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54
| type
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7
| title
stringclasses 393
values | time-limit
stringclasses 7
values | memory-limit
stringclasses 6
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stringlengths 0
2.97k
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stringlengths 4
1.87k
| output-specification
stringlengths 4
1.12k
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7
| demo-output
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| points
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1.7B
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int64 8
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values | verdict
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stringclasses 9
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int64 1
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int64 15
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61.4k
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stringlengths 297
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stringlengths 25
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| score
float64 2.82
3.99
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
180
|
C
|
Letter
|
PROGRAMMING
| 1,400
|
[
"dp"
] | null | null |
Patrick has just finished writing a message to his sweetheart Stacey when he noticed that the message didn't look fancy. Patrick was nervous while writing the message, so some of the letters there were lowercase and some of them were uppercase.
Patrick believes that a message is fancy if any uppercase letter stands to the left of any lowercase one. In other words, this rule describes the strings where first go zero or more uppercase letters, and then β zero or more lowercase letters.
To make the message fancy, Patrick can erase some letter and add the same letter in the same place in the opposite case (that is, he can replace an uppercase letter with the lowercase one and vice versa). Patrick got interested in the following question: what minimum number of actions do we need to make a message fancy? Changing a letter's case in the message counts as one action. Patrick cannot perform any other actions.
|
The only line of the input contains a non-empty string consisting of uppercase and lowercase letters. The string's length does not exceed 105.
|
Print a single number β the least number of actions needed to make the message fancy.
|
[
"PRuvetSTAaYA\n",
"OYPROSTIYAOPECHATALSYAPRIVETSTASYA\n",
"helloworld\n"
] |
[
"5\n",
"0\n",
"0\n"
] |
none
| 0
|
[
{
"input": "PRuvetSTAaYA",
"output": "5"
},
{
"input": "OYPROSTIYAOPECHATALSYAPRIVETSTASYA",
"output": "0"
},
{
"input": "helloworld",
"output": "0"
},
{
"input": "P",
"output": "0"
},
{
"input": "t",
"output": "0"
},
{
"input": "XdJ",
"output": "1"
},
{
"input": "FSFlNEelYY",
"output": "3"
},
{
"input": "lgtyasficu",
"output": "0"
},
{
"input": "WYKUDTDDBT",
"output": "0"
},
{
"input": "yysxwlyqboatikfnpxczmpijziiojbvadlfozjqldssffcxdegyxfrvohoxvgsrvlzjlkcuffoeisrpvagxtbkapkpzcafadzzjd",
"output": "0"
},
{
"input": "mnAkOBuKxaiJwXhKnlcCvjxYXGXDoIqfUYkiLrdSYWhMemgWFzsgpoKOtHqooxbLYFuABWQSXuHdbyPVWyrkeEfqOsnEBikiqhfu",
"output": "43"
},
{
"input": "MMVESdOCALHJCTBTUWWQRGUUVTTTABKKAAdIINAdKLRLLVLODHDXDPMcQfUhPNHFBJSDRGsHZNORSCPNvKOOIuZnZAmTPUCoPNlR",
"output": "13"
},
{
"input": "MMbJIBhgFXPVpdQHLkWJkAHFIfJSpITTCRzRCzvRPRYECCheOknfINZWuKATDBOrEVKTcWXiYPjtzQMGUSGPNTMCUrvYCSWQHqAi",
"output": "27"
},
{
"input": "ZnqXEBOABXVbHRFFqDLWpWBBLqZIagmbRAYMDKJAYyFQXnPYwvDyRXKHLQnMLHGEOnRKmXtGqyQkoBCSmgKPcEIEugwJyxDOtTVn",
"output": "31"
}
] | 1,575,117,428
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 40
| 310
| 2,355,200
|
s = input()
up_cnt = 0
low_cnt = 0
for i in s:
if i.isupper():
up_cnt += 1
else:
low_cnt += 1
cur_up = 0
cur_low = 0
ans = 1000000000
cost = up_cnt
for i in s:
if i.isupper():
cur_up += 1
else:
cur_low += 1
a = cur_low + up_cnt - cur_up
if a < ans:
ans = a
print(min(ans, up_cnt))
|
Title: Letter
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Patrick has just finished writing a message to his sweetheart Stacey when he noticed that the message didn't look fancy. Patrick was nervous while writing the message, so some of the letters there were lowercase and some of them were uppercase.
Patrick believes that a message is fancy if any uppercase letter stands to the left of any lowercase one. In other words, this rule describes the strings where first go zero or more uppercase letters, and then β zero or more lowercase letters.
To make the message fancy, Patrick can erase some letter and add the same letter in the same place in the opposite case (that is, he can replace an uppercase letter with the lowercase one and vice versa). Patrick got interested in the following question: what minimum number of actions do we need to make a message fancy? Changing a letter's case in the message counts as one action. Patrick cannot perform any other actions.
Input Specification:
The only line of the input contains a non-empty string consisting of uppercase and lowercase letters. The string's length does not exceed 105.
Output Specification:
Print a single number β the least number of actions needed to make the message fancy.
Demo Input:
['PRuvetSTAaYA\n', 'OYPROSTIYAOPECHATALSYAPRIVETSTASYA\n', 'helloworld\n']
Demo Output:
['5\n', '0\n', '0\n']
Note:
none
|
```python
s = input()
up_cnt = 0
low_cnt = 0
for i in s:
if i.isupper():
up_cnt += 1
else:
low_cnt += 1
cur_up = 0
cur_low = 0
ans = 1000000000
cost = up_cnt
for i in s:
if i.isupper():
cur_up += 1
else:
cur_low += 1
a = cur_low + up_cnt - cur_up
if a < ans:
ans = a
print(min(ans, up_cnt))
```
| 3
|
|
911
|
A
|
Nearest Minimums
|
PROGRAMMING
| 1,100
|
[
"implementation"
] | null | null |
You are given an array of *n* integer numbers *a*0,<=*a*1,<=...,<=*a**n*<=-<=1. Find the distance between two closest (nearest) minimums in it. It is guaranteed that in the array a minimum occurs at least two times.
|
The first line contains positive integer *n* (2<=β€<=*n*<=β€<=105) β size of the given array. The second line contains *n* integers *a*0,<=*a*1,<=...,<=*a**n*<=-<=1 (1<=β€<=*a**i*<=β€<=109) β elements of the array. It is guaranteed that in the array a minimum occurs at least two times.
|
Print the only number β distance between two nearest minimums in the array.
|
[
"2\n3 3\n",
"3\n5 6 5\n",
"9\n2 1 3 5 4 1 2 3 1\n"
] |
[
"1\n",
"2\n",
"3\n"
] |
none
| 0
|
[
{
"input": "2\n3 3",
"output": "1"
},
{
"input": "3\n5 6 5",
"output": "2"
},
{
"input": "9\n2 1 3 5 4 1 2 3 1",
"output": "3"
},
{
"input": "6\n4 6 7 8 6 4",
"output": "5"
},
{
"input": "2\n1000000000 1000000000",
"output": "1"
},
{
"input": "42\n1 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": "1"
},
{
"input": "2\n10000000 10000000",
"output": "1"
},
{
"input": "5\n100000000 100000001 100000000 100000001 100000000",
"output": "2"
},
{
"input": "9\n4 3 4 3 4 1 3 3 1",
"output": "3"
},
{
"input": "3\n10000000 1000000000 10000000",
"output": "2"
},
{
"input": "12\n5 6 6 5 6 1 9 9 9 9 9 1",
"output": "6"
},
{
"input": "5\n5 5 1 2 1",
"output": "2"
},
{
"input": "5\n2 2 1 3 1",
"output": "2"
},
{
"input": "3\n1000000000 1000000000 1000000000",
"output": "1"
},
{
"input": "3\n100000005 1000000000 100000005",
"output": "2"
},
{
"input": "5\n1 2 2 2 1",
"output": "4"
},
{
"input": "3\n10000 1000000 10000",
"output": "2"
},
{
"input": "3\n999999999 999999998 999999998",
"output": "1"
},
{
"input": "6\n2 1 1 2 3 4",
"output": "1"
},
{
"input": "4\n1000000000 900000000 900000000 1000000000",
"output": "1"
},
{
"input": "5\n7 7 2 7 2",
"output": "2"
},
{
"input": "6\n10 10 1 20 20 1",
"output": "3"
},
{
"input": "2\n999999999 999999999",
"output": "1"
},
{
"input": "10\n100000 100000 1 2 3 4 5 6 7 1",
"output": "7"
},
{
"input": "10\n3 3 1 2 2 1 10 10 10 10",
"output": "3"
},
{
"input": "5\n900000000 900000001 900000000 900000001 900000001",
"output": "2"
},
{
"input": "5\n3 3 2 5 2",
"output": "2"
},
{
"input": "2\n100000000 100000000",
"output": "1"
},
{
"input": "10\n10 15 10 2 54 54 54 54 2 10",
"output": "5"
},
{
"input": "2\n999999 999999",
"output": "1"
},
{
"input": "6\n1000000000 1000000000 1000000000 1000000000 1000000000 1000000000",
"output": "1"
},
{
"input": "5\n1000000000 100000000 1000000000 1000000000 100000000",
"output": "3"
},
{
"input": "4\n10 9 10 9",
"output": "2"
},
{
"input": "5\n1 3 2 3 1",
"output": "4"
},
{
"input": "5\n2 2 1 4 1",
"output": "2"
},
{
"input": "6\n1 2 2 2 2 1",
"output": "5"
},
{
"input": "7\n3 7 6 7 6 7 3",
"output": "6"
},
{
"input": "8\n1 2 2 2 2 1 2 2",
"output": "5"
},
{
"input": "10\n2 2 2 3 3 1 3 3 3 1",
"output": "4"
},
{
"input": "2\n88888888 88888888",
"output": "1"
},
{
"input": "3\n100000000 100000000 100000000",
"output": "1"
},
{
"input": "10\n1 3 2 4 5 5 4 3 2 1",
"output": "9"
},
{
"input": "5\n2 2 1 2 1",
"output": "2"
},
{
"input": "6\n900000005 900000000 900000001 900000000 900000001 900000001",
"output": "2"
},
{
"input": "5\n41 41 1 41 1",
"output": "2"
},
{
"input": "6\n5 5 1 3 3 1",
"output": "3"
},
{
"input": "8\n1 2 2 2 1 2 2 2",
"output": "4"
},
{
"input": "7\n6 6 6 6 1 8 1",
"output": "2"
},
{
"input": "3\n999999999 1000000000 999999999",
"output": "2"
},
{
"input": "5\n5 5 4 10 4",
"output": "2"
},
{
"input": "11\n2 2 3 4 1 5 3 4 2 5 1",
"output": "6"
},
{
"input": "5\n3 5 4 5 3",
"output": "4"
},
{
"input": "6\n6 6 6 6 1 1",
"output": "1"
},
{
"input": "7\n11 1 3 2 3 1 11",
"output": "4"
},
{
"input": "5\n3 3 1 2 1",
"output": "2"
},
{
"input": "5\n4 4 2 5 2",
"output": "2"
},
{
"input": "4\n10000099 10000567 10000099 10000234",
"output": "2"
},
{
"input": "4\n100000009 100000011 100000012 100000009",
"output": "3"
},
{
"input": "2\n1000000 1000000",
"output": "1"
},
{
"input": "2\n10000010 10000010",
"output": "1"
},
{
"input": "10\n1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000",
"output": "1"
},
{
"input": "8\n2 6 2 8 1 9 8 1",
"output": "3"
},
{
"input": "5\n7 7 1 8 1",
"output": "2"
},
{
"input": "7\n1 3 2 3 2 3 1",
"output": "6"
},
{
"input": "7\n2 3 2 1 3 4 1",
"output": "3"
},
{
"input": "5\n1000000000 999999999 1000000000 1000000000 999999999",
"output": "3"
},
{
"input": "4\n1000000000 1000000000 1000000000 1000000000",
"output": "1"
},
{
"input": "5\n5 5 3 5 3",
"output": "2"
},
{
"input": "6\n2 3 3 3 3 2",
"output": "5"
},
{
"input": "4\n1 1 2 2",
"output": "1"
},
{
"input": "5\n1 1 2 2 2",
"output": "1"
},
{
"input": "6\n2 1 1 2 2 2",
"output": "1"
},
{
"input": "5\n1000000000 1000000000 100000000 1000000000 100000000",
"output": "2"
},
{
"input": "7\n2 2 1 1 2 2 2",
"output": "1"
},
{
"input": "8\n2 2 2 1 1 2 2 2",
"output": "1"
},
{
"input": "10\n2 2 2 2 2 1 1 2 2 2",
"output": "1"
},
{
"input": "11\n2 2 2 2 2 2 1 1 2 2 2",
"output": "1"
},
{
"input": "12\n2 2 2 2 2 2 2 1 1 2 2 2",
"output": "1"
},
{
"input": "13\n2 2 2 2 2 2 2 2 1 1 2 2 2",
"output": "1"
},
{
"input": "14\n2 2 2 2 2 2 2 2 2 1 1 2 2 2",
"output": "1"
},
{
"input": "15\n2 2 2 2 2 2 2 2 2 2 1 1 2 2 2",
"output": "1"
},
{
"input": "16\n2 2 2 2 2 2 2 2 2 2 2 1 1 2 2 2",
"output": "1"
},
{
"input": "17\n2 2 2 2 2 2 2 2 2 2 2 2 1 1 2 2 2",
"output": "1"
},
{
"input": "18\n2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 2 2 2",
"output": "1"
},
{
"input": "19\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 2 2 2",
"output": "1"
},
{
"input": "20\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 2 2 2",
"output": "1"
},
{
"input": "4\n1000000000 100000000 100000000 1000000000",
"output": "1"
},
{
"input": "21\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 2 2 2",
"output": "1"
},
{
"input": "4\n1 2 3 1",
"output": "3"
},
{
"input": "8\n5 5 5 5 3 5 5 3",
"output": "3"
},
{
"input": "7\n2 3 2 1 4 4 1",
"output": "3"
},
{
"input": "6\n3 3 1 2 4 1",
"output": "3"
},
{
"input": "3\n2 1 1",
"output": "1"
},
{
"input": "5\n3 3 2 8 2",
"output": "2"
},
{
"input": "5\n1 2 1 2 2",
"output": "2"
},
{
"input": "4\n1 2 1 2",
"output": "2"
},
{
"input": "5\n3 1 1 3 2",
"output": "1"
},
{
"input": "4\n1 1 2 1",
"output": "1"
},
{
"input": "4\n2 2 1 1",
"output": "1"
},
{
"input": "5\n1 2 2 1 2",
"output": "3"
},
{
"input": "7\n2 1 2 1 1 2 1",
"output": "1"
},
{
"input": "9\n200000 500000 500000 500000 200000 500000 500000 500000 500000",
"output": "4"
},
{
"input": "3\n1 1 2",
"output": "1"
},
{
"input": "85\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 1",
"output": "84"
},
{
"input": "5\n1000000000 1000000000 999999999 1000000000 999999999",
"output": "2"
},
{
"input": "5\n2 1 2 2 1",
"output": "3"
},
{
"input": "3\n1 1 1",
"output": "1"
},
{
"input": "4\n1 2 1 1",
"output": "1"
},
{
"input": "6\n1 3 4 2 4 1",
"output": "5"
},
{
"input": "9\n2 2 5 1 6 8 7 9 1",
"output": "5"
},
{
"input": "10\n1000000000 1000000000 1000000000 999999999 1000000000 1000000000 1000000000 1000000000 1000000000 999999999",
"output": "6"
},
{
"input": "7\n3 3 1 2 4 1 2",
"output": "3"
},
{
"input": "7\n3 3 1 2 3 4 1",
"output": "4"
},
{
"input": "8\n10 5 10 1 10 10 10 1",
"output": "4"
}
] | 1,589,237,582
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 140
| 187
| 9,523,200
|
_=int(input())
A=list(map(int,input().split()))
xmin=min(*A)
index=[]
for i in range(len(A)):
if A[i]==xmin:
index.append(i)
m=[]
for i in range(len(index)-1):
m.append(index[i+1]-index[i])
print(min(m))
|
Title: Nearest Minimums
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given an array of *n* integer numbers *a*0,<=*a*1,<=...,<=*a**n*<=-<=1. Find the distance between two closest (nearest) minimums in it. It is guaranteed that in the array a minimum occurs at least two times.
Input Specification:
The first line contains positive integer *n* (2<=β€<=*n*<=β€<=105) β size of the given array. The second line contains *n* integers *a*0,<=*a*1,<=...,<=*a**n*<=-<=1 (1<=β€<=*a**i*<=β€<=109) β elements of the array. It is guaranteed that in the array a minimum occurs at least two times.
Output Specification:
Print the only number β distance between two nearest minimums in the array.
Demo Input:
['2\n3 3\n', '3\n5 6 5\n', '9\n2 1 3 5 4 1 2 3 1\n']
Demo Output:
['1\n', '2\n', '3\n']
Note:
none
|
```python
_=int(input())
A=list(map(int,input().split()))
xmin=min(*A)
index=[]
for i in range(len(A)):
if A[i]==xmin:
index.append(i)
m=[]
for i in range(len(index)-1):
m.append(index[i+1]-index[i])
print(min(m))
```
| 3
|
|
779
|
A
|
Pupils Redistribution
|
PROGRAMMING
| 1,000
|
[
"constructive algorithms",
"math"
] | null | null |
In Berland each high school student is characterized by academic performance β integer value between 1 and 5.
In high school 0xFF there are two groups of pupils: the group *A* and the group *B*. Each group consists of exactly *n* students. An academic performance of each student is known β integer value between 1 and 5.
The school director wants to redistribute students between groups so that each of the two groups has the same number of students whose academic performance is equal to 1, the same number of students whose academic performance is 2 and so on. In other words, the purpose of the school director is to change the composition of groups, so that for each value of academic performance the numbers of students in both groups are equal.
To achieve this, there is a plan to produce a series of exchanges of students between groups. During the single exchange the director selects one student from the class *A* and one student of class *B*. After that, they both change their groups.
Print the least number of exchanges, in order to achieve the desired equal numbers of students for each academic performance.
|
The first line of the input contains integer number *n* (1<=β€<=*n*<=β€<=100) β number of students in both groups.
The second line contains sequence of integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=β€<=*a**i*<=β€<=5), where *a**i* is academic performance of the *i*-th student of the group *A*.
The third line contains sequence of integer numbers *b*1,<=*b*2,<=...,<=*b**n* (1<=β€<=*b**i*<=β€<=5), where *b**i* is academic performance of the *i*-th student of the group *B*.
|
Print the required minimum number of exchanges or -1, if the desired distribution of students can not be obtained.
|
[
"4\n5 4 4 4\n5 5 4 5\n",
"6\n1 1 1 1 1 1\n5 5 5 5 5 5\n",
"1\n5\n3\n",
"9\n3 2 5 5 2 3 3 3 2\n4 1 4 1 1 2 4 4 1\n"
] |
[
"1\n",
"3\n",
"-1\n",
"4\n"
] |
none
| 500
|
[
{
"input": "4\n5 4 4 4\n5 5 4 5",
"output": "1"
},
{
"input": "6\n1 1 1 1 1 1\n5 5 5 5 5 5",
"output": "3"
},
{
"input": "1\n5\n3",
"output": "-1"
},
{
"input": "9\n3 2 5 5 2 3 3 3 2\n4 1 4 1 1 2 4 4 1",
"output": "4"
},
{
"input": "1\n1\n2",
"output": "-1"
},
{
"input": "1\n1\n1",
"output": "0"
},
{
"input": "8\n1 1 2 2 3 3 4 4\n4 4 5 5 1 1 1 1",
"output": "2"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1 1\n2 2 2 2 2 2 2 2 2 2",
"output": "5"
},
{
"input": "100\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5",
"output": "0"
},
{
"input": "2\n1 1\n1 1",
"output": "0"
},
{
"input": "2\n1 2\n1 1",
"output": "-1"
},
{
"input": "2\n2 2\n1 1",
"output": "1"
},
{
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},
{
"input": "6\n2 2 3 3 4 4\n2 3 4 5 5 5",
"output": "-1"
},
{
"input": "6\n1 1 2 2 3 4\n3 3 4 4 1 2",
"output": "-1"
},
{
"input": "4\n1 2 3 3\n3 3 3 3",
"output": "-1"
},
{
"input": "3\n1 2 3\n3 3 3",
"output": "-1"
},
{
"input": "5\n3 3 3 2 2\n2 2 2 3 3",
"output": "-1"
},
{
"input": "10\n1 2 3 4 1 2 3 4 1 2\n1 2 3 4 1 2 3 4 3 4",
"output": "-1"
},
{
"input": "2\n2 2\n1 3",
"output": "-1"
},
{
"input": "3\n1 2 3\n1 1 4",
"output": "-1"
},
{
"input": "4\n3 4 4 4\n3 3 4 4",
"output": "-1"
}
] | 1,488,108,543
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 81
| 62
| 4,608,000
|
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
aa = [0]*6
for v in a: aa[v] += 1
bb = [0]*6
for v in b: bb[v] += 1
res = sum(map(lambda x, y: max(x-y, 0), aa, bb))
if any(map(lambda x, y: (x-y) & 1, aa, bb)):
print(-1)
else:
print(res//2)
|
Title: Pupils Redistribution
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In Berland each high school student is characterized by academic performance β integer value between 1 and 5.
In high school 0xFF there are two groups of pupils: the group *A* and the group *B*. Each group consists of exactly *n* students. An academic performance of each student is known β integer value between 1 and 5.
The school director wants to redistribute students between groups so that each of the two groups has the same number of students whose academic performance is equal to 1, the same number of students whose academic performance is 2 and so on. In other words, the purpose of the school director is to change the composition of groups, so that for each value of academic performance the numbers of students in both groups are equal.
To achieve this, there is a plan to produce a series of exchanges of students between groups. During the single exchange the director selects one student from the class *A* and one student of class *B*. After that, they both change their groups.
Print the least number of exchanges, in order to achieve the desired equal numbers of students for each academic performance.
Input Specification:
The first line of the input contains integer number *n* (1<=β€<=*n*<=β€<=100) β number of students in both groups.
The second line contains sequence of integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=β€<=*a**i*<=β€<=5), where *a**i* is academic performance of the *i*-th student of the group *A*.
The third line contains sequence of integer numbers *b*1,<=*b*2,<=...,<=*b**n* (1<=β€<=*b**i*<=β€<=5), where *b**i* is academic performance of the *i*-th student of the group *B*.
Output Specification:
Print the required minimum number of exchanges or -1, if the desired distribution of students can not be obtained.
Demo Input:
['4\n5 4 4 4\n5 5 4 5\n', '6\n1 1 1 1 1 1\n5 5 5 5 5 5\n', '1\n5\n3\n', '9\n3 2 5 5 2 3 3 3 2\n4 1 4 1 1 2 4 4 1\n']
Demo Output:
['1\n', '3\n', '-1\n', '4\n']
Note:
none
|
```python
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
aa = [0]*6
for v in a: aa[v] += 1
bb = [0]*6
for v in b: bb[v] += 1
res = sum(map(lambda x, y: max(x-y, 0), aa, bb))
if any(map(lambda x, y: (x-y) & 1, aa, bb)):
print(-1)
else:
print(res//2)
```
| 3
|
|
808
|
C
|
Tea Party
|
PROGRAMMING
| 1,400
|
[
"constructive algorithms",
"greedy",
"sortings"
] | null | null |
Polycarp invited all his friends to the tea party to celebrate the holiday. He has *n* cups, one for each of his *n* friends, with volumes *a*1,<=*a*2,<=...,<=*a**n*. His teapot stores *w* milliliters of tea (*w*<=β€<=*a*1<=+<=*a*2<=+<=...<=+<=*a**n*). Polycarp wants to pour tea in cups in such a way that:
- Every cup will contain tea for at least half of its volume - Every cup will contain integer number of milliliters of tea - All the tea from the teapot will be poured into cups - All friends will be satisfied.
Friend with cup *i* won't be satisfied, if there exists such cup *j* that cup *i* contains less tea than cup *j* but *a**i*<=><=*a**j*.
For each cup output how many milliliters of tea should be poured in it. If it's impossible to pour all the tea and satisfy all conditions then output -1.
|
The first line contains two integer numbers *n* and *w* (1<=β€<=*n*<=β€<=100, ).
The second line contains *n* numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=β€<=*a**i*<=β€<=100).
|
Output how many milliliters of tea every cup should contain. If there are multiple answers, print any of them.
If it's impossible to pour all the tea and satisfy all conditions then output -1.
|
[
"2 10\n8 7\n",
"4 4\n1 1 1 1\n",
"3 10\n9 8 10\n"
] |
[
"6 4 \n",
"1 1 1 1 \n",
"-1\n"
] |
In the third example you should pour to the first cup at least 5 milliliters, to the second one at least 4, to the third one at least 5. It sums up to 14, which is greater than 10 milliliters available.
| 0
|
[
{
"input": "2 10\n8 7",
"output": "6 4 "
},
{
"input": "4 4\n1 1 1 1",
"output": "1 1 1 1 "
},
{
"input": "3 10\n9 8 10",
"output": "-1"
},
{
"input": "1 1\n1",
"output": "1 "
},
{
"input": "1 1\n2",
"output": "1 "
},
{
"input": "1 10\n20",
"output": "10 "
},
{
"input": "3 10\n8 4 8",
"output": "4 2 4 "
},
{
"input": "3 100\n37 26 37",
"output": "37 26 37 "
},
{
"input": "3 60\n43 23 24",
"output": "36 12 12 "
},
{
"input": "20 14\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "-1"
},
{
"input": "20 8\n1 2 1 2 1 1 1 2 1 1 1 2 1 1 2 1 1 1 2 2",
"output": "-1"
},
{
"input": "50 1113\n25 21 23 37 28 23 19 25 5 12 3 11 46 50 13 50 7 1 8 40 4 6 34 27 11 39 45 31 10 12 48 2 19 37 47 45 30 24 21 42 36 14 31 30 31 50 6 3 33 49",
"output": "13 11 12 37 28 12 10 18 3 6 2 6 46 50 7 50 4 1 4 40 2 3 34 27 6 39 45 31 5 6 48 1 10 37 47 45 30 12 11 42 36 7 31 30 31 50 3 2 33 49 "
},
{
"input": "50 440\n14 69 33 38 83 65 21 66 89 3 93 60 31 16 61 20 42 64 13 1 50 50 74 58 67 61 52 22 69 68 18 33 28 59 4 8 96 32 84 85 87 87 61 89 2 47 15 64 88 18",
"output": "-1"
},
{
"input": "100 640\n82 51 81 14 37 17 78 92 64 15 8 86 89 8 87 77 66 10 15 12 100 25 92 47 21 78 20 63 13 49 41 36 41 79 16 87 87 69 3 76 80 60 100 49 70 59 72 8 38 71 45 97 71 14 76 54 81 4 59 46 39 29 92 3 49 22 53 99 59 52 74 31 92 43 42 23 44 9 82 47 7 40 12 9 3 55 37 85 46 22 84 52 98 41 21 77 63 17 62 91",
"output": "-1"
},
{
"input": "100 82\n1 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 1",
"output": "-1"
},
{
"input": "100 55\n1 1 1 1 2 1 1 1 1 1 2 2 1 1 2 1 2 1 1 1 2 1 1 2 1 2 1 1 2 2 2 1 1 2 1 1 1 2 2 2 1 1 1 2 1 2 2 1 2 1 1 2 2 1 2 1 2 1 2 2 1 1 1 2 1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 1 1 1 1 2 2 2 2 2 2 2 1 1 1 2 1 2 1",
"output": "-1"
},
{
"input": "30 50\n3 1 2 4 1 2 2 4 3 4 4 3 3 3 3 5 3 2 5 4 3 3 5 3 3 5 4 5 3 5",
"output": "-1"
},
{
"input": "40 100\n3 3 3 3 4 1 1 1 1 1 2 2 1 3 1 2 3 2 1 2 2 2 1 4 2 2 3 3 3 2 4 6 4 4 3 2 2 2 4 5",
"output": "3 3 3 3 4 1 1 1 1 1 2 2 1 3 1 2 3 2 1 2 2 2 1 4 2 2 3 3 3 2 4 6 4 4 3 2 2 2 4 5 "
},
{
"input": "100 10000\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": "100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 "
},
{
"input": "2 5\n3 4",
"output": "2 3 "
},
{
"input": "2 6\n2 6",
"output": "1 5 "
},
{
"input": "23 855\n5 63 94 57 38 84 77 79 83 36 47 31 60 79 75 48 88 17 46 33 23 15 27",
"output": "3 32 94 29 19 84 39 72 83 18 24 16 30 79 38 24 88 9 23 17 12 8 14 "
},
{
"input": "52 2615\n73 78 70 92 94 74 46 19 55 20 70 3 1 42 68 10 66 80 1 31 65 19 73 74 56 35 53 38 92 35 65 81 6 98 74 51 27 49 76 19 86 76 5 60 14 75 64 99 43 7 36 79",
"output": "73 78 70 92 94 74 46 10 55 10 70 2 1 42 68 5 66 80 1 16 65 10 73 74 56 18 53 38 92 30 65 81 3 98 74 51 14 49 76 10 86 76 3 60 7 75 64 99 43 4 36 79 "
},
{
"input": "11 287\n34 30 69 86 22 53 11 91 62 44 5",
"output": "17 15 35 43 11 27 6 77 31 22 3 "
},
{
"input": "55 1645\n60 53 21 20 87 48 10 21 76 35 52 41 82 86 93 11 93 86 34 15 37 63 57 3 57 57 32 8 55 25 29 38 46 22 13 87 27 35 40 83 5 7 6 18 88 25 4 59 95 62 31 93 98 50 62",
"output": "30 27 11 10 82 24 5 11 38 18 26 21 41 43 93 6 93 43 17 8 19 32 29 2 29 29 16 4 28 13 15 19 23 11 7 87 14 18 20 42 3 4 3 9 88 13 2 30 95 31 16 93 98 25 31 "
},
{
"input": "71 3512\n97 46 76 95 81 96 99 83 10 50 19 18 73 5 41 60 12 73 60 31 21 64 88 61 43 57 61 19 75 35 41 85 12 59 32 47 37 43 35 92 90 47 3 98 21 18 61 79 39 86 74 8 52 33 39 27 93 54 35 38 96 36 83 51 97 10 8 66 75 87 68",
"output": "97 46 76 95 81 96 99 83 5 50 10 9 73 3 41 60 6 73 60 16 11 64 88 61 43 57 61 10 75 18 41 85 6 59 16 47 19 43 18 92 90 47 2 98 11 9 61 79 20 86 74 4 52 17 21 14 93 54 18 19 96 18 83 51 97 5 4 66 75 87 68 "
},
{
"input": "100 2633\n99 50 64 81 75 73 26 31 31 36 95 12 100 2 70 72 78 56 76 23 94 8 91 1 39 82 97 67 64 25 71 90 48 34 31 46 64 37 46 50 99 93 14 56 1 89 95 89 50 52 12 58 43 65 45 88 90 14 38 19 6 15 91 67 43 48 82 20 11 48 33 20 39 52 73 5 25 84 26 54 42 56 10 28 9 63 60 98 30 1 25 74 86 56 85 9 12 94 80 95",
"output": "50 25 32 41 38 37 13 16 16 18 48 6 61 1 35 36 39 28 38 12 47 4 46 1 20 41 49 34 32 13 36 45 24 17 16 23 32 19 23 25 50 47 7 28 1 45 48 45 25 26 6 29 22 33 23 44 45 7 19 10 3 8 46 34 22 24 41 10 6 24 17 10 20 26 37 3 13 42 13 27 21 28 5 14 5 32 30 49 15 1 13 37 43 28 43 5 6 47 40 48 "
},
{
"input": "71 1899\n23 55 58 87 69 85 100 21 19 72 81 68 20 25 29 92 18 74 89 70 53 7 78 57 41 79 64 87 63 76 95 84 1 28 32 1 79 34 77 17 71 61 35 31 62 92 69 99 60 26 2 18 61 9 27 77 82 6 30 65 52 3 51 43 13 77 41 59 19 29 86",
"output": "12 28 29 44 35 43 95 11 10 36 41 34 10 13 15 46 9 37 45 35 27 4 39 29 21 40 32 44 32 38 48 42 1 14 16 1 40 17 39 9 36 31 18 16 31 46 35 50 30 13 1 9 31 5 14 39 41 3 15 33 26 2 26 22 7 39 21 30 10 15 43 "
},
{
"input": "10 21\n3 3 3 3 4 3 3 3 3 3",
"output": "2 2 2 2 3 2 2 2 2 2 "
}
] | 1,638,711,156
| 1,056
|
PyPy 3
|
OK
|
TESTS
| 29
| 93
| 0
|
n, w = map(int, input().split())
A = list(map(int, input().split()))
B = []
for i, a in enumerate(A):
B.append((a, i))
B.sort(key=lambda x: x[0])
m = 0
ans = [0]*n
for a, i in B:
ans[i] = (a+1)//2
m += (a+1)//2
if m > w:
print(-1)
exit()
r = w-m
for j in reversed(range(n)):
if r == 0:
break
a, i = B[j]
if a-ans[i] <= r:
r -= a-ans[i]
ans[i] = a
else:
ans[i] += r
r = 0
if r:
print(-1)
else:
print(*ans)
|
Title: Tea Party
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp invited all his friends to the tea party to celebrate the holiday. He has *n* cups, one for each of his *n* friends, with volumes *a*1,<=*a*2,<=...,<=*a**n*. His teapot stores *w* milliliters of tea (*w*<=β€<=*a*1<=+<=*a*2<=+<=...<=+<=*a**n*). Polycarp wants to pour tea in cups in such a way that:
- Every cup will contain tea for at least half of its volume - Every cup will contain integer number of milliliters of tea - All the tea from the teapot will be poured into cups - All friends will be satisfied.
Friend with cup *i* won't be satisfied, if there exists such cup *j* that cup *i* contains less tea than cup *j* but *a**i*<=><=*a**j*.
For each cup output how many milliliters of tea should be poured in it. If it's impossible to pour all the tea and satisfy all conditions then output -1.
Input Specification:
The first line contains two integer numbers *n* and *w* (1<=β€<=*n*<=β€<=100, ).
The second line contains *n* numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=β€<=*a**i*<=β€<=100).
Output Specification:
Output how many milliliters of tea every cup should contain. If there are multiple answers, print any of them.
If it's impossible to pour all the tea and satisfy all conditions then output -1.
Demo Input:
['2 10\n8 7\n', '4 4\n1 1 1 1\n', '3 10\n9 8 10\n']
Demo Output:
['6 4 \n', '1 1 1 1 \n', '-1\n']
Note:
In the third example you should pour to the first cup at least 5 milliliters, to the second one at least 4, to the third one at least 5. It sums up to 14, which is greater than 10 milliliters available.
|
```python
n, w = map(int, input().split())
A = list(map(int, input().split()))
B = []
for i, a in enumerate(A):
B.append((a, i))
B.sort(key=lambda x: x[0])
m = 0
ans = [0]*n
for a, i in B:
ans[i] = (a+1)//2
m += (a+1)//2
if m > w:
print(-1)
exit()
r = w-m
for j in reversed(range(n)):
if r == 0:
break
a, i = B[j]
if a-ans[i] <= r:
r -= a-ans[i]
ans[i] = a
else:
ans[i] += r
r = 0
if r:
print(-1)
else:
print(*ans)
```
| 3
|
|
181
|
A
|
Series of Crimes
|
PROGRAMMING
| 800
|
[
"brute force",
"geometry",
"implementation"
] | null | null |
The Berland capital is shaken with three bold crimes committed by the Pihsters, a notorious criminal gang.
The Berland capital's map is represented by an *n*<=Γ<=*m* rectangular table. Each cell of the table on the map represents some districts of the capital.
The capital's main detective Polycarpus took a map and marked there the districts where the first three robberies had been committed as asterisks. Deduction tells Polycarpus that the fourth robbery will be committed in such district, that all four robbed districts will form the vertices of some rectangle, parallel to the sides of the map.
Polycarpus is good at deduction but he's hopeless at math. So he asked you to find the district where the fourth robbery will be committed.
|
The first line contains two space-separated integers *n* and *m* (2<=β€<=*n*,<=*m*<=β€<=100) β the number of rows and columns in the table, correspondingly.
Each of the next *n* lines contains *m* characters β the description of the capital's map. Each character can either be a "." (dot), or an "*" (asterisk). A character equals "*" if the corresponding district has been robbed. Otherwise, it equals ".".
It is guaranteed that the map has exactly three characters "*" and we can always find the fourth district that meets the problem requirements.
|
Print two integers β the number of the row and the number of the column of the city district that is the fourth one to be robbed. The rows are numbered starting from one from top to bottom and the columns are numbered starting from one from left to right.
|
[
"3 2\n.*\n..\n**\n",
"3 3\n*.*\n*..\n...\n"
] |
[
"1 1\n",
"2 3\n"
] |
none
| 500
|
[
{
"input": "3 2\n.*\n..\n**",
"output": "1 1"
},
{
"input": "2 5\n*....\n*...*",
"output": "1 5"
},
{
"input": "7 2\n..\n**\n..\n..\n..\n..\n.*",
"output": "7 1"
},
{
"input": "7 2\n*.\n..\n..\n..\n..\n..\n**",
"output": "1 2"
},
{
"input": "2 10\n*......*..\n.......*..",
"output": "2 1"
},
{
"input": "10 3\n*..\n...\n...\n...\n...\n...\n...\n...\n...\n**.",
"output": "1 2"
},
{
"input": "100 2\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..\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..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n**",
"output": "1 2"
},
{
"input": "99 3\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...\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...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n*..",
"output": "99 2"
},
{
"input": "98 3\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...\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...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n.**",
"output": "2 3"
},
{
"input": "99 3\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...\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...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n*..",
"output": "99 2"
},
{
"input": "98 3\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...\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...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n.**",
"output": "2 3"
},
{
"input": "2 100\n...................................................................................................*\n*..................................................................................................*",
"output": "1 1"
},
{
"input": "2 2\n*.\n**",
"output": "1 2"
},
{
"input": "2 2\n**\n.*",
"output": "2 1"
},
{
"input": "2 3\n*.*\n..*",
"output": "2 1"
}
] | 1,661,187,562
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 36
| 92
| 0
|
L = [int(h) for h in input().split(' ')]
n = L[0]; m = L[1]
if ((n >= 2) and (n <= 100) and (m >= 2) and (m <= 100)):
N = M = list()
for i in range(0, n):
R = list()
s = input()
for j in range(0, len(s)):
R = R + [s[j]]
for k in range(0, len(R)):
if (R[k] == '*'):
N = N + [i + 1]
M = M + [k + 1]
En = set(N); Em = set(M)
for p in range(0, len(N)):
for qn in En:
if (N.count(qn) == 1):
x = qn
for qm in Em:
if (M.count(qm) == 1):
y = qm
print(x, ' ', y)
|
Title: Series of Crimes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Berland capital is shaken with three bold crimes committed by the Pihsters, a notorious criminal gang.
The Berland capital's map is represented by an *n*<=Γ<=*m* rectangular table. Each cell of the table on the map represents some districts of the capital.
The capital's main detective Polycarpus took a map and marked there the districts where the first three robberies had been committed as asterisks. Deduction tells Polycarpus that the fourth robbery will be committed in such district, that all four robbed districts will form the vertices of some rectangle, parallel to the sides of the map.
Polycarpus is good at deduction but he's hopeless at math. So he asked you to find the district where the fourth robbery will be committed.
Input Specification:
The first line contains two space-separated integers *n* and *m* (2<=β€<=*n*,<=*m*<=β€<=100) β the number of rows and columns in the table, correspondingly.
Each of the next *n* lines contains *m* characters β the description of the capital's map. Each character can either be a "." (dot), or an "*" (asterisk). A character equals "*" if the corresponding district has been robbed. Otherwise, it equals ".".
It is guaranteed that the map has exactly three characters "*" and we can always find the fourth district that meets the problem requirements.
Output Specification:
Print two integers β the number of the row and the number of the column of the city district that is the fourth one to be robbed. The rows are numbered starting from one from top to bottom and the columns are numbered starting from one from left to right.
Demo Input:
['3 2\n.*\n..\n**\n', '3 3\n*.*\n*..\n...\n']
Demo Output:
['1 1\n', '2 3\n']
Note:
none
|
```python
L = [int(h) for h in input().split(' ')]
n = L[0]; m = L[1]
if ((n >= 2) and (n <= 100) and (m >= 2) and (m <= 100)):
N = M = list()
for i in range(0, n):
R = list()
s = input()
for j in range(0, len(s)):
R = R + [s[j]]
for k in range(0, len(R)):
if (R[k] == '*'):
N = N + [i + 1]
M = M + [k + 1]
En = set(N); Em = set(M)
for p in range(0, len(N)):
for qn in En:
if (N.count(qn) == 1):
x = qn
for qm in Em:
if (M.count(qm) == 1):
y = qm
print(x, ' ', y)
```
| 3
|
|
650
|
A
|
Watchmen
|
PROGRAMMING
| 1,400
|
[
"data structures",
"geometry",
"math"
] | null | null |
Watchmen are in a danger and Doctor Manhattan together with his friend Daniel Dreiberg should warn them as soon as possible. There are *n* watchmen on a plane, the *i*-th watchman is located at point (*x**i*,<=*y**i*).
They need to arrange a plan, but there are some difficulties on their way. As you know, Doctor Manhattan considers the distance between watchmen *i* and *j* to be |*x**i*<=-<=*x**j*|<=+<=|*y**i*<=-<=*y**j*|. Daniel, as an ordinary person, calculates the distance using the formula .
The success of the operation relies on the number of pairs (*i*,<=*j*) (1<=β€<=*i*<=<<=*j*<=β€<=*n*), such that the distance between watchman *i* and watchmen *j* calculated by Doctor Manhattan is equal to the distance between them calculated by Daniel. You were asked to compute the number of such pairs.
|
The first line of the input contains the single integer *n* (1<=β€<=*n*<=β€<=200<=000)Β β the number of watchmen.
Each of the following *n* lines contains two integers *x**i* and *y**i* (|*x**i*|,<=|*y**i*|<=β€<=109).
Some positions may coincide.
|
Print the number of pairs of watchmen such that the distance between them calculated by Doctor Manhattan is equal to the distance calculated by Daniel.
|
[
"3\n1 1\n7 5\n1 5\n",
"6\n0 0\n0 1\n0 2\n-1 1\n0 1\n1 1\n"
] |
[
"2\n",
"11\n"
] |
In the first sample, the distance between watchman 1 and watchman 2 is equal to |1β-β7|β+β|1β-β5|β=β10 for Doctor Manhattan and <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/bcb5b7064b5f02088da0fdcf677e6fda495dd0df.png" style="max-width: 100.0%;max-height: 100.0%;"/> for Daniel. For pairs (1,β1), (1,β5) and (7,β5), (1,β5) Doctor Manhattan and Daniel will calculate the same distances.
| 500
|
[
{
"input": "3\n1 1\n7 5\n1 5",
"output": "2"
},
{
"input": "6\n0 0\n0 1\n0 2\n-1 1\n0 1\n1 1",
"output": "11"
},
{
"input": "10\n46 -55\n46 45\n46 45\n83 -55\n46 45\n83 -55\n46 45\n83 45\n83 45\n46 -55",
"output": "33"
},
{
"input": "1\n-5 -90",
"output": "0"
},
{
"input": "2\n315 845\n-669 -762",
"output": "0"
},
{
"input": "3\n8911 7861\n-6888 7861\n8911 7861",
"output": "3"
},
{
"input": "2\n-1 1000000000\n0 -1",
"output": "0"
},
{
"input": "2\n1000000000 0\n-7 1",
"output": "0"
},
{
"input": "2\n1 4\n2 1",
"output": "0"
},
{
"input": "2\n1 0\n0 2333333",
"output": "0"
},
{
"input": "2\n2 1\n1 2",
"output": "0"
},
{
"input": "2\n1 1000000000\n2 -1000000000",
"output": "0"
},
{
"input": "2\n0 1000000000\n1 -7",
"output": "0"
},
{
"input": "2\n1 0\n0 19990213",
"output": "0"
}
] | 1,612,437,564
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 71
| 2,215
| 34,816,000
|
""" import sys
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w') """
q = {}
w = {}
e = {}
r = 0
for _ in range(int(input())):
x, y = map(int, input().split())
a, s, d = q.get(x, 0), w.get(y, 0), e.get((x, y), 0)
r += a + s - d
q[x], w[y], e[x, y] = a + 1, s + 1, d + 1
print(r)
|
Title: Watchmen
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Watchmen are in a danger and Doctor Manhattan together with his friend Daniel Dreiberg should warn them as soon as possible. There are *n* watchmen on a plane, the *i*-th watchman is located at point (*x**i*,<=*y**i*).
They need to arrange a plan, but there are some difficulties on their way. As you know, Doctor Manhattan considers the distance between watchmen *i* and *j* to be |*x**i*<=-<=*x**j*|<=+<=|*y**i*<=-<=*y**j*|. Daniel, as an ordinary person, calculates the distance using the formula .
The success of the operation relies on the number of pairs (*i*,<=*j*) (1<=β€<=*i*<=<<=*j*<=β€<=*n*), such that the distance between watchman *i* and watchmen *j* calculated by Doctor Manhattan is equal to the distance between them calculated by Daniel. You were asked to compute the number of such pairs.
Input Specification:
The first line of the input contains the single integer *n* (1<=β€<=*n*<=β€<=200<=000)Β β the number of watchmen.
Each of the following *n* lines contains two integers *x**i* and *y**i* (|*x**i*|,<=|*y**i*|<=β€<=109).
Some positions may coincide.
Output Specification:
Print the number of pairs of watchmen such that the distance between them calculated by Doctor Manhattan is equal to the distance calculated by Daniel.
Demo Input:
['3\n1 1\n7 5\n1 5\n', '6\n0 0\n0 1\n0 2\n-1 1\n0 1\n1 1\n']
Demo Output:
['2\n', '11\n']
Note:
In the first sample, the distance between watchman 1 and watchman 2 is equal to |1β-β7|β+β|1β-β5|β=β10 for Doctor Manhattan and <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/bcb5b7064b5f02088da0fdcf677e6fda495dd0df.png" style="max-width: 100.0%;max-height: 100.0%;"/> for Daniel. For pairs (1,β1), (1,β5) and (7,β5), (1,β5) Doctor Manhattan and Daniel will calculate the same distances.
|
```python
""" import sys
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w') """
q = {}
w = {}
e = {}
r = 0
for _ in range(int(input())):
x, y = map(int, input().split())
a, s, d = q.get(x, 0), w.get(y, 0), e.get((x, y), 0)
r += a + s - d
q[x], w[y], e[x, y] = a + 1, s + 1, d + 1
print(r)
```
| 3
|
|
202
|
A
|
LLPS
|
PROGRAMMING
| 800
|
[
"binary search",
"bitmasks",
"brute force",
"greedy",
"implementation",
"strings"
] | null | null |
This problem's actual name, "Lexicographically Largest Palindromic Subsequence" is too long to fit into the page headline.
You are given string *s* consisting of lowercase English letters only. Find its lexicographically largest palindromic subsequence.
We'll call a non-empty string *s*[*p*1*p*2... *p**k*] = *s**p*1*s**p*2... *s**p**k* (1 <=β€<= *p*1<=<<=*p*2<=<<=...<=<<=*p**k* <=β€<= |*s*|) a subsequence of string *s* = *s*1*s*2... *s*|*s*|, where |*s*| is the length of string *s*. For example, strings "abcb", "b" and "abacaba" are subsequences of string "abacaba".
String *x* = *x*1*x*2... *x*|*x*| is lexicographically larger than string *y* = *y*1*y*2... *y*|*y*| if either |*x*| > |*y*| and *x*1<==<=*y*1, *x*2<==<=*y*2, ...,<=*x*|*y*|<==<=*y*|*y*|, or there exists such number *r* (*r*<=<<=|*x*|, *r*<=<<=|*y*|) that *x*1<==<=*y*1, *x*2<==<=*y*2, ..., *x**r*<==<=*y**r* and *x**r*<=<=+<=<=1<=><=*y**r*<=<=+<=<=1. Characters in the strings are compared according to their ASCII codes. For example, string "ranger" is lexicographically larger than string "racecar" and string "poster" is lexicographically larger than string "post".
String *s* = *s*1*s*2... *s*|*s*| is a palindrome if it matches string *rev*(*s*) = *s*|*s*|*s*|*s*|<=-<=1... *s*1. In other words, a string is a palindrome if it reads the same way from left to right and from right to left. For example, palindromic strings are "racecar", "refer" and "z".
|
The only input line contains a non-empty string *s* consisting of lowercase English letters only. Its length does not exceed 10.
|
Print the lexicographically largest palindromic subsequence of string *s*.
|
[
"radar\n",
"bowwowwow\n",
"codeforces\n",
"mississipp\n"
] |
[
"rr\n",
"wwwww\n",
"s\n",
"ssss\n"
] |
Among all distinct subsequences of string "radar" the following ones are palindromes: "a", "d", "r", "aa", "rr", "ada", "rar", "rdr", "raar" and "radar". The lexicographically largest of them is "rr".
| 500
|
[
{
"input": "radar",
"output": "rr"
},
{
"input": "bowwowwow",
"output": "wwwww"
},
{
"input": "codeforces",
"output": "s"
},
{
"input": "mississipp",
"output": "ssss"
},
{
"input": "tourist",
"output": "u"
},
{
"input": "romka",
"output": "r"
},
{
"input": "helloworld",
"output": "w"
},
{
"input": "zzzzzzzazz",
"output": "zzzzzzzzz"
},
{
"input": "testcase",
"output": "tt"
},
{
"input": "hahahahaha",
"output": "hhhhh"
},
{
"input": "abbbbbbbbb",
"output": "bbbbbbbbb"
},
{
"input": "zaz",
"output": "zz"
},
{
"input": "aza",
"output": "z"
},
{
"input": "dcbaedcba",
"output": "e"
},
{
"input": "abcdeabcd",
"output": "e"
},
{
"input": "edcbabcde",
"output": "ee"
},
{
"input": "aaaaaaaaab",
"output": "b"
},
{
"input": "testzzzzzz",
"output": "zzzzzz"
},
{
"input": "zzzzzzwait",
"output": "zzzzzz"
},
{
"input": "rrrrrqponm",
"output": "rrrrr"
},
{
"input": "zzyzyy",
"output": "zzz"
},
{
"input": "aababb",
"output": "bbb"
},
{
"input": "zanzibar",
"output": "zz"
},
{
"input": "hhgfedcbaa",
"output": "hh"
},
{
"input": "aabcdefghh",
"output": "hh"
},
{
"input": "aruaru",
"output": "uu"
},
{
"input": "uraura",
"output": "uu"
},
{
"input": "aru",
"output": "u"
},
{
"input": "aburvabur",
"output": "v"
},
{
"input": "ura",
"output": "u"
},
{
"input": "eurottat",
"output": "u"
},
{
"input": "referee",
"output": "rr"
},
{
"input": "joking",
"output": "o"
},
{
"input": "seriously",
"output": "y"
},
{
"input": "sets",
"output": "t"
},
{
"input": "test",
"output": "tt"
},
{
"input": "klmgameklm",
"output": "mmm"
},
{
"input": "dfkjafdkdd",
"output": "kk"
},
{
"input": "zzzzzzzzzz",
"output": "zzzzzzzzzz"
},
{
"input": "aaaaaaaaaa",
"output": "aaaaaaaaaa"
},
{
"input": "buzz",
"output": "zz"
},
{
"input": "b",
"output": "b"
},
{
"input": "y",
"output": "y"
},
{
"input": "yy",
"output": "yy"
},
{
"input": "zz",
"output": "zz"
},
{
"input": "bb",
"output": "bb"
},
{
"input": "aa",
"output": "aa"
},
{
"input": "a",
"output": "a"
},
{
"input": "z",
"output": "z"
},
{
"input": "zz",
"output": "zz"
},
{
"input": "zzazazzzaz",
"output": "zzzzzzz"
},
{
"input": "hellhellhe",
"output": "llll"
},
{
"input": "hellohello",
"output": "oo"
},
{
"input": "refer",
"output": "rr"
}
] | 1,658,052,044
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 54
| 124
| 0
|
s = input()
M = sorted(list(s))[-1]
ans = ""
for c in s:
if c == M:
ans += M
print(ans)
|
Title: LLPS
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
This problem's actual name, "Lexicographically Largest Palindromic Subsequence" is too long to fit into the page headline.
You are given string *s* consisting of lowercase English letters only. Find its lexicographically largest palindromic subsequence.
We'll call a non-empty string *s*[*p*1*p*2... *p**k*] = *s**p*1*s**p*2... *s**p**k* (1 <=β€<= *p*1<=<<=*p*2<=<<=...<=<<=*p**k* <=β€<= |*s*|) a subsequence of string *s* = *s*1*s*2... *s*|*s*|, where |*s*| is the length of string *s*. For example, strings "abcb", "b" and "abacaba" are subsequences of string "abacaba".
String *x* = *x*1*x*2... *x*|*x*| is lexicographically larger than string *y* = *y*1*y*2... *y*|*y*| if either |*x*| > |*y*| and *x*1<==<=*y*1, *x*2<==<=*y*2, ...,<=*x*|*y*|<==<=*y*|*y*|, or there exists such number *r* (*r*<=<<=|*x*|, *r*<=<<=|*y*|) that *x*1<==<=*y*1, *x*2<==<=*y*2, ..., *x**r*<==<=*y**r* and *x**r*<=<=+<=<=1<=><=*y**r*<=<=+<=<=1. Characters in the strings are compared according to their ASCII codes. For example, string "ranger" is lexicographically larger than string "racecar" and string "poster" is lexicographically larger than string "post".
String *s* = *s*1*s*2... *s*|*s*| is a palindrome if it matches string *rev*(*s*) = *s*|*s*|*s*|*s*|<=-<=1... *s*1. In other words, a string is a palindrome if it reads the same way from left to right and from right to left. For example, palindromic strings are "racecar", "refer" and "z".
Input Specification:
The only input line contains a non-empty string *s* consisting of lowercase English letters only. Its length does not exceed 10.
Output Specification:
Print the lexicographically largest palindromic subsequence of string *s*.
Demo Input:
['radar\n', 'bowwowwow\n', 'codeforces\n', 'mississipp\n']
Demo Output:
['rr\n', 'wwwww\n', 's\n', 'ssss\n']
Note:
Among all distinct subsequences of string "radar" the following ones are palindromes: "a", "d", "r", "aa", "rr", "ada", "rar", "rdr", "raar" and "radar". The lexicographically largest of them is "rr".
|
```python
s = input()
M = sorted(list(s))[-1]
ans = ""
for c in s:
if c == M:
ans += M
print(ans)
```
| 3
|
|
1,005
|
B
|
Delete from the Left
|
PROGRAMMING
| 900
|
[
"brute force",
"implementation",
"strings"
] | null | null |
You are given two strings $s$ and $t$. In a single move, you can choose any of two strings and delete the first (that is, the leftmost) character. After a move, the length of the string decreases by $1$. You can't choose a string if it is empty.
For example:
- by applying a move to the string "where", the result is the string "here", - by applying a move to the string "a", the result is an empty string "".
You are required to make two given strings equal using the fewest number of moves. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the initial strings.
Write a program that finds the minimum number of moves to make two given strings $s$ and $t$ equal.
|
The first line of the input contains $s$. In the second line of the input contains $t$. Both strings consist only of lowercase Latin letters. The number of letters in each string is between 1 and $2\cdot10^5$, inclusive.
|
Output the fewest number of moves required. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the given strings.
|
[
"test\nwest\n",
"codeforces\nyes\n",
"test\nyes\n",
"b\nab\n"
] |
[
"2\n",
"9\n",
"7\n",
"1\n"
] |
In the first example, you should apply the move once to the first string and apply the move once to the second string. As a result, both strings will be equal to "est".
In the second example, the move should be applied to the string "codeforces" $8$ times. As a result, the string becomes "codeforces" $\to$ "es". The move should be applied to the string "yes" once. The result is the same string "yes" $\to$ "es".
In the third example, you can make the strings equal only by completely deleting them. That is, in the end, both strings will be empty.
In the fourth example, the first character of the second string should be deleted.
| 0
|
[
{
"input": "test\nwest",
"output": "2"
},
{
"input": "codeforces\nyes",
"output": "9"
},
{
"input": "test\nyes",
"output": "7"
},
{
"input": "b\nab",
"output": "1"
},
{
"input": "z\nz",
"output": "0"
},
{
"input": "abacabadabacaba\nabacabadacaba",
"output": "18"
},
{
"input": "z\na",
"output": "2"
},
{
"input": "ama\nsama",
"output": "1"
},
{
"input": "aaabc\nbc",
"output": "3"
},
{
"input": "lxuyspmieeuyxluv\ndfwxqvavzxctblaa",
"output": "32"
},
{
"input": "ahmedaly\naly",
"output": "5"
},
{
"input": "aa\na",
"output": "1"
},
{
"input": "aa\nba",
"output": "2"
},
{
"input": "cccba\nba",
"output": "3"
},
{
"input": "a\nab",
"output": "3"
},
{
"input": "dadc\ndddc",
"output": "4"
},
{
"input": "aacaaab\nb",
"output": "6"
},
{
"input": "dddc\ncdcde",
"output": "9"
},
{
"input": "bbdab\ndab",
"output": "2"
},
{
"input": "a\naa",
"output": "1"
},
{
"input": "aa\nab",
"output": "4"
},
{
"input": "codeforces\nces",
"output": "7"
}
] | 1,641,832,999
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 28
| 155
| 716,800
|
a = input()
b = input()
c = len(a)
d = len(b)
while c * d * (a[c - 1] == b[d - 1]):
c = c - 1
d = d - 1
print(c + d)
|
Title: Delete from the Left
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given two strings $s$ and $t$. In a single move, you can choose any of two strings and delete the first (that is, the leftmost) character. After a move, the length of the string decreases by $1$. You can't choose a string if it is empty.
For example:
- by applying a move to the string "where", the result is the string "here", - by applying a move to the string "a", the result is an empty string "".
You are required to make two given strings equal using the fewest number of moves. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the initial strings.
Write a program that finds the minimum number of moves to make two given strings $s$ and $t$ equal.
Input Specification:
The first line of the input contains $s$. In the second line of the input contains $t$. Both strings consist only of lowercase Latin letters. The number of letters in each string is between 1 and $2\cdot10^5$, inclusive.
Output Specification:
Output the fewest number of moves required. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the given strings.
Demo Input:
['test\nwest\n', 'codeforces\nyes\n', 'test\nyes\n', 'b\nab\n']
Demo Output:
['2\n', '9\n', '7\n', '1\n']
Note:
In the first example, you should apply the move once to the first string and apply the move once to the second string. As a result, both strings will be equal to "est".
In the second example, the move should be applied to the string "codeforces" $8$ times. As a result, the string becomes "codeforces" $\to$ "es". The move should be applied to the string "yes" once. The result is the same string "yes" $\to$ "es".
In the third example, you can make the strings equal only by completely deleting them. That is, in the end, both strings will be empty.
In the fourth example, the first character of the second string should be deleted.
|
```python
a = input()
b = input()
c = len(a)
d = len(b)
while c * d * (a[c - 1] == b[d - 1]):
c = c - 1
d = d - 1
print(c + d)
```
| 3
|
|
825
|
A
|
Binary Protocol
|
PROGRAMMING
| 1,100
|
[
"implementation"
] | null | null |
Polycarp has just invented a new binary protocol for data transmission. He is encoding positive integer decimal number to binary string using following algorithm:
- Each digit is represented with number of '1' characters equal to the value of that digit (for 0 it is zero ones). - Digits are written one by one in order corresponding to number and separated by single '0' character.
Though Polycarp learnt how to encode the numbers, he has no idea how to decode them back. Help him calculate the decoded number.
|
The first line contains one integer number *n* (1<=β€<=*n*<=β€<=89) β length of the string *s*.
The second line contains string *s* β sequence of '0' and '1' characters, number in its encoded format. It is guaranteed that the number corresponding to the string is positive and doesn't exceed 109. The string always starts with '1'.
|
Print the decoded number.
|
[
"3\n111\n",
"9\n110011101\n"
] |
[
"3\n",
"2031\n"
] |
none
| 0
|
[
{
"input": "3\n111",
"output": "3"
},
{
"input": "9\n110011101",
"output": "2031"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "3\n100",
"output": "100"
},
{
"input": "5\n10001",
"output": "1001"
},
{
"input": "14\n11001100011000",
"output": "202002000"
},
{
"input": "31\n1000011111111100011110111111111",
"output": "100090049"
},
{
"input": "53\n10110111011110111110111111011111110111111110111111111",
"output": "123456789"
},
{
"input": "89\n11111111101111111110111111111011111111101111111110111111111011111111101111111110111111111",
"output": "999999999"
},
{
"input": "10\n1000000000",
"output": "1000000000"
},
{
"input": "2\n10",
"output": "10"
},
{
"input": "4\n1110",
"output": "30"
},
{
"input": "8\n10101010",
"output": "11110"
}
] | 1,500,217,994
| 494
|
Python 3
|
OK
|
TESTS
| 13
| 62
| 4,608,000
|
length = input()
s = input()
s = s.split('0')
result = ""
for i in s:
result += str(len(i))
print(result)
|
Title: Binary Protocol
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp has just invented a new binary protocol for data transmission. He is encoding positive integer decimal number to binary string using following algorithm:
- Each digit is represented with number of '1' characters equal to the value of that digit (for 0 it is zero ones). - Digits are written one by one in order corresponding to number and separated by single '0' character.
Though Polycarp learnt how to encode the numbers, he has no idea how to decode them back. Help him calculate the decoded number.
Input Specification:
The first line contains one integer number *n* (1<=β€<=*n*<=β€<=89) β length of the string *s*.
The second line contains string *s* β sequence of '0' and '1' characters, number in its encoded format. It is guaranteed that the number corresponding to the string is positive and doesn't exceed 109. The string always starts with '1'.
Output Specification:
Print the decoded number.
Demo Input:
['3\n111\n', '9\n110011101\n']
Demo Output:
['3\n', '2031\n']
Note:
none
|
```python
length = input()
s = input()
s = s.split('0')
result = ""
for i in s:
result += str(len(i))
print(result)
```
| 3
|
|
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 stringΒ β 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'Β β $26$ 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 integersΒ β $n$ and $k$ ($1 \le k \le n \le 50$)Β β 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 integerΒ β 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"
}
] | 1,637,078,073
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 29
| 77
| 0
|
import math
import sys
input = sys.stdin.readline
output = sys.stdout.write
def inList():
return(list(map(int,input().split())))
def inVar():
return map(int,input().split())
def main():
n , k = inVar()
s = list(input())
s.pop()
mp = {}
for i in range(0 , 27):
mp[i] = 0
for char in s:
mp[ord(char) - ord('a') + 1] = 1
ans = 0
cur = -2
while k:
flag = 0
for i in range(27):
if mp[i] == 1 and cur + 1 < i:
mp[i] = 0
cur = i
ans += i
flag = 1
break
if flag == 0:
break
k -= 1
if k != 0:
ans = -1
print(ans)
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 stringΒ β 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'Β β $26$ 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 integersΒ β $n$ and $k$ ($1 \le k \le n \le 50$)Β β 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 integerΒ β 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
import math
import sys
input = sys.stdin.readline
output = sys.stdout.write
def inList():
return(list(map(int,input().split())))
def inVar():
return map(int,input().split())
def main():
n , k = inVar()
s = list(input())
s.pop()
mp = {}
for i in range(0 , 27):
mp[i] = 0
for char in s:
mp[ord(char) - ord('a') + 1] = 1
ans = 0
cur = -2
while k:
flag = 0
for i in range(27):
if mp[i] == 1 and cur + 1 < i:
mp[i] = 0
cur = i
ans += i
flag = 1
break
if flag == 0:
break
k -= 1
if k != 0:
ans = -1
print(ans)
main()
```
| 3
|
|
583
|
B
|
Robot's Task
|
PROGRAMMING
| 1,200
|
[
"greedy",
"implementation"
] | null | null |
Robot Doc is located in the hall, with *n* computers stand in a line, numbered from left to right from 1 to *n*. Each computer contains exactly one piece of information, each of which Doc wants to get eventually. The computers are equipped with a security system, so to crack the *i*-th of them, the robot needs to collect at least *a**i* any pieces of information from the other computers. Doc can hack the computer only if he is right next to it.
The robot is assembled using modern technologies and can move along the line of computers in either of the two possible directions, but the change of direction requires a large amount of resources from Doc. Tell the minimum number of changes of direction, which the robot will have to make to collect all *n* parts of information if initially it is next to computer with number 1.
It is guaranteed that there exists at least one sequence of the robot's actions, which leads to the collection of all information. Initially Doc doesn't have any pieces of information.
|
The first line contains number *n* (1<=β€<=*n*<=β€<=1000). The second line contains *n* non-negative integers *a*1,<=*a*2,<=...,<=*a**n* (0<=β€<=*a**i*<=<<=*n*), separated by a space. It is guaranteed that there exists a way for robot to collect all pieces of the information.
|
Print a single number β the minimum number of changes in direction that the robot will have to make in order to collect all *n* parts of information.
|
[
"3\n0 2 0\n",
"5\n4 2 3 0 1\n",
"7\n0 3 1 0 5 2 6\n"
] |
[
"1\n",
"3\n",
"2\n"
] |
In the first sample you can assemble all the pieces of information in the optimal manner by assembling first the piece of information in the first computer, then in the third one, then change direction and move to the second one, and then, having 2 pieces of information, collect the last piece.
In the second sample to collect all the pieces of information in the optimal manner, Doc can go to the fourth computer and get the piece of information, then go to the fifth computer with one piece and get another one, then go to the second computer in the same manner, then to the third one and finally, to the first one. Changes of direction will take place before moving from the fifth to the second computer, then from the second to the third computer, then from the third to the first computer.
In the third sample the optimal order of collecting parts from computers can look like that: 1->3->4->6->2->5->7.
| 1,000
|
[
{
"input": "3\n0 2 0",
"output": "1"
},
{
"input": "5\n4 2 3 0 1",
"output": "3"
},
{
"input": "7\n0 3 1 0 5 2 6",
"output": "2"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "2\n0 1",
"output": "0"
},
{
"input": "10\n0 0 0 0 0 0 0 0 0 0",
"output": "0"
},
{
"input": "3\n0 2 1",
"output": "1"
},
{
"input": "10\n7 1 9 3 5 8 6 0 2 4",
"output": "9"
},
{
"input": "10\n1 3 5 7 9 8 6 4 2 0",
"output": "9"
},
{
"input": "10\n5 0 0 1 3 2 2 2 5 7",
"output": "1"
},
{
"input": "10\n8 6 5 3 9 7 1 4 2 0",
"output": "8"
},
{
"input": "10\n1 2 4 5 0 1 3 7 1 4",
"output": "2"
},
{
"input": "10\n3 4 8 9 5 1 2 0 6 7",
"output": "6"
},
{
"input": "10\n2 2 0 0 6 2 9 0 2 0",
"output": "2"
},
{
"input": "10\n1 7 5 3 2 6 0 8 4 9",
"output": "8"
},
{
"input": "9\n1 3 8 6 2 4 5 0 7",
"output": "7"
},
{
"input": "9\n1 3 5 7 8 6 4 2 0",
"output": "8"
},
{
"input": "9\n2 4 3 1 3 0 5 4 3",
"output": "3"
},
{
"input": "9\n3 5 6 8 7 0 4 2 1",
"output": "5"
},
{
"input": "9\n2 0 8 1 0 3 0 5 3",
"output": "2"
},
{
"input": "9\n6 2 3 7 4 8 5 1 0",
"output": "4"
},
{
"input": "9\n3 1 5 6 0 3 2 0 0",
"output": "2"
},
{
"input": "9\n2 6 4 1 0 8 5 3 7",
"output": "7"
},
{
"input": "100\n27 20 18 78 93 38 56 2 48 75 36 88 96 57 69 10 25 74 68 86 65 85 66 14 22 12 43 80 99 34 42 63 61 71 77 15 37 54 21 59 23 94 28 30 50 84 62 76 47 16 26 64 82 92 72 53 17 11 41 91 35 83 79 95 67 13 1 7 3 4 73 90 8 19 33 58 98 32 39 45 87 52 60 46 6 44 49 70 51 9 5 29 31 24 40 97 81 0 89 55",
"output": "69"
},
{
"input": "100\n1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97 99 98 96 94 92 90 88 86 84 82 80 78 76 74 72 70 68 66 64 62 60 58 56 54 52 50 48 46 44 42 40 38 36 34 32 30 28 26 24 22 20 18 16 14 12 10 8 6 4 2 0",
"output": "99"
},
{
"input": "100\n13 89 81 0 62 1 59 92 29 13 1 37 2 8 53 15 20 34 12 70 0 85 97 55 84 60 37 54 14 65 22 69 30 22 95 44 59 85 50 80 9 71 91 93 74 21 11 78 28 21 40 81 76 24 26 60 48 85 61 68 89 76 46 73 34 52 98 29 4 38 94 51 5 55 6 27 74 27 38 37 82 70 44 89 51 59 30 37 15 55 63 78 42 39 71 43 4 10 2 13",
"output": "21"
},
{
"input": "100\n1 3 5 7 58 11 13 15 17 19 45 23 25 27 29 31 33 35 37 39 41 43 21 47 49 51 53 55 57 59 61 63 65 67 69 71 73 75 77 81 79 83 85 87 89 91 93 95 97 48 98 96 94 92 90 88 44 84 82 80 78 76 74 72 70 68 66 64 62 60 9 56 54 52 50 99 46 86 42 40 38 36 34 32 30 28 26 24 22 20 18 16 14 12 10 8 6 4 2 0",
"output": "96"
},
{
"input": "100\n32 47 74 8 14 4 12 68 18 0 44 80 14 38 6 57 4 72 69 3 21 78 74 22 39 32 58 63 34 33 23 6 39 11 6 12 18 4 0 11 20 28 16 1 22 12 57 55 13 48 43 1 50 18 87 6 11 45 38 67 37 14 7 56 6 41 1 55 5 73 78 64 38 18 38 8 37 0 18 61 37 58 58 62 86 5 0 2 15 43 34 61 2 21 15 9 69 1 11 24",
"output": "4"
},
{
"input": "100\n40 3 55 7 6 77 13 46 17 64 21 54 25 27 91 41 1 15 37 82 23 43 42 47 26 95 53 5 11 59 61 9 78 67 69 58 73 0 36 79 60 83 2 87 63 33 71 89 97 99 98 93 56 92 19 88 86 84 39 28 65 20 34 76 51 94 66 12 62 49 96 72 24 52 48 50 44 35 74 31 38 57 81 32 22 80 70 29 30 18 68 16 14 90 10 8 85 4 45 75",
"output": "75"
},
{
"input": "100\n34 16 42 21 84 27 11 7 82 16 95 39 36 64 26 0 38 37 2 2 16 56 16 61 55 42 26 5 61 8 30 20 19 15 9 78 5 34 15 0 3 17 36 36 1 5 4 26 18 0 14 25 7 5 91 7 43 26 79 37 17 27 40 55 66 7 0 2 16 23 68 35 2 5 9 21 1 7 2 9 4 3 22 15 27 6 0 47 5 0 12 9 20 55 36 10 6 8 5 1",
"output": "3"
},
{
"input": "100\n35 53 87 49 13 24 93 20 5 11 31 32 40 52 96 46 1 25 66 69 28 88 84 82 70 9 75 39 26 21 18 29 23 57 90 16 48 22 95 0 58 43 7 73 8 62 63 30 64 92 79 3 6 94 34 12 76 99 67 55 56 97 14 91 68 36 44 78 41 71 86 89 47 74 4 45 98 37 80 33 83 27 42 59 72 54 17 60 51 81 15 77 65 50 10 85 61 19 38 2",
"output": "67"
},
{
"input": "99\n89 96 56 31 32 14 9 66 87 34 69 5 92 54 41 52 46 30 22 26 16 18 20 68 62 73 90 43 79 33 58 98 37 45 10 78 94 51 19 0 91 39 28 47 17 86 3 61 77 7 15 64 55 83 65 71 97 88 6 48 24 11 8 42 81 4 63 93 50 74 35 12 95 27 53 82 29 85 84 60 72 40 36 57 23 13 38 59 49 1 75 44 76 2 21 25 70 80 67",
"output": "75"
},
{
"input": "99\n1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97 98 96 94 92 90 88 86 84 82 80 78 76 74 72 70 68 66 64 62 60 58 56 54 52 50 48 46 44 42 40 38 36 34 32 30 28 26 24 22 20 18 16 14 12 10 8 6 4 2 0",
"output": "98"
},
{
"input": "99\n82 7 6 77 17 28 90 3 68 12 63 60 24 20 4 81 71 85 57 45 11 84 3 91 49 34 89 82 0 50 48 88 36 76 36 5 62 48 20 2 20 45 69 27 37 62 42 31 57 51 92 84 89 25 7 62 12 23 23 56 30 90 27 10 77 58 48 38 56 68 57 15 33 1 34 67 16 47 75 70 69 28 38 16 5 61 85 76 44 90 37 22 77 94 55 1 97 8 69",
"output": "22"
},
{
"input": "99\n1 51 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 42 43 45 47 49 3 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97 98 96 94 92 90 88 86 84 82 80 8 76 74 72 70 68 66 22 62 60 58 56 54 52 0 48 46 44 41 40 38 36 34 32 30 28 26 24 64 20 18 16 14 12 10 78 6 4 2 50",
"output": "96"
},
{
"input": "99\n22 3 19 13 65 87 28 17 41 40 31 21 8 37 29 65 65 53 16 33 13 5 76 4 72 9 2 76 57 72 50 15 75 0 30 13 83 36 12 31 49 51 65 22 48 31 60 15 2 17 6 1 8 0 1 63 3 16 7 7 2 1 47 28 26 21 2 36 1 5 20 25 44 0 2 39 46 30 33 11 15 34 34 4 84 52 0 39 7 3 17 15 6 38 52 64 26 1 0",
"output": "3"
},
{
"input": "99\n24 87 25 82 97 11 37 15 23 19 34 17 76 13 45 89 33 1 27 78 63 43 54 47 49 2 42 41 75 83 61 90 65 67 21 71 60 57 77 62 81 58 85 69 3 91 68 55 72 93 29 94 66 16 88 86 84 53 14 39 35 44 9 70 80 92 56 79 74 5 64 31 52 50 48 46 51 59 40 38 36 96 32 30 28 95 7 22 20 18 26 73 12 10 8 6 4 98 0",
"output": "74"
},
{
"input": "99\n22 14 0 44 6 17 6 6 37 45 0 48 19 8 57 8 10 0 3 12 25 2 5 53 9 49 15 6 38 14 9 40 38 22 27 12 64 10 11 35 89 19 46 39 12 24 48 0 52 1 27 27 24 4 64 24 5 0 67 3 5 39 0 1 13 37 2 8 46 1 28 70 6 79 14 15 33 6 7 34 6 18 4 71 1 55 33 71 18 11 47 23 72 53 65 32 2 7 28",
"output": "3"
},
{
"input": "99\n28 59 73 89 52 27 0 20 36 12 83 95 31 24 54 94 49 14 51 34 50 93 13 1 2 68 63 48 41 81 23 43 18 9 16 38 33 60 62 3 40 85 72 69 90 98 11 37 22 44 35 6 21 39 82 10 64 66 96 42 74 30 8 67 97 46 84 32 17 57 75 71 5 26 4 55 58 29 7 15 45 19 92 91 78 65 88 25 86 80 77 87 79 53 47 70 56 76 61",
"output": "63"
}
] | 1,626,622,787
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 56
| 202
| 6,963,200
|
from sys import *
input = lambda:stdin.readline()
int_arr = lambda : list(map(int,stdin.readline().strip().split()))
str_arr = lambda :list(map(str,stdin.readline().split()))
get_str = lambda : map(str,stdin.readline().strip().split())
get_int = lambda: map(int,stdin.readline().strip().split())
get_float = lambda : map(float,stdin.readline().strip().split())
mod = 1000000007
setrecursionlimit(1000)
n = int(input())
arr = int_arr()
ct = 0
cond = 0
tot = -1
while True:
if len(set(arr)) == 1 and arr[0] == -1:
break
if cond % 2 == 0:
cond = 1
tot += 1
for i in range(n):
if arr[i] != -1 and arr[i] <= ct:
ct += 1
arr[i] = -1
last = i
else:
cond = 2
tot += 1
for i in range(last,-1,-1):
if arr[i] != -1 and arr[i] <= ct:
ct += 1
arr[i] = -1
print(tot)
|
Title: Robot's Task
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Robot Doc is located in the hall, with *n* computers stand in a line, numbered from left to right from 1 to *n*. Each computer contains exactly one piece of information, each of which Doc wants to get eventually. The computers are equipped with a security system, so to crack the *i*-th of them, the robot needs to collect at least *a**i* any pieces of information from the other computers. Doc can hack the computer only if he is right next to it.
The robot is assembled using modern technologies and can move along the line of computers in either of the two possible directions, but the change of direction requires a large amount of resources from Doc. Tell the minimum number of changes of direction, which the robot will have to make to collect all *n* parts of information if initially it is next to computer with number 1.
It is guaranteed that there exists at least one sequence of the robot's actions, which leads to the collection of all information. Initially Doc doesn't have any pieces of information.
Input Specification:
The first line contains number *n* (1<=β€<=*n*<=β€<=1000). The second line contains *n* non-negative integers *a*1,<=*a*2,<=...,<=*a**n* (0<=β€<=*a**i*<=<<=*n*), separated by a space. It is guaranteed that there exists a way for robot to collect all pieces of the information.
Output Specification:
Print a single number β the minimum number of changes in direction that the robot will have to make in order to collect all *n* parts of information.
Demo Input:
['3\n0 2 0\n', '5\n4 2 3 0 1\n', '7\n0 3 1 0 5 2 6\n']
Demo Output:
['1\n', '3\n', '2\n']
Note:
In the first sample you can assemble all the pieces of information in the optimal manner by assembling first the piece of information in the first computer, then in the third one, then change direction and move to the second one, and then, having 2 pieces of information, collect the last piece.
In the second sample to collect all the pieces of information in the optimal manner, Doc can go to the fourth computer and get the piece of information, then go to the fifth computer with one piece and get another one, then go to the second computer in the same manner, then to the third one and finally, to the first one. Changes of direction will take place before moving from the fifth to the second computer, then from the second to the third computer, then from the third to the first computer.
In the third sample the optimal order of collecting parts from computers can look like that: 1->3->4->6->2->5->7.
|
```python
from sys import *
input = lambda:stdin.readline()
int_arr = lambda : list(map(int,stdin.readline().strip().split()))
str_arr = lambda :list(map(str,stdin.readline().split()))
get_str = lambda : map(str,stdin.readline().strip().split())
get_int = lambda: map(int,stdin.readline().strip().split())
get_float = lambda : map(float,stdin.readline().strip().split())
mod = 1000000007
setrecursionlimit(1000)
n = int(input())
arr = int_arr()
ct = 0
cond = 0
tot = -1
while True:
if len(set(arr)) == 1 and arr[0] == -1:
break
if cond % 2 == 0:
cond = 1
tot += 1
for i in range(n):
if arr[i] != -1 and arr[i] <= ct:
ct += 1
arr[i] = -1
last = i
else:
cond = 2
tot += 1
for i in range(last,-1,-1):
if arr[i] != -1 and arr[i] <= ct:
ct += 1
arr[i] = -1
print(tot)
```
| 3
|
|
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" β 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<=β€<=*n*<=β€<=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<=β€<=*x**i*,<=*y**i*,<=*z**i*<=β€<=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"
}
] | 1,669,146,966
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 81
| 92
| 0
|
t=int(input())
x=[]
y=[]
z=[]
for i in range(0,t,1):
a,b,c=map(int,input().split())
x.append(a)
y.append(b)
z.append(c)
p=sum(x)
q=sum(y)
z=sum(z)
if(p==0 and q==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" β 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<=β€<=*n*<=β€<=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<=β€<=*x**i*,<=*y**i*,<=*z**i*<=β€<=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
t=int(input())
x=[]
y=[]
z=[]
for i in range(0,t,1):
a,b,c=map(int,input().split())
x.append(a)
y.append(b)
z.append(c)
p=sum(x)
q=sum(y)
z=sum(z)
if(p==0 and q==0 and z==0):
print("YES")
else:
print("NO")
```
| 3.977
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
Vlad likes to eat in cafes very much. During his life, he has visited cafes *n* times. Unfortunately, Vlad started to feel that his last visits are not any different from each other. To fix that Vlad had a small research.
First of all, Vlad assigned individual indices to all cafes. Then, he wrote down indices of cafes he visited in a row, in order of visiting them. Now, Vlad wants to find such a cafe that his last visit to that cafe was before his last visits to every other cafe. In other words, he wants to find such a cafe that he hasn't been there for as long as possible. Help Vlad to find that cafe.
|
In first line there is one integer *n* (1<=β€<=*n*<=β€<=2Β·105)Β β number of cafes indices written by Vlad.
In second line, *n* numbers *a*1,<=*a*2,<=...,<=*a**n* (0<=β€<=*a**i*<=β€<=2Β·105) are writtenΒ β indices of cafes in order of being visited by Vlad. Vlad could visit some cafes more than once. Note that in numeration, some indices could be omitted.
|
Print one integerΒ β index of the cafe that Vlad hasn't visited for as long as possible.
|
[
"5\n1 3 2 1 2\n",
"6\n2 1 2 2 4 1\n"
] |
[
"3\n",
"2\n"
] |
In first test, there are three cafes, and the last visits to cafes with indices 1 and 2 were after the last visit to cafe with index 3; so this cafe is the answer.
