water and pours all water from it into the cup. If the cup overfills, then we assume that this player lost.
As soon as Sereja's friends heard of the game, they wanted to play it. Sereja, on the other hand, wanted to find out whether his friends can play the game in such a way that there are no losers. You are given the volumes of all mugs and the cup. Also, you know that Sereja has (*n*<=-<=1) friends. Determine if Sereja's friends can play the game so that nobody loses.
|
The first line contains integers *n* and *s* (2<=≤<=*n*<=≤<=100; 1<=≤<=*s*<=≤<=1000) — the number of mugs and the volume of the cup. The next line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=10). Number *a**i* means the volume of the *i*-th mug.
|
In a single line, print "YES" (without the quotes) if his friends can play in the described manner, and "NO" (without the quotes) otherwise.
|
[
"3 4\n1 1 1\n",
"3 4\n3 1 3\n",
"3 4\n4 4 4\n"
] |
[
"YES\n",
"YES\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "3 4\n1 1 1",
"output": "YES"
},
{
"input": "3 4\n3 1 3",
"output": "YES"
},
{
"input": "3 4\n4 4 4",
"output": "NO"
},
{
"input": "2 1\n1 10",
"output": "YES"
},
{
"input": "3 12\n5 6 6",
"output": "YES"
},
{
"input": "4 10\n6 3 8 7",
"output": "NO"
},
{
"input": "5 16\n3 3 2 7 9",
"output": "YES"
},
{
"input": "6 38\n9 10 3 8 10 6",
"output": "YES"
},
{
"input": "7 12\n4 4 5 2 2 4 9",
"output": "NO"
},
{
"input": "8 15\n8 10 4 2 10 9 7 6",
"output": "NO"
},
{
"input": "9 22\n1 3 5 9 7 6 1 10 1",
"output": "NO"
},
{
"input": "10 30\n9 10 4 5 5 7 1 7 7 2",
"output": "NO"
},
{
"input": "38 83\n9 9 3 10 2 4 6 10 9 5 1 8 7 4 7 2 6 5 3 1 10 8 4 8 3 7 1 2 7 6 8 6 5 2 3 1 1 2",
"output": "NO"
},
{
"input": "84 212\n6 2 3 1 2 7 5 1 7 2 9 10 9 5 2 5 4 10 9 9 1 9 8 8 9 4 9 4 8 2 1 8 4 5 10 7 6 2 1 10 10 7 9 4 5 9 5 10 10 3 6 6 4 4 4 8 5 4 9 1 9 9 1 7 9 2 10 9 10 8 3 3 9 3 9 10 1 8 9 2 6 9 7 2",
"output": "NO"
},
{
"input": "8 50\n8 8 8 4 4 6 10 10",
"output": "YES"
},
{
"input": "7 24\n1 4 9 1 2 3 6",
"output": "YES"
},
{
"input": "47 262\n3 7 6 4 10 3 5 7 2 9 3 2 2 10 8 7 3 10 6 3 1 1 4 10 2 9 2 10 6 4 3 6 3 6 9 7 8 8 3 3 10 5 2 10 7 10 9",
"output": "YES"
},
{
"input": "42 227\n3 6 1 9 4 10 4 10 7 8 10 10 8 7 10 4 6 8 7 7 6 9 3 6 5 5 2 7 2 7 4 4 6 6 4 3 9 3 6 4 7 2",
"output": "NO"
},
{
"input": "97 65\n3 10 2 6 1 4 7 5 10 3 10 4 5 5 1 6 10 7 4 5 3 9 9 8 6 9 2 3 6 8 5 5 5 5 5 3 10 4 1 8 8 9 8 4 1 4 9 3 6 3 1 4 8 3 10 8 6 4 5 4 3 2 2 4 3 6 4 6 2 3 3 3 7 5 1 8 1 4 5 1 1 6 4 2 1 7 8 6 1 1 5 6 5 10 6 7 5",
"output": "NO"
},
{
"input": "94 279\n2 5 9 5 10 3 1 8 1 7 1 8 1 6 7 8 4 9 5 10 3 7 6 8 8 5 6 8 10 9 4 1 3 3 4 7 8 2 6 6 5 1 3 7 1 7 2 2 2 8 4 1 1 5 9 4 1 2 3 10 1 4 9 9 6 8 8 1 9 10 4 1 8 5 8 9 4 8 2 1 1 9 4 5 6 1 2 5 6 7 3 1 4 6",
"output": "NO"
},
{
"input": "58 70\n8 2 10 2 7 3 8 3 8 7 6 2 4 10 10 6 10 3 7 6 4 3 5 5 5 3 8 10 3 4 8 4 2 6 8 9 6 9 4 3 5 2 2 6 10 6 2 1 7 5 6 4 1 9 10 2 4 5",
"output": "NO"
},
{
"input": "6 14\n3 9 2 1 4 2",
"output": "YES"
},
{
"input": "78 400\n5 9 3 4 7 4 1 4 6 3 9 1 8 3 3 6 10 2 1 9 6 1 8 10 1 6 4 5 2 1 5 9 6 10 3 6 5 2 4 10 6 9 3 8 10 7 2 8 8 2 10 1 4 5 2 8 6 4 4 3 5 2 3 10 1 9 8 5 6 7 9 1 8 8 5 4 2 4",
"output": "YES"
},
{
"input": "41 181\n5 3 10 4 2 5 9 3 1 6 6 10 4 3 9 8 5 9 2 5 4 6 6 3 7 9 10 3 10 6 10 5 6 1 6 9 9 1 2 4 3",
"output": "NO"
},
{
"input": "2 4\n4 4",
"output": "YES"
},
{
"input": "29 71\n4 8 9 4 8 10 4 10 2 9 3 9 1 2 9 5 9 7 1 10 4 1 1 9 8 7 4 6 7",
"output": "NO"
},
{
"input": "49 272\n4 10 8 7 5 6 9 7 2 6 6 2 10 7 5 6 5 3 6 4 3 7 9 3 7 7 4 10 5 6 7 3 6 4 6 7 7 2 5 5 7 3 7 9 3 6 6 2 1",
"output": "YES"
},
{
"input": "91 486\n1 3 5 4 4 7 3 9 3 4 5 4 5 4 7 9 5 8 4 10 9 1 1 9 9 1 6 2 5 4 7 4 10 3 2 10 9 3 4 5 1 3 4 2 10 9 10 9 10 2 4 6 2 5 3 6 4 9 10 3 9 8 1 2 5 9 2 10 4 6 10 8 10 9 1 2 5 8 6 6 6 1 10 3 9 3 5 6 1 5 5",
"output": "YES"
},
{
"input": "80 78\n1 9 4 9 8 3 7 10 4 9 2 1 4 4 9 5 9 1 2 6 5 2 4 8 4 6 9 6 7 10 1 9 10 4 7 1 7 10 8 9 10 5 2 6 7 7 7 7 7 8 2 5 1 7 2 3 2 5 10 6 3 4 5 2 6 3 4 2 7 9 9 3 8 8 2 3 7 1 5 10",
"output": "NO"
},
{
"input": "53 245\n5 6 9 9 2 3 2 5 10 9 3 5 6 3 10 10 9 4 9 7 10 9 7 7 3 4 9 3 7 3 8 6 8 9 3 8 9 1 3 1 9 10 3 9 3 1 6 6 3 8 7 8 9",
"output": "NO"
},
{
"input": "50 271\n6 9 10 1 1 1 8 3 6 6 3 2 5 9 7 5 7 9 10 9 4 6 6 2 6 6 9 5 1 6 5 8 3 2 5 10 10 1 4 1 4 6 1 8 7 8 9 4 7 5",
"output": "YES"
},
{
"input": "38 214\n5 8 4 5 1 9 9 2 6 3 4 3 5 7 7 7 3 10 1 5 10 4 2 2 10 10 6 6 6 7 1 6 10 5 7 4 5 10",
"output": "YES"
},
{
"input": "100 1\n10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10",
"output": "NO"
},
{
"input": "100 100\n10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10",
"output": "NO"
},
{
"input": "100 989\n10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10",
"output": "NO"
},
{
"input": "100 990\n10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10",
"output": "YES"
},
{
"input": "100 1000\n10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10",
"output": "YES"
},
{
"input": "2 1\n1 1",
"output": "YES"
},
{
"input": "2 1\n2 2",
"output": "NO"
},
{
"input": "2 1000\n1 1",
"output": "YES"
},
{
"input": "10 89\n10 10 10 10 10 10 10 10 10 10",
"output": "NO"
},
{
"input": "10 44\n1 10 2 3 4 5 6 7 8 9",
"output": "NO"
}
] | 1,572,452,650
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 7
| 108
| 0
|
n, m = map(int, input().split())
s = list(map(int, input().split()))
total = 0
t = False
for i in range(len(s)-1):
total += s[i]
if total > m:
print("NO")
t = True
break
if t == False:
print("YES")
|
Title: Sereja and Mugs
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sereja showed an interesting game to his friends. The game goes like that. Initially, there is a table with an empty cup and *n* water mugs on it. Then all players take turns to move. During a move, a player takes a non-empty mug of water and pours all water from it into the cup. If the cup overfills, then we assume that this player lost.
As soon as Sereja's friends heard of the game, they wanted to play it. Sereja, on the other hand, wanted to find out whether his friends can play the game in such a way that there are no losers. You are given the volumes of all mugs and the cup. Also, you know that Sereja has (*n*<=-<=1) friends. Determine if Sereja's friends can play the game so that nobody loses.
Input Specification:
The first line contains integers *n* and *s* (2<=≤<=*n*<=≤<=100; 1<=≤<=*s*<=≤<=1000) — the number of mugs and the volume of the cup. The next line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=10). Number *a**i* means the volume of the *i*-th mug.
Output Specification:
In a single line, print "YES" (without the quotes) if his friends can play in the described manner, and "NO" (without the quotes) otherwise.
Demo Input:
['3 4\n1 1 1\n', '3 4\n3 1 3\n', '3 4\n4 4 4\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
none
|
```python
n, m = map(int, input().split())
s = list(map(int, input().split()))
total = 0
t = False
for i in range(len(s)-1):
total += s[i]
if total > m:
print("NO")
t = True
break
if t == False:
print("YES")
```
| 0
|
|
919
|
D
|
Substring
|
PROGRAMMING
| 1,700
|
[
"dfs and similar",
"dp",
"graphs"
] | null | null |
You are given a graph with $n$ nodes and $m$ directed edges. One lowercase letter is assigned to each node. We define a path's value as the number of the most frequently occurring letter. For example, if letters on a path are "abaca", then the value of that path is $3$. Your task is find a path whose value is the largest.
|
The first line contains two positive integers $n, m$ ($1 \leq n, m \leq 300\,000$), denoting that the graph has $n$ nodes and $m$ directed edges.
The second line contains a string $s$ with only lowercase English letters. The $i$-th character is the letter assigned to the $i$-th node.
Then $m$ lines follow. Each line contains two integers $x, y$ ($1 \leq x, y \leq n$), describing a directed edge from $x$ to $y$. Note that $x$ can be equal to $y$ and there can be multiple edges between $x$ and $y$. Also the graph can be not connected.
|
Output a single line with a single integer denoting the largest value. If the value can be arbitrarily large, output -1 instead.
|
[
"5 4\nabaca\n1 2\n1 3\n3 4\n4 5\n",
"6 6\nxzyabc\n1 2\n3 1\n2 3\n5 4\n4 3\n6 4\n",
"10 14\nxzyzyzyzqx\n1 2\n2 4\n3 5\n4 5\n2 6\n6 8\n6 5\n2 10\n3 9\n10 9\n4 6\n1 10\n2 8\n3 7\n"
] |
[
"3\n",
"-1\n",
"4\n"
] |
In the first sample, the path with largest value is $1 \to 3 \to 4 \to 5$. The value is $3$ because the letter 'a' appears $3$ times.
| 1,500
|
[
{
"input": "5 4\nabaca\n1 2\n1 3\n3 4\n4 5",
"output": "3"
},
{
"input": "6 6\nxzyabc\n1 2\n3 1\n2 3\n5 4\n4 3\n6 4",
"output": "-1"
},
{
"input": "10 14\nxzyzyzyzqx\n1 2\n2 4\n3 5\n4 5\n2 6\n6 8\n6 5\n2 10\n3 9\n10 9\n4 6\n1 10\n2 8\n3 7",
"output": "4"
},
{
"input": "1 1\nf\n1 1",
"output": "-1"
},
{
"input": "10 50\nebibwbjihv\n1 10\n1 2\n5 4\n1 8\n9 7\n5 6\n1 8\n8 7\n2 6\n5 4\n1 9\n3 2\n8 3\n5 6\n5 9\n2 4\n2 7\n3 9\n1 2\n1 7\n1 10\n3 7\n1 8\n3 10\n8 6\n1 7\n10 6\n1 6\n5 8\n1 5\n2 10\n3 9\n5 8\n8 3\n3 7\n5 2\n1 10\n1 4\n5 3\n3 2\n1 2\n5 8\n10 4\n2 10\n8 2\n1 9\n1 8\n1 2\n3 4\n1 8",
"output": "2"
},
{
"input": "13 37\ndwpzcppjmhkmz\n2 6\n3 6\n6 7\n6 7\n6 7\n6 7\n6 8\n6 8\n6 8\n6 8\n4 6\n4 6\n5 6\n4 6\n4 6\n6 9\n6 9\n6 10\n6 10\n6 10\n6 10\n4 6\n1 6\n1 6\n10 11\n6 11\n1 6\n6 12\n6 12\n6 12\n6 13\n6 13\n6 13\n6 13\n3 6\n2 6\n2 6",
"output": "3"
},
{
"input": "5 8\ntetqw\n2 1\n4 4\n5 5\n5 2\n4 5\n1 5\n1 5\n1 1",
"output": "-1"
},
{
"input": "5 8\nreeet\n4 3\n2 5\n4 2\n2 4\n4 2\n5 2\n3 3\n3 4",
"output": "-1"
}
] | 1,676,989,850
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
from collections import defaultdict
x=list(map(int,input().split()))
n=x[0]
m=x[1]
s=input()
graph=defaultdict(list)
for i in range(m):
y=list(map(int,input().split()))
graph[y[0]].append(y[1])
x={}
max=-1
for i in s:
if i in x:
x[i]+=1
if max< x[i]:
max=x[i]
else:
x[i]=1
if max<=1:
return -1
else:
return max
|
Title: Substring
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a graph with $n$ nodes and $m$ directed edges. One lowercase letter is assigned to each node. We define a path's value as the number of the most frequently occurring letter. For example, if letters on a path are "abaca", then the value of that path is $3$. Your task is find a path whose value is the largest.
Input Specification:
The first line contains two positive integers $n, m$ ($1 \leq n, m \leq 300\,000$), denoting that the graph has $n$ nodes and $m$ directed edges.
The second line contains a string $s$ with only lowercase English letters. The $i$-th character is the letter assigned to the $i$-th node.
Then $m$ lines follow. Each line contains two integers $x, y$ ($1 \leq x, y \leq n$), describing a directed edge from $x$ to $y$. Note that $x$ can be equal to $y$ and there can be multiple edges between $x$ and $y$. Also the graph can be not connected.
Output Specification:
Output a single line with a single integer denoting the largest value. If the value can be arbitrarily large, output -1 instead.
Demo Input:
['5 4\nabaca\n1 2\n1 3\n3 4\n4 5\n', '6 6\nxzyabc\n1 2\n3 1\n2 3\n5 4\n4 3\n6 4\n', '10 14\nxzyzyzyzqx\n1 2\n2 4\n3 5\n4 5\n2 6\n6 8\n6 5\n2 10\n3 9\n10 9\n4 6\n1 10\n2 8\n3 7\n']
Demo Output:
['3\n', '-1\n', '4\n']
Note:
In the first sample, the path with largest value is $1 \to 3 \to 4 \to 5$. The value is $3$ because the letter 'a' appears $3$ times.
|
```python
from collections import defaultdict
x=list(map(int,input().split()))
n=x[0]
m=x[1]
s=input()
graph=defaultdict(list)
for i in range(m):
y=list(map(int,input().split()))
graph[y[0]].append(y[1])
x={}
max=-1
for i in s:
if i in x:
x[i]+=1
if max< x[i]:
max=x[i]
else:
x[i]=1
if max<=1:
return -1
else:
return max
```
| -1
|
|
336
|
A
|
Vasily the Bear and Triangle
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] | null | null |
Vasily the bear has a favorite rectangle, it has one vertex at point (0,<=0), and the opposite vertex at point (*x*,<=*y*). Of course, the sides of Vasya's favorite rectangle are parallel to the coordinate axes.
Vasya also loves triangles, if the triangles have one vertex at point *B*<==<=(0,<=0). That's why today he asks you to find two points *A*<==<=(*x*1,<=*y*1) and *C*<==<=(*x*2,<=*y*2), such that the following conditions hold:
- the coordinates of points: *x*1, *x*2, *y*1, *y*2 are integers. Besides, the following inequation holds: *x*1<=<<=*x*2; - the triangle formed by point *A*, *B* and *C* is rectangular and isosceles ( is right); - all points of the favorite rectangle are located inside or on the border of triangle *ABC*; - the area of triangle *ABC* is as small as possible.