In second test case, there are also three cafes, but with indices 1, 2 and 4. Cafes with indices 1 and 4 were visited after the last visit of cafe with index 2, so the answer is 2. Note that Vlad could omit some numbers while numerating the cafes.
| 0
|
[
{
"input": "5\n1 3 2 1 2",
"output": "3"
},
{
"input": "6\n2 1 2 2 4 1",
"output": "2"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "1\n200000",
"output": "200000"
},
{
"input": "2\n2018 2017",
"output": "2018"
},
{
"input": "5\n100 1000 1000 1000 1000",
"output": "100"
},
{
"input": "8\n200000 1 200000 1 200000 1 200000 2",
"output": "1"
},
{
"input": "5\n20000 1 2 3 4",
"output": "20000"
},
{
"input": "2\n200000 1",
"output": "200000"
},
{
"input": "3\n2 100 2",
"output": "100"
},
{
"input": "2\n10 1",
"output": "10"
},
{
"input": "11\n1 1 1 1 1 1 1 1 1 1 1",
"output": "1"
},
{
"input": "3\n5 5 5",
"output": "5"
},
{
"input": "2\n5 1",
"output": "5"
}
] | 1,530,557,351
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 33
| 436
| 23,654,400
|
n=int(input())
l=list(map(int,input().split()))
d1,d2={},{}
for i in l:
try:
d1[i]+=1
except:
d1.update({i:0})
d1[i]+=1
x=list(d1.keys())
f=len(x)-1
if f==0:
print(l[0])
else:
s=set()
for i in range(n-1,-1,-1):
s.add(l[i])
if len(s)==f:
break
print(list(set(x)-s)[0])
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vlad likes to eat in cafes very much. During his life, he has visited cafes *n* times. Unfortunately, Vlad started to feel that his last visits are not any different from each other. To fix that Vlad had a small research.
First of all, Vlad assigned individual indices to all cafes. Then, he wrote down indices of cafes he visited in a row, in order of visiting them. Now, Vlad wants to find such a cafe that his last visit to that cafe was before his last visits to every other cafe. In other words, he wants to find such a cafe that he hasn't been there for as long as possible. Help Vlad to find that cafe.
Input Specification:
In first line there is one integer *n* (1<=β€<=*n*<=β€<=2Β·105)Β β number of cafes indices written by Vlad.
In second line, *n* numbers *a*1,<=*a*2,<=...,<=*a**n* (0<=β€<=*a**i*<=β€<=2Β·105) are writtenΒ β indices of cafes in order of being visited by Vlad. Vlad could visit some cafes more than once. Note that in numeration, some indices could be omitted.
Output Specification:
Print one integerΒ β index of the cafe that Vlad hasn't visited for as long as possible.
Demo Input:
['5\n1 3 2 1 2\n', '6\n2 1 2 2 4 1\n']
Demo Output:
['3\n', '2\n']
Note:
In first test, there are three cafes, and the last visits to cafes with indices 1 and 2 were after the last visit to cafe with index 3; so this cafe is the answer.
In second test case, there are also three cafes, but with indices 1, 2 and 4. Cafes with indices 1 and 4 were visited after the last visit of cafe with index 2, so the answer is 2. Note that Vlad could omit some numbers while numerating the cafes.
|
```python
n=int(input())
l=list(map(int,input().split()))
d1,d2={},{}
for i in l:
try:
d1[i]+=1
except:
d1.update({i:0})
d1[i]+=1
x=list(d1.keys())
f=len(x)-1
if f==0:
print(l[0])
else:
s=set()
for i in range(n-1,-1,-1):
s.add(l[i])
if len(s)==f:
break
print(list(set(x)-s)[0])
```
| 3
|
|
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 β (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 β 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",
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},
{
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{
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},
{
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},
{
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"output": "01000111100111001011110010100011111111110010101100001101"
},
{
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"output": "110001010001000011000101110101000100001011111001011001001"
},
{
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},
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"output": "01110110101110100100110011010000001000101100101111000111011"
},
{
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"output": "111100101000000011101011011001110010101111000110010010000000"
},
{
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},
{
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},
{
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},
{
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"output": "0010100110110100111100100100101101010100100111011010001001010101"
},
{
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"output": "11010110111100101111101001100001110100010110010110110111100110100"
},
{
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"output": "111111010011011100101110100110111111111001111110011010111111110000"
},
{
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"output": "1010101010100010001001001001100000111000010010010100010011000100000"
},
{
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"output": "00011111011111001000011100010011100011010100101011011000001001111110"
},
{
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"output": "001111000011001110100111010101111111011100110011001010010010000111011"
},
{
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"output": "0110001100110100010000110111000010011010011000011001010011010100010100"
},
{
"input": "00000111110010110001110110001010010101000111011001111111100110011110010\n00010111110100000100110101000010010001100001100011100000001100010100010",
"output": "00010000000110110101000011001000000100100110111010011111101010001010000"
},
{
"input": "100101011100101101000011010001011001101110101110001100010001010111001110\n100001111100101011011111110000001111000111001011111110000010101110111001",
"output": "000100100000000110011100100001010110101001100101110010010011111001110111"
},
{
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"output": "1000111100010011010110011101000000101010101100011111100001101111001010010"
},
{
"input": "01101101010011110101100001110101111011100010000010001101111000011110111111\n00101111001101001100111010000101110000100101101111100111101110010100011011",
"output": "01000010011110111001011011110000001011000111101101101010010110001010100100"
},
{
"input": "101100101100011001101111110110110010100110110010100001110010110011001101011\n000001011010101011110011111101001110000111000010001101000010010000010001101",
"output": "101101110110110010011100001011111100100001110000101100110000100011011100110"
},
{
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"output": "1110111111110010111000001100101010101011010100101010100101100011001001111111"
},
{
"input": "00101101010000000101011001101011001100010001100000101011101110000001111001000\n10010110010111000000101101000011101011001010000011011101101011010000000011111",
"output": "10111011000111000101110100101000100111011011100011110110000101010001111010111"
},
{
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"output": "110100100110000101010010011010011001100110000111010000010100001011110111111101"
},
{
"input": "0110001101100100001111110101101000100101010010101010011001101001001101110000000\n0111011000000010010111011110010000000001000110001000011001101000000001110100111",
"output": "0001010101100110011000101011111000100100010100100010000000000001001100000100111"
},
{
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"output": "00001000101011011011110011001111010110100010101011000011110001101011110010001001"
},
{
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"output": "010111111011000000100001100111101000001010100010001001100101110101001010000111011"
},
{
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"output": "0100101010111101000000010111101001101101010010000110001100110111110001000100000101"
},
{
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"output": "00001011101110000000011010111101011101110001011110010100010001001000010110111101101"
},
{
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"output": "100110111101100101110001010001000000100000011111101101001101001101111011011010100001"
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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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},
{
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"output": "010010111000010101000111111110111011001101010000000011010101010101000110111110101001010011001101011"
},
{
"input": "1010000011010110011101001101110001110010000111011101110010110110111111001001110100101100010101010001\n0111101000111100101100000101111010100100001000011101010100110011100011010011010101000100101011100011",
"output": "1101101011101010110001001000001011010110001111000000100110000101011100011010100001101000111110110010"
},
{
"input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001\n1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111010",
"output": "1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111011"
},
{
"input": "0\n0",
"output": "0"
},
{
"input": "0\n1",
"output": "1"
}
] | 1,697,107,978
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 102
| 46
| 0
|
a = str(input())
b = str(input())
n = 0
for i in a:
if i != b[n]:
print(1, end='')
if i == b[n]:
print(0, end='')
n += 1
|
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 β (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 β 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 = str(input())
b = str(input())
n = 0
for i in a:
if i != b[n]:
print(1, end='')
if i == b[n]:
print(0, end='')
n += 1
```
| 3.9885
|
760
|
A
|
Petr and a calendar
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
Petr wants to make a calendar for current month. For this purpose he draws a table in which columns correspond to weeks (a week is seven consequent days from Monday to Sunday), rows correspond to weekdays, and cells contain dates. For example, a calendar for January 2017 should look like on the picture:
Petr wants to know how many columns his table should have given the month and the weekday of the first date of that month? Assume that the year is non-leap.
|
The only line contain two integers *m* and *d* (1<=β€<=*m*<=β€<=12, 1<=β€<=*d*<=β€<=7)Β β the number of month (January is the first month, December is the twelfth) and the weekday of the first date of this month (1 is Monday, 7 is Sunday).
|
Print single integer: the number of columns the table should have.
|
[
"1 7\n",
"1 1\n",
"11 6\n"
] |
[
"6\n",
"5\n",
"5\n"
] |
The first example corresponds to the January 2017 shown on the picture in the statements.
In the second example 1-st January is Monday, so the whole month fits into 5 columns.
In the third example 1-st November is Saturday and 5 columns is enough.
| 500
|
[
{
"input": "1 7",
"output": "6"
},
{
"input": "1 1",
"output": "5"
},
{
"input": "11 6",
"output": "5"
},
{
"input": "2 7",
"output": "5"
},
{
"input": "2 1",
"output": "4"
},
{
"input": "8 6",
"output": "6"
},
{
"input": "1 1",
"output": "5"
},
{
"input": "1 2",
"output": "5"
},
{
"input": "1 3",
"output": "5"
},
{
"input": "1 4",
"output": "5"
},
{
"input": "1 5",
"output": "5"
},
{
"input": "1 6",
"output": "6"
},
{
"input": "1 7",
"output": "6"
},
{
"input": "2 1",
"output": "4"
},
{
"input": "2 2",
"output": "5"
},
{
"input": "2 3",
"output": "5"
},
{
"input": "2 4",
"output": "5"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "2 6",
"output": "5"
},
{
"input": "2 7",
"output": "5"
},
{
"input": "3 1",
"output": "5"
},
{
"input": "3 2",
"output": "5"
},
{
"input": "3 3",
"output": "5"
},
{
"input": "3 4",
"output": "5"
},
{
"input": "3 5",
"output": "5"
},
{
"input": "3 6",
"output": "6"
},
{
"input": "3 7",
"output": "6"
},
{
"input": "4 1",
"output": "5"
},
{
"input": "4 2",
"output": "5"
},
{
"input": "4 3",
"output": "5"
},
{
"input": "4 4",
"output": "5"
},
{
"input": "4 5",
"output": "5"
},
{
"input": "4 6",
"output": "5"
},
{
"input": "4 7",
"output": "6"
},
{
"input": "5 1",
"output": "5"
},
{
"input": "5 2",
"output": "5"
},
{
"input": "5 3",
"output": "5"
},
{
"input": "5 4",
"output": "5"
},
{
"input": "5 5",
"output": "5"
},
{
"input": "5 6",
"output": "6"
},
{
"input": "5 7",
"output": "6"
},
{
"input": "6 1",
"output": "5"
},
{
"input": "6 2",
"output": "5"
},
{
"input": "6 3",
"output": "5"
},
{
"input": "6 4",
"output": "5"
},
{
"input": "6 5",
"output": "5"
},
{
"input": "6 6",
"output": "5"
},
{
"input": "6 7",
"output": "6"
},
{
"input": "7 1",
"output": "5"
},
{
"input": "7 2",
"output": "5"
},
{
"input": "7 3",
"output": "5"
},
{
"input": "7 4",
"output": "5"
},
{
"input": "7 5",
"output": "5"
},
{
"input": "7 6",
"output": "6"
},
{
"input": "7 7",
"output": "6"
},
{
"input": "8 1",
"output": "5"
},
{
"input": "8 2",
"output": "5"
},
{
"input": "8 3",
"output": "5"
},
{
"input": "8 4",
"output": "5"
},
{
"input": "8 5",
"output": "5"
},
{
"input": "8 6",
"output": "6"
},
{
"input": "8 7",
"output": "6"
},
{
"input": "9 1",
"output": "5"
},
{
"input": "9 2",
"output": "5"
},
{
"input": "9 3",
"output": "5"
},
{
"input": "9 4",
"output": "5"
},
{
"input": "9 5",
"output": "5"
},
{
"input": "9 6",
"output": "5"
},
{
"input": "9 7",
"output": "6"
},
{
"input": "10 1",
"output": "5"
},
{
"input": "10 2",
"output": "5"
},
{
"input": "10 3",
"output": "5"
},
{
"input": "10 4",
"output": "5"
},
{
"input": "10 5",
"output": "5"
},
{
"input": "10 6",
"output": "6"
},
{
"input": "10 7",
"output": "6"
},
{
"input": "11 1",
"output": "5"
},
{
"input": "11 2",
"output": "5"
},
{
"input": "11 3",
"output": "5"
},
{
"input": "11 4",
"output": "5"
},
{
"input": "11 5",
"output": "5"
},
{
"input": "11 6",
"output": "5"
},
{
"input": "11 7",
"output": "6"
},
{
"input": "12 1",
"output": "5"
},
{
"input": "12 2",
"output": "5"
},
{
"input": "12 3",
"output": "5"
},
{
"input": "12 4",
"output": "5"
},
{
"input": "12 5",
"output": "5"
},
{
"input": "12 6",
"output": "6"
},
{
"input": "12 7",
"output": "6"
},
{
"input": "1 4",
"output": "5"
},
{
"input": "1 5",
"output": "5"
},
{
"input": "9 7",
"output": "6"
},
{
"input": "2 6",
"output": "5"
},
{
"input": "1 6",
"output": "6"
},
{
"input": "2 2",
"output": "5"
},
{
"input": "4 7",
"output": "6"
},
{
"input": "12 6",
"output": "6"
},
{
"input": "12 3",
"output": "5"
},
{
"input": "3 6",
"output": "6"
},
{
"input": "9 6",
"output": "5"
},
{
"input": "7 6",
"output": "6"
},
{
"input": "11 7",
"output": "6"
},
{
"input": "6 6",
"output": "5"
}
] | 1,685,201,702
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 104
| 46
| 0
|
from math import ceil
m, d = map(int,input().split())
a = [31,28,31,30,31,30,31,31,30,31,30,31]
n = a[m-1]
print(ceil((n-(8-d))/7)+1)
|
Title: Petr and a calendar
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petr wants to make a calendar for current month. For this purpose he draws a table in which columns correspond to weeks (a week is seven consequent days from Monday to Sunday), rows correspond to weekdays, and cells contain dates. For example, a calendar for January 2017 should look like on the picture:
Petr wants to know how many columns his table should have given the month and the weekday of the first date of that month? Assume that the year is non-leap.
Input Specification:
The only line contain two integers *m* and *d* (1<=β€<=*m*<=β€<=12, 1<=β€<=*d*<=β€<=7)Β β the number of month (January is the first month, December is the twelfth) and the weekday of the first date of this month (1 is Monday, 7 is Sunday).
Output Specification:
Print single integer: the number of columns the table should have.
Demo Input:
['1 7\n', '1 1\n', '11 6\n']
Demo Output:
['6\n', '5\n', '5\n']
Note:
The first example corresponds to the January 2017 shown on the picture in the statements.
In the second example 1-st January is Monday, so the whole month fits into 5 columns.
In the third example 1-st November is Saturday and 5 columns is enough.
|
```python
from math import ceil
m, d = map(int,input().split())
a = [31,28,31,30,31,30,31,31,30,31,30,31]
n = a[m-1]
print(ceil((n-(8-d))/7)+1)
```
| 3
|
|
894
|
A
|
QAQ
|
PROGRAMMING
| 800
|
[
"brute force",
"dp"
] | null | null |
"QAQ" is a word to denote an expression of crying. Imagine "Q" as eyes with tears and "A" as a mouth.
Now Diamond has given Bort a string consisting of only uppercase English letters of length *n*. There is a great number of "QAQ" in the string (Diamond is so cute!).
Bort wants to know how many subsequences "QAQ" are in the string Diamond has given. Note that the letters "QAQ" don't have to be consecutive, but the order of letters should be exact.
|
The only line contains a string of length *n* (1<=β€<=*n*<=β€<=100). It's guaranteed that the string only contains uppercase English letters.
|
Print a single integerΒ β the number of subsequences "QAQ" in the string.
|
[
"QAQAQYSYIOIWIN\n",
"QAQQQZZYNOIWIN\n"
] |
[
"4\n",
"3\n"
] |
In the first example there are 4 subsequences "QAQ": "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN".
| 500
|
[
{
"input": "QAQAQYSYIOIWIN",
"output": "4"
},
{
"input": "QAQQQZZYNOIWIN",
"output": "3"
},
{
"input": "QA",
"output": "0"
},
{
"input": "IAQVAQZLQBQVQFTQQQADAQJA",
"output": "24"
},
{
"input": "QQAAQASGAYAAAAKAKAQIQEAQAIAAIAQQQQQ",
"output": "378"
},
{
"input": "AMVFNFJIAVNQJWIVONQOAOOQSNQSONOASONAONQINAONAOIQONANOIQOANOQINAONOQINAONOXJCOIAQOAOQAQAQAQAQWWWAQQAQ",
"output": "1077"
},
{
"input": "AAQQAXBQQBQQXBNQRJAQKQNAQNQVDQASAGGANQQQQTJFFQQQTQQA",
"output": "568"
},
{
"input": "KAZXAVLPJQBQVQQQQQAPAQQGQTQVZQAAAOYA",
"output": "70"
},
{
"input": "W",
"output": "0"
},
{
"input": "DBA",
"output": "0"
},
{
"input": "RQAWNACASAAKAGAAAAQ",
"output": "10"
},
{
"input": "QJAWZAAOAAGIAAAAAOQATASQAEAAAAQFQQHPA",
"output": "111"
},
{
"input": "QQKWQAQAAAAAAAAGAAVAQUEQQUMQMAQQQNQLAMAAAUAEAAEMAAA",
"output": "411"
},
{
"input": "QQUMQAYAUAAGWAAAQSDAVAAQAAAASKQJJQQQQMAWAYYAAAAAAEAJAXWQQ",
"output": "625"
},
{
"input": "QORZOYAQ",
"output": "1"
},
{
"input": "QCQAQAGAWAQQQAQAVQAQQQQAQAQQQAQAAATQAAVAAAQQQQAAAUUQAQQNQQWQQWAQAAQQKQYAQAAQQQAAQRAQQQWBQQQQAPBAQGQA",
"output": "13174"
},
{
"input": "QQAQQAKQFAQLQAAWAMQAZQAJQAAQQOACQQAAAYANAQAQQAQAAQQAOBQQJQAQAQAQQQAAAAABQQQAVNZAQQQQAMQQAFAAEAQAQHQT",
"output": "10420"
},
{
"input": "AQEGQHQQKQAQQPQKAQQQAAAAQQQAQEQAAQAAQAQFSLAAQQAQOQQAVQAAAPQQAWAQAQAFQAXAQQQQTRLOQAQQJQNQXQQQQSQVDQQQ",
"output": "12488"
},
{
"input": "QNQKQQQLASQBAVQQQQAAQQOQRJQQAQQQEQZUOANAADAAQQJAQAQARAAAQQQEQBHTQAAQAAAAQQMKQQQIAOJJQQAQAAADADQUQQQA",
"output": "9114"
},
{
"input": "QQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQ",
"output": "35937"
},
{
"input": "AMQQAAQAAQAAAAAAQQQBOAAANAAKQJCYQAE",
"output": "254"
},
{
"input": "AYQBAEQGAQEOAKGIXLQJAIAKQAAAQPUAJAKAATFWQQAOQQQUFQYAQQMQHOKAAJXGFCARAQSATHAUQQAATQJJQDQRAANQQAE",
"output": "2174"
},
{
"input": "AAQXAAQAYQAAAAGAQHVQYAGIVACADFAAQAAAAQZAAQMAKZAADQAQDAAQDAAAMQQOXYAQQQAKQBAAQQKAXQBJZDDLAAHQQ",
"output": "2962"
},
{
"input": "AYQQYAVAMNIAUAAKBBQVACWKTQSAQZAAQAAASZJAWBCAALAARHACQAKQQAQAARPAQAAQAQAAZQUSHQAMFVFZQQQQSAQQXAA",
"output": "2482"
},
{
"input": "LQMAQQARQAQBJQQQAGAAZQQXALQQAARQAQQQQAAQQAQQQAQQCAQQAQQAYQQQRAAZATQALYQQAAHHAAQHAAAAAAAAQQMAAQNAKQ",
"output": "7768"
},
{
"input": "MAQQWAQOYQMAAAQAQPQZAOAAQAUAQNAAQAAAITQSAQAKAQKAQQWSQAAQQAGUCDQMQWKQUXKWQQAAQQAAQQZQDQQQAABXQUUXQOA",
"output": "5422"
},
{
"input": "QTAAQDAQXAQQJQQQGAAAQQQQSBQZKAQQAQQQQEAQNUQBZCQLYQZQEQQAAQHQVAORKQVAQYQNASZQAARZAAGAAAAOQDCQ",
"output": "3024"
},
{
"input": "QQWAQQGQQUZQQQLZAAQYQXQVAQFQUAQZUQZZQUKBHSHTQYLQAOQXAQQGAQQTQOAQARQADAJRAAQPQAQQUQAUAMAUVQAAAQQAWQ",
"output": "4527"
},
{
"input": "QQAAQQAQVAQZQQQQAOEAQZPQIBQZACQQAFQQLAAQDATZQANHKYQQAQTAAFQRQAIQAJPWQAQTEIRXAEQQAYWAAAUKQQAQAQQQSQQH",
"output": "6416"
},
{
"input": "AQQQQAQAAQQAQAQAAAAAAAAAQAQAAAAAQAQAQQQAQQQAAAQQQAAAAAAAQAAAAQQQQQQQAQQQQAQAAAQAAAAAQAQAAAAAQAQAAAA",
"output": "14270"
},
{
"input": "AQQQQAQAAQQAQAQAAAAAAAAAQAQAAAAAQAQAQQQAQQQAAAQQQAAAAAAAQAAAAQQQQQQQAQQQQAQAAAQAAAAAQAQAAAAAQ",
"output": "13136"
},
{
"input": "AQQQQAQAAQQAQAQAAAAAAAAAQAQAAAAAQAQAQQQAQQQAAAQQQAAAAAAAQAAAAQQQQQQQAQQQQAQAAAQAAAAAQAQAAAAAQAQAA",
"output": "14270"
},
{
"input": "AQQQQAQAAQQAQAQAAAAAAAAAQAQAAAAAQAQAQQQAQQQAAAQQQAAAAAAAQAAAAQQQQQQQAQQQQAQAAAQAAAAAQAQAAAAAQQAA",
"output": "14231"
},
{
"input": "QQQQQAQAAQQAQAQAAAAAAAAAQAQAAAAAQAQAQQQAQQQAAAQQQAAAAAAAQAAAAQQQQQQQAQQQQAQAAAQAAAAAQAQAAAAAQAQAA",
"output": "15296"
},
{
"input": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQ",
"output": "0"
},
{
"input": "QQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA",
"output": "0"
},
{
"input": "QQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQ",
"output": "0"
},
{
"input": "QAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQA",
"output": "20825"
},
{
"input": "AQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQ",
"output": "20825"
},
{
"input": "Q",
"output": "0"
},
{
"input": "A",
"output": "0"
},
{
"input": "FFF",
"output": "0"
},
{
"input": "AAAAAA",
"output": "0"
}
] | 1,614,586,143
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 43
| 77
| 307,200
|
import sys
word = str(sys.stdin.readline().strip())
letters = []
cnt = 0
for i in word:
if i == 'A' or i == 'Q':
letters.append(i)
for i in range(len(letters)):
if letters[i] == 'Q':
for j in range(i, len(letters)):
if letters[j] == 'A':
for k in range(j, len(letters)):
if letters[k] == 'Q':
cnt += 1
print(cnt)
|
Title: QAQ
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
"QAQ" is a word to denote an expression of crying. Imagine "Q" as eyes with tears and "A" as a mouth.
Now Diamond has given Bort a string consisting of only uppercase English letters of length *n*. There is a great number of "QAQ" in the string (Diamond is so cute!).
Bort wants to know how many subsequences "QAQ" are in the string Diamond has given. Note that the letters "QAQ" don't have to be consecutive, but the order of letters should be exact.
Input Specification:
The only line contains a string of length *n* (1<=β€<=*n*<=β€<=100). It's guaranteed that the string only contains uppercase English letters.
Output Specification:
Print a single integerΒ β the number of subsequences "QAQ" in the string.
Demo Input:
['QAQAQYSYIOIWIN\n', 'QAQQQZZYNOIWIN\n']
Demo Output:
['4\n', '3\n']
Note:
In the first example there are 4 subsequences "QAQ": "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN".
|
```python
import sys
word = str(sys.stdin.readline().strip())
letters = []
cnt = 0
for i in word:
if i == 'A' or i == 'Q':
letters.append(i)
for i in range(len(letters)):
if letters[i] == 'Q':
for j in range(i, len(letters)):
if letters[j] == 'A':
for k in range(j, len(letters)):
if letters[k] == 'Q':
cnt += 1
print(cnt)
```
| 3
|
|
109
|
A
|
Lucky Sum of Digits
|
PROGRAMMING
| 1,000
|
[
"brute force",
"implementation"
] |
A. Lucky Sum of Digits
|
2
|
256
|
Petya loves lucky numbers. We all know that 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.
Petya wonders eagerly what minimum lucky number has the sum of digits equal to *n*. Help him cope with the task.
|
The single line contains an integer *n* (1<=β€<=*n*<=β€<=106) β the sum of digits of the required lucky number.
|
Print on the single line the result β the minimum lucky number, whose sum of digits equals *n*. If such number does not exist, print -1.
|
[
"11\n",
"10\n"
] |
[
"47\n",
"-1\n"
] |
none
| 500
|
[
{
"input": "11",
"output": "47"
},
{
"input": "10",
"output": "-1"
},
{
"input": "64",
"output": "4477777777"
},
{
"input": "1",
"output": "-1"
},
{
"input": "4",
"output": "4"
},
{
"input": "7",
"output": "7"
},
{
"input": "12",
"output": "444"
},
{
"input": "1000000",
"output": "4477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "47",
"output": "44477777"
},
{
"input": "100",
"output": "4444777777777777"
},
{
"input": "700",
"output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777"
},
{
"input": "485",
"output": "44447777777777777777777777777777777777777777777777777777777777777777777"
},
{
"input": "111",
"output": "444447777777777777"
},
{
"input": "85",
"output": "4477777777777"
},
{
"input": "114",
"output": "444477777777777777"
},
{
"input": "474",
"output": "444777777777777777777777777777777777777777777777777777777777777777777"
},
{
"input": "74",
"output": "47777777777"
},
{
"input": "1000",
"output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777"
},
{
"input": "1024",
"output": "4444777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777"
},
{
"input": "4444",
"output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "45784",
"output": "4777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "10000",
"output": "4777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "9854",
"output": "4447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "186",
"output": "477777777777777777777777777"
},
{
"input": "10416",
"output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "10417",
"output": "4477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "3840",
"output": "4777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "100000",
"output": "4447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "9876",
"output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "99999",
"output": "4777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "777777",
"output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "854759",
"output": "4444447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "11000",
"output": "4444447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "18951",
"output": "4444777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "999999",
"output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "888887",
"output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "999998",
"output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "40008",
"output": "4444447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "10691",
"output": "4444777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "5",
"output": "-1"
},
{
"input": "6",
"output": "-1"
},
{
"input": "9",
"output": "-1"
},
{
"input": "8",
"output": "44"
},
{
"input": "2",
"output": "-1"
},
{
"input": "3",
"output": "-1"
},
{
"input": "999997",
"output": "4447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "999996",
"output": "4777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "999990",
"output": "4447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "999980",
"output": "4444777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "800000",
"output": "4447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "980000",
"output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
}
] | 1,585,363,623
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 51
| 342
| 307,200
|
n = input()
n = int(n)
four_amount = seven_amount = current_seven = 0
k = n
while k >= 0:
if k%4 == 0:
four_amount=k//4
seven_amount = current_seven
current_seven += 1
k = n - 7*current_seven
if four_amount == 0 and seven_amount == 0:
print(-1)
else:
print('4'*four_amount + '7'*seven_amount)
|
Title: Lucky Sum of Digits
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Petya loves lucky numbers. We all know that 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.
Petya wonders eagerly what minimum lucky number has the sum of digits equal to *n*. Help him cope with the task.
Input Specification:
The single line contains an integer *n* (1<=β€<=*n*<=β€<=106) β the sum of digits of the required lucky number.
Output Specification:
Print on the single line the result β the minimum lucky number, whose sum of digits equals *n*. If such number does not exist, print -1.
Demo Input:
['11\n', '10\n']
Demo Output:
['47\n', '-1\n']
Note:
none
|
```python
n = input()
n = int(n)
four_amount = seven_amount = current_seven = 0
k = n
while k >= 0:
if k%4 == 0:
four_amount=k//4
seven_amount = current_seven
current_seven += 1
k = n - 7*current_seven
if four_amount == 0 and seven_amount == 0:
print(-1)
else:
print('4'*four_amount + '7'*seven_amount)
```
| 3.913928
|
344
|
A
|
Magnets
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.
Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own.
Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed.
|
The first line of the input contains an integer *n* (1<=β€<=*n*<=β€<=100000) β the number of magnets. Then *n* lines follow. The *i*-th line (1<=β€<=*i*<=β€<=*n*) contains either characters "01", if Mike put the *i*-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position.
|
On the single line of the output print the number of groups of magnets.
|
[
"6\n10\n10\n10\n01\n10\n10\n",
"4\n01\n01\n10\n10\n"
] |
[
"3\n",
"2\n"
] |
The first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.
The second testcase has two groups, each consisting of two magnets.
| 500
|
[
{
"input": "6\n10\n10\n10\n01\n10\n10",
"output": "3"
},
{
"input": "4\n01\n01\n10\n10",
"output": "2"
},
{
"input": "1\n10",
"output": "1"
},
{
"input": "2\n01\n10",
"output": "2"
},
{
"input": "2\n10\n10",
"output": "1"
},
{
"input": "3\n10\n01\n10",
"output": "3"
},
{
"input": "1\n01",
"output": "1"
},
{
"input": "2\n01\n01",
"output": "1"
},
{
"input": "2\n10\n01",
"output": "2"
},
{
"input": "3\n01\n01\n01",
"output": "1"
},
{
"input": "3\n10\n10\n01",
"output": "2"
},
{
"input": "3\n01\n10\n10",
"output": "2"
},
{
"input": "115\n10\n10\n10\n10\n01\n01\n10\n10\n10\n01\n01\n10\n01\n01\n10\n10\n10\n01\n10\n01\n10\n10\n01\n01\n10\n10\n10\n10\n01\n10\n01\n01\n10\n10\n10\n10\n01\n10\n10\n10\n01\n10\n01\n10\n10\n10\n10\n01\n01\n01\n10\n10\n01\n01\n01\n10\n10\n01\n10\n01\n01\n01\n01\n10\n10\n01\n10\n01\n01\n01\n01\n01\n10\n01\n10\n10\n01\n01\n01\n10\n01\n01\n10\n10\n01\n01\n01\n01\n01\n10\n01\n10\n01\n10\n01\n01\n01\n10\n01\n10\n10\n01\n10\n10\n01\n01\n01\n10\n10\n10\n10\n10\n10\n10\n10",
"output": "55"
}
] | 1,694,357,660
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 31
| 310
| 7,680,000
|
n=int(input())
magnets=[input() for i in range(n)]
groups=1
for i in range(1,n):
if magnets[i]!=magnets[i-1]:
groups+=1
print(groups)
|
Title: Magnets
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.
Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own.
Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed.
Input Specification:
The first line of the input contains an integer *n* (1<=β€<=*n*<=β€<=100000) β the number of magnets. Then *n* lines follow. The *i*-th line (1<=β€<=*i*<=β€<=*n*) contains either characters "01", if Mike put the *i*-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position.
Output Specification:
On the single line of the output print the number of groups of magnets.
Demo Input:
['6\n10\n10\n10\n01\n10\n10\n', '4\n01\n01\n10\n10\n']
Demo Output:
['3\n', '2\n']
Note:
The first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.
The second testcase has two groups, each consisting of two magnets.
|
```python
n=int(input())
magnets=[input() for i in range(n)]
groups=1
for i in range(1,n):
if magnets[i]!=magnets[i-1]:
groups+=1
print(groups)
```
| 3
|
|
629
|
A
|
Far Relativeβs Birthday Cake
|
PROGRAMMING
| 800
|
[
"brute force",
"combinatorics",
"constructive algorithms",
"implementation"
] | null | null |
Door's family is going celebrate Famil Doors's birthday party. They love Famil Door so they are planning to make his birthday cake weird!
The cake is a *n*<=Γ<=*n* square consisting of equal squares with side length 1. Each square is either empty or consists of a single chocolate. They bought the cake and randomly started to put the chocolates on the cake. The value of Famil Door's happiness will be equal to the number of pairs of cells with chocolates that are in the same row or in the same column of the cake. Famil Doors's family is wondering what is the amount of happiness of Famil going to be?
Please, note that any pair can be counted no more than once, as two different cells can't share both the same row and the same column.
|
In the first line of the input, you are given a single integer *n* (1<=β€<=*n*<=β€<=100)Β β the length of the side of the cake.
Then follow *n* lines, each containing *n* characters. Empty cells are denoted with '.', while cells that contain chocolates are denoted by 'C'.
|
Print the value of Famil Door's happiness, i.e. the number of pairs of chocolate pieces that share the same row or the same column.
|
[
"3\n.CC\nC..\nC.C\n",
"4\nCC..\nC..C\n.CC.\n.CC.\n"
] |
[
"4\n",
"9\n"
] |
If we number rows from top to bottom and columns from left to right, then, pieces that share the same row in the first sample are:
1. (1,β2) and (1,β3) 1. (3,β1) and (3,β3) 1. (2,β1) and (3,β1) 1. (1,β3) and (3,β3)
| 500
|
[
{
"input": "3\n.CC\nC..\nC.C",
"output": "4"
},
{
"input": "4\nCC..\nC..C\n.CC.\n.CC.",
"output": "9"
},
{
"input": "5\n.CCCC\nCCCCC\n.CCC.\nCC...\n.CC.C",
"output": "46"
},
{
"input": "7\n.CC..CC\nCC.C..C\nC.C..C.\nC...C.C\nCCC.CCC\n.CC...C\n.C.CCC.",
"output": "84"
},
{
"input": "8\n..C....C\nC.CCC.CC\n.C..C.CC\nCC......\nC..C..CC\nC.C...C.\nC.C..C..\nC...C.C.",
"output": "80"
},
{
"input": "9\n.C...CCCC\nC.CCCC...\n....C..CC\n.CC.CCC..\n.C.C..CC.\nC...C.CCC\nCCC.C...C\nCCCC....C\n..C..C..C",
"output": "144"
},
{
"input": "10\n..C..C.C..\n..CC..C.CC\n.C.C...C.C\n..C.CC..CC\n....C..C.C\n...C..C..C\nCC.CC....C\n..CCCC.C.C\n..CC.CCC..\nCCCC..C.CC",
"output": "190"
},
{
"input": "11\nC.CC...C.CC\nCC.C....C.C\n.....C..CCC\n....C.CC.CC\nC..C..CC...\nC...C...C..\nCC..CCC.C.C\n..C.CC.C..C\nC...C.C..CC\n.C.C..CC..C\n.C.C.CC.C..",
"output": "228"
},
{
"input": "21\n...CCC.....CC..C..C.C\n..CCC...CC...CC.CCC.C\n....C.C.C..CCC..C.C.C\n....CCC..C..C.CC.CCC.\n...CCC.C..C.C.....CCC\n.CCC.....CCC..C...C.C\nCCCC.C...CCC.C...C.CC\nC..C...C.CCC..CC..C..\nC...CC..C.C.CC..C.CC.\nCC..CCCCCCCCC..C....C\n.C..CCCC.CCCC.CCC...C\nCCC...CCC...CCC.C..C.\n.CCCCCCCC.CCCC.CC.C..\n.C.C..C....C.CCCCCC.C\n...C...C.CCC.C.CC..C.\nCCC...CC..CC...C..C.C\n.CCCCC...C.C..C.CC.C.\n..CCC.C.C..CCC.CCC...\n..C..C.C.C.....CC.C..\n.CC.C...C.CCC.C....CC\n...C..CCCC.CCC....C..",
"output": "2103"
},
{
"input": "20\nC.C.CCC.C....C.CCCCC\nC.CC.C..CCC....CCCC.\n.CCC.CC...CC.CCCCCC.\n.C...CCCC..C....CCC.\n.C..CCCCCCC.C.C.....\nC....C.C..CCC.C..CCC\n...C.C.CC..CC..CC...\nC...CC.C.CCCCC....CC\n.CC.C.CCC....C.CCC.C\nCC...CC...CC..CC...C\nC.C..CC.C.CCCC.C.CC.\n..CCCCC.C.CCC..CCCC.\n....C..C..C.CC...C.C\nC..CCC..CC..C.CC..CC\n...CC......C.C..C.C.\nCC.CCCCC.CC.CC...C.C\n.C.CC..CC..CCC.C.CCC\nC..C.CC....C....C...\n..CCC..CCC...CC..C.C\n.C.CCC.CCCCCCCCC..CC",
"output": "2071"
},
{
"input": "17\nCCC..C.C....C.C.C\n.C.CC.CC...CC..C.\n.CCCC.CC.C..CCC.C\n...CCC.CC.CCC.C.C\nCCCCCCCC..C.CC.CC\n...C..C....C.CC.C\nCC....CCC...C.CC.\n.CC.C.CC..C......\n.CCCCC.C.CC.CCCCC\n..CCCC...C..CC..C\nC.CC.C.CC..C.C.C.\nC..C..C..CCC.C...\n.C..CCCC..C......\n.CC.C...C..CC.CC.\nC..C....CC...CC..\nC.CC.CC..C.C..C..\nCCCC...C.C..CCCC.",
"output": "1160"
},
{
"input": "15\nCCCC.C..CCC....\nCCCCCC.CC.....C\n...C.CC.C.C.CC.\nCCCCCCC..C..C..\nC..CCC..C.CCCC.\n.CC..C.C.C.CC.C\n.C.C..C..C.C..C\n...C...C..CCCC.\n.....C.C..CC...\nCC.C.C..CC.C..C\n..CCCCC..CCC...\nCC.CC.C..CC.CCC\n..CCC...CC.C..C\nCC..C.C..CCC..C\n.C.C....CCC...C",
"output": "789"
},
{
"input": "1\n.",
"output": "0"
},
{
"input": "3\n.CC\nC..\nC.C",
"output": "4"
},
{
"input": "13\nC.C...C.C.C..\nCC.CCCC.CC..C\n.C.CCCCC.CC..\nCCCC..C...C..\n...CC.C.C...C\n.CC.CCC...CC.\nCC.CCCCCC....\n.C...C..CC..C\nCCCC.CC...C..\n.C.CCC..C.CC.\n..C...CC..C.C\n..C.CCC..CC.C\n.C...CCC.CC.C",
"output": "529"
},
{
"input": "16\n.C.C.C.C.C...C.C\n..C..C.CCCCCC...\n..C.C.C.C..C..C.\n.CC....C.CCC..C.\n.C.CCC..C....CCC\nCC..C.CC..C.C.CC\n...C..C..CC..CC.\n.CCC..C.CC.C.C..\n.CC.C..........C\nC...C....CC..C..\nC.CCC.C..C..C...\n.CCCCCCCCCCCC..C\n..C.C.CC.CC.CCC.\nCC..C.C....C..CC\nC.CCC..C..C.C.CC\n.C.CCC.CC..CCC.C",
"output": "874"
},
{
"input": "2\nCC\nCC",
"output": "4"
},
{
"input": "3\nC..\nC..\nC..",
"output": "3"
}
] | 1,593,285,065
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 48
| 109
| 6,963,200
|
def pairs(n):
return n*(n-1)//2
n = int(input())
table = []
count_col = [0 for i in range(n)]
for i in range(n):
table.append([x for x in input()])
for j in range(n):
if table[-1][j] == 'C':
count_col[j] +=1
count_row = [row.count('C') for row in table]
happiness = [pairs(n) for n in count_row+count_col if n>=2]
print(sum(happiness))
|
Title: Far Relativeβs Birthday Cake
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Door's family is going celebrate Famil Doors's birthday party. They love Famil Door so they are planning to make his birthday cake weird!
The cake is a *n*<=Γ<=*n* square consisting of equal squares with side length 1. Each square is either empty or consists of a single chocolate. They bought the cake and randomly started to put the chocolates on the cake. The value of Famil Door's happiness will be equal to the number of pairs of cells with chocolates that are in the same row or in the same column of the cake. Famil Doors's family is wondering what is the amount of happiness of Famil going to be?
Please, note that any pair can be counted no more than once, as two different cells can't share both the same row and the same column.
Input Specification:
In the first line of the input, you are given a single integer *n* (1<=β€<=*n*<=β€<=100)Β β the length of the side of the cake.
Then follow *n* lines, each containing *n* characters. Empty cells are denoted with '.', while cells that contain chocolates are denoted by 'C'.
Output Specification:
Print the value of Famil Door's happiness, i.e. the number of pairs of chocolate pieces that share the same row or the same column.
Demo Input:
['3\n.CC\nC..\nC.C\n', '4\nCC..\nC..C\n.CC.\n.CC.\n']
Demo Output:
['4\n', '9\n']
Note:
If we number rows from top to bottom and columns from left to right, then, pieces that share the same row in the first sample are:
1. (1,β2) and (1,β3) 1. (3,β1) and (3,β3) 1. (2,β1) and (3,β1) 1. (1,β3) and (3,β3)
|
```python
def pairs(n):
return n*(n-1)//2
n = int(input())
table = []
count_col = [0 for i in range(n)]
for i in range(n):
table.append([x for x in input()])
for j in range(n):
if table[-1][j] == 'C':
count_col[j] +=1
count_row = [row.count('C') for row in table]
happiness = [pairs(n) for n in count_row+count_col if n>=2]
print(sum(happiness))
```
| 3
|
|
276
|
B
|
Little Girl and Game
|
PROGRAMMING
| 1,300
|
[
"games",
"greedy"
] | null | null |
The Little Girl loves problems on games very much. Here's one of them.
Two players have got a string *s*, consisting of lowercase English letters. They play a game that is described by the following rules:
- The players move in turns; In one move the player can remove an arbitrary letter from string *s*. - If the player before his turn can reorder the letters in string *s* so as to get a palindrome, this player wins. A palindrome is a string that reads the same both ways (from left to right, and vice versa). For example, string "abba" is a palindrome and string "abc" isn't.
Determine which player will win, provided that both sides play optimally well β the one who moves first or the one who moves second.
|
The input contains a single line, containing string *s* (1<=β€<=|*s*|<=<=β€<=<=103). String *s* consists of lowercase English letters.
|
In a single line print word "First" if the first player wins (provided that both players play optimally well). Otherwise, print word "Second". Print the words without the quotes.
|
[
"aba\n",
"abca\n"
] |
[
"First\n",
"Second\n"
] |
none
| 1,000
|
[
{
"input": "aba",
"output": "First"
},
{
"input": "abca",
"output": "Second"
},
{
"input": "aabb",
"output": "First"
},
{
"input": "ctjxzuimsxnarlciuynqeoqmmbqtagszuo",
"output": "Second"
},
{
"input": "gevqgtaorjixsxnbcoybr",
"output": "First"
},
{
"input": "xvhtcbtouuddhylxhplgjxwlo",
"output": "First"
},
{
"input": "knaxhkbokmtfvnjvlsbrfoefpjpkqwlumeqqbeohodnwevhllkylposdpjuoizyunuxivzrjofiyxxiliuwhkjqpkqxukxroivfhikxjdtwcqngqswptdwrywxszxrqojjphzwzxqftnfhkapeejdgckfyrxtpuipfljsjwgpjfatmxpylpnerllshuvkbomlpghjrxcgxvktgeyuhrcwgvdmppqnkdmjtxukzlzqhfbgrishuhkyggkpstvqabpxoqjuovwjwcmazmvpfpnljdgpokpatjnvwacotkvxheorzbsrazldsquijzkmtmqahakjrjvzkquvayxpqrmqqcknilpqpjapagezonfpz",
"output": "Second"
},
{
"input": "desktciwoidfuswycratvovutcgjrcyzmilsmadzaegseetexygedzxdmorxzxgiqhcuppshcsjcozkopebegfmxzxxagzwoymlghgjexcgfojychyt",
"output": "First"
},
{
"input": "gfhuidxgxpxduqrfnqrnefgtyxgmrtehmddjkddwdiayyilaknxhlxszeslnsjpcrwnoqubmbpcehiftteirkfvbtfyibiikdaxmondnawtvqccctdxrjcfxqwqhvvrqmhqflbzskrayvruqvqijrmikucwzodxvufwxpxxjxlifdjzxrttjzatafkbzsjupsiefmipdufqltedjlytphzppoevxawjdhbxgennevbvdgpoeihasycctyddenzypoprchkoioouhcexjqwjflxvkgpgjatstlmledxasecfhwvabzwviywsiaryqrxyeceefblherqjevdzkfxslqiytwzz",
"output": "First"
},
{
"input": "fezzkpyctjvvqtncmmjsitrxaliyhirspnjjngvzdoudrkkvvdiwcwtcxobpobzukegtcrwsgxxzlcphdxkbxdximqbycaicfdeqlvzboptfimkzvjzdsvahorqqhcirpkhtwjkplitpacpkpbhnxtoxuoqsxcxnhtrmzvexmpvlethbkvmlzftimjnidrzvcunbpysvukzgwghjmwrvstsunaocnoqohcsggtrwxiworkliqejajewbrtdwgnyynpupbrrvtfqtlaaq",
"output": "Second"
},
{
"input": "tsvxmeixijyavdalmrvscwohzubhhgsocdvnjmjtctojbxxpezzbgfltixwgzmkfwdnlhidhrdgyajggmrvmwaoydodjmzqvgabyszfqcuhwdncyfqvmackvijgpjyiauxljvvwgiofdxccwmybdfcfcrqppbvbagmnvvvhngxauwbpourviyfokwjweypzzrrzjcmddnpoaqgqfgglssjnlshrerfffmrwhapzknxveiqixflykjbnpivogtdpyjakwrdoklsbvbkjhdojfnuwbpcfdycwxecysbyjfvoykxsxgg",
"output": "First"
},
{
"input": "upgqmhfmfnodsyosgqswugfvpdxhtkxvhlsxrjiqlojchoddxkpsamwmuvopdbncymcgrkurwlxerexgswricuqxhvqvgekeofkgqabypamozmyjyfvpifsaotnyzqydcenphcsmplekinwkmwzpjnlapfdbhxjdcnarlgkfgxzfbpgsuxqfyhnxjhtojrlnprnxprfbkkcyriqztjeeepkzgzcaiutvbqqofyhddfebozhvtvrigtidxqmydjxegxipakzjcnenjkdroyjmxugj",
"output": "Second"
},
{
"input": "aaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbbbbbbbbccccccccccccccccccccddddddddddeeeeeeeeeeffffgggghhhhiiiijjjjqqqqwwwweeeerrrrttttyyyyuuuuiiiiooooppppaaaassssddddffffgggghhhhjjjjkkkkllllzzzzxxxxccccvvvvbbbbnnnnmmmm",
"output": "First"
},
{
"input": "vnvtvnxjrtffdhrfvczzoyeokjabxcilmmsrhwuakghvuabcmfpmblyroodmhfivmhqoiqhapoglwaluewhqkunzitmvijaictjdncivccedfpaezcnpwemlohbhjjlqsonuclaumgbzjamsrhuzqdqtitygggsnruuccdtxkgbdd",
"output": "First"
},
{
"input": "vqdtkbvlbdyndheoiiwqhnvcmmhnhsmwwrvesnpdfxvprqbwzbodoihrywagphlsrcbtnvppjsquuuzkjazaenienjiyctyajsqdfsdiedzugkymgzllvpxfetkwfabbiotjcknzdwsvmbbuqrxrulvgljagvxdmfsqtcczhifhoghqgffkbviphbabwiaqburerfkbqfjbptkwlahysrrfwjbqfnrgnsnsukqqcxxwqtuhvdzqmpfwrbqzdwxcaifuyhvojgurmchh",
"output": "First"
},
{
"input": "hxueikegwnrctlciwguepdsgupguykrntbszeqzzbpdlouwnmqgzcxejidstxyxhdlnttnibxstduwiflouzfswfikdudkazoefawm",
"output": "Second"
},
{
"input": "ershkhsywqftixappwqzoojtnamvqjbyfauvuubwpctspioqusnnivwsiyszfhlrskbswaiaczurygcioonjcndntwvrlaejyrghfnecltqytfmkvjxuujifgtujrqsisdawpwgttxynewiqhdhronamabysvpxankxeybcjqttbqnciwuqiehzyfjoedaradqnfthuuwrezwrkjiytpgwfwbslawbiezdbdltenjlaygwaxddplgseiaojndqjcopvolqbvnacuvfvirzbrnlnyjixngeevcggmirzatenjihpgnyfjhgsjgzepohbyhmzbatfwuorwutavlqsogrvcjpqziuifrhurq",
"output": "First"
},
{
"input": "qilwpsuxogazrfgfznngwklnioueuccyjfatjoizcctgsweitzofwkyjustizbopzwtaqxbtovkdrxeplukrcuozhpymldstbbfynkgsmafigetvzkxloxqtphvtwkgfjkiczttcsxkjpsoutdpzxytrsqgjtbdljjrbmkudrkodfvcwkcuggbsthxdyogeeyfuyhmnwgyuatfkvchavpzadfacckdurlbqjkthqbnirzzbpusxcenkpgtizayjmsahvobobudfeaewcqmrlxxnocqzmkessnguxkiccrxyvnxxlqnqfwuzmupk",
"output": "First"
},
{
"input": "opfokvwzpllctflkphutcrkferbjyyrasqqkrcvoymyrxwaudgsugcqveccymdplxmtlzfoptmrapfeizpnnhbzlkuyznwacnswibxhqunazbhdvrlidghisuqunstbuevjzimvlfvopgqxrvahhngnaumgywscfrfwfpnfxwhfrelbunmedvkssykwjyartxjiplerntzkpiiaalijiwhyuhxlvhxpkgfypvrpqqsacuwocdampnnhvibsbolyduvscsjfayxpldyvqzjbqojjxdvxtctwtifcdfcaoonzbgegyllngwvjivtxeezoabhsihcuvgvdsgjtzbzwovjshvwrljkxavowovzrfpdufpogdtujkerdorpboufrxhyswirjehgsxthlvjufvahdpeajidqipwaxokbeycrievgcfkyuyavaq",
"output": "Second"
},
{
"input": "a",
"output": "First"
},
{
"input": "ab",
"output": "Second"
},
{
"input": "abacaba",
"output": "First"
},
{
"input": "abazaba",
"output": "First"
},
{
"input": "zz",
"output": "First"
},
{
"input": "aassddxyz",
"output": "First"
},
{
"input": "abcdefghijklmnopqrstuvwxyz",
"output": "Second"
},
{
"input": "aabc",
"output": "Second"
},
{
"input": "abcabc",
"output": "First"
},
{
"input": "aaabbbccdd",
"output": "Second"
},
{
"input": "aabbcccc",
"output": "First"
}
] | 1,659,964,778
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 51
| 156
| 0
|
s=input();c = 0
for i in s:c += s.count(i) % 2
if c==0 or c%2:
print("First")
else:
print("Second")
|
Title: Little Girl and Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Little Girl loves problems on games very much. Here's one of them.
Two players have got a string *s*, consisting of lowercase English letters. They play a game that is described by the following rules:
- The players move in turns; In one move the player can remove an arbitrary letter from string *s*. - If the player before his turn can reorder the letters in string *s* so as to get a palindrome, this player wins. A palindrome is a string that reads the same both ways (from left to right, and vice versa). For example, string "abba" is a palindrome and string "abc" isn't.
Determine which player will win, provided that both sides play optimally well β the one who moves first or the one who moves second.
Input Specification:
The input contains a single line, containing string *s* (1<=β€<=|*s*|<=<=β€<=<=103). String *s* consists of lowercase English letters.
Output Specification:
In a single line print word "First" if the first player wins (provided that both players play optimally well). Otherwise, print word "Second". Print the words without the quotes.
Demo Input:
['aba\n', 'abca\n']
Demo Output:
['First\n', 'Second\n']
Note:
none
|
```python
s=input();c = 0
for i in s:c += s.count(i) % 2
if c==0 or c%2:
print("First")
else:
print("Second")
```
| 3
|
|
42
|
A
|
Guilty --- to the kitchen!
|
PROGRAMMING
| 1,400
|
[
"greedy",
"implementation"
] |
A. Guilty β to the kitchen!
|
2
|
256
|
It's a very unfortunate day for Volodya today. He got bad mark in algebra and was therefore forced to do some work in the kitchen, namely to cook borscht (traditional Russian soup). This should also improve his algebra skills.
According to the borscht recipe it consists of *n* ingredients that have to be mixed in proportion litres (thus, there should be *a*1<=Β·*x*,<=...,<=*a**n*<=Β·*x* litres of corresponding ingredients mixed for some non-negative *x*). In the kitchen Volodya found out that he has *b*1,<=...,<=*b**n* litres of these ingredients at his disposal correspondingly. In order to correct his algebra mistakes he ought to cook as much soup as possible in a *V* litres volume pan (which means the amount of soup cooked can be between 0 and *V* litres). What is the volume of borscht Volodya will cook ultimately?
|
The first line of the input contains two space-separated integers *n* and *V* (1<=β€<=*n*<=β€<=20,<=1<=β€<=*V*<=β€<=10000). The next line contains *n* space-separated integers *a**i* (1<=β€<=*a**i*<=β€<=100). Finally, the last line contains *n* space-separated integers *b**i* (0<=β€<=*b**i*<=β€<=100).
|
Your program should output just one real number β the volume of soup that Volodya will cook. Your answer must have a relative or absolute error less than 10<=-<=4.
|
[
"1 100\n1\n40\n",
"2 100\n1 1\n25 30\n",
"2 100\n1 1\n60 60\n"
] |
[
"40.0\n",
"50.0\n",
"100.0\n"
] |
none
| 500
|
[
{
"input": "1 100\n1\n40",
"output": "40.0"
},
{
"input": "2 100\n1 1\n25 30",
"output": "50.0"
},
{
"input": "2 100\n1 1\n60 60",
"output": "100.0"
},
{
"input": "2 100\n1 1\n50 50",
"output": "100.0"
},
{
"input": "2 100\n1 2\n33 66",
"output": "99.0"
},
{
"input": "3 10000\n1 1 1\n100 0 100",
"output": "0.0"
},
{
"input": "7 5100\n21 93 52 80 5 46 20\n79 37 74 54 22 15 90",
"output": "103.3695652173913"
},
{
"input": "10 2707\n80 91 41 99 99 48 81 25 80 17\n88 79 64 78 4 54 38 92 77 61",
"output": "26.70707070707071"
},
{
"input": "19 8111\n44 75 80 69 90 64 58 8 93 50 44 39 7 25 14 52 32 26 26\n38 57 38 23 73 24 4 49 0 34 96 93 14 26 29 89 54 12 24",
"output": "0.0"
},
{
"input": "5 1121\n14 37 91 35 71\n17 87 48 91 13",
"output": "45.40845070422535"
},
{
"input": "4 6054\n72 21 14 49\n43 53 42 55",
"output": "93.16666666666666"
},
{
"input": "6 8692\n20 61 56 4 78 76\n73 83 97 45 16 7",
"output": "27.171052631578945"
},
{
"input": "9 5583\n73 31 18 36 38 99 34 50 69\n48 24 75 78 75 69 13 74 3",
"output": "19.478260869565215"
},
{
"input": "1 5215\n24\n85",
"output": "85.0"
},
{
"input": "15 9559\n55 13 69 16 15 34 89 30 56 64 74 100 72 71 20\n40 73 29 12 31 5 59 5 90 13 32 75 99 7 44",
"output": "76.70422535211266"
},
{
"input": "13 2530\n83 59 19 69 8 81 99 74 14 75 61 13 36\n26 36 77 44 10 8 8 16 81 61 29 81 50",
"output": "55.83838383838385"
},
{
"input": "4 7672\n42 34 57 72\n56 7 24 24",
"output": "42.205882352941174"
},
{
"input": "17 6030\n100 77 5 87 28 50 51 64 45 79 60 80 49 20 25 91 64\n12 13 58 55 3 59 8 62 69 38 69 27 50 39 5 41 30",
"output": "104.46428571428571"
},
{
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"output": "0.0"
},
{
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"output": "209.72093023255815"
},
{
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"output": "38.0"
},
{
"input": "10 9501\n39 67 33 71 89 69 5 90 7 48\n89 91 8 68 7 54 61 66 53 51",
"output": "40.74157303370786"
},
{
"input": "1 1966\n85\n99",
"output": "99.0"
},
{
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"output": "112.64516129032259"
},
{
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"output": "180.22500000000002"
},
{
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"output": "35.0"
},
{
"input": "13 5115\n13 51 17 24 52 4 33 4 94 17 54 82 77\n40 34 90 29 81 24 38 74 28 81 14 40 24",
"output": "135.33333333333334"
},
{
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"output": "107.2258064516129"
},
{
"input": "2 775\n13 39\n76 35",
"output": "46.66666666666667"
},
{
"input": "7 8690\n73 93 32 47 80 82 97\n49 49 90 43 89 43 67",
"output": "264.2926829268293"
},
{
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"output": "175.6451612903226"
},
{
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"output": "36.317460317460316"
},
{
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"output": "113.26881720430106"
},
{
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"output": "25.685714285714283"
},
{
"input": "4 575\n24 23 16 64\n85 100 14 13",
"output": "25.796875"
},
{
"input": "9 423\n28 88 41 71 99 24 35 68 90\n7 76 44 27 64 52 92 81 98",
"output": "136.0"
},
{
"input": "2 1437\n66 58\n44 8",
"output": "17.10344827586207"
},
{
"input": "18 4733\n78 53 33 72 38 76 43 51 94 18 22 21 65 60 5 71 88 40\n5 78 50 43 81 44 10 18 23 51 52 31 10 55 63 46 82 92",
"output": "59.48717948717948"
},
{
"input": "16 7170\n17 1 48 51 28 16 41 14 59 93 25 76 46 69 74 41\n54 53 41 25 50 42 37 20 11 35 90 96 78 3 20 38",
"output": "30.391304347826086"
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{
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"output": "89.99999999999999"
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{
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"output": "96.0"
},
{
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"output": "39.45454545454545"
},
{
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"output": "135.8181818181818"
},
{
"input": "9 3269\n79 88 15 74 92 33 68 64 45\n55 84 75 50 68 32 41 82 42",
"output": "336.44117647058823"
},
{
"input": "6 1007\n93 23 35 15 25 6\n58 24 11 99 23 47",
"output": "61.91428571428571"
},
{
"input": "11 710\n2 49 56 33 79 69 64 62 64 9 87\n94 34 90 3 13 67 76 80 69 19 41",
"output": "52.18181818181819"
},
{
"input": "18 9292\n15 97 47 88 15 7 15 86 52 40 16 97 2 80 64 37 88 15\n39 47 94 12 34 17 45 39 98 99 19 8 94 50 87 68 31 6",
"output": "71.01030927835052"
},
{
"input": "11 3753\n78 75 17 65 97 36 79 56 97 62 43\n18 41 17 47 14 40 7 57 58 24 98",
"output": "62.46835443037974"
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{
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"output": "15.295454545454545"
},
{
"input": "20 1479\n69 30 15 62 81 24 5 16 25 65 47 23 62 51 87 50 6 44 88 61\n57 47 76 68 7 57 44 98 24 44 1 79 67 31 72 83 36 65 83 42",
"output": "19.382978723404253"
},
{
"input": "17 3856\n50 59 100 50 80 77 58 86 95 87 30 41 11 99 33 27 75\n47 47 39 62 58 91 55 18 65 47 8 97 31 80 61 87 66",
"output": "221.4418604651163"
},
{
"input": "9 2382\n84 51 95 66 34 77 96 9 57\n3 94 56 22 61 50 23 83 45",
"output": "20.32142857142857"
},
{
"input": "14 1751\n33 82 63 35 67 78 47 27 43 96 58 95 39 29\n42 7 15 83 95 91 60 3 85 39 7 56 39 4",
"output": "67.60975609756098"
},
{
"input": "6 8371\n34 11 24 95 62 32\n98 50 58 46 49 93",
"output": "124.92631578947369"
},
{
"input": "2 5181\n4 1\n6 33",
"output": "7.5"
},
{
"input": "9 632\n51 64 25 25 60 71 56 3 31\n70 28 76 84 86 33 77 11 69",
"output": "168.875"
},
{
"input": "3 2102\n76 15 85\n25 95 80",
"output": "57.89473684210526"
},
{
"input": "5 5005\n5 53 65 52 99\n21 49 9 3 66",
"output": "15.807692307692308"
},
{
"input": "17 8971\n54 62 7 47 48 70 78 96 91 34 84 23 72 75 72 60 21\n4 26 6 41 28 45 70 61 6 75 74 46 17 46 34 27 10",
"output": "65.53846153846153"
},
{
"input": "15 5527\n22 49 56 95 86 23 15 74 38 65 52 92 88 49 54\n33 61 71 95 69 31 30 0 1 93 66 48 65 92 11",
"output": "0.0"
},
{
"input": "20 3696\n87 22 21 83 95 31 28 96 71 25 56 40 70 79 46 87 19 19 34 25\n70 44 34 11 2 1 59 22 46 28 3 53 52 71 34 47 65 71 76 30",
"output": "21.768421052631577"
},
{
"input": "8 5540\n5 9 88 1 74 52 32 79\n17 48 99 33 68 28 2 58",
"output": "21.25"
},
{
"input": "15 303\n33 15 28 14 97 33 77 69 41 76 54 97 11 1 1\n83 70 63 11 71 10 48 65 5 5 82 2 6 79 19",
"output": "13.340206185567009"
},
{
"input": "10 9401\n4 53 39 66 52 42 65 39 1 76\n9 34 16 56 78 14 43 49 95 42",
"output": "145.66666666666666"
},
{
"input": "2 9083\n77 33\n22 22",
"output": "31.42857142857143"
},
{
"input": "16 8826\n29 21 40 93 48 49 43 96 60 68 66 5 96 49 84 44\n94 1 79 12 76 65 99 53 37 39 3 76 15 81 51 91",
"output": "40.5"
},
{
"input": "4 9426\n95 48 98 92\n65 40 43 90",
"output": "146.1122448979592"
},
{
"input": "13 175\n46 77 14 16 84 80 81 36 71 13 87 69 8\n54 46 69 59 30 72 83 97 83 96 43 94 84",
"output": "175.0"
},
{
"input": "13 5023\n11 30 92 40 26 77 33 94 71 2 70 97 50\n32 46 51 14 63 76 34 19 13 34 40 91 23",
"output": "126.88732394366197"
},
{
"input": "18 9978\n26 3 87 84 97 53 70 97 37 57 78 23 34 40 81 62 21 92\n56 73 0 79 93 14 17 80 0 20 3 81 22 71 7 82 71 81",
"output": "0.0"
},
{
"input": "14 8481\n64 2 90 76 49 30 88 32 98 64 20 85 40 35\n55 84 75 43 36 13 67 75 100 19 22 7 5 58",
"output": "63.65882352941177"
},
{
"input": "2 1674\n77 23\n23 25",
"output": "29.87012987012987"
},
{
"input": "10 2112\n45 11 32 14 82 30 34 11 42 56\n18 9 84 99 82 43 61 84 14 70",
"output": "119.0"
},
{
"input": "6 2006\n62 4 3 71 61 10\n37 45 61 84 24 15",
"output": "83.01639344262294"
},
{
"input": "8 3954\n80 77 64 1 50 21 89 26\n30 82 17 20 67 21 31 99",
"output": "108.375"
},
{
"input": "18 7253\n64 77 92 9 32 66 23 34 10 71 8 7 83 9 52 97 29 65\n46 90 65 43 44 63 7 38 38 20 62 9 53 39 17 13 5 90",
"output": "110.96907216494844"
},
{
"input": "4 4384\n42 41 85 79\n29 67 52 55",
"output": "151.1058823529412"
},
{
"input": "7 529\n77 18 67 64 43 51 30\n35 87 17 52 1 97 84",
"output": "8.139534883720929"
},
{
"input": "16 2915\n39 39 81 44 23 47 43 56 7 38 10 100 5 34 87 14\n10 96 34 20 62 88 46 38 29 35 2 43 26 55 31 63",
"output": "133.4"
},
{
"input": "14 6488\n53 41 36 28 17 15 63 33 75 40 85 88 90 100\n7 35 83 2 48 76 93 2 69 56 59 7 25 24",
"output": "46.30303030303031"
}
] | 1,669,773,563
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 80
| 92
| 0
|
n,V=map(int,input().split(" "))
alist=[int(x) for x in input().split(" ")]
blist=[int(x) for x in input().split(" ")]
#empty one item from blist
ans=0
minrate=100000
for i in range(n):
#empty ith item from blist,see if it works
rate=blist[i]/alist[i] # how many food should cook
minrate=min(minrate,rate)
total=0
for i in range(n):
total+=alist[i]*minrate
print(min(V,total))
|
Title: Guilty --- to the kitchen!
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
It's a very unfortunate day for Volodya today. He got bad mark in algebra and was therefore forced to do some work in the kitchen, namely to cook borscht (traditional Russian soup). This should also improve his algebra skills.
According to the borscht recipe it consists of *n* ingredients that have to be mixed in proportion litres (thus, there should be *a*1<=Β·*x*,<=...,<=*a**n*<=Β·*x* litres of corresponding ingredients mixed for some non-negative *x*). In the kitchen Volodya found out that he has *b*1,<=...,<=*b**n* litres of these ingredients at his disposal correspondingly. In order to correct his algebra mistakes he ought to cook as much soup as possible in a *V* litres volume pan (which means the amount of soup cooked can be between 0 and *V* litres). What is the volume of borscht Volodya will cook ultimately?
Input Specification:
The first line of the input contains two space-separated integers *n* and *V* (1<=β€<=*n*<=β€<=20,<=1<=β€<=*V*<=β€<=10000). The next line contains *n* space-separated integers *a**i* (1<=β€<=*a**i*<=β€<=100). Finally, the last line contains *n* space-separated integers *b**i* (0<=β€<=*b**i*<=β€<=100).
Output Specification:
Your program should output just one real number β the volume of soup that Volodya will cook. Your answer must have a relative or absolute error less than 10<=-<=4.
Demo Input:
['1 100\n1\n40\n', '2 100\n1 1\n25 30\n', '2 100\n1 1\n60 60\n']
Demo Output:
['40.0\n', '50.0\n', '100.0\n']
Note:
none
|
```python
n,V=map(int,input().split(" "))
alist=[int(x) for x in input().split(" ")]
blist=[int(x) for x in input().split(" ")]
#empty one item from blist
ans=0
minrate=100000
for i in range(n):
#empty ith item from blist,see if it works
rate=blist[i]/alist[i] # how many food should cook
minrate=min(minrate,rate)
total=0
for i in range(n):
total+=alist[i]*minrate
print(min(V,total))
```
| 3.977
|
79
|
B
|
Colorful Field
|
PROGRAMMING
| 1,400
|
[
"implementation",
"sortings"
] |
B. Colorful Field
|
2
|
256
|
Fox Ciel saw a large field while she was on a bus. The field was a *n*<=Γ<=*m* rectangle divided into 1<=Γ<=1 cells. Some cells were wasteland, and other each cell contained crop plants: either carrots or kiwis or grapes.
After seeing the field carefully, Ciel found that the crop plants of each cell were planted in following procedure:
- Assume that the rows are numbered 1 to *n* from top to bottom and the columns are numbered 1 to *m* from left to right, and a cell in row *i* and column *j* is represented as (*i*,<=*j*). - First, each field is either cultivated or waste. Crop plants will be planted in the cultivated cells in the order of (1,<=1)<=β<=...<=β<=(1,<=*m*)<=β<=(2,<=1)<=β<=...<=β<=(2,<=*m*)<=β<=...<=β<=(*n*,<=1)<=β<=...<=β<=(*n*,<=*m*). Waste cells will be ignored. - Crop plants (either carrots or kiwis or grapes) will be planted in each cell one after another cyclically. Carrots will be planted in the first cell, then kiwis in the second one, grapes in the third one, carrots in the forth one, kiwis in the fifth one, and so on.
The following figure will show you the example of this procedure. Here, a white square represents a cultivated cell, and a black square represents a waste cell.
Now she is wondering how to determine the crop plants in some certain cells.
|
In the first line there are four positive integers *n*,<=*m*,<=*k*,<=*t* (1<=β€<=*n*<=β€<=4Β·104,<=1<=β€<=*m*<=β€<=4Β·104,<=1<=β€<=*k*<=β€<=103,<=1<=β€<=*t*<=β€<=103), each of which represents the height of the field, the width of the field, the number of waste cells and the number of queries that ask the kind of crop plants in a certain cell.
Following each *k* lines contains two integers *a*,<=*b* (1<=β€<=*a*<=β€<=*n*,<=1<=β€<=*b*<=β€<=*m*), which denotes a cell (*a*,<=*b*) is waste. It is guaranteed that the same cell will not appear twice in this section.
Following each *t* lines contains two integers *i*,<=*j* (1<=β€<=*i*<=β€<=*n*,<=1<=β€<=*j*<=β€<=*m*), which is a query that asks you the kind of crop plants of a cell (*i*,<=*j*).
|
For each query, if the cell is waste, print Waste. Otherwise, print the name of crop plants in the cell: either Carrots or Kiwis or Grapes.
|
[
"4 5 5 6\n4 3\n1 3\n3 3\n2 5\n3 2\n1 3\n1 4\n2 3\n2 4\n1 1\n1 1\n"
] |
[
"Waste\nGrapes\nCarrots\nKiwis\nCarrots\nCarrots\n"
] |
The sample corresponds to the figure in the statement.
| 1,000
|
[
{
"input": "4 5 5 6\n4 3\n1 3\n3 3\n2 5\n3 2\n1 3\n1 4\n2 3\n2 4\n1 1\n1 1",
"output": "Waste\nGrapes\nCarrots\nKiwis\nCarrots\nCarrots"
},
{
"input": "2 3 2 2\n1 1\n2 2\n2 1\n2 2",
"output": "Grapes\nWaste"
},
{
"input": "31 31 31 4\n4 9\n16 27\n11 29\n8 28\n11 2\n10 7\n22 6\n1 25\n14 8\n9 7\n9 1\n2 3\n5 2\n21 16\n20 19\n23 14\n27 6\n25 21\n14 1\n18 14\n7 2\n19 12\n30 27\n4 27\n24 12\n25 20\n26 22\n21 17\n11 6\n5 28\n28 24\n17 30\n2 5\n30 10\n4 21",
"output": "Kiwis\nCarrots\nGrapes\nGrapes"
},
{
"input": "39898 39898 3 1\n4567 8901\n12345 23456\n24680 35679\n29292 12121",
"output": "Grapes"
},
{
"input": "1 1 1 1\n1 1\n1 1",
"output": "Waste"
},
{
"input": "1 2 1 2\n1 1\n1 2\n1 1",
"output": "Carrots\nWaste"
},
{
"input": "1 6 3 12\n1 2\n1 4\n1 6\n1 1\n1 2\n1 6\n1 2\n1 5\n1 3\n1 4\n1 5\n1 5\n1 4\n1 6\n1 3",
"output": "Carrots\nWaste\nWaste\nWaste\nGrapes\nKiwis\nWaste\nGrapes\nGrapes\nWaste\nWaste\nKiwis"
},
{
"input": "1 10 1 3\n1 5\n1 5\n1 5\n1 5",
"output": "Waste\nWaste\nWaste"
},
{
"input": "5 5 1 3\n2 2\n2 2\n2 2\n2 2",
"output": "Waste\nWaste\nWaste"
},
{
"input": "3 3 3 5\n1 1\n2 2\n3 3\n1 1\n2 2\n2 2\n2 2\n3 3",
"output": "Waste\nWaste\nWaste\nWaste\nWaste"
},
{
"input": "1 10 2 4\n1 9\n1 5\n1 5\n1 6\n1 9\n1 10",
"output": "Waste\nKiwis\nWaste\nKiwis"
},
{
"input": "5 2 2 1\n3 2\n4 2\n5 2",
"output": "Kiwis"
}
] | 1,661,604,714
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 29
| 466
| 0
|
FRUITS = ['Carrots', 'Kiwis', 'Grapes']
rn, cn, wn, qn = map(int, input().split())
waste_locations = []
for _ in range(wn):
r, c = map(int, input().split())
order = (r - 1) * cn + c - 1
waste_locations.append(order)
# waste_locations = sorted(waste_locations)
for _ in range(qn):
r, c = map(int, input().split())
fruit_order = (r - 1) * cn + c - 1
if fruit_order in waste_locations:
print("Waste")
else:
previous_waste = 0
for location in waste_locations:
if location < fruit_order:
previous_waste += 1
print(FRUITS[(fruit_order - previous_waste) % 3])
|
Title: Colorful Field
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Fox Ciel saw a large field while she was on a bus. The field was a *n*<=Γ<=*m* rectangle divided into 1<=Γ<=1 cells. Some cells were wasteland, and other each cell contained crop plants: either carrots or kiwis or grapes.
After seeing the field carefully, Ciel found that the crop plants of each cell were planted in following procedure:
- Assume that the rows are numbered 1 to *n* from top to bottom and the columns are numbered 1 to *m* from left to right, and a cell in row *i* and column *j* is represented as (*i*,<=*j*). - First, each field is either cultivated or waste. Crop plants will be planted in the cultivated cells in the order of (1,<=1)<=β<=...<=β<=(1,<=*m*)<=β<=(2,<=1)<=β<=...<=β<=(2,<=*m*)<=β<=...<=β<=(*n*,<=1)<=β<=...<=β<=(*n*,<=*m*). Waste cells will be ignored. - Crop plants (either carrots or kiwis or grapes) will be planted in each cell one after another cyclically. Carrots will be planted in the first cell, then kiwis in the second one, grapes in the third one, carrots in the forth one, kiwis in the fifth one, and so on.
The following figure will show you the example of this procedure. Here, a white square represents a cultivated cell, and a black square represents a waste cell.
Now she is wondering how to determine the crop plants in some certain cells.
Input Specification:
In the first line there are four positive integers *n*,<=*m*,<=*k*,<=*t* (1<=β€<=*n*<=β€<=4Β·104,<=1<=β€<=*m*<=β€<=4Β·104,<=1<=β€<=*k*<=β€<=103,<=1<=β€<=*t*<=β€<=103), each of which represents the height of the field, the width of the field, the number of waste cells and the number of queries that ask the kind of crop plants in a certain cell.
Following each *k* lines contains two integers *a*,<=*b* (1<=β€<=*a*<=β€<=*n*,<=1<=β€<=*b*<=β€<=*m*), which denotes a cell (*a*,<=*b*) is waste. It is guaranteed that the same cell will not appear twice in this section.
Following each *t* lines contains two integers *i*,<=*j* (1<=β€<=*i*<=β€<=*n*,<=1<=β€<=*j*<=β€<=*m*), which is a query that asks you the kind of crop plants of a cell (*i*,<=*j*).
Output Specification:
For each query, if the cell is waste, print Waste. Otherwise, print the name of crop plants in the cell: either Carrots or Kiwis or Grapes.
Demo Input:
['4 5 5 6\n4 3\n1 3\n3 3\n2 5\n3 2\n1 3\n1 4\n2 3\n2 4\n1 1\n1 1\n']
Demo Output:
['Waste\nGrapes\nCarrots\nKiwis\nCarrots\nCarrots\n']
Note:
The sample corresponds to the figure in the statement.
|
```python
FRUITS = ['Carrots', 'Kiwis', 'Grapes']
rn, cn, wn, qn = map(int, input().split())
waste_locations = []
for _ in range(wn):
r, c = map(int, input().split())
order = (r - 1) * cn + c - 1
waste_locations.append(order)
# waste_locations = sorted(waste_locations)
for _ in range(qn):
r, c = map(int, input().split())
fruit_order = (r - 1) * cn + c - 1
if fruit_order in waste_locations:
print("Waste")
else:
previous_waste = 0
for location in waste_locations:
if location < fruit_order:
previous_waste += 1
print(FRUITS[(fruit_order - previous_waste) % 3])
```
| 3.8835
|
296
|
A
|
Yaroslav and Permutations
|
PROGRAMMING
| 1,100
|
[
"greedy",
"math"
] | null | null |
Yaroslav has an array that consists of *n* integers. In one second Yaroslav can swap two neighboring array elements. Now Yaroslav is wondering if he can obtain an array where any two neighboring elements would be distinct in a finite time.
Help Yaroslav.
|
The first line contains integer *n* (1<=β€<=*n*<=β€<=100) β the number of elements in the array. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=β€<=*a**i*<=β€<=1000) β the array elements.
|
In the single line print "YES" (without the quotes) if Yaroslav can obtain the array he needs, and "NO" (without the quotes) otherwise.
|
[
"1\n1\n",
"3\n1 1 2\n",
"4\n7 7 7 7\n"
] |
[
"YES\n",
"YES\n",
"NO\n"
] |
In the first sample the initial array fits well.
In the second sample Yaroslav can get array: 1, 2, 1. He can swap the last and the second last elements to obtain it.
In the third sample Yarosav can't get the array he needs.
| 500
|
[
{
"input": "1\n1",
"output": "YES"
},
{
"input": "3\n1 1 2",
"output": "YES"
},
{
"input": "4\n7 7 7 7",
"output": "NO"
},
{
"input": "4\n479 170 465 146",
"output": "YES"
},
{
"input": "5\n996 437 605 996 293",
"output": "YES"
},
{
"input": "6\n727 539 896 668 36 896",
"output": "YES"
},
{
"input": "7\n674 712 674 674 674 674 674",
"output": "NO"
},
{
"input": "8\n742 742 742 742 742 289 742 742",
"output": "NO"
},
{
"input": "9\n730 351 806 806 806 630 85 757 967",
"output": "YES"
},
{
"input": "10\n324 539 83 440 834 640 440 440 440 440",
"output": "YES"
},
{
"input": "7\n925 830 925 98 987 162 356",
"output": "YES"
},
{
"input": "68\n575 32 53 351 151 942 725 967 431 108 192 8 338 458 288 754 384 946 910 210 759 222 589 423 947 507 31 414 169 901 592 763 656 411 360 625 538 549 484 596 42 603 351 292 837 375 21 597 22 349 200 669 485 282 735 54 1000 419 939 901 789 128 468 729 894 649 484 808",
"output": "YES"
},
{
"input": "22\n618 814 515 310 617 936 452 601 250 520 557 799 304 225 9 845 610 990 703 196 486 94",
"output": "YES"
},
{
"input": "44\n459 581 449 449 449 449 449 449 449 623 449 449 449 449 449 449 449 449 889 449 203 273 329 449 449 449 449 449 449 845 882 323 22 449 449 893 449 449 449 449 449 870 449 402",
"output": "NO"
},
{
"input": "90\n424 3 586 183 286 89 427 618 758 833 933 170 155 722 190 977 330 369 693 426 556 435 550 442 513 146 61 719 754 140 424 280 997 688 530 550 438 867 950 194 196 298 417 287 106 489 283 456 735 115 702 317 672 787 264 314 356 186 54 913 809 833 946 314 757 322 559 647 983 482 145 197 223 130 162 536 451 174 467 45 660 293 440 254 25 155 511 746 650 187",
"output": "YES"
},
{
"input": "14\n959 203 478 315 788 788 373 834 488 519 774 764 193 103",
"output": "YES"
},
{
"input": "81\n544 528 528 528 528 4 506 528 32 528 528 528 528 528 528 528 528 975 528 528 528 528 528 528 528 528 528 528 528 528 528 20 528 528 528 528 528 528 528 528 852 528 528 120 528 528 61 11 528 528 528 228 528 165 883 528 488 475 628 528 528 528 528 528 528 597 528 528 528 528 528 528 528 528 528 528 528 412 528 521 925",
"output": "NO"
},
{
"input": "89\n354 356 352 355 355 355 352 354 354 352 355 356 355 352 354 356 354 355 355 354 353 352 352 355 355 356 352 352 353 356 352 353 354 352 355 352 353 353 353 354 353 354 354 353 356 353 353 354 354 354 354 353 352 353 355 356 356 352 356 354 353 352 355 354 356 356 356 354 354 356 354 355 354 355 353 352 354 355 352 355 355 354 356 353 353 352 356 352 353",
"output": "YES"
},
{
"input": "71\n284 284 285 285 285 284 285 284 284 285 284 285 284 284 285 284 285 285 285 285 284 284 285 285 284 284 284 285 284 285 284 285 285 284 284 284 285 284 284 285 285 285 284 284 285 284 285 285 284 285 285 284 285 284 284 284 285 285 284 285 284 285 285 285 285 284 284 285 285 284 285",
"output": "NO"
},
{
"input": "28\n602 216 214 825 814 760 814 28 76 814 814 288 814 814 222 707 11 490 814 543 914 705 814 751 976 814 814 99",
"output": "YES"
},
{
"input": "48\n546 547 914 263 986 945 914 914 509 871 324 914 153 571 914 914 914 528 970 566 544 914 914 914 410 914 914 589 609 222 914 889 691 844 621 68 914 36 914 39 630 749 914 258 945 914 727 26",
"output": "YES"
},
{
"input": "56\n516 76 516 197 516 427 174 516 706 813 94 37 516 815 516 516 937 483 16 516 842 516 638 691 516 635 516 516 453 263 516 516 635 257 125 214 29 81 516 51 362 516 677 516 903 516 949 654 221 924 516 879 516 516 972 516",
"output": "YES"
},
{
"input": "46\n314 723 314 314 314 235 314 314 314 314 270 314 59 972 314 216 816 40 314 314 314 314 314 314 314 381 314 314 314 314 314 314 314 789 314 957 114 942 314 314 29 314 314 72 314 314",
"output": "NO"
},
{
"input": "72\n169 169 169 599 694 81 250 529 865 406 817 169 667 169 965 169 169 663 65 169 903 169 942 763 169 807 169 603 169 169 13 169 169 810 169 291 169 169 169 169 169 169 169 713 169 440 169 169 169 169 169 480 169 169 867 169 169 169 169 169 169 169 169 393 169 169 459 169 99 169 601 800",
"output": "NO"
},
{
"input": "100\n317 316 317 316 317 316 317 316 317 316 316 317 317 316 317 316 316 316 317 316 317 317 316 317 316 316 316 316 316 316 317 316 317 317 317 317 317 317 316 316 316 317 316 317 316 317 316 317 317 316 317 316 317 317 316 317 316 317 316 317 316 316 316 317 317 317 317 317 316 317 317 316 316 316 316 317 317 316 317 316 316 316 316 316 316 317 316 316 317 317 317 317 317 317 317 317 317 316 316 317",
"output": "NO"
},
{
"input": "100\n510 510 510 162 969 32 510 511 510 510 911 183 496 875 903 461 510 510 123 578 510 510 510 510 510 755 510 673 510 510 763 510 510 909 510 435 487 959 807 510 368 788 557 448 284 332 510 949 510 510 777 112 857 926 487 510 510 510 678 510 510 197 829 427 698 704 409 509 510 238 314 851 510 651 510 455 682 510 714 635 973 510 443 878 510 510 510 591 510 24 596 510 43 183 510 510 671 652 214 784",
"output": "YES"
},
{
"input": "100\n476 477 474 476 476 475 473 476 474 475 473 477 476 476 474 476 474 475 476 477 473 473 473 474 474 476 473 473 476 476 475 476 473 474 473 473 477 475 475 475 476 475 477 477 477 476 475 475 475 473 476 477 475 476 477 473 474 477 473 475 476 476 474 477 476 474 473 477 473 475 477 473 476 474 477 473 475 477 473 476 476 475 476 475 474 473 477 473 475 473 477 473 473 474 475 473 477 476 477 474",
"output": "YES"
},
{
"input": "100\n498 498 498 498 498 499 498 499 499 499 498 498 498 498 499 498 499 499 498 499 498 498 498 499 499 499 498 498 499 499 498 498 498 499 498 499 498 498 498 499 498 499 498 498 498 498 499 498 498 499 498 498 499 498 499 499 498 499 499 499 498 498 498 498 499 498 499 498 499 499 499 499 498 498 499 499 498 499 499 498 498 499 499 498 498 499 499 499 498 498 499 498 498 498 499 499 499 498 498 499",
"output": "NO"
},
{
"input": "100\n858 53 816 816 816 816 816 816 816 181 816 816 816 816 579 879 816 948 171 816 816 150 866 816 816 816 897 816 816 816 816 816 816 706 816 539 816 816 816 816 816 816 423 487 816 615 254 816 816 816 816 83 816 816 816 816 816 816 816 816 816 816 816 136 775 999 816 816 816 644 816 816 816 816 927 816 802 816 856 816 816 816 816 816 816 816 816 816 816 700 816 816 816 816 982 477 816 891 806 816",
"output": "NO"
},
{
"input": "100\n167 169 169 167 169 169 167 167 167 167 168 166 170 170 169 170 170 170 169 168 166 167 170 169 167 169 168 169 166 170 166 167 170 166 166 167 169 166 166 169 166 167 168 168 170 167 168 166 168 170 167 168 167 169 169 166 168 167 170 168 167 169 168 169 166 168 168 169 169 166 170 168 167 169 170 168 167 169 168 167 168 168 166 169 170 170 166 166 167 170 167 168 167 167 169 169 166 166 169 167",
"output": "YES"
},
{
"input": "100\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000",
"output": "NO"
},
{
"input": "99\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000",
"output": "NO"
},
{
"input": "100\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 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": "YES"
},
{
"input": "99\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 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": "YES"
},
{
"input": "2\n1 1",
"output": "NO"
},
{
"input": "1\n1000",
"output": "YES"
},
{
"input": "12\n2 2 4 4 4 4 6 6 6 6 6 6",
"output": "YES"
}
] | 1,633,453,472
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 37
| 154
| 6,758,400
|
n = int(input())
m = input().split()
possible = True
for i in set(m):
if m.count(i) >= n/2 + 1:
possible = False
break
print('YES' if possible else 'NO')
|
Title: Yaroslav and Permutations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Yaroslav has an array that consists of *n* integers. In one second Yaroslav can swap two neighboring array elements. Now Yaroslav is wondering if he can obtain an array where any two neighboring elements would be distinct in a finite time.
Help Yaroslav.
Input Specification:
The first line contains integer *n* (1<=β€<=*n*<=β€<=100) β the number of elements in the array. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=β€<=*a**i*<=β€<=1000) β the array elements.
Output Specification:
In the single line print "YES" (without the quotes) if Yaroslav can obtain the array he needs, and "NO" (without the quotes) otherwise.
Demo Input:
['1\n1\n', '3\n1 1 2\n', '4\n7 7 7 7\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
In the first sample the initial array fits well.
In the second sample Yaroslav can get array: 1, 2, 1. He can swap the last and the second last elements to obtain it.
In the third sample Yarosav can't get the array he needs.
|
```python
n = int(input())
m = input().split()
possible = True
for i in set(m):
if m.count(i) >= n/2 + 1:
possible = False
break
print('YES' if possible else 'NO')
```
| 3
|
|
208
|
A
|
Dubstep
|
PROGRAMMING
| 900
|
[
"strings"
] | null | null |
Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them.
Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club.
For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX".
Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song.
|
The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word.
|
Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space.
|
[
"WUBWUBABCWUB\n",
"WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n"
] |
[
"ABC ",
"WE ARE THE CHAMPIONS MY FRIEND "
] |
In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya.
In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" β between them Vasya added two "WUB".
| 500
|
[
{
"input": "WUBWUBABCWUB",
"output": "ABC "
},
{
"input": "WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB",
"output": "WE ARE THE CHAMPIONS MY FRIEND "
},
{
"input": "WUBWUBWUBSR",
"output": "SR "
},
{
"input": "RWUBWUBWUBLWUB",
"output": "R L "
},
{
"input": "ZJWUBWUBWUBJWUBWUBWUBL",
"output": "ZJ J L "
},
{
"input": "CWUBBWUBWUBWUBEWUBWUBWUBQWUBWUBWUB",
"output": "C B E Q "
},
{
"input": "WUBJKDWUBWUBWBIRAQKFWUBWUBYEWUBWUBWUBWVWUBWUB",
"output": "JKD WBIRAQKF YE WV "
},
{
"input": "WUBKSDHEMIXUJWUBWUBRWUBWUBWUBSWUBWUBWUBHWUBWUBWUB",
"output": "KSDHEMIXUJ R S H "
},
{
"input": "OGWUBWUBWUBXWUBWUBWUBIWUBWUBWUBKOWUBWUB",
"output": "OG X I KO "
},
{
"input": "QWUBQQWUBWUBWUBIWUBWUBWWWUBWUBWUBJOPJPBRH",
"output": "Q QQ I WW JOPJPBRH "
},
{
"input": "VSRNVEATZTLGQRFEGBFPWUBWUBWUBAJWUBWUBWUBPQCHNWUBCWUB",
"output": "VSRNVEATZTLGQRFEGBFP AJ PQCHN C "
},
{
"input": "WUBWUBEWUBWUBWUBIQMJNIQWUBWUBWUBGZZBQZAUHYPWUBWUBWUBPMRWUBWUBWUBDCV",
"output": "E IQMJNIQ GZZBQZAUHYP PMR DCV "
},
{
"input": "WUBWUBWUBFVWUBWUBWUBBPSWUBWUBWUBRXNETCJWUBWUBWUBJDMBHWUBWUBWUBBWUBWUBVWUBWUBB",
"output": "FV BPS RXNETCJ JDMBH B V B "
},
{
"input": "WUBWUBWUBFBQWUBWUBWUBIDFSYWUBWUBWUBCTWDMWUBWUBWUBSXOWUBWUBWUBQIWUBWUBWUBL",
"output": "FBQ IDFSY CTWDM SXO QI L "
},
{
"input": "IWUBWUBQLHDWUBYIIKZDFQWUBWUBWUBCXWUBWUBUWUBWUBWUBKWUBWUBWUBNL",
"output": "I QLHD YIIKZDFQ CX U K NL "
},
{
"input": "KWUBUPDYXGOKUWUBWUBWUBAGOAHWUBIZDWUBWUBWUBIYWUBWUBWUBVWUBWUBWUBPWUBWUBWUBE",
"output": "K UPDYXGOKU AGOAH IZD IY V P E "
},
{
"input": "WUBWUBOWUBWUBWUBIPVCQAFWYWUBWUBWUBQWUBWUBWUBXHDKCPYKCTWWYWUBWUBWUBVWUBWUBWUBFZWUBWUB",
"output": "O IPVCQAFWY Q XHDKCPYKCTWWY V FZ "
},
{
"input": "PAMJGYWUBWUBWUBXGPQMWUBWUBWUBTKGSXUYWUBWUBWUBEWUBWUBWUBNWUBWUBWUBHWUBWUBWUBEWUBWUB",
"output": "PAMJGY XGPQM TKGSXUY E N H E "
},
{
"input": "WUBYYRTSMNWUWUBWUBWUBCWUBWUBWUBCWUBWUBWUBFSYUINDWOBVWUBWUBWUBFWUBWUBWUBAUWUBWUBWUBVWUBWUBWUBJB",
"output": "YYRTSMNWU C C FSYUINDWOBV F AU V JB "
},
{
"input": "WUBWUBYGPYEYBNRTFKOQCWUBWUBWUBUYGRTQEGWLFYWUBWUBWUBFVWUBHPWUBWUBWUBXZQWUBWUBWUBZDWUBWUBWUBM",
"output": "YGPYEYBNRTFKOQC UYGRTQEGWLFY FV HP XZQ ZD M "
},
{
"input": "WUBZVMJWUBWUBWUBFOIMJQWKNZUBOFOFYCCWUBWUBWUBAUWWUBRDRADWUBWUBWUBCHQVWUBWUBWUBKFTWUBWUBWUBW",
"output": "ZVMJ FOIMJQWKNZUBOFOFYCC AUW RDRAD CHQV KFT W "
},
{
"input": "WUBWUBZBKOKHQLGKRVIMZQMQNRWUBWUBWUBDACWUBWUBNZHFJMPEYKRVSWUBWUBWUBPPHGAVVPRZWUBWUBWUBQWUBWUBAWUBG",
"output": "ZBKOKHQLGKRVIMZQMQNR DAC NZHFJMPEYKRVS PPHGAVVPRZ Q A G "
},
{
"input": "WUBWUBJWUBWUBWUBNFLWUBWUBWUBGECAWUBYFKBYJWTGBYHVSSNTINKWSINWSMAWUBWUBWUBFWUBWUBWUBOVWUBWUBLPWUBWUBWUBN",
"output": "J NFL GECA YFKBYJWTGBYHVSSNTINKWSINWSMA F OV LP N "
},
{
"input": "WUBWUBLCWUBWUBWUBZGEQUEATJVIXETVTWUBWUBWUBEXMGWUBWUBWUBRSWUBWUBWUBVWUBWUBWUBTAWUBWUBWUBCWUBWUBWUBQG",
"output": "LC ZGEQUEATJVIXETVT EXMG RS V TA C QG "
},
{
"input": "WUBMPWUBWUBWUBORWUBWUBDLGKWUBWUBWUBVVZQCAAKVJTIKWUBWUBWUBTJLUBZJCILQDIFVZWUBWUBYXWUBWUBWUBQWUBWUBWUBLWUB",
"output": "MP OR DLGK VVZQCAAKVJTIK TJLUBZJCILQDIFVZ YX Q L "
},
{
"input": "WUBNXOLIBKEGXNWUBWUBWUBUWUBGITCNMDQFUAOVLWUBWUBWUBAIJDJZJHFMPVTPOXHPWUBWUBWUBISCIOWUBWUBWUBGWUBWUBWUBUWUB",
"output": "NXOLIBKEGXN U GITCNMDQFUAOVL AIJDJZJHFMPVTPOXHP ISCIO G U "
},
{
"input": "WUBWUBNMMWCZOLYPNBELIYVDNHJUNINWUBWUBWUBDXLHYOWUBWUBWUBOJXUWUBWUBWUBRFHTGJCEFHCGWARGWUBWUBWUBJKWUBWUBSJWUBWUB",
"output": "NMMWCZOLYPNBELIYVDNHJUNIN DXLHYO OJXU RFHTGJCEFHCGWARG JK SJ "
},
{
"input": "SGWLYSAUJOJBNOXNWUBWUBWUBBOSSFWKXPDPDCQEWUBWUBWUBDIRZINODWUBWUBWUBWWUBWUBWUBPPHWUBWUBWUBRWUBWUBWUBQWUBWUBWUBJWUB",
"output": "SGWLYSAUJOJBNOXN BOSSFWKXPDPDCQE DIRZINOD W PPH R Q J "
},
{
"input": "TOWUBWUBWUBGBTBNWUBWUBWUBJVIOJBIZFUUYHUAIEBQLQXPQKZJMPTCWBKPOSAWUBWUBWUBSWUBWUBWUBTOLVXWUBWUBWUBNHWUBWUBWUBO",
"output": "TO GBTBN JVIOJBIZFUUYHUAIEBQLQXPQKZJMPTCWBKPOSA S TOLVX NH O "
},
{
"input": "WUBWUBWSPLAYSZSAUDSWUBWUBWUBUWUBWUBWUBKRWUBWUBWUBRSOKQMZFIYZQUWUBWUBWUBELSHUWUBWUBWUBUKHWUBWUBWUBQXEUHQWUBWUBWUBBWUBWUBWUBR",
"output": "WSPLAYSZSAUDS U KR RSOKQMZFIYZQU ELSHU UKH QXEUHQ B R "
},
{
"input": "WUBXEMWWVUHLSUUGRWUBWUBWUBAWUBXEGILZUNKWUBWUBWUBJDHHKSWUBWUBWUBDTSUYSJHWUBWUBWUBPXFWUBMOHNJWUBWUBWUBZFXVMDWUBWUBWUBZMWUBWUB",
"output": "XEMWWVUHLSUUGR A XEGILZUNK JDHHKS DTSUYSJH PXF MOHNJ ZFXVMD ZM "
},
{
"input": "BMBWUBWUBWUBOQKWUBWUBWUBPITCIHXHCKLRQRUGXJWUBWUBWUBVWUBWUBWUBJCWUBWUBWUBQJPWUBWUBWUBBWUBWUBWUBBMYGIZOOXWUBWUBWUBTAGWUBWUBHWUB",
"output": "BMB OQK PITCIHXHCKLRQRUGXJ V JC QJP B BMYGIZOOX TAG H "
},
{
"input": "CBZNWUBWUBWUBNHWUBWUBWUBYQSYWUBWUBWUBMWUBWUBWUBXRHBTMWUBWUBWUBPCRCWUBWUBWUBTZUYLYOWUBWUBWUBCYGCWUBWUBWUBCLJWUBWUBWUBSWUBWUBWUB",
"output": "CBZN NH YQSY M XRHBTM PCRC TZUYLYO CYGC CLJ S "
},
{
"input": "DPDWUBWUBWUBEUQKWPUHLTLNXHAEKGWUBRRFYCAYZFJDCJLXBAWUBWUBWUBHJWUBOJWUBWUBWUBNHBJEYFWUBWUBWUBRWUBWUBWUBSWUBWWUBWUBWUBXDWUBWUBWUBJWUB",
"output": "DPD EUQKWPUHLTLNXHAEKG RRFYCAYZFJDCJLXBA HJ OJ NHBJEYF R S W XD J "
},
{
"input": "WUBWUBWUBISERPQITVIYERSCNWUBWUBWUBQWUBWUBWUBDGSDIPWUBWUBWUBCAHKDZWEXBIBJVVSKKVQJWUBWUBWUBKIWUBWUBWUBCWUBWUBWUBAWUBWUBWUBPWUBWUBWUBHWUBWUBWUBF",
"output": "ISERPQITVIYERSCN Q DGSDIP CAHKDZWEXBIBJVVSKKVQJ KI C A P H F "
},
{
"input": "WUBWUBWUBIWUBWUBLIKNQVWUBWUBWUBPWUBWUBWUBHWUBWUBWUBMWUBWUBWUBDPRSWUBWUBWUBBSAGYLQEENWXXVWUBWUBWUBXMHOWUBWUBWUBUWUBWUBWUBYRYWUBWUBWUBCWUBWUBWUBY",
"output": "I LIKNQV P H M DPRS BSAGYLQEENWXXV XMHO U YRY C Y "
},
{
"input": "WUBWUBWUBMWUBWUBWUBQWUBWUBWUBITCFEYEWUBWUBWUBHEUWGNDFNZGWKLJWUBWUBWUBMZPWUBWUBWUBUWUBWUBWUBBWUBWUBWUBDTJWUBHZVIWUBWUBWUBPWUBFNHHWUBWUBWUBVTOWUB",
"output": "M Q ITCFEYE HEUWGNDFNZGWKLJ MZP U B DTJ HZVI P FNHH VTO "
},
{
"input": "WUBWUBNDNRFHYJAAUULLHRRDEDHYFSRXJWUBWUBWUBMUJVDTIRSGYZAVWKRGIFWUBWUBWUBHMZWUBWUBWUBVAIWUBWUBWUBDDKJXPZRGWUBWUBWUBSGXWUBWUBWUBIFKWUBWUBWUBUWUBWUBWUBW",
"output": "NDNRFHYJAAUULLHRRDEDHYFSRXJ MUJVDTIRSGYZAVWKRGIF HMZ VAI DDKJXPZRG SGX IFK U W "
},
{
"input": "WUBOJMWRSLAXXHQRTPMJNCMPGWUBWUBWUBNYGMZIXNLAKSQYWDWUBWUBWUBXNIWUBWUBWUBFWUBWUBWUBXMBWUBWUBWUBIWUBWUBWUBINWUBWUBWUBWDWUBWUBWUBDDWUBWUBWUBD",
"output": "OJMWRSLAXXHQRTPMJNCMPG NYGMZIXNLAKSQYWD XNI F XMB I IN WD DD D "
},
{
"input": "WUBWUBWUBREHMWUBWUBWUBXWUBWUBWUBQASNWUBWUBWUBNLSMHLCMTICWUBWUBWUBVAWUBWUBWUBHNWUBWUBWUBNWUBWUBWUBUEXLSFOEULBWUBWUBWUBXWUBWUBWUBJWUBWUBWUBQWUBWUBWUBAWUBWUB",
"output": "REHM X QASN NLSMHLCMTIC VA HN N UEXLSFOEULB X J Q A "
},
{
"input": "WUBWUBWUBSTEZTZEFFIWUBWUBWUBSWUBWUBWUBCWUBFWUBHRJPVWUBWUBWUBDYJUWUBWUBWUBPWYDKCWUBWUBWUBCWUBWUBWUBUUEOGCVHHBWUBWUBWUBEXLWUBWUBWUBVCYWUBWUBWUBMWUBWUBWUBYWUB",
"output": "STEZTZEFFI S C F HRJPV DYJU PWYDKC C UUEOGCVHHB EXL VCY M Y "
},
{
"input": "WPPNMSQOQIWUBWUBWUBPNQXWUBWUBWUBHWUBWUBWUBNFLWUBWUBWUBGWSGAHVJFNUWUBWUBWUBFWUBWUBWUBWCMLRICFSCQQQTNBWUBWUBWUBSWUBWUBWUBKGWUBWUBWUBCWUBWUBWUBBMWUBWUBWUBRWUBWUB",
"output": "WPPNMSQOQI PNQX H NFL GWSGAHVJFNU F WCMLRICFSCQQQTNB S KG C BM R "
},
{
"input": "YZJOOYITZRARKVFYWUBWUBRZQGWUBWUBWUBUOQWUBWUBWUBIWUBWUBWUBNKVDTBOLETKZISTWUBWUBWUBWLWUBQQFMMGSONZMAWUBZWUBWUBWUBQZUXGCWUBWUBWUBIRZWUBWUBWUBLTTVTLCWUBWUBWUBY",
"output": "YZJOOYITZRARKVFY RZQG UOQ I NKVDTBOLETKZIST WL QQFMMGSONZMA Z QZUXGC IRZ LTTVTLC Y "
},
{
"input": "WUBCAXNCKFBVZLGCBWCOAWVWOFKZVQYLVTWUBWUBWUBNLGWUBWUBWUBAMGDZBDHZMRMQMDLIRMIWUBWUBWUBGAJSHTBSWUBWUBWUBCXWUBWUBWUBYWUBZLXAWWUBWUBWUBOHWUBWUBWUBZWUBWUBWUBGBWUBWUBWUBE",
"output": "CAXNCKFBVZLGCBWCOAWVWOFKZVQYLVT NLG AMGDZBDHZMRMQMDLIRMI GAJSHTBS CX Y ZLXAW OH Z GB E "
},
{
"input": "WUBWUBCHXSOWTSQWUBWUBWUBCYUZBPBWUBWUBWUBSGWUBWUBWKWORLRRLQYUUFDNWUBWUBWUBYYGOJNEVEMWUBWUBWUBRWUBWUBWUBQWUBWUBWUBIHCKWUBWUBWUBKTWUBWUBWUBRGSNTGGWUBWUBWUBXCXWUBWUBWUBS",
"output": "CHXSOWTSQ CYUZBPB SG WKWORLRRLQYUUFDN YYGOJNEVEM R Q IHCK KT RGSNTGG XCX S "
},
{
"input": "WUBWUBWUBHJHMSBURXTHXWSCHNAIJOWBHLZGJZDHEDSPWBWACCGQWUBWUBWUBXTZKGIITWUBWUBWUBAWUBWUBWUBVNCXPUBCQWUBWUBWUBIDPNAWUBWUBWUBOWUBWUBWUBYGFWUBWUBWUBMQOWUBWUBWUBKWUBWUBWUBAZVWUBWUBWUBEP",
"output": "HJHMSBURXTHXWSCHNAIJOWBHLZGJZDHEDSPWBWACCGQ XTZKGIIT A VNCXPUBCQ IDPNA O YGF MQO K AZV EP "
},
{
"input": "WUBKYDZOYWZSNGMKJSWAXFDFLTHDHEOGTDBNZMSMKZTVWUBWUBWUBLRMIIWUBWUBWUBGWUBWUBWUBADPSWUBWUBWUBANBWUBWUBPCWUBWUBWUBPWUBWUBWUBGPVNLSWIRFORYGAABUXMWUBWUBWUBOWUBWUBWUBNWUBWUBWUBYWUBWUB",
"output": "KYDZOYWZSNGMKJSWAXFDFLTHDHEOGTDBNZMSMKZTV LRMII G ADPS ANB PC P GPVNLSWIRFORYGAABUXM O N Y "
},
{
"input": "REWUBWUBWUBJDWUBWUBWUBNWUBWUBWUBTWWUBWUBWUBWZDOCKKWUBWUBWUBLDPOVBFRCFWUBWUBAKZIBQKEUAZEEWUBWUBWUBLQYPNPFWUBYEWUBWUBWUBFWUBWUBWUBBPWUBWUBWUBAWWUBWUBWUBQWUBWUBWUBBRWUBWUBWUBXJL",
"output": "RE JD N TW WZDOCKK LDPOVBFRCF AKZIBQKEUAZEE LQYPNPF YE F BP AW Q BR XJL "
},
{
"input": "CUFGJDXGMWUBWUBWUBOMWUBWUBWUBSIEWUBWUBWUBJJWKNOWUBWUBWUBYBHVNRNORGYWUBWUBWUBOAGCAWUBWUBWUBSBLBKTPFKPBIWUBWUBWUBJBWUBWUBWUBRMFCJPGWUBWUBWUBDWUBWUBWUBOJOWUBWUBWUBZPWUBWUBWUBMWUBRWUBWUBWUBFXWWUBWUBWUBO",
"output": "CUFGJDXGM OM SIE JJWKNO YBHVNRNORGY OAGCA SBLBKTPFKPBI JB RMFCJPG D OJO ZP M R FXW O "
},
{
"input": "WUBJZGAEXFMFEWMAKGQLUWUBWUBWUBICYTPQWGENELVYWANKUOJYWUBWUBWUBGWUBWUBWUBHYCJVLPHTUPNEGKCDGQWUBWUBWUBOFWUBWUBWUBCPGSOGZBRPRPVJJEWUBWUBWUBDQBCWUBWUBWUBHWUBWUBWUBMHOHYBMATWUBWUBWUBVWUBWUBWUBSWUBWUBWUBKOWU",
"output": "JZGAEXFMFEWMAKGQLU ICYTPQWGENELVYWANKUOJY G HYCJVLPHTUPNEGKCDGQ OF CPGSOGZBRPRPVJJE DQBC H MHOHYBMAT V S KOWU "
},
{
"input": "A",
"output": "A "
},
{
"input": "WUBA",
"output": "A "
},
{
"input": "WUBWUBA",
"output": "A "
},
{
"input": "AWUBWUBWUB",
"output": "A "
},
{
"input": "AWUBBWUBCWUBD",
"output": "A B C D "
},
{
"input": "WUBWWUBWUBWUBUWUBWUBBWUB",
"output": "W U B "
},
{
"input": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA",
"output": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA "
},
{
"input": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAWUBAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA",
"output": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA "
},
{
"input": "WUWUBBWWUBUB",
"output": "WU BW UB "
},
{
"input": "WUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUABWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUB",
"output": "WUAB "
},
{
"input": "ZWUB",
"output": "Z "
},
{
"input": "WU",
"output": "WU "
},
{
"input": "UB",
"output": "UB "
},
{
"input": "U",
"output": "U "
},
{
"input": "WUBW",
"output": "W "
},
{
"input": "WUBWU",
"output": "WU "
},
{
"input": "WUWUB",
"output": "WU "
},
{
"input": "UBWUB",
"output": "UB "
},
{
"input": "WUWUBUBWUBUWUB",
"output": "WU UB U "
},
{
"input": "WUBWWUBAWUB",
"output": "W A "
},
{
"input": "WUUUUU",
"output": "WUUUUU "
}
] | 1,677,542,171
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 71
| 124
| 0
|
string = input()
print(string.replace("WUB", ' ').strip())
|
Title: Dubstep
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them.
Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club.
For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX".
Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song.
Input Specification:
The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word.
Output Specification:
Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space.
Demo Input:
['WUBWUBABCWUB\n', 'WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n']
Demo Output:
['ABC ', 'WE ARE THE CHAMPIONS MY FRIEND ']
Note:
In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya.
In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" β between them Vasya added two "WUB".
|
```python
string = input()
print(string.replace("WUB", ' ').strip())
```
| 3
|
|
705
|
A
|
Hulk
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Dr. Bruce Banner hates his enemies (like others don't). As we all know, he can barely talk when he turns into the incredible Hulk. That's why he asked you to help him to express his feelings.
Hulk likes the Inception so much, and like that his feelings are complicated. They have *n* layers. The first layer is hate, second one is love, third one is hate and so on...
For example if *n*<==<=1, then his feeling is "I hate it" or if *n*<==<=2 it's "I hate that I love it", and if *n*<==<=3 it's "I hate that I love that I hate it" and so on.
Please help Dr. Banner.
|
The only line of the input contains a single integer *n* (1<=β€<=*n*<=β€<=100)Β β the number of layers of love and hate.
|
Print Dr.Banner's feeling in one line.
|
[
"1\n",
"2\n",
"3\n"
] |
[
"I hate it\n",
"I hate that I love it\n",
"I hate that I love that I hate it\n"
] |
none
| 500
|
[
{
"input": "1",
"output": "I hate it"
},
{
"input": "2",
"output": "I hate that I love it"
},
{
"input": "3",
"output": "I hate that I love that I hate it"
},
{
"input": "4",
"output": "I hate that I love that I hate that I love it"
},
{
"input": "5",
"output": "I hate that I love that I hate that I love that I hate it"
},
{
"input": "10",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love it"
},
{
"input": "18",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love it"
},
{
"input": "19",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate it"
},
{
"input": "76",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate ..."
},
{
"input": "77",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate ..."
},
{
"input": "81",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate ..."
},
{
"input": "82",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate ..."
},
{
"input": "85",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate ..."
},
{
"input": "86",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate ..."
},
{
"input": "100",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate ..."
},
{
"input": "99",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate ..."
},
{
"input": "98",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate ..."
},
{
"input": "33",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate it"
},
{
"input": "34",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love it"
},
{
"input": "57",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate ..."
},
{
"input": "100",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate ..."
}
] | 1,695,227,278
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 21
| 78
| 0
|
V=''
T=int(input())
for i in range(1,T+1):
if i%2==0:
V=V+'I love'
if i==T:
V=V+' it'
else:
V=V+' that '
else:
V=V+'I hate'
if i==T:
V=V+' it'
else:
V=V+' that '
print(V)
|
Title: Hulk
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dr. Bruce Banner hates his enemies (like others don't). As we all know, he can barely talk when he turns into the incredible Hulk. That's why he asked you to help him to express his feelings.
Hulk likes the Inception so much, and like that his feelings are complicated. They have *n* layers. The first layer is hate, second one is love, third one is hate and so on...
For example if *n*<==<=1, then his feeling is "I hate it" or if *n*<==<=2 it's "I hate that I love it", and if *n*<==<=3 it's "I hate that I love that I hate it" and so on.
Please help Dr. Banner.
Input Specification:
The only line of the input contains a single integer *n* (1<=β€<=*n*<=β€<=100)Β β the number of layers of love and hate.
Output Specification:
Print Dr.Banner's feeling in one line.
Demo Input:
['1\n', '2\n', '3\n']
Demo Output:
['I hate it\n', 'I hate that I love it\n', 'I hate that I love that I hate it\n']
Note:
none
|
```python
V=''
T=int(input())
for i in range(1,T+1):
if i%2==0:
V=V+'I love'
if i==T:
V=V+' it'
else:
V=V+' that '
else:
V=V+'I hate'
if i==T:
V=V+' it'
else:
V=V+' that '
print(V)
```
| 3
|
|
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<=β€<=*n*<=β€<=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 β 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"
}
] | 1,649,727,311
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 41
| 46
| 0
|
n=int(input())-1
while n>4:
n=n-5>>1
print("SLPRHheeaoeonjwlnneadaysror hdnd"[n::5])
|
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<=β€<=*n*<=β€<=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 β 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())-1
while n>4:
n=n-5>>1
print("SLPRHheeaoeonjwlnneadaysror hdnd"[n::5])
```
| 3.977
|
492
|
B
|
Vanya and Lanterns
|
PROGRAMMING
| 1,200
|
[
"binary search",
"implementation",
"math",
"sortings"
] | null | null |
Vanya walks late at night along a straight street of length *l*, lit by *n* lanterns. Consider the coordinate system with the beginning of the street corresponding to the point 0, and its end corresponding to the point *l*. Then the *i*-th lantern is at the point *a**i*. The lantern lights all points of the street that are at the distance of at most *d* from it, where *d* is some positive number, common for all lanterns.
Vanya wonders: what is the minimum light radius *d* should the lanterns have to light the whole street?
|
The first line contains two integers *n*, *l* (1<=β€<=*n*<=β€<=1000, 1<=β€<=*l*<=β€<=109)Β β the number of lanterns and the length of the street respectively.
The next line contains *n* integers *a**i* (0<=β€<=*a**i*<=β€<=*l*). Multiple lanterns can be located at the same point. The lanterns may be located at the ends of the street.
|
Print the minimum light radius *d*, needed to light the whole street. The answer will be considered correct if its absolute or relative error doesn't exceed 10<=-<=9.
|
[
"7 15\n15 5 3 7 9 14 0\n",
"2 5\n2 5\n"
] |
[
"2.5000000000\n",
"2.0000000000\n"
] |
Consider the second sample. At *d*β=β2 the first lantern will light the segment [0,β4] of the street, and the second lantern will light segment [3,β5]. Thus, the whole street will be lit.
| 1,000
|
[
{
"input": "7 15\n15 5 3 7 9 14 0",
"output": "2.5000000000"
},
{
"input": "2 5\n2 5",
"output": "2.0000000000"
},
{
"input": "46 615683844\n431749087 271781274 274974690 324606253 480870261 401650581 13285442 478090364 266585394 425024433 588791449 492057200 391293435 563090494 317950 173675329 473068378 356306865 311731938 192959832 321180686 141984626 578985584 512026637 175885185 590844074 47103801 212211134 330150 509886963 565955809 315640375 612907074 500474373 524310737 568681652 315339618 478782781 518873818 271322031 74600969 539099112 85129347 222068995 106014720 77282307",
"output": "22258199.5000000000"
},
{
"input": "2 1000000000\n0 1000000000",
"output": "500000000.0000000000"
},
{
"input": "2 555\n200 300",
"output": "255.0000000000"
},
{
"input": "1 1\n1",
"output": "1.0000000000"
},
{
"input": "1 1\n0",
"output": "1.0000000000"
},
{
"input": "1 1000000000\n0",
"output": "1000000000.0000000000"
},
{
"input": "1 1000000000\n1000000000",
"output": "1000000000.0000000000"
},
{
"input": "1 999999999\n499999999",
"output": "500000000.0000000000"
},
{
"input": "4 700\n0 250 475 700",
"output": "125.0000000000"
},
{
"input": "4 700\n0 225 450 700",
"output": "125.0000000000"
},
{
"input": "5 1000\n0 250 500 750 1000",
"output": "125.0000000000"
},
{
"input": "5 2000\n1001 1001 1001 1001 1001",
"output": "1001.0000000000"
},
{
"input": "1 999\n501",
"output": "501.0000000000"
},
{
"input": "2 1000\n400 600",
"output": "400.0000000000"
},
{
"input": "4 15\n1 3 9 11",
"output": "4.0000000000"
},
{
"input": "1 5\n2",
"output": "3.0000000000"
}
] | 1,686,033,683
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 32
| 62
| 0
|
n,l = map(int,input().split())
a = list(map(int,input().split()))
a = sorted(a)
ma = -1
for i in range(1,n):
if (a[i] - a[i - 1])/2 > ma:
ma = (a[i] - a[i - 1])/2
if a[-1] != l:
if l - a[-1] > ma:
ma = l - a[-1]
if a[0] != 0:
if a[0] > ma:
ma = a[0]
print("%.10f" % ma)
|
Title: Vanya and Lanterns
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vanya walks late at night along a straight street of length *l*, lit by *n* lanterns. Consider the coordinate system with the beginning of the street corresponding to the point 0, and its end corresponding to the point *l*. Then the *i*-th lantern is at the point *a**i*. The lantern lights all points of the street that are at the distance of at most *d* from it, where *d* is some positive number, common for all lanterns.
Vanya wonders: what is the minimum light radius *d* should the lanterns have to light the whole street?
Input Specification:
The first line contains two integers *n*, *l* (1<=β€<=*n*<=β€<=1000, 1<=β€<=*l*<=β€<=109)Β β the number of lanterns and the length of the street respectively.
The next line contains *n* integers *a**i* (0<=β€<=*a**i*<=β€<=*l*). Multiple lanterns can be located at the same point. The lanterns may be located at the ends of the street.
Output Specification:
Print the minimum light radius *d*, needed to light the whole street. The answer will be considered correct if its absolute or relative error doesn't exceed 10<=-<=9.
Demo Input:
['7 15\n15 5 3 7 9 14 0\n', '2 5\n2 5\n']
Demo Output:
['2.5000000000\n', '2.0000000000\n']
Note:
Consider the second sample. At *d*β=β2 the first lantern will light the segment [0,β4] of the street, and the second lantern will light segment [3,β5]. Thus, the whole street will be lit.
|
```python
n,l = map(int,input().split())
a = list(map(int,input().split()))
a = sorted(a)
ma = -1
for i in range(1,n):
if (a[i] - a[i - 1])/2 > ma:
ma = (a[i] - a[i - 1])/2
if a[-1] != l:
if l - a[-1] > ma:
ma = l - a[-1]
if a[0] != 0:
if a[0] > ma:
ma = a[0]
print("%.10f" % ma)
```
| 3
|
|
723
|
A
|
The New Year: Meeting Friends
|
PROGRAMMING
| 800
|
[
"implementation",
"math",
"sortings"
] | null | null |
There are three friend living on the straight line *Ox* in Lineland. The first friend lives at the point *x*1, the second friend lives at the point *x*2, and the third friend lives at the point *x*3. They plan to celebrate the New Year together, so they need to meet at one point. What is the minimum total distance they have to travel in order to meet at some point and celebrate the New Year?
It's guaranteed that the optimal answer is always integer.
|
The first line of the input contains three distinct integers *x*1, *x*2 and *x*3 (1<=β€<=*x*1,<=*x*2,<=*x*3<=β€<=100)Β β the coordinates of the houses of the first, the second and the third friends respectively.
|
Print one integerΒ β the minimum total distance the friends need to travel in order to meet together.
|
[
"7 1 4\n",
"30 20 10\n"
] |
[
"6\n",
"20\n"
] |
In the first sample, friends should meet at the point 4. Thus, the first friend has to travel the distance of 3 (from the point 7 to the point 4), the second friend also has to travel the distance of 3 (from the point 1 to the point 4), while the third friend should not go anywhere because he lives at the point 4.
| 500
|
[
{
"input": "7 1 4",
"output": "6"
},
{
"input": "30 20 10",
"output": "20"
},
{
"input": "1 4 100",
"output": "99"
},
{
"input": "100 1 91",
"output": "99"
},
{
"input": "1 45 100",
"output": "99"
},
{
"input": "1 2 3",
"output": "2"
},
{
"input": "71 85 88",
"output": "17"
},
{
"input": "30 38 99",
"output": "69"
},
{
"input": "23 82 95",
"output": "72"
},
{
"input": "22 41 47",
"output": "25"
},
{
"input": "9 94 77",
"output": "85"
},
{
"input": "1 53 51",
"output": "52"
},
{
"input": "25 97 93",
"output": "72"
},
{
"input": "42 53 51",
"output": "11"
},
{
"input": "81 96 94",
"output": "15"
},
{
"input": "21 5 93",
"output": "88"
},
{
"input": "50 13 75",
"output": "62"
},
{
"input": "41 28 98",
"output": "70"
},
{
"input": "69 46 82",
"output": "36"
},
{
"input": "87 28 89",
"output": "61"
},
{
"input": "44 45 40",
"output": "5"
},
{
"input": "86 97 68",
"output": "29"
},
{
"input": "43 92 30",
"output": "62"
},
{
"input": "16 70 1",
"output": "69"
},
{
"input": "40 46 19",
"output": "27"
},
{
"input": "71 38 56",
"output": "33"
},
{
"input": "82 21 80",
"output": "61"
},
{
"input": "75 8 35",
"output": "67"
},
{
"input": "75 24 28",
"output": "51"
},
{
"input": "78 23 56",
"output": "55"
},
{
"input": "85 31 10",
"output": "75"
},
{
"input": "76 50 9",
"output": "67"
},
{
"input": "95 37 34",
"output": "61"
},
{
"input": "84 61 35",
"output": "49"
},
{
"input": "87 85 37",
"output": "50"
},
{
"input": "1 3 2",
"output": "2"
},
{
"input": "4 2 6",
"output": "4"
},
{
"input": "6 9 3",
"output": "6"
},
{
"input": "12 4 8",
"output": "8"
},
{
"input": "15 10 5",
"output": "10"
},
{
"input": "1 50 17",
"output": "49"
},
{
"input": "10 5 15",
"output": "10"
},
{
"input": "8 1 9",
"output": "8"
},
{
"input": "3 5 4",
"output": "2"
},
{
"input": "2 1 3",
"output": "2"
},
{
"input": "1 8 2",
"output": "7"
},
{
"input": "1 100 2",
"output": "99"
},
{
"input": "1 4 6",
"output": "5"
}
] | 1,677,677,785
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 48
| 93
| 0
|
distance=list(map(int,input().split()))
distance.sort()
first=distance[1]-distance[0]
second=distance[2]-distance[1]
total=first+second
print(total)
|
Title: The New Year: Meeting Friends
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are three friend living on the straight line *Ox* in Lineland. The first friend lives at the point *x*1, the second friend lives at the point *x*2, and the third friend lives at the point *x*3. They plan to celebrate the New Year together, so they need to meet at one point. What is the minimum total distance they have to travel in order to meet at some point and celebrate the New Year?
It's guaranteed that the optimal answer is always integer.
Input Specification:
The first line of the input contains three distinct integers *x*1, *x*2 and *x*3 (1<=β€<=*x*1,<=*x*2,<=*x*3<=β€<=100)Β β the coordinates of the houses of the first, the second and the third friends respectively.
Output Specification:
Print one integerΒ β the minimum total distance the friends need to travel in order to meet together.
Demo Input:
['7 1 4\n', '30 20 10\n']
Demo Output:
['6\n', '20\n']
Note:
In the first sample, friends should meet at the point 4. Thus, the first friend has to travel the distance of 3 (from the point 7 to the point 4), the second friend also has to travel the distance of 3 (from the point 1 to the point 4), while the third friend should not go anywhere because he lives at the point 4.
|
```python
distance=list(map(int,input().split()))
distance.sort()
first=distance[1]-distance[0]
second=distance[2]-distance[1]
total=first+second
print(total)
```
| 3
|
|
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<=β€<=*n*<=β€<=100) β amount of numbers in the sequence. The second line contains *n* space-separated integer numbers β 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"
}
] | 1,618,752,663
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 31
| 124
| 0
|
n=int(input())
l=list(map(int,input().split()))
l.sort()
a=l[0]
for i in range(n):
if l[i]!=a:
print(l[i])
break
if i==n-1 and l[i]==a:
print('NO')
|
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<=β€<=*n*<=β€<=100) β amount of numbers in the sequence. The second line contains *n* space-separated integer numbers β 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())
l=list(map(int,input().split()))
l.sort()
a=l[0]
for i in range(n):
if l[i]!=a:
print(l[i])
break
if i==n-1 and l[i]==a:
print('NO')
```
| 3.969
|
913
|
C
|
Party Lemonade
|
PROGRAMMING
| 1,600
|
[
"bitmasks",
"dp",
"greedy"
] | null | null |
A New Year party is not a New Year party without lemonade! As usual, you are expecting a lot of guests, and buying lemonade has already become a pleasant necessity.
Your favorite store sells lemonade in bottles of *n* different volumes at different costs. A single bottle of type *i* has volume 2*i*<=-<=1 liters and costs *c**i* roubles. The number of bottles of each type in the store can be considered infinite.
You want to buy at least *L* liters of lemonade. How many roubles do you have to spend?
|
The first line contains two integers *n* and *L* (1<=β€<=*n*<=β€<=30; 1<=β€<=*L*<=β€<=109)Β β the number of types of bottles in the store and the required amount of lemonade in liters, respectively.
The second line contains *n* integers *c*1,<=*c*2,<=...,<=*c**n* (1<=β€<=*c**i*<=β€<=109)Β β the costs of bottles of different types.
|
Output a single integerΒ β the smallest number of roubles you have to pay in order to buy at least *L* liters of lemonade.
|
[
"4 12\n20 30 70 90\n",
"4 3\n10000 1000 100 10\n",
"4 3\n10 100 1000 10000\n",
"5 787787787\n123456789 234567890 345678901 456789012 987654321\n"
] |
[
"150\n",
"10\n",
"30\n",
"44981600785557577\n"
] |
In the first example you should buy one 8-liter bottle for 90 roubles and two 2-liter bottles for 30 roubles each. In total you'll get 12 liters of lemonade for just 150 roubles.
In the second example, even though you need only 3 liters, it's cheaper to buy a single 8-liter bottle for 10 roubles.
In the third example it's best to buy three 1-liter bottles for 10 roubles each, getting three liters for 30 roubles.
| 1,000
|
[
{
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"output": "150"
},
{
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"output": "10"
},
{
"input": "4 3\n10 100 1000 10000",
"output": "30"
},
{
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"output": "44981600785557577"
},
{
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"output": "371343078"
},
{
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{
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"output": "1000000000000000000"
},
{
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"output": "1"
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{
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{
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{
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"output": "1000000000"
},
{
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{
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"output": "1000000000"
},
{
"input": "4 8\n8 4 4 1",
"output": "1"
},
{
"input": "2 3\n10 1",
"output": "2"
},
{
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{
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"output": "1000000000000000000"
},
{
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"output": "246183685"
},
{
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"output": "10"
},
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"output": "12"
},
{
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"output": "134217728000000000"
},
{
"input": "1 536870912\n1000000000",
"output": "536870912000000000"
},
{
"input": "5 5\n34 22 21 20 30",
"output": "20"
},
{
"input": "1 787787787\n1",
"output": "787787787"
},
{
"input": "7 7\n34 22 21 20 30 20 20",
"output": "20"
},
{
"input": "5 5\n34 22 21 25 30",
"output": "25"
},
{
"input": "5 787787787\n123456789 234567890 345678901 456789012 1",
"output": "49236737"
},
{
"input": "6 6\n34 22 21 25 30 35",
"output": "25"
}
] | 1,665,918,956
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 86
| 46
| 0
|
n, l = map(int, input().split())
a = list(map(int, input().split()))
for i in range(0, n-1):
a[i+1] = min(a[i+1], 2*a[i])
s = 0
ans = 100**100
for i in range(n-1, -1, -1):
d = l // (1<<i)
s += d * a[i]
l -= d<<i;
ans = min(ans, s+(l>0)*a[i])
print(ans)
|
Title: Party Lemonade
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A New Year party is not a New Year party without lemonade! As usual, you are expecting a lot of guests, and buying lemonade has already become a pleasant necessity.
Your favorite store sells lemonade in bottles of *n* different volumes at different costs. A single bottle of type *i* has volume 2*i*<=-<=1 liters and costs *c**i* roubles. The number of bottles of each type in the store can be considered infinite.
You want to buy at least *L* liters of lemonade. How many roubles do you have to spend?
Input Specification:
The first line contains two integers *n* and *L* (1<=β€<=*n*<=β€<=30; 1<=β€<=*L*<=β€<=109)Β β the number of types of bottles in the store and the required amount of lemonade in liters, respectively.
The second line contains *n* integers *c*1,<=*c*2,<=...,<=*c**n* (1<=β€<=*c**i*<=β€<=109)Β β the costs of bottles of different types.
Output Specification:
Output a single integerΒ β the smallest number of roubles you have to pay in order to buy at least *L* liters of lemonade.
Demo Input:
['4 12\n20 30 70 90\n', '4 3\n10000 1000 100 10\n', '4 3\n10 100 1000 10000\n', '5 787787787\n123456789 234567890 345678901 456789012 987654321\n']
Demo Output:
['150\n', '10\n', '30\n', '44981600785557577\n']
Note:
In the first example you should buy one 8-liter bottle for 90 roubles and two 2-liter bottles for 30 roubles each. In total you'll get 12 liters of lemonade for just 150 roubles.
In the second example, even though you need only 3 liters, it's cheaper to buy a single 8-liter bottle for 10 roubles.
In the third example it's best to buy three 1-liter bottles for 10 roubles each, getting three liters for 30 roubles.
|
```python
n, l = map(int, input().split())
a = list(map(int, input().split()))
for i in range(0, n-1):
a[i+1] = min(a[i+1], 2*a[i])
s = 0
ans = 100**100
for i in range(n-1, -1, -1):
d = l // (1<<i)
s += d * a[i]
l -= d<<i;
ans = min(ans, s+(l>0)*a[i])
print(ans)
```
| 3
|
|
214
|
A
|
System of Equations
|
PROGRAMMING
| 800
|
[
"brute force"
] | null | null |
Furik loves math lessons very much, so he doesn't attend them, unlike Rubik. But now Furik wants to get a good mark for math. For that Ms. Ivanova, his math teacher, gave him a new task. Furik solved the task immediately. Can you?
You are given a system of equations:
You should count, how many there are pairs of integers (*a*,<=*b*) (0<=β€<=*a*,<=*b*) which satisfy the system.
|
A single line contains two integers *n*,<=*m* (1<=β€<=*n*,<=*m*<=β€<=1000) β the parameters of the system. The numbers on the line are separated by a space.
|
On a single line print the answer to the problem.
|
[
"9 3\n",
"14 28\n",
"4 20\n"
] |
[
"1\n",
"1\n",
"0\n"
] |
In the first sample the suitable pair is integers (3,β0). In the second sample the suitable pair is integers (3,β5). In the third sample there is no suitable pair.
| 500
|
[
{
"input": "9 3",
"output": "1"
},
{
"input": "14 28",
"output": "1"
},
{
"input": "4 20",
"output": "0"
},
{
"input": "18 198",
"output": "1"
},
{
"input": "22 326",
"output": "1"
},
{
"input": "26 104",
"output": "1"
},
{
"input": "14 10",
"output": "0"
},
{
"input": "8 20",
"output": "0"
},
{
"input": "2 8",
"output": "0"
},
{
"input": "20 11",
"output": "0"
},
{
"input": "57 447",
"output": "1"
},
{
"input": "1 1",
"output": "2"
},
{
"input": "66 296",
"output": "1"
},
{
"input": "75 683",
"output": "1"
},
{
"input": "227 975",
"output": "1"
},
{
"input": "247 499",
"output": "1"
},
{
"input": "266 116",
"output": "1"
},
{
"input": "286 916",
"output": "1"
},
{
"input": "307 341",
"output": "1"
},
{
"input": "451 121",
"output": "1"
},
{
"input": "471 921",
"output": "1"
},
{
"input": "502 346",
"output": "1"
},
{
"input": "535 59",
"output": "1"
},
{
"input": "555 699",
"output": "1"
},
{
"input": "747 351",
"output": "1"
},
{
"input": "790 64",
"output": "1"
},
{
"input": "810 704",
"output": "1"
},
{
"input": "855 225",
"output": "1"
},
{
"input": "902 34",
"output": "1"
},
{
"input": "922 514",
"output": "1"
},
{
"input": "971 131",
"output": "1"
},
{
"input": "991 931",
"output": "1"
},
{
"input": "840 780",
"output": "0"
},
{
"input": "102 595",
"output": "0"
},
{
"input": "139 433",
"output": "0"
},
{
"input": "968 288",
"output": "0"
},
{
"input": "563 354",
"output": "0"
},
{
"input": "994 975",
"output": "0"
},
{
"input": "456 221",
"output": "0"
},
{
"input": "205 210",
"output": "0"
},
{
"input": "1 11",
"output": "0"
},
{
"input": "1000 1000",
"output": "0"
},
{
"input": "3 3",
"output": "0"
},
{
"input": "11 99",
"output": "0"
},
{
"input": "2 2",
"output": "1"
},
{
"input": "11 1",
"output": "0"
},
{
"input": "6 6",
"output": "1"
},
{
"input": "100 452",
"output": "0"
},
{
"input": "420 380",
"output": "0"
},
{
"input": "31 31",
"output": "0"
},
{
"input": "2 6",
"output": "0"
},
{
"input": "112 134",
"output": "0"
},
{
"input": "13 13",
"output": "0"
},
{
"input": "1 571",
"output": "0"
}
] | 1,676,222,735
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 54
| 124
| 1,945,600
|
k=input().split(" ")
n,m = int(k[0]), int(k[1])
count = 0
for a in range(n + 1):
for b in range(m + 1):
if a**2 + b == n and b**2 + a == m:
count += 1
print(count)
|
Title: System of Equations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Furik loves math lessons very much, so he doesn't attend them, unlike Rubik. But now Furik wants to get a good mark for math. For that Ms. Ivanova, his math teacher, gave him a new task. Furik solved the task immediately. Can you?
You are given a system of equations:
You should count, how many there are pairs of integers (*a*,<=*b*) (0<=β€<=*a*,<=*b*) which satisfy the system.
Input Specification:
A single line contains two integers *n*,<=*m* (1<=β€<=*n*,<=*m*<=β€<=1000) β the parameters of the system. The numbers on the line are separated by a space.
Output Specification:
On a single line print the answer to the problem.
Demo Input:
['9 3\n', '14 28\n', '4 20\n']
Demo Output:
['1\n', '1\n', '0\n']
Note:
In the first sample the suitable pair is integers (3,β0). In the second sample the suitable pair is integers (3,β5). In the third sample there is no suitable pair.
|
```python
k=input().split(" ")
n,m = int(k[0]), int(k[1])
count = 0
for a in range(n + 1):
for b in range(m + 1):
if a**2 + b == n and b**2 + a == m:
count += 1
print(count)
```
| 3
|
|
51
|
A
|
Cheaterius's Problem
|
PROGRAMMING
| 1,300
|
[
"implementation"
] |
A. Cheaterius's Problem
|
2
|
256
|
Cheaterius is a famous in all the Berland astrologist, magician and wizard, and he also is a liar and a cheater. One of his latest inventions is Cheaterius' amulets! They bring luck and wealth, but are rather expensive. Cheaterius makes them himself. The technology of their making is kept secret. But we know that throughout long nights Cheaterius glues together domino pairs with super glue to get squares 2<=Γ<=2 which are the Cheaterius' magic amulets!
After a hard night Cheaterius made *n* amulets. Everyone of them represents a square 2<=Γ<=2, every quarter contains 1 to 6 dots. Now he wants sort them into piles, every pile must contain similar amulets. Two amulets are called similar if they can be rotated by 90, 180 or 270 degrees so that the following condition is met: the numbers of dots in the corresponding quarters should be the same. It is forbidden to turn over the amulets.
Write a program that by the given amulets will find the number of piles on Cheaterius' desk.
|
The first line contains an integer *n* (1<=β€<=*n*<=β€<=1000), where *n* is the number of amulets. Then the amulet's descriptions are contained. Every description occupies two lines and contains two numbers (from 1 to 6) in each line. Between every pair of amulets the line "**" is located.
|
Print the required number of piles.
|
[
"4\n31\n23\n**\n31\n23\n**\n13\n32\n**\n32\n13\n",
"4\n51\n26\n**\n54\n35\n**\n25\n61\n**\n45\n53\n"
] |
[
"1\n",
"2\n"
] |
none
| 500
|
[
{
"input": "4\n31\n23\n**\n31\n23\n**\n13\n32\n**\n32\n13",
"output": "1"
},
{
"input": "4\n51\n26\n**\n54\n35\n**\n25\n61\n**\n45\n53",
"output": "2"
},
{
"input": "4\n56\n61\n**\n31\n31\n**\n33\n11\n**\n11\n33",
"output": "2"
},
{
"input": "4\n36\n44\n**\n32\n46\n**\n66\n41\n**\n64\n34",
"output": "3"
},
{
"input": "3\n63\n63\n**\n66\n33\n**\n36\n36",
"output": "1"
},
{
"input": "3\n11\n54\n**\n42\n63\n**\n42\n63",
"output": "2"
},
{
"input": "3\n14\n54\n**\n45\n41\n**\n12\n22",
"output": "2"
},
{
"input": "7\n21\n33\n**\n33\n12\n**\n32\n31\n**\n21\n33\n**\n33\n12\n**\n32\n31\n**\n13\n23",
"output": "1"
}
] | 1,670,938,318
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 20
| 92
| 0
|
n = int(input())
piles = []
for i in range(n-1):
line1 = input()
line2 = input()
seperator = input()
line1 = [int(x) for x in str(line1)]
line2 = [int(x) for x in str(line2)]
line1.append(line2[1])
line1.append(line2[0])
amulet = line1
amulet90 = amulet.copy()
amulet90.insert(0, amulet[-1])
amulet90.pop()
amulet180 = amulet90.copy()
amulet180.insert(0, amulet90[-1])
amulet180.pop()
amulet270 = amulet180.copy()
amulet270.insert(0, amulet180[-1])
amulet270.pop()
if amulet in piles:
pass
elif amulet90 in piles:
pass
elif amulet180 in piles:
pass
elif amulet270 in piles:
pass
else:
piles.append(amulet)
line1 = input()
line2 = input()
line1 = [int(x) for x in str(line1)]
line2 = [int(x) for x in str(line2)]
line1.append(line2[1])
line1.append(line2[0])
amulet = line1
amulet90 = amulet.copy()
amulet90.insert(0, amulet[-1])
amulet90.pop()
amulet180 = amulet90.copy()
amulet180.insert(0, amulet90[-1])
amulet180.pop()
amulet270 = amulet180.copy()
amulet270.insert(0, amulet180[-1])
amulet270.pop()
if amulet in piles:
pass
elif amulet90 in piles:
pass
elif amulet180 in piles:
pass
elif amulet270 in piles:
pass
else:
piles.append(amulet)
print(len(piles))
|
Title: Cheaterius's Problem
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Cheaterius is a famous in all the Berland astrologist, magician and wizard, and he also is a liar and a cheater. One of his latest inventions is Cheaterius' amulets! They bring luck and wealth, but are rather expensive. Cheaterius makes them himself. The technology of their making is kept secret. But we know that throughout long nights Cheaterius glues together domino pairs with super glue to get squares 2<=Γ<=2 which are the Cheaterius' magic amulets!
After a hard night Cheaterius made *n* amulets. Everyone of them represents a square 2<=Γ<=2, every quarter contains 1 to 6 dots. Now he wants sort them into piles, every pile must contain similar amulets. Two amulets are called similar if they can be rotated by 90, 180 or 270 degrees so that the following condition is met: the numbers of dots in the corresponding quarters should be the same. It is forbidden to turn over the amulets.
Write a program that by the given amulets will find the number of piles on Cheaterius' desk.
Input Specification:
The first line contains an integer *n* (1<=β€<=*n*<=β€<=1000), where *n* is the number of amulets. Then the amulet's descriptions are contained. Every description occupies two lines and contains two numbers (from 1 to 6) in each line. Between every pair of amulets the line "**" is located.
Output Specification:
Print the required number of piles.
Demo Input:
['4\n31\n23\n**\n31\n23\n**\n13\n32\n**\n32\n13\n', '4\n51\n26\n**\n54\n35\n**\n25\n61\n**\n45\n53\n']
Demo Output:
['1\n', '2\n']
Note:
none
|
```python
n = int(input())
piles = []
for i in range(n-1):
line1 = input()
line2 = input()
seperator = input()
line1 = [int(x) for x in str(line1)]
line2 = [int(x) for x in str(line2)]
line1.append(line2[1])
line1.append(line2[0])
amulet = line1
amulet90 = amulet.copy()
amulet90.insert(0, amulet[-1])
amulet90.pop()
amulet180 = amulet90.copy()
amulet180.insert(0, amulet90[-1])
amulet180.pop()
amulet270 = amulet180.copy()
amulet270.insert(0, amulet180[-1])
amulet270.pop()
if amulet in piles:
pass
elif amulet90 in piles:
pass
elif amulet180 in piles:
pass
elif amulet270 in piles:
pass
else:
piles.append(amulet)
line1 = input()
line2 = input()
line1 = [int(x) for x in str(line1)]
line2 = [int(x) for x in str(line2)]
line1.append(line2[1])
line1.append(line2[0])
amulet = line1
amulet90 = amulet.copy()
amulet90.insert(0, amulet[-1])
amulet90.pop()
amulet180 = amulet90.copy()
amulet180.insert(0, amulet90[-1])
amulet180.pop()
amulet270 = amulet180.copy()
amulet270.insert(0, amulet180[-1])
amulet270.pop()
if amulet in piles:
pass
elif amulet90 in piles:
pass
elif amulet180 in piles:
pass
elif amulet270 in piles:
pass
else:
piles.append(amulet)
print(len(piles))
```
| 3.977
|
435
|
B
|
Pasha Maximizes
|
PROGRAMMING
| 1,400
|
[
"greedy"
] | null | null |
Pasha has a positive integer *a* without leading zeroes. Today he decided that the number is too small and he should make it larger. Unfortunately, the only operation Pasha can do is to swap two adjacent decimal digits of the integer.
Help Pasha count the maximum number he can get if he has the time to make at most *k* swaps.
|
The single line contains two integers *a* and *k* (1<=β€<=*a*<=β€<=1018;Β 0<=β€<=*k*<=β€<=100).
|
Print the maximum number that Pasha can get if he makes at most *k* swaps.
|
[
"1990 1\n",
"300 0\n",
"1034 2\n",
"9090000078001234 6\n"
] |
[
"9190\n",
"300\n",
"3104\n",
"9907000008001234\n"
] |
none
| 1,000
|
[
{
"input": "1990 1",
"output": "9190"
},
{
"input": "300 0",
"output": "300"
},
{
"input": "1034 2",
"output": "3104"
},
{
"input": "9090000078001234 6",
"output": "9907000008001234"
},
{
"input": "1234 3",
"output": "4123"
},
{
"input": "5 100",
"output": "5"
},
{
"input": "1234 5",
"output": "4312"
},
{
"input": "1234 6",
"output": "4321"
},
{
"input": "9022 2",
"output": "9220"
},
{
"input": "66838 4",
"output": "86863"
},
{
"input": "39940894417248510 10",
"output": "99984304417248510"
},
{
"input": "5314 4",
"output": "5431"
},
{
"input": "1026 9",
"output": "6210"
},
{
"input": "4529 8",
"output": "9542"
},
{
"input": "83811284 3",
"output": "88321184"
},
{
"input": "92153348 6",
"output": "98215334"
},
{
"input": "5846059 3",
"output": "8654059"
},
{
"input": "521325125110071928 4",
"output": "552132125110071928"
},
{
"input": "39940894417248510 10",
"output": "99984304417248510"
},
{
"input": "77172428736634377 29",
"output": "87777764122363437"
},
{
"input": "337775999910796051 37",
"output": "999997733751076051"
},
{
"input": "116995340392134308 27",
"output": "999654331120134308"
},
{
"input": "10120921290110921 20",
"output": "99221010120110921"
},
{
"input": "929201010190831892 30",
"output": "999928201010103182"
},
{
"input": "111111111111111119 8",
"output": "111111111911111111"
},
{
"input": "219810011901120912 100",
"output": "999822211111110000"
},
{
"input": "191919191919119911 100",
"output": "999999991111111111"
},
{
"input": "801211288881101019 22",
"output": "982111028888110101"
},
{
"input": "619911311932347059 3",
"output": "969111311932347059"
},
{
"input": "620737553540689123 2",
"output": "672037553540689123"
},
{
"input": "621563797296514835 3",
"output": "662153797296514835"
},
{
"input": "915277434701161 9",
"output": "977541234701161"
},
{
"input": "15603712376708 28",
"output": "87761503123670"
},
{
"input": "784069392990841 0",
"output": "784069392990841"
},
{
"input": "787464780004 2",
"output": "877644780004"
},
{
"input": "74604713975 29",
"output": "97776544310"
},
{
"input": "901000000954321789 5",
"output": "910009000054321789"
},
{
"input": "901000000954321789 10",
"output": "991000000504321789"
},
{
"input": "901000000954321789 28",
"output": "999100050000432178"
},
{
"input": "901000000954321789 40",
"output": "999810000050043217"
},
{
"input": "901000000954321789 70",
"output": "999875410000300021"
},
{
"input": "1234567891234567 99",
"output": "9877665544332211"
},
{
"input": "123456789123456789 100",
"output": "998877665544213123"
},
{
"input": "12345670123456789 100",
"output": "98776655443322101"
},
{
"input": "12 100",
"output": "21"
},
{
"input": "11223344556677889 47",
"output": "98821213344556677"
}
] | 1,536,137,173
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 46
| 124
| 0
|
a, k = [i for i in input().split()]
k = int(k)
a = list(a)
a = [int(i) for i in a] + [-1]
count = 1
while k > 0 and count < len(a):
_max = max(a[count:min(len(a), k + count)])
ind = a[count:min(len(a), k + count)].index(_max) + count
if _max > a[count - 1]:
k = k - ind + count - 1
a.pop(ind)
a.insert(count - 1, _max)
count +=1
print(*a[:-1], sep='')
|
Title: Pasha Maximizes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pasha has a positive integer *a* without leading zeroes. Today he decided that the number is too small and he should make it larger. Unfortunately, the only operation Pasha can do is to swap two adjacent decimal digits of the integer.
Help Pasha count the maximum number he can get if he has the time to make at most *k* swaps.
Input Specification:
The single line contains two integers *a* and *k* (1<=β€<=*a*<=β€<=1018;Β 0<=β€<=*k*<=β€<=100).
Output Specification:
Print the maximum number that Pasha can get if he makes at most *k* swaps.
Demo Input:
['1990 1\n', '300 0\n', '1034 2\n', '9090000078001234 6\n']
Demo Output:
['9190\n', '300\n', '3104\n', '9907000008001234\n']
Note:
none
|
```python
a, k = [i for i in input().split()]
k = int(k)
a = list(a)
a = [int(i) for i in a] + [-1]
count = 1
while k > 0 and count < len(a):
_max = max(a[count:min(len(a), k + count)])
ind = a[count:min(len(a), k + count)].index(_max) + count
if _max > a[count - 1]:
k = k - ind + count - 1
a.pop(ind)
a.insert(count - 1, _max)
count +=1
print(*a[:-1], sep='')
```
| 3
|
|
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" β 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<=β€<=*n*<=β€<=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<=β€<=*x**i*,<=*y**i*,<=*z**i*<=β€<=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"
}
] | 1,690,657,122
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 81
| 92
| 0
|
n = int(input())
arr = [0,0,0]
for _ in range(n):
arr1 = list(map(int,input().split()))
for i in range(3): arr[i] += arr1[i]
print("YES" if arr == [0,0,0] else "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" β 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<=β€<=*n*<=β€<=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<=β€<=*x**i*,<=*y**i*,<=*z**i*<=β€<=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 = int(input())
arr = [0,0,0]
for _ in range(n):
arr1 = list(map(int,input().split()))
for i in range(3): arr[i] += arr1[i]
print("YES" if arr == [0,0,0] else "NO")
```
| 3.977
|
645
|
B
|
Mischievous Mess Makers
|
PROGRAMMING
| 1,200
|
[
"greedy",
"math"
] | null | null |
It is a balmy spring afternoon, and Farmer John's *n* cows are ruminating about link-cut cacti in their stalls. The cows, labeled 1 through *n*, are arranged so that the *i*-th cow occupies the *i*-th stall from the left. However, Elsie, after realizing that she will forever live in the shadows beyond Bessie's limelight, has formed the Mischievous Mess Makers and is plotting to disrupt this beautiful pastoral rhythm. While Farmer John takes his *k* minute long nap, Elsie and the Mess Makers plan to repeatedly choose two distinct stalls and swap the cows occupying those stalls, making no more than one swap each minute.
Being the meticulous pranksters that they are, the Mischievous Mess Makers would like to know the maximum messiness attainable in the *k* minutes that they have. We denote as *p**i* the label of the cow in the *i*-th stall. The messiness of an arrangement of cows is defined as the number of pairs (*i*,<=*j*) such that *i*<=<<=*j* and *p**i*<=><=*p**j*.
|
The first line of the input contains two integers *n* and *k* (1<=β€<=*n*,<=*k*<=β€<=100<=000)Β β the number of cows and the length of Farmer John's nap, respectively.
|
Output a single integer, the maximum messiness that the Mischievous Mess Makers can achieve by performing no more than *k* swaps.
|
[
"5 2\n",
"1 10\n"
] |
[
"10\n",
"0\n"
] |
In the first sample, the Mischievous Mess Makers can swap the cows in the stalls 1 and 5 during the first minute, then the cows in stalls 2 and 4 during the second minute. This reverses the arrangement of cows, giving us a total messiness of 10.
In the second sample, there is only one cow, so the maximum possible messiness is 0.
| 1,000
|
[
{
"input": "5 2",
"output": "10"
},
{
"input": "1 10",
"output": "0"
},
{
"input": "100000 2",
"output": "399990"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "8 3",
"output": "27"
},
{
"input": "7 1",
"output": "11"
},
{
"input": "100000 40000",
"output": "4799960000"
},
{
"input": "1 1000",
"output": "0"
},
{
"input": "100 45",
"output": "4905"
},
{
"input": "9 2",
"output": "26"
},
{
"input": "456 78",
"output": "58890"
},
{
"input": "100000 50000",
"output": "4999950000"
},
{
"input": "100000 50001",
"output": "4999950000"
},
{
"input": "100000 50002",
"output": "4999950000"
},
{
"input": "100000 50003",
"output": "4999950000"
},
{
"input": "100000 49998",
"output": "4999949994"
},
{
"input": "100000 49997",
"output": "4999949985"
},
{
"input": "99999 49998",
"output": "4999849998"
},
{
"input": "99999 49997",
"output": "4999849991"
},
{
"input": "99999 49996",
"output": "4999849980"
},
{
"input": "99999 50000",
"output": "4999850001"
},
{
"input": "99999 50001",
"output": "4999850001"
},
{
"input": "99999 50002",
"output": "4999850001"
},
{
"input": "30062 9",
"output": "540945"
},
{
"input": "13486 3",
"output": "80895"
},
{
"input": "29614 7",
"output": "414491"
},
{
"input": "13038 8",
"output": "208472"
},
{
"input": "96462 6",
"output": "1157466"
},
{
"input": "22599 93799",
"output": "255346101"
},
{
"input": "421 36817",
"output": "88410"
},
{
"input": "72859 65869",
"output": "2654180511"
},
{
"input": "37916 5241",
"output": "342494109"
},
{
"input": "47066 12852",
"output": "879423804"
},
{
"input": "84032 21951",
"output": "2725458111"
},
{
"input": "70454 75240",
"output": "2481847831"
},
{
"input": "86946 63967",
"output": "3779759985"
},
{
"input": "71128 11076",
"output": "1330260828"
},
{
"input": "46111 64940",
"output": "1063089105"
},
{
"input": "46111 64940",
"output": "1063089105"
},
{
"input": "56500 84184",
"output": "1596096750"
},
{
"input": "60108 83701",
"output": "1806455778"
},
{
"input": "1 2",
"output": "0"
},
{
"input": "1 3",
"output": "0"
},
{
"input": "1 4",
"output": "0"
},
{
"input": "1 5",
"output": "0"
},
{
"input": "1 6",
"output": "0"
},
{
"input": "2 1",
"output": "1"
},
{
"input": "2 2",
"output": "1"
},
{
"input": "2 3",
"output": "1"
},
{
"input": "2 4",
"output": "1"
},
{
"input": "2 5",
"output": "1"
},
{
"input": "3 1",
"output": "3"
},
{
"input": "3 2",
"output": "3"
},
{
"input": "3 3",
"output": "3"
},
{
"input": "3 4",
"output": "3"
},
{
"input": "3 5",
"output": "3"
},
{
"input": "4 1",
"output": "5"
},
{
"input": "4 2",
"output": "6"
},
{
"input": "4 3",
"output": "6"
},
{
"input": "4 4",
"output": "6"
},
{
"input": "4 5",
"output": "6"
},
{
"input": "5 1",
"output": "7"
},
{
"input": "5 3",
"output": "10"
},
{
"input": "5 4",
"output": "10"
},
{
"input": "5 5",
"output": "10"
},
{
"input": "6 1",
"output": "9"
},
{
"input": "6 2",
"output": "14"
},
{
"input": "6 3",
"output": "15"
},
{
"input": "7 2",
"output": "18"
},
{
"input": "7 3",
"output": "21"
},
{
"input": "7 4",
"output": "21"
},
{
"input": "10 2",
"output": "30"
},
{
"input": "60982 2",
"output": "243918"
},
{
"input": "23426 23",
"output": "1076515"
},
{
"input": "444 3",
"output": "2643"
},
{
"input": "18187 433",
"output": "15374531"
},
{
"input": "6895 3544",
"output": "23767065"
},
{
"input": "56204 22352",
"output": "1513297456"
},
{
"input": "41977 5207",
"output": "382917573"
},
{
"input": "78147 2321",
"output": "351981971"
},
{
"input": "99742 62198",
"output": "4974183411"
},
{
"input": "72099 38339",
"output": "2599096851"
},
{
"input": "82532 4838",
"output": "751762306"
},
{
"input": "79410 33144",
"output": "3066847464"
},
{
"input": "11021 3389",
"output": "51726307"
},
{
"input": "66900 7572",
"output": "898455660"
},
{
"input": "99999 49999",
"output": "4999850001"
},
{
"input": "100000 49999",
"output": "4999949999"
},
{
"input": "100000 100000",
"output": "4999950000"
},
{
"input": "100000 1",
"output": "199997"
},
{
"input": "4 100",
"output": "6"
},
{
"input": "100000 1234",
"output": "243753254"
}
] | 1,458,323,855
| 4,955
|
Python 3
|
OK
|
TESTS
| 92
| 93
| 0
|
n, k = map(int, input().split())
h = k * 2
n -= 1
ans = 0
for i in range(h):
if n == 0:
break
ans += n
n -= 1
print(ans)
|
Title: Mischievous Mess Makers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It is a balmy spring afternoon, and Farmer John's *n* cows are ruminating about link-cut cacti in their stalls. The cows, labeled 1 through *n*, are arranged so that the *i*-th cow occupies the *i*-th stall from the left. However, Elsie, after realizing that she will forever live in the shadows beyond Bessie's limelight, has formed the Mischievous Mess Makers and is plotting to disrupt this beautiful pastoral rhythm. While Farmer John takes his *k* minute long nap, Elsie and the Mess Makers plan to repeatedly choose two distinct stalls and swap the cows occupying those stalls, making no more than one swap each minute.
Being the meticulous pranksters that they are, the Mischievous Mess Makers would like to know the maximum messiness attainable in the *k* minutes that they have. We denote as *p**i* the label of the cow in the *i*-th stall. The messiness of an arrangement of cows is defined as the number of pairs (*i*,<=*j*) such that *i*<=<<=*j* and *p**i*<=><=*p**j*.
Input Specification:
The first line of the input contains two integers *n* and *k* (1<=β€<=*n*,<=*k*<=β€<=100<=000)Β β the number of cows and the length of Farmer John's nap, respectively.
Output Specification:
Output a single integer, the maximum messiness that the Mischievous Mess Makers can achieve by performing no more than *k* swaps.
Demo Input:
['5 2\n', '1 10\n']
Demo Output:
['10\n', '0\n']
Note:
In the first sample, the Mischievous Mess Makers can swap the cows in the stalls 1 and 5 during the first minute, then the cows in stalls 2 and 4 during the second minute. This reverses the arrangement of cows, giving us a total messiness of 10.
In the second sample, there is only one cow, so the maximum possible messiness is 0.
|
```python
n, k = map(int, input().split())
h = k * 2
n -= 1
ans = 0
for i in range(h):
if n == 0:
break
ans += n
n -= 1
print(ans)
```
| 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 "internationalizationΒ» 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<=β€<=*n*<=β€<=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"
}
] | 1,678,555,717
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 20
| 46
| 0
|
t = int(input())
for i in range(t):
s = input()
if len(s)<=10:
print(s)
else:
n = len(s)-2
new = s[0]+str(n)+s[-1]
print(new)
|
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 "internationalizationΒ» 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<=β€<=*n*<=β€<=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
t = int(input())
for i in range(t):
s = input()
if len(s)<=10:
print(s)
else:
n = len(s)-2
new = s[0]+str(n)+s[-1]
print(new)
```
| 3.977
|
224
|
A
|
Parallelepiped
|
PROGRAMMING
| 1,100
|
[
"brute force",
"geometry",
"math"
] | null | null |
You've got a rectangular parallelepiped with integer edge lengths. You know the areas of its three faces that have a common vertex. Your task is to find the sum of lengths of all 12 edges of this parallelepiped.
|
The first and the single line contains three space-separated integers β the areas of the parallelepiped's faces. The area's values are positive (<=><=0) and do not exceed 104. It is guaranteed that there exists at least one parallelepiped that satisfies the problem statement.
|
Print a single number β the sum of all edges of the parallelepiped.
|
[
"1 1 1\n",
"4 6 6\n"
] |
[
"12\n",
"28\n"
] |
In the first sample the parallelepiped has sizes 1βΓβ1βΓβ1, in the second oneΒ β 2βΓβ2βΓβ3.
| 500
|
[
{
"input": "1 1 1",
"output": "12"
},
{
"input": "4 6 6",
"output": "28"
},
{
"input": "20 10 50",
"output": "68"
},
{
"input": "9 4 36",
"output": "56"
},
{
"input": "324 9 36",
"output": "184"
},
{
"input": "1333 93 129",
"output": "308"
},
{
"input": "1022 584 112",
"output": "380"
},
{
"input": "66 174 319",
"output": "184"
},
{
"input": "912 276 1748",
"output": "444"
},
{
"input": "65 156 60",
"output": "120"
},
{
"input": "1 10000 10000",
"output": "40008"
},
{
"input": "1485 55 27",
"output": "332"
},
{
"input": "152 108 4104",
"output": "528"
},
{
"input": "1656 6900 1350",
"output": "740"
},
{
"input": "12 14 42",
"output": "60"
},
{
"input": "615 18 1230",
"output": "856"
},
{
"input": "680 60 408",
"output": "336"
},
{
"input": "644 966 6",
"output": "1308"
},
{
"input": "1 432 432",
"output": "1736"
},
{
"input": "2239 2239 1",
"output": "8964"
},
{
"input": "4106 8212 2",
"output": "16436"
},
{
"input": "10000 10000 10000",
"output": "1200"
},
{
"input": "3623 3623 1",
"output": "14500"
},
{
"input": "9801 9801 9801",
"output": "1188"
},
{
"input": "10000 1 10000",
"output": "40008"
},
{
"input": "9 9 9",
"output": "36"
},
{
"input": "9801 9702 9702",
"output": "1184"
}
] | 1,622,977,373
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 27
| 498
| 409,600
|
from sys import *
'''sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w') '''
from collections import defaultdict as dd
from math import *
from bisect import *
#sys.setrecursionlimit(10 ** 8)
def sinp():
return input()
def inp():
return int(sinp())
def minp():
return map(int, sinp().split())
def linp():
return list(minp())
def strl():
return list(sinp())
def pr(x):
print(x)
mod = int(1e9+7)
a, b, c = minp()
x = sqrt(a * b * c // b ** 2)
y = sqrt(a * b * c // c ** 2)
z = sqrt(a * b * c // a ** 2)
pr(4 * int((x + y + z)))
|
Title: Parallelepiped
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got a rectangular parallelepiped with integer edge lengths. You know the areas of its three faces that have a common vertex. Your task is to find the sum of lengths of all 12 edges of this parallelepiped.
Input Specification:
The first and the single line contains three space-separated integers β the areas of the parallelepiped's faces. The area's values are positive (<=><=0) and do not exceed 104. It is guaranteed that there exists at least one parallelepiped that satisfies the problem statement.
Output Specification:
Print a single number β the sum of all edges of the parallelepiped.
Demo Input:
['1 1 1\n', '4 6 6\n']
Demo Output:
['12\n', '28\n']
Note:
In the first sample the parallelepiped has sizes 1βΓβ1βΓβ1, in the second oneΒ β 2βΓβ2βΓβ3.
|
```python
from sys import *
'''sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w') '''
from collections import defaultdict as dd
from math import *
from bisect import *
#sys.setrecursionlimit(10 ** 8)
def sinp():
return input()
def inp():
return int(sinp())
def minp():
return map(int, sinp().split())
def linp():
return list(minp())
def strl():
return list(sinp())
def pr(x):
print(x)
mod = int(1e9+7)
a, b, c = minp()
x = sqrt(a * b * c // b ** 2)
y = sqrt(a * b * c // c ** 2)
z = sqrt(a * b * c // a ** 2)
pr(4 * int((x + y + z)))
```
| 3
|
|
116
|
A
|
Tram
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Linear Kingdom has exactly one tram line. It has *n* stops, numbered from 1 to *n* in the order of tram's movement. At the *i*-th stop *a**i* passengers exit the tram, while *b**i* passengers enter it. The tram is empty before it arrives at the first stop. Also, when the tram arrives at the last stop, all passengers exit so that it becomes empty.
Your task is to calculate the tram's minimum capacity such that the number of people inside the tram at any time never exceeds this capacity. Note that at each stop all exiting passengers exit before any entering passenger enters the tram.
|
The first line contains a single number *n* (2<=β€<=*n*<=β€<=1000) β the number of the tram's stops.
Then *n* lines follow, each contains two integers *a**i* and *b**i* (0<=β€<=*a**i*,<=*b**i*<=β€<=1000) β the number of passengers that exits the tram at the *i*-th stop, and the number of passengers that enter the tram at the *i*-th stop. The stops are given from the first to the last stop in the order of tram's movement.