Help the bear, find the required points. It is not so hard to proof that these points are unique.
|
The first line contains two integers *x*,<=*y* (<=-<=109<=≤<=*x*,<=*y*<=≤<=109,<=*x*<=≠<=0,<=*y*<=≠<=0).
|
Print in the single line four integers *x*1,<=*y*1,<=*x*2,<=*y*2 — the coordinates of the required points.
|
[
"10 5\n",
"-10 5\n"
] |
[
"0 15 15 0\n",
"-15 0 0 15\n"
] |
<img class="tex-graphics" src="https://espresso.codeforces.com/a9ea2088c4294ce8f23801562fda36b830df2c3f.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Figure to the first sample
| 500
|
[
{
"input": "10 5",
"output": "0 15 15 0"
},
{
"input": "-10 5",
"output": "-15 0 0 15"
},
{
"input": "20 -10",
"output": "0 -30 30 0"
},
{
"input": "-10 -1000000000",
"output": "-1000000010 0 0 -1000000010"
},
{
"input": "-1000000000 -1000000000",
"output": "-2000000000 0 0 -2000000000"
},
{
"input": "1000000000 1000000000",
"output": "0 2000000000 2000000000 0"
},
{
"input": "-123131 3123141",
"output": "-3246272 0 0 3246272"
},
{
"input": "-23423 -243242423",
"output": "-243265846 0 0 -243265846"
},
{
"input": "123112 4560954",
"output": "0 4684066 4684066 0"
},
{
"input": "1321 -23131",
"output": "0 -24452 24452 0"
},
{
"input": "1000000000 999999999",
"output": "0 1999999999 1999999999 0"
},
{
"input": "54543 432423",
"output": "0 486966 486966 0"
},
{
"input": "1 1",
"output": "0 2 2 0"
},
{
"input": "-1 -1",
"output": "-2 0 0 -2"
},
{
"input": "-1 1",
"output": "-2 0 0 2"
},
{
"input": "1 -1",
"output": "0 -2 2 0"
},
{
"input": "42 -2",
"output": "0 -44 44 0"
},
{
"input": "2 -435",
"output": "0 -437 437 0"
},
{
"input": "76 -76",
"output": "0 -152 152 0"
},
{
"input": "1000000000 1",
"output": "0 1000000001 1000000001 0"
},
{
"input": "1000000000 -1",
"output": "0 -1000000001 1000000001 0"
},
{
"input": "-1000000000 1",
"output": "-1000000001 0 0 1000000001"
},
{
"input": "-1000000000 -1",
"output": "-1000000001 0 0 -1000000001"
},
{
"input": "1000000000 -999999999",
"output": "0 -1999999999 1999999999 0"
},
{
"input": "-1000000000 999999999",
"output": "-1999999999 0 0 1999999999"
},
{
"input": "-1000000000 -999999999",
"output": "-1999999999 0 0 -1999999999"
},
{
"input": "999999999 1000000000",
"output": "0 1999999999 1999999999 0"
},
{
"input": "-999999999 1000000000",
"output": "-1999999999 0 0 1999999999"
},
{
"input": "999999999 -1000000000",
"output": "0 -1999999999 1999999999 0"
},
{
"input": "-999999999 -1000000000",
"output": "-1999999999 0 0 -1999999999"
}
] | 1,603,186,268
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 218
| 0
|
x,y=map(int,input().split())
q=abs(x)+abs(y)
if x>0and y>0:print(0,q,q,0)
elif x>0:print(0,q,-q,0)
elif y>0:print(-q,0,0,q)
else:print(-q,0,0-q)
|
Title: Vasily the Bear and Triangle
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasily the bear has a favorite rectangle, it has one vertex at point (0,<=0), and the opposite vertex at point (*x*,<=*y*). Of course, the sides of Vasya's favorite rectangle are parallel to the coordinate axes.
Vasya also loves triangles, if the triangles have one vertex at point *B*<==<=(0,<=0). That's why today he asks you to find two points *A*<==<=(*x*1,<=*y*1) and *C*<==<=(*x*2,<=*y*2), such that the following conditions hold:
- the coordinates of points: *x*1, *x*2, *y*1, *y*2 are integers. Besides, the following inequation holds: *x*1<=<<=*x*2; - the triangle formed by point *A*, *B* and *C* is rectangular and isosceles ( is right); - all points of the favorite rectangle are located inside or on the border of triangle *ABC*; - the area of triangle *ABC* is as small as possible.
Help the bear, find the required points. It is not so hard to proof that these points are unique.
Input Specification:
The first line contains two integers *x*,<=*y* (<=-<=109<=≤<=*x*,<=*y*<=≤<=109,<=*x*<=≠<=0,<=*y*<=≠<=0).
Output Specification:
Print in the single line four integers *x*1,<=*y*1,<=*x*2,<=*y*2 — the coordinates of the required points.
Demo Input:
['10 5\n', '-10 5\n']
Demo Output:
['0 15 15 0\n', '-15 0 0 15\n']
Note:
<img class="tex-graphics" src="https://espresso.codeforces.com/a9ea2088c4294ce8f23801562fda36b830df2c3f.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Figure to the first sample
|
```python
x,y=map(int,input().split())
q=abs(x)+abs(y)
if x>0and y>0:print(0,q,q,0)
elif x>0:print(0,q,-q,0)
elif y>0:print(-q,0,0,q)
else:print(-q,0,0-q)
```
| 0
|
|
71
|
A
|
Way Too Long Words
|
PROGRAMMING
| 800
|
[
"strings"
] |
A. Way Too Long Words
|
1
|
256
|
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
|
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
|
[
"4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n"
] |
[
"word\nl10n\ni18n\np43s\n"
] |
none
| 500
|
[
{
"input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis",
"output": "word\nl10n\ni18n\np43s"
},
{
"input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm",
"output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m"
},
{
"input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd",
"output": "n20n\nj27b\nk15d"
},
{
"input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt",
"output": "t41t"
},
{
"input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings",
"output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz",
"output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz"
},
{
"input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij",
"output": "a98j"
},
{
"input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco",
"output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco"
},
{
"input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt",
"output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t"
},
{
"input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr",
"output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr"
}
] | 1,688,282,221
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
a = input()
x = len(a)
if x>=10:
li = []
li.append(a[0])
li.append(x-2)
li.append(a[-1])
print(''.join(map(str,li)))
else:
print(a)
|
Title: Way Too Long Words
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output Specification:
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
Demo Input:
['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n']
Demo Output:
['word\nl10n\ni18n\np43s\n']
Note:
none
|
```python
a = input()
x = len(a)
if x>=10:
li = []
li.append(a[0])
li.append(x-2)
li.append(a[-1])
print(''.join(map(str,li)))
else:
print(a)
```
| 0
|
16
|
A
|
Flag
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Flag
|
2
|
64
|
According to a new ISO standard, a flag of every country should have a chequered field *n*<=×<=*m*, each square should be of one of 10 colours, and the flag should be «striped»: each horizontal row of the flag should contain squares of the same colour, and the colours of adjacent horizontal rows should be different. Berland's government asked you to find out whether their flag meets the new ISO standard.
|
The first line of the input contains numbers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100), *n* — the amount of rows, *m* — the amount of columns on the flag of Berland. Then there follows the description of the flag: each of the following *n* lines contain *m* characters. Each character is a digit between 0 and 9, and stands for the colour of the corresponding square.
|
Output YES, if the flag meets the new ISO standard, and NO otherwise.
|
[
"3 3\n000\n111\n222\n",
"3 3\n000\n000\n111\n",
"3 3\n000\n111\n002\n"
] |
[
"YES\n",
"NO\n",
"NO\n"
] |
none
| 0
|
[
{
"input": "3 3\n000\n111\n222",
"output": "YES"
},
{
"input": "3 3\n000\n000\n111",
"output": "NO"
},
{
"input": "3 3\n000\n111\n002",
"output": "NO"
},
{
"input": "10 10\n2222222222\n5555555555\n0000000000\n4444444444\n1111111111\n3333333393\n3333333333\n5555555555\n0000000000\n8888888888",
"output": "NO"
},
{
"input": "10 13\n4442444444444\n8888888888888\n6666666666666\n0000000000000\n3333333333333\n4444444444444\n7777777777777\n8388888888888\n1111111111111\n5555555555555",
"output": "NO"
},
{
"input": "10 8\n33333333\n44444444\n11111115\n81888888\n44444444\n11111111\n66666666\n33330333\n33333333\n33333333",
"output": "NO"
},
{
"input": "5 5\n88888\n44444\n66666\n55555\n88888",
"output": "YES"
},
{
"input": "20 19\n1111111111111111111\n5555555555555555555\n0000000000000000000\n3333333333333333333\n1111111111111111111\n2222222222222222222\n4444444444444444444\n5555555555555555555\n0000000000000000000\n4444444444444444444\n0000000000000000000\n5555555555555555555\n7777777777777777777\n9999999999999999999\n2222222222222222222\n4444444444444444444\n1111111111111111111\n6666666666666666666\n7777777777777777777\n2222222222222222222",
"output": "YES"
},
{
"input": "1 100\n8888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888",
"output": "YES"
},
{
"input": "100 1\n5\n7\n9\n4\n7\n2\n5\n1\n6\n7\n2\n7\n6\n8\n7\n4\n0\n2\n9\n8\n9\n1\n6\n4\n3\n4\n7\n1\n9\n3\n0\n8\n3\n1\n7\n5\n3\n9\n5\n1\n3\n5\n8\n1\n9\n3\n9\n0\n6\n0\n7\n6\n5\n2\n8\n3\n7\n6\n5\n1\n8\n3\n6\n9\n6\n0\n5\n8\n5\n2\n9\n1\n0\n1\n8\n3\n2\n1\n0\n3\n9\n0\n5\n1\n0\n4\n9\n3\n0\n4\n8\n4\n8\n6\n3\n0\n4\n6\n8\n4",
"output": "YES"
},
{
"input": "1 1\n2",
"output": "YES"
},
{
"input": "1 100\n1111111111111111111111111111111111111111111111111111111111111111111111111111181111111111111111111111",
"output": "NO"
},
{
"input": "100 1\n3\n6\n4\n3\n0\n2\n8\n7\n3\n2\n1\n7\n1\n3\n2\n3\n6\n9\n0\n8\n5\n9\n7\n9\n2\n1\n4\n5\n1\n9\n2\n5\n1\n4\n6\n4\n9\n1\n0\n2\n1\n4\n7\n1\n4\n8\n0\n9\n2\n1\n6\n2\n8\n6\n9\n5\n8\n6\n4\n5\n9\n2\n7\n4\n1\n5\n8\n0\n9\n5\n4\n6\n5\n0\n6\n3\n6\n9\n7\n2\n0\n9\n7\n3\n2\n4\n9\n4\n7\n1\n2\n3\n1\n7\n9\n1\n9\n0\n4\n0",
"output": "YES"
}
] | 1,696,600,136
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 4
| 62
| 0
|
n, m = map(int, input().split())
b = 0
d = 0
flag = True
a = ''
for i in range(m):
s = input()
if d==0:
b = s[0]
if s.count(s[0])!=n:
a = 'NO'
flag = False
if d>=1:
if s[0]==b:
a = 'NO'
flag = False
b = s[0]
d+=1
if flag:
a = 'YES'
print(a)
|
Title: Flag
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
According to a new ISO standard, a flag of every country should have a chequered field *n*<=×<=*m*, each square should be of one of 10 colours, and the flag should be «striped»: each horizontal row of the flag should contain squares of the same colour, and the colours of adjacent horizontal rows should be different. Berland's government asked you to find out whether their flag meets the new ISO standard.
Input Specification:
The first line of the input contains numbers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100), *n* — the amount of rows, *m* — the amount of columns on the flag of Berland. Then there follows the description of the flag: each of the following *n* lines contain *m* characters. Each character is a digit between 0 and 9, and stands for the colour of the corresponding square.
Output Specification:
Output YES, if the flag meets the new ISO standard, and NO otherwise.
Demo Input:
['3 3\n000\n111\n222\n', '3 3\n000\n000\n111\n', '3 3\n000\n111\n002\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none
|
```python
n, m = map(int, input().split())
b = 0
d = 0
flag = True
a = ''
for i in range(m):
s = input()
if d==0:
b = s[0]
if s.count(s[0])!=n:
a = 'NO'
flag = False
if d>=1:
if s[0]==b:
a = 'NO'
flag = False
b = s[0]
d+=1
if flag:
a = 'YES'
print(a)
```
| -1
|
2
|
B
|
The least round way
|
PROGRAMMING
| 2,000
|
[
"dp",
"math"
] |
B. The least round way
|
2
|
64
|
There is a square matrix *n*<=×<=*n*, consisting of non-negative integer numbers. You should find such a way on it that
- starts in the upper left cell of the matrix; - each following cell is to the right or down from the current cell; - the way ends in the bottom right cell.
Moreover, if we multiply together all the numbers along the way, the result should be the least "round". In other words, it should end in the least possible number of zeros.
|
The first line contains an integer number *n* (2<=≤<=*n*<=≤<=1000), *n* is the size of the matrix. Then follow *n* lines containing the matrix elements (non-negative integer numbers not exceeding 109).
|
In the first line print the least number of trailing zeros. In the second line print the correspondent way itself.
|
[
"3\n1 2 3\n4 5 6\n7 8 9\n"
] |
[
"0\nDDRR\n"
] |
none
| 0
|
[
{
"input": "3\n1 2 3\n4 5 6\n7 8 9",
"output": "0\nDDRR"
},
{
"input": "2\n7 6\n3 8",
"output": "0\nDR"
},
{
"input": "3\n4 10 5\n10 9 4\n6 5 3",
"output": "1\nDRRD"
},
{
"input": "4\n1 1 9 9\n3 4 7 3\n7 9 1 7\n1 7 1 5",
"output": "0\nDDDRRR"
},
{
"input": "5\n8 3 2 1 4\n3 7 2 4 8\n9 2 8 9 10\n2 3 6 10 1\n8 2 2 8 4",
"output": "0\nDDDDRRRR"
},
{
"input": "6\n5 5 4 10 5 5\n7 10 8 7 6 6\n7 1 7 9 7 8\n5 5 3 3 10 9\n5 8 10 6 3 8\n3 10 5 4 3 4",
"output": "1\nDDRRDRDDRR"
},
{
"input": "7\n2 9 8 2 7 4 8\n9 5 4 4 8 5 3\n5 7 2 10 8 1 8\n2 7 10 7 5 7 7\n9 2 7 6 4 8 4\n7 2 4 7 4 1 8\n9 5 3 10 1 6 2",
"output": "0\nRRDRRDRDDDDR"
},
{
"input": "8\n1 1 10 1 8 4 8 7\n9 3 3 2 2 6 2 4\n7 4 3 5 10 3 5 1\n8 4 4 10 4 5 9 4\n5 5 5 2 6 7 1 8\n4 10 1 3 2 4 8 3\n8 1 10 2 8 2 2 4\n2 10 6 8 10 2 8 4",
"output": "0\nDRRRRRRRDDDDDD"
},
{
"input": "9\n8 3 3 3 10 3 10 5 6\n2 1 6 1 8 1 9 1 6\n6 1 5 4 2 2 10 4 9\n1 9 1 3 10 6 10 5 5\n1 10 5 4 7 2 5 9 10\n6 6 1 3 1 9 4 9 9\n5 3 7 6 4 6 2 10 2\n9 3 3 10 5 6 7 6 4\n4 9 6 7 4 3 7 6 5",
"output": "1\nDDDDDRDDDRRRRRRR"
},
{
"input": "10\n10 8 6 5 9 8 2 5 3 2\n3 1 8 6 8 10 5 5 7 8\n5 9 7 7 4 9 7 2 5 2\n5 9 9 5 4 2 6 6 8 1\n10 6 9 9 10 5 6 3 5 9\n6 7 10 3 1 4 3 6 7 6\n10 10 4 7 4 4 8 6 7 4\n10 5 8 2 2 7 4 4 1 4\n8 4 6 10 10 6 1 3 3 1\n9 9 7 2 9 5 1 8 6 3",
"output": "1\nDRDDDRRDDDRRDRDRRR"
}
] | 1,610,089,417
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 15
| 2,000
| 11,878,400
|
import math
# misnomer
def tuple_sum(a1, a2):
if a1 == -1 or a2 == -1:
return -1
else:
return a1 + a2
def factorize(num, k):
if num==0:
return -1
i = 0
while True:
if num%k:
break
num = num // k
i += 1
return i
def countzero(a):
if a == -1:
return 1
else:
return a
def solve(block, n, k):
# create dp arr
dp = [[factorize(a, k) for a in row] for row in block]
for i in range(n):
for j in range(n):
if i==0 and j==0:
#dp[i][j] = factorize(block[0][0], k)
pass
elif i==0 and not j==0:
c = dp[i][j-1]
dp[i][j] = tuple_sum(c, dp[i][j])
elif not i==0 and j==0:
c = dp[i-1][j]
dp[i][j] = tuple_sum(c, dp[i][j])
else:
c1 = dp[i][j-1]
c2 = dp[i-1][j]
if c1 < c2:
dp[i][j] = tuple_sum(c1, dp[i][j])
else:
dp[i][j] = tuple_sum(c2, dp[i][j])
#get path
path=""
start = [0, 0]
cur = [n-1, n-1]
while not cur == start:
c1 = math.inf
c2 = math.inf
if cur[1] > 0:
c1 = dp[cur[0]][cur[1]-1]
if cur[0] > 0:
c2 = dp[cur[0]-1][cur[1]]
if c1 < c2:
path = "R" + path
cur[1] -= 1
else:
path = "D" + path
cur[0] -= 1
return (countzero(dp[n-1][n-1]), path)
n = int(input())
block = []
for i in range(n):
row = input().split()
row = list(map(int, row))
block.append(row)
(a, p) = min(solve(block, n, 2), solve(block, n, 5))
print(a)
print(p)
|
Title: The least round way
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
There is a square matrix *n*<=×<=*n*, consisting of non-negative integer numbers. You should find such a way on it that
- starts in the upper left cell of the matrix; - each following cell is to the right or down from the current cell; - the way ends in the bottom right cell.