- The number of people who exit at a given stop does not exceed the total number of people in the tram immediately before it arrives at the stop. More formally, . This particularly means that *a*1<==<=0. - At the last stop, all the passengers exit the tram and it becomes empty. More formally, . - No passenger will enter the train at the last stop. That is, *b**n*<==<=0.
|
Print a single integer denoting the minimum possible capacity of the tram (0 is allowed).
|
[
"4\n0 3\n2 5\n4 2\n4 0\n"
] |
[
"6\n"
] |
For the first example, a capacity of 6 is sufficient:
- At the first stop, the number of passengers inside the tram before arriving is 0. Then, 3 passengers enter the tram, and the number of passengers inside the tram becomes 3. - At the second stop, 2 passengers exit the tram (1 passenger remains inside). Then, 5 passengers enter the tram. There are 6 passengers inside the tram now. - At the third stop, 4 passengers exit the tram (2 passengers remain inside). Then, 2 passengers enter the tram. There are 4 passengers inside the tram now. - Finally, all the remaining passengers inside the tram exit the tram at the last stop. There are no passenger inside the tram now, which is in line with the constraints.
Since the number of passengers inside the tram never exceeds 6, a capacity of 6 is sufficient. Furthermore it is not possible for the tram to have a capacity less than 6. Hence, 6 is the correct answer.
| 500
|
[
{
"input": "4\n0 3\n2 5\n4 2\n4 0",
"output": "6"
},
{
"input": "5\n0 4\n4 6\n6 5\n5 4\n4 0",
"output": "6"
},
{
"input": "10\n0 5\n1 7\n10 8\n5 3\n0 5\n3 3\n8 8\n0 6\n10 1\n9 0",
"output": "18"
},
{
"input": "3\n0 1\n1 1\n1 0",
"output": "1"
},
{
"input": "4\n0 1\n0 1\n1 0\n1 0",
"output": "2"
},
{
"input": "3\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "3\n0 1000\n1000 1000\n1000 0",
"output": "1000"
},
{
"input": "5\n0 73\n73 189\n189 766\n766 0\n0 0",
"output": "766"
},
{
"input": "5\n0 0\n0 0\n0 0\n0 1\n1 0",
"output": "1"
},
{
"input": "5\n0 917\n917 923\n904 992\n1000 0\n11 0",
"output": "1011"
},
{
"input": "5\n0 1\n1 2\n2 1\n1 2\n2 0",
"output": "2"
},
{
"input": "5\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "20\n0 7\n2 1\n2 2\n5 7\n2 6\n6 10\n2 4\n0 4\n7 4\n8 0\n10 6\n2 1\n6 1\n1 7\n0 3\n8 7\n6 3\n6 3\n1 1\n3 0",
"output": "22"
},
{
"input": "5\n0 1000\n1000 1000\n1000 1000\n1000 1000\n1000 0",
"output": "1000"
},
{
"input": "10\n0 592\n258 598\n389 203\n249 836\n196 635\n478 482\n994 987\n1000 0\n769 0\n0 0",
"output": "1776"
},
{
"input": "10\n0 1\n1 0\n0 0\n0 0\n0 0\n0 1\n1 1\n0 1\n1 0\n1 0",
"output": "2"
},
{
"input": "10\n0 926\n926 938\n938 931\n931 964\n937 989\n983 936\n908 949\n997 932\n945 988\n988 0",
"output": "1016"
},
{
"input": "10\n0 1\n1 2\n1 2\n2 2\n2 2\n2 2\n1 1\n1 1\n2 1\n2 0",
"output": "3"
},
{
"input": "10\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "10\n0 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 0",
"output": "1000"
},
{
"input": "50\n0 332\n332 268\n268 56\n56 711\n420 180\n160 834\n149 341\n373 777\n763 93\n994 407\n86 803\n700 132\n471 608\n429 467\n75 5\n638 305\n405 853\n316 478\n643 163\n18 131\n648 241\n241 766\n316 847\n640 380\n923 759\n789 41\n125 421\n421 9\n9 388\n388 829\n408 108\n462 856\n816 411\n518 688\n290 7\n405 912\n397 772\n396 652\n394 146\n27 648\n462 617\n514 433\n780 35\n710 705\n460 390\n194 508\n643 56\n172 469\n1000 0\n194 0",
"output": "2071"
},
{
"input": "50\n0 0\n0 1\n1 1\n0 1\n0 0\n1 0\n0 0\n1 0\n0 0\n0 0\n0 0\n0 0\n0 1\n0 0\n0 0\n0 1\n1 0\n0 1\n0 0\n1 1\n1 0\n0 1\n0 0\n1 1\n0 1\n1 0\n1 1\n1 0\n0 0\n1 1\n1 0\n0 1\n0 0\n0 1\n1 1\n1 1\n1 1\n1 0\n1 1\n1 0\n0 1\n1 0\n0 0\n0 1\n1 1\n1 1\n0 1\n0 0\n1 0\n1 0",
"output": "3"
},
{
"input": "50\n0 926\n926 971\n915 980\n920 965\n954 944\n928 952\n955 980\n916 980\n906 935\n944 913\n905 923\n912 922\n965 934\n912 900\n946 930\n931 983\n979 905\n925 969\n924 926\n910 914\n921 977\n934 979\n962 986\n942 909\n976 903\n982 982\n991 941\n954 929\n902 980\n947 983\n919 924\n917 943\n916 905\n907 913\n964 977\n984 904\n905 999\n950 970\n986 906\n993 970\n960 994\n963 983\n918 986\n980 900\n931 986\n993 997\n941 909\n907 909\n1000 0\n278 0",
"output": "1329"
},
{
"input": "2\n0 863\n863 0",
"output": "863"
},
{
"input": "50\n0 1\n1 2\n2 2\n1 1\n1 1\n1 2\n1 2\n1 1\n1 2\n1 1\n1 1\n1 2\n1 2\n1 1\n2 1\n2 2\n1 2\n2 2\n1 2\n2 1\n2 1\n2 2\n2 1\n1 2\n1 2\n2 1\n1 1\n2 2\n1 1\n2 1\n2 2\n2 1\n1 2\n2 2\n1 2\n1 1\n1 1\n2 1\n2 1\n2 2\n2 1\n2 1\n1 2\n1 2\n1 2\n1 2\n2 0\n2 0\n2 0\n0 0",
"output": "8"
},
{
"input": "50\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\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\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\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "100\n0 1\n0 0\n0 0\n1 0\n0 0\n0 1\n0 1\n1 1\n0 0\n0 0\n1 1\n0 0\n1 1\n0 1\n1 1\n0 1\n1 1\n1 0\n1 0\n0 0\n1 0\n0 1\n1 0\n0 0\n0 0\n1 1\n1 1\n0 1\n0 0\n1 0\n1 1\n0 1\n1 0\n1 1\n0 1\n1 1\n1 0\n0 0\n0 0\n0 1\n0 0\n0 1\n1 1\n0 0\n1 1\n1 1\n0 0\n0 1\n1 0\n0 1\n0 0\n0 1\n0 1\n1 1\n1 1\n1 1\n0 0\n0 0\n1 1\n0 1\n0 1\n1 0\n0 0\n0 0\n1 1\n0 1\n0 1\n1 1\n1 1\n0 1\n1 1\n1 1\n0 0\n1 0\n0 1\n0 0\n0 0\n1 1\n1 1\n1 1\n1 1\n0 1\n1 0\n1 0\n1 0\n1 0\n1 0\n0 0\n1 0\n1 0\n0 0\n1 0\n0 0\n0 1\n1 0\n0 1\n1 0\n1 0\n1 0\n1 0",
"output": "11"
},
{
"input": "100\n0 2\n1 2\n2 1\n1 2\n1 2\n2 1\n2 2\n1 1\n1 1\n2 1\n1 2\n2 1\n1 2\n2 2\n2 2\n2 2\n1 2\n2 2\n2 1\n1 1\n1 1\n1 1\n2 2\n1 2\n2 2\n1 1\n1 1\n1 1\n1 1\n2 2\n1 2\n2 1\n1 1\n2 2\n1 1\n2 1\n1 1\n2 2\n2 1\n1 2\n1 1\n1 2\n2 1\n2 2\n1 1\n2 1\n1 1\n2 1\n1 1\n1 2\n2 2\n2 2\n1 1\n2 2\n1 2\n2 1\n2 1\n1 1\n1 1\n1 2\n1 2\n1 1\n1 1\n2 1\n1 2\n1 2\n2 1\n2 2\n2 2\n2 2\n2 1\n2 2\n1 1\n1 2\n1 2\n1 1\n2 2\n2 2\n1 1\n2 1\n1 1\n1 2\n1 2\n1 2\n1 1\n1 1\n2 2\n1 2\n2 1\n2 1\n2 1\n1 2\n1 2\n1 1\n2 2\n1 2\n2 0\n2 0\n2 0\n1 0",
"output": "7"
},
{
"input": "100\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\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\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\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\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\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\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "2\n0 1\n1 0",
"output": "1"
},
{
"input": "2\n0 900\n900 0",
"output": "900"
},
{
"input": "2\n0 1\n1 0",
"output": "1"
},
{
"input": "2\n0 0\n0 0",
"output": "0"
},
{
"input": "2\n0 1000\n1000 0",
"output": "1000"
},
{
"input": "3\n0 802\n175 188\n815 0",
"output": "815"
},
{
"input": "3\n0 910\n910 976\n976 0",
"output": "976"
},
{
"input": "3\n0 2\n2 1\n1 0",
"output": "2"
}
] | 1,691,348,350
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 71
| 92
| 0
|
s = []
s1 = 0
for _ in range(int(input())) :
a , b = map(int,input().split())
s1 += b
s1 -= a
s.append(s1)
print(max(s))
|
Title: Tram
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Linear Kingdom has exactly one tram line. It has *n* stops, numbered from 1 to *n* in the order of tram's movement. At the *i*-th stop *a**i* passengers exit the tram, while *b**i* passengers enter it. The tram is empty before it arrives at the first stop. Also, when the tram arrives at the last stop, all passengers exit so that it becomes empty.
Your task is to calculate the tram's minimum capacity such that the number of people inside the tram at any time never exceeds this capacity. Note that at each stop all exiting passengers exit before any entering passenger enters the tram.
Input Specification:
The first line contains a single number *n* (2<=β€<=*n*<=β€<=1000) β the number of the tram's stops.
Then *n* lines follow, each contains two integers *a**i* and *b**i* (0<=β€<=*a**i*,<=*b**i*<=β€<=1000) β the number of passengers that exits the tram at the *i*-th stop, and the number of passengers that enter the tram at the *i*-th stop. The stops are given from the first to the last stop in the order of tram's movement.
- The number of people who exit at a given stop does not exceed the total number of people in the tram immediately before it arrives at the stop. More formally, . This particularly means that *a*1<==<=0. - At the last stop, all the passengers exit the tram and it becomes empty. More formally, . - No passenger will enter the train at the last stop. That is, *b**n*<==<=0.
Output Specification:
Print a single integer denoting the minimum possible capacity of the tram (0 is allowed).
Demo Input:
['4\n0 3\n2 5\n4 2\n4 0\n']
Demo Output:
['6\n']
Note:
For the first example, a capacity of 6 is sufficient:
- At the first stop, the number of passengers inside the tram before arriving is 0. Then, 3 passengers enter the tram, and the number of passengers inside the tram becomes 3. - At the second stop, 2 passengers exit the tram (1 passenger remains inside). Then, 5 passengers enter the tram. There are 6 passengers inside the tram now. - At the third stop, 4 passengers exit the tram (2 passengers remain inside). Then, 2 passengers enter the tram. There are 4 passengers inside the tram now. - Finally, all the remaining passengers inside the tram exit the tram at the last stop. There are no passenger inside the tram now, which is in line with the constraints.
Since the number of passengers inside the tram never exceeds 6, a capacity of 6 is sufficient. Furthermore it is not possible for the tram to have a capacity less than 6. Hence, 6 is the correct answer.
|
```python
s = []
s1 = 0
for _ in range(int(input())) :
a , b = map(int,input().split())
s1 += b
s1 -= a
s.append(s1)
print(max(s))
```
| 3
|
|
59
|
A
|
Word
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] |
A. Word
|
2
|
256
|
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP β with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
|
The first line contains a word *s* β it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
|
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
|
[
"HoUse\n",
"ViP\n",
"maTRIx\n"
] |
[
"house\n",
"VIP\n",
"matrix\n"
] |
none
| 500
|
[
{
"input": "HoUse",
"output": "house"
},
{
"input": "ViP",
"output": "VIP"
},
{
"input": "maTRIx",
"output": "matrix"
},
{
"input": "BNHWpnpawg",
"output": "bnhwpnpawg"
},
{
"input": "VTYGP",
"output": "VTYGP"
},
{
"input": "CHNenu",
"output": "chnenu"
},
{
"input": "ERPZGrodyu",
"output": "erpzgrodyu"
},
{
"input": "KSXBXWpebh",
"output": "KSXBXWPEBH"
},
{
"input": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv",
"output": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv"
},
{
"input": "Amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd",
"output": "amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd"
},
{
"input": "ISAGFJFARYFBLOPQDSHWGMCNKMFTLVFUGNJEWGWNBLXUIATXEkqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv",
"output": "isagfjfaryfblopqdshwgmcnkmftlvfugnjewgwnblxuiatxekqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv"
},
{
"input": "XHRPXZEGHSOCJPICUIXSKFUZUPYTSGJSDIYBCMNMNBPNDBXLXBzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg",
"output": "xhrpxzeghsocjpicuixskfuzupytsgjsdiybcmnmnbpndbxlxbzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg"
},
{
"input": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGAdkcetqjljtmttlonpekcovdzebzdkzggwfsxhapmjkdbuceak",
"output": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGADKCETQJLJTMTTLONPEKCOVDZEBZDKZGGWFSXHAPMJKDBUCEAK"
},
{
"input": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFw",
"output": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFW"
},
{
"input": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB",
"output": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB"
},
{
"input": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge",
"output": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge"
},
{
"input": "Ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw",
"output": "ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw"
},
{
"input": "YQOMLKYAORUQQUCQZCDYMIVDHGWZFFRMUVTAWCHERFPMNRYRIkgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks",
"output": "yqomlkyaoruqqucqzcdymivdhgwzffrmuvtawcherfpmnryrikgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks"
},
{
"input": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJcuusigdqfkumewjtdyitveeiaybwrhomrwmpdipjwiuxfnwuz",
"output": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJCUUSIGDQFKUMEWJTDYITVEEIAYBWRHOMRWMPDIPJWIUXFNWUZ"
},
{
"input": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWuckzcpxosodcjaaakvlxpbiigsiauviilylnnqlyucziihqg",
"output": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWUCKZCPXOSODCJAAAKVLXPBIIGSIAUVIILYLNNQLYUCZIIHQG"
},
{
"input": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO",
"output": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO"
},
{
"input": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDd",
"output": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDD"
},
{
"input": "EBWOVSNLYTWWXrnovgugogtjahnmatomubebyxakas",
"output": "ebwovsnlytwwxrnovgugogtjahnmatomubebyxakas"
},
{
"input": "AORNNDKTRLRVGDPXJKXFTPFpopuzrlqumrxssskvbm",
"output": "AORNNDKTRLRVGDPXJKXFTPFPOPUZRLQUMRXSSSKVBM"
},
{
"input": "DBBZJCWQEVGYHEEUHJAWRHBDPsnplijbwtdibqrrheebfxqcvn",
"output": "dbbzjcwqevgyheeuhjawrhbdpsnplijbwtdibqrrheebfxqcvn"
},
{
"input": "t",
"output": "t"
},
{
"input": "N",
"output": "N"
},
{
"input": "kv",
"output": "kv"
},
{
"input": "Ur",
"output": "ur"
},
{
"input": "CN",
"output": "CN"
}
] | 1,521,888,147
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 186
| 7,372,800
|
s = input()
c, m = 0, 0
for i in s:
if 65<=ord(i)<=90:
c+=1
elif 97<=ord(i)<=122:
m+=1
if c>m:
print(s.upper())
else:
print(s.lower())
|
Title: Word
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP β with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
Input Specification:
The first line contains a word *s* β it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Output Specification:
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
Demo Input:
['HoUse\n', 'ViP\n', 'maTRIx\n']
Demo Output:
['house\n', 'VIP\n', 'matrix\n']
Note:
none
|
```python
s = input()
c, m = 0, 0
for i in s:
if 65<=ord(i)<=90:
c+=1
elif 97<=ord(i)<=122:
m+=1
if c>m:
print(s.upper())
else:
print(s.lower())
```
| 3.939767
|
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<=β€<=*n*<=β€<=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"
}
] | 1,574,447,906
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 33
| 156
| 0
|
import math
n = int(input())
# lower_pow = int(math.log10(n))
# answer = 0
# print(lower_pow)
# for i in range(0, lower_pow):
# answer = 2*answer + 2
# print(answer)
ans = 0
aux = 1
a = str(n)
for i in range(len(a)-1, -1, -1):
if(a[i] == '7'):
ans = ans + 2**aux
else:
ans = ans + 2**aux - 2**(aux-1)
aux = aux + 1
print(ans)
|
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<=β€<=*n*<=β€<=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
import math
n = int(input())
# lower_pow = int(math.log10(n))
# answer = 0
# print(lower_pow)
# for i in range(0, lower_pow):
# answer = 2*answer + 2
# print(answer)
ans = 0
aux = 1
a = str(n)
for i in range(len(a)-1, -1, -1):
if(a[i] == '7'):
ans = ans + 2**aux
else:
ans = ans + 2**aux - 2**(aux-1)
aux = aux + 1
print(ans)
```
| 3
|
|
799
|
A
|
Carrot Cakes
|
PROGRAMMING
| 1,100
|
[
"brute force",
"implementation"
] | null | null |
In some game by Playrix it takes *t* minutes for an oven to bake *k* carrot cakes, all cakes are ready at the same moment *t* minutes after they started baking. Arkady needs at least *n* cakes to complete a task, but he currently don't have any. However, he has infinitely many ingredients and one oven. Moreover, Arkady can build one more similar oven to make the process faster, it would take *d* minutes to build the oven. While the new oven is being built, only old one can bake cakes, after the new oven is built, both ovens bake simultaneously. Arkady can't build more than one oven.
Determine if it is reasonable to build the second oven, i.e. will it decrease the minimum time needed to get *n* cakes or not. If the time needed with the second oven is the same as with one oven, then it is unreasonable.
|
The only line contains four integers *n*, *t*, *k*, *d* (1<=β€<=*n*,<=*t*,<=*k*,<=*d*<=β€<=1<=000)Β β the number of cakes needed, the time needed for one oven to bake *k* cakes, the number of cakes baked at the same time, the time needed to build the second oven.
|
If it is reasonable to build the second oven, print "YES". Otherwise print "NO".
|
[
"8 6 4 5\n",
"8 6 4 6\n",
"10 3 11 4\n",
"4 2 1 4\n"
] |
[
"YES\n",
"NO\n",
"NO\n",
"YES\n"
] |
In the first example it is possible to get 8 cakes in 12 minutes using one oven. The second oven can be built in 5 minutes, so after 6 minutes the first oven bakes 4 cakes, the second oven bakes 4 more ovens after 11 minutes. Thus, it is reasonable to build the second oven.
In the second example it doesn't matter whether we build the second oven or not, thus it takes 12 minutes to bake 8 cakes in both cases. Thus, it is unreasonable to build the second oven.
In the third example the first oven bakes 11 cakes in 3 minutes, that is more than needed 10. It is unreasonable to build the second oven, because its building takes more time that baking the needed number of cakes using the only oven.
| 500
|
[
{
"input": "8 6 4 5",
"output": "YES"
},
{
"input": "8 6 4 6",
"output": "NO"
},
{
"input": "10 3 11 4",
"output": "NO"
},
{
"input": "4 2 1 4",
"output": "YES"
},
{
"input": "28 17 16 26",
"output": "NO"
},
{
"input": "60 69 9 438",
"output": "NO"
},
{
"input": "599 97 54 992",
"output": "YES"
},
{
"input": "11 22 18 17",
"output": "NO"
},
{
"input": "1 13 22 11",
"output": "NO"
},
{
"input": "1 1 1 1",
"output": "NO"
},
{
"input": "3 1 1 1",
"output": "YES"
},
{
"input": "1000 1000 1000 1000",
"output": "NO"
},
{
"input": "1000 1000 1 1",
"output": "YES"
},
{
"input": "1000 1000 1 400",
"output": "YES"
},
{
"input": "1000 1000 1 1000",
"output": "YES"
},
{
"input": "1000 1000 1 999",
"output": "YES"
},
{
"input": "53 11 3 166",
"output": "YES"
},
{
"input": "313 2 3 385",
"output": "NO"
},
{
"input": "214 9 9 412",
"output": "NO"
},
{
"input": "349 9 5 268",
"output": "YES"
},
{
"input": "611 16 8 153",
"output": "YES"
},
{
"input": "877 13 3 191",
"output": "YES"
},
{
"input": "340 9 9 10",
"output": "YES"
},
{
"input": "31 8 2 205",
"output": "NO"
},
{
"input": "519 3 2 148",
"output": "YES"
},
{
"input": "882 2 21 219",
"output": "NO"
},
{
"input": "982 13 5 198",
"output": "YES"
},
{
"input": "428 13 6 272",
"output": "YES"
},
{
"input": "436 16 14 26",
"output": "YES"
},
{
"input": "628 10 9 386",
"output": "YES"
},
{
"input": "77 33 18 31",
"output": "YES"
},
{
"input": "527 36 4 8",
"output": "YES"
},
{
"input": "128 18 2 169",
"output": "YES"
},
{
"input": "904 4 2 288",
"output": "YES"
},
{
"input": "986 4 3 25",
"output": "YES"
},
{
"input": "134 8 22 162",
"output": "NO"
},
{
"input": "942 42 3 69",
"output": "YES"
},
{
"input": "894 4 9 4",
"output": "YES"
},
{
"input": "953 8 10 312",
"output": "YES"
},
{
"input": "43 8 1 121",
"output": "YES"
},
{
"input": "12 13 19 273",
"output": "NO"
},
{
"input": "204 45 10 871",
"output": "YES"
},
{
"input": "342 69 50 425",
"output": "NO"
},
{
"input": "982 93 99 875",
"output": "NO"
},
{
"input": "283 21 39 132",
"output": "YES"
},
{
"input": "1000 45 83 686",
"output": "NO"
},
{
"input": "246 69 36 432",
"output": "NO"
},
{
"input": "607 93 76 689",
"output": "NO"
},
{
"input": "503 21 24 435",
"output": "NO"
},
{
"input": "1000 45 65 989",
"output": "NO"
},
{
"input": "30 21 2 250",
"output": "YES"
},
{
"input": "1000 49 50 995",
"output": "NO"
},
{
"input": "383 69 95 253",
"output": "YES"
},
{
"input": "393 98 35 999",
"output": "YES"
},
{
"input": "1000 22 79 552",
"output": "NO"
},
{
"input": "268 294 268 154",
"output": "NO"
},
{
"input": "963 465 706 146",
"output": "YES"
},
{
"input": "304 635 304 257",
"output": "NO"
},
{
"input": "4 2 1 6",
"output": "NO"
},
{
"input": "1 51 10 50",
"output": "NO"
},
{
"input": "5 5 4 4",
"output": "YES"
},
{
"input": "3 2 1 1",
"output": "YES"
},
{
"input": "3 4 3 3",
"output": "NO"
},
{
"input": "7 3 4 1",
"output": "YES"
},
{
"input": "101 10 1 1000",
"output": "NO"
},
{
"input": "5 1 1 1",
"output": "YES"
},
{
"input": "5 10 5 5",
"output": "NO"
},
{
"input": "19 1 7 1",
"output": "YES"
},
{
"input": "763 572 745 262",
"output": "YES"
},
{
"input": "1 2 1 1",
"output": "NO"
},
{
"input": "5 1 1 3",
"output": "YES"
},
{
"input": "170 725 479 359",
"output": "NO"
},
{
"input": "6 2 1 7",
"output": "YES"
},
{
"input": "6 2 5 1",
"output": "YES"
},
{
"input": "1 2 2 1",
"output": "NO"
},
{
"input": "24 2 8 3",
"output": "YES"
},
{
"input": "7 3 3 3",
"output": "YES"
},
{
"input": "5 2 2 2",
"output": "YES"
},
{
"input": "3 2 1 2",
"output": "YES"
},
{
"input": "1000 2 200 8",
"output": "NO"
},
{
"input": "3 100 2 100",
"output": "NO"
},
{
"input": "2 999 1 1000",
"output": "NO"
},
{
"input": "2 1 1 1",
"output": "NO"
},
{
"input": "2 3 5 1",
"output": "NO"
},
{
"input": "100 1 5 1",
"output": "YES"
},
{
"input": "7 2 3 3",
"output": "YES"
},
{
"input": "4 1 1 3",
"output": "NO"
},
{
"input": "3 2 2 1",
"output": "YES"
},
{
"input": "1 1 1 2",
"output": "NO"
},
{
"input": "91 8 7 13",
"output": "YES"
},
{
"input": "3 1 2 1",
"output": "NO"
},
{
"input": "5 3 2 3",
"output": "YES"
},
{
"input": "9 6 6 3",
"output": "YES"
}
] | 1,658,162,660
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 93
| 187
| 0
|
class Solution():
def is_worth_build_the_second_oven():
number_of_cakes, time_to_bake, \
cakes_baked_at_the_same_time, \
time_to_build_second_oven = map(int, input().split())
time = 0
second_oven_builded = 0
cakes_baked = 0
cakes_baked2 = 0
while cakes_baked < number_of_cakes and cakes_baked2 < number_of_cakes:
time += 1
if time % time_to_bake == 0:
cakes_baked += cakes_baked_at_the_same_time
cakes_baked2 += cakes_baked_at_the_same_time
if second_oven_builded > 0:
cakes_baked2 += cakes_baked_at_the_same_time
if time % time_to_build_second_oven == 0:
second_oven_builded = 1
if cakes_baked >= number_of_cakes:
return "NO"
else:
return "YES"
if __name__ == '__main__':
print(Solution.is_worth_build_the_second_oven())
|
Title: Carrot Cakes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In some game by Playrix it takes *t* minutes for an oven to bake *k* carrot cakes, all cakes are ready at the same moment *t* minutes after they started baking. Arkady needs at least *n* cakes to complete a task, but he currently don't have any. However, he has infinitely many ingredients and one oven. Moreover, Arkady can build one more similar oven to make the process faster, it would take *d* minutes to build the oven. While the new oven is being built, only old one can bake cakes, after the new oven is built, both ovens bake simultaneously. Arkady can't build more than one oven.
Determine if it is reasonable to build the second oven, i.e. will it decrease the minimum time needed to get *n* cakes or not. If the time needed with the second oven is the same as with one oven, then it is unreasonable.
Input Specification:
The only line contains four integers *n*, *t*, *k*, *d* (1<=β€<=*n*,<=*t*,<=*k*,<=*d*<=β€<=1<=000)Β β the number of cakes needed, the time needed for one oven to bake *k* cakes, the number of cakes baked at the same time, the time needed to build the second oven.
Output Specification:
If it is reasonable to build the second oven, print "YES". Otherwise print "NO".
Demo Input:
['8 6 4 5\n', '8 6 4 6\n', '10 3 11 4\n', '4 2 1 4\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n', 'YES\n']
Note:
In the first example it is possible to get 8 cakes in 12 minutes using one oven. The second oven can be built in 5 minutes, so after 6 minutes the first oven bakes 4 cakes, the second oven bakes 4 more ovens after 11 minutes. Thus, it is reasonable to build the second oven.
In the second example it doesn't matter whether we build the second oven or not, thus it takes 12 minutes to bake 8 cakes in both cases. Thus, it is unreasonable to build the second oven.
In the third example the first oven bakes 11 cakes in 3 minutes, that is more than needed 10. It is unreasonable to build the second oven, because its building takes more time that baking the needed number of cakes using the only oven.
|
```python
class Solution():
def is_worth_build_the_second_oven():
number_of_cakes, time_to_bake, \
cakes_baked_at_the_same_time, \
time_to_build_second_oven = map(int, input().split())
time = 0
second_oven_builded = 0
cakes_baked = 0
cakes_baked2 = 0
while cakes_baked < number_of_cakes and cakes_baked2 < number_of_cakes:
time += 1
if time % time_to_bake == 0:
cakes_baked += cakes_baked_at_the_same_time
cakes_baked2 += cakes_baked_at_the_same_time
if second_oven_builded > 0:
cakes_baked2 += cakes_baked_at_the_same_time
if time % time_to_build_second_oven == 0:
second_oven_builded = 1
if cakes_baked >= number_of_cakes:
return "NO"
else:
return "YES"
if __name__ == '__main__':
print(Solution.is_worth_build_the_second_oven())
```
| 3
|
|
408
|
A
|
Line to Cashier
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Little Vasya went to the supermarket to get some groceries. He walked about the supermarket for a long time and got a basket full of products. Now he needs to choose the cashier to pay for the products.
There are *n* cashiers at the exit from the supermarket. At the moment the queue for the *i*-th cashier already has *k**i* people. The *j*-th person standing in the queue to the *i*-th cashier has *m**i*,<=*j* items in the basket. Vasya knows that:
- the cashier needs 5 seconds to scan one item; - after the cashier scans each item of some customer, he needs 15 seconds to take the customer's money and give him the change.
Of course, Vasya wants to select a queue so that he can leave the supermarket as soon as possible. Help him write a program that displays the minimum number of seconds after which Vasya can get to one of the cashiers.
|
The first line contains integer *n* (1<=β€<=*n*<=β€<=100)Β β the number of cashes in the shop. The second line contains *n* space-separated integers: *k*1,<=*k*2,<=...,<=*k**n* (1<=β€<=*k**i*<=β€<=100), where *k**i* is the number of people in the queue to the *i*-th cashier.
The *i*-th of the next *n* lines contains *k**i* space-separated integers: *m**i*,<=1,<=*m**i*,<=2,<=...,<=*m**i*,<=*k**i* (1<=β€<=*m**i*,<=*j*<=β€<=100)Β β the number of products the *j*-th person in the queue for the *i*-th cash has.
|
Print a single integer β the minimum number of seconds Vasya needs to get to the cashier.
|
[
"1\n1\n1\n",
"4\n1 4 3 2\n100\n1 2 2 3\n1 9 1\n7 8\n"
] |
[
"20\n",
"100\n"
] |
In the second test sample, if Vasya goes to the first queue, he gets to the cashier in 100Β·5β+β15β=β515 seconds. But if he chooses the second queue, he will need 1Β·5β+β2Β·5β+β2Β·5β+β3Β·5β+β4Β·15β=β100 seconds. He will need 1Β·5β+β9Β·5β+β1Β·5β+β3Β·15β=β100 seconds for the third one and 7Β·5β+β8Β·5β+β2Β·15β=β105 seconds for the fourth one. Thus, Vasya gets to the cashier quicker if he chooses the second or the third queue.
| 500
|
[
{
"input": "1\n1\n1",
"output": "20"
},
{
"input": "4\n1 4 3 2\n100\n1 2 2 3\n1 9 1\n7 8",
"output": "100"
},
{
"input": "4\n5 4 5 5\n3 1 3 1 2\n3 1 1 3\n1 1 1 2 2\n2 2 1 1 3",
"output": "100"
},
{
"input": "5\n5 3 6 6 4\n7 5 3 3 9\n6 8 2\n1 10 8 5 9 2\n9 7 8 5 9 10\n9 8 3 3",
"output": "125"
},
{
"input": "5\n10 10 10 10 10\n6 7 8 6 8 5 9 8 10 5\n9 6 9 8 7 8 8 10 8 5\n8 7 7 8 7 5 6 8 9 5\n6 5 10 5 5 10 7 8 5 5\n10 9 8 7 6 9 7 9 6 5",
"output": "480"
},
{
"input": "10\n9 10 10 10 9 5 9 7 8 7\n11 6 10 4 4 15 7 15 5\n3 9 11 12 11 1 13 13 1 5\n6 15 9 12 3 2 8 12 11 10\n7 1 1 6 10 2 6 1 14 2\n8 14 2 3 6 1 14 1 12\n6 10 9 3 5\n13 12 12 7 13 4 4 8 10\n5 6 4 3 14 9 13\n8 12 1 5 7 4 13 1\n1 9 5 3 5 1 4",
"output": "240"
},
{
"input": "10\n5 5 5 5 5 5 5 5 5 5\n5 5 4 5 4\n6 5 7 7 6\n5 4 4 5 5\n4 4 5 5 5\n7 6 4 5 7\n4 6 5 4 5\n6 6 7 6 6\n4 5 4 4 7\n7 5 4 4 5\n6 6 7 4 4",
"output": "190"
},
{
"input": "1\n1\n100",
"output": "515"
},
{
"input": "1\n90\n90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90",
"output": "41850"
}
] | 1,396,766,898
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 20
| 77
| 0
|
N = int(input())
kassa = list(map(int, input().split()))
i = 0
people = list([0] * N)
while i != N:
people[i] = list(map(int, input().split()))
i += 1
min_sec = 100000000000
for human in people:
sec = 0
for tovar in human:
sec += tovar * 5 + 15
if min_sec > sec:
min_sec = sec
print(min_sec)
|
Title: Line to Cashier
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Vasya went to the supermarket to get some groceries. He walked about the supermarket for a long time and got a basket full of products. Now he needs to choose the cashier to pay for the products.
There are *n* cashiers at the exit from the supermarket. At the moment the queue for the *i*-th cashier already has *k**i* people. The *j*-th person standing in the queue to the *i*-th cashier has *m**i*,<=*j* items in the basket. Vasya knows that:
- the cashier needs 5 seconds to scan one item; - after the cashier scans each item of some customer, he needs 15 seconds to take the customer's money and give him the change.
Of course, Vasya wants to select a queue so that he can leave the supermarket as soon as possible. Help him write a program that displays the minimum number of seconds after which Vasya can get to one of the cashiers.
Input Specification:
The first line contains integer *n* (1<=β€<=*n*<=β€<=100)Β β the number of cashes in the shop. The second line contains *n* space-separated integers: *k*1,<=*k*2,<=...,<=*k**n* (1<=β€<=*k**i*<=β€<=100), where *k**i* is the number of people in the queue to the *i*-th cashier.
The *i*-th of the next *n* lines contains *k**i* space-separated integers: *m**i*,<=1,<=*m**i*,<=2,<=...,<=*m**i*,<=*k**i* (1<=β€<=*m**i*,<=*j*<=β€<=100)Β β the number of products the *j*-th person in the queue for the *i*-th cash has.
Output Specification:
Print a single integer β the minimum number of seconds Vasya needs to get to the cashier.
Demo Input:
['1\n1\n1\n', '4\n1 4 3 2\n100\n1 2 2 3\n1 9 1\n7 8\n']
Demo Output:
['20\n', '100\n']
Note:
In the second test sample, if Vasya goes to the first queue, he gets to the cashier in 100Β·5β+β15β=β515 seconds. But if he chooses the second queue, he will need 1Β·5β+β2Β·5β+β2Β·5β+β3Β·5β+β4Β·15β=β100 seconds. He will need 1Β·5β+β9Β·5β+β1Β·5β+β3Β·15β=β100 seconds for the third one and 7Β·5β+β8Β·5β+β2Β·15β=β105 seconds for the fourth one. Thus, Vasya gets to the cashier quicker if he chooses the second or the third queue.
|
```python
N = int(input())
kassa = list(map(int, input().split()))
i = 0
people = list([0] * N)
while i != N:
people[i] = list(map(int, input().split()))
i += 1
min_sec = 100000000000
for human in people:
sec = 0
for tovar in human:
sec += tovar * 5 + 15
if min_sec > sec:
min_sec = sec
print(min_sec)
```
| 3
|
|
276
|
C
|
Little Girl and Maximum Sum
|
PROGRAMMING
| 1,500
|
[
"data structures",
"greedy",
"implementation",
"sortings"
] | null | null |
The little girl loves the problems on array queries very much.
One day she came across a rather well-known problem: you've got an array of $n$ elements (the elements of the array are indexed starting from 1); also, there are $q$ queries, each one is defined by a pair of integers $l_i$, $r_i$ $(1 \le l_i \le r_i \le n)$. You need to find for each query the sum of elements of the array with indexes from $l_i$ to $r_i$, inclusive.
The little girl found the problem rather boring. She decided to reorder the array elements before replying to the queries in a way that makes the sum of query replies maximum possible. Your task is to find the value of this maximum sum.
|
The first line contains two space-separated integers $n$ ($1 \le n \le 2\cdot10^5$) and $q$ ($1 \le q \le 2\cdot10^5$) β the number of elements in the array and the number of queries, correspondingly.
The next line contains $n$ space-separated integers $a_i$ ($1 \le a_i \le 2\cdot10^5$) β the array elements.
Each of the following $q$ lines contains two space-separated integers $l_i$ and $r_i$ ($1 \le l_i \le r_i \le n$) β the $i$-th query.
|
In a single line print, a single integer β the maximum sum of query replies after the array elements are reordered.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
|
[
"3 3\n5 3 2\n1 2\n2 3\n1 3\n",
"5 3\n5 2 4 1 3\n1 5\n2 3\n2 3\n"
] |
[
"25\n",
"33\n"
] |
none
| 1,500
|
[
{
"input": "3 3\n5 3 2\n1 2\n2 3\n1 3",
"output": "25"
},
{
"input": "5 3\n5 2 4 1 3\n1 5\n2 3\n2 3",
"output": "33"
},
{
"input": "34 21\n23 38 16 49 44 50 48 34 33 19 18 31 11 15 20 47 44 30 39 33 45 46 1 13 27 16 31 36 17 23 38 5 30 16\n8 16\n14 27\n8 26\n1 8\n5 6\n23 28\n4 33\n13 30\n12 30\n11 30\n9 21\n1 14\n15 22\n4 11\n5 24\n8 20\n17 33\n6 9\n3 14\n25 34\n10 17",
"output": "9382"
},
{
"input": "16 13\n40 32 15 16 35 36 45 23 30 42 25 8 29 21 39 23\n2 9\n3 11\n8 9\n4 14\n1 6\n5 10\n5 14\n5 11\n13 13\n2 8\n9 16\n6 10\n7 8",
"output": "2838"
},
{
"input": "31 48\n45 19 16 42 38 18 50 7 28 40 39 25 45 14 36 18 27 30 16 4 22 6 1 23 16 47 14 35 27 47 2\n6 16\n11 28\n4 30\n25 26\n11 30\n5 9\n4 17\n15 17\n10 25\n15 26\n1 3\n9 26\n8 29\n16 30\n5 24\n27 30\n9 10\n22 29\n2 6\n15 24\n6 21\n19 21\n4 28\n1 7\n18 21\n10 22\n6 15\n14 28\n4 29\n12 13\n19 29\n5 6\n13 31\n21 27\n9 25\n6 18\n6 8\n28 30\n2 4\n15 21\n1 1\n8 30\n3 31\n11 27\n28 29\n6 22\n20 22\n9 25",
"output": "17471"
},
{
"input": "22 7\n44 41 40 41 37 42 46 16 50 47 30 7 32 6 20 14 47 25 11 1 35 12\n1 10\n10 20\n4 5\n1 8\n2 12\n8 8\n2 16",
"output": "2202"
},
{
"input": "3 2\n1 2 3\n2 3\n1 2",
"output": "9"
}
] | 1,654,779,468
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 63
| 701
| 13,926,400
|
n, q = map(int, input().split())
a = list(map(int, input().split()))
lst = [0]*(n+1)
for i in range(q):
l, r = map(int, input().split())
lst[l-1] += 1
lst[r] -= 1
p = [0]*n
t = 0
for i in range(n):
t += lst[i]
p[i] = t
p.sort(reverse = True)
a.sort(reverse = True)
ans = 0
for i in range(n):
ans += a[i]*p[i]
print(ans)
|
Title: Little Girl and Maximum Sum
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The little girl loves the problems on array queries very much.
One day she came across a rather well-known problem: you've got an array of $n$ elements (the elements of the array are indexed starting from 1); also, there are $q$ queries, each one is defined by a pair of integers $l_i$, $r_i$ $(1 \le l_i \le r_i \le n)$. You need to find for each query the sum of elements of the array with indexes from $l_i$ to $r_i$, inclusive.
The little girl found the problem rather boring. She decided to reorder the array elements before replying to the queries in a way that makes the sum of query replies maximum possible. Your task is to find the value of this maximum sum.
Input Specification:
The first line contains two space-separated integers $n$ ($1 \le n \le 2\cdot10^5$) and $q$ ($1 \le q \le 2\cdot10^5$) β the number of elements in the array and the number of queries, correspondingly.
The next line contains $n$ space-separated integers $a_i$ ($1 \le a_i \le 2\cdot10^5$) β the array elements.
Each of the following $q$ lines contains two space-separated integers $l_i$ and $r_i$ ($1 \le l_i \le r_i \le n$) β the $i$-th query.
Output Specification:
In a single line print, a single integer β the maximum sum of query replies after the array elements are reordered.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Demo Input:
['3 3\n5 3 2\n1 2\n2 3\n1 3\n', '5 3\n5 2 4 1 3\n1 5\n2 3\n2 3\n']
Demo Output:
['25\n', '33\n']
Note:
none
|
```python
n, q = map(int, input().split())
a = list(map(int, input().split()))
lst = [0]*(n+1)
for i in range(q):
l, r = map(int, input().split())
lst[l-1] += 1
lst[r] -= 1
p = [0]*n
t = 0
for i in range(n):
t += lst[i]
p[i] = t
p.sort(reverse = True)
a.sort(reverse = True)
ans = 0
for i in range(n):
ans += a[i]*p[i]
print(ans)
```
| 3
|
|
379
|
A
|
New Year Candles
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles.
Vasily has *a* candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make *b* went out candles into a new candle. As a result, this new candle can be used like any other new candle.
Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number.
|
The single line contains two integers, *a* and *b* (1<=β€<=*a*<=β€<=1000;Β 2<=β€<=*b*<=β€<=1000).
|
Print a single integer β the number of hours Vasily can light up the room for.
|
[
"4 2\n",
"6 3\n"
] |
[
"7\n",
"8\n"
] |
Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours.
| 500
|
[
{
"input": "4 2",
"output": "7"
},
{
"input": "6 3",
"output": "8"
},
{
"input": "1000 1000",
"output": "1001"
},
{
"input": "123 5",
"output": "153"
},
{
"input": "1000 2",
"output": "1999"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "1 3",
"output": "1"
},
{
"input": "1 4",
"output": "1"
},
{
"input": "2 2",
"output": "3"
},
{
"input": "3 2",
"output": "5"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "999 2",
"output": "1997"
},
{
"input": "1000 3",
"output": "1499"
},
{
"input": "1000 4",
"output": "1333"
},
{
"input": "1 1000",
"output": "1"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "10 4",
"output": "13"
},
{
"input": "4 3",
"output": "5"
},
{
"input": "91 5",
"output": "113"
},
{
"input": "777 17",
"output": "825"
},
{
"input": "100 4",
"output": "133"
},
{
"input": "5 3",
"output": "7"
},
{
"input": "6 4",
"output": "7"
},
{
"input": "26 8",
"output": "29"
},
{
"input": "9 4",
"output": "11"
},
{
"input": "20 3",
"output": "29"
},
{
"input": "17 3",
"output": "25"
}
] | 1,649,662,824
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 32
| 46
| 0
|
import math
(a,b) = map(int,input().split())
print(a + (a-1)//(b-1))
|
Title: New Year Candles
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles.
Vasily has *a* candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make *b* went out candles into a new candle. As a result, this new candle can be used like any other new candle.
Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number.
Input Specification:
The single line contains two integers, *a* and *b* (1<=β€<=*a*<=β€<=1000;Β 2<=β€<=*b*<=β€<=1000).
Output Specification:
Print a single integer β the number of hours Vasily can light up the room for.
Demo Input:
['4 2\n', '6 3\n']
Demo Output:
['7\n', '8\n']
Note:
Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours.
|
```python
import math
(a,b) = map(int,input().split())
print(a + (a-1)//(b-1))
```
| 3
|
|
1
|
A
|
Theatre Square
|
PROGRAMMING
| 1,000
|
[
"math"
] |
A. Theatre Square
|
1
|
256
|
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=Γ<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=Γ<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
|
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=β€<=<=*n*,<=*m*,<=*a*<=β€<=109).
|
Write the needed number of flagstones.
|
[
"6 6 4\n"
] |
[
"4\n"
] |
none
| 0
|
[
{
"input": "6 6 4",
"output": "4"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "2 1 1",
"output": "2"
},
{
"input": "1 2 1",
"output": "2"
},
{
"input": "2 2 1",
"output": "4"
},
{
"input": "2 1 2",
"output": "1"
},
{
"input": "1 1 3",
"output": "1"
},
{
"input": "2 3 4",
"output": "1"
},
{
"input": "1000000000 1000000000 1",
"output": "1000000000000000000"
},
{
"input": "12 13 4",
"output": "12"
},
{
"input": "222 332 5",
"output": "3015"
},
{
"input": "1000 1000 10",
"output": "10000"
},
{
"input": "1001 1000 10",
"output": "10100"
},
{
"input": "100 10001 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000 999999999",
"output": "4"
},
{
"input": "1000000000 1000000000 192",
"output": "27126743055556"
},
{
"input": "1000000000 987654321 1",
"output": "987654321000000000"
},
{
"input": "456784567 1000000000 51",
"output": "175618850864484"
},
{
"input": "39916800 134217728 40320",
"output": "3295710"
}
] | 1,692,860,876
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 20
| 46
| 0
|
n, m, a = map(int, input().split()) # Read n, m, and a
# Calculate the number of flagstones needed for each dimension
flagstones_n = (n + a - 1) // a # Round up division
flagstones_m = (m + a - 1) // a # Round up division
# Calculate the total number of flagstones needed
total_flagstones = flagstones_n * flagstones_m
print(total_flagstones) # Print the result
|
Title: Theatre Square
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=Γ<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=Γ<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
Input Specification:
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=β€<=<=*n*,<=*m*,<=*a*<=β€<=109).
Output Specification:
Write the needed number of flagstones.
Demo Input:
['6 6 4\n']
Demo Output:
['4\n']
Note:
none
|
```python
n, m, a = map(int, input().split()) # Read n, m, and a
# Calculate the number of flagstones needed for each dimension
flagstones_n = (n + a - 1) // a # Round up division
flagstones_m = (m + a - 1) // a # Round up division
# Calculate the total number of flagstones needed
total_flagstones = flagstones_n * flagstones_m
print(total_flagstones) # Print the result
```
| 3.977
|
932
|
A
|
Palindromic Supersequence
|
PROGRAMMING
| 800
|
[
"constructive algorithms"
] | null | null |
You are given a string *A*. Find a string *B*, where *B* is a palindrome and *A* is a subsequence of *B*.
A subsequence of a string is a string that can be derived from it by deleting some (not necessarily consecutive) characters without changing the order of the remaining characters. For example, "cotst" is a subsequence of "contest".
A palindrome is a string that reads the same forward or backward.
The length of string *B* should be at most 104. It is guaranteed that there always exists such string.
You do not need to find the shortest answer, the only restriction is that the length of string *B* should not exceed 104.
|
First line contains a string *A* (1<=β€<=|*A*|<=β€<=103) consisting of lowercase Latin letters, where |*A*| is a length of *A*.
|
Output single line containing *B* consisting of only lowercase Latin letters. You do not need to find the shortest answer, the only restriction is that the length of string *B* should not exceed 104. If there are many possible *B*, print any of them.
|
[
"aba\n",
"ab\n"
] |
[
"aba",
"aabaa"
] |
In the first example, "aba" is a subsequence of "aba" which is a palindrome.
In the second example, "ab" is a subsequence of "aabaa" which is a palindrome.
| 500
|
[
{
"input": "aba",
"output": "abaaba"
},
{
"input": "ab",
"output": "abba"
},
{
"input": "krnyoixirslfszfqivgkaflgkctvbvksipwomqxlyqxhlbceuhbjbfnhofcgpgwdseffycthmlpcqejgskwjkbkbbmifnurnwyhevsoqzmtvzgfiqajfrgyuzxnrtxectcnlyoisbglpdbjbslxlpoymrcxmdtqhcnlvtqdwftuzgbdxsyscwbrguostbelnvtaqdmkmihmoxqtqlxvlsssisvqvvzotoyqryuyqwoknnqcqggysrqpkrccvyhxsjmhoqoyocwcriplarjoyiqrmmpmueqbsbljddwrumauczfziodpudheexalbwpiypmdjlmwtgdrzhpxneofhqzjdmurgvmrwdotuwyknlrbvuvtnhiouvqitgyfgfieonbaapyhwpcrmehxcpkijzfiayfvoxkpa",
"output": "krnyoixirslfszfqivgkaflgkctvbvksipwomqxlyqxhlbceuhbjbfnhofcgpgwdseffycthmlpcqejgskwjkbkbbmifnurnwyhevsoqzmtvzgfiqajfrgyuzxnrtxectcnlyoisbglpdbjbslxlpoymrcxmdtqhcnlvtqdwftuzgbdxsyscwbrguostbelnvtaqdmkmihmoxqtqlxvlsssisvqvvzotoyqryuyqwoknnqcqggysrqpkrccvyhxsjmhoqoyocwcriplarjoyiqrmmpmueqbsbljddwrumauczfziodpudheexalbwpiypmdjlmwtgdrzhpxneofhqzjdmurgvmrwdotuwyknlrbvuvtnhiouvqitgyfgfieonbaapyhwpcrmehxcpkijzfiayfvoxkpaapkxovfyaifzjikpcxhemrcpwhypaabnoeifgfygtiqvuoihntvuvbrlnkywutodwrmvgrumdjzqhfoenxphzrdgtwmljdm..."
},
{
"input": "mgrfmzxqpejcixxppqgvuawutgrmezjkteofjbnrvzzkvjtacfxjjokisavsgrslryxfqgrmdsqwptajbqzvethuljbdatxghfzqrwvfgakwmoawlzqjypmhllbbuuhbpriqsnibywlgjlxowyzagrfnqafvcqwktkcjwejevzbnxhsfmwojshcdypnvbuhhuzqmgovmvgwiizatoxgblyudipahfbkewmuneoqhjmbpdtwnznblwvtjrniwlbyblhppndspojrouffazpoxtqdfpjuhitvijrohavpqatofxwmksvjcvhdecxwwmosqiczjpkfafqlboxosnjgzgdraehzdltthemeusxhiiimrdrugabnxwsygsktkcslhjebfexucsyvlwrptebkjhefsvfrmcqqdlanbetrgzwylizmrystvpgrkhlicfadco",
"output": "mgrfmzxqpejcixxppqgvuawutgrmezjkteofjbnrvzzkvjtacfxjjokisavsgrslryxfqgrmdsqwptajbqzvethuljbdatxghfzqrwvfgakwmoawlzqjypmhllbbuuhbpriqsnibywlgjlxowyzagrfnqafvcqwktkcjwejevzbnxhsfmwojshcdypnvbuhhuzqmgovmvgwiizatoxgblyudipahfbkewmuneoqhjmbpdtwnznblwvtjrniwlbyblhppndspojrouffazpoxtqdfpjuhitvijrohavpqatofxwmksvjcvhdecxwwmosqiczjpkfafqlboxosnjgzgdraehzdltthemeusxhiiimrdrugabnxwsygsktkcslhjebfexucsyvlwrptebkjhefsvfrmcqqdlanbetrgzwylizmrystvpgrkhlicfadcoocdafcilhkrgpvtsyrmzilywzgrtebnaldqqcmrfvsfehjkbetprwlvyscuxef..."
},
{
"input": "hdmasfcjuigrwjchmjslmpynewnzpphmudzcbxzdexjuhktdtcoibzvevsmwaxakrtdfoivkvoooypyemiidadquqepxwqkesdnakxkbzrcjkgvwwxtqxvfpxcwitljyehldgsjytmekimkkndjvnzqtjykiymkmdzpwakxdtkzcqcatlevppgfhyykgmipuodjrnfjzhcmjdbzvhywprbwdcfxiffpzbjbmbyijkqnosslqbfvvicxvoeuzruraetglthgourzhfpnubzvblfzmmbgepjjyshchthulxar",
"output": "hdmasfcjuigrwjchmjslmpynewnzpphmudzcbxzdexjuhktdtcoibzvevsmwaxakrtdfoivkvoooypyemiidadquqepxwqkesdnakxkbzrcjkgvwwxtqxvfpxcwitljyehldgsjytmekimkkndjvnzqtjykiymkmdzpwakxdtkzcqcatlevppgfhyykgmipuodjrnfjzhcmjdbzvhywprbwdcfxiffpzbjbmbyijkqnosslqbfvvicxvoeuzruraetglthgourzhfpnubzvblfzmmbgepjjyshchthulxarraxluhthchsyjjpegbmmzflbvzbunpfhzruoghtlgtearurzueovxcivvfbqlssonqkjiybmbjbzpffixfcdwbrpwyhvzbdjmchzjfnrjdoupimgkyyhfgppveltacqczktdxkawpzdmkmyikyjtqznvjdnkkmikemtyjsgdlheyjltiwcxpfvxqtxwwvgkjcrzbkxkandsekqwxpequ..."
},
{
"input": "fggbyzobbmxtwdajawqdywnppflkkmtxzjvxopqvliwdwhzepcuiwelhbuotlkvesexnwkytonfrpqcxzzqzdvsmbsjcxxeugavekozfjlolrtqgwzqxsfgrnvrgfrqpixhsskbpzghndesvwptpvvkasfalzsetopervpwzmkgpcexqnvtnoulprwnowmsorscecvvvrjfwumcjqyrounqsgdruxttvtmrkivtxauhosokdiahsyrftzsgvgyveqwkzhqstbgywrvmsgfcfyuxpphvmyydzpohgdicoxbtjnsbyhoidnkrialowvlvmjpxcfeygqzphmbcjkupojsmmuqlydixbaluwezvnfasjfxilbyllwyipsmovdzosuwotcxerzcfuvxprtziseshjfcosalyqglpotxvxaanpocypsiyazsejjoximnbvqucftuvdksaxutvjeunodbipsumlaymjnzljurefjg",
"output": "fggbyzobbmxtwdajawqdywnppflkkmtxzjvxopqvliwdwhzepcuiwelhbuotlkvesexnwkytonfrpqcxzzqzdvsmbsjcxxeugavekozfjlolrtqgwzqxsfgrnvrgfrqpixhsskbpzghndesvwptpvvkasfalzsetopervpwzmkgpcexqnvtnoulprwnowmsorscecvvvrjfwumcjqyrounqsgdruxttvtmrkivtxauhosokdiahsyrftzsgvgyveqwkzhqstbgywrvmsgfcfyuxpphvmyydzpohgdicoxbtjnsbyhoidnkrialowvlvmjpxcfeygqzphmbcjkupojsmmuqlydixbaluwezvnfasjfxilbyllwyipsmovdzosuwotcxerzcfuvxprtziseshjfcosalyqglpotxvxaanpocypsiyazsejjoximnbvqucftuvdksaxutvjeunodbipsumlaymjnzljurefjggjferujlznjmyalmuspib..."
},
{
"input": "qyyxqkbxsvfnjzttdqmpzinbdgayllxpfrpopwciejjjzadguurnnhvixgueukugkkjyghxknedojvmdrskswiotgatsajowionuiumuhyggjuoympuxyfahwftwufvocdguxmxabbxnfviscxtilzzauizsgugwcqtbqgoosefhkumhodwpgolfdkbuiwlzjydonwbgyzzrjwxnceltqgqelrrljmzdbftmaogiuosaqhngmdzxzlmyrwefzhqawmkdckfnyyjgdjgadtfjvrkdwysqofcgyqrnyzutycvspzbjmmesobvhshtqlrytztyieknnkporrbcmlopgtknlmsstzkigreqwgsvagmvbrvwypoxttmzzsgm",
"output": "qyyxqkbxsvfnjzttdqmpzinbdgayllxpfrpopwciejjjzadguurnnhvixgueukugkkjyghxknedojvmdrskswiotgatsajowionuiumuhyggjuoympuxyfahwftwufvocdguxmxabbxnfviscxtilzzauizsgugwcqtbqgoosefhkumhodwpgolfdkbuiwlzjydonwbgyzzrjwxnceltqgqelrrljmzdbftmaogiuosaqhngmdzxzlmyrwefzhqawmkdckfnyyjgdjgadtfjvrkdwysqofcgyqrnyzutycvspzbjmmesobvhshtqlrytztyieknnkporrbcmlopgtknlmsstzkigreqwgsvagmvbrvwypoxttmzzsgmmgszzmttxopywvrbvmgavsgwqergikztssmlnktgpolmcbrropknnkeiytztyrlqthshvbosemmjbzpsvcytuzynrqygcfoqsywdkrvjftdagjdgjyynfkcdkmwaqhzfewry..."
},
{
"input": "scvlhflaqvniyiyofonowwcuqajuwscdrzhbvasymvqfnthzvtjcfuaftrbjghhvslcohwpxkggrbtatjtgehuqtorwinwvrtdldyoeeozxwippuahgkuehvsmyqtodqvlufqqmqautaqirvwzvtodzxtgxiinubhrbeoiybidutrqamsdnasctxatzkvkjkrmavdravnsxyngjlugwftmhmcvvxdbfndurrbmcpuoigjpssqcortmqoqttrabhoqvopjkxvpbqdqsilvlplhgqazauyvnodsxtwnomlinjpozwhrgrkqwmlwcwdkxjxjftexiavwrejvdjcfptterblxysjcheesyqsbgdrzjxbfjqgjgmvccqcyj",
"output": "scvlhflaqvniyiyofonowwcuqajuwscdrzhbvasymvqfnthzvtjcfuaftrbjghhvslcohwpxkggrbtatjtgehuqtorwinwvrtdldyoeeozxwippuahgkuehvsmyqtodqvlufqqmqautaqirvwzvtodzxtgxiinubhrbeoiybidutrqamsdnasctxatzkvkjkrmavdravnsxyngjlugwftmhmcvvxdbfndurrbmcpuoigjpssqcortmqoqttrabhoqvopjkxvpbqdqsilvlplhgqazauyvnodsxtwnomlinjpozwhrgrkqwmlwcwdkxjxjftexiavwrejvdjcfptterblxysjcheesyqsbgdrzjxbfjqgjgmvccqcyjjycqccvmgjgqjfbxjzrdgbsqyseehcjsyxlbrettpfcjdvjerwvaixetfjxjxkdwcwlmwqkrgrhwzopjnilmonwtxsdonvyuazaqghlplvlisqdqbpvxkjpovqohbarttqoqm..."
},
{
"input": "oohkqxxtvxzmvfjjxyjwlbqmeqwwlienzkdbhswgfbkhfygltsucdijozwaiewpixapyazfztksjeoqjugjfhdbqzuezbuajfvvffkwprroyivfoocvslejffgxuiofisenroxoeixmdbzonmreikpflciwsbafrdqfvdfojgoziiibqhwwsvhnzmptgirqqulkgmyzrfekzqqujmdumxkudsgexisupedisgmdgebvlvrpyfrbrqjknrxyzfpwmsxjxismgd",
"output": "oohkqxxtvxzmvfjjxyjwlbqmeqwwlienzkdbhswgfbkhfygltsucdijozwaiewpixapyazfztksjeoqjugjfhdbqzuezbuajfvvffkwprroyivfoocvslejffgxuiofisenroxoeixmdbzonmreikpflciwsbafrdqfvdfojgoziiibqhwwsvhnzmptgirqqulkgmyzrfekzqqujmdumxkudsgexisupedisgmdgebvlvrpyfrbrqjknrxyzfpwmsxjxismgddgmsixjxsmwpfzyxrnkjqrbrfyprvlvbegdmgsidepusixegsdukxmudmjuqqzkefrzymgkluqqrigtpmznhvswwhqbiiizogjofdvfqdrfabswiclfpkiermnozbdmxieoxornesifoiuxgffjelsvcoofviyorrpwkffvvfjaubzeuzqbdhfjgujqoejsktzfzaypaxipweiawzojidcustlgyfhkbfgwshbdkzneilwwqemqblw..."
},
{
"input": "gilhoixzjgidfanqrmekjelnvicpuujlpxittgadgrhqallnkjlemwazntwfywjnrxdkgrnczlwzjyeyfktduzdjnivcldjjarfzmmdbyytvipbbnjqolfnlqjpidotxxfobgtgpvjmpddcyddwdcjsxxumuoyznhpvpqccgqnuouzojntanfwctthcgynrukcvshsuuqrxfdvqqggaatwytikkitywtaaggqqvdfxrquushsvckurnygchttcwfnatnjozuounqgccqpvphnzyoumuxxsjcdwddycddpmjvpgtgbofxxtodipjqlnfloqjnbbpivtyybdmmzfrajjdlcvinjdzudtkfyeyjzwlzcnrgkdxrnjwyfwtnzawmeljknllaqhrgdagttixpljuupcivnlejkemrqnafdigjzxiohlig",
"output": "gilhoixzjgidfanqrmekjelnvicpuujlpxittgadgrhqallnkjlemwazntwfywjnrxdkgrnczlwzjyeyfktduzdjnivcldjjarfzmmdbyytvipbbnjqolfnlqjpidotxxfobgtgpvjmpddcyddwdcjsxxumuoyznhpvpqccgqnuouzojntanfwctthcgynrukcvshsuuqrxfdvqqggaatwytikkitywtaaggqqvdfxrquushsvckurnygchttcwfnatnjozuounqgccqpvphnzyoumuxxsjcdwddycddpmjvpgtgbofxxtodipjqlnfloqjnbbpivtyybdmmzfrajjdlcvinjdzudtkfyeyjzwlzcnrgkdxrnjwyfwtnzawmeljknllaqhrgdagttixpljuupcivnlejkemrqnafdigjzxiohliggilhoixzjgidfanqrmekjelnvicpuujlpxittgadgrhqallnkjlemwazntwfywjnrxdkgrnczlw..."
},
{
"input": "abcab",
"output": "abcabbacba"
},
{
"input": "baaaaaaa",
"output": "baaaaaaaaaaaaaab"
},
{
"input": "baaaaaa",
"output": "baaaaaaaaaaaab"
},
{
"input": "baaaaaaaaa",
"output": "baaaaaaaaaaaaaaaaaab"
},
{
"input": "baaaaaaaa",
"output": "baaaaaaaaaaaaaaaab"
}
] | 1,518,981,150
| 1,230
|
Python 3
|
OK
|
TESTS
| 48
| 61
| 5,632,000
|
#author @karngyan
s=input()
s=s+s[-1:-len(s)-1:-1]
print(s)
|
Title: Palindromic Supersequence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a string *A*. Find a string *B*, where *B* is a palindrome and *A* is a subsequence of *B*.
A subsequence of a string is a string that can be derived from it by deleting some (not necessarily consecutive) characters without changing the order of the remaining characters. For example, "cotst" is a subsequence of "contest".
A palindrome is a string that reads the same forward or backward.
The length of string *B* should be at most 104. It is guaranteed that there always exists such string.
You do not need to find the shortest answer, the only restriction is that the length of string *B* should not exceed 104.
Input Specification:
First line contains a string *A* (1<=β€<=|*A*|<=β€<=103) consisting of lowercase Latin letters, where |*A*| is a length of *A*.
Output Specification:
Output single line containing *B* consisting of only lowercase Latin letters. You do not need to find the shortest answer, the only restriction is that the length of string *B* should not exceed 104. If there are many possible *B*, print any of them.
Demo Input:
['aba\n', 'ab\n']
Demo Output:
['aba', 'aabaa']
Note:
In the first example, "aba" is a subsequence of "aba" which is a palindrome.
In the second example, "ab" is a subsequence of "aabaa" which is a palindrome.
|
```python
#author @karngyan
s=input()
s=s+s[-1:-len(s)-1:-1]
print(s)
```
| 3
|
|
854
|
A
|
Fraction
|
PROGRAMMING
| 800
|
[
"brute force",
"constructive algorithms",
"math"
] | null | null |
Petya is a big fan of mathematics, especially its part related to fractions. Recently he learned that a fraction is called proper iff its numerator is smaller than its denominator (*a*<=<<=*b*) and that the fraction is called irreducible if its numerator and its denominator are coprime (they do not have positive common divisors except 1).
During his free time, Petya thinks about proper irreducible fractions and converts them to decimals using the calculator. One day he mistakenly pressed addition button (<=+<=) instead of division button (Γ·) and got sum of numerator and denominator that was equal to *n* instead of the expected decimal notation.
Petya wanted to restore the original fraction, but soon he realized that it might not be done uniquely. That's why he decided to determine maximum possible proper irreducible fraction such that sum of its numerator and denominator equals *n*. Help Petya deal with this problem.
|
In the only line of input there is an integer *n* (3<=β€<=*n*<=β€<=1000), the sum of numerator and denominator of the fraction.
|
Output two space-separated positive integers *a* and *b*, numerator and denominator of the maximum possible proper irreducible fraction satisfying the given sum.
|
[
"3\n",
"4\n",
"12\n"
] |
[
"1 2\n",
"1 3\n",
"5 7\n"
] |
none
| 500
|
[
{
"input": "3",
"output": "1 2"
},
{
"input": "4",
"output": "1 3"
},
{
"input": "12",
"output": "5 7"
},
{
"input": "34",
"output": "15 19"
},
{
"input": "13",
"output": "6 7"
},
{
"input": "11",
"output": "5 6"
},
{
"input": "24",
"output": "11 13"
},
{
"input": "17",
"output": "8 9"
},
{
"input": "10",
"output": "3 7"
},
{
"input": "69",
"output": "34 35"
},
{
"input": "100",
"output": "49 51"
},
{
"input": "57",
"output": "28 29"
},
{
"input": "1000",
"output": "499 501"
},
{
"input": "999",
"output": "499 500"
},
{
"input": "998",
"output": "497 501"
},
{
"input": "997",
"output": "498 499"
},
{
"input": "996",
"output": "497 499"
},
{
"input": "995",
"output": "497 498"
},
{
"input": "994",
"output": "495 499"
},
{
"input": "5",
"output": "2 3"
},
{
"input": "6",
"output": "1 5"
},
{
"input": "8",
"output": "3 5"
},
{
"input": "9",
"output": "4 5"
},
{
"input": "423",
"output": "211 212"
},
{
"input": "876",
"output": "437 439"
},
{
"input": "29",
"output": "14 15"
}
] | 1,505,057,511
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 26
| 62
| 0
|
n=int(input())
if n%2:print(int(n/2),int(n/2)+1)
elif int(n/2)%2:print(int(n/2)-2,int(n/2)+2)
else:print(int(n/2)-1,int(n/2)+1)
|
Title: Fraction
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya is a big fan of mathematics, especially its part related to fractions. Recently he learned that a fraction is called proper iff its numerator is smaller than its denominator (*a*<=<<=*b*) and that the fraction is called irreducible if its numerator and its denominator are coprime (they do not have positive common divisors except 1).
During his free time, Petya thinks about proper irreducible fractions and converts them to decimals using the calculator. One day he mistakenly pressed addition button (<=+<=) instead of division button (Γ·) and got sum of numerator and denominator that was equal to *n* instead of the expected decimal notation.
Petya wanted to restore the original fraction, but soon he realized that it might not be done uniquely. That's why he decided to determine maximum possible proper irreducible fraction such that sum of its numerator and denominator equals *n*. Help Petya deal with this problem.
Input Specification:
In the only line of input there is an integer *n* (3<=β€<=*n*<=β€<=1000), the sum of numerator and denominator of the fraction.
Output Specification:
Output two space-separated positive integers *a* and *b*, numerator and denominator of the maximum possible proper irreducible fraction satisfying the given sum.
Demo Input:
['3\n', '4\n', '12\n']
Demo Output:
['1 2\n', '1 3\n', '5 7\n']
Note:
none
|
```python
n=int(input())
if n%2:print(int(n/2),int(n/2)+1)
elif int(n/2)%2:print(int(n/2)-2,int(n/2)+2)
else:print(int(n/2)-1,int(n/2)+1)
```
| 3
|
|
102
|
B
|
Sum of Digits
|
PROGRAMMING
| 1,000
|
[
"implementation"
] |
B. Sum of Digits
|
2
|
265
|
Having watched the last Harry Potter film, little Gerald also decided to practice magic. He found in his father's magical book a spell that turns any number in the sum of its digits. At the moment Gerald learned that, he came across a number *n*. How many times can Gerald put a spell on it until the number becomes one-digit?
|
The first line contains the only integer *n* (0<=β€<=*n*<=β€<=10100000). It is guaranteed that *n* doesn't contain any leading zeroes.
|
Print the number of times a number can be replaced by the sum of its digits until it only contains one digit.
|
[
"0\n",
"10\n",
"991\n"
] |
[
"0\n",
"1\n",
"3\n"
] |
In the first sample the number already is one-digit β Herald can't cast a spell.
The second test contains number 10. After one casting of a spell it becomes 1, and here the process is completed. Thus, Gerald can only cast the spell once.
The third test contains number 991. As one casts a spell the following transformations take place: 991βββ19βββ10βββ1. After three transformations the number becomes one-digit.
| 1,000
|
[
{
"input": "0",
"output": "0"
},
{
"input": "10",
"output": "1"
},
{
"input": "991",
"output": "3"
},
{
"input": "99",
"output": "2"
},
{
"input": "100",
"output": "1"
},
{
"input": "123456789",
"output": "2"
},
{
"input": "32",
"output": "1"
},
{
"input": "86",
"output": "2"
},
{
"input": "2",
"output": "0"
},
{
"input": "8",
"output": "0"
},
{
"input": "34",
"output": "1"
},
{
"input": "13",
"output": "1"
},
{
"input": "28",
"output": "2"
},
{
"input": "23",
"output": "1"
},
{
"input": "57",
"output": "2"
},
{
"input": "29",
"output": "2"
},
{
"input": "353869285",
"output": "3"
},
{
"input": "549548646",
"output": "2"
},
{
"input": "858893424",
"output": "2"
},
{
"input": "644818852",
"output": "3"
},
{
"input": "360322525",
"output": "3"
},
{
"input": "122937520",
"output": "2"
},
{
"input": "288403032",
"output": "2"
},
{
"input": "677257481",
"output": "3"
},
{
"input": "58059561",
"output": "3"
},
{
"input": "211288847",
"output": "2"
},
{
"input": "339900034079539584",
"output": "2"
},
{
"input": "784084029310269952",
"output": "3"
},
{
"input": "774730557264864000",
"output": "3"
},
{
"input": "478233807148352256",
"output": "3"
},
{
"input": "165646874056812544",
"output": "3"
},
{
"input": "477533739511673792",
"output": "3"
},
{
"input": "660119944377002240",
"output": "3"
},
{
"input": "448375218506078080",
"output": "3"
},
{
"input": "77213334785795616",
"output": "3"
},
{
"input": "165767221702271872",
"output": "3"
},
{
"input": "1",
"output": "0"
},
{
"input": "5",
"output": "0"
},
{
"input": "8",
"output": "0"
},
{
"input": "156161456146164104103460452045416165146141414651641564105461465156445621465146514562146145465165145145614561465145614561",
"output": "3"
},
{
"input": "9",
"output": "0"
}
] | 1,653,054,544
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 51
| 124
| 1,638,400
|
nos = input()
po = 0
while True:
if len(nos) == 1:
print(po)
break
else:
nos = str(sum(list(map(int,list(nos)))))
po+=1
|
Title: Sum of Digits
Time Limit: 2 seconds
Memory Limit: 265 megabytes
Problem Description:
Having watched the last Harry Potter film, little Gerald also decided to practice magic. He found in his father's magical book a spell that turns any number in the sum of its digits. At the moment Gerald learned that, he came across a number *n*. How many times can Gerald put a spell on it until the number becomes one-digit?
Input Specification:
The first line contains the only integer *n* (0<=β€<=*n*<=β€<=10100000). It is guaranteed that *n* doesn't contain any leading zeroes.
Output Specification:
Print the number of times a number can be replaced by the sum of its digits until it only contains one digit.
Demo Input:
['0\n', '10\n', '991\n']
Demo Output:
['0\n', '1\n', '3\n']
Note:
In the first sample the number already is one-digit β Herald can't cast a spell.
The second test contains number 10. After one casting of a spell it becomes 1, and here the process is completed. Thus, Gerald can only cast the spell once.
The third test contains number 991. As one casts a spell the following transformations take place: 991βββ19βββ10βββ1. After three transformations the number becomes one-digit.
|
```python
nos = input()
po = 0
while True:
if len(nos) == 1:
print(po)
break
else:
nos = str(sum(list(map(int,list(nos)))))
po+=1
```
| 3.966052
|
808
|
A
|
Lucky Year
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Apart from having lots of holidays throughout the year, residents of Berland also have whole lucky years. Year is considered lucky if it has no more than 1 non-zero digit in its number. So years 100, 40000, 5 are lucky and 12, 3001 and 12345 are not.
You are given current year in Berland. Your task is to find how long will residents of Berland wait till the next lucky year.
|
The first line contains integer number *n* (1<=β€<=*n*<=β€<=109) β current year in Berland.
|
Output amount of years from the current year to the next lucky one.
|
[
"4\n",
"201\n",
"4000\n"
] |
[
"1\n",
"99\n",
"1000\n"
] |
In the first example next lucky year is 5. In the second one β 300. In the third β 5000.
| 0
|
[
{
"input": "4",
"output": "1"
},
{
"input": "201",
"output": "99"
},
{
"input": "4000",
"output": "1000"
},
{
"input": "9",
"output": "1"
},
{
"input": "10",
"output": "10"
},
{
"input": "1",
"output": "1"
},
{
"input": "100000000",
"output": "100000000"
},
{
"input": "900000000",
"output": "100000000"
},
{
"input": "999999999",
"output": "1"
},
{
"input": "1000000000",
"output": "1000000000"
},
{
"input": "9999999",
"output": "1"
},
{
"input": "100000001",
"output": "99999999"
},
{
"input": "3660",
"output": "340"
},
{
"input": "21",
"output": "9"
},
{
"input": "900000001",
"output": "99999999"
},
{
"input": "62911",
"output": "7089"
},
{
"input": "11",
"output": "9"
},
{
"input": "940302010",
"output": "59697990"
},
{
"input": "91",
"output": "9"
},
{
"input": "101",
"output": "99"
},
{
"input": "1090",
"output": "910"
},
{
"input": "987654321",
"output": "12345679"
},
{
"input": "703450474",
"output": "96549526"
},
{
"input": "1091",
"output": "909"
},
{
"input": "89",
"output": "1"
},
{
"input": "109",
"output": "91"
},
{
"input": "190",
"output": "10"
},
{
"input": "19",
"output": "1"
},
{
"input": "8",
"output": "1"
},
{
"input": "482",
"output": "18"
},
{
"input": "1",
"output": "1"
},
{
"input": "2",
"output": "1"
},
{
"input": "3",
"output": "1"
},
{
"input": "4",
"output": "1"
},
{
"input": "5",
"output": "1"
},
{
"input": "6",
"output": "1"
},
{
"input": "7",
"output": "1"
},
{
"input": "8",
"output": "1"
},
{
"input": "9",
"output": "1"
},
{
"input": "10",
"output": "10"
},
{
"input": "11",
"output": "9"
},
{
"input": "12",
"output": "8"
},
{
"input": "13",
"output": "7"
},
{
"input": "14",
"output": "6"
},
{
"input": "15",
"output": "5"
},
{
"input": "16",
"output": "4"
},
{
"input": "17",
"output": "3"
},
{
"input": "18",
"output": "2"
},
{
"input": "19",
"output": "1"
},
{
"input": "20",
"output": "10"
},
{
"input": "21",
"output": "9"
},
{
"input": "22",
"output": "8"
},
{
"input": "23",
"output": "7"
},
{
"input": "24",
"output": "6"
},
{
"input": "25",
"output": "5"
},
{
"input": "26",
"output": "4"
},
{
"input": "27",
"output": "3"
},
{
"input": "28",
"output": "2"
},
{
"input": "29",
"output": "1"
},
{
"input": "30",
"output": "10"
},
{
"input": "31",
"output": "9"
},
{
"input": "32",
"output": "8"
},
{
"input": "33",
"output": "7"
},
{
"input": "34",
"output": "6"
},
{
"input": "35",
"output": "5"
},
{
"input": "36",
"output": "4"
},
{
"input": "37",
"output": "3"
},
{
"input": "38",
"output": "2"
},
{
"input": "39",
"output": "1"
},
{
"input": "40",
"output": "10"
},
{
"input": "41",
"output": "9"
},
{
"input": "42",
"output": "8"
},
{
"input": "43",
"output": "7"
},
{
"input": "44",
"output": "6"
},
{
"input": "45",
"output": "5"
},
{
"input": "46",
"output": "4"
},
{
"input": "47",
"output": "3"
},
{
"input": "48",
"output": "2"
},
{
"input": "49",
"output": "1"
},
{
"input": "50",
"output": "10"
},
{
"input": "51",
"output": "9"
},
{
"input": "52",
"output": "8"
},
{
"input": "53",
"output": "7"
},
{
"input": "54",
"output": "6"
},
{
"input": "55",
"output": "5"
},
{
"input": "56",
"output": "4"
},
{
"input": "57",
"output": "3"
},
{
"input": "58",
"output": "2"
},
{
"input": "59",
"output": "1"
},
{
"input": "60",
"output": "10"
},
{
"input": "61",
"output": "9"
},
{
"input": "62",
"output": "8"
},
{
"input": "63",
"output": "7"
},
{
"input": "64",
"output": "6"
},
{
"input": "65",
"output": "5"
},
{
"input": "66",
"output": "4"
},
{
"input": "67",
"output": "3"
},
{
"input": "68",
"output": "2"
},
{
"input": "69",
"output": "1"
},
{
"input": "70",
"output": "10"
},
{
"input": "71",
"output": "9"
},
{
"input": "72",
"output": "8"
},
{
"input": "73",
"output": "7"
},
{
"input": "74",
"output": "6"
},
{
"input": "75",
"output": "5"
},
{
"input": "76",
"output": "4"
},
{
"input": "77",
"output": "3"
},
{
"input": "78",
"output": "2"
},
{
"input": "79",
"output": "1"
},
{
"input": "80",
"output": "10"
},
{
"input": "81",
"output": "9"
},
{
"input": "82",
"output": "8"
},
{
"input": "83",
"output": "7"
},
{
"input": "84",
"output": "6"
},
{
"input": "85",
"output": "5"
},
{
"input": "86",
"output": "4"
},
{
"input": "87",
"output": "3"
},
{
"input": "88",
"output": "2"
},
{
"input": "89",
"output": "1"
},
{
"input": "90",
"output": "10"
},
{
"input": "91",
"output": "9"
},
{
"input": "92",
"output": "8"
},
{
"input": "93",
"output": "7"
},
{
"input": "94",
"output": "6"
},
{
"input": "95",
"output": "5"
},
{
"input": "96",
"output": "4"
},
{
"input": "97",
"output": "3"
},
{
"input": "98",
"output": "2"
},
{
"input": "99",
"output": "1"
},
{
"input": "100",
"output": "100"
},
{
"input": "100",
"output": "100"
},
{
"input": "100",
"output": "100"
},
{
"input": "1000",
"output": "1000"
},
{
"input": "1000",
"output": "1000"
},
{
"input": "1000",
"output": "1000"
},
{
"input": "10000",
"output": "10000"
},
{
"input": "10000",
"output": "10000"
},
{
"input": "101",
"output": "99"
},
{
"input": "110",
"output": "90"
},
{
"input": "1001",
"output": "999"
},
{
"input": "1100",
"output": "900"
},
{
"input": "1010",
"output": "990"
},
{
"input": "10010",
"output": "9990"
},
{
"input": "10100",
"output": "9900"
},
{
"input": "102",
"output": "98"
},
{
"input": "120",
"output": "80"
},
{
"input": "1002",
"output": "998"
},
{
"input": "1200",
"output": "800"
},
{
"input": "1020",
"output": "980"
},
{
"input": "10020",
"output": "9980"
},
{
"input": "10200",
"output": "9800"
},
{
"input": "108",
"output": "92"
},
{
"input": "180",
"output": "20"
},
{
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"output": "992"
},
{
"input": "1800",
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},
{
"input": "1080",
"output": "920"
},
{
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"output": "9920"
},
{
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"output": "9200"
},
{
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"output": "91"
},
{
"input": "190",
"output": "10"
},
{
"input": "1009",
"output": "991"
},
{
"input": "1900",
"output": "100"
},
{
"input": "1090",
"output": "910"
},
{
"input": "10090",
"output": "9910"
},
{
"input": "10900",
"output": "9100"
},
{
"input": "200",
"output": "100"
},
{
"input": "200",
"output": "100"
},
{
"input": "2000",
"output": "1000"
},
{
"input": "2000",
"output": "1000"
},
{
"input": "2000",
"output": "1000"
},
{
"input": "20000",
"output": "10000"
},
{
"input": "20000",
"output": "10000"
},
{
"input": "201",
"output": "99"
},
{
"input": "210",
"output": "90"
},
{
"input": "2001",
"output": "999"
},
{
"input": "2100",
"output": "900"
},
{
"input": "2010",
"output": "990"
},
{
"input": "20010",
"output": "9990"
},
{
"input": "20100",
"output": "9900"
},
{
"input": "202",
"output": "98"
},
{
"input": "220",
"output": "80"
},
{
"input": "2002",
"output": "998"
},
{
"input": "2200",
"output": "800"
},
{
"input": "2020",
"output": "980"
},
{
"input": "20020",
"output": "9980"
},
{
"input": "20200",
"output": "9800"
},
{
"input": "208",
"output": "92"
},
{
"input": "280",
"output": "20"
},
{
"input": "2008",
"output": "992"
},
{
"input": "2800",
"output": "200"
},
{
"input": "2080",
"output": "920"
},
{
"input": "20080",
"output": "9920"
},
{
"input": "20800",
"output": "9200"
},
{
"input": "209",
"output": "91"
},
{
"input": "290",
"output": "10"
},
{
"input": "2009",
"output": "991"
},
{
"input": "2900",
"output": "100"
},
{
"input": "2090",
"output": "910"
},
{
"input": "20090",
"output": "9910"
},
{
"input": "20900",
"output": "9100"
},
{
"input": "800",
"output": "100"
},
{
"input": "800",
"output": "100"
},
{
"input": "8000",
"output": "1000"
},
{
"input": "8000",
"output": "1000"
},
{
"input": "8000",
"output": "1000"
},
{
"input": "80000",
"output": "10000"
},
{
"input": "80000",
"output": "10000"
},
{
"input": "801",
"output": "99"
},
{
"input": "810",
"output": "90"
},
{
"input": "8001",
"output": "999"
},
{
"input": "8100",
"output": "900"
},
{
"input": "8010",
"output": "990"
},
{
"input": "80010",
"output": "9990"
},
{
"input": "80100",
"output": "9900"
},
{
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"output": "98"
},
{
"input": "820",
"output": "80"
},
{
"input": "8002",
"output": "998"
},
{
"input": "8200",
"output": "800"
},
{
"input": "8020",
"output": "980"
},
{
"input": "80020",
"output": "9980"
},
{
"input": "80200",
"output": "9800"
},
{
"input": "808",
"output": "92"
},
{
"input": "880",
"output": "20"
},
{
"input": "8008",
"output": "992"
},
{
"input": "8800",
"output": "200"
},
{
"input": "8080",
"output": "920"
},
{
"input": "80080",
"output": "9920"
},
{
"input": "80800",
"output": "9200"
},
{
"input": "809",
"output": "91"
},
{
"input": "890",
"output": "10"
},
{
"input": "8009",
"output": "991"
},
{
"input": "8900",
"output": "100"
},
{
"input": "8090",
"output": "910"
},
{
"input": "80090",
"output": "9910"
},
{
"input": "80900",
"output": "9100"
},
{
"input": "900",
"output": "100"
},
{
"input": "900",
"output": "100"
},
{
"input": "9000",
"output": "1000"
},
{
"input": "9000",
"output": "1000"
},
{
"input": "9000",
"output": "1000"
},
{
"input": "90000",
"output": "10000"
},
{
"input": "90000",
"output": "10000"
},
{
"input": "901",
"output": "99"
},
{
"input": "910",
"output": "90"
},
{
"input": "9001",
"output": "999"
},
{
"input": "9100",
"output": "900"
},
{
"input": "9010",
"output": "990"
},
{
"input": "90010",
"output": "9990"
},
{
"input": "90100",
"output": "9900"
},
{
"input": "902",
"output": "98"
},
{
"input": "920",
"output": "80"
},
{
"input": "9002",
"output": "998"
},
{
"input": "9200",
"output": "800"
},
{
"input": "9020",
"output": "980"
},
{
"input": "90020",
"output": "9980"
},
{
"input": "90200",
"output": "9800"
},
{
"input": "908",
"output": "92"
},
{
"input": "980",
"output": "20"
},
{
"input": "9008",
"output": "992"
},
{
"input": "9800",
"output": "200"
},
{
"input": "9080",
"output": "920"
},
{
"input": "90080",
"output": "9920"
},
{
"input": "90800",
"output": "9200"
},
{
"input": "909",
"output": "91"
},
{
"input": "990",
"output": "10"
},
{
"input": "9009",
"output": "991"
},
{
"input": "9900",
"output": "100"
},
{
"input": "9090",
"output": "910"
},
{
"input": "90090",
"output": "9910"
},
{
"input": "90900",
"output": "9100"
},
{
"input": "92651241",
"output": "7348759"
}
] | 1,500,370,624
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 271
| 62
| 4,608,000
|
a=input()
l=len(a)
if l!=1:
b=10**(l-1)
print(b-int(a[1:]))
else:print(1)
|
Title: Lucky Year
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Apart from having lots of holidays throughout the year, residents of Berland also have whole lucky years. Year is considered lucky if it has no more than 1 non-zero digit in its number. So years 100, 40000, 5 are lucky and 12, 3001 and 12345 are not.
You are given current year in Berland. Your task is to find how long will residents of Berland wait till the next lucky year.
Input Specification:
The first line contains integer number *n* (1<=β€<=*n*<=β€<=109) β current year in Berland.
Output Specification:
Output amount of years from the current year to the next lucky one.
Demo Input:
['4\n', '201\n', '4000\n']
Demo Output:
['1\n', '99\n', '1000\n']
Note:
In the first example next lucky year is 5. In the second one β 300. In the third β 5000.
|
```python
a=input()
l=len(a)
if l!=1:
b=10**(l-1)
print(b-int(a[1:]))
else:print(1)
```
| 3
|
|
276
|
A
|
Lunch Rush
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Having written another programming contest, three Rabbits decided to grab some lunch. The coach gave the team exactly *k* time units for the lunch break.
The Rabbits have a list of *n* restaurants to lunch in: the *i*-th restaurant is characterized by two integers *f**i* and *t**i*. Value *t**i* shows the time the Rabbits need to lunch in the *i*-th restaurant. If time *t**i* exceeds the time *k* that the coach has given for the lunch break, then the Rabbits' joy from lunching in this restaurant will equal *f**i*<=-<=(*t**i*<=-<=*k*). Otherwise, the Rabbits get exactly *f**i* units of joy.
Your task is to find the value of the maximum joy the Rabbits can get from the lunch, depending on the restaurant. The Rabbits must choose exactly one restaurant to lunch in. Note that the joy value isn't necessarily a positive value.
|
The first line contains two space-separated integers β *n* (1<=β€<=*n*<=β€<=104) and *k* (1<=β€<=*k*<=β€<=109) β the number of restaurants in the Rabbits' list and the time the coach has given them to lunch, correspondingly. Each of the next *n* lines contains two space-separated integers β *f**i* (1<=β€<=*f**i*<=β€<=109) and *t**i* (1<=β€<=*t**i*<=β€<=109) β the characteristics of the *i*-th restaurant.
|
In a single line print a single integer β the maximum joy value that the Rabbits will get from the lunch.
|
[
"2 5\n3 3\n4 5\n",
"4 6\n5 8\n3 6\n2 3\n2 2\n",
"1 5\n1 7\n"
] |
[
"4\n",
"3\n",
"-1\n"
] |
none
| 500
|
[
{
"input": "2 5\n3 3\n4 5",
"output": "4"
},
{
"input": "4 6\n5 8\n3 6\n2 3\n2 2",
"output": "3"
},
{
"input": "1 5\n1 7",
"output": "-1"
},
{
"input": "4 9\n10 13\n4 18\n13 3\n10 6",
"output": "13"
},
{
"input": "1 1\n1 1000000000",
"output": "-999999998"
},
{
"input": "1 1\n1000000000 1000000000",
"output": "1"
},
{
"input": "1 1\n1000000000 1",
"output": "1000000000"
},
{
"input": "2 3\n1000000000 1\n2 2",
"output": "1000000000"
},
{
"input": "2 5\n1 7\n1 1000000000",
"output": "-1"
}
] | 1,674,938,405
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 35
| 280
| 5,529,600
|
n, k=map(int, input().split())
ans = []
for _ in range(n):
f, t=map(int, input().split())
d = abs(t-k)
e = f-d
if t<=k:
ans.append(f)
else:
ans.append(e)
ans.sort(reverse=True)
print(ans[0])
|
Title: Lunch Rush
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Having written another programming contest, three Rabbits decided to grab some lunch. The coach gave the team exactly *k* time units for the lunch break.
The Rabbits have a list of *n* restaurants to lunch in: the *i*-th restaurant is characterized by two integers *f**i* and *t**i*. Value *t**i* shows the time the Rabbits need to lunch in the *i*-th restaurant. If time *t**i* exceeds the time *k* that the coach has given for the lunch break, then the Rabbits' joy from lunching in this restaurant will equal *f**i*<=-<=(*t**i*<=-<=*k*). Otherwise, the Rabbits get exactly *f**i* units of joy.
Your task is to find the value of the maximum joy the Rabbits can get from the lunch, depending on the restaurant. The Rabbits must choose exactly one restaurant to lunch in. Note that the joy value isn't necessarily a positive value.
Input Specification:
The first line contains two space-separated integers β *n* (1<=β€<=*n*<=β€<=104) and *k* (1<=β€<=*k*<=β€<=109) β the number of restaurants in the Rabbits' list and the time the coach has given them to lunch, correspondingly. Each of the next *n* lines contains two space-separated integers β *f**i* (1<=β€<=*f**i*<=β€<=109) and *t**i* (1<=β€<=*t**i*<=β€<=109) β the characteristics of the *i*-th restaurant.
Output Specification:
In a single line print a single integer β the maximum joy value that the Rabbits will get from the lunch.
Demo Input:
['2 5\n3 3\n4 5\n', '4 6\n5 8\n3 6\n2 3\n2 2\n', '1 5\n1 7\n']
Demo Output:
['4\n', '3\n', '-1\n']
Note:
none
|
```python
n, k=map(int, input().split())
ans = []
for _ in range(n):
f, t=map(int, input().split())
d = abs(t-k)
e = f-d
if t<=k:
ans.append(f)
else:
ans.append(e)
ans.sort(reverse=True)
print(ans[0])
```
| 3
|
|
461
|
A
|
Appleman and Toastman
|
PROGRAMMING
| 1,200
|
[
"greedy",
"sortings"
] | null | null |
Appleman and Toastman play a game. Initially Appleman gives one group of *n* numbers to the Toastman, then they start to complete the following tasks:
- Each time Toastman gets a group of numbers, he sums up all the numbers and adds this sum to the score. Then he gives the group to the Appleman. - Each time Appleman gets a group consisting of a single number, he throws this group out. Each time Appleman gets a group consisting of more than one number, he splits the group into two non-empty groups (he can do it in any way) and gives each of them to Toastman.
After guys complete all the tasks they look at the score value. What is the maximum possible value of score they can get?
|
The first line contains a single integer *n* (1<=β€<=*n*<=β€<=3Β·105). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=β€<=*a**i*<=β€<=106) β the initial group that is given to Toastman.
|
Print a single integer β the largest possible score.
|
[
"3\n3 1 5\n",
"1\n10\n"
] |
[
"26\n",
"10\n"
] |
Consider the following situation in the first example. Initially Toastman gets group [3, 1, 5] and adds 9 to the score, then he give the group to Appleman. Appleman splits group [3, 1, 5] into two groups: [3, 5] and [1]. Both of them should be given to Toastman. When Toastman receives group [1], he adds 1 to score and gives the group to Appleman (he will throw it out). When Toastman receives group [3, 5], he adds 8 to the score and gives the group to Appleman. Appleman splits [3, 5] in the only possible way: [5] and [3]. Then he gives both groups to Toastman. When Toastman receives [5], he adds 5 to the score and gives the group to Appleman (he will throws it out). When Toastman receives [3], he adds 3 to the score and gives the group to Appleman (he will throws it out). Finally Toastman have added 9 + 1 + 8 + 5 + 3 = 26 to the score. This is the optimal sequence of actions.
| 500
|
[
{
"input": "3\n3 1 5",
"output": "26"
},
{
"input": "1\n10",
"output": "10"
},
{
"input": "10\n8 10 2 5 6 2 4 7 2 1",
"output": "376"
},
{
"input": "10\n171308 397870 724672 431255 228496 892002 542924 718337 888642 161821",
"output": "40204082"
},
{
"input": "10\n1 2 2 2 4 5 6 7 8 10",
"output": "376"
},
{
"input": "10\n161821 171308 228496 397870 431255 542924 718337 724672 888642 892002",
"output": "40204082"
},
{
"input": "1\n397870",
"output": "397870"
},
{
"input": "1\n1000000",
"output": "1000000"
},
{
"input": "10\n10 8 7 6 5 4 2 2 2 1",
"output": "376"
},
{
"input": "10\n892002 888642 724672 718337 542924 431255 397870 228496 171308 161821",
"output": "40204082"
},
{
"input": "10\n5 2 6 10 10 10 10 2 2 5",
"output": "485"
},
{
"input": "10\n431255 724672 228496 397870 397870 397870 397870 724672 888642 431255",
"output": "36742665"
},
{
"input": "10\n2 2 2 5 5 6 10 10 10 10",
"output": "485"
},
{
"input": "10\n228496 397870 397870 397870 397870 431255 431255 724672 724672 888642",
"output": "36742665"
},
{
"input": "10\n10 10 10 10 6 5 5 2 2 2",
"output": "485"
},
{
"input": "10\n888642 724672 724672 431255 431255 397870 397870 397870 397870 228496",
"output": "36742665"
},
{
"input": "10\n10 10 10 10 10 10 10 10 10 10",
"output": "640"
},
{
"input": "10\n1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000",
"output": "64000000"
},
{
"input": "1\n397870",
"output": "397870"
},
{
"input": "2\n1 2",
"output": "6"
},
{
"input": "2\n2 3",
"output": "10"
},
{
"input": "2\n1 1",
"output": "4"
}
] | 1,450,011,648
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 36
| 436
| 21,401,600
|
#@author: xyj
n=int(input())
a=[int(i) for i in input().split()]
a=sorted(a)
ans=sum(a)
for i in range(len(a)-1):
a[i]*=(i+1)
a[n-1]*=n-1
ans+=sum(a)
print(ans)
|
Title: Appleman and Toastman
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Appleman and Toastman play a game. Initially Appleman gives one group of *n* numbers to the Toastman, then they start to complete the following tasks:
- Each time Toastman gets a group of numbers, he sums up all the numbers and adds this sum to the score. Then he gives the group to the Appleman. - Each time Appleman gets a group consisting of a single number, he throws this group out. Each time Appleman gets a group consisting of more than one number, he splits the group into two non-empty groups (he can do it in any way) and gives each of them to Toastman.
After guys complete all the tasks they look at the score value. What is the maximum possible value of score they can get?
Input Specification:
The first line contains a single integer *n* (1<=β€<=*n*<=β€<=3Β·105). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=β€<=*a**i*<=β€<=106) β the initial group that is given to Toastman.
Output Specification:
Print a single integer β the largest possible score.
Demo Input:
['3\n3 1 5\n', '1\n10\n']
Demo Output:
['26\n', '10\n']
Note:
Consider the following situation in the first example. Initially Toastman gets group [3, 1, 5] and adds 9 to the score, then he give the group to Appleman. Appleman splits group [3, 1, 5] into two groups: [3, 5] and [1]. Both of them should be given to Toastman. When Toastman receives group [1], he adds 1 to score and gives the group to Appleman (he will throw it out). When Toastman receives group [3, 5], he adds 8 to the score and gives the group to Appleman. Appleman splits [3, 5] in the only possible way: [5] and [3]. Then he gives both groups to Toastman. When Toastman receives [5], he adds 5 to the score and gives the group to Appleman (he will throws it out). When Toastman receives [3], he adds 3 to the score and gives the group to Appleman (he will throws it out). Finally Toastman have added 9 + 1 + 8 + 5 + 3 = 26 to the score. This is the optimal sequence of actions.
|
```python
#@author: xyj
n=int(input())
a=[int(i) for i in input().split()]
a=sorted(a)
ans=sum(a)
for i in range(len(a)-1):
a[i]*=(i+1)
a[n-1]*=n-1
ans+=sum(a)
print(ans)
```
| 3
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
Santa Claus is the first who came to the Christmas Olympiad, and he is going to be the first to take his place at a desk! In the classroom there are *n* lanes of *m* desks each, and there are two working places at each of the desks. The lanes are numbered from 1 to *n* from the left to the right, the desks in a lane are numbered from 1 to *m* starting from the blackboard. Note that the lanes go perpendicularly to the blackboard, not along it (see picture).
The organizers numbered all the working places from 1 to 2*nm*. The places are numbered by lanes (i.Β e. all the places of the first lane go first, then all the places of the second lane, and so on), in a lane the places are numbered starting from the nearest to the blackboard (i.Β e. from the first desk in the lane), at each desk, the place on the left is numbered before the place on the right.
Santa Clause knows that his place has number *k*. Help him to determine at which lane at which desk he should sit, and whether his place is on the left or on the right!
|
The only line contains three integers *n*, *m* and *k* (1<=β€<=*n*,<=*m*<=β€<=10<=000, 1<=β€<=*k*<=β€<=2*nm*)Β β the number of lanes, the number of desks in each lane and the number of Santa Claus' place.
|
Print two integers: the number of lane *r*, the number of desk *d*, and a character *s*, which stands for the side of the desk Santa Claus. The character *s* should be "L", if Santa Clause should sit on the left, and "R" if his place is on the right.
|
[
"4 3 9\n",
"4 3 24\n",
"2 4 4\n"
] |
[
"2 2 L\n",
"4 3 R\n",
"1 2 R\n"
] |
The first and the second samples are shown on the picture. The green place corresponds to Santa Claus' place in the first example, the blue place corresponds to Santa Claus' place in the second example.
In the third sample there are two lanes with four desks in each, and Santa Claus has the fourth place. Thus, his place is in the first lane at the second desk on the right.
| 0
|
[
{
"input": "4 3 9",
"output": "2 2 L"
},
{
"input": "4 3 24",
"output": "4 3 R"
},
{
"input": "2 4 4",
"output": "1 2 R"
},
{
"input": "3 10 24",
"output": "2 2 R"
},
{
"input": "10 3 59",
"output": "10 3 L"
},
{
"input": "10000 10000 160845880",
"output": "8043 2940 R"
},
{
"input": "1 1 1",
"output": "1 1 L"
},
{
"input": "1 1 2",
"output": "1 1 R"
},
{
"input": "1 10000 1",
"output": "1 1 L"
},
{
"input": "1 10000 20000",
"output": "1 10000 R"
},
{
"input": "10000 1 1",
"output": "1 1 L"
},
{
"input": "10000 1 10000",
"output": "5000 1 R"
},
{
"input": "10000 1 20000",
"output": "10000 1 R"
},
{
"input": "3 2 1",
"output": "1 1 L"
},
{
"input": "3 2 2",
"output": "1 1 R"
},
{
"input": "3 2 3",
"output": "1 2 L"
},
{
"input": "3 2 4",
"output": "1 2 R"
},
{
"input": "3 2 5",
"output": "2 1 L"
},
{
"input": "3 2 6",
"output": "2 1 R"
},
{
"input": "3 2 7",
"output": "2 2 L"
},
{
"input": "3 2 8",
"output": "2 2 R"
},
{
"input": "3 2 9",
"output": "3 1 L"
},
{
"input": "3 2 10",
"output": "3 1 R"
},
{
"input": "3 2 11",
"output": "3 2 L"
},
{
"input": "3 2 12",
"output": "3 2 R"
},
{
"input": "300 2000 1068628",
"output": "268 314 R"
},
{
"input": "300 2000 584756",
"output": "147 378 R"
},
{
"input": "300 2000 268181",
"output": "68 91 L"
},
{
"input": "10000 9999 186450844",
"output": "9324 4745 R"
},
{
"input": "10000 9999 197114268",
"output": "9857 6990 R"
},
{
"input": "10000 9999 112390396",
"output": "5621 818 R"
},
{
"input": "10000 10000 1",
"output": "1 1 L"
},
{
"input": "10000 10000 2",
"output": "1 1 R"
},
{
"input": "10000 10000 100000001",
"output": "5001 1 L"
},
{
"input": "10000 10000 199999999",
"output": "10000 10000 L"
},
{
"input": "10000 10000 200000000",
"output": "10000 10000 R"
},
{
"input": "1 2 1",
"output": "1 1 L"
},
{
"input": "1 2 2",
"output": "1 1 R"
},
{
"input": "1 2 3",
"output": "1 2 L"
},
{
"input": "1 2 4",
"output": "1 2 R"
},
{
"input": "2 1 1",
"output": "1 1 L"
},
{
"input": "2 1 2",
"output": "1 1 R"
},
{
"input": "2 1 3",
"output": "2 1 L"
},
{
"input": "2 1 4",
"output": "2 1 R"
},
{
"input": "4 3 7",
"output": "2 1 L"
},
{
"input": "1 1 1",
"output": "1 1 L"
}
] | 1,482,657,503
| 803
|
PyPy 3
|
OK
|
TESTS
| 46
| 139
| 23,347,200
|
from math import ceil
n, m, k = map(int, input().split())
r = ceil(k / (m * 2))
c = k % (m * 2)
c = m * 2 if c == 0 else c
print(r, end = ' ')
print(ceil(c / 2), end = ' ')
print('L' if c % 2 else 'R')
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Santa Claus is the first who came to the Christmas Olympiad, and he is going to be the first to take his place at a desk! In the classroom there are *n* lanes of *m* desks each, and there are two working places at each of the desks. The lanes are numbered from 1 to *n* from the left to the right, the desks in a lane are numbered from 1 to *m* starting from the blackboard. Note that the lanes go perpendicularly to the blackboard, not along it (see picture).
The organizers numbered all the working places from 1 to 2*nm*. The places are numbered by lanes (i.Β e. all the places of the first lane go first, then all the places of the second lane, and so on), in a lane the places are numbered starting from the nearest to the blackboard (i.Β e. from the first desk in the lane), at each desk, the place on the left is numbered before the place on the right.
Santa Clause knows that his place has number *k*. Help him to determine at which lane at which desk he should sit, and whether his place is on the left or on the right!
Input Specification:
The only line contains three integers *n*, *m* and *k* (1<=β€<=*n*,<=*m*<=β€<=10<=000, 1<=β€<=*k*<=β€<=2*nm*)Β β the number of lanes, the number of desks in each lane and the number of Santa Claus' place.
Output Specification:
Print two integers: the number of lane *r*, the number of desk *d*, and a character *s*, which stands for the side of the desk Santa Claus. The character *s* should be "L", if Santa Clause should sit on the left, and "R" if his place is on the right.
Demo Input:
['4 3 9\n', '4 3 24\n', '2 4 4\n']
Demo Output:
['2 2 L\n', '4 3 R\n', '1 2 R\n']
Note:
The first and the second samples are shown on the picture. The green place corresponds to Santa Claus' place in the first example, the blue place corresponds to Santa Claus' place in the second example.
In the third sample there are two lanes with four desks in each, and Santa Claus has the fourth place. Thus, his place is in the first lane at the second desk on the right.
|
```python
from math import ceil
n, m, k = map(int, input().split())
r = ceil(k / (m * 2))
c = k % (m * 2)
c = m * 2 if c == 0 else c
print(r, end = ' ')
print(ceil(c / 2), end = ' ')
print('L' if c % 2 else 'R')
```
| 3
|
|
804
|
A
|
Find Amir
|
PROGRAMMING
| 1,000
|
[
"constructive algorithms",
"greedy",
"math"
] | null | null |
A few years ago Sajjad left his school and register to another one due to security reasons. Now he wishes to find Amir, one of his schoolmates and good friends.
There are *n* schools numerated from 1 to *n*. One can travel between each pair of them, to do so, he needs to buy a ticket. The ticker between schools *i* and *j* costs and can be used multiple times. Help Sajjad to find the minimum cost he needs to pay for tickets to visit all schools. He can start and finish in any school.
|
The first line contains a single integer *n* (1<=β€<=*n*<=β€<=105)Β β the number of schools.
|
Print single integer: the minimum cost of tickets needed to visit all schools.
|
[
"2\n",
"10\n"
] |
[
"0\n",
"4\n"
] |
In the first example we can buy a ticket between the schools that costs <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c67d72010e0498bfd065a6a38fdeaec90358507b.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
| 500
|
[
{
"input": "2",
"output": "0"
},
{
"input": "10",
"output": "4"
},
{
"input": "43670",
"output": "21834"
},
{
"input": "4217",
"output": "2108"
},
{
"input": "17879",
"output": "8939"
},
{
"input": "31809",
"output": "15904"
},
{
"input": "40873",
"output": "20436"
},
{
"input": "77859",
"output": "38929"
},
{
"input": "53022",
"output": "26510"
},
{
"input": "79227",
"output": "39613"
},
{
"input": "100000",
"output": "49999"
},
{
"input": "82801",
"output": "41400"
},
{
"input": "5188",
"output": "2593"
},
{
"input": "86539",
"output": "43269"
},
{
"input": "12802",
"output": "6400"
},
{
"input": "20289",
"output": "10144"
},
{
"input": "32866",
"output": "16432"
},
{
"input": "33377",
"output": "16688"
},
{
"input": "31775",
"output": "15887"
},
{
"input": "60397",
"output": "30198"
},
{
"input": "100000",
"output": "49999"
},
{
"input": "99999",
"output": "49999"
},
{
"input": "99998",
"output": "49998"
},
{
"input": "99997",
"output": "49998"
},
{
"input": "99996",
"output": "49997"
},
{
"input": "1",
"output": "0"
},
{
"input": "2",
"output": "0"
},
{
"input": "3",
"output": "1"
},
{
"input": "4",
"output": "1"
},
{
"input": "1",
"output": "0"
},
{
"input": "3",
"output": "1"
}
] | 1,666,873,602
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 31
| 46
| 0
|
"""
https://codeforces.com/problemset/problem/804/A
"""
ecoles = int(input())
if ecoles % 2 == 0:
print(ecoles // 2 - 1)
else:
print(ecoles // 2)
|
Title: Find Amir
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A few years ago Sajjad left his school and register to another one due to security reasons. Now he wishes to find Amir, one of his schoolmates and good friends.
There are *n* schools numerated from 1 to *n*. One can travel between each pair of them, to do so, he needs to buy a ticket. The ticker between schools *i* and *j* costs and can be used multiple times. Help Sajjad to find the minimum cost he needs to pay for tickets to visit all schools. He can start and finish in any school.
Input Specification:
The first line contains a single integer *n* (1<=β€<=*n*<=β€<=105)Β β the number of schools.
Output Specification:
Print single integer: the minimum cost of tickets needed to visit all schools.
Demo Input:
['2\n', '10\n']
Demo Output:
['0\n', '4\n']
Note:
In the first example we can buy a ticket between the schools that costs <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c67d72010e0498bfd065a6a38fdeaec90358507b.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
|
```python
"""
https://codeforces.com/problemset/problem/804/A
"""
ecoles = int(input())
if ecoles % 2 == 0:
print(ecoles // 2 - 1)
else:
print(ecoles // 2)
```
| 3
|
|
1
|
A
|
Theatre Square
|
PROGRAMMING
| 1,000
|
[
"math"
] |
A. Theatre Square
|
1
|
256
|
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=Γ<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=Γ<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
|
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=β€<=<=*n*,<=*m*,<=*a*<=β€<=109).
|
Write the needed number of flagstones.
|
[
"6 6 4\n"
] |
[
"4\n"
] |
none
| 0
|
[
{
"input": "6 6 4",
"output": "4"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "2 1 1",
"output": "2"
},
{
"input": "1 2 1",
"output": "2"
},
{
"input": "2 2 1",
"output": "4"
},
{
"input": "2 1 2",
"output": "1"
},
{
"input": "1 1 3",
"output": "1"
},
{
"input": "2 3 4",
"output": "1"
},
{
"input": "1000000000 1000000000 1",
"output": "1000000000000000000"
},
{
"input": "12 13 4",
"output": "12"
},
{
"input": "222 332 5",
"output": "3015"
},
{
"input": "1000 1000 10",
"output": "10000"
},
{
"input": "1001 1000 10",
"output": "10100"
},
{
"input": "100 10001 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000 999999999",
"output": "4"
},
{
"input": "1000000000 1000000000 192",
"output": "27126743055556"
},
{
"input": "1000000000 987654321 1",
"output": "987654321000000000"
},
{
"input": "456784567 1000000000 51",
"output": "175618850864484"
},
{
"input": "39916800 134217728 40320",
"output": "3295710"
}
] | 1,692,377,832
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 20
| 31
| 0
|
import math
def main():
n, m, a = map(int, input().split())
flagstones_needed = math.ceil(n/a) * math.ceil(m/a)
print(flagstones_needed)
if __name__ == "__main__":
main()
|
Title: Theatre Square
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=Γ<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=Γ<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
Input Specification:
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=β€<=<=*n*,<=*m*,<=*a*<=β€<=109).
Output Specification:
Write the needed number of flagstones.
Demo Input:
['6 6 4\n']
Demo Output:
['4\n']
Note:
none
|
```python
import math
def main():
n, m, a = map(int, input().split())
flagstones_needed = math.ceil(n/a) * math.ceil(m/a)
print(flagstones_needed)
if __name__ == "__main__":
main()
```
| 3.9845
|
287
|
A
|
IQ Test
|
PROGRAMMING
| 1,100
|
[
"brute force",
"implementation"
] | null | null |
In the city of Ultima Thule job applicants are often offered an IQ test.
The test is as follows: the person gets a piece of squared paper with a 4<=Γ<=4 square painted on it. Some of the square's cells are painted black and others are painted white. Your task is to repaint at most one cell the other color so that the picture has a 2<=Γ<=2 square, completely consisting of cells of the same color. If the initial picture already has such a square, the person should just say so and the test will be completed.
Your task is to write a program that determines whether it is possible to pass the test. You cannot pass the test if either repainting any cell or no action doesn't result in a 2<=Γ<=2 square, consisting of cells of the same color.
|
Four lines contain four characters each: the *j*-th character of the *i*-th line equals "." if the cell in the *i*-th row and the *j*-th column of the square is painted white, and "#", if the cell is black.
|
Print "YES" (without the quotes), if the test can be passed and "NO" (without the quotes) otherwise.
|
[
"####\n.#..\n####\n....\n",
"####\n....\n####\n....\n"
] |
[
"YES\n",
"NO\n"
] |
In the first test sample it is enough to repaint the first cell in the second row. After such repainting the required 2βΓβ2 square is on the intersection of the 1-st and 2-nd row with the 1-st and 2-nd column.
| 500
|
[
{
"input": "###.\n...#\n###.\n...#",
"output": "NO"
},
{
"input": ".##.\n#..#\n.##.\n#..#",
"output": "NO"
},
{
"input": ".#.#\n#.#.\n.#.#\n#.#.",
"output": "NO"
},
{
"input": "##..\n..##\n##..\n..##",
"output": "NO"
},
{
"input": "#.#.\n#.#.\n.#.#\n.#.#",
"output": "NO"
},
{
"input": ".#.#\n#.#.\n#.#.\n#.#.",
"output": "NO"
},
{
"input": ".#.#\n#.#.\n#.#.\n.#.#",
"output": "NO"
},
{
"input": "#.#.\n#.#.\n#.#.\n#.#.",
"output": "NO"
},
{
"input": ".#.#\n.#.#\n.#.#\n.#.#",
"output": "NO"
},
{
"input": "#..#\n.##.\n####\n####",
"output": "YES"
},
{
"input": "#.#.\n.###\n#.#.\n.###",
"output": "YES"
},
{
"input": "#..#\n.##.\n.##.\n#..#",
"output": "YES"
},
{
"input": ".##.\n.#..\n##.#\n#..#",
"output": "YES"
},
{
"input": ".##.\n##..\n#..#\n..##",
"output": "YES"
},
{
"input": "##..\n##..\n..##\n..##",
"output": "YES"
},
{
"input": ".#..\n###.\n.#.#\n..#.",
"output": "YES"
},
{
"input": "###.\n###.\n...#\n...#",
"output": "YES"
},
{
"input": "#.##\n##.#\n#.##\n##.#",
"output": "YES"
},
{
"input": ".#.#\n#.#.\n.#.#\n#.##",
"output": "YES"
},
{
"input": "##..\n..##\n##..\n...#",
"output": "YES"
},
{
"input": ".#..\n..##\n##..\n..##",
"output": "YES"
},
{
"input": "##..\n...#\n##..\n...#",
"output": "YES"
},
{
"input": ".#..\n..#.\n.#..\n..#.",
"output": "YES"
},
{
"input": "....\n....\n....\n.#.#",
"output": "YES"
},
{
"input": "....\n....\n....\n...#",
"output": "YES"
}
] | 1,694,450,364
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 30
| 93
| 0
|
import sys
from os import path
if (path.exists('input.txt')):
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt','w')
def solve(arr):
for i in range (3):
for j in range (3):
s = arr[i][j] + arr[i][j+1] + arr[i+1][j] + arr[i+1][j+1]
if s.count('#')!=2: return True
return False
arr = []
for k in range (4):
arr.append(list(input()))
print('YES' if solve(arr) else 'NO')
|
Title: IQ Test
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In the city of Ultima Thule job applicants are often offered an IQ test.
The test is as follows: the person gets a piece of squared paper with a 4<=Γ<=4 square painted on it. Some of the square's cells are painted black and others are painted white. Your task is to repaint at most one cell the other color so that the picture has a 2<=Γ<=2 square, completely consisting of cells of the same color. If the initial picture already has such a square, the person should just say so and the test will be completed.
Your task is to write a program that determines whether it is possible to pass the test. You cannot pass the test if either repainting any cell or no action doesn't result in a 2<=Γ<=2 square, consisting of cells of the same color.
Input Specification:
Four lines contain four characters each: the *j*-th character of the *i*-th line equals "." if the cell in the *i*-th row and the *j*-th column of the square is painted white, and "#", if the cell is black.
Output Specification:
Print "YES" (without the quotes), if the test can be passed and "NO" (without the quotes) otherwise.
Demo Input:
['####\n.#..\n####\n....\n', '####\n....\n####\n....\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first test sample it is enough to repaint the first cell in the second row. After such repainting the required 2βΓβ2 square is on the intersection of the 1-st and 2-nd row with the 1-st and 2-nd column.
|
```python
import sys
from os import path
if (path.exists('input.txt')):
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt','w')
def solve(arr):
for i in range (3):
for j in range (3):
s = arr[i][j] + arr[i][j+1] + arr[i+1][j] + arr[i+1][j+1]
if s.count('#')!=2: return True
return False
arr = []
for k in range (4):
arr.append(list(input()))
print('YES' if solve(arr) else 'NO')
```
| 3
|
|
501
|
B
|
Misha and Changing Handles
|
PROGRAMMING
| 1,100
|
[
"data structures",
"dsu",
"strings"
] | null | null |
Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point.
Misha has a list of handle change requests. After completing the requests he wants to understand the relation between the original and the new handles of the users. Help him to do that.
|
The first line contains integer *q* (1<=β€<=*q*<=β€<=1000), the number of handle change requests.
Next *q* lines contain the descriptions of the requests, one per line.
Each query consists of two non-empty strings *old* and *new*, separated by a space. The strings consist of lowercase and uppercase Latin letters and digits. Strings *old* and *new* are distinct. The lengths of the strings do not exceed 20.
The requests are given chronologically. In other words, by the moment of a query there is a single person with handle *old*, and handle *new* is not used and has not been used by anyone.
|
In the first line output the integer *n* β the number of users that changed their handles at least once.
In the next *n* lines print the mapping between the old and the new handles of the users. Each of them must contain two strings, *old* and *new*, separated by a space, meaning that before the user had handle *old*, and after all the requests are completed, his handle is *new*. You may output lines in any order.
Each user who changes the handle must occur exactly once in this description.
|
[
"5\nMisha ILoveCodeforces\nVasya Petrov\nPetrov VasyaPetrov123\nILoveCodeforces MikeMirzayanov\nPetya Ivanov\n"
] |
[
"3\nPetya Ivanov\nMisha MikeMirzayanov\nVasya VasyaPetrov123\n"
] |
none
| 500
|
[
{
"input": "5\nMisha ILoveCodeforces\nVasya Petrov\nPetrov VasyaPetrov123\nILoveCodeforces MikeMirzayanov\nPetya Ivanov",
"output": "3\nPetya Ivanov\nMisha MikeMirzayanov\nVasya VasyaPetrov123"
},
{
"input": "1\nMisha Vasya",
"output": "1\nMisha Vasya"
},
{
"input": "10\na b\nb c\nc d\nd e\ne f\nf g\ng h\nh i\ni j\nj k",
"output": "1\na k"
},
{
"input": "5\n123abc abc123\nabc123 a1b2c3\na1b2c3 1A2B3C\n1 2\n2 Misha",
"output": "2\n123abc 1A2B3C\n1 Misha"
},
{
"input": "8\nM F\nS D\n1 2\nF G\n2 R\nD Q\nQ W\nW e",
"output": "3\nM G\n1 R\nS e"
},
{
"input": "17\nn5WhQ VCczxtxKwFio5U\nVCczxtxKwFio5U 1WMVGA17cd1LRcp4r\n1WMVGA17cd1LRcp4r SJl\nSJl D8bPUoIft5v1\nNAvvUgunbPZNCL9ZY2 jnLkarKYsotz\nD8bPUoIft5v1 DnDkHi7\njnLkarKYsotz GfjX109HSQ81gFEBJc\nGfjX109HSQ81gFEBJc kBJ0zrH78mveJ\nkBJ0zrH78mveJ 9DrAypYW\nDnDkHi7 3Wkho2PglMDaFQw\n3Wkho2PglMDaFQw pOqW\n9DrAypYW G3y0cXXGsWAh\npOqW yr1Ec\nG3y0cXXGsWAh HrmWWg5u4Hsy\nyr1Ec GkFeivXjQ01\nGkFeivXjQ01 mSsWgbCCZcotV4goiA\nHrmWWg5u4Hsy zkCmEV",
"output": "2\nn5WhQ mSsWgbCCZcotV4goiA\nNAvvUgunbPZNCL9ZY2 zkCmEV"
},
{
"input": "10\nH1nauWCJOImtVqXk gWPMQ9DHv5CtkYp9lwm9\nSEj 2knOMLyzr\n0v69ijnAc S7d7zGTjmlku01Gv\n2knOMLyzr otGmEd\nacwr3TfMV7oCIp RUSVFa9TIWlLsd7SB\nS7d7zGTjmlku01Gv Gd6ZufVmQnBpi\nS1 WOJLpk\nWOJLpk Gu\nRUSVFa9TIWlLsd7SB RFawatGnbVB\notGmEd OTB1zKiOI",
"output": "5\n0v69ijnAc Gd6ZufVmQnBpi\nS1 Gu\nSEj OTB1zKiOI\nacwr3TfMV7oCIp RFawatGnbVB\nH1nauWCJOImtVqXk gWPMQ9DHv5CtkYp9lwm9"
},
{
"input": "14\nTPdoztSZROpjZe z6F8bYFvnER4V5SP0n\n8Aa3PQY3hzHZTPEUz fhrZZPJ3iUS\nm9p888KaZAoQaO KNmdRSAlUVn8zXOM0\nAO s1VGWTCbHzM\ni 4F\nfhrZZPJ3iUS j0OVZQF6MvNcKN9xDZFJ\nDnlkXtaKNlYEI2ApBuwu DMA9i8ScKRxwhe72a3\nj0OVZQF6MvNcKN9xDZFJ DzjmeNqN0H4Teq0Awr\n4F wJcdxt1kwqfDeJ\nqxXlsa5t RHCL1K6aUyns\nr6WYbDaXt hEHw\nJ0Usg DKdKMFJ6tK8XA\nz6F8bYFvnER4V5SP0n 0alJ\nMijh2O6 qic8kXWuR6",
"output": "10\nTPdoztSZROpjZe 0alJ\nJ0Usg DKdKMFJ6tK8XA\nDnlkXtaKNlYEI2ApBuwu DMA9i8ScKRxwhe72a3\n8Aa3PQY3hzHZTPEUz DzjmeNqN0H4Teq0Awr\nm9p888KaZAoQaO KNmdRSAlUVn8zXOM0\nqxXlsa5t RHCL1K6aUyns\nr6WYbDaXt hEHw\nMijh2O6 qic8kXWuR6\nAO s1VGWTCbHzM\ni wJcdxt1kwqfDeJ"
},
{
"input": "14\nHAXRxayyf1Dj1F0mT hjR4A8IQMb0nyBtqG\nWNuMJa5Jg05qkqZOrL noNkWXrSidHGwxgbQ\nmOitVy6W52s0FENMz6 oLUkLNfojssvLvb1t\nhjR4A8IQMb0nyBtqG oA7uBFu4Oo\noA7uBFu4Oo M450\nM450 LXEzO4\noLUkLNfojssvLvb1t YG5\nnoNkWXrSidHGwxgbQ L\nL YBWzu4W\nYBWzu4W ML\nML scVZE9m8JnH\nLXEzO4 Ne0oBPY0Iy\nscVZE9m8JnH GXhznv\nYG5 UY08abilYF1LaXj49hQ",
"output": "3\nWNuMJa5Jg05qkqZOrL GXhznv\nHAXRxayyf1Dj1F0mT Ne0oBPY0Iy\nmOitVy6W52s0FENMz6 UY08abilYF1LaXj49hQ"
}
] | 1,684,530,746
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 49
| 109
| 6,041,600
|
import sys
import math
import bisect
import heapq
import itertools
from sys import stdin,stdout
from math import gcd,floor,sqrt,log
from collections import defaultdict, Counter, deque
from bisect import bisect_left,bisect_right, insort_left, insort_right
mod=1000000007
def get_ints(): return map(int, sys.stdin.readline().strip().split())
def get_list(): return list(map(int, sys.stdin.readline().strip().split()))
def get_string(): return sys.stdin.readline().strip()
def get_int(): return int(sys.stdin.readline().strip())
def get_list_strings(): return list(map(str, sys.stdin.readline().strip().split()))
def solve():
n = get_int()
# parent, size, distance
parent = {}
def find(member):
if member not in parent:
parent[member] = member
return member
if member == parent[member]:
return member
parent[member] = find(parent[member])
return parent[member]
def union(x,y):
xpar = find(x)
ypar = find(y)
parent[ypar] = xpar
def isConnected(x,y):
return find(x) == find(y)
for i in range(n):
inp = get_list_strings()
# print(inp, parent)
union(inp[0], inp[1])
parentset = defaultdict(list)
for key in parent.keys():
parentset[find(key)].append(key)
print(len(parentset))
for p in parentset.values():
print(p[0], p[-1])
if __name__ == "__main__":
solve()
|
Title: Misha and Changing Handles
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point.
Misha has a list of handle change requests. After completing the requests he wants to understand the relation between the original and the new handles of the users. Help him to do that.
Input Specification:
The first line contains integer *q* (1<=β€<=*q*<=β€<=1000), the number of handle change requests.
Next *q* lines contain the descriptions of the requests, one per line.
Each query consists of two non-empty strings *old* and *new*, separated by a space. The strings consist of lowercase and uppercase Latin letters and digits. Strings *old* and *new* are distinct. The lengths of the strings do not exceed 20.
The requests are given chronologically. In other words, by the moment of a query there is a single person with handle *old*, and handle *new* is not used and has not been used by anyone.
Output Specification:
In the first line output the integer *n* β the number of users that changed their handles at least once.
In the next *n* lines print the mapping between the old and the new handles of the users. Each of them must contain two strings, *old* and *new*, separated by a space, meaning that before the user had handle *old*, and after all the requests are completed, his handle is *new*. You may output lines in any order.
Each user who changes the handle must occur exactly once in this description.
Demo Input:
['5\nMisha ILoveCodeforces\nVasya Petrov\nPetrov VasyaPetrov123\nILoveCodeforces MikeMirzayanov\nPetya Ivanov\n']
Demo Output:
['3\nPetya Ivanov\nMisha MikeMirzayanov\nVasya VasyaPetrov123\n']
Note:
none
|
```python
import sys
import math
import bisect
import heapq
import itertools
from sys import stdin,stdout
from math import gcd,floor,sqrt,log
from collections import defaultdict, Counter, deque
from bisect import bisect_left,bisect_right, insort_left, insort_right
mod=1000000007
def get_ints(): return map(int, sys.stdin.readline().strip().split())
def get_list(): return list(map(int, sys.stdin.readline().strip().split()))
def get_string(): return sys.stdin.readline().strip()
def get_int(): return int(sys.stdin.readline().strip())
def get_list_strings(): return list(map(str, sys.stdin.readline().strip().split()))
def solve():
n = get_int()
# parent, size, distance
parent = {}
def find(member):
if member not in parent:
parent[member] = member
return member
if member == parent[member]:
return member
parent[member] = find(parent[member])
return parent[member]
def union(x,y):
xpar = find(x)
ypar = find(y)
parent[ypar] = xpar
def isConnected(x,y):
return find(x) == find(y)
for i in range(n):
inp = get_list_strings()
# print(inp, parent)
union(inp[0], inp[1])
parentset = defaultdict(list)
for key in parent.keys():
parentset[find(key)].append(key)
print(len(parentset))
for p in parentset.values():
print(p[0], p[-1])
if __name__ == "__main__":
solve()
```
| 3
|
|
984
|
A
|
Game
|
PROGRAMMING
| 800
|
[
"sortings"
] | null | null |
Two players play a game.
Initially there are $n$ integers $a_1, a_2, \ldots, a_n$ written on the board. Each turn a player selects one number and erases it from the board. This continues until there is only one number left on the board, i.Β e. $n - 1$ turns are made. The first player makes the first move, then players alternate turns.
The first player wants to minimize the last number that would be left on the board, while the second player wants to maximize it.
You want to know what number will be left on the board after $n - 1$ turns if both players make optimal moves.
|
The first line contains one integer $n$ ($1 \le n \le 1000$)Β β the number of numbers on the board.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$).
|
Print one number that will be left on the board.
|
[
"3\n2 1 3\n",
"3\n2 2 2\n"
] |
[
"2",
"2"
] |
In the first sample, the first player erases $3$ and the second erases $1$. $2$ is left on the board.
In the second sample, $2$ is left on the board regardless of the actions of the players.
| 500
|
[
{
"input": "3\n2 1 3",
"output": "2"
},
{
"input": "3\n2 2 2",
"output": "2"
},
{
"input": "9\n44 53 51 80 5 27 74 79 94",
"output": "53"
},
{
"input": "10\n38 82 23 37 96 4 81 60 67 86",
"output": "60"
},
{
"input": "10\n58 26 77 15 53 81 68 48 22 65",
"output": "53"
},
{
"input": "1\n124",
"output": "124"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "3\n1 1 1000",
"output": "1"
},
{
"input": "2\n322 322",
"output": "322"
},
{
"input": "3\n9 92 12",
"output": "12"
},
{
"input": "3\n1 2 2",
"output": "2"
}
] | 1,561,199,340
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 35
| 109
| 0
|
# import sys
# sys.stdin=open("input.in",'r')
# sys.stdout=open("ou.out",'w')
n=int(input())
a=list(map(int,input().split()))
a.sort()
if n%2==0:
x=n//2-1
else:
x=n//2
print(a[x])
|
Title: Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Two players play a game.
Initially there are $n$ integers $a_1, a_2, \ldots, a_n$ written on the board. Each turn a player selects one number and erases it from the board. This continues until there is only one number left on the board, i.Β e. $n - 1$ turns are made. The first player makes the first move, then players alternate turns.
The first player wants to minimize the last number that would be left on the board, while the second player wants to maximize it.
You want to know what number will be left on the board after $n - 1$ turns if both players make optimal moves.
Input Specification:
The first line contains one integer $n$ ($1 \le n \le 1000$)Β β the number of numbers on the board.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$).
Output Specification:
Print one number that will be left on the board.
Demo Input:
['3\n2 1 3\n', '3\n2 2 2\n']
Demo Output:
['2', '2']
Note:
In the first sample, the first player erases $3$ and the second erases $1$. $2$ is left on the board.
In the second sample, $2$ is left on the board regardless of the actions of the players.
|
```python
# import sys
# sys.stdin=open("input.in",'r')
# sys.stdout=open("ou.out",'w')
n=int(input())
a=list(map(int,input().split()))
a.sort()
if n%2==0:
x=n//2-1
else:
x=n//2
print(a[x])
```
| 3
|
|
915
|
A
|
Garden
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Luba thinks about watering her garden. The garden can be represented as a segment of length *k*. Luba has got *n* buckets, the *i*-th bucket allows her to water some continuous subsegment of garden of length exactly *a**i* each hour. Luba can't water any parts of the garden that were already watered, also she can't water the ground outside the garden.
Luba has to choose one of the buckets in order to water the garden as fast as possible (as mentioned above, each hour she will water some continuous subsegment of length *a**i* if she chooses the *i*-th bucket). Help her to determine the minimum number of hours she has to spend watering the garden. It is guaranteed that Luba can always choose a bucket so it is possible water the garden.
See the examples for better understanding.
|
The first line of input contains two integer numbers *n* and *k* (1<=β€<=*n*,<=*k*<=β€<=100) β the number of buckets and the length of the garden, respectively.
The second line of input contains *n* integer numbers *a**i* (1<=β€<=*a**i*<=β€<=100) β the length of the segment that can be watered by the *i*-th bucket in one hour.
It is guaranteed that there is at least one bucket such that it is possible to water the garden in integer number of hours using only this bucket.
|
Print one integer number β the minimum number of hours required to water the garden.
|
[
"3 6\n2 3 5\n",
"6 7\n1 2 3 4 5 6\n"
] |
[
"2\n",
"7\n"
] |
In the first test the best option is to choose the bucket that allows to water the segment of length 3. We can't choose the bucket that allows to water the segment of length 5 because then we can't water the whole garden.
In the second test we can choose only the bucket that allows us to water the segment of length 1.
| 0
|
[
{
"input": "3 6\n2 3 5",
"output": "2"
},
{
"input": "6 7\n1 2 3 4 5 6",
"output": "7"
},
{
"input": "5 97\n1 10 50 97 2",
"output": "1"
},
{
"input": "5 97\n1 10 50 100 2",
"output": "97"
},
{
"input": "100 100\n2 46 24 18 86 90 31 38 84 49 58 28 15 80 14 24 87 56 62 87 41 87 55 71 87 32 41 56 91 32 24 75 43 42 35 30 72 53 31 26 54 61 87 85 36 75 44 31 7 38 77 57 61 54 70 77 45 96 39 57 11 8 91 42 52 15 42 30 92 41 27 26 34 27 3 80 32 86 26 97 63 91 30 75 14 7 19 23 45 11 8 43 44 73 11 56 3 55 63 16",
"output": "50"
},
{
"input": "100 91\n13 13 62 96 74 47 81 46 78 21 20 42 4 73 25 30 76 74 58 28 25 52 42 48 74 40 82 9 25 29 17 22 46 64 57 95 81 39 47 86 40 95 97 35 31 98 45 98 47 78 52 63 58 14 89 97 17 95 28 22 20 36 68 38 95 16 2 26 54 47 42 31 31 81 21 21 65 40 82 53 60 71 75 33 96 98 6 22 95 12 5 48 18 27 58 62 5 96 36 75",
"output": "7"
},
{
"input": "8 8\n8 7 6 5 4 3 2 1",
"output": "1"
},
{
"input": "3 8\n4 3 2",
"output": "2"
},
{
"input": "3 8\n2 4 2",
"output": "2"
},
{
"input": "3 6\n1 3 2",
"output": "2"
},
{
"input": "3 6\n3 2 5",
"output": "2"
},
{
"input": "3 8\n4 2 1",
"output": "2"
},
{
"input": "5 6\n2 3 5 1 2",
"output": "2"
},
{
"input": "2 6\n5 3",
"output": "2"
},
{
"input": "4 12\n6 4 3 1",
"output": "2"
},
{
"input": "3 18\n1 9 6",
"output": "2"
},
{
"input": "3 9\n3 2 1",
"output": "3"
},
{
"input": "3 6\n5 3 2",
"output": "2"
},
{
"input": "2 10\n5 2",
"output": "2"
},
{
"input": "2 18\n6 3",
"output": "3"
},
{
"input": "4 12\n1 2 12 3",
"output": "1"
},
{
"input": "3 7\n3 2 1",
"output": "7"
},
{
"input": "3 6\n3 2 1",
"output": "2"
},
{
"input": "5 10\n5 4 3 2 1",
"output": "2"
},
{
"input": "5 16\n8 4 2 1 7",
"output": "2"
},
{
"input": "6 7\n6 5 4 3 7 1",
"output": "1"
},
{
"input": "2 6\n3 2",
"output": "2"
},
{
"input": "2 4\n4 1",
"output": "1"
},
{
"input": "6 8\n2 4 1 3 5 7",
"output": "2"
},
{
"input": "6 8\n6 5 4 3 2 1",
"output": "2"
},
{
"input": "6 15\n5 2 3 6 4 3",
"output": "3"
},
{
"input": "4 8\n2 4 8 1",
"output": "1"
},
{
"input": "2 5\n5 1",
"output": "1"
},
{
"input": "4 18\n3 1 1 2",
"output": "6"
},
{
"input": "2 1\n2 1",
"output": "1"
},
{
"input": "3 10\n2 10 5",
"output": "1"
},
{
"input": "5 12\n12 4 4 4 3",
"output": "1"
},
{
"input": "3 6\n6 3 2",
"output": "1"
},
{
"input": "2 2\n2 1",
"output": "1"
},
{
"input": "3 18\n1 9 3",
"output": "2"
},
{
"input": "3 8\n7 2 4",
"output": "2"
},
{
"input": "2 100\n99 1",
"output": "100"
},
{
"input": "4 12\n1 3 4 2",
"output": "3"
},
{
"input": "3 6\n2 3 1",
"output": "2"
},
{
"input": "4 6\n3 2 5 12",
"output": "2"
},
{
"input": "4 97\n97 1 50 10",
"output": "1"
},
{
"input": "3 12\n1 12 2",
"output": "1"
},
{
"input": "4 12\n1 4 3 2",
"output": "3"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "3 19\n7 1 1",
"output": "19"
},
{
"input": "5 12\n12 4 3 4 4",
"output": "1"
},
{
"input": "3 8\n8 4 2",
"output": "1"
},
{
"input": "3 3\n3 2 1",
"output": "1"
},
{
"input": "5 6\n3 2 4 2 2",
"output": "2"
},
{
"input": "2 16\n8 4",
"output": "2"
},
{
"input": "3 6\n10 2 3",
"output": "2"
},
{
"input": "5 3\n2 4 5 3 6",
"output": "1"
},
{
"input": "11 99\n1 2 3 6 5 4 7 8 99 33 66",
"output": "1"
},
{
"input": "3 12\n3 12 2",
"output": "1"
},
{
"input": "5 25\n24 5 15 25 23",
"output": "1"
},
{
"input": "2 4\n8 1",
"output": "4"
},
{
"input": "4 100\n2 50 4 1",
"output": "2"
},
{
"input": "3 28\n7 14 1",
"output": "2"
},
{
"input": "4 8\n2 8 4 1",
"output": "1"
},
{
"input": "4 6\n6 1 2 3",
"output": "1"
},
{
"input": "2 12\n4 3",
"output": "3"
},
{
"input": "4 12\n1 2 4 3",
"output": "3"
},
{
"input": "5 12\n2 3 12 6 4",
"output": "1"
},
{
"input": "4 4\n1 2 2 4",
"output": "1"
},
{
"input": "3 6\n2 3 2",
"output": "2"
},
{
"input": "4 21\n21 20 21 2",
"output": "1"
},
{
"input": "3 8\n3 4 2",
"output": "2"
},
{
"input": "1 25\n25",
"output": "1"
},
{
"input": "99 12\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",
"output": "1"
},
{
"input": "98 12\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",
"output": "1"
},
{
"input": "79 12\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",
"output": "1"
},
{
"input": "4 32\n1 1 1 1",
"output": "32"
},
{
"input": "1 100\n1",
"output": "100"
},
{
"input": "2 100\n7 1",
"output": "100"
},
{
"input": "7 24\n1 3 6 4 5 2 7",
"output": "4"
},
{
"input": "6 87\n1 2 8 4 5 7",
"output": "87"
},
{
"input": "1 88\n1",
"output": "88"
},
{
"input": "1 89\n1",
"output": "89"
}
] | 1,588,600,068
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 83
| 109
| 0
|
n, k = map(int, input().split())
num = [int(x) for x in input().split()]
res = []
for i in num:
if k % i == 0:
res.append(k // i)
print(min(res))
|
Title: Garden
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Luba thinks about watering her garden. The garden can be represented as a segment of length *k*. Luba has got *n* buckets, the *i*-th bucket allows her to water some continuous subsegment of garden of length exactly *a**i* each hour. Luba can't water any parts of the garden that were already watered, also she can't water the ground outside the garden.
Luba has to choose one of the buckets in order to water the garden as fast as possible (as mentioned above, each hour she will water some continuous subsegment of length *a**i* if she chooses the *i*-th bucket). Help her to determine the minimum number of hours she has to spend watering the garden. It is guaranteed that Luba can always choose a bucket so it is possible water the garden.
See the examples for better understanding.
Input Specification:
The first line of input contains two integer numbers *n* and *k* (1<=β€<=*n*,<=*k*<=β€<=100) β the number of buckets and the length of the garden, respectively.
The second line of input contains *n* integer numbers *a**i* (1<=β€<=*a**i*<=β€<=100) β the length of the segment that can be watered by the *i*-th bucket in one hour.
It is guaranteed that there is at least one bucket such that it is possible to water the garden in integer number of hours using only this bucket.
Output Specification:
Print one integer number β the minimum number of hours required to water the garden.
Demo Input:
['3 6\n2 3 5\n', '6 7\n1 2 3 4 5 6\n']
Demo Output:
['2\n', '7\n']
Note:
In the first test the best option is to choose the bucket that allows to water the segment of length 3. We can't choose the bucket that allows to water the segment of length 5 because then we can't water the whole garden.
In the second test we can choose only the bucket that allows us to water the segment of length 1.
|
```python
n, k = map(int, input().split())
num = [int(x) for x in input().split()]
res = []
for i in num:
if k % i == 0:
res.append(k // i)
print(min(res))
```
| 3
|
|
312
|
B
|
Archer
|
PROGRAMMING
| 1,300
|
[
"math",
"probabilities"
] | null | null |
SmallR is an archer. SmallR is taking a match of archer with Zanoes. They try to shoot in the target in turns, and SmallR shoots first. The probability of shooting the target each time is for SmallR while for Zanoes. The one who shoots in the target first should be the winner.
Output the probability that SmallR will win the match.
|
A single line contains four integers .
|
Print a single real number, the probability that SmallR will win the match.
The answer will be considered correct if the absolute or relative error doesn't exceed 10<=-<=6.
|
[
"1 2 1 2\n"
] |
[
"0.666666666667"
] |
none
| 1,000
|
[
{
"input": "1 2 1 2",
"output": "0.666666666667"
},
{
"input": "1 3 1 3",
"output": "0.600000000000"
},
{
"input": "1 3 2 3",
"output": "0.428571428571"
},
{
"input": "3 4 3 4",
"output": "0.800000000000"
},
{
"input": "1 2 10 11",
"output": "0.523809523810"
},
{
"input": "4 5 4 5",
"output": "0.833333333333"
},
{
"input": "466 701 95 721",
"output": "0.937693791148"
},
{
"input": "268 470 444 885",
"output": "0.725614009325"
},
{
"input": "632 916 713 821",
"output": "0.719292895126"
},
{
"input": "269 656 918 992",
"output": "0.428937461623"
},
{
"input": "71 657 187 695",
"output": "0.310488463257"
},
{
"input": "435 852 973 978",
"output": "0.511844133157"
},
{
"input": "518 816 243 359",
"output": "0.719734031025"
},
{
"input": "882 962 311 811",
"output": "0.966386645447"
},
{
"input": "684 774 580 736",
"output": "0.906051574446"
},
{
"input": "486 868 929 999",
"output": "0.577723252958"
},
{
"input": "132 359 996 998",
"output": "0.368154532345"
},
{
"input": "933 977 266 450",
"output": "0.972879407907"
},
{
"input": "298 833 615 872",
"output": "0.441270817024"
},
{
"input": "34 554 14 958",
"output": "0.817324099167"
},
{
"input": "836 934 800 905",
"output": "0.906105535462"
},
{
"input": "482 815 69 509",
"output": "0.914365577772"
},
{
"input": "284 423 137 521",
"output": "0.885974839378"
},
{
"input": "648 881 486 703",
"output": "0.800911421248"
},
{
"input": "450 885 755 836",
"output": "0.533901011176"
},
{
"input": "533 773 823 998",
"output": "0.729222130525"
},
{
"input": "897 957 92 898",
"output": "0.993193806364"
},
{
"input": "699 925 441 928",
"output": "0.866816866175"
},
{
"input": "64 704 148 603",
"output": "0.289486317811"
},
{
"input": "719 735 626 990",
"output": "0.986124079764"
},
{
"input": "1 1000 1 1000",
"output": "0.500250125063"
}
] | 1,517,808,647
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 31
| 62
| 5,632,000
|
import sys
a, b, c, d = [float(x) for x in sys.stdin.readline().strip().split(" ")]
f = (1 - (a/b)) * (1 - (c/d))
ans = (a/b) * (1 / (1 - f))
print(ans)
|
Title: Archer
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
SmallR is an archer. SmallR is taking a match of archer with Zanoes. They try to shoot in the target in turns, and SmallR shoots first. The probability of shooting the target each time is for SmallR while for Zanoes. The one who shoots in the target first should be the winner.
Output the probability that SmallR will win the match.
Input Specification:
A single line contains four integers .
Output Specification:
Print a single real number, the probability that SmallR will win the match.
The answer will be considered correct if the absolute or relative error doesn't exceed 10<=-<=6.
Demo Input:
['1 2 1 2\n']
Demo Output:
['0.666666666667']
Note:
none
|
```python
import sys
a, b, c, d = [float(x) for x in sys.stdin.readline().strip().split(" ")]
f = (1 - (a/b)) * (1 - (c/d))
ans = (a/b) * (1 / (1 - f))
print(ans)
```
| 3
|
|
6
|
C
|
Alice, Bob and Chocolate
|
PROGRAMMING
| 1,200
|
[
"greedy",
"two pointers"
] |
C. Alice, Bob and Chocolate
|
2
|
64
|
Alice and Bob like games. And now they are ready to start a new game. They have placed *n* chocolate bars in a line. Alice starts to eat chocolate bars one by one from left to right, and Bob β from right to left. For each chocololate bar the time, needed for the player to consume it, is known (Alice and Bob eat them with equal speed). When the player consumes a chocolate bar, he immediately starts with another. It is not allowed to eat two chocolate bars at the same time, to leave the bar unfinished and to make pauses. If both players start to eat the same bar simultaneously, Bob leaves it to Alice as a true gentleman.
How many bars each of the players will consume?
|
The first line contains one integer *n* (1<=β€<=*n*<=β€<=105) β the amount of bars on the table. The second line contains a sequence *t*1,<=*t*2,<=...,<=*t**n* (1<=β€<=*t**i*<=β€<=1000), where *t**i* is the time (in seconds) needed to consume the *i*-th bar (in the order from left to right).
|
Print two numbers *a* and *b*, where *a* is the amount of bars consumed by Alice, and *b* is the amount of bars consumed by Bob.
|
[
"5\n2 9 8 2 7\n"
] |
[
"2 3\n"
] |
none
| 0
|
[
{
"input": "5\n2 9 8 2 7",
"output": "2 3"
},
{
"input": "1\n1",
"output": "1 0"
},
{
"input": "2\n1 1",
"output": "1 1"
},
{
"input": "3\n1 1 1",
"output": "2 1"
},
{
"input": "4\n1 1 1 1",
"output": "2 2"
},
{
"input": "1\n6",
"output": "1 0"
},
{
"input": "2\n8 2",
"output": "1 1"
},
{
"input": "3\n6 8 4",
"output": "1 2"
},
{
"input": "4\n4 3 2 8",
"output": "3 1"
},
{
"input": "5\n1 9 8 2 7",
"output": "2 3"
},
{
"input": "6\n1 4 7 3 2 1",
"output": "3 3"
},
{
"input": "11\n7 9 1 8 7 1 7 1 4 8 1",
"output": "4 7"
},
{
"input": "12\n4 2 7 1 9 5 6 2 7 8 5 7",
"output": "7 5"
},
{
"input": "13\n9 9 9 9 9 9 9 9 9 9 9 9 9",
"output": "7 6"
}
] | 1,590,508,669
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 40
| 560
| 8,499,200
|
n=int(input())
l1=list(map(int,input().split()))
pre=[0]*n
suf=[0]*n
for i in range(n):
if(i==0):
pre[i]=l1[i]
else:
pre[i]=pre[i-1]+l1[i]
for i in range(n-1,-1,-1):
if(i==(n-1)):
suf[i]=l1[i]
else:
suf[i]=suf[i+1]+l1[i]
alice=0
bob=0
for i in range(n):
if(pre[i]<=suf[i]):
alice+=1
else:
bob+=1
print(alice,bob)
|
Title: Alice, Bob and Chocolate
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
Alice and Bob like games. And now they are ready to start a new game. They have placed *n* chocolate bars in a line. Alice starts to eat chocolate bars one by one from left to right, and Bob β from right to left. For each chocololate bar the time, needed for the player to consume it, is known (Alice and Bob eat them with equal speed). When the player consumes a chocolate bar, he immediately starts with another. It is not allowed to eat two chocolate bars at the same time, to leave the bar unfinished and to make pauses. If both players start to eat the same bar simultaneously, Bob leaves it to Alice as a true gentleman.
How many bars each of the players will consume?
Input Specification:
The first line contains one integer *n* (1<=β€<=*n*<=β€<=105) β the amount of bars on the table. The second line contains a sequence *t*1,<=*t*2,<=...,<=*t**n* (1<=β€<=*t**i*<=β€<=1000), where *t**i* is the time (in seconds) needed to consume the *i*-th bar (in the order from left to right).
Output Specification:
Print two numbers *a* and *b*, where *a* is the amount of bars consumed by Alice, and *b* is the amount of bars consumed by Bob.
Demo Input:
['5\n2 9 8 2 7\n']
Demo Output:
['2 3\n']
Note:
none
|
```python
n=int(input())
l1=list(map(int,input().split()))
pre=[0]*n
suf=[0]*n
for i in range(n):
if(i==0):
pre[i]=l1[i]
else:
pre[i]=pre[i-1]+l1[i]
for i in range(n-1,-1,-1):
if(i==(n-1)):
suf[i]=l1[i]
else:
suf[i]=suf[i+1]+l1[i]
alice=0
bob=0
for i in range(n):
if(pre[i]<=suf[i]):
alice+=1
else:
bob+=1
print(alice,bob)
```
| 3.796676
|
59
|
A
|
Word
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] |
A. Word
|
2
|
256
|
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP β with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
|
The first line contains a word *s* β it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
|
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
|
[
"HoUse\n",
"ViP\n",
"maTRIx\n"
] |
[
"house\n",
"VIP\n",
"matrix\n"
] |
none
| 500
|
[
{
"input": "HoUse",
"output": "house"
},
{
"input": "ViP",
"output": "VIP"
},
{
"input": "maTRIx",
"output": "matrix"
},
{
"input": "BNHWpnpawg",
"output": "bnhwpnpawg"
},
{
"input": "VTYGP",
"output": "VTYGP"
},
{
"input": "CHNenu",
"output": "chnenu"
},
{
"input": "ERPZGrodyu",
"output": "erpzgrodyu"
},
{
"input": "KSXBXWpebh",
"output": "KSXBXWPEBH"
},
{
"input": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv",
"output": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv"
},
{
"input": "Amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd",
"output": "amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd"
},
{
"input": "ISAGFJFARYFBLOPQDSHWGMCNKMFTLVFUGNJEWGWNBLXUIATXEkqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv",
"output": "isagfjfaryfblopqdshwgmcnkmftlvfugnjewgwnblxuiatxekqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv"
},
{
"input": "XHRPXZEGHSOCJPICUIXSKFUZUPYTSGJSDIYBCMNMNBPNDBXLXBzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg",
"output": "xhrpxzeghsocjpicuixskfuzupytsgjsdiybcmnmnbpndbxlxbzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg"
},
{
"input": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGAdkcetqjljtmttlonpekcovdzebzdkzggwfsxhapmjkdbuceak",
"output": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGADKCETQJLJTMTTLONPEKCOVDZEBZDKZGGWFSXHAPMJKDBUCEAK"
},
{
"input": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFw",
"output": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFW"
},
{
"input": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB",
"output": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB"
},
{
"input": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge",
"output": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge"
},
{
"input": "Ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw",
"output": "ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw"
},
{
"input": "YQOMLKYAORUQQUCQZCDYMIVDHGWZFFRMUVTAWCHERFPMNRYRIkgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks",
"output": "yqomlkyaoruqqucqzcdymivdhgwzffrmuvtawcherfpmnryrikgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks"
},
{
"input": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJcuusigdqfkumewjtdyitveeiaybwrhomrwmpdipjwiuxfnwuz",
"output": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJCUUSIGDQFKUMEWJTDYITVEEIAYBWRHOMRWMPDIPJWIUXFNWUZ"
},
{
"input": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWuckzcpxosodcjaaakvlxpbiigsiauviilylnnqlyucziihqg",
"output": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWUCKZCPXOSODCJAAAKVLXPBIIGSIAUVIILYLNNQLYUCZIIHQG"
},
{
"input": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO",
"output": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO"
},
{
"input": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDd",
"output": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDD"
},
{
"input": "EBWOVSNLYTWWXrnovgugogtjahnmatomubebyxakas",
"output": "ebwovsnlytwwxrnovgugogtjahnmatomubebyxakas"
},
{
"input": "AORNNDKTRLRVGDPXJKXFTPFpopuzrlqumrxssskvbm",
"output": "AORNNDKTRLRVGDPXJKXFTPFPOPUZRLQUMRXSSSKVBM"
},
{
"input": "DBBZJCWQEVGYHEEUHJAWRHBDPsnplijbwtdibqrrheebfxqcvn",
"output": "dbbzjcwqevgyheeuhjawrhbdpsnplijbwtdibqrrheebfxqcvn"
},
{
"input": "t",
"output": "t"
},
{
"input": "N",
"output": "N"
},
{
"input": "kv",
"output": "kv"
},
{
"input": "Ur",
"output": "ur"
},
{
"input": "CN",
"output": "CN"
}
] | 1,603,242,709
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 218
| 0
|
s = input()
small = 0
big = 0
for i in s:
if i >= 'a' and i <= 'z':
small += 1
else:
big += 1
if small >= big:
s = s.lower()
else:
s = s.upper()
print(s)
|
Title: Word
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP β with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
Input Specification:
The first line contains a word *s* β it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Output Specification:
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
Demo Input:
['HoUse\n', 'ViP\n', 'maTRIx\n']
Demo Output:
['house\n', 'VIP\n', 'matrix\n']
Note:
none
|
```python
s = input()
small = 0
big = 0
for i in s:
if i >= 'a' and i <= 'z':
small += 1
else:
big += 1
if small >= big:
s = s.lower()
else:
s = s.upper()
print(s)
```
| 3.9455
|
938
|
A
|
Word Correction
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Victor tries to write his own text editor, with word correction included. However, the rules of word correction are really strange.
Victor thinks that if a word contains two consecutive vowels, then it's kinda weird and it needs to be replaced. So the word corrector works in such a way: as long as there are two consecutive vowels in the word, it deletes the first vowel in a word such that there is another vowel right before it. If there are no two consecutive vowels in the word, it is considered to be correct.
You are given a word *s*. Can you predict what will it become after correction?
In this problem letters a, e, i, o, u and y are considered to be vowels.
|
The first line contains one integer *n* (1<=β€<=*n*<=β€<=100) β the number of letters in word *s* before the correction.
The second line contains a string *s* consisting of exactly *n* lowercase Latin letters β the word before the correction.
|
Output the word *s* after the correction.
|
[
"5\nweird\n",
"4\nword\n",
"5\naaeaa\n"
] |
[
"werd\n",
"word\n",
"a\n"
] |
Explanations of the examples:
1. There is only one replace: weird <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> werd;1. No replace needed since there are no two consecutive vowels;1. aaeaa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> aeaa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> aaa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> aa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> a.
| 0
|
[
{
"input": "5\nweird",
"output": "werd"
},
{
"input": "4\nword",
"output": "word"
},
{
"input": "5\naaeaa",
"output": "a"
},
{
"input": "100\naaaaabbbbboyoyoyoyoyacadabbbbbiuiufgiuiuaahjabbbklboyoyoyoyoyaaaaabbbbbiuiuiuiuiuaaaaabbbbbeyiyuyzyw",
"output": "abbbbbocadabbbbbifgihjabbbklbobbbbbibbbbbezyw"
},
{
"input": "69\nbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb",
"output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb"
},
{
"input": "12\nmmmmmmmmmmmm",
"output": "mmmmmmmmmmmm"
},
{
"input": "18\nyaywptqwuyiqypwoyw",
"output": "ywptqwuqypwow"
},
{
"input": "85\nbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb",
"output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb"
},
{
"input": "13\nmmmmmmmmmmmmm",
"output": "mmmmmmmmmmmmm"
},
{
"input": "10\nmmmmmmmmmm",
"output": "mmmmmmmmmm"
},
{
"input": "11\nmmmmmmmmmmm",
"output": "mmmmmmmmmmm"
},
{
"input": "15\nmmmmmmmmmmmmmmm",
"output": "mmmmmmmmmmmmmmm"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "14\nmmmmmmmmmmmmmm",
"output": "mmmmmmmmmmmmmm"
},
{
"input": "33\nmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm",
"output": "mmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm"
},
{
"input": "79\nbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb",
"output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb"
},
{
"input": "90\nbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb",
"output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb"
},
{
"input": "2\naa",
"output": "a"
},
{
"input": "18\niuiuqpyyaoaetiwliu",
"output": "iqpytiwli"
},
{
"input": "5\nxxxxx",
"output": "xxxxx"
},
{
"input": "6\nxxxahg",
"output": "xxxahg"
},
{
"input": "3\nzcv",
"output": "zcv"
},
{
"input": "4\naepo",
"output": "apo"
},
{
"input": "5\nqqqqq",
"output": "qqqqq"
},
{
"input": "6\naaaaaa",
"output": "a"
},
{
"input": "4\naeta",
"output": "ata"
},
{
"input": "20\nttyttlwaoieulyiluuri",
"output": "ttyttlwalyluri"
},
{
"input": "1\nb",
"output": "b"
},
{
"input": "3\nanc",
"output": "anc"
},
{
"input": "1\ne",
"output": "e"
},
{
"input": "3\naie",
"output": "a"
},
{
"input": "3\nvio",
"output": "vi"
},
{
"input": "2\nea",
"output": "e"
},
{
"input": "3\nuas",
"output": "us"
},
{
"input": "2\nba",
"output": "ba"
},
{
"input": "2\naq",
"output": "aq"
},
{
"input": "2\nya",
"output": "y"
},
{
"input": "2\nou",
"output": "o"
},
{
"input": "2\nbb",
"output": "bb"
},
{
"input": "7\nayylmao",
"output": "alma"
},
{
"input": "2\nab",
"output": "ab"
},
{
"input": "19\nyuouiyaoiiweqrryqqp",
"output": "yweqrryqqp"
},
{
"input": "25\niqypwqpriiioetiuqqqttouei",
"output": "iqypwqpritiqqqtto"
},
{
"input": "100\naaaaabbbbboyoyoyoyoyacadabbbbbiuiufgiuiuaahjabbbklboyoyoyoyoyaaaaabbbbbiuiuiuiuiuaaaaabbbbbeyiyuyzyz",
"output": "abbbbbocadabbbbbifgihjabbbklbobbbbbibbbbbezyz"
},
{
"input": "17\naccccccccccccccca",
"output": "accccccccccccccca"
},
{
"input": "5\nababa",
"output": "ababa"
},
{
"input": "10\naaaaaaaaaa",
"output": "a"
},
{
"input": "22\naaaaabbbbboyoyoyoyoyac",
"output": "abbbbboc"
},
{
"input": "7\nmahmoud",
"output": "mahmod"
}
] | 1,522,107,484
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 49
| 108
| 7,372,800
|
n = int(input())
s = input()
vowels = ['a', 'e', 'i', 'o', 'u', 'y']
firstOvl = False
result = ""
for each in s:
if firstOvl is False and each in vowels:
result+= each
firstOvl = True
elif firstOvl is True and each not in vowels:
result+= each
firstOvl = False
elif firstOvl is True:
pass
else:
result+= each
firstOvl = False
print(result)
|
Title: Word Correction
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Victor tries to write his own text editor, with word correction included. However, the rules of word correction are really strange.
Victor thinks that if a word contains two consecutive vowels, then it's kinda weird and it needs to be replaced. So the word corrector works in such a way: as long as there are two consecutive vowels in the word, it deletes the first vowel in a word such that there is another vowel right before it. If there are no two consecutive vowels in the word, it is considered to be correct.
You are given a word *s*. Can you predict what will it become after correction?
In this problem letters a, e, i, o, u and y are considered to be vowels.
Input Specification:
The first line contains one integer *n* (1<=β€<=*n*<=β€<=100) β the number of letters in word *s* before the correction.
The second line contains a string *s* consisting of exactly *n* lowercase Latin letters β the word before the correction.
Output Specification:
Output the word *s* after the correction.
Demo Input:
['5\nweird\n', '4\nword\n', '5\naaeaa\n']
Demo Output:
['werd\n', 'word\n', 'a\n']
Note:
Explanations of the examples:
1. There is only one replace: weird <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> werd;1. No replace needed since there are no two consecutive vowels;1. aaeaa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> aeaa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> aaa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> aa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> a.
|
```python
n = int(input())
s = input()
vowels = ['a', 'e', 'i', 'o', 'u', 'y']
firstOvl = False
result = ""
for each in s:
if firstOvl is False and each in vowels:
result+= each
firstOvl = True
elif firstOvl is True and each not in vowels:
result+= each
firstOvl = False
elif firstOvl is True:
pass
else:
result+= each
firstOvl = False
print(result)
```
| 3
|
|
49
|
D
|
Game
|
PROGRAMMING
| 1,800
|
[
"brute force",
"dp",
"implementation"
] |
D. Game
|
2
|
256
|
Vasya and Petya have invented a new game. Vasya takes a stripe consisting of 1<=Γ<=*n* square and paints the squares black and white. After that Petya can start moves β during a move he may choose any two neighboring squares of one color and repaint these two squares any way he wants, perhaps in different colors. Petya can only repaint the squares in white and black colors. Petyaβs aim is to repaint the stripe so that no two neighboring squares were of one color. Help Petya, using the given initial coloring, find the minimum number of moves Petya needs to win.
|
The first line contains number *n* (1<=β€<=*n*<=β€<=1000) which represents the stripeβs length. The second line contains exactly *n* symbols β the lineβs initial coloring. 0 corresponds to a white square, 1 corresponds to a black one.
|
If Petya cannot win with such an initial coloring, print -1. Otherwise print the minimum number of moves Petya needs to win.
|
[
"6\n111010\n",
"5\n10001\n",
"7\n1100010\n",
"5\n00100\n"
] |
[
"1\n",
"1\n",
"2\n",
"2\n"
] |
In the first sample Petya can take squares 1 and 2. He repaints square 1 to black and square 2 to white.
In the second sample Petya can take squares 2 and 3. He repaints square 2 to white and square 3 to black.
| 2,000
|
[
{
"input": "6\n111010",
"output": "1"
},
{
"input": "5\n10001",
"output": "1"
},
{
"input": "7\n1100010",
"output": "2"
},
{
"input": "5\n00100",
"output": "2"
},
{
"input": "3\n101",
"output": "0"
},
{
"input": "6\n111111",
"output": "3"
},
{
"input": "6\n000000",
"output": "3"
},
{
"input": "10\n1000101001",
"output": "3"
},
{
"input": "100\n1001001000011010001101000011100101101110101001110110010001110011011100111000010010011011101000011101",
"output": "49"
},
{
"input": "100\n0000000000100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "49"
},
{
"input": "100\n1111111111111111111111111111111111111111111111111111111111111111111111111111101111111111111111111111",
"output": "49"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "2\n11",
"output": "1"
},
{
"input": "3\n111",
"output": "1"
},
{
"input": "3\n010",
"output": "0"
},
{
"input": "70\n0010011001010100000110011001011111101011010110110101110101111011101010",
"output": "32"
},
{
"input": "149\n11110101110111101111110110001111110101111011111111111111101111110000101101110110111101011111011111111000111011011110111111001011111111111010110111110",
"output": "73"
},
{
"input": "35\n11111111111111111111111111111111110",
"output": "17"
},
{
"input": "10\n1000000000",
"output": "4"
},
{
"input": "337\n0000000000000000000000000000000000000010000000000000000000000000000000010000000000001000000000100100000000000000000000000000000000000000000000000000000000100000000000000010000000000000000000000000000000000000000000000000000000000000000000000010000000000000000000000000000000000000001000100000000000000000000000000000000000100010000000100",
"output": "160"
},
{
"input": "7\n0000000",
"output": "3"
},
{
"input": "7\n0100011",
"output": "2"
},
{
"input": "4\n0000",
"output": "2"
},
{
"input": "18\n110100000000000000",
"output": "8"
},
{
"input": "17\n00100000000000000",
"output": "8"
},
{
"input": "16\n1101010010000000",
"output": "7"
},
{
"input": "15\n110101101111111",
"output": "7"
},
{
"input": "14\n00101111111111",
"output": "6"
},
{
"input": "8\n10011001",
"output": "4"
},
{
"input": "4\n1001",
"output": "2"
},
{
"input": "8\n00000110",
"output": "4"
},
{
"input": "8\n10011001",
"output": "4"
},
{
"input": "12\n101101010010",
"output": "6"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "6\n110011",
"output": "3"
},
{
"input": "23\n10010101010101010101001",
"output": "4"
},
{
"input": "14\n10101001101010",
"output": "2"
},
{
"input": "4\n0110",
"output": "2"
},
{
"input": "4\n1100",
"output": "2"
}
] | 1,685,773,280
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 61
| 124
| 0
|
n = int(input())
s = input()
ans = 0
# s = s.replace("1", "B").replace("0", "W")
for i in range(n):
if (i&1 and s[i] == "0") or (not i&1 and s[i] == "1"):
ans += 1
print(min(ans, n-ans))
|
Title: Game
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya and Petya have invented a new game. Vasya takes a stripe consisting of 1<=Γ<=*n* square and paints the squares black and white. After that Petya can start moves β during a move he may choose any two neighboring squares of one color and repaint these two squares any way he wants, perhaps in different colors. Petya can only repaint the squares in white and black colors. Petyaβs aim is to repaint the stripe so that no two neighboring squares were of one color. Help Petya, using the given initial coloring, find the minimum number of moves Petya needs to win.