Moreover, if we multiply together all the numbers along the way, the result should be the least "round". In other words, it should end in the least possible number of zeros.
Input Specification:
The first line contains an integer number *n* (2<=≤<=*n*<=≤<=1000), *n* is the size of the matrix. Then follow *n* lines containing the matrix elements (non-negative integer numbers not exceeding 109).
Output Specification:
In the first line print the least number of trailing zeros. In the second line print the correspondent way itself.
Demo Input:
['3\n1 2 3\n4 5 6\n7 8 9\n']
Demo Output:
['0\nDDRR\n']
Note:
none
|
```python
import math
# misnomer
def tuple_sum(a1, a2):
if a1 == -1 or a2 == -1:
return -1
else:
return a1 + a2
def factorize(num, k):
if num==0:
return -1
i = 0
while True:
if num%k:
break
num = num // k
i += 1
return i
def countzero(a):
if a == -1:
return 1
else:
return a
def solve(block, n, k):
# create dp arr
dp = [[factorize(a, k) for a in row] for row in block]
for i in range(n):
for j in range(n):
if i==0 and j==0:
#dp[i][j] = factorize(block[0][0], k)
pass
elif i==0 and not j==0:
c = dp[i][j-1]
dp[i][j] = tuple_sum(c, dp[i][j])
elif not i==0 and j==0:
c = dp[i-1][j]
dp[i][j] = tuple_sum(c, dp[i][j])
else:
c1 = dp[i][j-1]
c2 = dp[i-1][j]
if c1 < c2:
dp[i][j] = tuple_sum(c1, dp[i][j])
else:
dp[i][j] = tuple_sum(c2, dp[i][j])
#get path
path=""
start = [0, 0]
cur = [n-1, n-1]
while not cur == start:
c1 = math.inf
c2 = math.inf
if cur[1] > 0:
c1 = dp[cur[0]][cur[1]-1]
if cur[0] > 0:
c2 = dp[cur[0]-1][cur[1]]
if c1 < c2:
path = "R" + path
cur[1] -= 1
else:
path = "D" + path
cur[0] -= 1
return (countzero(dp[n-1][n-1]), path)
n = int(input())
block = []
for i in range(n):
row = input().split()
row = list(map(int, row))
block.append(row)
(a, p) = min(solve(block, n, 2), solve(block, n, 5))
print(a)
print(p)
```
| 0
|
83
|
A
|
Magical Array
|
PROGRAMMING
| 1,300
|
[
"math"
] |
A. Magical Array
|
2
|
256
|
Valery is very interested in magic. Magic attracts him so much that he sees it everywhere. He explains any strange and weird phenomenon through intervention of supernatural forces. But who would have thought that even in a regular array of numbers Valera manages to see something beautiful and magical.
Valera absolutely accidentally got a piece of ancient parchment on which an array of numbers was written. He immediately thought that the numbers in this array were not random. As a result of extensive research Valera worked out a wonderful property that a magical array should have: an array is defined as magic if its minimum and maximum coincide.
He decided to share this outstanding discovery with you, but he asks you for help in return. Despite the tremendous intelligence and wit, Valera counts very badly and so you will have to complete his work. All you have to do is count the number of magical subarrays of the original array of numbers, written on the parchment. Subarray is defined as non-empty sequence of consecutive elements.
|
The first line of the input data contains an integer *n* (1<=≤<=*n*<=≤<=105). The second line contains an array of original integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=109<=≤<=*a**i*<=≤<=109).
|
Print on the single line the answer to the problem: the amount of subarrays, which are magical.
Please do not use the %lld specificator to read or write 64-bit numbers in C++. It is recommended to use cin, cout streams (you can also use the %I64d specificator).
|
[
"4\n2 1 1 4\n",
"5\n-2 -2 -2 0 1\n"
] |
[
"5\n",
"8\n"
] |
Notes to sample tests:
Magical subarrays are shown with pairs of indices [a;b] of the beginning and the end.
In the first sample: [1;1], [2;2], [3;3], [4;4], [2;3].
In the second sample: [1;1], [2;2], [3;3], [4;4], [5;5], [1;2], [2;3], [1;3].
| 500
|
[
{
"input": "4\n2 1 1 4",
"output": "5"
},
{
"input": "5\n-2 -2 -2 0 1",
"output": "8"
},
{
"input": "1\n10",
"output": "1"
},
{
"input": "2\n5 6",
"output": "2"
},
{
"input": "5\n5 5 4 5 5",
"output": "7"
},
{
"input": "8\n1 2 0 0 0 0 3 3",
"output": "15"
},
{
"input": "12\n-4 3 3 2 3 3 3 -4 2 -4 -4 -4",
"output": "19"
},
{
"input": "10\n7 1 0 10 0 -5 -3 -2 0 0",
"output": "11"
},
{
"input": "20\n6 0 0 -3 1 -3 0 -8 1 3 5 2 -1 -5 -1 9 0 6 -2 4",
"output": "21"
},
{
"input": "100\n0 -18 -9 -15 3 16 -28 0 -28 0 28 -20 -9 9 -11 0 18 -15 -18 -26 0 -27 -25 -22 6 -5 8 14 -17 24 20 3 -6 24 -27 1 -23 0 4 12 -20 0 -10 30 22 -6 13 16 0 15 17 -8 -2 0 -5 13 11 23 -17 -29 10 15 -28 0 -23 4 20 17 -7 -5 -16 -17 16 2 20 19 -8 0 8 -5 12 0 0 -14 -15 -28 -10 20 0 8 -1 10 14 9 0 4 -16 15 13 -10",
"output": "101"
},
{
"input": "50\n2 0 2 0 0 0 0 -1 -2 -2 -2 1 1 2 2 0 2 0 2 -3 0 0 0 0 3 1 -2 0 -1 0 -2 3 -1 2 0 2 0 0 0 0 2 0 1 0 0 3 0 0 -2 0",
"output": "75"
},
{
"input": "2\n-510468670 0",
"output": "2"
},
{
"input": "150\n0 -2 1 -2 0 0 0 0 -2 0 -2 -1 0 0 2 0 1 -2 1 -1 0 0 0 2 -2 2 -1 0 0 0 -2 0 2 0 1 0 -2 0 -2 -1 -1 -2 -2 2 0 0 1 -2 -2 -1 -2 0 2 1 1 -1 1 0 -2 2 0 0 0 1 -1 0 -2 -1 0 -2 2 1 1 0 0 2 0 0 2 -1 0 0 2 0 2 0 -2 -1 1 -2 1 0 0 -2 -1 -1 0 0 2 -1 -1 -1 -1 -2 0 0 2 -1 -1 1 0 0 1 -1 0 0 -1 2 2 0 0 0 -1 -2 0 1 0 -1 0 -1 -1 0 2 0 1 2 0 0 2 0 0 1 2 0 2 -2 2 0 2 2",
"output": "196"
}
] | 1,657,914,131
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 50
| 154
| 14,131,200
|
import sys
input = sys.stdin.readline
n = int(input())
w = list(map(int, input().split()))
c = 0
d = 1
for i in range(n-1):
if w[i] == w[i+1]:
d += 1
else:
c += d*(d+1)//2
d = 1
c += d*(d+1)//2
print(c)
|
Title: Magical Array
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Valery is very interested in magic. Magic attracts him so much that he sees it everywhere. He explains any strange and weird phenomenon through intervention of supernatural forces. But who would have thought that even in a regular array of numbers Valera manages to see something beautiful and magical.
Valera absolutely accidentally got a piece of ancient parchment on which an array of numbers was written. He immediately thought that the numbers in this array were not random. As a result of extensive research Valera worked out a wonderful property that a magical array should have: an array is defined as magic if its minimum and maximum coincide.
He decided to share this outstanding discovery with you, but he asks you for help in return. Despite the tremendous intelligence and wit, Valera counts very badly and so you will have to complete his work. All you have to do is count the number of magical subarrays of the original array of numbers, written on the parchment. Subarray is defined as non-empty sequence of consecutive elements.
Input Specification:
The first line of the input data contains an integer *n* (1<=≤<=*n*<=≤<=105). The second line contains an array of original integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=109<=≤<=*a**i*<=≤<=109).
Output Specification:
Print on the single line the answer to the problem: the amount of subarrays, which are magical.
Please do not use the %lld specificator to read or write 64-bit numbers in C++. It is recommended to use cin, cout streams (you can also use the %I64d specificator).
Demo Input:
['4\n2 1 1 4\n', '5\n-2 -2 -2 0 1\n']
Demo Output:
['5\n', '8\n']
Note:
Notes to sample tests:
Magical subarrays are shown with pairs of indices [a;b] of the beginning and the end.
In the first sample: [1;1], [2;2], [3;3], [4;4], [2;3].
In the second sample: [1;1], [2;2], [3;3], [4;4], [5;5], [1;2], [2;3], [1;3].
|
```python
import sys
input = sys.stdin.readline
n = int(input())
w = list(map(int, input().split()))
c = 0
d = 1
for i in range(n-1):
if w[i] == w[i+1]:
d += 1
else:
c += d*(d+1)//2
d = 1
c += d*(d+1)//2
print(c)
```
| 3.935179
|
1,011
|
A
|
Stages
|
PROGRAMMING
| 900
|
[
"greedy",
"implementation",
"sortings"
] | null | null |
Natasha is going to fly to Mars. She needs to build a rocket, which consists of several stages in some order. Each of the stages is defined by a lowercase Latin letter. This way, the rocket can be described by the string — concatenation of letters, which correspond to the stages.
There are $n$ stages available. The rocket must contain exactly $k$ of them. Stages in the rocket should be ordered by their weight. So, after the stage with some letter can go only stage with a letter, which is at least two positions after in the alphabet (skipping one letter in between, or even more). For example, after letter 'c' can't go letters 'a', 'b', 'c' and 'd', but can go letters 'e', 'f', ..., 'z'.
For the rocket to fly as far as possible, its weight should be minimal. The weight of the rocket is equal to the sum of the weights of its stages. The weight of the stage is the number of its letter in the alphabet. For example, the stage 'a 'weighs one ton,' b 'weighs two tons, and' z' — $26$ tons.
Build the rocket with the minimal weight or determine, that it is impossible to build a rocket at all. Each stage can be used at most once.
|
The first line of input contains two integers — $n$ and $k$ ($1 \le k \le n \le 50$) – the number of available stages and the number of stages to use in the rocket.
The second line contains string $s$, which consists of exactly $n$ lowercase Latin letters. Each letter defines a new stage, which can be used to build the rocket. Each stage can be used at most once.
|
Print a single integer — the minimal total weight of the rocket or -1, if it is impossible to build the rocket at all.
|
[
"5 3\nxyabd\n",
"7 4\nproblem\n",
"2 2\nab\n",
"12 1\nabaabbaaabbb\n"
] |
[
"29",
"34",
"-1",
"1"
] |
In the first example, the following rockets satisfy the condition:
- "adx" (weight is $1+4+24=29$);- "ady" (weight is $1+4+25=30$);- "bdx" (weight is $2+4+24=30$);- "bdy" (weight is $2+4+25=31$).
Rocket "adx" has the minimal weight, so the answer is $29$.
In the second example, target rocket is "belo". Its weight is $2+5+12+15=34$.
In the third example, $n=k=2$, so the rocket must have both stages: 'a' and 'b'. This rocket doesn't satisfy the condition, because these letters are adjacent in the alphabet. Answer is -1.
| 500
|
[
{
"input": "5 3\nxyabd",
"output": "29"
},
{
"input": "7 4\nproblem",
"output": "34"
},
{
"input": "2 2\nab",
"output": "-1"
},
{
"input": "12 1\nabaabbaaabbb",
"output": "1"
},
{
"input": "50 13\nqwertyuiopasdfghjklzxcvbnmaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "169"
},
{
"input": "50 14\nqwertyuiopasdfghjklzxcvbnmaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "-1"
},
{
"input": "1 1\na",
"output": "1"
},
{
"input": "50 1\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "1"
},
{
"input": "50 2\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "-1"
},
{
"input": "13 13\nuwgmkyqeiaocs",
"output": "169"
},
{
"input": "13 13\nhzdxpbfvrltnj",
"output": "182"
},
{
"input": "1 1\nn",
"output": "14"
},
{
"input": "10 8\nsmzeblyjqw",
"output": "113"
},
{
"input": "20 20\ntzmvhskkyugkuuxpvtbh",
"output": "-1"
},
{
"input": "30 15\nwjzolzzkfulwgioksfxmcxmnnjtoav",
"output": "-1"
},
{
"input": "40 30\nxumfrflllrrgswehqtsskefixhcxjrxbjmrpsshv",
"output": "-1"
},
{
"input": "50 31\nahbyyoxltryqdmvenemaqnbakglgqolxnaifnqtoclnnqiabpz",
"output": "-1"
},
{
"input": "10 7\niuiukrxcml",
"output": "99"
},
{
"input": "38 2\nvjzarfykmrsrvwbwfwldsulhxtykmjbnwmdufa",
"output": "5"
},
{
"input": "12 6\nfwseyrarkwcd",
"output": "61"
},
{
"input": "2 2\nac",
"output": "4"
},
{
"input": "1 1\nc",
"output": "3"
},
{
"input": "2 2\nad",
"output": "5"
},
{
"input": "2 1\nac",
"output": "1"
},
{
"input": "4 3\nadjz",
"output": "15"
},
{
"input": "3 3\naoz",
"output": "42"
},
{
"input": "3 1\nzzz",
"output": "26"
},
{
"input": "2 1\nxz",
"output": "24"
},
{
"input": "5 1\naaddd",
"output": "1"
}
] | 1,644,582,916
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 29
| 77
| 0
|
def main():
n, k = list(map(int, input().split()))
string = "".join(sorted(input()))
answer_string = ""
previous_ord = -1
for i in range(n):
if previous_ord == -1:
answer_string += string[i]
previous_ord = ord(string[i])
else:
if ord(string[i]) - previous_ord > 1:
answer_string += string[i]
previous_ord = ord(string[i])
if len(answer_string) == k:
s = 0
for character in answer_string:
s += ord(character) - ord("a") + 1
print(s)
return
print(-1)
if __name__ == "__main__":
main()
|
Title: Stages
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Natasha is going to fly to Mars. She needs to build a rocket, which consists of several stages in some order. Each of the stages is defined by a lowercase Latin letter. This way, the rocket can be described by the string — concatenation of letters, which correspond to the stages.
There are $n$ stages available. The rocket must contain exactly $k$ of them. Stages in the rocket should be ordered by their weight. So, after the stage with some letter can go only stage with a letter, which is at least two positions after in the alphabet (skipping one letter in between, or even more). For example, after letter 'c' can't go letters 'a', 'b', 'c' and 'd', but can go letters 'e', 'f', ..., 'z'.
For the rocket to fly as far as possible, its weight should be minimal. The weight of the rocket is equal to the sum of the weights of its stages. The weight of the stage is the number of its letter in the alphabet. For example, the stage 'a 'weighs one ton,' b 'weighs two tons, and' z' — $26$ tons.
Build the rocket with the minimal weight or determine, that it is impossible to build a rocket at all. Each stage can be used at most once.
Input Specification:
The first line of input contains two integers — $n$ and $k$ ($1 \le k \le n \le 50$) – the number of available stages and the number of stages to use in the rocket.
The second line contains string $s$, which consists of exactly $n$ lowercase Latin letters. Each letter defines a new stage, which can be used to build the rocket. Each stage can be used at most once.
Output Specification:
Print a single integer — the minimal total weight of the rocket or -1, if it is impossible to build the rocket at all.
Demo Input:
['5 3\nxyabd\n', '7 4\nproblem\n', '2 2\nab\n', '12 1\nabaabbaaabbb\n']
Demo Output:
['29', '34', '-1', '1']
Note:
In the first example, the following rockets satisfy the condition:
- "adx" (weight is $1+4+24=29$);- "ady" (weight is $1+4+25=30$);- "bdx" (weight is $2+4+24=30$);- "bdy" (weight is $2+4+25=31$).
Rocket "adx" has the minimal weight, so the answer is $29$.
In the second example, target rocket is "belo". Its weight is $2+5+12+15=34$.
In the third example, $n=k=2$, so the rocket must have both stages: 'a' and 'b'. This rocket doesn't satisfy the condition, because these letters are adjacent in the alphabet. Answer is -1.
|
```python
def main():
n, k = list(map(int, input().split()))
string = "".join(sorted(input()))
answer_string = ""
previous_ord = -1
for i in range(n):
if previous_ord == -1:
answer_string += string[i]
previous_ord = ord(string[i])
else:
if ord(string[i]) - previous_ord > 1:
answer_string += string[i]
previous_ord = ord(string[i])
if len(answer_string) == k:
s = 0
for character in answer_string:
s += ord(character) - ord("a") + 1
print(s)
return
print(-1)
if __name__ == "__main__":
main()
```
| 3
|
|
136
|
A
|
Presents
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Little Petya very much likes gifts. Recently he has received a new laptop as a New Year gift from his mother. He immediately decided to give it to somebody else as what can be more pleasant than giving somebody gifts. And on this occasion he organized a New Year party at his place and invited *n* his friends there.
If there's one thing Petya likes more that receiving gifts, that's watching others giving gifts to somebody else. Thus, he safely hid the laptop until the next New Year and made up his mind to watch his friends exchanging gifts while he does not participate in the process. He numbered all his friends with integers from 1 to *n*. Petya remembered that a friend number *i* gave a gift to a friend number *p**i*. He also remembered that each of his friends received exactly one gift.
Now Petya wants to know for each friend *i* the number of a friend who has given him a gift.
|
The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the quantity of friends Petya invited to the party. The second line contains *n* space-separated integers: the *i*-th number is *p**i* — the number of a friend who gave a gift to friend number *i*. It is guaranteed that each friend received exactly one gift. It is possible that some friends do not share Petya's ideas of giving gifts to somebody else. Those friends gave the gifts to themselves.
|
Print *n* space-separated integers: the *i*-th number should equal the number of the friend who gave a gift to friend number *i*.
|
[
"4\n2 3 4 1\n",
"3\n1 3 2\n",
"2\n1 2\n"
] |
[
"4 1 2 3\n",
"1 3 2\n",
"1 2\n"
] |
none
| 500
|
[
{
"input": "4\n2 3 4 1",
"output": "4 1 2 3"
},
{
"input": "3\n1 3 2",
"output": "1 3 2"
},
{
"input": "2\n1 2",
"output": "1 2"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "10\n1 3 2 6 4 5 7 9 8 10",
"output": "1 3 2 5 6 4 7 9 8 10"
},
{
"input": "5\n5 4 3 2 1",
"output": "5 4 3 2 1"
},
{
"input": "20\n2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19"
},
{
"input": "21\n3 2 1 6 5 4 9 8 7 12 11 10 15 14 13 18 17 16 21 20 19",
"output": "3 2 1 6 5 4 9 8 7 12 11 10 15 14 13 18 17 16 21 20 19"
},
{
"input": "10\n3 4 5 6 7 8 9 10 1 2",
"output": "9 10 1 2 3 4 5 6 7 8"
},
{
"input": "8\n1 5 3 7 2 6 4 8",
"output": "1 5 3 7 2 6 4 8"
},
{
"input": "50\n49 22 4 2 20 46 7 32 5 19 48 24 26 15 45 21 44 11 50 43 39 17 31 1 42 34 3 27 36 25 12 30 13 33 28 35 18 6 8 37 38 14 10 9 29 16 40 23 41 47",
"output": "24 4 27 3 9 38 7 39 44 43 18 31 33 42 14 46 22 37 10 5 16 2 48 12 30 13 28 35 45 32 23 8 34 26 36 29 40 41 21 47 49 25 20 17 15 6 50 11 1 19"
},
{
"input": "34\n13 20 33 30 15 11 27 4 8 2 29 25 24 7 3 22 18 10 26 16 5 1 32 9 34 6 12 14 28 19 31 21 23 17",
"output": "22 10 15 8 21 26 14 9 24 18 6 27 1 28 5 20 34 17 30 2 32 16 33 13 12 19 7 29 11 4 31 23 3 25"
},
{
"input": "92\n23 1 6 4 84 54 44 76 63 34 61 20 48 13 28 78 26 46 90 72 24 55 91 89 53 38 82 5 79 92 29 32 15 64 11 88 60 70 7 66 18 59 8 57 19 16 42 21 80 71 62 27 75 86 36 9 83 73 74 50 43 31 56 30 17 33 40 81 49 12 10 41 22 77 25 68 51 2 47 3 58 69 87 67 39 37 35 65 14 45 52 85",
"output": "2 78 80 4 28 3 39 43 56 71 35 70 14 89 33 46 65 41 45 12 48 73 1 21 75 17 52 15 31 64 62 32 66 10 87 55 86 26 85 67 72 47 61 7 90 18 79 13 69 60 77 91 25 6 22 63 44 81 42 37 11 51 9 34 88 40 84 76 82 38 50 20 58 59 53 8 74 16 29 49 68 27 57 5 92 54 83 36 24 19 23 30"
},
{
"input": "49\n30 24 33 48 7 3 17 2 8 35 10 39 23 40 46 32 18 21 26 22 1 16 47 45 41 28 31 6 12 43 27 11 13 37 19 15 44 5 29 42 4 38 20 34 14 9 25 36 49",
"output": "21 8 6 41 38 28 5 9 46 11 32 29 33 45 36 22 7 17 35 43 18 20 13 2 47 19 31 26 39 1 27 16 3 44 10 48 34 42 12 14 25 40 30 37 24 15 23 4 49"
},
{
"input": "12\n3 8 7 4 6 5 2 1 11 9 10 12",
"output": "8 7 1 4 6 5 3 2 10 11 9 12"
},
{
"input": "78\n16 56 36 78 21 14 9 77 26 57 70 61 41 47 18 44 5 31 50 74 65 52 6 39 22 62 67 69 43 7 64 29 24 40 48 51 73 54 72 12 19 34 4 25 55 33 17 35 23 53 10 8 27 32 42 68 20 63 3 2 1 71 58 46 13 30 49 11 37 66 38 60 28 75 15 59 45 76",
"output": "61 60 59 43 17 23 30 52 7 51 68 40 65 6 75 1 47 15 41 57 5 25 49 33 44 9 53 73 32 66 18 54 46 42 48 3 69 71 24 34 13 55 29 16 77 64 14 35 67 19 36 22 50 38 45 2 10 63 76 72 12 26 58 31 21 70 27 56 28 11 62 39 37 20 74 78 8 4"
},
{
"input": "64\n64 57 40 3 15 8 62 18 33 59 51 19 22 13 4 37 47 45 50 35 63 11 58 42 46 21 7 2 41 48 32 23 28 38 17 12 24 27 49 31 60 6 30 25 61 52 26 54 9 14 29 20 44 39 55 10 34 16 5 56 1 36 53 43",
"output": "61 28 4 15 59 42 27 6 49 56 22 36 14 50 5 58 35 8 12 52 26 13 32 37 44 47 38 33 51 43 40 31 9 57 20 62 16 34 54 3 29 24 64 53 18 25 17 30 39 19 11 46 63 48 55 60 2 23 10 41 45 7 21 1"
},
{
"input": "49\n38 20 49 32 14 41 39 45 25 48 40 19 26 43 34 12 10 3 35 42 5 7 46 47 4 2 13 22 16 24 33 15 11 18 29 31 23 9 44 36 6 17 37 1 30 28 8 21 27",
"output": "44 26 18 25 21 41 22 47 38 17 33 16 27 5 32 29 42 34 12 2 48 28 37 30 9 13 49 46 35 45 36 4 31 15 19 40 43 1 7 11 6 20 14 39 8 23 24 10 3"
},
{
"input": "78\n17 50 30 48 33 12 42 4 18 53 76 67 38 3 20 72 51 55 60 63 46 10 57 45 54 32 24 62 8 11 35 44 65 74 58 28 2 6 56 52 39 23 47 49 61 1 66 41 15 77 7 27 78 13 14 34 5 31 37 21 40 16 29 69 59 43 64 36 70 19 25 73 71 75 9 68 26 22",
"output": "46 37 14 8 57 38 51 29 75 22 30 6 54 55 49 62 1 9 70 15 60 78 42 27 71 77 52 36 63 3 58 26 5 56 31 68 59 13 41 61 48 7 66 32 24 21 43 4 44 2 17 40 10 25 18 39 23 35 65 19 45 28 20 67 33 47 12 76 64 69 73 16 72 34 74 11 50 53"
},
{
"input": "29\n14 21 27 1 4 18 10 17 20 23 2 24 7 9 28 22 8 25 12 15 11 6 16 29 3 26 19 5 13",
"output": "4 11 25 5 28 22 13 17 14 7 21 19 29 1 20 23 8 6 27 9 2 16 10 12 18 26 3 15 24"
},
{
"input": "82\n6 1 10 75 28 66 61 81 78 63 17 19 58 34 49 12 67 50 41 44 3 15 59 38 51 72 36 11 46 29 18 64 27 23 13 53 56 68 2 25 47 40 69 54 42 5 60 55 4 16 24 79 57 20 7 73 32 80 76 52 82 37 26 31 65 8 39 62 33 71 30 9 77 43 48 74 70 22 14 45 35 21",
"output": "2 39 21 49 46 1 55 66 72 3 28 16 35 79 22 50 11 31 12 54 82 78 34 51 40 63 33 5 30 71 64 57 69 14 81 27 62 24 67 42 19 45 74 20 80 29 41 75 15 18 25 60 36 44 48 37 53 13 23 47 7 68 10 32 65 6 17 38 43 77 70 26 56 76 4 59 73 9 52 58 8 61"
},
{
"input": "82\n74 18 15 69 71 77 19 26 80 20 66 7 30 82 22 48 21 44 52 65 64 61 35 49 12 8 53 81 54 16 11 9 40 46 13 1 29 58 5 41 55 4 78 60 6 51 56 2 38 36 34 62 63 25 17 67 45 14 32 37 75 79 10 47 27 39 31 68 59 24 50 43 72 70 42 28 76 23 57 3 73 33",
"output": "36 48 80 42 39 45 12 26 32 63 31 25 35 58 3 30 55 2 7 10 17 15 78 70 54 8 65 76 37 13 67 59 82 51 23 50 60 49 66 33 40 75 72 18 57 34 64 16 24 71 46 19 27 29 41 47 79 38 69 44 22 52 53 21 20 11 56 68 4 74 5 73 81 1 61 77 6 43 62 9 28 14"
},
{
"input": "45\n2 32 34 13 3 15 16 33 22 12 31 38 42 14 27 7 36 8 4 19 45 41 5 35 10 11 39 20 29 44 17 9 6 40 37 28 25 21 1 30 24 18 43 26 23",
"output": "39 1 5 19 23 33 16 18 32 25 26 10 4 14 6 7 31 42 20 28 38 9 45 41 37 44 15 36 29 40 11 2 8 3 24 17 35 12 27 34 22 13 43 30 21"
},
{
"input": "45\n4 32 33 39 43 21 22 35 45 7 14 5 16 9 42 31 24 36 17 29 41 25 37 34 27 20 11 44 3 13 19 2 1 10 26 30 38 18 6 8 15 23 40 28 12",
"output": "33 32 29 1 12 39 10 40 14 34 27 45 30 11 41 13 19 38 31 26 6 7 42 17 22 35 25 44 20 36 16 2 3 24 8 18 23 37 4 43 21 15 5 28 9"
},
{
"input": "74\n48 72 40 67 17 4 27 53 11 32 25 9 74 2 41 24 56 22 14 21 33 5 18 55 20 7 29 36 69 13 52 19 38 30 68 59 66 34 63 6 47 45 54 44 62 12 50 71 16 10 8 64 57 73 46 26 49 42 3 23 35 1 61 39 70 60 65 43 15 28 37 51 58 31",
"output": "62 14 59 6 22 40 26 51 12 50 9 46 30 19 69 49 5 23 32 25 20 18 60 16 11 56 7 70 27 34 74 10 21 38 61 28 71 33 64 3 15 58 68 44 42 55 41 1 57 47 72 31 8 43 24 17 53 73 36 66 63 45 39 52 67 37 4 35 29 65 48 2 54 13"
},
{
"input": "47\n9 26 27 10 6 34 28 42 39 22 45 21 11 43 14 47 38 15 40 32 46 1 36 29 17 25 2 23 31 5 24 4 7 8 12 19 16 44 37 20 18 33 30 13 35 41 3",
"output": "22 27 47 32 30 5 33 34 1 4 13 35 44 15 18 37 25 41 36 40 12 10 28 31 26 2 3 7 24 43 29 20 42 6 45 23 39 17 9 19 46 8 14 38 11 21 16"
},
{
"input": "49\n14 38 6 29 9 49 36 43 47 3 44 20 34 15 7 11 1 28 12 40 16 37 31 10 42 41 33 21 18 30 5 27 17 35 25 26 45 19 2 13 23 32 4 22 46 48 24 39 8",
"output": "17 39 10 43 31 3 15 49 5 24 16 19 40 1 14 21 33 29 38 12 28 44 41 47 35 36 32 18 4 30 23 42 27 13 34 7 22 2 48 20 26 25 8 11 37 45 9 46 6"
},
{
"input": "100\n78 56 31 91 90 95 16 65 58 77 37 89 33 61 10 76 62 47 35 67 69 7 63 83 22 25 49 8 12 30 39 44 57 64 48 42 32 11 70 43 55 50 99 24 85 73 45 14 54 21 98 84 74 2 26 18 9 36 80 53 75 46 66 86 59 93 87 68 94 13 72 28 79 88 92 29 52 82 34 97 19 38 1 41 27 4 40 5 96 100 51 6 20 23 81 15 17 3 60 71",
"output": "83 54 98 86 88 92 22 28 57 15 38 29 70 48 96 7 97 56 81 93 50 25 94 44 26 55 85 72 76 30 3 37 13 79 19 58 11 82 31 87 84 36 40 32 47 62 18 35 27 42 91 77 60 49 41 2 33 9 65 99 14 17 23 34 8 63 20 68 21 39 100 71 46 53 61 16 10 1 73 59 95 78 24 52 45 64 67 74 12 5 4 75 66 69 6 89 80 51 43 90"
},
{
"input": "22\n12 8 11 2 16 7 13 6 22 21 20 10 4 14 18 1 5 15 3 19 17 9",
"output": "16 4 19 13 17 8 6 2 22 12 3 1 7 14 18 5 21 15 20 11 10 9"
},
{
"input": "72\n16 11 49 51 3 27 60 55 23 40 66 7 53 70 13 5 15 32 18 72 33 30 8 31 46 12 28 67 25 38 50 22 69 34 71 52 58 39 24 35 42 9 41 26 62 1 63 65 36 64 68 61 37 14 45 47 6 57 54 20 17 2 56 59 29 10 4 48 21 43 19 44",
"output": "46 62 5 67 16 57 12 23 42 66 2 26 15 54 17 1 61 19 71 60 69 32 9 39 29 44 6 27 65 22 24 18 21 34 40 49 53 30 38 10 43 41 70 72 55 25 56 68 3 31 4 36 13 59 8 63 58 37 64 7 52 45 47 50 48 11 28 51 33 14 35 20"
},
{
"input": "63\n21 56 11 10 62 24 20 42 28 52 38 2 37 43 48 22 7 8 40 14 13 46 53 1 23 4 60 63 51 36 25 12 39 32 49 16 58 44 31 61 33 50 55 54 45 6 47 41 9 57 30 29 26 18 19 27 15 34 3 35 59 5 17",
"output": "24 12 59 26 62 46 17 18 49 4 3 32 21 20 57 36 63 54 55 7 1 16 25 6 31 53 56 9 52 51 39 34 41 58 60 30 13 11 33 19 48 8 14 38 45 22 47 15 35 42 29 10 23 44 43 2 50 37 61 27 40 5 28"
},
{
"input": "18\n2 16 8 4 18 12 3 6 5 9 10 15 11 17 14 13 1 7",
"output": "17 1 7 4 9 8 18 3 10 11 13 6 16 15 12 2 14 5"
},
{
"input": "47\n6 9 10 41 25 3 4 37 20 1 36 22 29 27 11 24 43 31 12 17 34 42 38 39 13 2 7 21 18 5 15 35 44 26 33 46 19 40 30 14 28 23 47 32 45 8 16",
"output": "10 26 6 7 30 1 27 46 2 3 15 19 25 40 31 47 20 29 37 9 28 12 42 16 5 34 14 41 13 39 18 44 35 21 32 11 8 23 24 38 4 22 17 33 45 36 43"
},
{
"input": "96\n41 91 48 88 29 57 1 19 44 43 37 5 10 75 25 63 30 78 76 53 8 92 18 70 39 17 49 60 9 16 3 34 86 59 23 79 55 45 72 51 28 33 96 40 26 54 6 32 89 61 85 74 7 82 52 31 64 66 94 95 11 22 2 73 35 13 42 71 14 47 84 69 50 67 58 12 77 46 38 68 15 36 20 93 27 90 83 56 87 4 21 24 81 62 80 65",
"output": "7 63 31 90 12 47 53 21 29 13 61 76 66 69 81 30 26 23 8 83 91 62 35 92 15 45 85 41 5 17 56 48 42 32 65 82 11 79 25 44 1 67 10 9 38 78 70 3 27 73 40 55 20 46 37 88 6 75 34 28 50 94 16 57 96 58 74 80 72 24 68 39 64 52 14 19 77 18 36 95 93 54 87 71 51 33 89 4 49 86 2 22 84 59 60 43"
},
{
"input": "73\n67 24 39 22 23 20 48 34 42 40 19 70 65 69 64 21 53 11 59 15 26 10 30 33 72 29 55 25 56 71 8 9 57 49 41 61 13 12 6 27 66 36 47 50 73 60 2 37 7 4 51 17 1 46 14 62 35 3 45 63 43 58 54 32 31 5 28 44 18 52 68 38 16",
"output": "53 47 58 50 66 39 49 31 32 22 18 38 37 55 20 73 52 69 11 6 16 4 5 2 28 21 40 67 26 23 65 64 24 8 57 42 48 72 3 10 35 9 61 68 59 54 43 7 34 44 51 70 17 63 27 29 33 62 19 46 36 56 60 15 13 41 1 71 14 12 30 25 45"
},
{
"input": "81\n25 2 78 40 12 80 69 13 49 43 17 33 23 54 32 61 77 66 27 71 24 26 42 55 60 9 5 30 7 37 45 63 53 11 38 44 68 34 28 52 67 22 57 46 47 50 8 16 79 62 4 36 20 14 73 64 6 76 35 74 58 10 29 81 59 31 19 1 75 39 70 18 41 21 72 65 3 48 15 56 51",
"output": "68 2 77 51 27 57 29 47 26 62 34 5 8 54 79 48 11 72 67 53 74 42 13 21 1 22 19 39 63 28 66 15 12 38 59 52 30 35 70 4 73 23 10 36 31 44 45 78 9 46 81 40 33 14 24 80 43 61 65 25 16 50 32 56 76 18 41 37 7 71 20 75 55 60 69 58 17 3 49 6 64"
},
{
"input": "12\n12 3 1 5 11 6 7 10 2 8 9 4",
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},
{