Input Specification:
The first line contains number *n* (1<=β€<=*n*<=β€<=1000) which represents the stripeβs length. The second line contains exactly *n* symbols β the lineβs initial coloring. 0 corresponds to a white square, 1 corresponds to a black one.
Output Specification:
If Petya cannot win with such an initial coloring, print -1. Otherwise print the minimum number of moves Petya needs to win.
Demo Input:
['6\n111010\n', '5\n10001\n', '7\n1100010\n', '5\n00100\n']
Demo Output:
['1\n', '1\n', '2\n', '2\n']
Note:
In the first sample Petya can take squares 1 and 2. He repaints square 1 to black and square 2 to white.
In the second sample Petya can take squares 2 and 3. He repaints square 2 to white and square 3 to black.
|
```python
n = int(input())
s = input()
ans = 0
# s = s.replace("1", "B").replace("0", "W")
for i in range(n):
if (i&1 and s[i] == "0") or (not i&1 and s[i] == "1"):
ans += 1
print(min(ans, n-ans))
```
| 3.969
|
237
|
A
|
Free Cash
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
Valera runs a 24/7 fast food cafe. He magically learned that next day *n* people will visit his cafe. For each person we know the arrival time: the *i*-th person comes exactly at *h**i* hours *m**i* minutes. The cafe spends less than a minute to serve each client, but if a client comes in and sees that there is no free cash, than he doesn't want to wait and leaves the cafe immediately.
Valera is very greedy, so he wants to serve all *n* customers next day (and get more profit). However, for that he needs to ensure that at each moment of time the number of working cashes is no less than the number of clients in the cafe.
Help Valera count the minimum number of cashes to work at his cafe next day, so that they can serve all visitors.
|
The first line contains a single integer *n* (1<=β€<=*n*<=β€<=105), that is the number of cafe visitors.
Each of the following *n* lines has two space-separated integers *h**i* and *m**i* (0<=β€<=*h**i*<=β€<=23;Β 0<=β€<=*m**i*<=β€<=59), representing the time when the *i*-th person comes into the cafe.
Note that the time is given in the chronological order. All time is given within one 24-hour period.
|
Print a single integer β the minimum number of cashes, needed to serve all clients next day.
|
[
"4\n8 0\n8 10\n8 10\n8 45\n",
"3\n0 12\n10 11\n22 22\n"
] |
[
"2\n",
"1\n"
] |
In the first sample it is not enough one cash to serve all clients, because two visitors will come into cafe in 8:10. Therefore, if there will be one cash in cafe, then one customer will be served by it, and another one will not wait and will go away.
In the second sample all visitors will come in different times, so it will be enough one cash.
| 500
|
[
{
"input": "4\n8 0\n8 10\n8 10\n8 45",
"output": "2"
},
{
"input": "3\n0 12\n10 11\n22 22",
"output": "1"
},
{
"input": "5\n12 8\n15 27\n15 27\n16 2\n19 52",
"output": "2"
},
{
"input": "7\n5 6\n7 34\n7 34\n7 34\n12 29\n15 19\n20 23",
"output": "3"
},
{
"input": "8\n0 36\n4 7\n4 7\n4 7\n11 46\n12 4\n15 39\n18 6",
"output": "3"
},
{
"input": "20\n4 12\n4 21\n4 27\n4 56\n5 55\n7 56\n11 28\n11 36\n14 58\n15 59\n16 8\n17 12\n17 23\n17 23\n17 23\n17 23\n17 23\n17 23\n20 50\n22 32",
"output": "6"
},
{
"input": "10\n1 30\n1 30\n1 30\n1 30\n1 30\n1 30\n1 30\n1 30\n1 30\n1 30",
"output": "10"
},
{
"input": "50\n0 23\n1 21\n2 8\n2 45\n3 1\n4 19\n4 37\n7 7\n7 40\n8 43\n9 51\n10 13\n11 2\n11 19\n11 30\n12 37\n12 37\n12 37\n12 37\n12 37\n12 37\n12 37\n12 37\n12 54\n13 32\n13 42\n14 29\n14 34\n14 48\n15 0\n15 27\n16 22\n16 31\n17 25\n17 26\n17 33\n18 14\n18 16\n18 20\n19 0\n19 5\n19 56\n20 22\n21 26\n22 0\n22 10\n22 11\n22 36\n23 17\n23 20",
"output": "8"
},
{
"input": "10\n0 39\n1 35\n1 49\n1 51\n5 24\n7 40\n7 56\n16 42\n23 33\n23 49",
"output": "1"
},
{
"input": "15\n0 16\n6 15\n8 2\n8 6\n8 7\n10 1\n10 1\n10 3\n10 12\n13 5\n14 16\n14 16\n14 16\n14 16\n14 16",
"output": "5"
},
{
"input": "2\n0 24\n1 0",
"output": "1"
},
{
"input": "1\n0 0",
"output": "1"
},
{
"input": "1\n1 5",
"output": "1"
},
{
"input": "1\n1 1",
"output": "1"
},
{
"input": "3\n0 0\n0 0\n0 0",
"output": "3"
},
{
"input": "1\n5 0",
"output": "1"
},
{
"input": "5\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "5"
},
{
"input": "1\n10 10",
"output": "1"
},
{
"input": "1\n8 0",
"output": "1"
},
{
"input": "10\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "10"
},
{
"input": "2\n0 0\n0 1",
"output": "1"
},
{
"input": "2\n8 5\n8 5",
"output": "2"
}
] | 1,631,621,324
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 67
| 1,744
| 30,310,400
|
n = int(input())
time = []
for i in range(n):
h, m = [int(i) for i in input().split()]
time.append((h, m))
prev = (-1, -1)
res = 0
count = 0
for i in time:
if i == prev:
count += 1
else:
count = 1
prev = i
res = max(res, count)
print(res)
|
Title: Free Cash
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera runs a 24/7 fast food cafe. He magically learned that next day *n* people will visit his cafe. For each person we know the arrival time: the *i*-th person comes exactly at *h**i* hours *m**i* minutes. The cafe spends less than a minute to serve each client, but if a client comes in and sees that there is no free cash, than he doesn't want to wait and leaves the cafe immediately.
Valera is very greedy, so he wants to serve all *n* customers next day (and get more profit). However, for that he needs to ensure that at each moment of time the number of working cashes is no less than the number of clients in the cafe.
Help Valera count the minimum number of cashes to work at his cafe next day, so that they can serve all visitors.
Input Specification:
The first line contains a single integer *n* (1<=β€<=*n*<=β€<=105), that is the number of cafe visitors.
Each of the following *n* lines has two space-separated integers *h**i* and *m**i* (0<=β€<=*h**i*<=β€<=23;Β 0<=β€<=*m**i*<=β€<=59), representing the time when the *i*-th person comes into the cafe.
Note that the time is given in the chronological order. All time is given within one 24-hour period.
Output Specification:
Print a single integer β the minimum number of cashes, needed to serve all clients next day.
Demo Input:
['4\n8 0\n8 10\n8 10\n8 45\n', '3\n0 12\n10 11\n22 22\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample it is not enough one cash to serve all clients, because two visitors will come into cafe in 8:10. Therefore, if there will be one cash in cafe, then one customer will be served by it, and another one will not wait and will go away.
In the second sample all visitors will come in different times, so it will be enough one cash.
|
```python
n = int(input())
time = []
for i in range(n):
h, m = [int(i) for i in input().split()]
time.append((h, m))
prev = (-1, -1)
res = 0
count = 0
for i in time:
if i == prev:
count += 1
else:
count = 1
prev = i
res = max(res, count)
print(res)
```
| 3
|
|
404
|
A
|
Valera and X
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
Valera is a little boy. Yesterday he got a huge Math hometask at school, so Valera didn't have enough time to properly learn the English alphabet for his English lesson. Unfortunately, the English teacher decided to have a test on alphabet today. At the test Valera got a square piece of squared paper. The length of the side equals *n* squares (*n* is an odd number) and each unit square contains some small letter of the English alphabet.
Valera needs to know if the letters written on the square piece of paper form letter "X". Valera's teacher thinks that the letters on the piece of paper form an "X", if:
- on both diagonals of the square paper all letters are the same; - all other squares of the paper (they are not on the diagonals) contain the same letter that is different from the letters on the diagonals.
Help Valera, write the program that completes the described task for him.
|
The first line contains integer *n* (3<=β€<=*n*<=<<=300; *n* is odd). Each of the next *n* lines contains *n* small English letters β the description of Valera's paper.
|
Print string "YES", if the letters on the paper form letter "X". Otherwise, print string "NO". Print the strings without quotes.
|
[
"5\nxooox\noxoxo\nsoxoo\noxoxo\nxooox\n",
"3\nwsw\nsws\nwsw\n",
"3\nxpx\npxp\nxpe\n"
] |
[
"NO\n",
"YES\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "5\nxooox\noxoxo\nsoxoo\noxoxo\nxooox",
"output": "NO"
},
{
"input": "3\nwsw\nsws\nwsw",
"output": "YES"
},
{
"input": "3\nxpx\npxp\nxpe",
"output": "NO"
},
{
"input": "5\nliiil\nilili\niilii\nilili\nliiil",
"output": "YES"
},
{
"input": "7\nbwccccb\nckcccbj\nccbcbcc\ncccbccc\nccbcbcc\ncbcccbc\nbccccdt",
"output": "NO"
},
{
"input": "13\nsooooooooooos\nosoooooooooso\noosooooooosoo\nooosooooosooo\noooosooosoooo\nooooososooooo\noooooosoooooo\nooooososooooo\noooosooosoooo\nooosooooosooo\noosooooooosoo\nosoooooooooso\nsooooooooooos",
"output": "YES"
},
{
"input": "3\naaa\naaa\naaa",
"output": "NO"
},
{
"input": "3\naca\noec\nzba",
"output": "NO"
},
{
"input": "15\nrxeeeeeeeeeeeer\nereeeeeeeeeeere\needeeeeeeeeeoee\neeereeeeeeeewee\neeeereeeeebeeee\nqeeeereeejedyee\neeeeeerereeeeee\neeeeeeereeeeeee\neeeeeerereeeeze\neeeeereeereeeee\neeeereeeeegeeee\neeereeeeeeereee\neereeeeeeqeeved\ncreeeeeeceeeere\nreeerneeeeeeeer",
"output": "NO"
},
{
"input": "5\nxxxxx\nxxxxx\nxxxxx\nxxxxx\nxxxxx",
"output": "NO"
},
{
"input": "5\nxxxxx\nxxxxx\nxoxxx\nxxxxx\nxxxxx",
"output": "NO"
},
{
"input": "5\noxxxo\nxoxox\nxxxxx\nxoxox\noxxxo",
"output": "NO"
},
{
"input": "5\noxxxo\nxoxox\nxxoox\nxoxox\noxxxo",
"output": "NO"
},
{
"input": "5\noxxxo\nxoxox\nxxaxx\nxoxox\noxxxo",
"output": "NO"
},
{
"input": "5\noxxxo\nxoxox\noxoxx\nxoxox\noxxxo",
"output": "NO"
},
{
"input": "3\nxxx\naxa\nxax",
"output": "NO"
},
{
"input": "3\nxax\naxx\nxax",
"output": "NO"
},
{
"input": "3\nxax\naxa\nxxx",
"output": "NO"
},
{
"input": "3\nxax\nxxa\nxax",
"output": "NO"
},
{
"input": "3\nxax\naaa\nxax",
"output": "NO"
},
{
"input": "3\naax\naxa\nxax",
"output": "NO"
},
{
"input": "3\nxaa\naxa\nxax",
"output": "NO"
},
{
"input": "3\nxax\naxa\naax",
"output": "NO"
},
{
"input": "3\nxax\naxa\nxaa",
"output": "NO"
},
{
"input": "3\nxfx\naxa\nxax",
"output": "NO"
},
{
"input": "3\nxax\nafa\nxax",
"output": "NO"
},
{
"input": "3\nxax\naxa\nxaf",
"output": "NO"
},
{
"input": "3\nxox\nxxx\nxxx",
"output": "NO"
},
{
"input": "3\naxa\naax\nxxa",
"output": "NO"
},
{
"input": "3\nxox\noxx\nxox",
"output": "NO"
},
{
"input": "3\nxox\nooo\nxox",
"output": "NO"
},
{
"input": "3\naaa\naab\nbbb",
"output": "NO"
},
{
"input": "3\nxxx\nsxs\nxsx",
"output": "NO"
},
{
"input": "5\nabbba\nbabab\nbbbbb\nbaaab\nabbba",
"output": "NO"
},
{
"input": "5\nabaaa\nbbbbb\nbbabb\nbabab\nabbba",
"output": "NO"
},
{
"input": "5\nxoxox\noxoxo\nooxoo\noxoxo\nxooox",
"output": "NO"
},
{
"input": "3\nxox\noxx\nxxx",
"output": "NO"
},
{
"input": "5\nxoooo\noxooo\nooxoo\noooxo\noooox",
"output": "NO"
},
{
"input": "5\nxoooo\noxoxx\nooxoo\noxoxo\noxoox",
"output": "NO"
},
{
"input": "3\naaa\nbab\naba",
"output": "NO"
}
] | 1,588,124,961
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 47
| 124
| 1,024,000
|
j = int(input())
m=[]
r2 = []
r3 = []
for i in range(0, j):
m.append(input().strip())
# print(m)
for a in range(0, j):
for b in range(0, j):
if a == b or b== (j - a - 1):
r2.append(m[a][b])
else:
r3.append(m[a][b])
# print(r2, r3)
r2 = list(set(r2))
r3 = list(set(r3))
if len(r2)== 1 and len(r3) ==1 and r2 != r3:
print("YES")
else:
print("NO")
|
Title: Valera and X
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera is a little boy. Yesterday he got a huge Math hometask at school, so Valera didn't have enough time to properly learn the English alphabet for his English lesson. Unfortunately, the English teacher decided to have a test on alphabet today. At the test Valera got a square piece of squared paper. The length of the side equals *n* squares (*n* is an odd number) and each unit square contains some small letter of the English alphabet.
Valera needs to know if the letters written on the square piece of paper form letter "X". Valera's teacher thinks that the letters on the piece of paper form an "X", if:
- on both diagonals of the square paper all letters are the same; - all other squares of the paper (they are not on the diagonals) contain the same letter that is different from the letters on the diagonals.
Help Valera, write the program that completes the described task for him.
Input Specification:
The first line contains integer *n* (3<=β€<=*n*<=<<=300; *n* is odd). Each of the next *n* lines contains *n* small English letters β the description of Valera's paper.
Output Specification:
Print string "YES", if the letters on the paper form letter "X". Otherwise, print string "NO". Print the strings without quotes.
Demo Input:
['5\nxooox\noxoxo\nsoxoo\noxoxo\nxooox\n', '3\nwsw\nsws\nwsw\n', '3\nxpx\npxp\nxpe\n']
Demo Output:
['NO\n', 'YES\n', 'NO\n']
Note:
none
|
```python
j = int(input())
m=[]
r2 = []
r3 = []
for i in range(0, j):
m.append(input().strip())
# print(m)
for a in range(0, j):
for b in range(0, j):
if a == b or b== (j - a - 1):
r2.append(m[a][b])
else:
r3.append(m[a][b])
# print(r2, r3)
r2 = list(set(r2))
r3 = list(set(r3))
if len(r2)== 1 and len(r3) ==1 and r2 != r3:
print("YES")
else:
print("NO")
```
| 3
|
|
155
|
A
|
I_love_\%username\%
|
PROGRAMMING
| 800
|
[
"brute force"
] | null | null |
Vasya adores sport programming. He can't write programs but he loves to watch the contests' progress. Vasya even has a favorite coder and Vasya pays special attention to him.
One day Vasya decided to collect the results of all contests where his favorite coder participated and track the progress of his coolness. For each contest where this coder participated, he wrote out a single non-negative number β the number of points his favorite coder earned in the contest. Vasya wrote out the points for the contest in the order, in which the contests run (naturally, no two contests ran simultaneously).
Vasya considers a coder's performance in a contest amazing in two situations: he can break either his best or his worst performance record. First, it is amazing if during the contest the coder earns strictly more points that he earned on each past contest. Second, it is amazing if during the contest the coder earns strictly less points that he earned on each past contest. A coder's first contest isn't considered amazing. Now he wants to count the number of amazing performances the coder had throughout his whole history of participating in contests. But the list of earned points turned out long and Vasya can't code... That's why he asks you to help him.
|
The first line contains the single integer *n* (1<=β€<=*n*<=β€<=1000) β the number of contests where the coder participated.
The next line contains *n* space-separated non-negative integer numbers β they are the points which the coder has earned. The points are given in the chronological order. All points do not exceed 10000.
|
Print the single number β the number of amazing performances the coder has had during his whole history of participating in the contests.
|
[
"5\n100 50 200 150 200\n",
"10\n4664 6496 5814 7010 5762 5736 6944 4850 3698 7242\n"
] |
[
"2\n",
"4\n"
] |
In the first sample the performances number 2 and 3 are amazing.
In the second sample the performances number 2, 4, 9 and 10 are amazing.
| 500
|
[
{
"input": "5\n100 50 200 150 200",
"output": "2"
},
{
"input": "10\n4664 6496 5814 7010 5762 5736 6944 4850 3698 7242",
"output": "4"
},
{
"input": "1\n6",
"output": "0"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "5\n100 36 53 7 81",
"output": "2"
},
{
"input": "5\n7 36 53 81 100",
"output": "4"
},
{
"input": "5\n100 81 53 36 7",
"output": "4"
},
{
"input": "10\n8 6 3 4 9 10 7 7 1 3",
"output": "5"
},
{
"input": "10\n1627 1675 1488 1390 1812 1137 1746 1324 1952 1862",
"output": "6"
},
{
"input": "10\n1 3 3 4 6 7 7 8 9 10",
"output": "7"
},
{
"input": "10\n1952 1862 1812 1746 1675 1627 1488 1390 1324 1137",
"output": "9"
},
{
"input": "25\n1448 4549 2310 2725 2091 3509 1565 2475 2232 3989 4231 779 2967 2702 608 3739 721 1552 2767 530 3114 665 1940 48 4198",
"output": "5"
},
{
"input": "33\n1097 1132 1091 1104 1049 1038 1023 1080 1104 1029 1035 1061 1049 1060 1088 1106 1105 1087 1063 1076 1054 1103 1047 1041 1028 1120 1126 1063 1117 1110 1044 1093 1101",
"output": "5"
},
{
"input": "34\n821 5536 2491 6074 7216 9885 764 1603 778 8736 8987 771 617 1587 8943 7922 439 7367 4115 8886 7878 6899 8811 5752 3184 3401 9760 9400 8995 4681 1323 6637 6554 6498",
"output": "7"
},
{
"input": "68\n6764 6877 6950 6768 6839 6755 6726 6778 6699 6805 6777 6985 6821 6801 6791 6805 6940 6761 6677 6999 6911 6699 6959 6933 6903 6843 6972 6717 6997 6756 6789 6668 6735 6852 6735 6880 6723 6834 6810 6694 6780 6679 6698 6857 6826 6896 6979 6968 6957 6988 6960 6700 6919 6892 6984 6685 6813 6678 6715 6857 6976 6902 6780 6686 6777 6686 6842 6679",
"output": "9"
},
{
"input": "60\n9000 9014 9034 9081 9131 9162 9174 9199 9202 9220 9221 9223 9229 9235 9251 9260 9268 9269 9270 9298 9307 9309 9313 9323 9386 9399 9407 9495 9497 9529 9531 9544 9614 9615 9627 9627 9643 9654 9656 9657 9685 9699 9701 9736 9745 9758 9799 9827 9843 9845 9854 9854 9885 9891 9896 9913 9942 9963 9986 9992",
"output": "57"
},
{
"input": "100\n7 61 12 52 41 16 34 99 30 44 48 89 31 54 21 1 48 52 61 15 35 87 21 76 64 92 44 81 16 93 84 92 32 15 68 76 53 39 26 4 11 26 7 4 99 99 61 65 55 85 65 67 47 39 2 74 63 49 98 87 5 94 22 30 25 42 31 84 49 23 89 60 16 26 92 27 9 57 75 61 94 35 83 47 99 100 63 24 91 88 79 10 15 45 22 64 3 11 89 83",
"output": "4"
},
{
"input": "100\n9999 9999 9999 9998 9998 9998 9997 9996 9996 9995 9993 9993 9991 9990 9989 9986 9984 9984 9983 9981 9981 9980 9980 9980 9979 9977 9977 9977 9977 9977 9976 9976 9975 9975 9973 9972 9972 9972 9972 9971 9969 9969 9969 9968 9967 9965 9962 9962 9962 9962 9960 9958 9958 9955 9954 9954 9954 9954 9950 9949 9949 9947 9943 9943 9940 9940 9939 9937 9937 9937 9936 9935 9932 9932 9931 9929 9927 9927 9926 9923 9919 9919 9918 9918 9918 9917 9917 9916 9916 9915 9914 9911 9910 9907 9907 9906 9905 9905 9904 9901",
"output": "57"
},
{
"input": "100\n158 159 157 156 155 154 160 153 161 152 162 163 151 164 150 165 149 148 147 166 146 167 145 144 143 142 168 141 169 170 140 139 138 137 171 136 135 134 172 173 174 133 175 132 177 178 131 179 180 130 129 181 128 127 182 126 183 125 124 184 123 122 121 120 119 118 117 185 186 187 188 189 116 190 115 191 192 193 114 113 112 111 110 109 108 194 195 107 106 105 196 197 104 198 199 103 102 200 101 100",
"output": "99"
},
{
"input": "2\n0 10000",
"output": "1"
},
{
"input": "2\n5 5",
"output": "0"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "2\n10 10",
"output": "0"
},
{
"input": "1\n0",
"output": "0"
}
] | 1,684,015,489
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 37
| 92
| 0
|
n = int(input())
l = input()
performances = []
for i in l.split():
performances.append(int(i))
count = 0
max = performances[0]
min = performances[0]
performances.remove(performances[0])
for performa in performances:
if performa > max:
count += 1
max = performa
elif performa < min:
count += 1
min = performa
print(count)
|
Title: I_love_\%username\%
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya adores sport programming. He can't write programs but he loves to watch the contests' progress. Vasya even has a favorite coder and Vasya pays special attention to him.
One day Vasya decided to collect the results of all contests where his favorite coder participated and track the progress of his coolness. For each contest where this coder participated, he wrote out a single non-negative number β the number of points his favorite coder earned in the contest. Vasya wrote out the points for the contest in the order, in which the contests run (naturally, no two contests ran simultaneously).
Vasya considers a coder's performance in a contest amazing in two situations: he can break either his best or his worst performance record. First, it is amazing if during the contest the coder earns strictly more points that he earned on each past contest. Second, it is amazing if during the contest the coder earns strictly less points that he earned on each past contest. A coder's first contest isn't considered amazing. Now he wants to count the number of amazing performances the coder had throughout his whole history of participating in contests. But the list of earned points turned out long and Vasya can't code... That's why he asks you to help him.
Input Specification:
The first line contains the single integer *n* (1<=β€<=*n*<=β€<=1000) β the number of contests where the coder participated.
The next line contains *n* space-separated non-negative integer numbers β they are the points which the coder has earned. The points are given in the chronological order. All points do not exceed 10000.
Output Specification:
Print the single number β the number of amazing performances the coder has had during his whole history of participating in the contests.
Demo Input:
['5\n100 50 200 150 200\n', '10\n4664 6496 5814 7010 5762 5736 6944 4850 3698 7242\n']
Demo Output:
['2\n', '4\n']
Note:
In the first sample the performances number 2 and 3 are amazing.
In the second sample the performances number 2, 4, 9 and 10 are amazing.
|
```python
n = int(input())
l = input()
performances = []
for i in l.split():
performances.append(int(i))
count = 0
max = performances[0]
min = performances[0]
performances.remove(performances[0])
for performa in performances:
if performa > max:
count += 1
max = performa
elif performa < min:
count += 1
min = performa
print(count)
```
| 3
|
|
546
|
A
|
Soldier and Bananas
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation",
"math"
] | null | null |
A soldier wants to buy *w* bananas in the shop. He has to pay *k* dollars for the first banana, 2*k* dollars for the second one and so on (in other words, he has to pay *i*Β·*k* dollars for the *i*-th banana).
He has *n* dollars. How many dollars does he have to borrow from his friend soldier to buy *w* bananas?
|
The first line contains three positive integers *k*,<=*n*,<=*w* (1<=<=β€<=<=*k*,<=*w*<=<=β€<=<=1000, 0<=β€<=*n*<=β€<=109), the cost of the first banana, initial number of dollars the soldier has and number of bananas he wants.
|
Output one integer β the amount of dollars that the soldier must borrow from his friend. If he doesn't have to borrow money, output 0.
|
[
"3 17 4\n"
] |
[
"13"
] |
none
| 500
|
[
{
"input": "3 17 4",
"output": "13"
},
{
"input": "1 2 1",
"output": "0"
},
{
"input": "1 1 1",
"output": "0"
},
{
"input": "1 5 6",
"output": "16"
},
{
"input": "1 1000000000 1",
"output": "0"
},
{
"input": "1000 0 1000",
"output": "500500000"
},
{
"input": "859 453892 543",
"output": "126416972"
},
{
"input": "1000 1000000000 1000",
"output": "0"
},
{
"input": "1000 500500000 1000",
"output": "0"
},
{
"input": "1000 500500001 1000",
"output": "0"
},
{
"input": "1000 500499999 1000",
"output": "1"
},
{
"input": "634 87973 214",
"output": "14497197"
},
{
"input": "432 10000 241",
"output": "12587552"
},
{
"input": "111 111111111 111",
"output": "0"
},
{
"input": "20 43 3",
"output": "77"
}
] | 1,698,077,131
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 15
| 46
| 0
|
k, n, w= map(int, input().split())
#count all price
price = 0
for i in range(1, w +1):
price += i * k
#borrow
borrow = max(0, price - n )
print(borrow)
|
Title: Soldier and Bananas
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A soldier wants to buy *w* bananas in the shop. He has to pay *k* dollars for the first banana, 2*k* dollars for the second one and so on (in other words, he has to pay *i*Β·*k* dollars for the *i*-th banana).
He has *n* dollars. How many dollars does he have to borrow from his friend soldier to buy *w* bananas?
Input Specification:
The first line contains three positive integers *k*,<=*n*,<=*w* (1<=<=β€<=<=*k*,<=*w*<=<=β€<=<=1000, 0<=β€<=*n*<=β€<=109), the cost of the first banana, initial number of dollars the soldier has and number of bananas he wants.
Output Specification:
Output one integer β the amount of dollars that the soldier must borrow from his friend. If he doesn't have to borrow money, output 0.
Demo Input:
['3 17 4\n']
Demo Output:
['13']
Note:
none
|
```python
k, n, w= map(int, input().split())
#count all price
price = 0
for i in range(1, w +1):
price += i * k
#borrow
borrow = max(0, price - n )
print(borrow)
```
| 3
|
|
218
|
A
|
Mountain Scenery
|
PROGRAMMING
| 1,100
|
[
"brute force",
"constructive algorithms",
"implementation"
] | null | null |
Little Bolek has found a picture with *n* mountain peaks painted on it. The *n* painted peaks are represented by a non-closed polyline, consisting of 2*n* segments. The segments go through 2*n*<=+<=1 points with coordinates (1,<=*y*1), (2,<=*y*2), ..., (2*n*<=+<=1,<=*y*2*n*<=+<=1), with the *i*-th segment connecting the point (*i*,<=*y**i*) and the point (*i*<=+<=1,<=*y**i*<=+<=1). For any even *i* (2<=β€<=*i*<=β€<=2*n*) the following condition holds: *y**i*<=-<=1<=<<=*y**i* and *y**i*<=><=*y**i*<=+<=1.
We shall call a vertex of a polyline with an even *x* coordinate a mountain peak.
Bolek fancied a little mischief. He chose exactly *k* mountain peaks, rubbed out the segments that went through those peaks and increased each peak's height by one (that is, he increased the *y* coordinate of the corresponding points). Then he painted the missing segments to get a new picture of mountain peaks. Let us denote the points through which the new polyline passes on Bolek's new picture as (1,<=*r*1), (2,<=*r*2), ..., (2*n*<=+<=1,<=*r*2*n*<=+<=1).
Given Bolek's final picture, restore the initial one.
|
The first line contains two space-separated integers *n* and *k* (1<=β€<=*k*<=β€<=*n*<=β€<=100). The next line contains 2*n*<=+<=1 space-separated integers *r*1,<=*r*2,<=...,<=*r*2*n*<=+<=1 (0<=β€<=*r**i*<=β€<=100) β the *y* coordinates of the polyline vertices on Bolek's picture.
It is guaranteed that we can obtain the given picture after performing the described actions on some picture of mountain peaks.
|
Print 2*n*<=+<=1 integers *y*1,<=*y*2,<=...,<=*y*2*n*<=+<=1 β the *y* coordinates of the vertices of the polyline on the initial picture. If there are multiple answers, output any one of them.
|
[
"3 2\n0 5 3 5 1 5 2\n",
"1 1\n0 2 0\n"
] |
[
"0 5 3 4 1 4 2 \n",
"0 1 0 \n"
] |
none
| 500
|
[
{
"input": "3 2\n0 5 3 5 1 5 2",
"output": "0 5 3 4 1 4 2 "
},
{
"input": "1 1\n0 2 0",
"output": "0 1 0 "
},
{
"input": "1 1\n1 100 0",
"output": "1 99 0 "
},
{
"input": "3 1\n0 1 0 1 0 2 0",
"output": "0 1 0 1 0 1 0 "
},
{
"input": "3 1\n0 1 0 2 0 1 0",
"output": "0 1 0 1 0 1 0 "
},
{
"input": "3 3\n0 100 35 67 40 60 3",
"output": "0 99 35 66 40 59 3 "
},
{
"input": "7 3\n1 2 1 3 1 2 1 2 1 3 1 3 1 2 1",
"output": "1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 "
},
{
"input": "100 100\n1 3 1 3 1 3 0 2 0 3 1 3 1 3 1 3 0 3 1 3 0 2 0 2 0 3 0 2 0 2 0 3 1 3 1 3 1 3 1 3 0 2 0 3 1 3 0 2 0 2 0 2 0 2 0 2 0 3 0 3 0 3 0 3 0 2 0 3 1 3 1 3 1 3 0 3 0 2 0 2 0 2 0 2 0 3 0 3 1 3 0 3 1 3 1 3 0 3 1 3 0 3 1 3 1 3 0 3 1 3 0 3 1 3 0 2 0 3 1 3 0 3 1 3 0 2 0 3 1 3 0 3 0 2 0 3 1 3 0 3 0 3 0 2 0 2 0 2 0 3 0 3 1 3 1 3 0 3 1 3 1 3 1 3 0 2 0 3 0 2 0 3 1 3 0 3 0 3 1 3 0 2 0 3 0 2 0 2 0 2 0 2 0 3 1 3 0 3 1 3 1",
"output": "1 2 1 2 1 2 0 1 0 2 1 2 1 2 1 2 0 2 1 2 0 1 0 1 0 2 0 1 0 1 0 2 1 2 1 2 1 2 1 2 0 1 0 2 1 2 0 1 0 1 0 1 0 1 0 1 0 2 0 2 0 2 0 2 0 1 0 2 1 2 1 2 1 2 0 2 0 1 0 1 0 1 0 1 0 2 0 2 1 2 0 2 1 2 1 2 0 2 1 2 0 2 1 2 1 2 0 2 1 2 0 2 1 2 0 1 0 2 1 2 0 2 1 2 0 1 0 2 1 2 0 2 0 1 0 2 1 2 0 2 0 2 0 1 0 1 0 1 0 2 0 2 1 2 1 2 0 2 1 2 1 2 1 2 0 1 0 2 0 1 0 2 1 2 0 2 0 2 1 2 0 1 0 2 0 1 0 1 0 1 0 1 0 2 1 2 0 2 1 2 1 "
},
{
"input": "30 20\n1 3 1 3 0 2 0 4 1 3 0 3 1 3 1 4 2 3 1 2 0 4 2 4 0 4 1 3 0 4 1 4 2 4 2 4 0 3 1 2 1 4 0 3 0 4 1 3 1 4 1 3 0 1 0 4 0 3 2 3 1",
"output": "1 3 1 3 0 2 0 4 1 2 0 2 1 2 1 3 2 3 1 2 0 3 2 3 0 3 1 2 0 3 1 3 2 3 2 3 0 2 1 2 1 3 0 2 0 3 1 2 1 3 1 2 0 1 0 3 0 3 2 3 1 "
},
{
"input": "10 6\n0 5 2 4 1 5 2 5 2 4 2 5 3 5 0 2 0 1 0 1 0",
"output": "0 5 2 4 1 4 2 4 2 3 2 4 3 4 0 1 0 1 0 1 0 "
},
{
"input": "11 6\n3 5 1 4 3 5 0 2 0 2 0 4 0 3 0 4 1 5 2 4 0 4 0",
"output": "3 5 1 4 3 5 0 2 0 2 0 3 0 2 0 3 1 4 2 3 0 3 0 "
},
{
"input": "12 6\n1 2 1 5 0 2 0 4 1 3 1 4 2 4 0 4 0 4 2 4 0 4 0 5 3",
"output": "1 2 1 5 0 2 0 4 1 3 1 4 2 3 0 3 0 3 2 3 0 3 0 4 3 "
},
{
"input": "13 6\n3 5 2 5 0 3 0 1 0 2 0 1 0 1 0 2 1 4 3 5 1 3 1 3 2 3 1",
"output": "3 4 2 4 0 2 0 1 0 1 0 1 0 1 0 2 1 4 3 4 1 2 1 3 2 3 1 "
},
{
"input": "24 7\n3 4 2 4 1 4 3 4 3 5 1 3 1 3 0 3 0 3 1 4 0 3 0 1 0 1 0 3 2 3 2 3 1 2 1 3 2 5 1 3 0 1 0 2 0 3 1 3 1",
"output": "3 4 2 4 1 4 3 4 3 5 1 3 1 3 0 3 0 3 1 3 0 2 0 1 0 1 0 3 2 3 2 3 1 2 1 3 2 4 1 2 0 1 0 1 0 2 1 2 1 "
},
{
"input": "25 8\n3 5 2 4 2 4 0 1 0 1 0 1 0 2 1 5 2 4 2 4 2 3 1 2 0 1 0 2 0 3 2 5 3 5 0 4 2 3 2 4 1 4 0 4 1 4 0 1 0 4 2",
"output": "3 5 2 4 2 4 0 1 0 1 0 1 0 2 1 5 2 4 2 4 2 3 1 2 0 1 0 2 0 3 2 4 3 4 0 3 2 3 2 3 1 3 0 3 1 3 0 1 0 3 2 "
},
{
"input": "26 9\n3 4 2 3 1 3 1 3 2 4 0 1 0 2 1 3 1 3 0 5 1 4 3 5 0 5 2 3 0 3 1 4 1 3 1 4 2 3 1 4 3 4 1 3 2 4 1 3 2 5 1 2 0",
"output": "3 4 2 3 1 3 1 3 2 4 0 1 0 2 1 3 1 3 0 4 1 4 3 4 0 4 2 3 0 2 1 3 1 2 1 3 2 3 1 4 3 4 1 3 2 3 1 3 2 4 1 2 0 "
},
{
"input": "27 10\n3 5 3 5 3 4 1 3 1 3 1 3 2 3 2 3 2 4 2 3 0 4 2 5 3 4 3 4 1 5 3 4 1 2 1 5 0 3 0 5 0 5 3 4 0 1 0 2 0 2 1 4 0 2 1",
"output": "3 5 3 5 3 4 1 3 1 3 1 3 2 3 2 3 2 3 2 3 0 3 2 4 3 4 3 4 1 4 3 4 1 2 1 4 0 2 0 4 0 4 3 4 0 1 0 1 0 2 1 3 0 2 1 "
},
{
"input": "40 1\n0 2 1 2 0 2 1 2 1 2 1 2 1 2 1 3 0 1 0 1 0 1 0 2 0 2 1 2 0 2 1 2 1 2 1 2 1 2 0 2 1 2 1 2 0 1 0 2 0 2 0 1 0 1 0 1 0 1 0 1 0 2 0 2 0 2 0 1 0 2 0 1 0 2 0 1 0 2 1 2 0",
"output": "0 2 1 2 0 2 1 2 1 2 1 2 1 2 1 3 0 1 0 1 0 1 0 2 0 2 1 2 0 2 1 2 1 2 1 2 1 2 0 2 1 2 1 2 0 1 0 2 0 2 0 1 0 1 0 1 0 1 0 1 0 2 0 2 0 2 0 1 0 2 0 1 0 1 0 1 0 2 1 2 0 "
},
{
"input": "40 2\n0 3 1 2 1 2 0 1 0 2 1 3 0 2 0 3 0 3 0 1 0 2 0 3 1 2 0 2 1 2 0 2 0 1 0 1 0 2 0 2 1 3 0 2 0 1 0 1 0 1 0 3 1 3 1 2 1 2 0 3 0 1 0 3 0 2 1 2 0 1 0 2 0 3 1 2 1 3 1 3 0",
"output": "0 3 1 2 1 2 0 1 0 2 1 3 0 2 0 3 0 3 0 1 0 2 0 3 1 2 0 2 1 2 0 2 0 1 0 1 0 2 0 2 1 3 0 2 0 1 0 1 0 1 0 3 1 3 1 2 1 2 0 3 0 1 0 3 0 2 1 2 0 1 0 2 0 3 1 2 1 2 1 2 0 "
},
{
"input": "40 3\n1 3 1 2 0 4 1 2 0 1 0 1 0 3 0 3 2 3 0 3 1 3 0 4 1 3 2 3 0 2 1 3 0 2 0 1 0 3 1 3 2 3 2 3 0 1 0 2 0 1 0 1 0 3 1 3 0 3 1 3 1 2 0 1 0 3 0 2 0 3 0 1 0 2 0 3 1 2 0 3 0",
"output": "1 3 1 2 0 4 1 2 0 1 0 1 0 3 0 3 2 3 0 3 1 3 0 4 1 3 2 3 0 2 1 3 0 2 0 1 0 3 1 3 2 3 2 3 0 1 0 2 0 1 0 1 0 3 1 3 0 3 1 3 1 2 0 1 0 3 0 2 0 3 0 1 0 1 0 2 1 2 0 2 0 "
},
{
"input": "50 40\n1 4 2 4 1 2 1 4 1 4 2 3 1 2 1 4 1 3 0 2 1 4 0 1 0 3 1 3 1 3 0 4 2 4 2 4 2 4 2 4 2 4 2 4 0 4 1 3 1 3 0 4 1 4 2 3 2 3 0 3 0 3 0 4 1 4 1 3 1 4 1 3 0 4 0 3 0 2 0 2 0 4 1 4 0 2 0 4 1 4 0 3 0 2 1 3 0 2 0 4 0",
"output": "1 4 2 4 1 2 1 3 1 3 2 3 1 2 1 3 1 2 0 2 1 3 0 1 0 2 1 2 1 2 0 3 2 3 2 3 2 3 2 3 2 3 2 3 0 3 1 2 1 2 0 3 1 3 2 3 2 3 0 2 0 2 0 3 1 3 1 2 1 3 1 2 0 3 0 2 0 1 0 1 0 3 1 3 0 1 0 3 1 3 0 2 0 2 1 2 0 1 0 3 0 "
},
{
"input": "100 2\n1 3 1 2 1 3 2 3 1 3 1 3 1 3 1 2 0 3 0 2 0 3 2 3 0 3 1 2 1 2 0 3 0 1 0 1 0 3 2 3 1 2 0 1 0 2 0 1 0 2 1 3 1 2 1 3 2 3 1 3 1 2 0 3 2 3 0 2 1 3 1 2 0 3 2 3 1 3 2 3 0 4 0 3 0 1 0 3 0 1 0 1 0 2 0 2 1 3 1 2 1 2 0 2 0 1 0 2 0 2 1 3 1 3 2 3 0 2 1 2 0 3 0 1 0 2 0 3 2 3 1 3 0 3 1 2 0 1 0 3 0 1 0 1 0 1 0 2 0 1 0 2 1 2 1 2 1 3 0 1 0 2 1 3 0 2 1 3 0 2 1 2 0 3 1 3 1 3 0 2 1 2 1 3 0 2 1 3 2 3 1 2 0 3 1 2 0 3 1 2 0",
"output": "1 3 1 2 1 3 2 3 1 3 1 3 1 3 1 2 0 3 0 2 0 3 2 3 0 3 1 2 1 2 0 3 0 1 0 1 0 3 2 3 1 2 0 1 0 2 0 1 0 2 1 3 1 2 1 3 2 3 1 3 1 2 0 3 2 3 0 2 1 3 1 2 0 3 2 3 1 3 2 3 0 4 0 3 0 1 0 3 0 1 0 1 0 2 0 2 1 3 1 2 1 2 0 2 0 1 0 2 0 2 1 3 1 3 2 3 0 2 1 2 0 3 0 1 0 2 0 3 2 3 1 3 0 3 1 2 0 1 0 3 0 1 0 1 0 1 0 2 0 1 0 2 1 2 1 2 1 3 0 1 0 2 1 3 0 2 1 3 0 2 1 2 0 3 1 3 1 3 0 2 1 2 1 3 0 2 1 3 2 3 1 2 0 2 1 2 0 2 1 2 0 "
},
{
"input": "100 3\n0 2 1 2 0 1 0 1 0 3 0 2 1 3 1 3 2 3 0 2 0 1 0 2 0 1 0 3 2 3 2 3 1 2 1 3 1 2 1 3 2 3 2 3 0 3 2 3 2 3 2 3 0 2 0 3 0 3 2 3 2 3 2 3 2 3 0 3 0 1 0 2 1 3 0 2 1 2 0 3 2 3 2 3 1 3 0 3 1 3 0 3 0 1 0 1 0 2 0 2 1 2 0 3 1 3 0 3 2 3 2 3 2 3 2 3 0 1 0 1 0 1 0 2 1 2 0 2 1 3 2 3 0 1 0 1 0 1 0 1 0 2 0 1 0 3 1 2 1 2 1 3 1 2 0 3 0 2 1 2 1 3 2 3 1 3 2 3 0 1 0 1 0 1 0 1 0 3 0 1 0 2 1 2 0 3 1 3 2 3 0 3 1 2 1 3 1 3 1 3 0",
"output": "0 2 1 2 0 1 0 1 0 3 0 2 1 3 1 3 2 3 0 2 0 1 0 2 0 1 0 3 2 3 2 3 1 2 1 3 1 2 1 3 2 3 2 3 0 3 2 3 2 3 2 3 0 2 0 3 0 3 2 3 2 3 2 3 2 3 0 3 0 1 0 2 1 3 0 2 1 2 0 3 2 3 2 3 1 3 0 3 1 3 0 3 0 1 0 1 0 2 0 2 1 2 0 3 1 3 0 3 2 3 2 3 2 3 2 3 0 1 0 1 0 1 0 2 1 2 0 2 1 3 2 3 0 1 0 1 0 1 0 1 0 2 0 1 0 3 1 2 1 2 1 3 1 2 0 3 0 2 1 2 1 3 2 3 1 3 2 3 0 1 0 1 0 1 0 1 0 3 0 1 0 2 1 2 0 3 1 3 2 3 0 3 1 2 1 2 1 2 1 2 0 "
},
{
"input": "100 20\n0 1 0 3 0 3 2 3 2 4 0 2 0 3 1 3 0 2 0 2 0 3 0 1 0 3 2 4 0 1 0 2 0 2 1 2 1 4 2 4 1 2 0 1 0 2 1 3 0 2 1 3 2 3 1 2 0 2 1 4 0 3 0 2 0 1 0 1 0 1 0 2 1 3 2 3 2 3 2 3 0 1 0 1 0 4 2 3 2 3 0 3 1 2 0 2 0 2 1 3 2 3 1 4 0 1 0 2 1 2 0 2 0 3 2 3 0 2 0 2 1 4 2 3 1 3 0 3 0 2 0 2 1 2 1 3 0 3 1 2 1 3 1 3 1 2 1 2 0 2 1 3 0 2 0 3 0 1 0 3 0 3 0 1 0 4 1 3 0 1 0 1 0 2 1 2 0 2 1 4 1 3 0 2 1 3 1 3 1 3 0 3 0 2 0 1 0 2 1 2 1",
"output": "0 1 0 3 0 3 2 3 2 4 0 2 0 3 1 3 0 2 0 2 0 3 0 1 0 3 2 4 0 1 0 2 0 2 1 2 1 4 2 4 1 2 0 1 0 2 1 3 0 2 1 3 2 3 1 2 0 2 1 4 0 3 0 2 0 1 0 1 0 1 0 2 1 3 2 3 2 3 2 3 0 1 0 1 0 4 2 3 2 3 0 3 1 2 0 2 0 2 1 3 2 3 1 4 0 1 0 2 1 2 0 2 0 3 2 3 0 2 0 2 1 4 2 3 1 3 0 2 0 1 0 2 1 2 1 2 0 2 1 2 1 2 1 2 1 2 1 2 0 2 1 2 0 1 0 2 0 1 0 2 0 2 0 1 0 3 1 2 0 1 0 1 0 2 1 2 0 2 1 3 1 2 0 2 1 2 1 2 1 2 0 2 0 1 0 1 0 2 1 2 1 "
},
{
"input": "100 20\n2 3 0 4 0 1 0 6 3 4 3 6 4 6 0 9 0 6 2 7 3 8 7 10 2 9 3 9 5 6 5 10 3 7 1 5 2 8 3 7 2 3 1 6 0 8 3 8 0 4 1 8 3 7 1 9 5 9 5 8 7 8 5 6 5 8 1 9 8 9 8 10 7 10 5 8 6 10 2 6 3 9 2 6 3 10 5 9 3 10 1 3 2 11 8 9 8 10 1 8 7 11 0 9 5 8 4 5 0 7 3 7 5 9 5 10 1 7 1 9 1 6 3 8 2 4 1 4 2 6 0 4 2 4 2 7 6 9 0 1 0 4 0 4 0 9 2 7 6 7 2 8 0 8 2 7 5 10 1 2 0 2 0 4 3 5 4 7 0 10 2 10 3 6 3 7 1 4 0 9 1 4 3 8 1 10 1 10 0 3 2 5 3 9 0 7 4 5 0 1 0",
"output": "2 3 0 4 0 1 0 6 3 4 3 6 4 6 0 9 0 6 2 7 3 8 7 10 2 9 3 9 5 6 5 10 3 7 1 5 2 8 3 7 2 3 1 6 0 8 3 8 0 4 1 8 3 7 1 9 5 9 5 8 7 8 5 6 5 8 1 9 8 9 8 10 7 10 5 8 6 10 2 6 3 9 2 6 3 10 5 9 3 10 1 3 2 11 8 9 8 10 1 8 7 11 0 9 5 8 4 5 0 7 3 7 5 9 5 10 1 7 1 9 1 6 3 8 2 4 1 4 2 6 0 4 2 4 2 7 6 9 0 1 0 4 0 3 0 8 2 7 6 7 2 7 0 7 2 6 5 9 1 2 0 1 0 4 3 5 4 6 0 9 2 9 3 5 3 6 1 3 0 8 1 4 3 7 1 9 1 9 0 3 2 4 3 8 0 6 4 5 0 1 0 "
},
{
"input": "98 3\n1 2 1 2 0 2 0 2 1 2 0 1 0 2 1 2 0 2 1 2 1 2 0 1 0 2 1 2 1 2 0 2 1 2 0 2 0 2 0 1 0 1 0 1 0 2 1 3 1 2 1 2 1 2 1 2 1 2 1 2 0 2 0 2 1 2 1 2 0 2 1 2 0 1 0 1 0 1 0 1 0 2 0 1 0 2 0 2 1 2 1 2 1 2 0 1 0 1 0 1 0 2 1 2 0 2 1 2 0 2 0 1 0 2 1 2 0 1 0 2 1 2 1 2 1 2 0 2 1 2 1 2 1 2 0 2 1 2 1 2 0 1 0 2 0 2 0 1 0 2 0 2 0 1 0 1 0 1 0 2 0 2 1 2 0 1 0 2 0 2 0 1 0 2 1 2 1 2 1 2 0 2 1 2 1 2 1 2 0 1 0 1 0 2 0 2 0",
"output": "1 2 1 2 0 2 0 2 1 2 0 1 0 2 1 2 0 2 1 2 1 2 0 1 0 2 1 2 1 2 0 2 1 2 0 2 0 2 0 1 0 1 0 1 0 2 1 3 1 2 1 2 1 2 1 2 1 2 1 2 0 2 0 2 1 2 1 2 0 2 1 2 0 1 0 1 0 1 0 1 0 2 0 1 0 2 0 2 1 2 1 2 1 2 0 1 0 1 0 1 0 2 1 2 0 2 1 2 0 2 0 1 0 2 1 2 0 1 0 2 1 2 1 2 1 2 0 2 1 2 1 2 1 2 0 2 1 2 1 2 0 1 0 2 0 2 0 1 0 2 0 2 0 1 0 1 0 1 0 2 0 2 1 2 0 1 0 2 0 1 0 1 0 2 1 2 1 2 1 2 0 2 1 2 1 2 1 2 0 1 0 1 0 1 0 1 0 "
},
{
"input": "2 1\n0 2 1 4 1",
"output": "0 2 1 3 1 "
},
{
"input": "2 1\n0 2 1 5 1",
"output": "0 2 1 4 1 "
},
{
"input": "3 3\n1 12 9 11 6 8 1",
"output": "1 11 9 10 6 7 1 "
},
{
"input": "3 2\n0 7 4 7 1 3 2",
"output": "0 6 4 6 1 3 2 "
},
{
"input": "2 1\n1 3 2 4 1",
"output": "1 3 2 3 1 "
},
{
"input": "4 1\n5 6 5 6 5 6 1 3 1",
"output": "5 6 5 6 5 6 1 2 1 "
},
{
"input": "2 1\n0 2 1 3 0",
"output": "0 2 1 2 0 "
},
{
"input": "2 2\n98 100 1 7 2",
"output": "98 99 1 6 2 "
},
{
"input": "3 1\n8 10 9 10 3 5 1",
"output": "8 10 9 10 3 4 1 "
},
{
"input": "3 2\n0 4 3 5 2 5 2",
"output": "0 4 3 4 2 4 2 "
},
{
"input": "2 1\n4 5 2 4 2",
"output": "4 5 2 3 2 "
},
{
"input": "3 1\n0 2 1 2 0 2 0",
"output": "0 2 1 2 0 1 0 "
},
{
"input": "1 1\n5 7 2",
"output": "5 6 2 "
},
{
"input": "2 1\n3 4 1 3 1",
"output": "3 4 1 2 1 "
},
{
"input": "3 1\n0 4 3 5 0 5 0",
"output": "0 4 3 5 0 4 0 "
},
{
"input": "3 1\n1 3 2 3 1 3 1",
"output": "1 3 2 3 1 2 1 "
},
{
"input": "2 1\n0 8 7 100 0",
"output": "0 8 7 99 0 "
},
{
"input": "2 1\n1 3 2 5 1",
"output": "1 3 2 4 1 "
}
] | 1,595,931,885
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 48
| 280
| 20,172,800
|
import sys
import math
input = sys.stdin.readline
n, m = map(int, input().strip().split())
a = list(map(int, input().strip().split()))
for i in range(1, 2 * n):
if m and i % 2 and a[i] - 1 > a[i - 1] and a[i] - 1 > a[i + 1]:
a[i] -= 1
m -= 1
for i in range(0, 2 * n + 1): print(a[i], end = ' ')
|
Title: Mountain Scenery
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Bolek has found a picture with *n* mountain peaks painted on it. The *n* painted peaks are represented by a non-closed polyline, consisting of 2*n* segments. The segments go through 2*n*<=+<=1 points with coordinates (1,<=*y*1), (2,<=*y*2), ..., (2*n*<=+<=1,<=*y*2*n*<=+<=1), with the *i*-th segment connecting the point (*i*,<=*y**i*) and the point (*i*<=+<=1,<=*y**i*<=+<=1). For any even *i* (2<=β€<=*i*<=β€<=2*n*) the following condition holds: *y**i*<=-<=1<=<<=*y**i* and *y**i*<=><=*y**i*<=+<=1.
We shall call a vertex of a polyline with an even *x* coordinate a mountain peak.
Bolek fancied a little mischief. He chose exactly *k* mountain peaks, rubbed out the segments that went through those peaks and increased each peak's height by one (that is, he increased the *y* coordinate of the corresponding points). Then he painted the missing segments to get a new picture of mountain peaks. Let us denote the points through which the new polyline passes on Bolek's new picture as (1,<=*r*1), (2,<=*r*2), ..., (2*n*<=+<=1,<=*r*2*n*<=+<=1).
Given Bolek's final picture, restore the initial one.
Input Specification:
The first line contains two space-separated integers *n* and *k* (1<=β€<=*k*<=β€<=*n*<=β€<=100). The next line contains 2*n*<=+<=1 space-separated integers *r*1,<=*r*2,<=...,<=*r*2*n*<=+<=1 (0<=β€<=*r**i*<=β€<=100) β the *y* coordinates of the polyline vertices on Bolek's picture.
It is guaranteed that we can obtain the given picture after performing the described actions on some picture of mountain peaks.
Output Specification:
Print 2*n*<=+<=1 integers *y*1,<=*y*2,<=...,<=*y*2*n*<=+<=1 β the *y* coordinates of the vertices of the polyline on the initial picture. If there are multiple answers, output any one of them.
Demo Input:
['3 2\n0 5 3 5 1 5 2\n', '1 1\n0 2 0\n']
Demo Output:
['0 5 3 4 1 4 2 \n', '0 1 0 \n']
Note:
none
|
```python
import sys
import math
input = sys.stdin.readline
n, m = map(int, input().strip().split())
a = list(map(int, input().strip().split()))
for i in range(1, 2 * n):
if m and i % 2 and a[i] - 1 > a[i - 1] and a[i] - 1 > a[i + 1]:
a[i] -= 1
m -= 1
for i in range(0, 2 * n + 1): print(a[i], end = ' ')
```
| 3
|
|
831
|
B
|
Keyboard Layouts
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] | null | null |
There are two popular keyboard layouts in Berland, they differ only in letters positions. All the other keys are the same. In Berland they use alphabet with 26 letters which coincides with English alphabet.
You are given two strings consisting of 26 distinct letters each: all keys of the first and the second layouts in the same order.
You are also given some text consisting of small and capital English letters and digits. It is known that it was typed in the first layout, but the writer intended to type it in the second layout. Print the text if the same keys were pressed in the second layout.
Since all keys but letters are the same in both layouts, the capitalization of the letters should remain the same, as well as all other characters.
|
The first line contains a string of length 26 consisting of distinct lowercase English letters. This is the first layout.
The second line contains a string of length 26 consisting of distinct lowercase English letters. This is the second layout.
The third line contains a non-empty string *s* consisting of lowercase and uppercase English letters and digits. This is the text typed in the first layout. The length of *s* does not exceed 1000.
|
Print the text if the same keys were pressed in the second layout.
|
[
"qwertyuiopasdfghjklzxcvbnm\nveamhjsgqocnrbfxdtwkylupzi\nTwccpQZAvb2017\n",
"mnbvcxzlkjhgfdsapoiuytrewq\nasdfghjklqwertyuiopzxcvbnm\n7abaCABAABAcaba7\n"
] |
[
"HelloVKCup2017\n",
"7uduGUDUUDUgudu7\n"
] |
none
| 750
|
[
{
"input": "qwertyuiopasdfghjklzxcvbnm\nveamhjsgqocnrbfxdtwkylupzi\nTwccpQZAvb2017",
"output": "HelloVKCup2017"
},
{
"input": "mnbvcxzlkjhgfdsapoiuytrewq\nasdfghjklqwertyuiopzxcvbnm\n7abaCABAABAcaba7",
"output": "7uduGUDUUDUgudu7"
},
{
"input": "ayvguplhjsoiencbkxdrfwmqtz\nkhzvtbspcndierqumlojyagfwx\n3",
"output": "3"
},
{
"input": "oaihbljgekzsxucwnqyrvfdtmp\nwznqcfvrthjibokeglmudpayxs\ntZ8WI33UZZytE8A99EvJjck228LxUQtL5A8q7O217KrmdhpmdhN7JEdVXc8CRm07TFidlIou9AKW9cCl1c4289rfU87oXoSCwHpZO7ggC2GmmDl0KGuA2IimDco2iKaBKl46H089r2tw16mhzI44d2X6g3cnoD0OU5GvA8l89nhNpzTbY9FtZ2wE3Y2a5EC7zXryudTZhXFr9EEcX8P71fp6694aa02B4T0w1pDaVml8FM3N2qB78DBrS723Vpku105sbTJEdBpZu77b1C47DujdoR7rjm5k2nsaPBqX93EfhW95Mm0sBnFtgo12gS87jegSR5u88tM5l420dkt1l1b18UjatzU7P2i9KNJA528caiEpE3JtRw4m4TJ7M1zchxO53skt3Fqvxk2C51gD8XEY7YJC2xmTUqyEUFmPX581Gow2HWq4jaP8FK87",
"output": "yJ8EN33OJJmyT8Z99TdVvkh228FbOLyF5Z8l7W217HuxaqsxaqG7VTaDBk8KUx07YPnafNwo9ZHE9kKf1k4289upO87wBwIKeQsJW7rrK2RxxAf0HRoZ2NnxAkw2nHzCHf46Q089u2ye16xqjN44a2B6r3kgwA0WO5RdZ8f89gqGsjYcM9PyJ2eT3M2z5TK7jBumoaYJqBPu9TTkB8S71ps6694zz02C4Y0e1sAzDxf8PX3G2lC78ACuI723Dsho105icYVTaCsJo77c1K47AovawU7uvx5h2gizSClB93TpqE95Xx0iCgPyrw12rI87vtrIU5o88yX5f420ahy1f1c18OvzyjO7S2n9HGVZ528kznTsT3VyUe4x4YV7X1jkqbW53ihy3Pldbh2K51rA8BTM7MVK2bxYOlmTOPxSB581Rwe2QEl4vzS8PH87"
},
{
"input": "aymrnptzhklcbuxfdvjsgqweio\nwzsavqryltmjnfgcedxpiokbuh\nB5",
"output": "N5"
},
{
"input": "unbclszprgiqjodxeawkymvfth\ncxfwbdvuqlotkgparmhsyinjze\nk081O",
"output": "s081G"
},
{
"input": "evfsnczuiodgbhqmlypkjatxrw\nhvsockwjxtgreqmyanlzidpbuf\n306QMPpaqZ",
"output": "306MYLldmW"
},
{
"input": "pbfjtvryklwmuhxnqsoceiadgz\ntaipfdvlzemhjsnkwyocqgrxbu\nTm9H66Ux59PuGe3lEG94q18u11Dda6w59q1hAAIvHR1qquKI2Xf5ZFdKAPhcEnqKT6BF6Oh16P48YvrIKWGDlRcx9BZwwEF64o0As",
"output": "Fh9S66Jn59TjBq3eQB94w18j11Xxr6m59w1sRRGdSV1wwjZG2Ni5UIxZRTscQkwZF6AI6Os16T48LdvGZMBXeVcn9AUmmQI64o0Ry"
},
{
"input": "rtqgahmkeoldsiynjbuwpvcxfz\noxqiuwflvebnapyrmcghtkdjzs\nJqNskelr3FNjbDhfKPfPXxlqOw72p9BVBwf0tN8Ucs48Vlfjxqo9V3ruU5205UgTYi3JKFbW91NLQ1683315VJ4RSLFW7s26s6uZKs5cO2wAT4JS8rCytZVlPWXdNXaCTq06F1v1Fj2zq7DeJbBSfM5Eko6vBndR75d46mf5Pq7Ark9NARTtQ176ukljBdaqXRsYxrBYl7hda1V7sy38hfbjz59HYM9U55P9eh1CX7tUE44NFlQu7zSjSBHyS3Tte2XaXD3O470Q8U20p8W5rViIh8lsn2TvmcdFdxrF3Ye26J2ZK0BR3KShN597WSJmHJTl4ZZ88IMhzHi6vFyr7MuGYNFGebTB573e6Crwj8P18h344yd8sR2NPge36Y3QC8Y2uW577CO2w4fz",
"output": "MqRalvbo3ZRmcNwzLTzTJjbqEh72t9CKChz0xR8Gda48Kbzmjqe9K3ogG5205GiXYp3MLZcH91RBQ1683315KM4OABZH7a26a6gSLa5dE2hUX4MA8oDyxSKbTHJnRJuDXq06Z1k1Zm2sq7NvMcCAzF5Vle6kCrnO75n46fz5Tq7Uol9RUOXxQ176glbmCnuqJOaYjoCYb7wnu1K7ay38wzcms59WYF9G55T9vw1DJ7xGV44RZbQg7sAmACWyA3Xxv2JuJN3E470Q8G20t8H5oKpPw8bar2XkfdnZnjoZ3Yv26M2SL0CO3LAwR597HAMfWMXb4SS88PFwsWp6kZyo7FgIYRZIvcXC573v6Dohm8T18w344yn8aO2RTiv36Y3QD8Y2gH577DE2h4zs"
},
{
"input": "buneohqdgxjsafrmwtzickvlpy\nzblwamjxifyuqtnrgdkchpoves\n4RZf8YivG6414X1GdDfcCbc10GA0Wz8514LI9D647XzPb66UNh7lX1rDQv0hQvJ7aqhyh1Z39yABGKn24g185Y85ER5q9UqPFaQ2JeK97wHZ78CMSuU8Zf091mePl2OX61BLe5KdmUWodt4BXPiseOZkZ4SZ27qtBM4hT499mCirjy6nB0ZqjQie4Wr3uhW2mGqBlHyEZbW7A6QnsNX9d3j5aHQN0H6GF8J0365KWuAmcroutnJD6l6HI3kSSq17Sdo2htt9y967y8sc98ZAHbutH1m9MOVT1E9Mb5UIK3qNatk9A0m2i1fQl9A65204Q4z4O4rQf374YEq0s2sfmQNW9K7E1zSbj51sGINJVr5736Gw8aW6u9Cjr0sjffXctLopJ0YQ47xD1yEP6bB3odG7slgiM8hJ9BuwfGUwN8tbAgJU8wMI2L0P446MO",
"output": "4NKt8ScoI6414F1IxXthHzh10IQ0Gk8514VC9X647FkEz66BLm7vF1nXJo0mJoY7qjmsm1K39sQZIPl24i185S85WN5j9BjETqJ2YwP97gMK78HRUbB8Kt091rwEv2AF61ZVw5PxrBGaxd4ZFEcuwAKpK4UK27jdZR4mD499rHcnys6lZ0KjyJcw4Gn3bmG2rIjZvMsWKzG7Q6JluLF9x3y5qMJL0M6IT8Y0365PGbQrhnabdlYX6v6MC3pUUj17Uxa2mdd9s967s8uh98KQMzbdM1r9RAOD1W9Rz5BCP3jLqdp9Q0r2c1tJv9Q65204J4k4A4nJt374SWj0u2utrJLG9P7W1kUzy51uICLYOn5736Ig8qG6b9Hyn0uyttFhdVaeY0SJ47fX1sWE6zZ3axI7uvicR8mY9ZbgtIBgL8dzQiYB8gRC2V0E446RA"
},
{
"input": "qwertyuiopasdfghjklzxcvbnm\nqwertyuiopasdfghjklzxcvbnm\nqwertyuiopasdfghjklzxcvbnmPOIUYTREWQLKJHGFDSAMNBVCXZ12345678900987654321ASDFGHJKLqwertyuiopZXCVBNM",
"output": "qwertyuiopasdfghjklzxcvbnmPOIUYTREWQLKJHGFDSAMNBVCXZ12345678900987654321ASDFGHJKLqwertyuiopZXCVBNM"
},
{
"input": "qwertyuiopasdfghjklzxcvbnm\nmnbvcxzlkjhgfdsapoiuytrewq\nasdfghjklzxcvbnmqwertyuiopASDFGHJKLQWERTYUIOPZXCVBNM12345678900987654321QWSDFGVBNxcvghjkoWQEDFGHNJMzxcfghjkl",
"output": "hgfdsapoiuytrewqmnbvcxzlkjHGFDSAPOIMNBVCXZLKJUYTREWQ12345678900987654321MNGFDSREWytrsapokNMBFDSAWPQuytdsapoi"
}
] | 1,657,367,219
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 19
| 62
| 0
|
l1 = input()
l2 = input()
ques = input()
d = dict(zip(l1,l2))
ans = ""
for i in ques:
if i.isdigit():
ans = ans + i
elif i.isupper():
ans = ans + d[i.lower()].upper()
else:
ans = ans + d[i]
print(ans)
|
Title: Keyboard Layouts
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are two popular keyboard layouts in Berland, they differ only in letters positions. All the other keys are the same. In Berland they use alphabet with 26 letters which coincides with English alphabet.
You are given two strings consisting of 26 distinct letters each: all keys of the first and the second layouts in the same order.
You are also given some text consisting of small and capital English letters and digits. It is known that it was typed in the first layout, but the writer intended to type it in the second layout. Print the text if the same keys were pressed in the second layout.
Since all keys but letters are the same in both layouts, the capitalization of the letters should remain the same, as well as all other characters.
Input Specification:
The first line contains a string of length 26 consisting of distinct lowercase English letters. This is the first layout.
The second line contains a string of length 26 consisting of distinct lowercase English letters. This is the second layout.
The third line contains a non-empty string *s* consisting of lowercase and uppercase English letters and digits. This is the text typed in the first layout. The length of *s* does not exceed 1000.
Output Specification:
Print the text if the same keys were pressed in the second layout.
Demo Input:
['qwertyuiopasdfghjklzxcvbnm\nveamhjsgqocnrbfxdtwkylupzi\nTwccpQZAvb2017\n', 'mnbvcxzlkjhgfdsapoiuytrewq\nasdfghjklqwertyuiopzxcvbnm\n7abaCABAABAcaba7\n']
Demo Output:
['HelloVKCup2017\n', '7uduGUDUUDUgudu7\n']
Note:
none
|
```python
l1 = input()
l2 = input()
ques = input()
d = dict(zip(l1,l2))
ans = ""
for i in ques:
if i.isdigit():
ans = ans + i
elif i.isupper():
ans = ans + d[i.lower()].upper()
else:
ans = ans + d[i]
print(ans)
```
| 3
|
|
224
|
A
|
Parallelepiped
|
PROGRAMMING
| 1,100
|
[
"brute force",
"geometry",
"math"
] | null | null |
You've got a rectangular parallelepiped with integer edge lengths. You know the areas of its three faces that have a common vertex. Your task is to find the sum of lengths of all 12 edges of this parallelepiped.
|
The first and the single line contains three space-separated integers β the areas of the parallelepiped's faces. The area's values are positive (<=><=0) and do not exceed 104. It is guaranteed that there exists at least one parallelepiped that satisfies the problem statement.
|
Print a single number β the sum of all edges of the parallelepiped.
|
[
"1 1 1\n",
"4 6 6\n"
] |
[
"12\n",
"28\n"
] |
In the first sample the parallelepiped has sizes 1βΓβ1βΓβ1, in the second oneΒ β 2βΓβ2βΓβ3.
| 500
|
[
{
"input": "1 1 1",
"output": "12"
},
{
"input": "4 6 6",
"output": "28"
},
{
"input": "20 10 50",
"output": "68"
},
{
"input": "9 4 36",
"output": "56"
},
{
"input": "324 9 36",
"output": "184"
},
{
"input": "1333 93 129",
"output": "308"
},
{
"input": "1022 584 112",
"output": "380"
},
{
"input": "66 174 319",
"output": "184"
},
{
"input": "912 276 1748",
"output": "444"
},
{
"input": "65 156 60",
"output": "120"
},
{
"input": "1 10000 10000",
"output": "40008"
},
{
"input": "1485 55 27",
"output": "332"
},
{
"input": "152 108 4104",
"output": "528"
},
{
"input": "1656 6900 1350",
"output": "740"
},
{
"input": "12 14 42",
"output": "60"
},
{
"input": "615 18 1230",
"output": "856"
},
{
"input": "680 60 408",
"output": "336"
},
{
"input": "644 966 6",
"output": "1308"
},
{
"input": "1 432 432",
"output": "1736"
},
{
"input": "2239 2239 1",
"output": "8964"
},
{
"input": "4106 8212 2",
"output": "16436"
},
{
"input": "10000 10000 10000",
"output": "1200"
},
{
"input": "3623 3623 1",
"output": "14500"
},
{
"input": "9801 9801 9801",
"output": "1188"
},
{
"input": "10000 1 10000",
"output": "40008"
},
{
"input": "9 9 9",
"output": "36"
},
{
"input": "9801 9702 9702",
"output": "1184"
}
] | 1,657,948,229
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 27
| 124
| 0
|
x,y,z = map(int , input().split())
#>-> shivank singh
k = (x*y*z)**0.5
print(4*int(k//x+k//y+k//z))
|
Title: Parallelepiped
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got a rectangular parallelepiped with integer edge lengths. You know the areas of its three faces that have a common vertex. Your task is to find the sum of lengths of all 12 edges of this parallelepiped.
Input Specification:
The first and the single line contains three space-separated integers β the areas of the parallelepiped's faces. The area's values are positive (<=><=0) and do not exceed 104. It is guaranteed that there exists at least one parallelepiped that satisfies the problem statement.
Output Specification:
Print a single number β the sum of all edges of the parallelepiped.
Demo Input:
['1 1 1\n', '4 6 6\n']
Demo Output:
['12\n', '28\n']
Note:
In the first sample the parallelepiped has sizes 1βΓβ1βΓβ1, in the second oneΒ β 2βΓβ2βΓβ3.
|
```python
x,y,z = map(int , input().split())
#>-> shivank singh
k = (x*y*z)**0.5
print(4*int(k//x+k//y+k//z))
```
| 3
|
|
519
|
B
|
A and B and Compilation Errors
|
PROGRAMMING
| 1,100
|
[
"data structures",
"implementation",
"sortings"
] | null | null |
A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed *n* compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared β the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected?
|
The first line of the input contains integer *n* (3<=β€<=*n*<=β€<=105) β the initial number of compilation errors.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=β€<=*a**i*<=β€<=109) β the errors the compiler displayed for the first time.
The third line contains *n*<=-<=1 space-separated integers *b*1,<=*b*2,<=...,<=*b**n*<=-<=1 β the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains *n*<=-<=2 space-separated integers *Ρ*1,<=*Ρ*2,<=...,<=*Ρ**n*<=-<=2 β the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one.
|
Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively.
|
[
"5\n1 5 8 123 7\n123 7 5 1\n5 1 7\n",
"6\n1 4 3 3 5 7\n3 7 5 4 3\n4 3 7 5\n"
] |
[
"8\n123\n",
"1\n3\n"
] |
In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step.
| 1,000
|
[
{
"input": "5\n1 5 8 123 7\n123 7 5 1\n5 1 7",
"output": "8\n123"
},
{
"input": "6\n1 4 3 3 5 7\n3 7 5 4 3\n4 3 7 5",
"output": "1\n3"
},
{
"input": "3\n1 2 3\n3 2\n2",
"output": "1\n3"
},
{
"input": "10\n460626451 802090732 277246428 661369649 388684428 784303821 376287098 656422756 9301599 25720377\n277246428 388684428 661369649 460626451 656422756 802090732 9301599 784303821 376287098\n376287098 802090732 388684428 9301599 656422756 784303821 460626451 277246428",
"output": "25720377\n661369649"
},
{
"input": "3\n796067435 964699482 819602309\n964699482 796067435\n964699482",
"output": "819602309\n796067435"
},
{
"input": "3\n374054998 726316780 902899520\n902899520 726316780\n726316780",
"output": "374054998\n902899520"
},
{
"input": "3\n168638990 939116221 323703261\n168638990 323703261\n168638990",
"output": "939116221\n323703261"
},
{
"input": "3\n77 77 77\n77 77\n77",
"output": "77\n77"
},
{
"input": "3\n84 30 9\n9 84\n9",
"output": "30\n84"
},
{
"input": "6\n5 4 3 3 5 5\n3 5 5 4 3\n3 5 4 3",
"output": "5\n5"
},
{
"input": "4\n1 5 7 8\n1 5 7\n1 5",
"output": "8\n7"
}
] | 1,675,330,829
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 33
| 156
| 8,908,800
|
if __name__ == "__main__":
input()
outputs = [sum([int (x) for x in input().split()]), sum([int (x) for x in input().split()]), sum([int (x) for x in input().split()])]
errors = ""
for _ in range(1, 3):
errors+= str(outputs[_ - 1] - outputs[_]) + "\n"
print (errors)
|
Title: A and B and Compilation Errors
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed *n* compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared β the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected?
Input Specification:
The first line of the input contains integer *n* (3<=β€<=*n*<=β€<=105) β the initial number of compilation errors.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=β€<=*a**i*<=β€<=109) β the errors the compiler displayed for the first time.
The third line contains *n*<=-<=1 space-separated integers *b*1,<=*b*2,<=...,<=*b**n*<=-<=1 β the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains *n*<=-<=2 space-separated integers *Ρ*1,<=*Ρ*2,<=...,<=*Ρ**n*<=-<=2 β the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one.
Output Specification:
Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively.
Demo Input:
['5\n1 5 8 123 7\n123 7 5 1\n5 1 7\n', '6\n1 4 3 3 5 7\n3 7 5 4 3\n4 3 7 5\n']
Demo Output:
['8\n123\n', '1\n3\n']
Note:
In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step.
|
```python
if __name__ == "__main__":
input()
outputs = [sum([int (x) for x in input().split()]), sum([int (x) for x in input().split()]), sum([int (x) for x in input().split()])]
errors = ""
for _ in range(1, 3):
errors+= str(outputs[_ - 1] - outputs[_]) + "\n"
print (errors)
```
| 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<=β€<=*r*<=β€<=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<=β€<=*k*<=β€<=1000, 1<=β€<=*r*<=β€<=9)Β β 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 9Β·117β=β1053 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 2Β·15β=β30 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"
}
] | 1,690,498,189
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 54
| 46
| 0
|
x1, x2 = [int(x) for x in input().split()]
q = x1
while (str(q)[-1] != '0') and (str(q)[-1] != str(x2)):
q += x1
else:
print(q // x1)
|
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<=β€<=*r*<=β€<=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<=β€<=*k*<=β€<=1000, 1<=β€<=*r*<=β€<=9)Β β 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 9Β·117β=β1053 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 2Β·15β=β30 burles. It is obvious that he can pay this sum without any change.
|
```python
x1, x2 = [int(x) for x in input().split()]
q = x1
while (str(q)[-1] != '0') and (str(q)[-1] != str(x2)):
q += x1
else:
print(q // x1)
```
| 3
|
|
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<=β€<=*n*<=β€<=2000) β the number of employees.
The next *n* lines contain the integers *p**i* (1<=β€<=*p**i*<=β€<=*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*<=β <=*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"
}
] | 1,690,779,121
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 106
| 218
| 4,198,400
|
n = int(input())
c,s = [int(input()) for _ in range(n)],0
for i in range(n):
d = 0
while i>=0:
i=c[i]-1
d+=1
s = max(d,s)
print(s)
|
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<=β€<=*n*<=β€<=2000) β the number of employees.
The next *n* lines contain the integers *p**i* (1<=β€<=*p**i*<=β€<=*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*<=β <=*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
n = int(input())
c,s = [int(input()) for _ in range(n)],0
for i in range(n):
d = 0
while i>=0:
i=c[i]-1
d+=1
s = max(d,s)
print(s)
```
| 3
|
|
32
|
B
|
Borze
|
PROGRAMMING
| 800
|
[
"expression parsing",
"implementation"
] |
B. Borze
|
2
|
256
|
Ternary numeric notation is quite popular in Berland. To telegraph the ternary number the Borze alphabet is used. Digit 0 is transmitted as Β«.Β», 1 as Β«-.Β» and 2 as Β«--Β». You are to decode the Borze code, i.e. to find out the ternary number given its representation in Borze alphabet.
|
The first line contains a number in Borze code. The length of the string is between 1 and 200 characters. It's guaranteed that the given string is a valid Borze code of some ternary number (this number can have leading zeroes).
|
Output the decoded ternary number. It can have leading zeroes.
|
[
".-.--\n",
"--.\n",
"-..-.--\n"
] |
[
"012",
"20",
"1012"
] |
none
| 1,000
|
[
{
"input": ".-.--",
"output": "012"
},
{
"input": "--.",
"output": "20"
},
{
"input": "-..-.--",
"output": "1012"
},
{
"input": "---..",
"output": "210"
},
{
"input": "..--.---..",
"output": "0020210"
},
{
"input": "-.....----.",
"output": "10000220"
},
{
"input": ".",
"output": "0"
},
{
"input": "-.",
"output": "1"
},
{
"input": "--",
"output": "2"
},
{
"input": "..",
"output": "00"
},
{
"input": "--.",
"output": "20"
},
{
"input": ".--.",
"output": "020"
},
{
"input": ".-.-..",
"output": "0110"
},
{
"input": "----.-.",
"output": "2201"
},
{
"input": "-..--.-.",
"output": "10201"
},
{
"input": "..--..--.",
"output": "0020020"
},
{
"input": "-.-.---.--..-..-.-.-..-..-.--.",
"output": "112120010111010120"
},
{
"input": "---.-.-.------..-..-..-..-.-..-.--.-.-..-.-.-----..-.-.",
"output": "21112220010101011012011011221011"
},
{
"input": "-.-..--.-.-.-.-.-..-.-.-.---------.--.---..--...--.-----.-.-.-...--.-.-.---.------.--..-.--.-----.-...-..------",
"output": "11020111110111222212021020002022111100201121222020012022110010222"
},
{
"input": "-.-..-.--.---..---.-..---.-...-.-.----..-.---.-.---..-.--.---.-.-------.---.--....----.-.---.---.---.----.-----..---.-.-.-.-----.--.-------.-..",
"output": "110120210211021100112200121121012021122212120000220121212122022102111122120222110"
},
{
"input": ".-..-.-.---.-----.--.---...-.--.-.-....-..",
"output": "01011212212021001201100010"
},
{
"input": ".------.-.---..--...-..-..-.-.-.--.--.-..-.--...-.-.---.-.-.------..--..-.---..----.-..-.--.---.-.----.-.---...-.-.-.-----.-.-.---.---.-.....-.-...-----.-...-.---.-..-.-----.--...---.-.-..-.--.-.---..",
"output": "022201210200010101112020101200011211122200200121022010120211220121001112211121211000011002211001211012212000211101201210"
},
{
"input": ".-.--.---.-----.-.-----.-.-..-----..-..----..--.-.--.----..---.---..-.-.-----..-------.----..----.-..---...-----..-..-----...-..-.-.-----....---..---..-.-----...-.--...--.-.---.-.-.-.-.-...---..----.",
"output": "01202122112211102210102200201202200212101122102221220022010210022101022100101122100021021012210012000201211111100210220"
},
{
"input": "..-.-.-.---.-.-.-..-.-..-.-.---.-------.---..-----.---....-.---.--.--.-.---.---------.-..---.-.-.--..---.---.-.---.-.-..-.-..-.-.-.----.--.-....--------.-.---..----.------.-.-.--.--.-----.-----.----",
"output": "0011121111011011212221210221210001212020121222211021112002121121110110111220201000222201210220222011202022122122"
},
{
"input": "-..-------.------.-..--.-.-..--.-.-..-----..-.-.-..-..-..--.---..-----..---..-..--.-..-.-.---...-.....-------.---.-----.-...-.-...-.-.---.---.-----.--.--...-.--..-.-..-...-.-.-.-.---..---.-..-.-.-.-..",
"output": "102221222010201102011022101110101020210221021010201011210010000222121221100110011212122120200012001101001111210211011110"
},
{
"input": ".-.----.-.--..-.-.-.-..----..-.-...--.-.---.---.-------..-.--..-......--.------.--.----.--...-.--.--..-----..-.....--.--.-.-.------..--------.----------..-.---.----.---.-..--..-.....-..------.--.",
"output": "012201200111102200110020121212221012001000002022202022020001202002210100002020112220022220222220012122021102001000010222020"
},
{
"input": "------.-----.-....--.-.----.-.---.-.-..---.-.---.-----..-...-.-.---..-.-.-..-.-.-...-.-.-.----..--.------.----.-..-.--...-.-------...-.-..-.-.--.--.---..--..--------.--.-.-.---.-.-.-...----.--..-.--..",
"output": "222022110002012201211102112122101001121011101110011122002022202201012000122210011011202021020022220201121111002202001200"
},
{
"input": "-.---...----...--.--...-.--.----",
"output": "121002200020200012022"
},
{
"input": "--.--.--.---.--.-.---.-.-..-..--.-..---.-.....-..---.-----.--...-.-.-------.-.--.-.----.-..-.------.",
"output": "202020212012111010201021100001021221200011222112012201012220"
}
] | 1,620,401,162
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 124
| 6,758,400
|
borze = input()
code = ""
i = 0
while i < len(borze):
if borze[i] == ".":
code += "0"
if borze[i] == "-":
if borze[i+1] == ".":
code += "1"
else:
code += "2"
i += 2
continue
i += 1
print(code)
|
Title: Borze
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Ternary numeric notation is quite popular in Berland. To telegraph the ternary number the Borze alphabet is used. Digit 0 is transmitted as Β«.Β», 1 as Β«-.Β» and 2 as Β«--Β». You are to decode the Borze code, i.e. to find out the ternary number given its representation in Borze alphabet.
Input Specification:
The first line contains a number in Borze code. The length of the string is between 1 and 200 characters. It's guaranteed that the given string is a valid Borze code of some ternary number (this number can have leading zeroes).
Output Specification:
Output the decoded ternary number. It can have leading zeroes.
Demo Input:
['.-.--\n', '--.\n', '-..-.--\n']
Demo Output:
['012', '20', '1012']
Note:
none
|
```python
borze = input()
code = ""
i = 0
while i < len(borze):
if borze[i] == ".":
code += "0"
if borze[i] == "-":
if borze[i+1] == ".":
code += "1"
else:
code += "2"
i += 2
continue
i += 1
print(code)
```
| 3.956411
|
263
|
A
|
Beautiful Matrix
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
You've got a 5<=Γ<=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<=β€<=*i*<=<<=5). 1. Swap two neighboring matrix columns, that is, columns with indexes *j* and *j*<=+<=1 for some integer *j* (1<=β€<=*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 β 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"
}
] | 1,694,882,976
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 25
| 92
| 0
|
# COMPLEJIDAD O(r*c)
x = [list(map(int, input().split())) for _ in range (5)]
for r in range (5):
for c in range (5):
if x [r][c] == 1:
rx, cx = r,c
distanciaR = abs(rx - 2)
distanciaC = abs (cx - 2)
distanciaT = distanciaC + distanciaR
print (distanciaT)
|
Title: Beautiful Matrix
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got a 5<=Γ<=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<=β€<=*i*<=<<=5). 1. Swap two neighboring matrix columns, that is, columns with indexes *j* and *j*<=+<=1 for some integer *j* (1<=β€<=*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 β 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
# COMPLEJIDAD O(r*c)
x = [list(map(int, input().split())) for _ in range (5)]
for r in range (5):
for c in range (5):
if x [r][c] == 1:
rx, cx = r,c
distanciaR = abs(rx - 2)
distanciaC = abs (cx - 2)
distanciaT = distanciaC + distanciaR
print (distanciaT)
```
| 3
|
|
499
|
B
|
Lecture
|
PROGRAMMING
| 1,000
|
[
"implementation",
"strings"
] | null | null |
You have a new professor of graph theory and he speaks very quickly. You come up with the following plan to keep up with his lecture and make notes.
You know two languages, and the professor is giving the lecture in the first one. The words in both languages consist of lowercase English characters, each language consists of several words. For each language, all words are distinct, i.e. they are spelled differently. Moreover, the words of these languages have a one-to-one correspondence, that is, for each word in each language, there exists exactly one word in the other language having has the same meaning.
You can write down every word the professor says in either the first language or the second language. Of course, during the lecture you write down each word in the language in which the word is shorter. In case of equal lengths of the corresponding words you prefer the word of the first language.
You are given the text of the lecture the professor is going to read. Find out how the lecture will be recorded in your notes.
|
The first line contains two integers, *n* and *m* (1<=β€<=*n*<=β€<=3000, 1<=β€<=*m*<=β€<=3000) β the number of words in the professor's lecture and the number of words in each of these languages.
The following *m* lines contain the words. The *i*-th line contains two strings *a**i*, *b**i* meaning that the word *a**i* belongs to the first language, the word *b**i* belongs to the second language, and these two words have the same meaning. It is guaranteed that no word occurs in both languages, and each word occurs in its language exactly once.
The next line contains *n* space-separated strings *c*1,<=*c*2,<=...,<=*c**n* β the text of the lecture. It is guaranteed that each of the strings *c**i* belongs to the set of strings {*a*1,<=*a*2,<=... *a**m*}.
All the strings in the input are non-empty, each consisting of no more than 10 lowercase English letters.
|
Output exactly *n* words: how you will record the lecture in your notebook. Output the words of the lecture in the same order as in the input.
|
[
"4 3\ncodeforces codesecrof\ncontest round\nletter message\ncodeforces contest letter contest\n",
"5 3\njoll wuqrd\neuzf un\nhbnyiyc rsoqqveh\nhbnyiyc joll joll euzf joll\n"
] |
[
"codeforces round letter round\n",
"hbnyiyc joll joll un joll\n"
] |
none
| 500
|
[
{
"input": "4 3\ncodeforces codesecrof\ncontest round\nletter message\ncodeforces contest letter contest",
"output": "codeforces round letter round"
},
{
"input": "5 3\njoll wuqrd\neuzf un\nhbnyiyc rsoqqveh\nhbnyiyc joll joll euzf joll",
"output": "hbnyiyc joll joll un joll"
},
{
"input": "5 5\nqueyqj f\nb vn\ntabzvk qpfzoqx\nytnyonoc hnxsd\njpggvr lchinjmt\nqueyqj jpggvr b ytnyonoc b",
"output": "f jpggvr b hnxsd b"
},
{
"input": "10 22\nazbrll oen\ngh vdyayei\njphveblohx vfglv\nmfyxib jepnvhcuwo\nrpikazqj uam\nl rx\nokjenof qpnyi\nj tixqrno\nod itozmfct\nikkdxmirx ev\nqexftojc p\nkdazb zjs\nmbk ykvqjrxaxu\nhbcwhouzq pwt\nmirpsz zfaegpl\nuhkkvcj rlvwj\nef iqnnwtolrc\npjzfcpmeag ecdayth\nsa qcthz\ncbfhfxi qrnbvdryz\nwqel tj\natx smkbid\nef hbcwhouzq cbfhfxi hbcwhouzq mirpsz cbfhfxi cbfhfxi okjenof pjzfcpmeag kdazb",
"output": "ef pwt cbfhfxi pwt mirpsz cbfhfxi cbfhfxi qpnyi ecdayth zjs"
},
{
"input": "1 1\namit am\namit",
"output": "am"
},
{
"input": "1 1\na c\na",
"output": "a"
}
] | 1,681,741,814
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 37
| 202
| 5,632,000
|
import sys
input = sys.stdin.readline
n, m = list(map(int, input().split()))
d = {}
for i in range(m):
a,b = list(map(str, input().split()))
if len(a) <= len(b):
d[a] = a;
else:
d[a] = b;
lecture = list(map(str, input().split()))
for i in range(len(lecture)):
if i != len(lecture) - 1:
print(d[lecture[i]], end=' ')
else:
print(d[lecture[i]])
|
Title: Lecture
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have a new professor of graph theory and he speaks very quickly. You come up with the following plan to keep up with his lecture and make notes.
You know two languages, and the professor is giving the lecture in the first one. The words in both languages consist of lowercase English characters, each language consists of several words. For each language, all words are distinct, i.e. they are spelled differently. Moreover, the words of these languages have a one-to-one correspondence, that is, for each word in each language, there exists exactly one word in the other language having has the same meaning.
You can write down every word the professor says in either the first language or the second language. Of course, during the lecture you write down each word in the language in which the word is shorter. In case of equal lengths of the corresponding words you prefer the word of the first language.
You are given the text of the lecture the professor is going to read. Find out how the lecture will be recorded in your notes.
Input Specification:
The first line contains two integers, *n* and *m* (1<=β€<=*n*<=β€<=3000, 1<=β€<=*m*<=β€<=3000) β the number of words in the professor's lecture and the number of words in each of these languages.
The following *m* lines contain the words. The *i*-th line contains two strings *a**i*, *b**i* meaning that the word *a**i* belongs to the first language, the word *b**i* belongs to the second language, and these two words have the same meaning. It is guaranteed that no word occurs in both languages, and each word occurs in its language exactly once.
The next line contains *n* space-separated strings *c*1,<=*c*2,<=...,<=*c**n* β the text of the lecture. It is guaranteed that each of the strings *c**i* belongs to the set of strings {*a*1,<=*a*2,<=... *a**m*}.
All the strings in the input are non-empty, each consisting of no more than 10 lowercase English letters.
Output Specification:
Output exactly *n* words: how you will record the lecture in your notebook. Output the words of the lecture in the same order as in the input.
Demo Input:
['4 3\ncodeforces codesecrof\ncontest round\nletter message\ncodeforces contest letter contest\n', '5 3\njoll wuqrd\neuzf un\nhbnyiyc rsoqqveh\nhbnyiyc joll joll euzf joll\n']
Demo Output:
['codeforces round letter round\n', 'hbnyiyc joll joll un joll\n']
Note:
none
|
```python
import sys
input = sys.stdin.readline
n, m = list(map(int, input().split()))
d = {}
for i in range(m):
a,b = list(map(str, input().split()))
if len(a) <= len(b):
d[a] = a;
else:
d[a] = b;
lecture = list(map(str, input().split()))
for i in range(len(lecture)):
if i != len(lecture) - 1:
print(d[lecture[i]], end=' ')
else:
print(d[lecture[i]])
```
| 3
|
|
78
|
A
|
Haiku
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] |
A. Haiku
|
2
|
256
|
Haiku is a genre of Japanese traditional poetry.
A haiku poem consists of 17 syllables split into three phrases, containing 5, 7 and 5 syllables correspondingly (the first phrase should contain exactly 5 syllables, the second phrase should contain exactly 7 syllables, and the third phrase should contain exactly 5 syllables). A haiku masterpiece contains a description of a moment in those three phrases. Every word is important in a small poem, which is why haiku are rich with symbols. Each word has a special meaning, a special role. The main principle of haiku is to say much using a few words.
To simplify the matter, in the given problem we will consider that the number of syllable in the phrase is equal to the number of vowel letters there. Only the following letters are regarded as vowel letters: "a", "e", "i", "o" and "u".
Three phases from a certain poem are given. Determine whether it is haiku or not.
|
The input data consists of three lines. The length of each line is between 1 and 100, inclusive. The *i*-th line contains the *i*-th phrase of the poem. Each phrase consists of one or more words, which are separated by one or more spaces. A word is a non-empty sequence of lowercase Latin letters. Leading and/or trailing spaces in phrases are allowed. Every phrase has at least one non-space character. See the example for clarification.
|
Print "YES" (without the quotes) if the poem is a haiku. Otherwise, print "NO" (also without the quotes).
|
[
"on codeforces \nbeta round is running\n a rustling of keys \n",
"how many gallons\nof edo s rain did you drink\n cuckoo\n"
] |
[
"YES",
"NO"
] |
none
| 500
|
[
{
"input": "on codeforces \nbeta round is running\n a rustling of keys ",
"output": "YES"
},
{
"input": "how many gallons\nof edo s rain did you drink\n cuckoo",
"output": "NO"
},
{
"input": " hatsu shigure\n saru mo komino wo\nhoshige nari",
"output": "YES"
},
{
"input": "o vetus stagnum\n rana de ripa salit\n ac sonant aquae",
"output": "NO"
},
{
"input": " furuike ya\nkawazu tobikomu\nmizu no oto ",
"output": "YES"
},
{
"input": " noch da leich\na stamperl zum aufwaerma\n da pfarrer kimmt a ",
"output": "NO"
},
{
"input": " sommerfuglene \n hvorfor bruge mange ord\n et kan gore det",
"output": "YES"
},
{
"input": " ab der mittagszeit\n ist es etwas schattiger\n ein wolkenhimmel",
"output": "NO"
},
{
"input": "tornando a vederli\ni fiori di ciliegio la sera\nson divenuti frutti",
"output": "NO"
},
{
"input": "kutaburete\nyado karu koro ya\nfuji no hana",
"output": "YES"
},
{
"input": " beginnings of poetry\n the rice planting songs \n of the interior",
"output": "NO"
},
{
"input": " door zomerregens\n zijn de kraanvogelpoten\n korter geworden",
"output": "NO"
},
{
"input": " derevo na srub\na ptitsi bezzabotno\n gnezdishko tam vyut",
"output": "YES"
},
{
"input": "writing in the dark\nunaware that my pen\nhas run out of ink",
"output": "NO"
},
{
"input": "kusaaiu\nuieueua\nuo efaa",
"output": "YES"
},
{
"input": "v\nh\np",
"output": "NO"
},
{
"input": "i\ni\nu",
"output": "NO"
},
{
"input": "awmio eoj\nabdoolceegood\nwaadeuoy",
"output": "YES"
},
{
"input": "xzpnhhnqsjpxdboqojixmofawhdjcfbscq\nfoparnxnbzbveycoltwdrfbwwsuobyoz hfbrszy\nimtqryscsahrxpic agfjh wvpmczjjdrnwj mcggxcdo",
"output": "YES"
},
{
"input": "wxjcvccp cppwsjpzbd dhizbcnnllckybrnfyamhgkvkjtxxfzzzuyczmhedhztugpbgpvgh\nmdewztdoycbpxtp bsiw hknggnggykdkrlihvsaykzfiiw\ndewdztnngpsnn lfwfbvnwwmxoojknygqb hfe ibsrxsxr",
"output": "YES"
},
{
"input": "nbmtgyyfuxdvrhuhuhpcfywzrbclp znvxw synxmzymyxcntmhrjriqgdjh xkjckydbzjbvtjurnf\nhhnhxdknvamywhsrkprofnyzlcgtdyzzjdsfxyddvilnzjziz qmwfdvzckgcbrrxplxnxf mpxwxyrpesnewjrx ajxlfj\nvcczq hddzd cvefmhxwxxyqcwkr fdsndckmesqeq zyjbwbnbyhybd cta nsxzidl jpcvtzkldwd",
"output": "YES"
},
{
"input": "rvwdsgdsrutgjwscxz pkd qtpmfbqsmctuevxdj kjzknzghdvxzlaljcntg jxhvzn yciktbsbyscfypx x xhkxnfpdp\nwdfhvqgxbcts mnrwbr iqttsvigwdgvlxwhsmnyxnttedonxcfrtmdjjmacvqtkbmsnwwvvrlxwvtggeowtgsqld qj\nvsxcdhbzktrxbywpdvstr meykarwtkbm pkkbhvwvelclfmpngzxdmblhcvf qmabmweldplmczgbqgzbqnhvcdpnpjtch ",
"output": "YES"
},
{
"input": "brydyfsmtzzkpdsqvvztmprhqzbzqvgsblnz naait tdtiprjsttwusdykndwcccxfmzmrmfmzjywkpgbfnjpypgcbcfpsyfj k\nucwdfkfyxxxht lxvnovqnnsqutjsyagrplb jhvtwdptrwcqrovncdvqljjlrpxcfbxqgsfylbgmcjpvpl ccbcybmigpmjrxpu\nfgwtpcjeywgnxgbttgx htntpbk tkkpwbgxwtbxvcpkqbzetjdkcwad tftnjdxxjdvbpfibvxuglvx llyhgjvggtw jtjyphs",
"output": "YES"
},
{
"input": "nyc aqgqzjjlj mswgmjfcxlqdscheskchlzljlsbhyn iobxymwzykrsnljj\nnnebeaoiraga\nqpjximoqzswhyyszhzzrhfwhf iyxysdtcpmikkwpugwlxlhqfkn",
"output": "NO"
},
{
"input": "lzrkztgfe mlcnq ay ydmdzxh cdgcghxnkdgmgfzgahdjjmqkpdbskreswpnblnrc fmkwziiqrbskp\np oukeaz gvvy kghtrjlczyl qeqhgfgfej\nwfolhkmktvsjnrpzfxcxzqmfidtlzmuhxac wsncjgmkckrywvxmnjdpjpfydhk qlmdwphcvyngansqhl",
"output": "NO"
},
{
"input": "yxcboqmpwoevrdhvpxfzqmammak\njmhphkxppkqkszhqqtkvflarsxzla pbxlnnnafqbsnmznfj qmhoktgzix qpmrgzxqvmjxhskkksrtryehfnmrt dtzcvnvwp\nscwymuecjxhw rdgsffqywwhjpjbfcvcrnisfqllnbplpadfklayjguyvtrzhwblftclfmsr",
"output": "NO"
},
{
"input": "qfdwsr jsbrpfmn znplcx nhlselflytndzmgxqpgwhpi ghvbbxrkjdirfghcybhkkqdzmyacvrrcgsneyjlgzfvdmxyjmph\nylxlyrzs drbktzsniwcbahjkgohcghoaczsmtzhuwdryjwdijmxkmbmxv yyfrokdnsx\nyw xtwyzqlfxwxghugoyscqlx pljtz aldfskvxlsxqgbihzndhxkswkxqpwnfcxzfyvncstfpqf",
"output": "NO"
},
{
"input": "g rguhqhcrzmuqthtmwzhfyhpmqzzosa\nmhjimzvchkhejh irvzejhtjgaujkqfxhpdqjnxr dvqallgssktqvsxi\npcwbliftjcvuzrsqiswohi",
"output": "NO"
},
{
"input": " ngxtlq iehiise vgffqcpnmsoqzyseuqqtggokymol zn\nvjdjljazeujwoubkcvtsbepooxqzrueaauokhepiquuopfild\ngoabauauaeotoieufueeknudiilupouaiaexcoapapu",
"output": "NO"
},
{
"input": "ycnvnnqk mhrmhctpkfbc qbyvtjznmndqjzgbcxmvrpkfcll zwspfptmbxgrdv dsgkk nfytsqjrnfbhh pzdldzymvkdxxwh\nvnhjfwgdnyjptsmblyxmpzylsbjlmtkkwjcbqwjctqvrlqqkdsrktxlnslspvnn mdgsmzblhbnvpczmqkcffwhwljqkzmk hxcm\nrghnjvzcpprrgmtgytpkzyc mrdnnhpkwypwqbtzjyfwvrdwyjltbzxtbstzs xdjzdmx yjsqtzlrnvyssvglsdjrmsrfrcdpqt",
"output": "NO"
},
{
"input": "ioeeaioeiuoeaeieuuieooaouiuouiioaueeaiaiuoaoiioeeaauooiuuieeuaeeoauieeaiuoieiaieuoauaaoioooieueueuai\nuooaoeeaoiuuoeioaoouaououoeioiaeueoioaiouaeaoioiuuaueeuaiuoiueoiuaoeeieeouaeeaeeieioeoiiieuuueuuieuo\naeeouieeieoueaioeoioooiouaeeeiaaioueauaueiouuuaieuuioiaeiueauueaoieauauoeueuaiueuuaueeoueauaeaoieeoo",
"output": "NO"
},
{
"input": "mkgycgmrqtmlwj vddlbsgdptyfrcj nmrddjchydvbsyhgzytfzvwbhqmmhkpsyrcnknpsqrr wcnf wfvgskrnsmnrcqgcnc q\npn cclcyjjdp rqkpgyrvyyd hhynrochfa pthyffqvskazfshvzcmvhmsrxvquzkdvivlbfypwbgltfbjmhzljvgzmxjlrjdjq\nauuouieuaaauoooeoiuoaueuueauiuuuiaeuwuoiaeoeaieauieouoaeiiieiioiuuuiuauioiiuoeauueieaueiaioeuaoaaau",
"output": "NO"
},
{
"input": "ab\na\na",
"output": "NO"
},
{
"input": "on codeforcs\nbeta round is runnineg\na rustling of keys",
"output": "NO"
},
{
"input": "coooooooooooooooood\nd\nd",
"output": "NO"
},
{
"input": "on codeforces \nbeta round is runninog\n a rustling of keys",
"output": "NO"
},
{
"input": "aaaaaaaaaaaaaaa\na\na",
"output": "NO"
},
{
"input": "aaaaaccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc\naaaaaaaccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc\naaaaaccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc",
"output": "YES"
},
{
"input": "cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc\ncccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc\ncccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc",
"output": "NO"
},
{
"input": "aaaaaa\naaaaaa\naaaaa",
"output": "NO"
}
] | 1,566,996,914
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 40
| 218
| 0
|
#!/usr/bin/python3
is_haiku = True
for x in range(0, 3):
phrase = input()
phrase = phrase.strip()
# count the number of vowels in the phrase
vowel_cnt = phrase.count('a') + phrase.count('e') + phrase.count('i') + phrase.count('o') + phrase.count('u')
if (x == 0 and vowel_cnt != 5) or (x == 1 and vowel_cnt != 7) or (x == 2 and vowel_cnt != 5):
is_haiku = False
break
if is_haiku:
print ("YES")
else:
print ("NO")
|
Title: Haiku
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Haiku is a genre of Japanese traditional poetry.
A haiku poem consists of 17 syllables split into three phrases, containing 5, 7 and 5 syllables correspondingly (the first phrase should contain exactly 5 syllables, the second phrase should contain exactly 7 syllables, and the third phrase should contain exactly 5 syllables). A haiku masterpiece contains a description of a moment in those three phrases. Every word is important in a small poem, which is why haiku are rich with symbols. Each word has a special meaning, a special role. The main principle of haiku is to say much using a few words.
To simplify the matter, in the given problem we will consider that the number of syllable in the phrase is equal to the number of vowel letters there. Only the following letters are regarded as vowel letters: "a", "e", "i", "o" and "u".
Three phases from a certain poem are given. Determine whether it is haiku or not.
Input Specification:
The input data consists of three lines. The length of each line is between 1 and 100, inclusive. The *i*-th line contains the *i*-th phrase of the poem. Each phrase consists of one or more words, which are separated by one or more spaces. A word is a non-empty sequence of lowercase Latin letters. Leading and/or trailing spaces in phrases are allowed. Every phrase has at least one non-space character. See the example for clarification.
Output Specification:
Print "YES" (without the quotes) if the poem is a haiku. Otherwise, print "NO" (also without the quotes).
Demo Input:
['on codeforces \nbeta round is running\n a rustling of keys \n', 'how many gallons\nof edo s rain did you drink\n cuckoo\n']
Demo Output:
['YES', 'NO']
Note:
none
|
```python
#!/usr/bin/python3
is_haiku = True
for x in range(0, 3):
phrase = input()
phrase = phrase.strip()
# count the number of vowels in the phrase
vowel_cnt = phrase.count('a') + phrase.count('e') + phrase.count('i') + phrase.count('o') + phrase.count('u')
if (x == 0 and vowel_cnt != 5) or (x == 1 and vowel_cnt != 7) or (x == 2 and vowel_cnt != 5):
is_haiku = False
break
if is_haiku:
print ("YES")
else:
print ("NO")
```
| 3.9455
|
151
|
A
|
Soft Drinking
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
This winter is so cold in Nvodsk! A group of *n* friends decided to buy *k* bottles of a soft drink called "Take-It-Light" to warm up a bit. Each bottle has *l* milliliters of the drink. Also they bought *c* limes and cut each of them into *d* slices. After that they found *p* grams of salt.
To make a toast, each friend needs *nl* milliliters of the drink, a slice of lime and *np* grams of salt. The friends want to make as many toasts as they can, provided they all drink the same amount. How many toasts can each friend make?
|
The first and only line contains positive integers *n*, *k*, *l*, *c*, *d*, *p*, *nl*, *np*, not exceeding 1000 and no less than 1. The numbers are separated by exactly one space.
|
Print a single integer β the number of toasts each friend can make.
|
[
"3 4 5 10 8 100 3 1\n",
"5 100 10 1 19 90 4 3\n",
"10 1000 1000 25 23 1 50 1\n"
] |
[
"2\n",
"3\n",
"0\n"
] |
A comment to the first sample:
Overall the friends have 4β*β5β=β20 milliliters of the drink, it is enough to make 20β/β3β=β6 toasts. The limes are enough for 10β*β8β=β80 toasts and the salt is enough for 100β/β1β=β100 toasts. However, there are 3 friends in the group, so the answer is *min*(6,β80,β100)β/β3β=β2.
| 500
|
[
{
"input": "3 4 5 10 8 100 3 1",
"output": "2"
},
{
"input": "5 100 10 1 19 90 4 3",
"output": "3"
},
{
"input": "10 1000 1000 25 23 1 50 1",
"output": "0"
},
{
"input": "1 7 4 5 5 8 3 2",
"output": "4"
},
{
"input": "2 3 3 5 5 10 1 3",
"output": "1"
},
{
"input": "2 6 4 5 6 5 1 3",
"output": "0"
},
{
"input": "1 7 3 5 3 6 2 1",
"output": "6"
},
{
"input": "2 4 5 4 5 7 3 2",
"output": "1"
},
{
"input": "2 3 6 5 7 8 2 1",
"output": "4"
},
{
"input": "1 4 5 5 3 10 3 1",
"output": "6"
},
{
"input": "1 4 6 7 3 5 1 3",
"output": "1"
},
{
"input": "1 6 5 5 5 8 3 1",
"output": "8"
},
{
"input": "1 7 5 3 3 9 2 1",
"output": "9"
},
{
"input": "3 5 3 7 6 10 3 1",
"output": "1"
},
{
"input": "3 6 3 5 3 6 3 1",
"output": "2"
},
{
"input": "1 7 5 5 5 5 2 2",
"output": "2"
},
{
"input": "2 5 3 5 6 9 2 1",
"output": "3"
},
{
"input": "3 4 3 5 3 6 2 1",
"output": "2"
},
{
"input": "1 5 5 4 7 6 3 1",
"output": "6"
},
{
"input": "2 3 7 6 5 9 3 1",
"output": "3"
},
{
"input": "2 6 5 3 3 8 1 1",
"output": "4"
},
{
"input": "2 4 7 3 4 10 2 1",
"output": "5"
},
{
"input": "1 1000 1000 1000 1000 1000 1 1",
"output": "1000"
},
{
"input": "17 1000 1000 1000 1000 1000 3 7",
"output": "8"
},
{
"input": "115 1000 1000 1000 1000 1000 17 15",
"output": "0"
},
{
"input": "1 587 981 1 2 1 1 1",
"output": "1"
},
{
"input": "1 1 2 1 2 2 1 1",
"output": "2"
}
] | 1,676,301,598
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 27
| 92
| 0
|
# import math
# import re
# import random
def solve():
t=1
# t = int(input())
while(t):
t-=1
#n = int(input())
n, k, l, c, d, salt, nl, np = [int(x) for x in input().split()]
# s = input()
slice_of_lime = c * d
mili_drink = k * l
toast_by_lime = int(slice_of_lime/1)
toast_by_drinks = int(mili_drink/nl)
toast_by_salt = int(salt/np)
print(int(min(toast_by_drinks, toast_by_lime, toast_by_salt)/n))
solve()
|
Title: Soft Drinking
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
This winter is so cold in Nvodsk! A group of *n* friends decided to buy *k* bottles of a soft drink called "Take-It-Light" to warm up a bit. Each bottle has *l* milliliters of the drink. Also they bought *c* limes and cut each of them into *d* slices. After that they found *p* grams of salt.
To make a toast, each friend needs *nl* milliliters of the drink, a slice of lime and *np* grams of salt. The friends want to make as many toasts as they can, provided they all drink the same amount. How many toasts can each friend make?
Input Specification:
The first and only line contains positive integers *n*, *k*, *l*, *c*, *d*, *p*, *nl*, *np*, not exceeding 1000 and no less than 1. The numbers are separated by exactly one space.
Output Specification:
Print a single integer β the number of toasts each friend can make.
Demo Input:
['3 4 5 10 8 100 3 1\n', '5 100 10 1 19 90 4 3\n', '10 1000 1000 25 23 1 50 1\n']
Demo Output:
['2\n', '3\n', '0\n']
Note:
A comment to the first sample:
Overall the friends have 4β*β5β=β20 milliliters of the drink, it is enough to make 20β/β3β=β6 toasts. The limes are enough for 10β*β8β=β80 toasts and the salt is enough for 100β/β1β=β100 toasts. However, there are 3 friends in the group, so the answer is *min*(6,β80,β100)β/β3β=β2.
|
```python
# import math
# import re
# import random
def solve():
t=1
# t = int(input())
while(t):
t-=1
#n = int(input())
n, k, l, c, d, salt, nl, np = [int(x) for x in input().split()]
# s = input()
slice_of_lime = c * d
mili_drink = k * l
toast_by_lime = int(slice_of_lime/1)
toast_by_drinks = int(mili_drink/nl)
toast_by_salt = int(salt/np)
print(int(min(toast_by_drinks, toast_by_lime, toast_by_salt)/n))
solve()
```
| 3
|
|
460
|
A
|
Vasya and Socks
|
PROGRAMMING
| 900
|
[
"brute force",
"implementation",
"math"
] | null | null |
Vasya has *n* pairs of socks. In the morning of each day Vasya has to put on a pair of socks before he goes to school. When he comes home in the evening, Vasya takes off the used socks and throws them away. Every *m*-th day (at days with numbers *m*,<=2*m*,<=3*m*,<=...) mom buys a pair of socks to Vasya. She does it late in the evening, so that Vasya cannot put on a new pair of socks before the next day. How many consecutive days pass until Vasya runs out of socks?
|
The single line contains two integers *n* and *m* (1<=β€<=*n*<=β€<=100;Β 2<=β€<=*m*<=β€<=100), separated by a space.
|
Print a single integer β the answer to the problem.
|
[
"2 2\n",
"9 3\n"
] |
[
"3\n",
"13\n"
] |
In the first sample Vasya spends the first two days wearing the socks that he had initially. Then on day three he puts on the socks that were bought on day two.
In the second sample Vasya spends the first nine days wearing the socks that he had initially. Then he spends three days wearing the socks that were bought on the third, sixth and ninth days. Than he spends another day wearing the socks that were bought on the twelfth day.
| 500
|
[
{
"input": "2 2",
"output": "3"
},
{
"input": "9 3",
"output": "13"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "2 3",
"output": "2"
},
{
"input": "1 99",
"output": "1"
},
{
"input": "4 4",
"output": "5"
},
{
"input": "10 2",
"output": "19"
},
{
"input": "10 9",
"output": "11"
},
{
"input": "100 100",
"output": "101"
},
{
"input": "2 27",
"output": "2"
},
{
"input": "99 100",
"output": "99"
},
{
"input": "99 2",
"output": "197"
},
{
"input": "100 3",
"output": "149"
},
{
"input": "98 3",
"output": "146"
},
{
"input": "4 4",
"output": "5"
},
{
"input": "100 2",
"output": "199"
},
{
"input": "62 4",
"output": "82"
},
{
"input": "99 10",
"output": "109"
},
{
"input": "100 5",
"output": "124"
},
{
"input": "80 80",
"output": "81"
},
{
"input": "95 16",
"output": "101"
},
{
"input": "75 16",
"output": "79"
},
{
"input": "99 74",
"output": "100"
},
{
"input": "20 21",
"output": "20"
},
{
"input": "52 96",
"output": "52"
},
{
"input": "24 5",
"output": "29"
}
] | 1,638,991,244
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 26
| 62
| 0
|
n, m = input().split()
n = int(n)
m = int(m)
rooz = 0
tedad_joorab = n
while tedad_joorab != 0:
rooz += 1
tedad_joorab -= 1
if rooz % m == 0:
tedad_joorab += 1
print(rooz)
|
Title: Vasya and Socks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has *n* pairs of socks. In the morning of each day Vasya has to put on a pair of socks before he goes to school. When he comes home in the evening, Vasya takes off the used socks and throws them away. Every *m*-th day (at days with numbers *m*,<=2*m*,<=3*m*,<=...) mom buys a pair of socks to Vasya. She does it late in the evening, so that Vasya cannot put on a new pair of socks before the next day. How many consecutive days pass until Vasya runs out of socks?
Input Specification:
The single line contains two integers *n* and *m* (1<=β€<=*n*<=β€<=100;Β 2<=β€<=*m*<=β€<=100), separated by a space.
Output Specification:
Print a single integer β the answer to the problem.
Demo Input:
['2 2\n', '9 3\n']
Demo Output:
['3\n', '13\n']
Note:
In the first sample Vasya spends the first two days wearing the socks that he had initially. Then on day three he puts on the socks that were bought on day two.
In the second sample Vasya spends the first nine days wearing the socks that he had initially. Then he spends three days wearing the socks that were bought on the third, sixth and ninth days. Than he spends another day wearing the socks that were bought on the twelfth day.
|
```python
n, m = input().split()
n = int(n)
m = int(m)
rooz = 0
tedad_joorab = n
while tedad_joorab != 0:
rooz += 1
tedad_joorab -= 1
if rooz % m == 0:
tedad_joorab += 1
print(rooz)
```
| 3
|
|
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*<=β€<=*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<=β€<=*n*,<=*m*<=β€<=100). The second line contains *n* space-separated positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=β€<=*a**i*<=β€<=100) β 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<=β€<=*b**i*<=β€<=100) β 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"
}
] | 1,380,642,306
| 906
|
Python 3
|
OK
|
TESTS
| 45
| 92
| 0
|
a, b = map(int, input().split())
P = list(map(int, input().split()))
flag = 0
M = P[0]
m = P[0]
for i in range(a):
M = max(M, P[i])
m = min(m, P[i])
s = max(M, 2 * m)
A = list(map(int, input().split()))
for i in range(b):
if A[i] <= s:
flag = 1
if flag == 1:
print(-1)
else:
print(s)
|
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*<=β€<=*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<=β€<=*n*,<=*m*<=β€<=100). The second line contains *n* space-separated positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=β€<=*a**i*<=β€<=100) β 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<=β€<=*b**i*<=β€<=100) β 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
a, b = map(int, input().split())
P = list(map(int, input().split()))
flag = 0
M = P[0]
m = P[0]
for i in range(a):
M = max(M, P[i])
m = min(m, P[i])
s = max(M, 2 * m)
A = list(map(int, input().split()))
for i in range(b):
if A[i] <= s:
flag = 1
if flag == 1:
print(-1)
else:
print(s)
```
| 3
|
|
381
|
A
|
Sereja and Dima
|
PROGRAMMING
| 800
|
[
"greedy",
"implementation",
"two pointers"
] | null | null |
Sereja and Dima play a game. The rules of the game are very simple. The players have *n* cards in a row. Each card contains a number, all numbers on the cards are distinct. The players take turns, Sereja moves first. During his turn a player can take one card: either the leftmost card in a row, or the rightmost one. The game ends when there is no more cards. The player who has the maximum sum of numbers on his cards by the end of the game, wins.
Sereja and Dima are being greedy. Each of them chooses the card with the larger number during his move.
Inna is a friend of Sereja and Dima. She knows which strategy the guys are using, so she wants to determine the final score, given the initial state of the game. Help her.
|
The first line contains integer *n* (1<=β€<=*n*<=β€<=1000) β the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000.
|
On a single line, print two integers. The first number is the number of Sereja's points at the end of the game, the second number is the number of Dima's points at the end of the game.
|
[
"4\n4 1 2 10\n",
"7\n1 2 3 4 5 6 7\n"
] |
[
"12 5\n",
"16 12\n"
] |
In the first sample Sereja will take cards with numbers 10 and 2, so Sereja's sum is 12. Dima will take cards with numbers 4 and 1, so Dima's sum is 5.
| 500
|
[
{
"input": "4\n4 1 2 10",
"output": "12 5"
},
{
"input": "7\n1 2 3 4 5 6 7",
"output": "16 12"
},
{
"input": "42\n15 29 37 22 16 5 26 31 6 32 19 3 45 36 33 14 25 20 48 7 42 11 24 28 9 18 8 21 47 17 38 40 44 4 35 1 43 39 41 27 12 13",
"output": "613 418"
},
{
"input": "43\n32 1 15 48 38 26 25 14 20 44 11 30 3 42 49 19 18 46 5 45 10 23 34 9 29 41 2 52 6 17 35 4 50 22 33 51 7 28 47 13 39 37 24",
"output": "644 500"
},
{
"input": "1\n3",
"output": "3 0"
},
{
"input": "45\n553 40 94 225 415 471 126 190 647 394 515 303 189 159 308 6 139 132 326 78 455 75 85 295 135 613 360 614 351 228 578 259 258 591 444 29 33 463 561 174 368 183 140 168 646",
"output": "6848 6568"
},
{
"input": "44\n849 373 112 307 479 608 856 769 526 82 168 143 573 762 115 501 688 36 214 450 396 496 236 309 287 786 397 43 811 141 745 846 350 270 276 677 420 459 403 722 267 54 394 727",
"output": "9562 9561"
},
{
"input": "35\n10 15 18 1 28 16 2 33 6 22 23 4 9 25 35 8 7 26 3 20 30 14 31 19 27 32 11 5 29 24 21 34 13 17 12",
"output": "315 315"
},
{
"input": "17\n580 376 191 496 73 44 520 357 483 149 81 178 514 300 216 598 304",
"output": "3238 2222"
},
{
"input": "30\n334 443 223 424 168 549 189 303 429 559 516 220 459 134 344 346 316 446 209 148 487 526 69 286 102 366 518 280 392 325",
"output": "5246 4864"
},
{
"input": "95\n122 29 188 265 292 287 183 225 222 187 155 256 64 148 173 278 218 136 290 17 31 130 2 87 57 283 255 280 68 166 174 142 102 39 116 206 288 154 26 78 296 172 184 232 77 91 277 8 249 186 94 93 207 251 257 195 101 299 193 124 293 65 58 35 24 302 220 189 252 125 27 284 247 182 141 103 198 97 234 83 281 216 85 180 267 236 109 143 149 239 79 300 191 244 71",
"output": "8147 7807"
},
{
"input": "1\n1",
"output": "1 0"
}
] | 1,691,307,944
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 34
| 46
| 0
|
n=int(input(""))
l=list(map(int,input().split()))
i=0
j=len(l)-1
n=0
m=0
count=0
while i<=j:
if count%2==0:
if l[i]==max(l[i],l[j]):
n=n+l[i]
i=i+1
else:
n=n+l[j]
j=j-1
else:
if l[i]==max(l[i],l[j]):
m=m+l[i]
i=i+1
else:
m=m+l[j]
j=j-1
count=count+1
print(n,m)
|
Title: Sereja and Dima
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sereja and Dima play a game. The rules of the game are very simple. The players have *n* cards in a row. Each card contains a number, all numbers on the cards are distinct. The players take turns, Sereja moves first. During his turn a player can take one card: either the leftmost card in a row, or the rightmost one. The game ends when there is no more cards. The player who has the maximum sum of numbers on his cards by the end of the game, wins.
Sereja and Dima are being greedy. Each of them chooses the card with the larger number during his move.
Inna is a friend of Sereja and Dima. She knows which strategy the guys are using, so she wants to determine the final score, given the initial state of the game. Help her.
Input Specification:
The first line contains integer *n* (1<=β€<=*n*<=β€<=1000) β the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000.
Output Specification:
On a single line, print two integers. The first number is the number of Sereja's points at the end of the game, the second number is the number of Dima's points at the end of the game.
Demo Input:
['4\n4 1 2 10\n', '7\n1 2 3 4 5 6 7\n']
Demo Output:
['12 5\n', '16 12\n']
Note:
In the first sample Sereja will take cards with numbers 10 and 2, so Sereja's sum is 12. Dima will take cards with numbers 4 and 1, so Dima's sum is 5.
|
```python
n=int(input(""))
l=list(map(int,input().split()))
i=0
j=len(l)-1
n=0
m=0
count=0
while i<=j:
if count%2==0:
if l[i]==max(l[i],l[j]):
n=n+l[i]
i=i+1
else:
n=n+l[j]
j=j-1
else:
if l[i]==max(l[i],l[j]):
m=m+l[i]
i=i+1
else:
m=m+l[j]
j=j-1
count=count+1
print(n,m)
```
| 3
|
|
593
|
A
|
2Char
|
PROGRAMMING
| 1,200
|
[
"brute force",
"implementation"
] | null | null |
Andrew often reads articles in his favorite magazine 2Char. The main feature of these articles is that each of them uses at most two distinct letters. Andrew decided to send an article to the magazine, but as he hasn't written any article, he just decided to take a random one from magazine 26Char. However, before sending it to the magazine 2Char, he needs to adapt the text to the format of the journal. To do so, he removes some words from the chosen article, in such a way that the remaining text can be written using no more than two distinct letters.
Since the payment depends from the number of non-space characters in the article, Andrew wants to keep the words with the maximum total length.
|
The first line of the input contains number *n* (1<=β€<=*n*<=β€<=100)Β β the number of words in the article chosen by Andrew. Following are *n* lines, each of them contains one word. All the words consist only of small English letters and their total length doesn't exceed 1000. The words are not guaranteed to be distinct, in this case you are allowed to use a word in the article as many times as it appears in the input.
|
Print a single integerΒ β the maximum possible total length of words in Andrew's article.
|
[
"4\nabb\ncacc\naaa\nbbb\n",
"5\na\na\nbcbcb\ncdecdecdecdecdecde\naaaa\n"
] |
[
"9",
"6"
] |
In the first sample the optimal way to choose words is {'abb', 'aaa', 'bbb'}.
In the second sample the word 'cdecdecdecdecdecde' consists of three distinct letters, and thus cannot be used in the article. The optimal answer is {'a', 'a', 'aaaa'}.
| 250
|
[
{
"input": "4\nabb\ncacc\naaa\nbbb",
"output": "9"
},
{
"input": "5\na\na\nbcbcb\ncdecdecdecdecdecde\naaaa",
"output": "6"
},
{
"input": "1\na",
"output": "1"
},
{
"input": "2\nz\nz",
"output": "2"
},
{
"input": "5\nabcde\nfghij\nklmno\npqrst\nuvwxy",
"output": "0"
},
{
"input": "6\ngggggg\ngggggg\ngggggg\ngggggg\ngggggg\ngggggg",
"output": "36"
},
{
"input": "6\naaaaaa\naaaaaa\nbbbbbb\nbbbbbb\naaabbb\nababab",
"output": "36"
},
{
"input": "1\nabc",
"output": "0"
},
{
"input": "2\nabc\nbca",
"output": "0"
},
{
"input": "3\nab\nba\nzzz",
"output": "4"
},
{
"input": "3\nab\nba\nzzzzz",
"output": "5"
},
{
"input": "5\nzzz\nzzzz\nzz\nz\naaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "37"
},
{
"input": "26\nq\nw\ne\nr\nt\ny\nu\ni\no\np\na\ns\nd\nf\ng\nh\nj\nk\nl\nz\nx\nc\nv\nb\nn\nm",
"output": "2"
},
{
"input": "5\nzzz\nzzzz\nzz\nz\naaaaaaaaaaaaaaaaaaaaaaaaaaaf",
"output": "28"
},
{
"input": "7\npavel\nerika\nalexxxxxxx\ngracio\nzhenya\nsudarev\nchelyaba",
"output": "0"
},
{
"input": "31\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml",
"output": "0"
},
{
"input": "5\nzloyfreid\ngraciocode\nschooldiary\nkazakov\nevgesha",
"output": "0"
},
{
"input": "4\nurkop\nvisualac\ngnutl\nwtf",
"output": "0"
},
{
"input": "3\naa\nb\nccc",
"output": "5"
},
{
"input": "3\na\nbd\ncaaaaaaa",
"output": "9"
},
{
"input": "4\naa\nax\nay\nxxxx",
"output": "8"
},
{
"input": "5\nc\nbb\ne\ndd\nf",
"output": "4"
},
{
"input": "2\naaaaa\naaaaa",
"output": "10"
}
] | 1,446,671,611
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 49
| 124
| 2,560,000
|
import string
alphabet = string.ascii_lowercase
class Breakdown:
def __init__(self, word):
self.word = word
self.length = len(word)
self.count = { ch: word.count(ch) for ch in alphabet }
n = int(input())
breakdowns = [ Breakdown(input().strip()) for i in range(n) ]
best = 0
for i, a in enumerate(alphabet):
for j in range(i + 1, len(alphabet)):
b = alphabet[j]
count = 0
for breakdown in breakdowns:
if breakdown.count[a] + breakdown.count[b] == breakdown.length:
count += breakdown.length
best = max(best, count)
print(best)
|
Title: 2Char
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Andrew often reads articles in his favorite magazine 2Char. The main feature of these articles is that each of them uses at most two distinct letters. Andrew decided to send an article to the magazine, but as he hasn't written any article, he just decided to take a random one from magazine 26Char. However, before sending it to the magazine 2Char, he needs to adapt the text to the format of the journal. To do so, he removes some words from the chosen article, in such a way that the remaining text can be written using no more than two distinct letters.
Since the payment depends from the number of non-space characters in the article, Andrew wants to keep the words with the maximum total length.
Input Specification:
The first line of the input contains number *n* (1<=β€<=*n*<=β€<=100)Β β the number of words in the article chosen by Andrew. Following are *n* lines, each of them contains one word. All the words consist only of small English letters and their total length doesn't exceed 1000. The words are not guaranteed to be distinct, in this case you are allowed to use a word in the article as many times as it appears in the input.
Output Specification:
Print a single integerΒ β the maximum possible total length of words in Andrew's article.
Demo Input:
['4\nabb\ncacc\naaa\nbbb\n', '5\na\na\nbcbcb\ncdecdecdecdecdecde\naaaa\n']
Demo Output:
['9', '6']
Note:
In the first sample the optimal way to choose words is {'abb', 'aaa', 'bbb'}.
In the second sample the word 'cdecdecdecdecdecde' consists of three distinct letters, and thus cannot be used in the article. The optimal answer is {'a', 'a', 'aaaa'}.
|
```python
import string
alphabet = string.ascii_lowercase
class Breakdown:
def __init__(self, word):
self.word = word
self.length = len(word)
self.count = { ch: word.count(ch) for ch in alphabet }
n = int(input())
breakdowns = [ Breakdown(input().strip()) for i in range(n) ]
best = 0
for i, a in enumerate(alphabet):
for j in range(i + 1, len(alphabet)):
b = alphabet[j]
count = 0
for breakdown in breakdowns:
if breakdown.count[a] + breakdown.count[b] == breakdown.length:
count += breakdown.length
best = max(best, count)
print(best)
```
| 3
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
Andryusha is an orderly boy and likes to keep things in their place.
Today he faced a problem to put his socks in the wardrobe. He has *n* distinct pairs of socks which are initially in a bag. The pairs are numbered from 1 to *n*. Andryusha wants to put paired socks together and put them in the wardrobe. He takes the socks one by one from the bag, and for each sock he looks whether the pair of this sock has been already took out of the bag, or not. If not (that means the pair of this sock is still in the bag), he puts the current socks on the table in front of him. Otherwise, he puts both socks from the pair to the wardrobe.
Andryusha remembers the order in which he took the socks from the bag. Can you tell him what is the maximum number of socks that were on the table at the same time?
|
The first line contains the single integer *n* (1<=β€<=*n*<=β€<=105)Β β the number of sock pairs.
The second line contains 2*n* integers *x*1,<=*x*2,<=...,<=*x*2*n* (1<=β€<=*x**i*<=β€<=*n*), which describe the order in which Andryusha took the socks from the bag. More precisely, *x**i* means that the *i*-th sock Andryusha took out was from pair *x**i*.
It is guaranteed that Andryusha took exactly two socks of each pair.
|
Print single integerΒ β the maximum number of socks that were on the table at the same time.
|
[
"1\n1 1\n",
"3\n2 1 1 3 2 3\n"
] |
[
"1\n",
"2\n"
] |
In the first example Andryusha took a sock from the first pair and put it on the table. Then he took the next sock which is from the first pair as well, so he immediately puts both socks to the wardrobe. Thus, at most one sock was on the table at the same time.
In the second example Andryusha behaved as follows:
- Initially the table was empty, he took out a sock from pair 2 and put it on the table. - Sock (2) was on the table. Andryusha took out a sock from pair 1 and put it on the table. - Socks (1,β2) were on the table. Andryusha took out a sock from pair 1, and put this pair into the wardrobe. - Sock (2) was on the table. Andryusha took out a sock from pair 3 and put it on the table. - Socks (2,β3) were on the table. Andryusha took out a sock from pair 2, and put this pair into the wardrobe. - Sock (3) was on the table. Andryusha took out a sock from pair 3 and put this pair into the wardrobe.
| 0
|
[
{
"input": "1\n1 1",
"output": "1"
},
{
"input": "3\n2 1 1 3 2 3",
"output": "2"
},
{
"input": "5\n5 1 3 2 4 3 1 2 4 5",
"output": "5"
},
{
"input": "10\n4 2 6 3 4 8 7 1 1 5 2 10 6 8 3 5 10 9 9 7",
"output": "6"
},
{
"input": "50\n30 47 31 38 37 50 36 43 9 23 2 2 15 31 14 49 9 16 6 44 27 14 5 6 3 47 25 26 1 35 3 15 24 19 8 46 49 41 4 26 40 28 42 11 34 35 46 18 7 28 18 40 19 42 4 41 38 48 50 12 29 39 33 17 25 22 22 21 36 45 27 30 20 7 13 29 39 44 21 8 37 45 34 1 20 10 11 17 33 12 43 13 10 16 48 24 32 5 23 32",
"output": "25"
},
{
"input": "50\n1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 15 15 16 16 17 17 18 18 19 19 20 20 21 21 22 22 23 23 24 24 25 25 26 26 27 27 28 28 29 29 30 30 31 31 32 32 33 33 34 34 35 35 36 36 37 37 38 38 39 39 40 40 41 41 42 42 43 43 44 44 45 45 46 46 47 47 48 48 49 49 50 50",
"output": "1"
},
{
"input": "50\n50 50 49 49 48 48 47 47 46 46 45 45 44 44 43 43 42 42 41 41 40 40 39 39 38 38 37 37 36 36 35 35 34 34 33 33 32 32 31 31 30 30 29 29 28 28 27 27 26 26 25 25 24 24 23 23 22 22 21 21 20 20 19 19 18 18 17 17 16 16 15 15 14 14 13 13 12 12 11 11 10 10 9 9 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1",
"output": "1"
},
{
"input": "50\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 1 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",
"output": "50"
},
{
"input": "50\n50 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 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": "50"
},
{
"input": "10\n2 9 4 1 6 7 10 3 1 5 8 6 2 3 10 7 4 8 5 9",
"output": "9"
}
] | 1,537,956,347
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 56
| 249
| 14,028,800
|
class data:
def __init__(self,data):
self.data = data
def solve(self):
dict_1 ={}
count = solution = 0
for i in self.data:
if i in dict_1:
count = count-1
else:
dict_1[i]=1
count = count+1
solution = max(solution,count)
return solution
socks_pair = int(input())
order = list(map(int,input().strip().split()))
if 2*socks_pair == len(order):
sol = data(order)
print(sol.solve())
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Andryusha is an orderly boy and likes to keep things in their place.
Today he faced a problem to put his socks in the wardrobe. He has *n* distinct pairs of socks which are initially in a bag. The pairs are numbered from 1 to *n*. Andryusha wants to put paired socks together and put them in the wardrobe. He takes the socks one by one from the bag, and for each sock he looks whether the pair of this sock has been already took out of the bag, or not. If not (that means the pair of this sock is still in the bag), he puts the current socks on the table in front of him. Otherwise, he puts both socks from the pair to the wardrobe.
Andryusha remembers the order in which he took the socks from the bag. Can you tell him what is the maximum number of socks that were on the table at the same time?
Input Specification:
The first line contains the single integer *n* (1<=β€<=*n*<=β€<=105)Β β the number of sock pairs.
The second line contains 2*n* integers *x*1,<=*x*2,<=...,<=*x*2*n* (1<=β€<=*x**i*<=β€<=*n*), which describe the order in which Andryusha took the socks from the bag. More precisely, *x**i* means that the *i*-th sock Andryusha took out was from pair *x**i*.
It is guaranteed that Andryusha took exactly two socks of each pair.
Output Specification:
Print single integerΒ β the maximum number of socks that were on the table at the same time.
Demo Input:
['1\n1 1\n', '3\n2 1 1 3 2 3\n']
Demo Output:
['1\n', '2\n']
Note:
In the first example Andryusha took a sock from the first pair and put it on the table. Then he took the next sock which is from the first pair as well, so he immediately puts both socks to the wardrobe. Thus, at most one sock was on the table at the same time.
In the second example Andryusha behaved as follows:
- Initially the table was empty, he took out a sock from pair 2 and put it on the table. - Sock (2) was on the table. Andryusha took out a sock from pair 1 and put it on the table. - Socks (1,β2) were on the table. Andryusha took out a sock from pair 1, and put this pair into the wardrobe. - Sock (2) was on the table. Andryusha took out a sock from pair 3 and put it on the table. - Socks (2,β3) were on the table. Andryusha took out a sock from pair 2, and put this pair into the wardrobe. - Sock (3) was on the table. Andryusha took out a sock from pair 3 and put this pair into the wardrobe.
|
```python
class data:
def __init__(self,data):
self.data = data
def solve(self):
dict_1 ={}
count = solution = 0
for i in self.data:
if i in dict_1:
count = count-1
else:
dict_1[i]=1
count = count+1
solution = max(solution,count)
return solution
socks_pair = int(input())
order = list(map(int,input().strip().split()))
if 2*socks_pair == len(order):
sol = data(order)
print(sol.solve())
```
| 3
|
|
427
|
A
|
Police Recruits
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
The police department of your city has just started its journey. Initially, they donβt 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*Β (1<=β€<=*n*<=β€<=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"
}
] | 1,693,764,540
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 63
| 61
| 10,547,200
|
t=int(input())
l=list(map(int,input().split()))
u=0
p=0
for i in range(t):
if(l[i]==-1):
if(p!=0):
p=p-1
else:
u=u+1
else:
p=p+l[i]
print(u)
|
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 donβt 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*Β (1<=β€<=*n*<=β€<=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
t=int(input())
l=list(map(int,input().split()))
u=0
p=0
for i in range(t):
if(l[i]==-1):
if(p!=0):
p=p-1
else:
u=u+1
else:
p=p+l[i]
print(u)
```
| 3
|
|
721
|
A
|
One-dimensional Japanese Crossword
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Recently Adaltik discovered japanese crosswords. Japanese crossword is a picture, represented as a table sized *a*<=Γ<=*b* squares, and each square is colored white or black. There are integers to the left of the rows and to the top of the columns, encrypting the corresponding row or column. The number of integers represents how many groups of black squares there are in corresponding row or column, and the integers themselves represents the number of consecutive black squares in corresponding group (you can find more detailed explanation in Wikipedia [https://en.wikipedia.org/wiki/Japanese_crossword](https://en.wikipedia.org/wiki/Japanese_crossword)).
Adaltik decided that the general case of japanese crossword is too complicated and drew a row consisting of *n* squares (e.g. japanese crossword sized 1<=Γ<=*n*), which he wants to encrypt in the same way as in japanese crossword.
Help Adaltik find the numbers encrypting the row he drew.
|
The first line of the input contains a single integer *n* (1<=β€<=*n*<=β€<=100)Β β the length of the row. The second line of the input contains a single string consisting of *n* characters 'B' or 'W', ('B' corresponds to black square, 'W'Β β to white square in the row that Adaltik drew).
|
The first line should contain a single integer *k*Β β the number of integers encrypting the row, e.g. the number of groups of black squares in the row.
The second line should contain *k* integers, encrypting the row, e.g. corresponding to sizes of groups of consecutive black squares in the order from left to right.
|
[
"3\nBBW\n",
"5\nBWBWB\n",
"4\nWWWW\n",
"4\nBBBB\n",
"13\nWBBBBWWBWBBBW\n"
] |
[
"1\n2 ",
"3\n1 1 1 ",
"0\n",
"1\n4 ",
"3\n4 1 3 "
] |
The last sample case correspond to the picture in the statement.
| 500
|
[
{
"input": "3\nBBW",
"output": "1\n2 "
},
{
"input": "5\nBWBWB",
"output": "3\n1 1 1 "
},
{
"input": "4\nWWWW",
"output": "0"
},
{
"input": "4\nBBBB",
"output": "1\n4 "
},
{
"input": "13\nWBBBBWWBWBBBW",
"output": "3\n4 1 3 "
},
{
"input": "1\nB",
"output": "1\n1 "
},
{
"input": "2\nBB",
"output": "1\n2 "
},
{
"input": "100\nWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWB",
"output": "50\n1 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 "
},
{
"input": "1\nW",
"output": "0"
},
{
"input": "2\nWW",
"output": "0"
},
{
"input": "2\nWB",
"output": "1\n1 "
},
{
"input": "2\nBW",
"output": "1\n1 "
},
{
"input": "3\nBBB",
"output": "1\n3 "
},
{
"input": "3\nBWB",
"output": "2\n1 1 "
},
{
"input": "3\nWBB",
"output": "1\n2 "
},
{
"input": "3\nWWB",
"output": "1\n1 "
},
{
"input": "3\nWBW",
"output": "1\n1 "
},
{
"input": "3\nBWW",
"output": "1\n1 "
},
{
"input": "3\nWWW",
"output": "0"
},
{
"input": "100\nBBBWWWWWWBBWWBBWWWBBWBBBBBBBBBBBWBBBWBBWWWBBWWBBBWBWWBBBWWBBBWBBBBBWWWBWWBBWWWWWWBWBBWWBWWWBWBWWWWWB",
"output": "21\n3 2 2 2 11 3 2 2 3 1 3 3 5 1 2 1 2 1 1 1 1 "
},
{
"input": "5\nBBBWB",
"output": "2\n3 1 "
},
{
"input": "5\nBWWWB",
"output": "2\n1 1 "
},
{
"input": "5\nWWWWB",
"output": "1\n1 "
},
{
"input": "5\nBWWWW",
"output": "1\n1 "
},
{
"input": "5\nBBBWW",
"output": "1\n3 "
},
{
"input": "5\nWWBBB",
"output": "1\n3 "
},
{
"input": "10\nBBBBBWWBBB",
"output": "2\n5 3 "
},
{
"input": "10\nBBBBWBBWBB",
"output": "3\n4 2 2 "
},
{
"input": "20\nBBBBBWWBWBBWBWWBWBBB",
"output": "6\n5 1 2 1 1 3 "
},
{
"input": "20\nBBBWWWWBBWWWBWBWWBBB",
"output": "5\n3 2 1 1 3 "
},
{
"input": "20\nBBBBBBBBWBBBWBWBWBBB",
"output": "5\n8 3 1 1 3 "
},
{
"input": "20\nBBBWBWBWWWBBWWWWBWBB",
"output": "6\n3 1 1 2 1 2 "
},
{
"input": "40\nBBBBBBWWWWBWBWWWBWWWWWWWWWWWBBBBBBBBBBBB",
"output": "5\n6 1 1 1 12 "
},
{
"input": "40\nBBBBBWBWWWBBWWWBWBWWBBBBWWWWBWBWBBBBBBBB",
"output": "9\n5 1 2 1 1 4 1 1 8 "
},
{
"input": "50\nBBBBBBBBBBBWWWWBWBWWWWBBBBBBBBWWWWWWWBWWWWBWBBBBBB",
"output": "7\n11 1 1 8 1 1 6 "
},
{
"input": "50\nWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWW",
"output": "0"
},
{
"input": "50\nBBBBBWWWWWBWWWBWWWWWBWWWBWWWWWWBBWBBWWWWBWWWWWWWBW",
"output": "9\n5 1 1 1 1 2 2 1 1 "
},
{
"input": "50\nWWWWBWWBWWWWWWWWWWWWWWWWWWWWWWWWWBWBWBWWWWWWWBBBBB",
"output": "6\n1 1 1 1 1 5 "
},
{
"input": "50\nBBBBBWBWBWWBWBWWWWWWBWBWBWWWWWWWWWWWWWBWBWWWWBWWWB",
"output": "12\n5 1 1 1 1 1 1 1 1 1 1 1 "
},
{
"input": "50\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB",
"output": "1\n50 "
},
{
"input": "100\nBBBBBBBBBBBWBWWWWBWWBBWBBWWWWWWWWWWBWBWWBWWWWWWWWWWWBBBWWBBWWWWWBWBWWWWBWWWWWWWWWWWBWWWWWBBBBBBBBBBB",
"output": "15\n11 1 1 2 2 1 1 1 3 2 1 1 1 1 11 "
},
{
"input": "100\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB",
"output": "1\n100 "
},
{
"input": "100\nBBBBBBBBBBBBBBBBBBBBWBWBWWWWWBWWWWWWWWWWWWWWBBWWWBWWWWBWWBWWWWWWBWWWWWWWWWWWWWBWBBBBBBBBBBBBBBBBBBBB",
"output": "11\n20 1 1 1 2 1 1 1 1 1 20 "
},
{
"input": "100\nBBBBWWWWWWWWWWWWWWWWWWWWWWWWWBWBWWWWWBWBWWWWWWBBWWWWWWWWWWWWBWWWWBWWWWWWWWWWWWBWWWWWWWBWWWWWWWBBBBBB",
"output": "11\n4 1 1 1 1 2 1 1 1 1 6 "
},
{
"input": "5\nBWBWB",
"output": "3\n1 1 1 "
},
{
"input": "10\nWWBWWWBWBB",
"output": "3\n1 1 2 "
},
{
"input": "50\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB",
"output": "1\n50 "
},
{
"input": "50\nBBBBBBBBBBBBBBBBBWWBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB",
"output": "2\n17 31 "
},
{
"input": "100\nBBBBBBBBBBBBBBBBBBBBBBBBWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB",
"output": "2\n24 42 "
},
{
"input": "90\nWWBWWBWBBWBBWWBWBWBBBWBWBBBWBWBWBWBWBWBWBWBBBBBWBBWWWWBWBBWBWWBBBWBWBWWBWBWBWBWWWWWWBWBBBB",
"output": "30\n1 1 2 2 1 1 3 1 3 1 1 1 1 1 1 1 5 2 1 2 1 3 1 1 1 1 1 1 1 4 "
},
{
"input": "100\nBWWWBWBWBBBBBWBWWBWBWWWBWBWBWWBBWWBBBWBBBWWBWBWWBBBBWBWBBBWBWBBWWWWWWBWWBBBBWBWBWWBWBWWWBWBWWBWBWWWB",
"output": "31\n1 1 1 5 1 1 1 1 1 1 2 3 3 1 1 4 1 3 1 2 1 4 1 1 1 1 1 1 1 1 1 "
},
{
"input": "90\nWBWBBBBBBWWWBBWWBWWWBBWWBWWWBWBBWBWBBWWWWBWBWBBWBBWBWWWBBWBBWWWWBWBBWWWBBBWBBWBWBBBBWWBWWB",
"output": "25\n1 6 2 1 2 1 1 2 1 2 1 1 2 2 1 2 2 1 2 3 2 1 4 1 1 "
},
{
"input": "80\nBBWWBBBWBBWWWWBBWBWBBWWWWWBWBBWWBWBWBWBWBWWBWWBWWWBWBBWBBWBBWBBBWWBBBBBBBWBBBWBB",
"output": "23\n2 3 2 2 1 2 1 2 1 1 1 1 1 1 1 1 2 2 2 3 7 3 2 "
},
{
"input": "65\nWWWWBWWWBBBBBWWWWWWBBBWWBBBBWWWWWWWWBBBWWWWBWBWWBBWWWWBWWWBBWBBBB",
"output": "11\n1 5 3 4 3 1 1 2 1 2 4 "
}
] | 1,592,813,988
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 54
| 140
| 307,200
|
# One-dimensional Japanese Crossword
def black_white(s):
if s.count('B') == 0:
print('0')
return
if s.count('W') == 0:
print(1)
print(s.count('B'))
return
ans = []
count = 0
i = 0
while i < len(s):
if s[i] == 'B':
count += 1
else:
if count != 0:
ans.append(str(count))
count = 0
i += 1
if count != 0:
ans.append(str(count))
val = ' '.join(ans)
print(len(ans))
print(val)
n = int(input())
s = input()
black_white(s)
|
Title: One-dimensional Japanese Crossword
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently Adaltik discovered japanese crosswords. Japanese crossword is a picture, represented as a table sized *a*<=Γ<=*b* squares, and each square is colored white or black. There are integers to the left of the rows and to the top of the columns, encrypting the corresponding row or column. The number of integers represents how many groups of black squares there are in corresponding row or column, and the integers themselves represents the number of consecutive black squares in corresponding group (you can find more detailed explanation in Wikipedia [https://en.wikipedia.org/wiki/Japanese_crossword](https://en.wikipedia.org/wiki/Japanese_crossword)).
Adaltik decided that the general case of japanese crossword is too complicated and drew a row consisting of *n* squares (e.g. japanese crossword sized 1<=Γ<=*n*), which he wants to encrypt in the same way as in japanese crossword.
Help Adaltik find the numbers encrypting the row he drew.
Input Specification:
The first line of the input contains a single integer *n* (1<=β€<=*n*<=β€<=100)Β β the length of the row. The second line of the input contains a single string consisting of *n* characters 'B' or 'W', ('B' corresponds to black square, 'W'Β β to white square in the row that Adaltik drew).
Output Specification:
The first line should contain a single integer *k*Β β the number of integers encrypting the row, e.g. the number of groups of black squares in the row.
The second line should contain *k* integers, encrypting the row, e.g. corresponding to sizes of groups of consecutive black squares in the order from left to right.
Demo Input:
['3\nBBW\n', '5\nBWBWB\n', '4\nWWWW\n', '4\nBBBB\n', '13\nWBBBBWWBWBBBW\n']
Demo Output:
['1\n2 ', '3\n1 1 1 ', '0\n', '1\n4 ', '3\n4 1 3 ']
Note:
The last sample case correspond to the picture in the statement.
|
```python
# One-dimensional Japanese Crossword
def black_white(s):
if s.count('B') == 0:
print('0')
return
if s.count('W') == 0:
print(1)
print(s.count('B'))
return
ans = []
count = 0
i = 0
while i < len(s):
if s[i] == 'B':
count += 1
else:
if count != 0:
ans.append(str(count))
count = 0
i += 1
if count != 0:
ans.append(str(count))
val = ' '.join(ans)
print(len(ans))
print(val)
n = int(input())
s = input()
black_white(s)
```
| 3
|
|
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 β (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 β 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"
}
] | 1,649,980,577
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 102
| 62
| 0
|
n = input()
m = input()
op = []
for i in range(len(n)):
if n[i] != m[i]:
op.append('1')
else:
op.append('0')
print("".join(op))
|
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 β (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 β 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
n = input()
m = input()
op = []
for i in range(len(n)):
if n[i] != m[i]:
op.append('1')
else:
op.append('0')
print("".join(op))
```
| 3.9845
|
570
|
A
|
Elections
|
PROGRAMMING
| 1,100
|
[
"implementation"
] | null | null |
The country of Byalechinsk is running elections involving *n* candidates. The country consists of *m* cities. We know how many people in each city voted for each candidate.
The electoral system in the country is pretty unusual. At the first stage of elections the votes are counted for each city: it is assumed that in each city won the candidate who got the highest number of votes in this city, and if several candidates got the maximum number of votes, then the winner is the one with a smaller index.
At the second stage of elections the winner is determined by the same principle over the cities: the winner of the elections is the candidate who won in the maximum number of cities, and among those who got the maximum number of cities the winner is the one with a smaller index.
Determine who will win the elections.
|
The first line of the input contains two integers *n*, *m* (1<=β€<=*n*,<=*m*<=β€<=100) β the number of candidates and of cities, respectively.
Each of the next *m* lines contains *n* non-negative integers, the *j*-th number in the *i*-th line *a**ij* (1<=β€<=*j*<=β€<=*n*, 1<=β€<=*i*<=β€<=*m*, 0<=β€<=*a**ij*<=β€<=109) denotes the number of votes for candidate *j* in city *i*.
It is guaranteed that the total number of people in all the cities does not exceed 109.
|
Print a single number β the index of the candidate who won the elections. The candidates are indexed starting from one.
|
[
"3 3\n1 2 3\n2 3 1\n1 2 1\n",
"3 4\n10 10 3\n5 1 6\n2 2 2\n1 5 7\n"
] |
[
"2",
"1"
] |
Note to the first sample test. At the first stage city 1 chosen candidate 3, city 2 chosen candidate 2, city 3 chosen candidate 2. The winner is candidate 2, he gained 2 votes.
Note to the second sample test. At the first stage in city 1 candidates 1 and 2 got the same maximum number of votes, but candidate 1 has a smaller index, so the city chose candidate 1. City 2 chosen candidate 3. City 3 chosen candidate 1, due to the fact that everyone has the same number of votes, and 1 has the smallest index. City 4 chosen the candidate 3. On the second stage the same number of cities chose candidates 1 and 3. The winner is candidate 1, the one with the smaller index.
| 500
|
[
{
"input": "3 3\n1 2 3\n2 3 1\n1 2 1",
"output": "2"
},
{
"input": "3 4\n10 10 3\n5 1 6\n2 2 2\n1 5 7",
"output": "1"
},
{
"input": "1 3\n5\n3\n2",
"output": "1"
},
{
"input": "3 1\n1 2 3",
"output": "3"
},
{
"input": "3 1\n100 100 100",
"output": "1"
},
{
"input": "2 2\n1 2\n2 1",
"output": "1"
},
{
"input": "2 2\n2 1\n2 1",
"output": "1"
},
{
"input": "2 2\n1 2\n1 2",
"output": "2"
},
{
"input": "3 3\n0 0 0\n1 1 1\n2 2 2",
"output": "1"
},
{
"input": "1 1\n1000000000",
"output": "1"
},
{
"input": "5 5\n1 2 3 4 5\n2 3 4 5 6\n3 4 5 6 7\n4 5 6 7 8\n5 6 7 8 9",
"output": "5"
},
{
"input": "4 4\n1 3 1 3\n3 1 3 1\n2 0 0 2\n0 1 1 0",
"output": "1"
},
{
"input": "4 4\n1 4 1 3\n3 1 2 1\n1 0 0 2\n0 1 10 0",
"output": "1"
},
{
"input": "4 4\n1 4 1 300\n3 1 2 1\n5 0 0 2\n0 1 10 100",
"output": "1"
},
{
"input": "5 5\n15 45 15 300 10\n53 15 25 51 10\n5 50 50 2 10\n1000 1 10 100 10\n10 10 10 10 10",
"output": "1"
},
{
"input": "1 100\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1",
"output": "1"
},
{
"input": "100 1\n1 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 1",
"output": "1"
},
{
"input": "1 100\n859\n441\n272\n47\n355\n345\n612\n569\n545\n599\n410\n31\n720\n303\n58\n537\n561\n730\n288\n275\n446\n955\n195\n282\n153\n455\n996\n121\n267\n702\n769\n560\n353\n89\n990\n282\n801\n335\n573\n258\n722\n768\n324\n41\n249\n125\n557\n303\n664\n945\n156\n884\n985\n816\n433\n65\n976\n963\n85\n647\n46\n877\n665\n523\n714\n182\n377\n549\n994\n385\n184\n724\n447\n99\n766\n353\n494\n747\n324\n436\n915\n472\n879\n582\n928\n84\n627\n156\n972\n651\n159\n372\n70\n903\n590\n480\n184\n540\n270\n892",
"output": "1"
},
{
"input": "100 1\n439 158 619 538 187 153 973 781 610 475 94 947 449 531 220 51 788 118 189 501 54 434 465 902 280 635 688 214 737 327 682 690 683 519 261 923 254 388 529 659 662 276 376 735 976 664 521 285 42 147 187 259 407 977 879 465 522 17 550 701 114 921 577 265 668 812 232 267 135 371 586 201 608 373 771 358 101 412 195 582 199 758 507 882 16 484 11 712 916 699 783 618 405 124 904 257 606 610 230 718",
"output": "54"
},
{
"input": "1 99\n511\n642\n251\n30\n494\n128\n189\n324\n884\n656\n120\n616\n959\n328\n411\n933\n895\n350\n1\n838\n996\n761\n619\n131\n824\n751\n707\n688\n915\n115\n244\n476\n293\n986\n29\n787\n607\n259\n756\n864\n394\n465\n303\n387\n521\n582\n485\n355\n299\n997\n683\n472\n424\n948\n339\n383\n285\n957\n591\n203\n866\n79\n835\n980\n344\n493\n361\n159\n160\n947\n46\n362\n63\n553\n793\n754\n429\n494\n523\n227\n805\n313\n409\n243\n927\n350\n479\n971\n825\n460\n544\n235\n660\n327\n216\n729\n147\n671\n738",
"output": "1"
},
{
"input": "99 1\n50 287 266 159 551 198 689 418 809 43 691 367 160 664 86 805 461 55 127 950 576 351 721 493 972 560 934 885 492 92 321 759 767 989 883 7 127 413 404 604 80 645 666 874 371 718 893 158 722 198 563 293 134 255 742 913 252 378 859 721 502 251 839 284 133 209 962 514 773 124 205 903 785 859 911 93 861 786 747 213 690 69 942 697 211 203 284 961 351 137 962 952 408 249 238 850 944 40 346",
"output": "34"
},
{
"input": "100 1\n1 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 2",
"output": "100"
},
{
"input": "1 1\n0",
"output": "1"
},
{
"input": "2 1\n0 0",
"output": "1"
},
{
"input": "2 2\n0 0\n0 0",
"output": "1"
},
{
"input": "2 2\n1 2\n0 0",
"output": "1"
},
{
"input": "3 3\n0 0 0\n0 0 0\n0 0 0",
"output": "1"
},
{
"input": "2 3\n0 0\n0 0\n0 1",
"output": "1"
},
{
"input": "3 2\n1 1 3\n0 0 0",
"output": "1"
},
{
"input": "3 4\n1 10 3\n0 0 0\n0 0 0\n0 0 0",
"output": "1"
},
{
"input": "2 4\n2 1\n1 2\n0 0\n1 2",
"output": "1"
},
{
"input": "2 2\n0 1\n0 1",
"output": "2"
},
{
"input": "2 3\n1 2\n0 0\n2 1",
"output": "1"
},
{
"input": "2 2\n0 0\n4 5",
"output": "1"
},
{
"input": "3 2\n10 15 20\n0 0 0",
"output": "1"
},
{
"input": "3 4\n0 0 0\n0 0 0\n0 0 0\n1 2 3",
"output": "1"
},
{
"input": "3 3\n0 0 0\n0 0 0\n0 0 1",
"output": "1"
},
{
"input": "3 3\n0 0 0\n1 2 3\n1 3 2",
"output": "1"
},
{
"input": "3 1\n0 0 0",
"output": "1"
},
{
"input": "3 3\n0 0 1\n0 0 0\n0 0 0",
"output": "1"
}
] | 1,546,556,281
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 62
| 109
| 0
|
# http://codeforces.com/problemset/problem/570/A
# Elections
def first_that_equals(array, number):
for i in range(len(array)):
if array[i] == number:
return i
return -1
candidates, cities = list(map(int, input().split()))
candidates = [0 for n in range(candidates)]
for i in range(cities):
votes = list(map(int, input().split()))
winner = first_that_equals(votes, max(votes))
candidates[winner] += 1
print(first_that_equals(candidates, max(candidates)) + 1)
|
Title: Elections
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The country of Byalechinsk is running elections involving *n* candidates. The country consists of *m* cities. We know how many people in each city voted for each candidate.
The electoral system in the country is pretty unusual. At the first stage of elections the votes are counted for each city: it is assumed that in each city won the candidate who got the highest number of votes in this city, and if several candidates got the maximum number of votes, then the winner is the one with a smaller index.
At the second stage of elections the winner is determined by the same principle over the cities: the winner of the elections is the candidate who won in the maximum number of cities, and among those who got the maximum number of cities the winner is the one with a smaller index.
Determine who will win the elections.
Input Specification:
The first line of the input contains two integers *n*, *m* (1<=β€<=*n*,<=*m*<=β€<=100) β the number of candidates and of cities, respectively.
Each of the next *m* lines contains *n* non-negative integers, the *j*-th number in the *i*-th line *a**ij* (1<=β€<=*j*<=β€<=*n*, 1<=β€<=*i*<=β€<=*m*, 0<=β€<=*a**ij*<=β€<=109) denotes the number of votes for candidate *j* in city *i*.
It is guaranteed that the total number of people in all the cities does not exceed 109.
Output Specification:
Print a single number β the index of the candidate who won the elections. The candidates are indexed starting from one.
Demo Input:
['3 3\n1 2 3\n2 3 1\n1 2 1\n', '3 4\n10 10 3\n5 1 6\n2 2 2\n1 5 7\n']
Demo Output:
['2', '1']
Note:
Note to the first sample test. At the first stage city 1 chosen candidate 3, city 2 chosen candidate 2, city 3 chosen candidate 2. The winner is candidate 2, he gained 2 votes.
Note to the second sample test. At the first stage in city 1 candidates 1 and 2 got the same maximum number of votes, but candidate 1 has a smaller index, so the city chose candidate 1. City 2 chosen candidate 3. City 3 chosen candidate 1, due to the fact that everyone has the same number of votes, and 1 has the smallest index. City 4 chosen the candidate 3. On the second stage the same number of cities chose candidates 1 and 3. The winner is candidate 1, the one with the smaller index.
|
```python
# http://codeforces.com/problemset/problem/570/A
# Elections
def first_that_equals(array, number):
for i in range(len(array)):
if array[i] == number:
return i
return -1
candidates, cities = list(map(int, input().split()))
candidates = [0 for n in range(candidates)]
for i in range(cities):
votes = list(map(int, input().split()))
winner = first_that_equals(votes, max(votes))
candidates[winner] += 1
print(first_that_equals(candidates, max(candidates)) + 1)
```
| 3
|
|
334
|
A
|
Candy Bags
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
Gerald has *n* younger brothers and their number happens to be even. One day he bought *n*2 candy bags. One bag has one candy, one bag has two candies, one bag has three candies and so on. In fact, for each integer *k* from 1 to *n*2 he has exactly one bag with *k* candies.
Help him give *n* bags of candies to each brother so that all brothers got the same number of candies.
|
The single line contains a single integer *n* (*n* is even, 2<=β€<=*n*<=β€<=100) β the number of Gerald's brothers.
|
Let's assume that Gerald indexes his brothers with numbers from 1 to *n*. You need to print *n* lines, on the *i*-th line print *n* integers β the numbers of candies in the bags for the *i*-th brother. Naturally, all these numbers should be distinct and be within limits from 1 to *n*2. You can print the numbers in the lines in any order.
It is guaranteed that the solution exists at the given limits.
|
[
"2\n"
] |
[
"1 4\n2 3\n"
] |
The sample shows Gerald's actions if he has two brothers. In this case, his bags contain 1, 2, 3 and 4 candies. He can give the bags with 1 and 4 candies to one brother and the bags with 2 and 3 to the other brother.
| 500
|
[
{
"input": "2",
"output": "1 4\n2 3"
},
{
"input": "4",
"output": "1 16 2 15\n3 14 4 13\n5 12 6 11\n7 10 8 9"
},
{
"input": "6",
"output": "1 36 2 35 3 34\n4 33 5 32 6 31\n7 30 8 29 9 28\n10 27 11 26 12 25\n13 24 14 23 15 22\n16 21 17 20 18 19"
},
{
"input": "8",
"output": "1 64 2 63 3 62 4 61\n5 60 6 59 7 58 8 57\n9 56 10 55 11 54 12 53\n13 52 14 51 15 50 16 49\n17 48 18 47 19 46 20 45\n21 44 22 43 23 42 24 41\n25 40 26 39 27 38 28 37\n29 36 30 35 31 34 32 33"
},
{
"input": "10",
"output": "1 100 2 99 3 98 4 97 5 96\n6 95 7 94 8 93 9 92 10 91\n11 90 12 89 13 88 14 87 15 86\n16 85 17 84 18 83 19 82 20 81\n21 80 22 79 23 78 24 77 25 76\n26 75 27 74 28 73 29 72 30 71\n31 70 32 69 33 68 34 67 35 66\n36 65 37 64 38 63 39 62 40 61\n41 60 42 59 43 58 44 57 45 56\n46 55 47 54 48 53 49 52 50 51"
},
{
"input": "100",
"output": "1 10000 2 9999 3 9998 4 9997 5 9996 6 9995 7 9994 8 9993 9 9992 10 9991 11 9990 12 9989 13 9988 14 9987 15 9986 16 9985 17 9984 18 9983 19 9982 20 9981 21 9980 22 9979 23 9978 24 9977 25 9976 26 9975 27 9974 28 9973 29 9972 30 9971 31 9970 32 9969 33 9968 34 9967 35 9966 36 9965 37 9964 38 9963 39 9962 40 9961 41 9960 42 9959 43 9958 44 9957 45 9956 46 9955 47 9954 48 9953 49 9952 50 9951\n51 9950 52 9949 53 9948 54 9947 55 9946 56 9945 57 9944 58 9943 59 9942 60 9941 61 9940 62 9939 63 9938 64 9937 65 993..."
},
{
"input": "62",
"output": "1 3844 2 3843 3 3842 4 3841 5 3840 6 3839 7 3838 8 3837 9 3836 10 3835 11 3834 12 3833 13 3832 14 3831 15 3830 16 3829 17 3828 18 3827 19 3826 20 3825 21 3824 22 3823 23 3822 24 3821 25 3820 26 3819 27 3818 28 3817 29 3816 30 3815 31 3814\n32 3813 33 3812 34 3811 35 3810 36 3809 37 3808 38 3807 39 3806 40 3805 41 3804 42 3803 43 3802 44 3801 45 3800 46 3799 47 3798 48 3797 49 3796 50 3795 51 3794 52 3793 53 3792 54 3791 55 3790 56 3789 57 3788 58 3787 59 3786 60 3785 61 3784 62 3783\n63 3782 64 3781 65 378..."