"input": "47\n7 21 41 18 40 31 12 28 24 14 43 23 33 10 19 38 26 8 34 15 29 44 5 13 39 25 3 27 20 42 35 9 2 1 30 46 36 32 4 22 37 45 6 47 11 16 17",
"output": "34 33 27 39 23 43 1 18 32 14 45 7 24 10 20 46 47 4 15 29 2 40 12 9 26 17 28 8 21 35 6 38 13 19 31 37 41 16 25 5 3 30 11 22 42 36 44"
},
{
"input": "8\n1 3 5 2 4 8 6 7",
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},
{
"input": "38\n28 8 2 33 20 32 26 29 23 31 15 38 11 37 18 21 22 19 4 34 1 35 16 7 17 6 27 30 36 12 9 24 25 13 5 3 10 14",
"output": "21 3 36 19 35 26 24 2 31 37 13 30 34 38 11 23 25 15 18 5 16 17 9 32 33 7 27 1 8 28 10 6 4 20 22 29 14 12"
},
{
"input": "10\n2 9 4 6 10 1 7 5 3 8",
"output": "6 1 9 3 8 4 7 10 2 5"
},
{
"input": "23\n20 11 15 1 5 12 23 9 2 22 13 19 16 14 7 4 8 21 6 17 18 10 3",
"output": "4 9 23 16 5 19 15 17 8 22 2 6 11 14 3 13 20 21 12 1 18 10 7"
},
{
"input": "10\n2 4 9 3 6 8 10 5 1 7",
"output": "9 1 4 2 8 5 10 6 3 7"
},
{
"input": "55\n9 48 23 49 11 24 4 22 34 32 17 45 39 13 14 21 19 25 2 31 37 7 55 36 20 51 5 12 54 10 35 40 43 1 46 18 53 41 38 26 29 50 3 42 52 27 8 28 47 33 6 16 30 44 15",
"output": "34 19 43 7 27 51 22 47 1 30 5 28 14 15 55 52 11 36 17 25 16 8 3 6 18 40 46 48 41 53 20 10 50 9 31 24 21 39 13 32 38 44 33 54 12 35 49 2 4 42 26 45 37 29 23"
},
{
"input": "58\n49 13 12 54 2 38 56 11 33 25 26 19 28 8 23 41 20 36 46 55 15 35 9 7 32 37 58 6 3 14 47 31 40 30 53 44 4 50 29 34 10 43 39 57 5 22 27 45 51 42 24 16 18 21 52 17 48 1",
"output": "58 5 29 37 45 28 24 14 23 41 8 3 2 30 21 52 56 53 12 17 54 46 15 51 10 11 47 13 39 34 32 25 9 40 22 18 26 6 43 33 16 50 42 36 48 19 31 57 1 38 49 55 35 4 20 7 44 27"
},
{
"input": "34\n20 25 2 3 33 29 1 16 14 7 21 9 32 31 6 26 22 4 27 23 24 10 34 12 19 15 5 18 28 17 13 8 11 30",
"output": "7 3 4 18 27 15 10 32 12 22 33 24 31 9 26 8 30 28 25 1 11 17 20 21 2 16 19 29 6 34 14 13 5 23"
},
{
"input": "53\n47 29 46 25 23 13 7 31 33 4 38 11 35 16 42 14 15 43 34 39 28 18 6 45 30 1 40 20 2 37 5 32 24 12 44 26 27 3 19 51 36 21 22 9 10 50 41 48 49 53 8 17 52",
"output": "26 29 38 10 31 23 7 51 44 45 12 34 6 16 17 14 52 22 39 28 42 43 5 33 4 36 37 21 2 25 8 32 9 19 13 41 30 11 20 27 47 15 18 35 24 3 1 48 49 46 40 53 50"
},
{
"input": "99\n77 87 90 48 53 38 68 6 28 57 35 82 63 71 60 41 3 12 86 65 10 59 22 67 33 74 93 27 24 1 61 43 25 4 51 52 15 88 9 31 30 42 89 49 23 21 29 32 46 73 37 16 5 69 56 26 92 64 20 54 75 14 98 13 94 2 95 7 36 66 58 8 50 78 84 45 11 96 76 62 97 80 40 39 47 85 34 79 83 17 91 72 19 44 70 81 55 99 18",
"output": "30 66 17 34 53 8 68 72 39 21 77 18 64 62 37 52 90 99 93 59 46 23 45 29 33 56 28 9 47 41 40 48 25 87 11 69 51 6 84 83 16 42 32 94 76 49 85 4 44 73 35 36 5 60 97 55 10 71 22 15 31 80 13 58 20 70 24 7 54 95 14 92 50 26 61 79 1 74 88 82 96 12 89 75 86 19 2 38 43 3 91 57 27 65 67 78 81 63 98"
},
{
"input": "32\n17 29 2 6 30 8 26 7 1 27 10 9 13 24 31 21 15 19 22 18 4 11 25 28 32 3 23 12 5 14 20 16",
"output": "9 3 26 21 29 4 8 6 12 11 22 28 13 30 17 32 1 20 18 31 16 19 27 14 23 7 10 24 2 5 15 25"
},
{
"input": "65\n18 40 1 60 17 19 4 6 12 49 28 58 2 25 13 14 64 56 61 34 62 30 59 51 26 8 33 63 36 48 46 7 43 21 31 27 11 44 29 5 32 23 35 9 53 57 52 50 15 38 42 3 54 65 55 41 20 24 22 47 45 10 39 16 37",
"output": "3 13 52 7 40 8 32 26 44 62 37 9 15 16 49 64 5 1 6 57 34 59 42 58 14 25 36 11 39 22 35 41 27 20 43 29 65 50 63 2 56 51 33 38 61 31 60 30 10 48 24 47 45 53 55 18 46 12 23 4 19 21 28 17 54"
},
{
"input": "71\n35 50 55 58 25 32 26 40 63 34 44 53 24 18 37 7 64 27 56 65 1 19 2 43 42 14 57 47 22 13 59 61 39 67 30 45 54 38 33 48 6 5 3 69 36 21 41 4 16 46 20 17 15 12 10 70 68 23 60 31 52 29 66 28 51 49 62 11 8 9 71",
"output": "21 23 43 48 42 41 16 69 70 55 68 54 30 26 53 49 52 14 22 51 46 29 58 13 5 7 18 64 62 35 60 6 39 10 1 45 15 38 33 8 47 25 24 11 36 50 28 40 66 2 65 61 12 37 3 19 27 4 31 59 32 67 9 17 20 63 34 57 44 56 71"
},
{
"input": "74\n33 8 42 63 64 61 31 74 11 50 68 14 36 25 57 30 7 44 21 15 6 9 23 59 46 3 73 16 62 51 40 60 41 54 5 39 35 28 48 4 58 12 66 69 13 26 71 1 24 19 29 52 37 2 20 43 18 72 17 56 34 38 65 67 27 10 47 70 53 32 45 55 49 22",
"output": "48 54 26 40 35 21 17 2 22 66 9 42 45 12 20 28 59 57 50 55 19 74 23 49 14 46 65 38 51 16 7 70 1 61 37 13 53 62 36 31 33 3 56 18 71 25 67 39 73 10 30 52 69 34 72 60 15 41 24 32 6 29 4 5 63 43 64 11 44 68 47 58 27 8"
},
{
"input": "96\n78 10 82 46 38 91 77 69 2 27 58 80 79 44 59 41 6 31 76 11 42 48 51 37 19 87 43 25 52 32 1 39 63 29 21 65 53 74 92 16 15 95 90 83 30 73 71 5 50 17 96 33 86 60 67 64 20 26 61 40 55 88 94 93 9 72 47 57 14 45 22 3 54 68 13 24 4 7 56 81 89 70 49 8 84 28 18 62 35 36 75 23 66 85 34 12",
"output": "31 9 72 77 48 17 78 84 65 2 20 96 75 69 41 40 50 87 25 57 35 71 92 76 28 58 10 86 34 45 18 30 52 95 89 90 24 5 32 60 16 21 27 14 70 4 67 22 83 49 23 29 37 73 61 79 68 11 15 54 59 88 33 56 36 93 55 74 8 82 47 66 46 38 91 19 7 1 13 12 80 3 44 85 94 53 26 62 81 43 6 39 64 63 42 51"
},
{
"input": "7\n2 1 5 7 3 4 6",
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},
{
"input": "51\n8 33 37 2 16 22 24 30 4 9 5 15 27 3 18 39 31 26 10 17 46 41 25 14 6 1 29 48 36 20 51 49 21 43 19 13 38 50 47 34 11 23 28 12 42 7 32 40 44 45 35",
"output": "26 4 14 9 11 25 46 1 10 19 41 44 36 24 12 5 20 15 35 30 33 6 42 7 23 18 13 43 27 8 17 47 2 40 51 29 3 37 16 48 22 45 34 49 50 21 39 28 32 38 31"
},
{
"input": "27\n12 14 7 3 20 21 25 13 22 15 23 4 2 24 10 17 19 8 26 11 27 18 9 5 6 1 16",
"output": "26 13 4 12 24 25 3 18 23 15 20 1 8 2 10 27 16 22 17 5 6 9 11 14 7 19 21"
},
{
"input": "71\n51 13 20 48 54 23 24 64 14 62 71 67 57 53 3 30 55 43 33 25 39 40 66 6 46 18 5 19 61 16 32 68 70 41 60 44 29 49 27 69 50 38 10 17 45 56 9 21 26 63 28 35 7 59 1 65 2 15 8 11 12 34 37 47 58 22 31 4 36 42 52",
"output": "55 57 15 68 27 24 53 59 47 43 60 61 2 9 58 30 44 26 28 3 48 66 6 7 20 49 39 51 37 16 67 31 19 62 52 69 63 42 21 22 34 70 18 36 45 25 64 4 38 41 1 71 14 5 17 46 13 65 54 35 29 10 50 8 56 23 12 32 40 33 11"
},
{
"input": "9\n8 5 2 6 1 9 4 7 3",
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},
{
"input": "29\n10 24 11 5 26 25 2 9 22 15 8 14 29 21 4 1 23 17 3 12 13 16 18 28 19 20 7 6 27",
"output": "16 7 19 15 4 28 27 11 8 1 3 20 21 12 10 22 18 23 25 26 14 9 17 2 6 5 29 24 13"
},
{
"input": "60\n39 25 42 4 55 60 16 18 47 1 11 40 7 50 19 35 49 54 12 3 30 38 2 58 17 26 45 6 33 43 37 32 52 36 15 23 27 59 24 20 28 14 8 9 13 29 44 46 41 21 5 48 51 22 31 56 57 53 10 34",
"output": "10 23 20 4 51 28 13 43 44 59 11 19 45 42 35 7 25 8 15 40 50 54 36 39 2 26 37 41 46 21 55 32 29 60 16 34 31 22 1 12 49 3 30 47 27 48 9 52 17 14 53 33 58 18 5 56 57 24 38 6"
},
{
"input": "50\n37 45 22 5 12 21 28 24 18 47 20 25 8 50 14 2 34 43 11 16 49 41 48 1 19 31 39 46 32 23 15 42 3 35 38 30 44 26 10 9 40 36 7 17 33 4 27 6 13 29",
"output": "24 16 33 46 4 48 43 13 40 39 19 5 49 15 31 20 44 9 25 11 6 3 30 8 12 38 47 7 50 36 26 29 45 17 34 42 1 35 27 41 22 32 18 37 2 28 10 23 21 14"
},
{
"input": "30\n8 29 28 16 17 25 27 15 21 11 6 20 2 13 1 30 5 4 24 10 14 3 23 18 26 9 12 22 19 7",
"output": "15 13 22 18 17 11 30 1 26 20 10 27 14 21 8 4 5 24 29 12 9 28 23 19 6 25 7 3 2 16"
},
{
"input": "46\n15 2 44 43 38 19 31 42 4 37 29 30 24 45 27 41 8 20 33 7 35 3 18 46 36 26 1 28 21 40 16 22 32 11 14 13 12 9 25 39 10 6 23 17 5 34",
"output": "27 2 22 9 45 42 20 17 38 41 34 37 36 35 1 31 44 23 6 18 29 32 43 13 39 26 15 28 11 12 7 33 19 46 21 25 10 5 40 30 16 8 4 3 14 24"
},
{
"input": "9\n4 8 6 5 3 9 2 7 1",
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},
{
"input": "46\n31 30 33 23 45 7 36 8 11 3 32 39 41 20 1 28 6 27 18 24 17 5 16 37 26 13 22 14 2 38 15 46 9 4 19 21 12 44 10 35 25 34 42 43 40 29",
"output": "15 29 10 34 22 17 6 8 33 39 9 37 26 28 31 23 21 19 35 14 36 27 4 20 41 25 18 16 46 2 1 11 3 42 40 7 24 30 12 45 13 43 44 38 5 32"
},
{
"input": "66\n27 12 37 48 46 21 34 58 38 28 66 2 64 32 44 31 13 36 40 15 19 11 22 5 30 29 6 7 61 39 20 42 23 54 51 33 50 9 60 8 57 45 49 10 62 41 59 3 55 63 52 24 25 26 43 56 65 4 16 14 1 35 18 17 53 47",
"output": "61 12 48 58 24 27 28 40 38 44 22 2 17 60 20 59 64 63 21 31 6 23 33 52 53 54 1 10 26 25 16 14 36 7 62 18 3 9 30 19 46 32 55 15 42 5 66 4 43 37 35 51 65 34 49 56 41 8 47 39 29 45 50 13 57 11"
},
{
"input": "13\n3 12 9 2 8 5 13 4 11 1 10 7 6",
"output": "10 4 1 8 6 13 12 5 3 11 9 2 7"
},
{
"input": "80\n21 25 56 50 20 61 7 74 51 69 8 2 46 57 45 71 14 52 17 43 9 30 70 78 31 10 38 13 23 15 37 79 6 16 77 73 80 4 49 48 18 28 26 58 33 41 64 22 54 72 59 60 40 63 53 27 1 5 75 67 62 34 19 39 68 65 44 55 3 32 11 42 76 12 35 47 66 36 24 29",
"output": "57 12 69 38 58 33 7 11 21 26 71 74 28 17 30 34 19 41 63 5 1 48 29 79 2 43 56 42 80 22 25 70 45 62 75 78 31 27 64 53 46 72 20 67 15 13 76 40 39 4 9 18 55 49 68 3 14 44 51 52 6 61 54 47 66 77 60 65 10 23 16 50 36 8 59 73 35 24 32 37"
},
{
"input": "63\n9 49 53 25 40 46 43 51 54 22 58 16 23 26 10 47 5 27 2 8 61 59 19 35 63 56 28 20 34 4 62 38 6 55 36 31 57 15 29 33 1 48 50 37 7 30 18 42 32 52 12 41 14 21 45 11 24 17 39 13 44 60 3",
"output": "41 19 63 30 17 33 45 20 1 15 56 51 60 53 38 12 58 47 23 28 54 10 13 57 4 14 18 27 39 46 36 49 40 29 24 35 44 32 59 5 52 48 7 61 55 6 16 42 2 43 8 50 3 9 34 26 37 11 22 62 21 31 25"
},
{
"input": "26\n11 4 19 13 17 9 2 24 6 5 22 23 14 15 3 25 16 8 18 10 21 1 12 26 7 20",
"output": "22 7 15 2 10 9 25 18 6 20 1 23 4 13 14 17 5 19 3 26 21 11 12 8 16 24"
},
{
"input": "69\n40 22 11 66 4 27 31 29 64 53 37 55 51 2 7 36 18 52 6 1 30 21 17 20 14 9 59 62 49 68 3 50 65 57 44 5 67 46 33 13 34 15 24 48 63 58 38 25 41 35 16 54 32 10 60 61 39 12 69 8 23 45 26 47 56 43 28 19 42",
"output": "20 14 31 5 36 19 15 60 26 54 3 58 40 25 42 51 23 17 68 24 22 2 61 43 48 63 6 67 8 21 7 53 39 41 50 16 11 47 57 1 49 69 66 35 62 38 64 44 29 32 13 18 10 52 12 65 34 46 27 55 56 28 45 9 33 4 37 30 59"
},
{
"input": "6\n4 3 6 5 1 2",
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},
{
"input": "9\n7 8 5 3 1 4 2 9 6",
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},
{
"input": "41\n27 24 16 30 25 8 32 2 26 20 39 33 41 22 40 14 36 9 28 4 34 11 31 23 19 18 17 35 3 10 6 13 5 15 29 38 7 21 1 12 37",
"output": "39 8 29 20 33 31 37 6 18 30 22 40 32 16 34 3 27 26 25 10 38 14 24 2 5 9 1 19 35 4 23 7 12 21 28 17 41 36 11 15 13"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "20\n2 6 4 18 7 10 17 13 16 8 14 9 20 5 19 12 1 3 15 11",
"output": "17 1 18 3 14 2 5 10 12 6 20 16 8 11 19 9 7 4 15 13"
},
{
"input": "2\n2 1",
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},
{
"input": "60\n2 4 31 51 11 7 34 20 3 14 18 23 48 54 15 36 38 60 49 40 5 33 41 26 55 58 10 8 13 9 27 30 37 1 21 59 44 57 35 19 46 43 42 45 12 22 39 32 24 16 6 56 53 52 25 17 47 29 50 28",
"output": "34 1 9 2 21 51 6 28 30 27 5 45 29 10 15 50 56 11 40 8 35 46 12 49 55 24 31 60 58 32 3 48 22 7 39 16 33 17 47 20 23 43 42 37 44 41 57 13 19 59 4 54 53 14 25 52 38 26 36 18"
},
{
"input": "14\n14 6 3 12 11 2 7 1 10 9 8 5 4 13",
"output": "8 6 3 13 12 2 7 11 10 9 5 4 14 1"
},
{
"input": "81\n13 43 79 8 7 21 73 46 63 4 62 78 56 11 70 68 61 53 60 49 16 27 59 47 69 5 22 44 77 57 52 48 1 9 72 81 28 55 58 33 51 18 31 17 41 20 42 3 32 54 19 2 75 34 64 10 65 50 30 29 67 12 71 66 74 15 26 23 6 38 25 35 37 24 80 76 40 45 39 36 14",
"output": "33 52 48 10 26 69 5 4 34 56 14 62 1 81 66 21 44 42 51 46 6 27 68 74 71 67 22 37 60 59 43 49 40 54 72 80 73 70 79 77 45 47 2 28 78 8 24 32 20 58 41 31 18 50 38 13 30 39 23 19 17 11 9 55 57 64 61 16 25 15 63 35 7 65 53 76 29 12 3 75 36"
},
{
"input": "42\n41 11 10 8 21 37 32 19 31 25 1 15 36 5 6 27 4 3 13 7 16 17 2 23 34 24 38 28 12 20 30 42 18 26 39 35 33 40 9 14 22 29",
"output": "11 23 18 17 14 15 20 4 39 3 2 29 19 40 12 21 22 33 8 30 5 41 24 26 10 34 16 28 42 31 9 7 37 25 36 13 6 27 35 38 1 32"
},
{
"input": "97\n20 6 76 42 4 18 35 59 39 63 27 7 66 47 61 52 15 36 88 93 19 33 10 92 1 34 46 86 78 57 51 94 77 29 26 73 41 2 58 97 43 65 17 74 21 49 25 3 91 82 95 12 96 13 84 90 69 24 72 37 16 55 54 71 64 62 48 89 11 70 80 67 30 40 44 85 53 83 79 9 56 45 75 87 22 14 81 68 8 38 60 50 28 23 31 32 5",
"output": "25 38 48 5 97 2 12 89 80 23 69 52 54 86 17 61 43 6 21 1 45 85 94 58 47 35 11 93 34 73 95 96 22 26 7 18 60 90 9 74 37 4 41 75 82 27 14 67 46 92 31 16 77 63 62 81 30 39 8 91 15 66 10 65 42 13 72 88 57 70 64 59 36 44 83 3 33 29 79 71 87 50 78 55 76 28 84 19 68 56 49 24 20 32 51 53 40"
},
{
"input": "62\n15 27 46 6 8 51 14 56 23 48 42 49 52 22 20 31 29 12 47 3 62 34 37 35 32 57 19 25 5 60 61 38 18 10 11 55 45 53 17 30 9 36 4 50 41 16 44 28 40 59 24 1 13 39 26 7 33 58 2 43 21 54",
"output": "52 59 20 43 29 4 56 5 41 34 35 18 53 7 1 46 39 33 27 15 61 14 9 51 28 55 2 48 17 40 16 25 57 22 24 42 23 32 54 49 45 11 60 47 37 3 19 10 12 44 6 13 38 62 36 8 26 58 50 30 31 21"
},
{
"input": "61\n35 27 4 61 52 32 41 46 14 37 17 54 55 31 11 26 44 49 15 30 9 50 45 39 7 38 53 3 58 40 13 56 18 19 28 6 43 5 21 42 20 34 2 25 36 12 33 57 16 60 1 8 59 10 22 23 24 48 51 47 29",
"output": "51 43 28 3 38 36 25 52 21 54 15 46 31 9 19 49 11 33 34 41 39 55 56 57 44 16 2 35 61 20 14 6 47 42 1 45 10 26 24 30 7 40 37 17 23 8 60 58 18 22 59 5 27 12 13 32 48 29 53 50 4"
},
{
"input": "59\n31 26 36 15 17 19 10 53 11 34 13 46 55 9 44 7 8 37 32 52 47 25 51 22 35 39 41 4 43 24 5 27 20 57 6 38 3 28 21 40 50 18 14 56 33 45 12 2 49 59 54 29 16 48 42 58 1 30 23",
"output": "57 48 37 28 31 35 16 17 14 7 9 47 11 43 4 53 5 42 6 33 39 24 59 30 22 2 32 38 52 58 1 19 45 10 25 3 18 36 26 40 27 55 29 15 46 12 21 54 49 41 23 20 8 51 13 44 34 56 50"
},
{
"input": "10\n2 10 7 4 1 5 8 6 3 9",
"output": "5 1 9 4 6 8 3 7 10 2"
},
{
"input": "14\n14 2 1 8 6 12 11 10 9 7 3 4 5 13",
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},
{
"input": "43\n28 38 15 14 31 42 27 30 19 33 43 26 22 29 18 32 3 13 1 8 35 34 4 12 11 17 41 21 5 25 39 37 20 23 7 24 16 10 40 9 6 36 2",
"output": "19 43 17 23 29 41 35 20 40 38 25 24 18 4 3 37 26 15 9 33 28 13 34 36 30 12 7 1 14 8 5 16 10 22 21 42 32 2 31 39 27 6 11"
},
{
"input": "86\n39 11 20 31 28 76 29 64 35 21 41 71 12 82 5 37 80 73 38 26 79 75 23 15 59 45 47 6 3 62 50 49 51 22 2 65 86 60 70 42 74 17 1 30 55 44 8 66 81 27 57 77 43 13 54 32 72 46 48 56 14 34 78 52 36 85 24 19 69 83 25 61 7 4 84 33 63 58 18 40 68 10 67 9 16 53",
"output": "43 35 29 74 15 28 73 47 84 82 2 13 54 61 24 85 42 79 68 3 10 34 23 67 71 20 50 5 7 44 4 56 76 62 9 65 16 19 1 80 11 40 53 46 26 58 27 59 32 31 33 64 86 55 45 60 51 78 25 38 72 30 77 8 36 48 83 81 69 39 12 57 18 41 22 6 52 63 21 17 49 14 70 75 66 37"
},
{
"input": "99\n65 78 56 98 33 24 61 40 29 93 1 64 57 22 25 52 67 95 50 3 31 15 90 68 71 83 38 36 6 46 89 26 4 87 14 88 72 37 23 43 63 12 80 96 5 34 73 86 9 48 92 62 99 10 16 20 66 27 28 2 82 70 30 94 49 8 84 69 18 60 58 59 44 39 21 7 91 76 54 19 75 85 74 47 55 32 97 77 51 13 35 79 45 42 11 41 17 81 53",
"output": "11 60 20 33 45 29 76 66 49 54 95 42 90 35 22 55 97 69 80 56 75 14 39 6 15 32 58 59 9 63 21 86 5 46 91 28 38 27 74 8 96 94 40 73 93 30 84 50 65 19 89 16 99 79 85 3 13 71 72 70 7 52 41 12 1 57 17 24 68 62 25 37 47 83 81 78 88 2 92 43 98 61 26 67 82 48 34 36 31 23 77 51 10 64 18 44 87 4 53"
},
{
"input": "100\n42 23 48 88 36 6 18 70 96 1 34 40 46 22 39 55 85 93 45 67 71 75 59 9 21 3 86 63 65 68 20 38 73 31 84 90 50 51 56 95 72 33 49 19 83 76 54 74 100 30 17 98 15 94 4 97 5 99 81 27 92 32 89 12 13 91 87 29 60 11 52 43 35 58 10 25 16 80 28 2 44 61 8 82 66 69 41 24 57 62 78 37 79 77 53 7 14 47 26 64",
"output": "10 80 26 55 57 6 96 83 24 75 70 64 65 97 53 77 51 7 44 31 25 14 2 88 76 99 60 79 68 50 34 62 42 11 73 5 92 32 15 12 87 1 72 81 19 13 98 3 43 37 38 71 95 47 16 39 89 74 23 69 82 90 28 100 29 85 20 30 86 8 21 41 33 48 22 46 94 91 93 78 59 84 45 35 17 27 67 4 63 36 66 61 18 54 40 9 56 52 58 49"
},
{
"input": "99\n8 68 94 75 71 60 57 58 6 11 5 48 65 41 49 12 46 72 95 59 13 70 74 7 84 62 17 36 55 76 38 79 2 85 23 10 32 99 87 50 83 28 54 91 53 51 1 3 97 81 21 89 93 78 61 26 82 96 4 98 25 40 31 44 24 47 30 52 14 16 39 27 9 29 45 18 67 63 37 43 90 66 19 69 88 22 92 77 34 42 73 80 56 64 20 35 15 33 86",
"output": "47 33 48 59 11 9 24 1 73 36 10 16 21 69 97 70 27 76 83 95 51 86 35 65 61 56 72 42 74 67 63 37 98 89 96 28 79 31 71 62 14 90 80 64 75 17 66 12 15 40 46 68 45 43 29 93 7 8 20 6 55 26 78 94 13 82 77 2 84 22 5 18 91 23 4 30 88 54 32 92 50 57 41 25 34 99 39 85 52 81 44 87 53 3 19 58 49 60 38"
},
{
"input": "99\n12 99 88 13 7 19 74 47 23 90 16 29 26 11 58 60 64 98 37 18 82 67 72 46 51 85 17 92 87 20 77 36 78 71 57 35 80 54 73 15 14 62 97 45 31 79 94 56 76 96 28 63 8 44 38 86 49 2 52 66 61 59 10 43 55 50 22 34 83 53 95 40 81 21 30 42 27 3 5 41 1 70 69 25 93 48 65 6 24 89 91 33 39 68 9 4 32 84 75",
"output": "81 58 78 96 79 88 5 53 95 63 14 1 4 41 40 11 27 20 6 30 74 67 9 89 84 13 77 51 12 75 45 97 92 68 36 32 19 55 93 72 80 76 64 54 44 24 8 86 57 66 25 59 70 38 65 48 35 15 62 16 61 42 52 17 87 60 22 94 83 82 34 23 39 7 99 49 31 33 46 37 73 21 69 98 26 56 29 3 90 10 91 28 85 47 71 50 43 18 2"
},
{
"input": "99\n20 79 26 75 99 69 98 47 93 62 18 42 43 38 90 66 67 8 13 84 76 58 81 60 64 46 56 23 78 17 86 36 19 52 85 39 48 27 96 49 37 95 5 31 10 24 12 1 80 35 92 33 16 68 57 54 32 29 45 88 72 77 4 87 97 89 59 3 21 22 61 94 83 15 44 34 70 91 55 9 51 50 73 11 14 6 40 7 63 25 2 82 41 65 28 74 71 30 53",
"output": "48 91 68 63 43 86 88 18 80 45 84 47 19 85 74 53 30 11 33 1 69 70 28 46 90 3 38 95 58 98 44 57 52 76 50 32 41 14 36 87 93 12 13 75 59 26 8 37 40 82 81 34 99 56 79 27 55 22 67 24 71 10 89 25 94 16 17 54 6 77 97 61 83 96 4 21 62 29 2 49 23 92 73 20 35 31 64 60 66 15 78 51 9 72 42 39 65 7 5"
},
{
"input": "99\n74 20 9 1 60 85 65 13 4 25 40 99 5 53 64 3 36 31 73 44 55 50 45 63 98 51 68 6 47 37 71 82 88 34 84 18 19 12 93 58 86 7 11 46 90 17 33 27 81 69 42 59 56 32 95 52 76 61 96 62 78 43 66 21 49 97 75 14 41 72 89 16 30 79 22 23 15 83 91 38 48 2 87 26 28 80 94 70 54 92 57 10 8 35 67 77 29 24 39",
"output": "4 82 16 9 13 28 42 93 3 92 43 38 8 68 77 72 46 36 37 2 64 75 76 98 10 84 48 85 97 73 18 54 47 34 94 17 30 80 99 11 69 51 62 20 23 44 29 81 65 22 26 56 14 89 21 53 91 40 52 5 58 60 24 15 7 63 95 27 50 88 31 70 19 1 67 57 96 61 74 86 49 32 78 35 6 41 83 33 71 45 79 90 39 87 55 59 66 25 12"
},
{
"input": "99\n50 94 2 18 69 90 59 83 75 68 77 97 39 78 25 7 16 9 49 4 42 89 44 48 17 96 61 70 3 10 5 81 56 57 88 6 98 1 46 67 92 37 11 30 85 41 8 36 51 29 20 71 19 79 74 93 43 34 55 40 38 21 64 63 32 24 72 14 12 86 82 15 65 23 66 22 28 53 13 26 95 99 91 52 76 27 60 45 47 33 73 84 31 35 54 80 58 62 87",
"output": "38 3 29 20 31 36 16 47 18 30 43 69 79 68 72 17 25 4 53 51 62 76 74 66 15 80 86 77 50 44 93 65 90 58 94 48 42 61 13 60 46 21 57 23 88 39 89 24 19 1 49 84 78 95 59 33 34 97 7 87 27 98 64 63 73 75 40 10 5 28 52 67 91 55 9 85 11 14 54 96 32 71 8 92 45 70 99 35 22 6 83 41 56 2 81 26 12 37 82"
},
{
"input": "99\n19 93 14 34 39 37 33 15 52 88 7 43 69 27 9 77 94 31 48 22 63 70 79 17 50 6 81 8 76 58 23 74 86 11 57 62 41 87 75 51 12 18 68 56 95 3 80 83 84 29 24 61 71 78 59 96 20 85 90 28 45 36 38 97 1 49 40 98 44 67 13 73 72 91 47 10 30 54 35 42 4 2 92 26 64 60 53 21 5 82 46 32 55 66 16 89 99 65 25",
"output": "65 82 46 81 89 26 11 28 15 76 34 41 71 3 8 95 24 42 1 57 88 20 31 51 99 84 14 60 50 77 18 92 7 4 79 62 6 63 5 67 37 80 12 69 61 91 75 19 66 25 40 9 87 78 93 44 35 30 55 86 52 36 21 85 98 94 70 43 13 22 53 73 72 32 39 29 16 54 23 47 27 90 48 49 58 33 38 10 96 59 74 83 2 17 45 56 64 68 97"
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{
"input": "99\n86 25 50 51 62 39 41 67 44 20 45 14 80 88 66 7 36 59 13 84 78 58 96 75 2 43 48 47 69 12 19 98 22 38 28 55 11 76 68 46 53 70 85 34 16 33 91 30 8 40 74 60 94 82 87 32 37 4 5 10 89 73 90 29 35 26 23 57 27 65 24 3 9 83 77 72 6 31 15 92 93 79 64 18 63 42 56 1 52 97 17 81 71 21 49 99 54 95 61",
"output": "88 25 72 58 59 77 16 49 73 60 37 30 19 12 79 45 91 84 31 10 94 33 67 71 2 66 69 35 64 48 78 56 46 44 65 17 57 34 6 50 7 86 26 9 11 40 28 27 95 3 4 89 41 97 36 87 68 22 18 52 99 5 85 83 70 15 8 39 29 42 93 76 62 51 24 38 75 21 82 13 92 54 74 20 43 1 55 14 61 63 47 80 81 53 98 23 90 32 96"
},
{
"input": "100\n66 44 99 15 43 79 28 33 88 90 49 68 82 38 9 74 4 58 29 81 31 94 10 42 89 21 63 40 62 61 18 6 84 72 48 25 67 69 71 85 98 34 83 70 65 78 91 77 93 41 23 24 87 11 55 12 59 73 36 97 7 14 26 39 30 27 45 20 50 17 53 2 57 47 95 56 75 19 37 96 16 35 8 3 76 60 13 86 5 32 64 80 46 51 54 100 1 22 52 92",
"output": "97 72 84 17 89 32 61 83 15 23 54 56 87 62 4 81 70 31 78 68 26 98 51 52 36 63 66 7 19 65 21 90 8 42 82 59 79 14 64 28 50 24 5 2 67 93 74 35 11 69 94 99 71 95 55 76 73 18 57 86 30 29 27 91 45 1 37 12 38 44 39 34 58 16 77 85 48 46 6 92 20 13 43 33 40 88 53 9 25 10 47 100 49 22 75 80 60 41 3 96"
},
{
"input": "99\n3 73 32 37 25 15 93 63 85 8 91 78 80 5 39 48 46 7 83 70 23 96 9 29 77 53 30 20 56 50 13 45 21 76 87 99 65 31 16 18 14 72 51 28 43 2 81 34 38 40 66 54 74 26 71 4 61 17 58 24 22 33 49 36 42 11 12 55 60 27 62 90 79 92 94 68 1 52 84 41 86 35 69 75 47 10 64 88 97 98 67 19 89 95 59 82 57 44 6",
"output": "77 46 1 56 14 99 18 10 23 86 66 67 31 41 6 39 58 40 92 28 33 61 21 60 5 54 70 44 24 27 38 3 62 48 82 64 4 49 15 50 80 65 45 98 32 17 85 16 63 30 43 78 26 52 68 29 97 59 95 69 57 71 8 87 37 51 91 76 83 20 55 42 2 53 84 34 25 12 73 13 47 96 19 79 9 81 35 88 93 72 11 74 7 75 94 22 89 90 36"
},
{
"input": "100\n100 99 98 97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 80 79 78 77 76 75 74 73 72 71 70 69 68 67 66 65 64 63 62 61 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1",
"output": "100 99 98 97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 80 79 78 77 76 75 74 73 72 71 70 69 68 67 66 65 64 63 62 61 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1"
}
] | 1,694,396,280
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 62
| 0
|
n = int(input())
sentence = list(map(int,input().split()))
word_list = [""]*int(len(sentence))
if n == int(len(sentence)):
for num in sentence:
word_list[num - 1] = int(sentence.index(num)) + 1
print(word_list)
|
Title: Presents
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Petya very much likes gifts. Recently he has received a new laptop as a New Year gift from his mother. He immediately decided to give it to somebody else as what can be more pleasant than giving somebody gifts. And on this occasion he organized a New Year party at his place and invited *n* his friends there.
If there's one thing Petya likes more that receiving gifts, that's watching others giving gifts to somebody else. Thus, he safely hid the laptop until the next New Year and made up his mind to watch his friends exchanging gifts while he does not participate in the process. He numbered all his friends with integers from 1 to *n*. Petya remembered that a friend number *i* gave a gift to a friend number *p**i*. He also remembered that each of his friends received exactly one gift.
Now Petya wants to know for each friend *i* the number of a friend who has given him a gift.
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the quantity of friends Petya invited to the party. The second line contains *n* space-separated integers: the *i*-th number is *p**i* — the number of a friend who gave a gift to friend number *i*. It is guaranteed that each friend received exactly one gift. It is possible that some friends do not share Petya's ideas of giving gifts to somebody else. Those friends gave the gifts to themselves.
Output Specification:
Print *n* space-separated integers: the *i*-th number should equal the number of the friend who gave a gift to friend number *i*.
Demo Input:
['4\n2 3 4 1\n', '3\n1 3 2\n', '2\n1 2\n']
Demo Output:
['4 1 2 3\n', '1 3 2\n', '1 2\n']
Note:
none
|
```python
n = int(input())
sentence = list(map(int,input().split()))
word_list = [""]*int(len(sentence))
if n == int(len(sentence)):
for num in sentence:
word_list[num - 1] = int(sentence.index(num)) + 1
print(word_list)
```
| 0
|
|
17
|
A
|
Noldbach problem
|
PROGRAMMING
| 1,000
|
[
"brute force",
"math",
"number theory"
] |
A. Noldbach problem
|
2
|
64
|
Nick is interested in prime numbers. Once he read about Goldbach problem. It states that every even integer greater than 2 can be expressed as the sum of two primes. That got Nick's attention and he decided to invent a problem of his own and call it Noldbach problem. Since Nick is interested only in prime numbers, Noldbach problem states that at least *k* prime numbers from 2 to *n* inclusively can be expressed as the sum of three integer numbers: two neighboring prime numbers and 1. For example, 19 = 7 + 11 + 1, or 13 = 5 + 7 + 1.
Two prime numbers are called neighboring if there are no other prime numbers between them.
You are to help Nick, and find out if he is right or wrong.
|
The first line of the input contains two integers *n* (2<=≤<=*n*<=≤<=1000) and *k* (0<=≤<=*k*<=≤<=1000).
|
Output YES if at least *k* prime numbers from 2 to *n* inclusively can be expressed as it was described above. Otherwise output NO.
|
[
"27 2\n",
"45 7\n"
] |
[
"YES",
"NO"
] |
In the first sample the answer is YES since at least two numbers can be expressed as it was described (for example, 13 and 19). In the second sample the answer is NO since it is impossible to express 7 prime numbers from 2 to 45 in the desired form.
| 0
|
[
{
"input": "27 2",
"output": "YES"
},
{
"input": "45 7",
"output": "NO"
},
{
"input": "2 0",
"output": "YES"
},
{
"input": "15 1",
"output": "YES"
},
{
"input": "17 1",
"output": "YES"
},
{
"input": "34 5",
"output": "NO"
},
{
"input": "37 4",
"output": "YES"
},
{
"input": "43 5",
"output": "YES"
},
{
"input": "47 7",
"output": "NO"
},
{
"input": "50 5",
"output": "YES"
},
{
"input": "57 6",
"output": "YES"
},
{
"input": "60 8",
"output": "NO"
},
{
"input": "62 7",
"output": "YES"
},
{
"input": "76 9",
"output": "NO"
},
{
"input": "69 7",
"output": "YES"
},
{
"input": "113 10",
"output": "YES"
},
{
"input": "141 11",
"output": "YES"
},
{
"input": "207 16",
"output": "NO"
},
{
"input": "231 18",
"output": "NO"
},
{
"input": "296 19",
"output": "YES"
},
{
"input": "332 20",
"output": "YES"
},
{
"input": "378 24",
"output": "NO"
},
{
"input": "428 23",
"output": "YES"
},
{
"input": "497 27",
"output": "NO"
},
{
"input": "640 32",
"output": "YES"
},
{
"input": "798 35",
"output": "YES"
},
{
"input": "802 35",
"output": "YES"
},
{
"input": "864 40",
"output": "NO"
},
{
"input": "953 44",
"output": "NO"
},
{
"input": "995 44",
"output": "YES"
},
{
"input": "1000 44",
"output": "YES"
},
{
"input": "1000 0",
"output": "YES"
},
{
"input": "1000 1000",
"output": "NO"
},
{
"input": "2 1000",
"output": "NO"
},
{
"input": "2 0",
"output": "YES"
}
] | 1,650,685,325
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 12
| 92
| 0
|
n,k = map(int, input().split())
prev_prime = 3
c = 0
def check_prime(n):
for i in range(2,int(n**0.5)+1):
if n%i == 0:
return False
return True
for i in range(3,n+1):
# find the next prime
if check_prime(i) == False:
continue
else:
next_prime = i
x = prev_prime + next_prime + 1
if x<=n:
if check_prime(x):
c += 1
if c>=k:
print('Yes')
else:
print('No')
|
Title: Noldbach problem
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
Nick is interested in prime numbers. Once he read about Goldbach problem. It states that every even integer greater than 2 can be expressed as the sum of two primes. That got Nick's attention and he decided to invent a problem of his own and call it Noldbach problem. Since Nick is interested only in prime numbers, Noldbach problem states that at least *k* prime numbers from 2 to *n* inclusively can be expressed as the sum of three integer numbers: two neighboring prime numbers and 1. For example, 19 = 7 + 11 + 1, or 13 = 5 + 7 + 1.
Two prime numbers are called neighboring if there are no other prime numbers between them.
You are to help Nick, and find out if he is right or wrong.
Input Specification:
The first line of the input contains two integers *n* (2<=≤<=*n*<=≤<=1000) and *k* (0<=≤<=*k*<=≤<=1000).
Output Specification:
Output YES if at least *k* prime numbers from 2 to *n* inclusively can be expressed as it was described above. Otherwise output NO.
Demo Input:
['27 2\n', '45 7\n']
Demo Output:
['YES', 'NO']
Note:
In the first sample the answer is YES since at least two numbers can be expressed as it was described (for example, 13 and 19). In the second sample the answer is NO since it is impossible to express 7 prime numbers from 2 to 45 in the desired form.
|
```python
n,k = map(int, input().split())
prev_prime = 3
c = 0
def check_prime(n):
for i in range(2,int(n**0.5)+1):
if n%i == 0:
return False
return True
for i in range(3,n+1):
# find the next prime
if check_prime(i) == False:
continue
else:
next_prime = i
x = prev_prime + next_prime + 1
if x<=n:
if check_prime(x):
c += 1
if c>=k:
print('Yes')
else:
print('No')
```
| 0
|
697
|
A
|
Pineapple Incident
|
PROGRAMMING
| 900
|
[
"implementation",
"math"
] | null | null |
Ted has a pineapple. This pineapple is able to bark like a bulldog! At time *t* (in seconds) it barks for the first time. Then every *s* seconds after it, it barks twice with 1 second interval. Thus it barks at times *t*, *t*<=+<=*s*, *t*<=+<=*s*<=+<=1, *t*<=+<=2*s*, *t*<=+<=2*s*<=+<=1, etc.
Barney woke up in the morning and wants to eat the pineapple, but he can't eat it when it's barking. Barney plans to eat it at time *x* (in seconds), so he asked you to tell him if it's gonna bark at that time.
|
The first and only line of input contains three integers *t*, *s* and *x* (0<=≤<=*t*,<=*x*<=≤<=109, 2<=≤<=*s*<=≤<=109) — the time the pineapple barks for the first time, the pineapple barking interval, and the time Barney wants to eat the pineapple respectively.
|
Print a single "YES" (without quotes) if the pineapple will bark at time *x* or a single "NO" (without quotes) otherwise in the only line of output.
|
[
"3 10 4\n",
"3 10 3\n",
"3 8 51\n",
"3 8 52\n"
] |
[
"NO\n",
"YES\n",
"YES\n",
"YES\n"
] |
In the first and the second sample cases pineapple will bark at moments 3, 13, 14, ..., so it won't bark at the moment 4 and will bark at the moment 3.
In the third and fourth sample cases pineapple will bark at moments 3, 11, 12, 19, 20, 27, 28, 35, 36, 43, 44, 51, 52, 59, ..., so it will bark at both moments 51 and 52.
| 500
|
[
{
"input": "3 10 4",
"output": "NO"
},
{
"input": "3 10 3",
"output": "YES"
},
{
"input": "3 8 51",
"output": "YES"
},
{
"input": "3 8 52",
"output": "YES"
},
{
"input": "456947336 740144 45",
"output": "NO"
},
{
"input": "33 232603 599417964",
"output": "YES"
},
{
"input": "4363010 696782227 701145238",
"output": "YES"
},
{
"input": "9295078 2 6",
"output": "NO"
},
{
"input": "76079 281367 119938421",
"output": "YES"
},
{
"input": "93647 7 451664565",
"output": "YES"
},
{
"input": "5 18553 10908",
"output": "NO"
},
{
"input": "6 52 30",
"output": "NO"
},
{
"input": "6431 855039 352662",
"output": "NO"
},
{
"input": "749399100 103031711 761562532",
"output": "NO"
},
{
"input": "21 65767 55245",
"output": "NO"
},
{
"input": "4796601 66897 4860613",
"output": "NO"
},
{
"input": "8 6728951 860676",
"output": "NO"
},
{
"input": "914016 6 914019",
"output": "NO"
},
{
"input": "60686899 78474 60704617",
"output": "NO"
},
{
"input": "3 743604 201724",
"output": "NO"
},
{
"input": "571128 973448796 10",
"output": "NO"
},
{
"input": "688051712 67 51",
"output": "NO"
},
{
"input": "74619 213344 6432326",
"output": "NO"
},
{
"input": "6947541 698167 6",
"output": "NO"
},
{
"input": "83 6 6772861",
"output": "NO"
},
{
"input": "251132 67561 135026988",
"output": "NO"
},
{
"input": "8897216 734348516 743245732",
"output": "YES"
},
{
"input": "50 64536 153660266",
"output": "YES"
},
{
"input": "876884 55420 971613604",
"output": "YES"
},
{
"input": "0 6906451 366041903",
"output": "YES"
},
{
"input": "11750 8 446010134",
"output": "YES"
},
{
"input": "582692707 66997 925047377",
"output": "YES"
},
{
"input": "11 957526890 957526901",
"output": "YES"
},
{
"input": "556888 514614196 515171084",
"output": "YES"
},
{
"input": "6 328006 584834704",
"output": "YES"
},
{
"input": "4567998 4 204966403",
"output": "YES"
},
{
"input": "60 317278 109460971",
"output": "YES"
},
{
"input": "906385 342131991 685170368",
"output": "YES"
},
{
"input": "1 38 902410512",
"output": "YES"
},
{
"input": "29318 787017 587931018",
"output": "YES"
},
{
"input": "351416375 243431 368213115",
"output": "YES"
},
{
"input": "54 197366062 197366117",
"output": "YES"
},
{
"input": "586389 79039 850729874",
"output": "YES"
},
{
"input": "723634470 2814619 940360134",
"output": "YES"
},
{
"input": "0 2 0",
"output": "YES"
},
{
"input": "0 2 1",
"output": "NO"
},
{
"input": "0 2 2",
"output": "YES"
},
{
"input": "0 2 3",
"output": "YES"
},
{
"input": "0 2 1000000000",
"output": "YES"
},
{
"input": "0 10 23",
"output": "NO"
},
{
"input": "0 2 999999999",
"output": "YES"
},
{
"input": "10 5 11",
"output": "NO"
},
{
"input": "1 2 1000000000",
"output": "YES"
},
{
"input": "1 10 20",
"output": "NO"
},
{
"input": "1 2 999999937",
"output": "YES"
},
{
"input": "10 3 5",
"output": "NO"
},
{
"input": "3 2 5",
"output": "YES"
},
{
"input": "0 4 0",
"output": "YES"
},
{
"input": "0 215 403",
"output": "NO"
},
{
"input": "5 2 10",
"output": "YES"
},
{
"input": "0 2 900000000",
"output": "YES"
},
{
"input": "0 79 4000",
"output": "NO"
},
{
"input": "5 1000 1000",
"output": "NO"
},
{
"input": "1 5 103",
"output": "NO"
},
{
"input": "5 2 6",
"output": "NO"
},
{
"input": "120 2 1000000000",
"output": "YES"
},
{
"input": "2 2 1000000000",
"output": "YES"
},
{
"input": "5 5 13",
"output": "NO"
},
{
"input": "10 5 15",
"output": "YES"
},
{
"input": "11 2 0",
"output": "NO"
},
{
"input": "3 8 53",
"output": "NO"
},
{
"input": "2 2 4",
"output": "YES"
},
{
"input": "4 4 0",
"output": "NO"
},
{
"input": "1 2 3",
"output": "YES"
},
{
"input": "5 3 9",
"output": "YES"
},
{
"input": "5 6 19",
"output": "NO"
},
{
"input": "3 10 125",
"output": "NO"
},
{
"input": "5 3 8",
"output": "YES"
},
{
"input": "6 3 9",
"output": "YES"
},
{
"input": "0 3 5",
"output": "NO"
},
{
"input": "5 3 300000035",
"output": "YES"
},
{
"input": "5 2 7",
"output": "YES"
},
{
"input": "1 5 6",
"output": "YES"
},
{
"input": "4 2 6",
"output": "YES"
},
{
"input": "0 3 999999998",
"output": "NO"
},
{
"input": "0 10001 0",
"output": "YES"
},
{
"input": "6 5 3",
"output": "NO"
},
{
"input": "1 5 1000000000",
"output": "NO"
},
{
"input": "1 3 6",
"output": "NO"
},
{
"input": "3 3 1000000000",
"output": "YES"
},
{
"input": "3 3 4",
"output": "NO"
},
{
"input": "3 3 5",
"output": "NO"
},
{
"input": "3 3 0",
"output": "NO"
},
{
"input": "1 2 4",
"output": "YES"
},
{
"input": "5 5 10",
"output": "YES"
}
] | 1,590,087,006
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 7
| 93
| 0
|
t,s,x=(int(i) for i in input().split())
ans="NO"
if (x-t)%s==0 or (x-t)%s==1 and x!=t+1 and x>t:
ans="YES"
print(ans)
|
Title: Pineapple Incident
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ted has a pineapple. This pineapple is able to bark like a bulldog! At time *t* (in seconds) it barks for the first time. Then every *s* seconds after it, it barks twice with 1 second interval. Thus it barks at times *t*, *t*<=+<=*s*, *t*<=+<=*s*<=+<=1, *t*<=+<=2*s*, *t*<=+<=2*s*<=+<=1, etc.
Barney woke up in the morning and wants to eat the pineapple, but he can't eat it when it's barking. Barney plans to eat it at time *x* (in seconds), so he asked you to tell him if it's gonna bark at that time.
Input Specification:
The first and only line of input contains three integers *t*, *s* and *x* (0<=≤<=*t*,<=*x*<=≤<=109, 2<=≤<=*s*<=≤<=109) — the time the pineapple barks for the first time, the pineapple barking interval, and the time Barney wants to eat the pineapple respectively.
Output Specification:
Print a single "YES" (without quotes) if the pineapple will bark at time *x* or a single "NO" (without quotes) otherwise in the only line of output.
Demo Input:
['3 10 4\n', '3 10 3\n', '3 8 51\n', '3 8 52\n']
Demo Output:
['NO\n', 'YES\n', 'YES\n', 'YES\n']
Note:
In the first and the second sample cases pineapple will bark at moments 3, 13, 14, ..., so it won't bark at the moment 4 and will bark at the moment 3.
In the third and fourth sample cases pineapple will bark at moments 3, 11, 12, 19, 20, 27, 28, 35, 36, 43, 44, 51, 52, 59, ..., so it will bark at both moments 51 and 52.
|
```python
t,s,x=(int(i) for i in input().split())
ans="NO"
if (x-t)%s==0 or (x-t)%s==1 and x!=t+1 and x>t:
ans="YES"
print(ans)
```
| 0
|
|
268
|
A
|
Games
|
PROGRAMMING
| 800
|
[
"brute force"
] | null | null |
Manao works on a sports TV. He's spent much time watching the football games of some country. After a while he began to notice different patterns. For example, each team has two sets of uniforms: home uniform and guest uniform. When a team plays a game at home, the players put on the home uniform. When a team plays as a guest on somebody else's stadium, the players put on the guest uniform. The only exception to that rule is: when the home uniform color of the host team matches the guests' uniform, the host team puts on its guest uniform as well. For each team the color of the home and guest uniform is different.
There are *n* teams taking part in the national championship. The championship consists of *n*·(*n*<=-<=1) games: each team invites each other team to its stadium. At this point Manao wondered: how many times during the championship is a host team going to put on the guest uniform? Note that the order of the games does not affect this number.
You know the colors of the home and guest uniform for each team. For simplicity, the colors are numbered by integers in such a way that no two distinct colors have the same number. Help Manao find the answer to his question.
|
The first line contains an integer *n* (2<=≤<=*n*<=≤<=30). Each of the following *n* lines contains a pair of distinct space-separated integers *h**i*, *a**i* (1<=≤<=*h**i*,<=*a**i*<=≤<=100) — the colors of the *i*-th team's home and guest uniforms, respectively.
|
In a single line print the number of games where the host team is going to play in the guest uniform.
|
[
"3\n1 2\n2 4\n3 4\n",
"4\n100 42\n42 100\n5 42\n100 5\n",
"2\n1 2\n1 2\n"
] |
[
"1\n",
"5\n",
"0\n"
] |
In the first test case the championship consists of 6 games. The only game with the event in question is the game between teams 2 and 1 on the stadium of team 2.
In the second test sample the host team will have to wear guest uniform in the games between teams: 1 and 2, 2 and 1, 2 and 3, 3 and 4, 4 and 2 (the host team is written first).
| 500
|
[
{
"input": "3\n1 2\n2 4\n3 4",
"output": "1"
},
{
"input": "4\n100 42\n42 100\n5 42\n100 5",
"output": "5"
},
{
"input": "2\n1 2\n1 2",
"output": "0"
},
{
"input": "7\n4 7\n52 55\n16 4\n55 4\n20 99\n3 4\n7 52",
"output": "6"
},
{
"input": "10\n68 42\n1 35\n25 70\n59 79\n65 63\n46 6\n28 82\n92 62\n43 96\n37 28",
"output": "1"
},
{
"input": "30\n10 39\n89 1\n78 58\n75 99\n36 13\n77 50\n6 97\n79 28\n27 52\n56 5\n93 96\n40 21\n33 74\n26 37\n53 59\n98 56\n61 65\n42 57\n9 7\n25 63\n74 34\n96 84\n95 47\n12 23\n34 21\n71 6\n27 13\n15 47\n64 14\n12 77",
"output": "6"
},
{
"input": "30\n46 100\n87 53\n34 84\n44 66\n23 20\n50 34\n90 66\n17 39\n13 22\n94 33\n92 46\n63 78\n26 48\n44 61\n3 19\n41 84\n62 31\n65 89\n23 28\n58 57\n19 85\n26 60\n75 66\n69 67\n76 15\n64 15\n36 72\n90 89\n42 69\n45 35",
"output": "4"
},
{
"input": "2\n46 6\n6 46",
"output": "2"
},
{
"input": "29\n8 18\n33 75\n69 22\n97 95\n1 97\n78 10\n88 18\n13 3\n19 64\n98 12\n79 92\n41 72\n69 15\n98 31\n57 74\n15 56\n36 37\n15 66\n63 100\n16 42\n47 56\n6 4\n73 15\n30 24\n27 71\n12 19\n88 69\n85 6\n50 11",
"output": "10"
},
{
"input": "23\n43 78\n31 28\n58 80\n66 63\n20 4\n51 95\n40 20\n50 14\n5 34\n36 39\n77 42\n64 97\n62 89\n16 56\n8 34\n58 16\n37 35\n37 66\n8 54\n50 36\n24 8\n68 48\n85 33",
"output": "6"
},
{
"input": "13\n76 58\n32 85\n99 79\n23 58\n96 59\n72 35\n53 43\n96 55\n41 78\n75 10\n28 11\n72 7\n52 73",
"output": "0"
},
{
"input": "18\n6 90\n70 79\n26 52\n67 81\n29 95\n41 32\n94 88\n18 58\n59 65\n51 56\n64 68\n34 2\n6 98\n95 82\n34 2\n40 98\n83 78\n29 2",
"output": "1"
},
{
"input": "18\n6 90\n100 79\n26 100\n67 100\n29 100\n100 32\n94 88\n18 58\n59 65\n51 56\n64 68\n34 2\n6 98\n95 82\n34 2\n40 98\n83 78\n29 100",
"output": "8"
},
{
"input": "30\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1",
"output": "450"
},
{
"input": "30\n100 99\n58 59\n56 57\n54 55\n52 53\n50 51\n48 49\n46 47\n44 45\n42 43\n40 41\n38 39\n36 37\n34 35\n32 33\n30 31\n28 29\n26 27\n24 25\n22 23\n20 21\n18 19\n16 17\n14 15\n12 13\n10 11\n8 9\n6 7\n4 5\n2 3",
"output": "0"
},
{
"input": "15\n9 3\n2 6\n7 6\n5 10\n9 5\n8 1\n10 5\n2 8\n4 5\n9 8\n5 3\n3 8\n9 8\n4 10\n8 5",
"output": "20"
},
{
"input": "15\n2 1\n1 2\n1 2\n1 2\n2 1\n2 1\n2 1\n1 2\n2 1\n2 1\n2 1\n1 2\n2 1\n2 1\n1 2",
"output": "108"
},
{
"input": "25\n2 1\n1 2\n1 2\n1 2\n2 1\n1 2\n1 2\n1 2\n2 1\n2 1\n2 1\n1 2\n1 2\n1 2\n2 1\n2 1\n2 1\n1 2\n2 1\n1 2\n2 1\n2 1\n2 1\n2 1\n1 2",
"output": "312"
},
{
"input": "25\n91 57\n2 73\n54 57\n2 57\n23 57\n2 6\n57 54\n57 23\n91 54\n91 23\n57 23\n91 57\n54 2\n6 91\n57 54\n2 57\n57 91\n73 91\n57 23\n91 57\n2 73\n91 2\n23 6\n2 73\n23 6",
"output": "96"
},
{
"input": "28\n31 66\n31 91\n91 31\n97 66\n31 66\n31 66\n66 91\n91 31\n97 31\n91 97\n97 31\n66 31\n66 97\n91 31\n31 66\n31 66\n66 31\n31 97\n66 97\n97 31\n31 91\n66 91\n91 66\n31 66\n91 66\n66 31\n66 31\n91 97",
"output": "210"
},
{
"input": "29\n78 27\n50 68\n24 26\n68 43\n38 78\n26 38\n78 28\n28 26\n27 24\n23 38\n24 26\n24 43\n61 50\n38 78\n27 23\n61 26\n27 28\n43 23\n28 78\n43 27\n43 78\n27 61\n28 38\n61 78\n50 26\n43 27\n26 78\n28 50\n43 78",
"output": "73"
},
{
"input": "29\n80 27\n69 80\n27 80\n69 80\n80 27\n80 27\n80 27\n80 69\n27 69\n80 69\n80 27\n27 69\n69 27\n80 69\n27 69\n69 80\n27 69\n80 69\n80 27\n69 27\n27 69\n27 80\n80 27\n69 80\n27 69\n80 69\n69 80\n69 80\n27 80",
"output": "277"
},
{
"input": "30\n19 71\n7 89\n89 71\n21 7\n19 21\n7 89\n19 71\n89 8\n89 21\n19 8\n21 7\n8 89\n19 89\n7 21\n19 8\n19 7\n7 19\n8 21\n71 21\n71 89\n7 19\n7 19\n21 7\n21 19\n21 19\n71 8\n21 8\n71 19\n19 71\n8 21",
"output": "154"
},
{
"input": "30\n44 17\n44 17\n44 17\n17 44\n44 17\n44 17\n17 44\n17 44\n17 44\n44 17\n44 17\n44 17\n44 17\n44 17\n17 44\n17 44\n17 44\n44 17\n44 17\n17 44\n44 17\n44 17\n44 17\n17 44\n17 44\n44 17\n17 44\n44 17\n44 17\n44 17",
"output": "418"
},
{
"input": "22\n78 92\n15 92\n92 78\n78 80\n92 16\n24 80\n92 16\n16 92\n78 16\n24 78\n80 78\n92 80\n16 80\n80 78\n15 78\n92 16\n24 15\n24 80\n80 16\n16 80\n92 80\n24 80",
"output": "74"
},
{
"input": "24\n9 83\n90 31\n83 3\n83 3\n21 31\n83 3\n32 31\n12 21\n31 21\n90 32\n32 21\n12 9\n12 31\n9 83\n83 12\n32 3\n32 83\n90 31\n9 32\n31 21\n83 90\n32 21\n21 3\n32 9",
"output": "59"
},
{
"input": "30\n67 21\n85 39\n85 87\n21 39\n66 85\n10 95\n10 21\n87 85\n82 21\n67 21\n95 10\n21 39\n82 21\n21 66\n66 39\n95 30\n67 85\n66 82\n85 82\n21 66\n10 39\n67 10\n21 85\n10 82\n85 95\n10 85\n21 39\n85 39\n39 10\n95 67",
"output": "100"
},
{
"input": "4\n8 7\n8 7\n7 8\n7 8",
"output": "8"
},
{
"input": "6\n1 2\n1 2\n1 2\n1 2\n1 2\n2 1",
"output": "10"
},
{
"input": "12\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1",
"output": "72"
},
{
"input": "4\n1 2\n1 2\n2 1\n2 1",
"output": "8"
}
] | 1,688,050,953
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 31
| 124
| 0
|
n = int(input())
homeColors = []
guestColors = []
for _ in range(n):
home, guest = map(int, input().split())
homeColors.append(home)
guestColors.append(guest)
count = 0
for i in range(n):
for j in range(n):
if homeColors[i] == guestColors[j]:
count += 1
print(count)
|
Title: Games
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Manao works on a sports TV. He's spent much time watching the football games of some country. After a while he began to notice different patterns. For example, each team has two sets of uniforms: home uniform and guest uniform. When a team plays a game at home, the players put on the home uniform. When a team plays as a guest on somebody else's stadium, the players put on the guest uniform. The only exception to that rule is: when the home uniform color of the host team matches the guests' uniform, the host team puts on its guest uniform as well. For each team the color of the home and guest uniform is different.
There are *n* teams taking part in the national championship. The championship consists of *n*·(*n*<=-<=1) games: each team invites each other team to its stadium. At this point Manao wondered: how many times during the championship is a host team going to put on the guest uniform? Note that the order of the games does not affect this number.
You know the colors of the home and guest uniform for each team. For simplicity, the colors are numbered by integers in such a way that no two distinct colors have the same number. Help Manao find the answer to his question.
Input Specification:
The first line contains an integer *n* (2<=≤<=*n*<=≤<=30). Each of the following *n* lines contains a pair of distinct space-separated integers *h**i*, *a**i* (1<=≤<=*h**i*,<=*a**i*<=≤<=100) — the colors of the *i*-th team's home and guest uniforms, respectively.
Output Specification:
In a single line print the number of games where the host team is going to play in the guest uniform.
Demo Input:
['3\n1 2\n2 4\n3 4\n', '4\n100 42\n42 100\n5 42\n100 5\n', '2\n1 2\n1 2\n']
Demo Output:
['1\n', '5\n', '0\n']
Note:
In the first test case the championship consists of 6 games. The only game with the event in question is the game between teams 2 and 1 on the stadium of team 2.
In the second test sample the host team will have to wear guest uniform in the games between teams: 1 and 2, 2 and 1, 2 and 3, 3 and 4, 4 and 2 (the host team is written first).
|
```python
n = int(input())
homeColors = []
guestColors = []
for _ in range(n):
home, guest = map(int, input().split())
homeColors.append(home)
guestColors.append(guest)
count = 0
for i in range(n):
for j in range(n):
if homeColors[i] == guestColors[j]:
count += 1
print(count)
```
| 3
|
|
159
|
B
|
Matchmaker
|
PROGRAMMING
| 1,100
|
[
"*special",
"greedy",
"sortings"
] | null | null |
Polycarpus has *n* markers and *m* marker caps. Each marker is described by two numbers: *x**i* is the color and *y**i* is the diameter. Correspondingly, each cap is described by two numbers: *a**j* is the color and *b**j* is the diameter. Cap (*a**j*,<=*b**j*) can close marker (*x**i*,<=*y**i*) only if their diameters match, that is, *b**j*<==<=*y**i*. Besides, a marker is considered to be beautifully closed, if the cap color and the marker color match, that is, *a**j*<==<=*x**i*.
Find the way to close the maximum number of markers. If there are several such ways, then choose the one that has the maximum number of beautifully closed markers.
|
The first input line contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=105) — the number of markers and the number of caps, correspondingly.
Next *n* lines describe the markers. The *i*-th line contains two space-separated integers *x**i*, *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=1000) — the *i*-th marker's color and diameter, correspondingly.
Next *m* lines describe the caps. The *j*-th line contains two space-separated integers *a**j*, *b**j* (1<=≤<=*a**j*,<=*b**j*<=≤<=1000) — the color and diameter of the *j*-th cap, correspondingly.
|
Print two space-separated integers *u*,<=*v*, where *u* is the number of closed markers and *v* is the number of beautifully closed markers in the sought optimal way. Remember that you have to find the way to close the maximum number of markers, and if there are several such ways, you should choose the one where the number of beautifully closed markers is maximum.
|
[
"3 4\n1 2\n3 4\n2 4\n5 4\n2 4\n1 1\n1 2\n",
"2 2\n1 2\n2 1\n3 4\n5 1\n"
] |
[
"3 2\n",
"1 0\n"
] |
In the first test sample the first marker should be closed by the fourth cap, the second marker should be closed by the first cap and the third marker should be closed by the second cap. Thus, three markers will be closed, and two of them will be beautifully closed — the first and the third markers.
| 1,000
|
[
{
"input": "3 4\n1 2\n3 4\n2 4\n5 4\n2 4\n1 1\n1 2",
"output": "3 2"
},
{
"input": "2 2\n1 2\n2 1\n3 4\n5 1",
"output": "1 0"
},
{
"input": "6 7\n2 1\n2 2\n2 1\n1 1\n2 1\n1 2\n2 2\n2 2\n2 2\n1 2\n2 2\n1 1\n1 2",
"output": "3 3"
},
{
"input": "6 7\n2 1\n1 1\n2 2\n1 2\n1 1\n1 2\n2 1\n1 1\n1 1\n1 1\n1 2\n2 2\n1 1",
"output": "5 5"
},
{
"input": "6 7\n2 1\n2 2\n1 1\n1 2\n2 2\n1 2\n1 2\n1 1\n1 2\n2 2\n2 2\n2 2\n1 1",
"output": "6 5"
},
{
"input": "6 7\n1 1\n1 1\n1 2\n1 2\n2 2\n1 2\n2 1\n2 1\n2 1\n2 1\n1 1\n1 1\n1 2",
"output": "3 3"
},
{
"input": "6 2\n1 1\n2 1\n1 1\n1 1\n1 2\n1 1\n2 2\n1 2",
"output": "1 1"
},
{
"input": "3 3\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1",
"output": "3 3"
},
{
"input": "1 1\n1 1\n1 1",
"output": "1 1"
},
{
"input": "1 1\n1 1\n2 2",
"output": "0 0"
},
{
"input": "1 1\n1 1\n1 2",
"output": "0 0"
},
{
"input": "1 1\n1 2\n2 2",
"output": "1 0"
},
{
"input": "1 5\n1 1\n2 2\n1 1\n2 1\n1 2\n1 1",
"output": "1 1"
}
] | 1,590,822,274
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 2
| 248
| 0
|
n,m=map(int,input().split())
x,y=[0]*1001,[0]*1001
l,d=[],[]
for i in range(n):
l.append(list(map(int,input().split())))
for i in range(m):
d.append(list(map(int,input().split())))
kk,mm=0,0
for i in range(m):
x[d[i][0]]+=1
y[d[i][1]]+=1
for i in range(n):
if(y[l[i][1]]>0):
kk+=1
y[l[i][1]]-=1
for i in range(n):
if(x[l[i][0]]>0):
mm+=1
x[l[i][0]]-=1
print(kk,mm)
|
Title: Matchmaker
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus has *n* markers and *m* marker caps. Each marker is described by two numbers: *x**i* is the color and *y**i* is the diameter. Correspondingly, each cap is described by two numbers: *a**j* is the color and *b**j* is the diameter. Cap (*a**j*,<=*b**j*) can close marker (*x**i*,<=*y**i*) only if their diameters match, that is, *b**j*<==<=*y**i*. Besides, a marker is considered to be beautifully closed, if the cap color and the marker color match, that is, *a**j*<==<=*x**i*.
Find the way to close the maximum number of markers. If there are several such ways, then choose the one that has the maximum number of beautifully closed markers.
Input Specification:
The first input line contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=105) — the number of markers and the number of caps, correspondingly.
Next *n* lines describe the markers. The *i*-th line contains two space-separated integers *x**i*, *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=1000) — the *i*-th marker's color and diameter, correspondingly.
Next *m* lines describe the caps. The *j*-th line contains two space-separated integers *a**j*, *b**j* (1<=≤<=*a**j*,<=*b**j*<=≤<=1000) — the color and diameter of the *j*-th cap, correspondingly.
Output Specification:
Print two space-separated integers *u*,<=*v*, where *u* is the number of closed markers and *v* is the number of beautifully closed markers in the sought optimal way. Remember that you have to find the way to close the maximum number of markers, and if there are several such ways, you should choose the one where the number of beautifully closed markers is maximum.
Demo Input:
['3 4\n1 2\n3 4\n2 4\n5 4\n2 4\n1 1\n1 2\n', '2 2\n1 2\n2 1\n3 4\n5 1\n']
Demo Output:
['3 2\n', '1 0\n']
Note:
In the first test sample the first marker should be closed by the fourth cap, the second marker should be closed by the first cap and the third marker should be closed by the second cap. Thus, three markers will be closed, and two of them will be beautifully closed — the first and the third markers.
|
```python
n,m=map(int,input().split())
x,y=[0]*1001,[0]*1001
l,d=[],[]
for i in range(n):
l.append(list(map(int,input().split())))
for i in range(m):
d.append(list(map(int,input().split())))
kk,mm=0,0
for i in range(m):
x[d[i][0]]+=1
y[d[i][1]]+=1
for i in range(n):
if(y[l[i][1]]>0):
kk+=1
y[l[i][1]]-=1
for i in range(n):
if(x[l[i][0]]>0):
mm+=1
x[l[i][0]]-=1
print(kk,mm)
```
| 0
|
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