},
{
"input": "66",
"output": "1 4356 2 4355 3 4354 4 4353 5 4352 6 4351 7 4350 8 4349 9 4348 10 4347 11 4346 12 4345 13 4344 14 4343 15 4342 16 4341 17 4340 18 4339 19 4338 20 4337 21 4336 22 4335 23 4334 24 4333 25 4332 26 4331 27 4330 28 4329 29 4328 30 4327 31 4326 32 4325 33 4324\n34 4323 35 4322 36 4321 37 4320 38 4319 39 4318 40 4317 41 4316 42 4315 43 4314 44 4313 45 4312 46 4311 47 4310 48 4309 49 4308 50 4307 51 4306 52 4305 53 4304 54 4303 55 4302 56 4301 57 4300 58 4299 59 4298 60 4297 61 4296 62 4295 63 4294 64 4293 65 4292..."
},
{
"input": "18",
"output": "1 324 2 323 3 322 4 321 5 320 6 319 7 318 8 317 9 316\n10 315 11 314 12 313 13 312 14 311 15 310 16 309 17 308 18 307\n19 306 20 305 21 304 22 303 23 302 24 301 25 300 26 299 27 298\n28 297 29 296 30 295 31 294 32 293 33 292 34 291 35 290 36 289\n37 288 38 287 39 286 40 285 41 284 42 283 43 282 44 281 45 280\n46 279 47 278 48 277 49 276 50 275 51 274 52 273 53 272 54 271\n55 270 56 269 57 268 58 267 59 266 60 265 61 264 62 263 63 262\n64 261 65 260 66 259 67 258 68 257 69 256 70 255 71 254 72 253\n73 252 7..."
},
{
"input": "68",
"output": "1 4624 2 4623 3 4622 4 4621 5 4620 6 4619 7 4618 8 4617 9 4616 10 4615 11 4614 12 4613 13 4612 14 4611 15 4610 16 4609 17 4608 18 4607 19 4606 20 4605 21 4604 22 4603 23 4602 24 4601 25 4600 26 4599 27 4598 28 4597 29 4596 30 4595 31 4594 32 4593 33 4592 34 4591\n35 4590 36 4589 37 4588 38 4587 39 4586 40 4585 41 4584 42 4583 43 4582 44 4581 45 4580 46 4579 47 4578 48 4577 49 4576 50 4575 51 4574 52 4573 53 4572 54 4571 55 4570 56 4569 57 4568 58 4567 59 4566 60 4565 61 4564 62 4563 63 4562 64 4561 65 4560..."
},
{
"input": "86",
"output": "1 7396 2 7395 3 7394 4 7393 5 7392 6 7391 7 7390 8 7389 9 7388 10 7387 11 7386 12 7385 13 7384 14 7383 15 7382 16 7381 17 7380 18 7379 19 7378 20 7377 21 7376 22 7375 23 7374 24 7373 25 7372 26 7371 27 7370 28 7369 29 7368 30 7367 31 7366 32 7365 33 7364 34 7363 35 7362 36 7361 37 7360 38 7359 39 7358 40 7357 41 7356 42 7355 43 7354\n44 7353 45 7352 46 7351 47 7350 48 7349 49 7348 50 7347 51 7346 52 7345 53 7344 54 7343 55 7342 56 7341 57 7340 58 7339 59 7338 60 7337 61 7336 62 7335 63 7334 64 7333 65 7332..."
},
{
"input": "96",
"output": "1 9216 2 9215 3 9214 4 9213 5 9212 6 9211 7 9210 8 9209 9 9208 10 9207 11 9206 12 9205 13 9204 14 9203 15 9202 16 9201 17 9200 18 9199 19 9198 20 9197 21 9196 22 9195 23 9194 24 9193 25 9192 26 9191 27 9190 28 9189 29 9188 30 9187 31 9186 32 9185 33 9184 34 9183 35 9182 36 9181 37 9180 38 9179 39 9178 40 9177 41 9176 42 9175 43 9174 44 9173 45 9172 46 9171 47 9170 48 9169\n49 9168 50 9167 51 9166 52 9165 53 9164 54 9163 55 9162 56 9161 57 9160 58 9159 59 9158 60 9157 61 9156 62 9155 63 9154 64 9153 65 9152..."
},
{
"input": "12",
"output": "1 144 2 143 3 142 4 141 5 140 6 139\n7 138 8 137 9 136 10 135 11 134 12 133\n13 132 14 131 15 130 16 129 17 128 18 127\n19 126 20 125 21 124 22 123 23 122 24 121\n25 120 26 119 27 118 28 117 29 116 30 115\n31 114 32 113 33 112 34 111 35 110 36 109\n37 108 38 107 39 106 40 105 41 104 42 103\n43 102 44 101 45 100 46 99 47 98 48 97\n49 96 50 95 51 94 52 93 53 92 54 91\n55 90 56 89 57 88 58 87 59 86 60 85\n61 84 62 83 63 82 64 81 65 80 66 79\n67 78 68 77 69 76 70 75 71 74 72 73"
},
{
"input": "88",
"output": "1 7744 2 7743 3 7742 4 7741 5 7740 6 7739 7 7738 8 7737 9 7736 10 7735 11 7734 12 7733 13 7732 14 7731 15 7730 16 7729 17 7728 18 7727 19 7726 20 7725 21 7724 22 7723 23 7722 24 7721 25 7720 26 7719 27 7718 28 7717 29 7716 30 7715 31 7714 32 7713 33 7712 34 7711 35 7710 36 7709 37 7708 38 7707 39 7706 40 7705 41 7704 42 7703 43 7702 44 7701\n45 7700 46 7699 47 7698 48 7697 49 7696 50 7695 51 7694 52 7693 53 7692 54 7691 55 7690 56 7689 57 7688 58 7687 59 7686 60 7685 61 7684 62 7683 63 7682 64 7681 65 7680..."
},
{
"input": "28",
"output": "1 784 2 783 3 782 4 781 5 780 6 779 7 778 8 777 9 776 10 775 11 774 12 773 13 772 14 771\n15 770 16 769 17 768 18 767 19 766 20 765 21 764 22 763 23 762 24 761 25 760 26 759 27 758 28 757\n29 756 30 755 31 754 32 753 33 752 34 751 35 750 36 749 37 748 38 747 39 746 40 745 41 744 42 743\n43 742 44 741 45 740 46 739 47 738 48 737 49 736 50 735 51 734 52 733 53 732 54 731 55 730 56 729\n57 728 58 727 59 726 60 725 61 724 62 723 63 722 64 721 65 720 66 719 67 718 68 717 69 716 70 715\n71 714 72 713 73 712 74 7..."
},
{
"input": "80",
"output": "1 6400 2 6399 3 6398 4 6397 5 6396 6 6395 7 6394 8 6393 9 6392 10 6391 11 6390 12 6389 13 6388 14 6387 15 6386 16 6385 17 6384 18 6383 19 6382 20 6381 21 6380 22 6379 23 6378 24 6377 25 6376 26 6375 27 6374 28 6373 29 6372 30 6371 31 6370 32 6369 33 6368 34 6367 35 6366 36 6365 37 6364 38 6363 39 6362 40 6361\n41 6360 42 6359 43 6358 44 6357 45 6356 46 6355 47 6354 48 6353 49 6352 50 6351 51 6350 52 6349 53 6348 54 6347 55 6346 56 6345 57 6344 58 6343 59 6342 60 6341 61 6340 62 6339 63 6338 64 6337 65 6336..."
},
{
"input": "48",
"output": "1 2304 2 2303 3 2302 4 2301 5 2300 6 2299 7 2298 8 2297 9 2296 10 2295 11 2294 12 2293 13 2292 14 2291 15 2290 16 2289 17 2288 18 2287 19 2286 20 2285 21 2284 22 2283 23 2282 24 2281\n25 2280 26 2279 27 2278 28 2277 29 2276 30 2275 31 2274 32 2273 33 2272 34 2271 35 2270 36 2269 37 2268 38 2267 39 2266 40 2265 41 2264 42 2263 43 2262 44 2261 45 2260 46 2259 47 2258 48 2257\n49 2256 50 2255 51 2254 52 2253 53 2252 54 2251 55 2250 56 2249 57 2248 58 2247 59 2246 60 2245 61 2244 62 2243 63 2242 64 2241 65 224..."
},
{
"input": "54",
"output": "1 2916 2 2915 3 2914 4 2913 5 2912 6 2911 7 2910 8 2909 9 2908 10 2907 11 2906 12 2905 13 2904 14 2903 15 2902 16 2901 17 2900 18 2899 19 2898 20 2897 21 2896 22 2895 23 2894 24 2893 25 2892 26 2891 27 2890\n28 2889 29 2888 30 2887 31 2886 32 2885 33 2884 34 2883 35 2882 36 2881 37 2880 38 2879 39 2878 40 2877 41 2876 42 2875 43 2874 44 2873 45 2872 46 2871 47 2870 48 2869 49 2868 50 2867 51 2866 52 2865 53 2864 54 2863\n55 2862 56 2861 57 2860 58 2859 59 2858 60 2857 61 2856 62 2855 63 2854 64 2853 65 285..."
},
{
"input": "58",
"output": "1 3364 2 3363 3 3362 4 3361 5 3360 6 3359 7 3358 8 3357 9 3356 10 3355 11 3354 12 3353 13 3352 14 3351 15 3350 16 3349 17 3348 18 3347 19 3346 20 3345 21 3344 22 3343 23 3342 24 3341 25 3340 26 3339 27 3338 28 3337 29 3336\n30 3335 31 3334 32 3333 33 3332 34 3331 35 3330 36 3329 37 3328 38 3327 39 3326 40 3325 41 3324 42 3323 43 3322 44 3321 45 3320 46 3319 47 3318 48 3317 49 3316 50 3315 51 3314 52 3313 53 3312 54 3311 55 3310 56 3309 57 3308 58 3307\n59 3306 60 3305 61 3304 62 3303 63 3302 64 3301 65 330..."
},
{
"input": "64",
"output": "1 4096 2 4095 3 4094 4 4093 5 4092 6 4091 7 4090 8 4089 9 4088 10 4087 11 4086 12 4085 13 4084 14 4083 15 4082 16 4081 17 4080 18 4079 19 4078 20 4077 21 4076 22 4075 23 4074 24 4073 25 4072 26 4071 27 4070 28 4069 29 4068 30 4067 31 4066 32 4065\n33 4064 34 4063 35 4062 36 4061 37 4060 38 4059 39 4058 40 4057 41 4056 42 4055 43 4054 44 4053 45 4052 46 4051 47 4050 48 4049 49 4048 50 4047 51 4046 52 4045 53 4044 54 4043 55 4042 56 4041 57 4040 58 4039 59 4038 60 4037 61 4036 62 4035 63 4034 64 4033\n65 403..."
},
{
"input": "44",
"output": "1 1936 2 1935 3 1934 4 1933 5 1932 6 1931 7 1930 8 1929 9 1928 10 1927 11 1926 12 1925 13 1924 14 1923 15 1922 16 1921 17 1920 18 1919 19 1918 20 1917 21 1916 22 1915\n23 1914 24 1913 25 1912 26 1911 27 1910 28 1909 29 1908 30 1907 31 1906 32 1905 33 1904 34 1903 35 1902 36 1901 37 1900 38 1899 39 1898 40 1897 41 1896 42 1895 43 1894 44 1893\n45 1892 46 1891 47 1890 48 1889 49 1888 50 1887 51 1886 52 1885 53 1884 54 1883 55 1882 56 1881 57 1880 58 1879 59 1878 60 1877 61 1876 62 1875 63 1874 64 1873 65 187..."
}
] | 1,502,439,603
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 21
| 184
| 5,324,800
|
n=int(input())
a=[int(i) for i in range(1,n*n+1)]
b=a[:round(n*n/2)]
c=a[round(n*n/2):]
c=c[::-1]
for i in range(n):
for i in range(round(n/2)):
print(b.pop(0),c.pop(0),end=' ')
print()
|
Title: Candy Bags
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Gerald has *n* younger brothers and their number happens to be even. One day he bought *n*2 candy bags. One bag has one candy, one bag has two candies, one bag has three candies and so on. In fact, for each integer *k* from 1 to *n*2 he has exactly one bag with *k* candies.
Help him give *n* bags of candies to each brother so that all brothers got the same number of candies.
Input Specification:
The single line contains a single integer *n* (*n* is even, 2<=β€<=*n*<=β€<=100) β the number of Gerald's brothers.
Output Specification:
Let's assume that Gerald indexes his brothers with numbers from 1 to *n*. You need to print *n* lines, on the *i*-th line print *n* integers β the numbers of candies in the bags for the *i*-th brother. Naturally, all these numbers should be distinct and be within limits from 1 to *n*2. You can print the numbers in the lines in any order.
It is guaranteed that the solution exists at the given limits.
Demo Input:
['2\n']
Demo Output:
['1 4\n2 3\n']
Note:
The sample shows Gerald's actions if he has two brothers. In this case, his bags contain 1, 2, 3 and 4 candies. He can give the bags with 1 and 4 candies to one brother and the bags with 2 and 3 to the other brother.
|
```python
n=int(input())
a=[int(i) for i in range(1,n*n+1)]
b=a[:round(n*n/2)]
c=a[round(n*n/2):]
c=c[::-1]
for i in range(n):
for i in range(round(n/2)):
print(b.pop(0),c.pop(0),end=' ')
print()
```
| 3
|
|
812
|
C
|
Sagheer and Nubian Market
|
PROGRAMMING
| 1,500
|
[
"binary search",
"sortings"
] | null | null |
On his trip to Luxor and Aswan, Sagheer went to a Nubian market to buy some souvenirs for his friends and relatives. The market has some strange rules. It contains *n* different items numbered from 1 to *n*. The *i*-th item has base cost *a**i* Egyptian pounds. If Sagheer buys *k* items with indices *x*1,<=*x*2,<=...,<=*x**k*, then the cost of item *x**j* is *a**x**j*<=+<=*x**j*Β·*k* for 1<=β€<=*j*<=β€<=*k*. In other words, the cost of an item is equal to its base cost in addition to its index multiplied by the factor *k*.
Sagheer wants to buy as many souvenirs as possible without paying more than *S* Egyptian pounds. Note that he cannot buy a souvenir more than once. If there are many ways to maximize the number of souvenirs, he will choose the way that will minimize the total cost. Can you help him with this task?
|
The first line contains two integers *n* and *S* (1<=β€<=*n*<=β€<=105 and 1<=β€<=*S*<=β€<=109)Β β the number of souvenirs in the market and Sagheer's budget.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=β€<=*a**i*<=β€<=105)Β β the base costs of the souvenirs.
|
On a single line, print two integers *k*, *T*Β β the maximum number of souvenirs Sagheer can buy and the minimum total cost to buy these *k* souvenirs.
|
[
"3 11\n2 3 5\n",
"4 100\n1 2 5 6\n",
"1 7\n7\n"
] |
[
"2 11\n",
"4 54\n",
"0 0\n"
] |
In the first example, he cannot take the three items because they will cost him [5,β9,β14] with total cost 28. If he decides to take only two items, then the costs will be [4,β7,β11]. So he can afford the first and second items.
In the second example, he can buy all items as they will cost him [5,β10,β17,β22].
In the third example, there is only one souvenir in the market which will cost him 8 pounds, so he cannot buy it.
| 1,500
|
[
{
"input": "3 11\n2 3 5",
"output": "2 11"
},
{
"input": "4 100\n1 2 5 6",
"output": "4 54"
},
{
"input": "1 7\n7",
"output": "0 0"
},
{
"input": "1 7\n5",
"output": "1 6"
},
{
"input": "1 1\n1",
"output": "0 0"
},
{
"input": "4 33\n4 3 2 1",
"output": "3 27"
},
{
"input": "86 96\n89 48 14 55 5 35 7 79 49 70 74 18 64 63 35 93 63 97 90 77 33 11 100 75 60 99 54 38 3 6 55 1 7 64 56 90 21 76 35 16 61 78 38 78 93 21 89 1 58 53 34 77 56 37 46 59 30 5 85 1 52 87 84 99 97 9 15 66 29 60 17 16 59 23 88 93 32 2 98 89 63 42 9 86 70 80",
"output": "3 71"
},
{
"input": "9 2727\n73 41 68 90 51 7 20 48 69",
"output": "9 872"
},
{
"input": "35 792600\n61 11 82 29 3 50 65 60 62 86 83 78 15 82 7 77 38 87 100 12 93 86 96 79 14 58 60 47 94 39 36 23 69 93 18",
"output": "35 24043"
},
{
"input": "63 47677090\n53 4 59 68 6 12 47 63 28 93 9 53 61 63 53 70 77 63 49 76 70 23 4 40 4 34 24 70 42 83 84 95 11 46 38 83 26 85 34 29 67 96 3 62 97 7 42 65 49 45 50 54 81 74 83 59 10 87 95 87 89 27 3",
"output": "63 130272"
},
{
"input": "88 631662736\n93 75 25 7 6 55 92 23 22 32 4 48 61 29 91 79 16 18 18 9 66 9 57 62 3 81 48 16 21 90 93 58 30 8 31 47 44 70 34 85 52 71 58 42 99 53 43 54 96 26 6 13 38 4 13 60 1 48 32 100 52 8 27 99 66 34 98 45 19 50 37 59 31 56 58 70 61 14 100 66 74 85 64 57 92 89 7 92",
"output": "88 348883"
},
{
"input": "12 12\n1232 1848 2048 4694 5121 3735 9968 4687 2040 6033 5839 2507",
"output": "0 0"
},
{
"input": "37 5271\n368 6194 4856 8534 944 4953 2085 5350 788 7772 9786 1321 4310 4453 7078 9912 5799 4066 5471 5079 5161 9773 1300 5474 1202 1353 9499 9694 9020 6332 595 7619 1271 7430 1199 3127 8867",
"output": "5 4252"
},
{
"input": "65 958484\n9597 1867 5346 637 6115 5833 3318 6059 4430 9169 8155 7895 3534 7962 9900 9495 5694 3461 5370 1945 1724 9264 3475 618 3421 551 8359 6889 1843 6716 9216 2356 1592 6265 2945 6496 4947 2840 9057 6141 887 4823 4004 8027 1993 1391 796 7059 5500 4369 4012 4983 6495 8990 3633 5439 421 1129 6970 8796 7826 1200 8741 6555 5037",
"output": "65 468998"
},
{
"input": "90 61394040\n2480 6212 4506 829 8191 797 5336 6722 3178 1007 5849 3061 3588 6684 5983 5452 7654 5321 660 2569 2809 2179 679 4858 6887 2580 6880 6120 4159 5542 4999 8703 2386 8221 7046 1229 1662 4542 7089 3548 4298 1973 1854 2473 5507 241 359 5248 7907 5201 9624 4596 1723 2622 4800 4716 693 961 7402 9004 7994 8048 6590 5866 7502 3304 4331 5218 6906 1016 5342 6644 2205 5823 8525 4839 1914 2651 3940 7751 3489 4178 7234 6640 7602 9765 8559 7819 5827 163",
"output": "90 795634"
},
{
"input": "14 891190480\n1424 3077 9632 6506 4568 9650 5534 1085 6934 9340 2867 367 7075 618",
"output": "14 70147"
},
{
"input": "39 43\n22166 81842 15513 80979 39645 60168 96994 13493 12904 79871 49910 45356 93691 51829 18226 34288 11525 41944 40433 67295 30123 1081 55623 22279 75814 82316 2963 39329 38223 8445 43202 61912 15122 86367 37200 68113 57194 38541 49641",
"output": "0 0"
},
{
"input": "67 8824\n75515 67590 86373 34191 3446 27408 31581 24727 40005 23718 39738 30960 4786 51040 32590 80454 14335 47173 20079 41204 67289 58347 88969 88396 37681 43963 13886 85690 12259 14732 42036 62620 15011 41890 20150 59469 62104 30136 47163 19790 25699 27453 36151 52914 52684 20503 78622 81082 94500 55756 94030 54764 72763 37830 13210 64559 53600 87998 80472 19001 83769 79700 88794 10161 99980 95184 74439",
"output": "2 8268"
},
{
"input": "16 56532535\n84567 85265 99012 62115 58908 10120 27355 90226 17191 35588 43807 6327 89157 71156 16253 22387",
"output": "16 821610"
},
{
"input": "42 818723640\n57019 99450 43192 25131 80678 41786 52490 34454 51620 56132 41773 94622 23949 11597 96944 51420 44164 85727 72959 26622 1838 36302 99335 35572 92734 10651 22293 97601 38634 24066 81870 10904 5825 71968 2377 61198 3104 4191 97851 99920 65295 1385",
"output": "42 2094569"
}
] | 1,680,186,373
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 57
| 186
| 19,763,200
|
n,budget = map(int,input().split())
arr = list(map(int,input().split()))
def check(x):
temp = [arr[i]+(mid*(i+1)) for i in range(n)]
temp.sort()
expense = sum(temp[:mid])
return [expense,expense<=budget]
l,r = 1,n
ans = 0
exp = 0
while(l<=r):
mid = (l+r)//2
price,status = check(mid)
if status:
if ans<mid:
exp = price
ans = mid
l = mid+1
else:
r = mid-1
print(ans,exp)
|
Title: Sagheer and Nubian Market
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
On his trip to Luxor and Aswan, Sagheer went to a Nubian market to buy some souvenirs for his friends and relatives. The market has some strange rules. It contains *n* different items numbered from 1 to *n*. The *i*-th item has base cost *a**i* Egyptian pounds. If Sagheer buys *k* items with indices *x*1,<=*x*2,<=...,<=*x**k*, then the cost of item *x**j* is *a**x**j*<=+<=*x**j*Β·*k* for 1<=β€<=*j*<=β€<=*k*. In other words, the cost of an item is equal to its base cost in addition to its index multiplied by the factor *k*.
Sagheer wants to buy as many souvenirs as possible without paying more than *S* Egyptian pounds. Note that he cannot buy a souvenir more than once. If there are many ways to maximize the number of souvenirs, he will choose the way that will minimize the total cost. Can you help him with this task?
Input Specification:
The first line contains two integers *n* and *S* (1<=β€<=*n*<=β€<=105 and 1<=β€<=*S*<=β€<=109)Β β the number of souvenirs in the market and Sagheer's budget.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=β€<=*a**i*<=β€<=105)Β β the base costs of the souvenirs.
Output Specification:
On a single line, print two integers *k*, *T*Β β the maximum number of souvenirs Sagheer can buy and the minimum total cost to buy these *k* souvenirs.
Demo Input:
['3 11\n2 3 5\n', '4 100\n1 2 5 6\n', '1 7\n7\n']
Demo Output:
['2 11\n', '4 54\n', '0 0\n']
Note:
In the first example, he cannot take the three items because they will cost him [5,β9,β14] with total cost 28. If he decides to take only two items, then the costs will be [4,β7,β11]. So he can afford the first and second items.
In the second example, he can buy all items as they will cost him [5,β10,β17,β22].
In the third example, there is only one souvenir in the market which will cost him 8 pounds, so he cannot buy it.
|
```python
n,budget = map(int,input().split())
arr = list(map(int,input().split()))
def check(x):
temp = [arr[i]+(mid*(i+1)) for i in range(n)]
temp.sort()
expense = sum(temp[:mid])
return [expense,expense<=budget]
l,r = 1,n
ans = 0
exp = 0
while(l<=r):
mid = (l+r)//2
price,status = check(mid)
if status:
if ans<mid:
exp = price
ans = mid
l = mid+1
else:
r = mid-1
print(ans,exp)
```
| 3
|
|
490
|
A
|
Team Olympiad
|
PROGRAMMING
| 800
|
[
"greedy",
"implementation",
"sortings"
] | null | null |
The School β0 of the capital of Berland has *n* children studying in it. All the children in this school are gifted: some of them are good at programming, some are good at maths, others are good at PE (Physical Education). Hence, for each child we know value *t**i*:
- *t**i*<==<=1, if the *i*-th child is good at programming, - *t**i*<==<=2, if the *i*-th child is good at maths, - *t**i*<==<=3, if the *i*-th child is good at PE
Each child happens to be good at exactly one of these three subjects.
The Team Scientific Decathlon Olympias requires teams of three students. The school teachers decided that the teams will be composed of three children that are good at different subjects. That is, each team must have one mathematician, one programmer and one sportsman. Of course, each child can be a member of no more than one team.
What is the maximum number of teams that the school will be able to present at the Olympiad? How should the teams be formed for that?
|
The first line contains integer *n* (1<=β€<=*n*<=β€<=5000) β the number of children in the school. The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=β€<=*t**i*<=β€<=3), where *t**i* describes the skill of the *i*-th child.
|
In the first line output integer *w* β the largest possible number of teams.
Then print *w* lines, containing three numbers in each line. Each triple represents the indexes of the children forming the team. You can print both the teams, and the numbers in the triplets in any order. The children are numbered from 1 to *n* in the order of their appearance in the input. Each child must participate in no more than one team. If there are several solutions, print any of them.
If no teams can be compiled, print the only line with value *w* equal to 0.
|
[
"7\n1 3 1 3 2 1 2\n",
"4\n2 1 1 2\n"
] |
[
"2\n3 5 2\n6 7 4\n",
"0\n"
] |
none
| 500
|
[
{
"input": "7\n1 3 1 3 2 1 2",
"output": "2\n3 5 2\n6 7 4"
},
{
"input": "4\n2 1 1 2",
"output": "0"
},
{
"input": "1\n2",
"output": "0"
},
{
"input": "2\n3 1",
"output": "0"
},
{
"input": "3\n2 1 2",
"output": "0"
},
{
"input": "3\n1 2 3",
"output": "1\n1 2 3"
},
{
"input": "12\n3 3 3 3 3 3 3 3 1 3 3 2",
"output": "1\n9 12 2"
},
{
"input": "60\n3 3 1 2 2 1 3 1 1 1 3 2 2 2 3 3 1 3 2 3 2 2 1 3 3 2 3 1 2 2 2 1 3 2 1 1 3 3 1 1 1 3 1 2 1 1 3 3 3 2 3 2 3 2 2 2 1 1 1 2",
"output": "20\n6 60 1\n17 44 20\n3 5 33\n36 21 42\n59 14 2\n58 26 49\n9 29 48\n23 19 24\n10 30 37\n41 54 15\n45 31 27\n57 55 38\n39 12 25\n35 34 11\n32 52 7\n8 50 18\n43 4 53\n46 56 51\n40 22 16\n28 13 47"
},
{
"input": "12\n3 1 1 1 1 1 1 2 1 1 1 1",
"output": "1\n3 8 1"
},
{
"input": "22\n2 2 2 2 2 2 2 2 2 2 3 2 2 2 2 2 2 1 2 2 2 2",
"output": "1\n18 2 11"
},
{
"input": "138\n2 3 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 3 2 2 2 1 2 3 2 2 2 3 1 3 2 3 2 3 2 2 2 2 3 2 2 2 2 2 1 2 2 3 2 2 3 2 1 2 2 2 2 2 3 1 2 2 2 2 2 3 2 2 3 2 2 2 2 2 1 1 2 3 2 2 2 2 3 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 3 2 3 2 2 2 1 2 2 2 1 2 2 2 2 1 2 2 2 2 1 3",
"output": "18\n13 91 84\n34 90 48\n11 39 77\n78 129 50\n137 68 119\n132 122 138\n19 12 96\n40 7 2\n22 88 69\n107 73 46\n115 15 52\n127 106 87\n93 92 66\n71 112 117\n63 124 42\n17 70 101\n109 121 57\n123 25 36"
},
{
"input": "203\n2 2 1 2 1 2 2 2 1 2 2 1 1 3 1 2 1 2 1 1 2 3 1 1 2 3 3 2 2 2 1 2 1 1 1 1 1 3 1 1 2 1 1 2 2 2 1 2 2 2 1 2 3 2 1 1 2 2 1 2 1 2 2 1 1 2 2 2 1 1 2 2 1 2 1 2 2 3 2 1 2 1 1 1 1 1 1 1 1 1 1 2 2 1 1 2 2 2 2 1 1 1 1 1 1 1 2 2 2 2 2 1 1 1 2 2 2 1 2 2 1 3 2 1 1 1 2 1 1 2 1 1 2 2 2 1 1 2 2 2 1 2 1 3 2 1 2 2 2 1 1 1 2 2 2 1 2 1 1 2 2 2 2 2 1 1 2 1 2 2 1 1 1 1 1 1 2 2 3 1 1 2 3 1 1 1 1 1 1 2 2 1 1 1 2 2 3 2 1 3 1 1 1",
"output": "13\n188 72 14\n137 4 197\n158 76 122\n152 142 26\n104 119 179\n40 63 38\n12 1 78\n17 30 27\n189 60 53\n166 190 144\n129 7 183\n83 41 22\n121 81 200"
},
{
"input": "220\n1 1 3 1 3 1 1 3 1 3 3 3 3 1 3 3 1 3 3 3 3 3 1 1 1 3 1 1 1 3 2 3 3 3 1 1 3 3 1 1 3 3 3 3 1 3 3 1 1 1 2 3 1 1 1 2 3 3 3 2 3 1 1 3 1 1 1 3 2 1 3 2 3 1 1 3 3 3 1 3 1 1 1 3 3 2 1 3 2 1 1 3 3 1 1 1 2 1 1 3 2 1 2 1 1 1 3 1 3 3 1 2 3 3 3 3 1 3 1 1 1 1 2 3 1 1 1 1 1 1 3 2 3 1 3 1 3 1 1 3 1 3 1 3 1 3 1 3 3 2 3 1 3 3 1 3 3 3 3 1 1 3 3 3 3 1 1 3 3 3 2 1 1 1 3 3 1 3 3 3 1 1 1 3 1 3 3 1 1 1 2 3 1 1 3 1 1 1 1 2 3 1 1 2 3 3 1 3 1 3 3 3 3 1 3 2 3 1 1 3",
"output": "20\n198 89 20\n141 56 131\n166 204 19\n160 132 142\n111 112 195\n45 216 92\n6 31 109\n14 150 170\n199 60 18\n173 123 140\n134 69 156\n82 191 85\n126 200 80\n24 97 46\n62 86 149\n214 101 26\n79 171 78\n125 72 118\n172 103 162\n219 51 64"
},
{
"input": "61\n2 3 1 3 2 2 2 3 1 3 2 3 1 2 1 1 2 2 2 2 3 2 3 1 2 1 3 1 3 2 1 1 3 2 1 3 3 3 1 3 3 1 1 3 1 3 2 2 1 2 2 2 1 3 2 3 1 3 3 1 1",
"output": "20\n9 55 2\n24 34 27\n3 5 37\n35 17 41\n61 11 4\n60 19 54\n15 20 59\n26 14 29\n16 22 38\n43 50 12\n49 25 36\n57 51 40\n39 6 33\n32 30 10\n31 48 8\n13 47 23\n45 1 58\n53 52 56\n42 18 21\n28 7 44"
},
{
"input": "5\n1 2 2 3 3",
"output": "1\n1 3 4"
}
] | 1,659,248,873
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 41
| 46
| 0
|
children_in_school = int(input())
skills = list(map(int, input().split(' ')))
programming = []
maths = []
pe = []
for i in range(children_in_school):
if skills[i] == 1:
programming.append(i + 1)
elif skills[i] == 2:
maths.append(i + 1)
else:
pe.append(i + 1)
number_of_teams = min(len(programming), len(maths), len(pe))
teams = []
if number_of_teams > 0:
for i in range(number_of_teams):
teams.append([programming[i], maths[i], pe[i]])
print(len(teams))
for team in teams:
print(*team)
|
Title: Team Olympiad
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The School β0 of the capital of Berland has *n* children studying in it. All the children in this school are gifted: some of them are good at programming, some are good at maths, others are good at PE (Physical Education). Hence, for each child we know value *t**i*:
- *t**i*<==<=1, if the *i*-th child is good at programming, - *t**i*<==<=2, if the *i*-th child is good at maths, - *t**i*<==<=3, if the *i*-th child is good at PE
Each child happens to be good at exactly one of these three subjects.
The Team Scientific Decathlon Olympias requires teams of three students. The school teachers decided that the teams will be composed of three children that are good at different subjects. That is, each team must have one mathematician, one programmer and one sportsman. Of course, each child can be a member of no more than one team.
What is the maximum number of teams that the school will be able to present at the Olympiad? How should the teams be formed for that?
Input Specification:
The first line contains integer *n* (1<=β€<=*n*<=β€<=5000) β the number of children in the school. The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=β€<=*t**i*<=β€<=3), where *t**i* describes the skill of the *i*-th child.
Output Specification:
In the first line output integer *w* β the largest possible number of teams.
Then print *w* lines, containing three numbers in each line. Each triple represents the indexes of the children forming the team. You can print both the teams, and the numbers in the triplets in any order. The children are numbered from 1 to *n* in the order of their appearance in the input. Each child must participate in no more than one team. If there are several solutions, print any of them.
If no teams can be compiled, print the only line with value *w* equal to 0.
Demo Input:
['7\n1 3 1 3 2 1 2\n', '4\n2 1 1 2\n']
Demo Output:
['2\n3 5 2\n6 7 4\n', '0\n']
Note:
none
|
```python
children_in_school = int(input())
skills = list(map(int, input().split(' ')))
programming = []
maths = []
pe = []
for i in range(children_in_school):
if skills[i] == 1:
programming.append(i + 1)
elif skills[i] == 2:
maths.append(i + 1)
else:
pe.append(i + 1)
number_of_teams = min(len(programming), len(maths), len(pe))
teams = []
if number_of_teams > 0:
for i in range(number_of_teams):
teams.append([programming[i], maths[i], pe[i]])
print(len(teams))
for team in teams:
print(*team)
```
| 3
|
|
306
|
A
|
Candies
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Polycarpus has got *n* candies and *m* friends (*n*<=β₯<=*m*). He wants to make a New Year present with candies to each friend. Polycarpus is planning to present all candies and he wants to do this in the fairest (that is, most equal) manner. He wants to choose such *a**i*, where *a**i* is the number of candies in the *i*-th friend's present, that the maximum *a**i* differs from the least *a**i* as little as possible.
For example, if *n* is divisible by *m*, then he is going to present the same number of candies to all his friends, that is, the maximum *a**i* won't differ from the minimum one.
|
The single line of the input contains a pair of space-separated positive integers *n*, *m* (1<=β€<=*n*,<=*m*<=β€<=100;*n*<=β₯<=*m*) β the number of candies and the number of Polycarpus's friends.
|
Print the required sequence *a*1,<=*a*2,<=...,<=*a**m*, where *a**i* is the number of candies in the *i*-th friend's present. All numbers *a**i* must be positive integers, total up to *n*, the maximum one should differ from the minimum one by the smallest possible value.
|
[
"12 3\n",
"15 4\n",
"18 7\n"
] |
[
"4 4 4 ",
"3 4 4 4 ",
"2 2 2 3 3 3 3 "
] |
Print *a*<sub class="lower-index">*i*</sub> in any order, separate the numbers by spaces.
| 500
|
[
{
"input": "12 4",
"output": "3 3 3 3 "
},
{
"input": "15 6",
"output": "2 2 2 3 3 3 "
},
{
"input": "18 8",
"output": "2 2 2 2 2 2 3 3 "
},
{
"input": "1 1",
"output": "1 "
},
{
"input": "2 1",
"output": "2 "
},
{
"input": "100 1",
"output": "100 "
},
{
"input": "100 100",
"output": "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 1 1 "
},
{
"input": "100 10",
"output": "10 10 10 10 10 10 10 10 10 10 "
},
{
"input": "64 8",
"output": "8 8 8 8 8 8 8 8 "
},
{
"input": "81 27",
"output": "3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 "
},
{
"input": "100 99",
"output": "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 2 "
},
{
"input": "100 51",
"output": "1 1 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": "100 49",
"output": "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 3 3 "
},
{
"input": "100 13",
"output": "7 7 7 7 8 8 8 8 8 8 8 8 8 "
},
{
"input": "97 11",
"output": "8 8 9 9 9 9 9 9 9 9 9 "
},
{
"input": "19 3",
"output": "6 6 7 "
},
{
"input": "99 10",
"output": "9 10 10 10 10 10 10 10 10 10 "
},
{
"input": "88 5",
"output": "17 17 18 18 18 "
},
{
"input": "87 34",
"output": "2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 "
},
{
"input": "97 55",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 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 "
}
] | 1,371,898,559
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 20
| 78
| 102,400
|
I=[int(i) for i in input().split()]
conf, friend = I[0], I[1]
answer=[]
ans=""
mod=conf%friend
for i in range(friend): answer.append(int((conf-mod)/friend))
for i in range(mod): answer[i]+=1
for i in range(friend): ans+=str(answer[i])+" "
print(ans[:-1])
|
Title: Candies
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus has got *n* candies and *m* friends (*n*<=β₯<=*m*). He wants to make a New Year present with candies to each friend. Polycarpus is planning to present all candies and he wants to do this in the fairest (that is, most equal) manner. He wants to choose such *a**i*, where *a**i* is the number of candies in the *i*-th friend's present, that the maximum *a**i* differs from the least *a**i* as little as possible.
For example, if *n* is divisible by *m*, then he is going to present the same number of candies to all his friends, that is, the maximum *a**i* won't differ from the minimum one.
Input Specification:
The single line of the input contains a pair of space-separated positive integers *n*, *m* (1<=β€<=*n*,<=*m*<=β€<=100;*n*<=β₯<=*m*) β the number of candies and the number of Polycarpus's friends.
Output Specification:
Print the required sequence *a*1,<=*a*2,<=...,<=*a**m*, where *a**i* is the number of candies in the *i*-th friend's present. All numbers *a**i* must be positive integers, total up to *n*, the maximum one should differ from the minimum one by the smallest possible value.
Demo Input:
['12 3\n', '15 4\n', '18 7\n']
Demo Output:
['4 4 4 ', '3 4 4 4 ', '2 2 2 3 3 3 3 ']
Note:
Print *a*<sub class="lower-index">*i*</sub> in any order, separate the numbers by spaces.
|
```python
I=[int(i) for i in input().split()]
conf, friend = I[0], I[1]
answer=[]
ans=""
mod=conf%friend
for i in range(friend): answer.append(int((conf-mod)/friend))
for i in range(mod): answer[i]+=1
for i in range(friend): ans+=str(answer[i])+" "
print(ans[:-1])
```
| 3
|
|
977
|
A
|
Wrong Subtraction
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Little girl Tanya is learning how to decrease a number by one, but she does it wrong with a number consisting of two or more digits. Tanya subtracts one from a number by the following algorithm:
- if the last digit of the number is non-zero, she decreases the number by one; - if the last digit of the number is zero, she divides the number by 10 (i.e. removes the last digit).
You are given an integer number $n$. Tanya will subtract one from it $k$ times. Your task is to print the result after all $k$ subtractions.
It is guaranteed that the result will be positive integer number.
|
The first line of the input contains two integer numbers $n$ and $k$ ($2 \le n \le 10^9$, $1 \le k \le 50$) β the number from which Tanya will subtract and the number of subtractions correspondingly.
|
Print one integer number β the result of the decreasing $n$ by one $k$ times.
It is guaranteed that the result will be positive integer number.
|
[
"512 4\n",
"1000000000 9\n"
] |
[
"50\n",
"1\n"
] |
The first example corresponds to the following sequence: $512 \rightarrow 511 \rightarrow 510 \rightarrow 51 \rightarrow 50$.
| 0
|
[
{
"input": "512 4",
"output": "50"
},
{
"input": "1000000000 9",
"output": "1"
},
{
"input": "131203 11",
"output": "12"
},
{
"input": "999999999 50",
"output": "9999"
},
{
"input": "999999999 49",
"output": "99990"
},
{
"input": "131203 9",
"output": "130"
},
{
"input": "900000000 16",
"output": "1"
},
{
"input": "909090909 50",
"output": "3"
},
{
"input": "1001 2",
"output": "100"
},
{
"input": "5 2",
"output": "3"
},
{
"input": "2 1",
"output": "1"
}
] | 1,696,078,611
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 11
| 62
| 0
|
a,b = stuff = map(int,input().split(" "))
for i in range(b):
if a % 10 == 0:
a /= 10
else: a -= 1
print(int(a))
|
Title: Wrong Subtraction
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little girl Tanya is learning how to decrease a number by one, but she does it wrong with a number consisting of two or more digits. Tanya subtracts one from a number by the following algorithm:
- if the last digit of the number is non-zero, she decreases the number by one; - if the last digit of the number is zero, she divides the number by 10 (i.e. removes the last digit).
You are given an integer number $n$. Tanya will subtract one from it $k$ times. Your task is to print the result after all $k$ subtractions.
It is guaranteed that the result will be positive integer number.
Input Specification:
The first line of the input contains two integer numbers $n$ and $k$ ($2 \le n \le 10^9$, $1 \le k \le 50$) β the number from which Tanya will subtract and the number of subtractions correspondingly.
Output Specification:
Print one integer number β the result of the decreasing $n$ by one $k$ times.
It is guaranteed that the result will be positive integer number.
Demo Input:
['512 4\n', '1000000000 9\n']
Demo Output:
['50\n', '1\n']
Note:
The first example corresponds to the following sequence: $512 \rightarrow 511 \rightarrow 510 \rightarrow 51 \rightarrow 50$.
|
```python
a,b = stuff = map(int,input().split(" "))
for i in range(b):
if a % 10 == 0:
a /= 10
else: a -= 1
print(int(a))
```
| 3
|
|
621
|
A
|
Wet Shark and Odd and Even
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Today, Wet Shark is given *n* integers. Using any of these integers no more than once, Wet Shark wants to get maximum possible even (divisible by 2) sum. Please, calculate this value for Wet Shark.
Note, that if Wet Shark uses no integers from the *n* integers, the sum is an even integer 0.
|
The first line of the input contains one integer, *n* (1<=β€<=*n*<=β€<=100<=000). The next line contains *n* space separated integers given to Wet Shark. Each of these integers is in range from 1 to 109, inclusive.
|
Print the maximum possible even sum that can be obtained if we use some of the given integers.
|
[
"3\n1 2 3\n",
"5\n999999999 999999999 999999999 999999999 999999999\n"
] |
[
"6",
"3999999996"
] |
In the first sample, we can simply take all three integers for a total sum of 6.
In the second sample Wet Shark should take any four out of five integers 999β999β999.
| 500
|
[
{
"input": "3\n1 2 3",
"output": "6"
},
{
"input": "5\n999999999 999999999 999999999 999999999 999999999",
"output": "3999999996"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "15\n39 52 88 78 46 95 84 98 55 3 68 42 6 18 98",
"output": "870"
},
{
"input": "15\n59 96 34 48 8 72 67 90 15 85 7 90 97 47 25",
"output": "840"
},
{
"input": "15\n87 37 91 29 58 45 51 74 70 71 47 38 91 89 44",
"output": "922"
},
{
"input": "15\n11 81 49 7 11 14 30 67 29 50 90 81 77 18 59",
"output": "674"
},
{
"input": "15\n39 21 95 89 73 90 9 55 85 32 30 21 68 59 82",
"output": "848"
},
{
"input": "15\n59 70 48 54 26 67 84 39 40 18 77 69 70 88 93",
"output": "902"
},
{
"input": "15\n87 22 98 32 88 36 72 31 100 97 17 16 60 22 20",
"output": "798"
},
{
"input": "15\n15 63 51 13 37 9 43 19 55 79 57 60 50 59 31",
"output": "632"
},
{
"input": "1\n4",
"output": "4"
},
{
"input": "2\n1 4",
"output": "4"
},
{
"input": "3\n1 2 4",
"output": "6"
},
{
"input": "2\n9 3",
"output": "12"
},
{
"input": "2\n1000000000 1001",
"output": "1000000000"
},
{
"input": "3\n1 8 4",
"output": "12"
},
{
"input": "3\n7 4 4",
"output": "8"
},
{
"input": "5\n2 3 4 5 3",
"output": "14"
},
{
"input": "2\n4 5",
"output": "4"
},
{
"input": "3\n2 4 5",
"output": "6"
},
{
"input": "3\n2 2 3",
"output": "4"
},
{
"input": "2\n2 3",
"output": "2"
},
{
"input": "4\n2 3 7 7",
"output": "16"
},
{
"input": "2\n999999999 2",
"output": "2"
},
{
"input": "2\n2 5",
"output": "2"
},
{
"input": "3\n5 3 1",
"output": "8"
},
{
"input": "4\n3 2 5 7",
"output": "14"
}
] | 1,620,353,112
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 87
| 140
| 14,950,400
|
# input
n = int(input())
nums = [int(i) for i in input().split()[:n]]
# code
least_odd = None
total = 0
for num in nums:
num_mod = num % 2
if num_mod is 1:
if least_odd is None:
least_odd = num
elif least_odd > num:
least_odd = num
total = total + num
if total % 2 is 1:
total = total - least_odd
print(total)
|
Title: Wet Shark and Odd and Even
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today, Wet Shark is given *n* integers. Using any of these integers no more than once, Wet Shark wants to get maximum possible even (divisible by 2) sum. Please, calculate this value for Wet Shark.
Note, that if Wet Shark uses no integers from the *n* integers, the sum is an even integer 0.
Input Specification:
The first line of the input contains one integer, *n* (1<=β€<=*n*<=β€<=100<=000). The next line contains *n* space separated integers given to Wet Shark. Each of these integers is in range from 1 to 109, inclusive.
Output Specification:
Print the maximum possible even sum that can be obtained if we use some of the given integers.
Demo Input:
['3\n1 2 3\n', '5\n999999999 999999999 999999999 999999999 999999999\n']
Demo Output:
['6', '3999999996']
Note:
In the first sample, we can simply take all three integers for a total sum of 6.
In the second sample Wet Shark should take any four out of five integers 999β999β999.
|
```python
# input
n = int(input())
nums = [int(i) for i in input().split()[:n]]
# code
least_odd = None
total = 0
for num in nums:
num_mod = num % 2
if num_mod is 1:
if least_odd is None:
least_odd = num
elif least_odd > num:
least_odd = num
total = total + num
if total % 2 is 1:
total = total - least_odd
print(total)
```
| 3
|
|
844
|
A
|
Diversity
|
PROGRAMMING
| 1,000
|
[
"greedy",
"implementation",
"strings"
] | null | null |
Calculate the minimum number of characters you need to change in the string *s*, so that it contains at least *k* different letters, or print that it is impossible.
String *s* consists only of lowercase Latin letters, and it is allowed to change characters only to lowercase Latin letters too.
|
First line of input contains string *s*, consisting only of lowercase Latin letters (1<=β€<=|*s*|<=β€<=1000, |*s*| denotes the length of *s*).
Second line of input contains integer *k* (1<=β€<=*k*<=β€<=26).
|
Print single line with a minimum number of necessary changes, or the word Β«impossibleΒ» (without quotes) if it is impossible.
|
[
"yandex\n6\n",
"yahoo\n5\n",
"google\n7\n"
] |
[
"0\n",
"1\n",
"impossible\n"
] |
In the first test case string contains 6 different letters, so we don't need to change anything.
In the second test case string contains 4 different letters: {'*a*',β'*h*',β'*o*',β'*y*'}. To get 5 different letters it is necessary to change one occurrence of '*o*' to some letter, which doesn't occur in the string, for example, {'*b*'}.
In the third test case, it is impossible to make 7 different letters because the length of the string is 6.
| 500
|
[
{
"input": "yandex\n6",
"output": "0"
},
{
"input": "yahoo\n5",
"output": "1"
},
{
"input": "google\n7",
"output": "impossible"
},
{
"input": "a\n1",
"output": "0"
},
{
"input": "z\n2",
"output": "impossible"
},
{
"input": "fwgfrwgkuwghfiruhewgirueguhergiqrbvgrgf\n26",
"output": "14"
},
{
"input": "nfevghreuoghrueighoqghbnebvnejbvnbgneluqe\n26",
"output": "12"
},
{
"input": "a\n3",
"output": "impossible"
},
{
"input": "smaxpqplaqqbxuqxalqmbmmgubbpspxhawbxsuqhhegpmmpebqmqpbbeplwaepxmsahuepuhuhwxeqmmlgqubuaxehwuwasgxpqmugbmuawuhwqlswllssueglbxepbmwgs\n1",
"output": "0"
},
{
"input": "cuguccgcugcugucgggggcgcgucgucugcuuuccccuugccg\n4",
"output": "1"
},
{
"input": "fcfccfcfccfcfcffcffffffcfccfccfcffccccfcffffccfccfcffcfcccccffcfffcccffcfccfffffcccfccffffffccfccccf\n20",
"output": "18"
},
{
"input": "swmkwaruyv\n5",
"output": "0"
},
{
"input": "tnbqpsuhkczmejirvyfdolxwga\n22",
"output": "0"
},
{
"input": "abcde\n3",
"output": "0"
},
{
"input": "abb\n1",
"output": "0"
},
{
"input": "aaaa\n1",
"output": "0"
},
{
"input": "abcde\n2",
"output": "0"
},
{
"input": "yandex\n4",
"output": "0"
},
{
"input": "aaabbbccc\n1",
"output": "0"
},
{
"input": "abcd\n2",
"output": "0"
},
{
"input": "asdfgh\n2",
"output": "0"
},
{
"input": "aab\n1",
"output": "0"
},
{
"input": "mynameissako\n5",
"output": "0"
},
{
"input": "abcde\n1",
"output": "0"
},
{
"input": "abcd\n3",
"output": "0"
},
{
"input": "abcdef\n2",
"output": "0"
},
{
"input": "abcdefg\n4",
"output": "0"
},
{
"input": "abc\n1",
"output": "0"
},
{
"input": "asdafjsgljdllgjdgkl\n5",
"output": "0"
},
{
"input": "yaay\n3",
"output": "1"
},
{
"input": "yaay\n4",
"output": "2"
},
{
"input": "zzzzzz\n2",
"output": "1"
}
] | 1,646,159,552
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 41
| 62
| 0
|
x = input()
s = len(set(list(x)))
n = int(input())
if len(x)<n:
print ('impossible')
else :
if s>=n:
print(0)
else:
print(n-s)
|
Title: Diversity
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Calculate the minimum number of characters you need to change in the string *s*, so that it contains at least *k* different letters, or print that it is impossible.
String *s* consists only of lowercase Latin letters, and it is allowed to change characters only to lowercase Latin letters too.
Input Specification:
First line of input contains string *s*, consisting only of lowercase Latin letters (1<=β€<=|*s*|<=β€<=1000, |*s*| denotes the length of *s*).
Second line of input contains integer *k* (1<=β€<=*k*<=β€<=26).
Output Specification:
Print single line with a minimum number of necessary changes, or the word Β«impossibleΒ» (without quotes) if it is impossible.
Demo Input:
['yandex\n6\n', 'yahoo\n5\n', 'google\n7\n']
Demo Output:
['0\n', '1\n', 'impossible\n']
Note:
In the first test case string contains 6 different letters, so we don't need to change anything.
In the second test case string contains 4 different letters: {'*a*',β'*h*',β'*o*',β'*y*'}. To get 5 different letters it is necessary to change one occurrence of '*o*' to some letter, which doesn't occur in the string, for example, {'*b*'}.
In the third test case, it is impossible to make 7 different letters because the length of the string is 6.
|
```python
x = input()
s = len(set(list(x)))
n = int(input())
if len(x)<n:
print ('impossible')
else :
if s>=n:
print(0)
else:
print(n-s)
```
| 3
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
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<=β€<=*n*<=β€<=105). The second line contains an array of original integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=109<=β€<=*a**i*<=β€<=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].
| 0
|
[
{
"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"
}
] | 1,682,616,594
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 50
| 372
| 11,161,600
|
n = (int)(input())
a = (list)(map(int , input().split()))
ans = 0
cnt = 1
for i in range(1 , n) :
if a[i] == a[i-1] : cnt += 1
else :
ans += cnt * (cnt + 1) // 2
cnt = 1
ans += cnt * (cnt + 1) // 2
print(ans)
|
Title: none
Time Limit: None seconds
Memory Limit: None 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<=β€<=*n*<=β€<=105). The second line contains an array of original integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=109<=β€<=*a**i*<=β€<=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
n = (int)(input())
a = (list)(map(int , input().split()))
ans = 0
cnt = 1
for i in range(1 , n) :
if a[i] == a[i-1] : cnt += 1
else :
ans += cnt * (cnt + 1) // 2
cnt = 1
ans += cnt * (cnt + 1) // 2
print(ans)
```
| 3
|
|
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<=β€<=*n*<=β€<=200<=000)Β β 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"
}
] | 1,666,875,223
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 95
| 1,731
| 28,876,800
|
n=int(input())
ar=[input() for i in range(n)]
ar=ar[::-1]
d={}
for i in ar:
if not i in d:
d[i]=1
print(i)
|
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<=β€<=*n*<=β€<=200<=000)Β β 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())
ar=[input() for i in range(n)]
ar=ar[::-1]
d={}
for i in ar:
if not i in d:
d[i]=1
print(i)
```
| 3
|
|
279
|
B
|
Books
|
PROGRAMMING
| 1,400
|
[
"binary search",
"brute force",
"implementation",
"two pointers"
] | null | null |
When Valera has got some free time, he goes to the library to read some books. Today he's got *t* free minutes to read. That's why Valera took *n* books in the library and for each book he estimated the time he is going to need to read it. Let's number the books by integers from 1 to *n*. Valera needs *a**i* minutes to read the *i*-th book.
Valera decided to choose an arbitrary book with number *i* and read the books one by one, starting from this book. In other words, he will first read book number *i*, then book number *i*<=+<=1, then book number *i*<=+<=2 and so on. He continues the process until he either runs out of the free time or finishes reading the *n*-th book. Valera reads each book up to the end, that is, he doesn't start reading the book if he doesn't have enough free time to finish reading it.
Print the maximum number of books Valera can read.
|
The first line contains two integers *n* and *t* (1<=β€<=*n*<=β€<=105;Β 1<=β€<=*t*<=β€<=109) β the number of books and the number of free minutes Valera's got. The second line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=β€<=*a**i*<=β€<=104), where number *a**i* shows the number of minutes that the boy needs to read the *i*-th book.
|
Print a single integer β the maximum number of books Valera can read.
|
[
"4 5\n3 1 2 1\n",
"3 3\n2 2 3\n"
] |
[
"3\n",
"1\n"
] |
none
| 1,000
|
[
{
"input": "4 5\n3 1 2 1",
"output": "3"
},
{
"input": "3 3\n2 2 3",
"output": "1"
},
{
"input": "1 3\n5",
"output": "0"
},
{
"input": "1 10\n4",
"output": "1"
},
{
"input": "2 10\n6 4",
"output": "2"
},
{
"input": "6 10\n2 3 4 2 1 1",
"output": "4"
},
{
"input": "7 13\n6 8 14 9 4 11 10",
"output": "2"
},
{
"input": "10 15\n10 9 1 1 5 10 5 3 7 2",
"output": "3"
},
{
"input": "20 30\n8 1 2 6 9 4 1 9 9 10 4 7 8 9 5 7 1 8 7 4",
"output": "6"
},
{
"input": "30 60\n16 13 22 38 13 35 17 17 20 38 12 19 9 22 20 3 35 34 34 21 35 40 22 3 27 19 12 4 8 19",
"output": "4"
},
{
"input": "100 100\n75 92 18 6 81 67 7 92 100 65 82 32 50 67 85 31 80 91 84 63 39 52 92 81 1 98 24 12 43 48 17 86 51 72 48 95 45 50 12 66 19 79 49 89 34 1 97 75 20 33 96 27 42 23 73 71 93 1 85 19 66 14 17 61 20 39 36 33 42 61 56 64 23 91 80 99 40 74 13 18 98 85 74 39 62 84 46 74 50 23 38 11 79 14 9 25 66 100 25 52",
"output": "3"
},
{
"input": "10 1\n4418 7528 8170 1736 1317 3205 8183 4995 8039 4708",
"output": "0"
},
{
"input": "50 2\n124 214 63 73 996 760 38 571 451 300 970 1 706 937 837 494 619 88 851 411 957 990 842 613 821 649 627 34 693 678 734 116 816 985 705 940 499 493 922 967 854 439 112 644 961 438 189 572 655 550",
"output": "1"
}
] | 1,695,112,780
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 38
| 186
| 13,824,000
|
#from submission id 29501657 cf
n, t = map(int, input().split())
a = list(map(int, input().split()))
i, j, s = 0, 0, 0
for j in range(len(a)):
s += a[j]
if s > t:
s -= a[i]
i += 1
print(j - i + 1)
|
Title: Books
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
When Valera has got some free time, he goes to the library to read some books. Today he's got *t* free minutes to read. That's why Valera took *n* books in the library and for each book he estimated the time he is going to need to read it. Let's number the books by integers from 1 to *n*. Valera needs *a**i* minutes to read the *i*-th book.
Valera decided to choose an arbitrary book with number *i* and read the books one by one, starting from this book. In other words, he will first read book number *i*, then book number *i*<=+<=1, then book number *i*<=+<=2 and so on. He continues the process until he either runs out of the free time or finishes reading the *n*-th book. Valera reads each book up to the end, that is, he doesn't start reading the book if he doesn't have enough free time to finish reading it.
Print the maximum number of books Valera can read.
Input Specification:
The first line contains two integers *n* and *t* (1<=β€<=*n*<=β€<=105;Β 1<=β€<=*t*<=β€<=109) β the number of books and the number of free minutes Valera's got. The second line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=β€<=*a**i*<=β€<=104), where number *a**i* shows the number of minutes that the boy needs to read the *i*-th book.
Output Specification:
Print a single integer β the maximum number of books Valera can read.
Demo Input:
['4 5\n3 1 2 1\n', '3 3\n2 2 3\n']
Demo Output:
['3\n', '1\n']
Note:
none
|
```python
#from submission id 29501657 cf
n, t = map(int, input().split())
a = list(map(int, input().split()))
i, j, s = 0, 0, 0
for j in range(len(a)):
s += a[j]
if s > t:
s -= a[i]
i += 1
print(j - i + 1)
```
| 3
|
|
839
|
A
|
Arya and Bran
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Bran and his older sister Arya are from the same house. Bran like candies so much, so Arya is going to give him some Candies.
At first, Arya and Bran have 0 Candies. There are *n* days, at the *i*-th day, Arya finds *a**i* candies in a box, that is given by the Many-Faced God. Every day she can give Bran at most 8 of her candies. If she don't give him the candies at the same day, they are saved for her and she can give them to him later.
Your task is to find the minimum number of days Arya needs to give Bran *k* candies before the end of the *n*-th day. Formally, you need to output the minimum day index to the end of which *k* candies will be given out (the days are indexed from 1 to *n*).
Print -1 if she can't give him *k* candies during *n* given days.
|
The first line contains two integers *n* and *k* (1<=β€<=*n*<=β€<=100, 1<=β€<=*k*<=β€<=10000).
The second line contains *n* integers *a*1,<=*a*2,<=*a*3,<=...,<=*a**n* (1<=β€<=*a**i*<=β€<=100).
|
If it is impossible for Arya to give Bran *k* candies within *n* days, print -1.
Otherwise print a single integerΒ β the minimum number of days Arya needs to give Bran *k* candies before the end of the *n*-th day.
|
[
"2 3\n1 2\n",
"3 17\n10 10 10\n",
"1 9\n10\n"
] |
[
"2",
"3",
"-1"
] |
In the first sample, Arya can give Bran 3 candies in 2 days.
In the second sample, Arya can give Bran 17 candies in 3 days, because she can give him at most 8 candies per day.
In the third sample, Arya can't give Bran 9 candies, because she can give him at most 8 candies per day and she must give him the candies within 1 day.
| 500
|
[
{
"input": "2 3\n1 2",
"output": "2"
},
{
"input": "3 17\n10 10 10",
"output": "3"
},
{
"input": "1 9\n10",
"output": "-1"
},
{
"input": "10 70\n6 5 2 3 3 2 1 4 3 2",
"output": "-1"
},
{
"input": "20 140\n40 4 81 40 10 54 34 50 84 60 16 1 90 78 38 93 99 60 81 99",
"output": "18"
},
{
"input": "30 133\n3 2 3 4 3 7 4 5 5 6 7 2 1 3 4 6 7 4 6 4 7 5 7 1 3 4 1 6 8 5",
"output": "30"
},
{
"input": "40 320\n70 79 21 64 95 36 63 29 66 89 30 34 100 76 42 12 4 56 80 78 83 1 39 9 34 45 6 71 27 31 55 52 72 71 38 21 43 83 48 47",
"output": "40"
},
{
"input": "50 300\n5 3 11 8 7 4 9 5 5 1 6 3 5 7 4 2 2 10 8 1 7 10 4 4 11 5 2 4 9 1 5 4 11 9 11 2 7 4 4 8 10 9 1 11 10 2 4 11 6 9",
"output": "-1"
},
{
"input": "37 30\n1 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": "30"
},
{
"input": "100 456\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": "57"
},
{
"input": "90 298\n94 90 98 94 93 90 99 98 90 96 93 96 92 92 97 98 94 94 96 100 93 96 95 98 94 91 95 95 94 90 93 96 93 100 99 98 94 95 98 91 91 98 97 100 98 93 92 93 91 100 92 97 95 95 97 94 98 97 99 100 90 96 93 100 95 99 92 100 99 91 97 99 98 93 90 93 97 95 94 96 90 100 94 93 91 92 97 97 97 100",
"output": "38"
},
{
"input": "7 43\n4 3 7 9 3 8 10",
"output": "-1"
},
{
"input": "99 585\n8 2 3 3 10 7 9 4 7 4 6 8 7 11 5 8 7 4 7 7 6 7 11 8 1 7 3 2 10 1 6 10 10 5 10 2 5 5 11 6 4 1 5 10 5 8 1 3 7 10 6 1 1 3 8 11 5 8 2 2 5 4 7 6 7 5 8 7 10 9 6 11 4 8 2 7 1 7 1 4 11 1 9 6 1 10 6 10 1 5 6 5 2 5 11 5 1 10 8",
"output": "-1"
},
{
"input": "30 177\n8 7 5 8 3 7 2 4 3 8 11 3 9 11 2 4 1 4 5 6 11 5 8 3 6 3 11 2 11 8",
"output": "-1"
},
{
"input": "19 129\n3 3 10 11 4 7 3 8 10 2 11 6 11 9 4 2 11 10 5",
"output": "-1"
},
{
"input": "100 100\n1 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 1",
"output": "100"
},
{
"input": "13 104\n94 55 20 96 86 76 13 71 13 1 32 76 69",
"output": "13"
},
{
"input": "85 680\n61 44 55 6 30 74 27 26 17 45 73 1 67 71 39 32 13 25 79 66 4 59 49 28 29 22 10 17 98 80 36 99 52 24 59 44 27 79 29 46 29 12 47 72 82 25 6 30 81 72 95 65 30 71 72 45 39 16 16 89 48 42 59 71 50 58 31 65 91 70 48 56 28 34 53 89 94 98 49 55 94 65 91 11 53",
"output": "85"
},
{
"input": "100 458\n3 6 4 1 8 4 1 5 4 4 5 8 4 4 6 6 5 1 2 2 2 1 7 1 1 2 6 5 7 8 3 3 8 3 7 5 7 6 6 2 4 2 2 1 1 8 6 1 5 3 3 4 1 4 6 8 5 4 8 5 4 5 5 1 3 1 6 7 6 2 7 3 4 8 1 8 6 7 1 2 4 6 7 4 8 8 8 4 8 7 5 2 8 4 2 5 6 8 8 5",
"output": "100"
},
{
"input": "98 430\n4 7 6 3 4 1 7 1 1 6 6 1 5 4 6 1 5 4 6 6 1 5 1 1 8 1 6 6 2 6 8 4 4 6 6 8 8 7 4 1 2 4 1 5 4 3 7 3 2 5 7 7 7 2 2 2 7 2 8 7 3 4 5 7 8 3 7 6 7 3 2 4 7 1 4 4 7 1 1 8 4 5 8 3 1 5 3 5 2 1 3 3 8 1 3 5 8 6",
"output": "98"
},
{
"input": "90 80\n6 1 7 1 1 8 6 6 6 1 5 4 2 2 8 4 8 7 7 2 5 7 7 8 5 5 6 3 3 8 3 5 6 3 4 2 6 5 5 3 3 3 8 6 6 1 8 3 6 5 4 8 5 4 3 7 1 3 2 3 3 7 7 7 3 5 2 6 2 3 6 4 6 5 5 3 2 1 1 7 3 3 4 3 4 2 1 2 3 1",
"output": "18"
},
{
"input": "89 99\n7 7 3 5 2 7 8 8 1 1 5 7 7 4 1 5 3 4 4 8 8 3 3 2 6 3 8 2 7 5 8 1 3 5 3 6 4 3 6 2 3 3 4 5 1 6 1 7 7 7 6 7 7 7 8 8 8 2 1 7 5 8 6 7 7 4 7 5 7 8 1 3 5 8 7 1 4 2 5 8 3 4 4 5 5 6 2 4 2",
"output": "21"
},
{
"input": "50 700\n4 3 2 8 8 5 5 3 3 4 7 2 6 6 3 3 8 4 2 4 8 6 5 4 5 4 5 8 6 5 4 7 2 4 1 6 2 6 8 6 2 5 8 1 3 8 3 8 4 1",
"output": "-1"
},
{
"input": "82 359\n95 98 95 90 90 96 91 94 93 99 100 100 92 99 96 94 99 90 94 96 91 91 90 93 97 96 90 94 97 99 93 90 99 98 96 100 93 97 100 91 100 92 93 100 92 90 90 94 99 95 100 98 99 96 94 96 96 99 99 91 97 100 95 100 99 91 94 91 98 98 100 97 93 93 96 97 94 94 92 100 91 91",
"output": "45"
},
{
"input": "60 500\n93 93 100 99 91 92 95 93 95 99 93 91 97 98 90 91 98 100 95 100 94 93 92 91 91 98 98 90 93 91 90 96 92 93 92 94 94 91 96 94 98 100 97 96 96 97 91 99 97 95 96 94 91 92 99 95 97 92 98 90",
"output": "-1"
},
{
"input": "98 776\n48 63 26 3 88 81 27 33 37 10 2 89 41 84 98 93 25 44 42 90 41 65 97 1 28 69 42 14 86 18 96 28 28 94 78 8 44 31 96 45 26 52 93 25 48 39 3 75 94 93 63 59 67 86 18 74 27 38 68 7 31 60 69 67 20 11 19 34 47 43 86 96 3 49 56 60 35 49 89 28 92 69 48 15 17 73 99 69 2 73 27 35 28 53 11 1 96 50",
"output": "97"
},
{
"input": "100 189\n15 14 32 65 28 96 33 93 48 28 57 20 32 20 90 42 57 53 18 58 94 21 27 29 37 22 94 45 67 60 83 23 20 23 35 93 3 42 6 46 68 46 34 25 17 16 50 5 49 91 23 76 69 100 58 68 81 32 88 41 64 29 37 13 95 25 6 59 74 58 31 35 16 80 13 80 10 59 85 18 16 70 51 40 44 28 8 76 8 87 53 86 28 100 2 73 14 100 52 9",
"output": "24"
},
{
"input": "99 167\n72 4 79 73 49 58 15 13 92 92 42 36 35 21 13 10 51 94 64 35 86 50 6 80 93 77 59 71 2 88 22 10 27 30 87 12 77 6 34 56 31 67 78 84 36 27 15 15 12 56 80 7 56 14 10 9 14 59 15 20 34 81 8 49 51 72 4 58 38 77 31 86 18 61 27 86 95 36 46 36 39 18 78 39 48 37 71 12 51 92 65 48 39 22 16 87 4 5 42",
"output": "21"
},
{
"input": "90 4\n48 4 4 78 39 3 85 29 69 52 70 39 11 98 42 56 65 98 77 24 61 31 6 59 60 62 84 46 67 59 15 44 99 23 12 74 2 48 84 60 51 28 17 90 10 82 3 43 50 100 45 57 57 95 53 71 20 74 52 46 64 59 72 33 74 16 44 44 80 71 83 1 70 59 61 6 82 69 81 45 88 28 17 24 22 25 53 97 1 100",
"output": "1"
},
{
"input": "30 102\n55 94 3 96 3 47 92 85 25 78 27 70 97 83 40 2 55 12 74 84 91 37 31 85 7 40 33 54 72 5",
"output": "13"
},
{
"input": "81 108\n61 59 40 100 8 75 5 74 87 12 6 23 98 26 59 68 27 4 98 79 14 44 4 11 89 77 29 90 33 3 43 1 87 91 28 24 4 84 75 7 37 46 15 46 8 87 68 66 5 21 36 62 77 74 91 95 88 28 12 48 18 93 14 51 33 5 99 62 99 38 49 15 56 87 52 64 69 46 41 12 92",
"output": "14"
},
{
"input": "2 16\n10 6",
"output": "2"
},
{
"input": "2 8\n7 8",
"output": "2"
},
{
"input": "2 9\n4 8",
"output": "2"
},
{
"input": "3 19\n9 9 1",
"output": "3"
},
{
"input": "4 32\n9 9 9 5",
"output": "4"
},
{
"input": "2 15\n14 1",
"output": "2"
},
{
"input": "2 3\n3 3",
"output": "1"
},
{
"input": "3 10\n10 1 1",
"output": "2"
},
{
"input": "12 20\n3 16 19 10 1 6 17 8 6 20 1 4",
"output": "4"
},
{
"input": "4 15\n14 3 3 3",
"output": "2"
},
{
"input": "5 40\n10 10 10 10 1",
"output": "5"
},
{
"input": "4 31\n9 9 8 5",
"output": "4"
},
{
"input": "4 31\n20 7 1 1",
"output": "-1"
},
{
"input": "2 10\n9 1",
"output": "2"
},
{
"input": "10 50\n100 10 1 1 1 1 1 1 1 1",
"output": "7"
},
{
"input": "2 11\n10 2",
"output": "2"
},
{
"input": "3 21\n10 10 1",
"output": "3"
},
{
"input": "2 2\n1 2",
"output": "2"
},
{
"input": "3 2\n1 8 8",
"output": "2"
},
{
"input": "2 11\n10 1",
"output": "2"
},
{
"input": "2 16\n12 4",
"output": "2"
},
{
"input": "3 11\n9 2 2",
"output": "2"
},
{
"input": "3 11\n4 3 4",
"output": "3"
},
{
"input": "2 13\n7 6",
"output": "2"
},
{
"input": "3 24\n14 3 4",
"output": "-1"
},
{
"input": "2 13\n10 3",
"output": "2"
},
{
"input": "3 11\n9 2 1",
"output": "2"
},
{
"input": "2 15\n12 3",
"output": "2"
},
{
"input": "2 14\n11 4",
"output": "2"
}
] | 1,577,818,458
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 60
| 140
| 0
|
n,k=map(int,input().split())
a=list(map(int,input().split()))
ans,res=0,0
for i in range(n):
if a[i]>8:
if i==n-1:
a[n-1]+=a[i]-8
else:
a[i+1]+=a[i]-8
a[i]=8
ans+=a[i]
res+=1
#print(ans)
if ans>=k:
print(res)
exit()
print(-1)
|
Title: Arya and Bran
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bran and his older sister Arya are from the same house. Bran like candies so much, so Arya is going to give him some Candies.
At first, Arya and Bran have 0 Candies. There are *n* days, at the *i*-th day, Arya finds *a**i* candies in a box, that is given by the Many-Faced God. Every day she can give Bran at most 8 of her candies. If she don't give him the candies at the same day, they are saved for her and she can give them to him later.
Your task is to find the minimum number of days Arya needs to give Bran *k* candies before the end of the *n*-th day. Formally, you need to output the minimum day index to the end of which *k* candies will be given out (the days are indexed from 1 to *n*).
Print -1 if she can't give him *k* candies during *n* given days.
Input Specification:
The first line contains two integers *n* and *k* (1<=β€<=*n*<=β€<=100, 1<=β€<=*k*<=β€<=10000).
The second line contains *n* integers *a*1,<=*a*2,<=*a*3,<=...,<=*a**n* (1<=β€<=*a**i*<=β€<=100).
Output Specification:
If it is impossible for Arya to give Bran *k* candies within *n* days, print -1.
Otherwise print a single integerΒ β the minimum number of days Arya needs to give Bran *k* candies before the end of the *n*-th day.
Demo Input:
['2 3\n1 2\n', '3 17\n10 10 10\n', '1 9\n10\n']
Demo Output:
['2', '3', '-1']
Note:
In the first sample, Arya can give Bran 3 candies in 2 days.
In the second sample, Arya can give Bran 17 candies in 3 days, because she can give him at most 8 candies per day.
In the third sample, Arya can't give Bran 9 candies, because she can give him at most 8 candies per day and she must give him the candies within 1 day.
|
```python
n,k=map(int,input().split())
a=list(map(int,input().split()))
ans,res=0,0
for i in range(n):
if a[i]>8:
if i==n-1:
a[n-1]+=a[i]-8
else:
a[i+1]+=a[i]-8
a[i]=8
ans+=a[i]
res+=1
#print(ans)
if ans>=k:
print(res)
exit()
print(-1)
```
| 3
|
|
805
|
A
|
Fake NP
|
PROGRAMMING
| 1,000
|
[
"greedy",
"math"
] | null | null |
Tavak and Seyyed are good friends. Seyyed is very funny and he told Tavak to solve the following problem instead of longest-path.
You are given *l* and *r*. For all integers from *l* to *r*, inclusive, we wrote down all of their integer divisors except 1. Find the integer that we wrote down the maximum number of times.
Solve the problem to show that it's not a NP problem.
|
The first line contains two integers *l* and *r* (2<=β€<=*l*<=β€<=*r*<=β€<=109).
|
Print single integer, the integer that appears maximum number of times in the divisors.
If there are multiple answers, print any of them.
|
[
"19 29\n",
"3 6\n"
] |
[
"2\n",
"3\n"
] |
Definition of a divisor: [https://www.mathsisfun.com/definitions/divisor-of-an-integer-.html](https://www.mathsisfun.com/definitions/divisor-of-an-integer-.html)
The first example: from 19 to 29 these numbers are divisible by 2: {20,β22,β24,β26,β28}.
The second example: from 3 to 6 these numbers are divisible by 3: {3,β6}.
| 500
|
[
{
"input": "19 29",
"output": "2"
},
{
"input": "3 6",
"output": "2"
},
{
"input": "39 91",
"output": "2"
},
{
"input": "76 134",
"output": "2"
},
{
"input": "93 95",
"output": "2"
},
{
"input": "17 35",
"output": "2"
},
{
"input": "94 95",
"output": "2"
},
{
"input": "51 52",
"output": "2"
},
{
"input": "47 52",
"output": "2"
},
{
"input": "38 98",
"output": "2"
},
{
"input": "30 37",
"output": "2"
},
{
"input": "56 92",
"output": "2"
},
{
"input": "900000000 1000000000",
"output": "2"
},
{
"input": "37622224 162971117",
"output": "2"
},
{
"input": "760632746 850720703",
"output": "2"
},
{
"input": "908580370 968054552",
"output": "2"
},
{
"input": "951594860 953554446",
"output": "2"
},
{
"input": "347877978 913527175",
"output": "2"
},
{
"input": "620769961 988145114",
"output": "2"
},
{
"input": "820844234 892579936",
"output": "2"
},
{
"input": "741254764 741254768",
"output": "2"
},
{
"input": "80270976 80270977",
"output": "2"
},
{
"input": "392602363 392602367",
"output": "2"
},
{
"input": "519002744 519002744",
"output": "519002744"
},
{
"input": "331900277 331900277",
"output": "331900277"
},
{
"input": "419873015 419873018",
"output": "2"
},
{
"input": "349533413 349533413",
"output": "349533413"
},
{
"input": "28829775 28829776",
"output": "2"
},
{
"input": "568814539 568814539",
"output": "568814539"
},
{
"input": "720270740 720270743",
"output": "2"
},
{
"input": "871232720 871232722",
"output": "2"
},
{
"input": "305693653 305693653",
"output": "305693653"
},
{
"input": "634097178 634097179",
"output": "2"
},
{
"input": "450868287 450868290",
"output": "2"
},
{
"input": "252662256 252662260",
"output": "2"
},
{
"input": "575062045 575062049",
"output": "2"
},
{
"input": "273072892 273072894",
"output": "2"
},
{
"input": "770439256 770439256",
"output": "770439256"
},
{
"input": "2 1000000000",
"output": "2"
},
{
"input": "6 8",
"output": "2"
},
{
"input": "2 879190747",
"output": "2"
},
{
"input": "5 5",
"output": "5"
},
{
"input": "999999937 999999937",
"output": "999999937"
},
{
"input": "3 3",
"output": "3"
},
{
"input": "5 100",
"output": "2"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "3 18",
"output": "2"
},
{
"input": "7 7",
"output": "7"
},
{
"input": "39916801 39916801",
"output": "39916801"
},
{
"input": "3 8",
"output": "2"
},
{
"input": "13 13",
"output": "13"
},
{
"input": "4 8",
"output": "2"
},
{
"input": "3 12",
"output": "2"
},
{
"input": "6 12",
"output": "2"
},
{
"input": "999999103 999999103",
"output": "999999103"
},
{
"input": "100000007 100000007",
"output": "100000007"
},
{
"input": "3 99",
"output": "2"
},
{
"input": "999999733 999999733",
"output": "999999733"
},
{
"input": "5 10",
"output": "2"
},
{
"input": "982451653 982451653",
"output": "982451653"
},
{
"input": "999900001 1000000000",
"output": "2"
},
{
"input": "999727999 999727999",
"output": "999727999"
},
{
"input": "2 999999999",
"output": "2"
},
{
"input": "242 244",
"output": "2"
},
{
"input": "3 10",
"output": "2"
},
{
"input": "15 27",
"output": "2"
},
{
"input": "998244353 998244353",
"output": "998244353"
},
{
"input": "5 15",
"output": "2"
},
{
"input": "999999797 999999797",
"output": "999999797"
},
{
"input": "2 3",
"output": "2"
},
{
"input": "999999929 999999929",
"output": "999999929"
},
{
"input": "3 111111",
"output": "2"
},
{
"input": "12 18",
"output": "2"
},
{
"input": "479001599 479001599",
"output": "479001599"
},
{
"input": "10000019 10000019",
"output": "10000019"
},
{
"input": "715827883 715827883",
"output": "715827883"
},
{
"input": "999992977 999992977",
"output": "999992977"
},
{
"input": "11 11",
"output": "11"
},
{
"input": "29 29",
"output": "29"
},
{
"input": "1000003 1000003",
"output": "1000003"
},
{
"input": "6 15",
"output": "2"
},
{
"input": "1200007 1200007",
"output": "1200007"
},
{
"input": "3 1000000000",
"output": "2"
},
{
"input": "990000023 990000023",
"output": "990000023"
},
{
"input": "1717 1717",
"output": "1717"
},
{
"input": "141650963 141650963",
"output": "141650963"
},
{
"input": "1002523 1002523",
"output": "1002523"
},
{
"input": "900000011 900000011",
"output": "900000011"
},
{
"input": "104729 104729",
"output": "104729"
},
{
"input": "4 12",
"output": "2"
},
{
"input": "100003 100003",
"output": "100003"
},
{
"input": "17 17",
"output": "17"
},
{
"input": "10 100",
"output": "2"
}
] | 1,638,985,544
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 93
| 124
| 512,000
|
a, b = list(map(int, input().split()))
if abs(a - b) > 1 or a % 2 == 0:
print(2)
else:
print(a)
|
Title: Fake NP
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Tavak and Seyyed are good friends. Seyyed is very funny and he told Tavak to solve the following problem instead of longest-path.
You are given *l* and *r*. For all integers from *l* to *r*, inclusive, we wrote down all of their integer divisors except 1. Find the integer that we wrote down the maximum number of times.
Solve the problem to show that it's not a NP problem.
Input Specification:
The first line contains two integers *l* and *r* (2<=β€<=*l*<=β€<=*r*<=β€<=109).
Output Specification:
Print single integer, the integer that appears maximum number of times in the divisors.
If there are multiple answers, print any of them.
Demo Input:
['19 29\n', '3 6\n']
Demo Output:
['2\n', '3\n']
Note:
Definition of a divisor: [https://www.mathsisfun.com/definitions/divisor-of-an-integer-.html](https://www.mathsisfun.com/definitions/divisor-of-an-integer-.html)
The first example: from 19 to 29 these numbers are divisible by 2: {20,β22,β24,β26,β28}.
The second example: from 3 to 6 these numbers are divisible by 3: {3,β6}.
|
```python
a, b = list(map(int, input().split()))
if abs(a - b) > 1 or a % 2 == 0:
print(2)
else:
print(a)
```
| 3
|
|
992
|
B
|
Nastya Studies Informatics
|
PROGRAMMING
| 1,600
|
[
"math",
"number theory"
] | null | null |
Today on Informatics class Nastya learned about GCD and LCM (see links below). Nastya is very intelligent, so she solved all the tasks momentarily and now suggests you to solve one of them as well.
We define a pair of integers (*a*,<=*b*) good, if *GCD*(*a*,<=*b*)<==<=*x* and *LCM*(*a*,<=*b*)<==<=*y*, where *GCD*(*a*,<=*b*) denotes the [greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor) of *a* and *b*, and *LCM*(*a*,<=*b*) denotes the [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) of *a* and *b*.
You are given two integers *x* and *y*. You are to find the number of good pairs of integers (*a*,<=*b*) such that *l*<=β€<=*a*,<=*b*<=β€<=*r*. Note that pairs (*a*,<=*b*) and (*b*,<=*a*) are considered different if *a*<=β <=*b*.
|
The only line contains four integers *l*,<=*r*,<=*x*,<=*y* (1<=β€<=*l*<=β€<=*r*<=β€<=109, 1<=β€<=*x*<=β€<=*y*<=β€<=109).
|
In the only line print the only integerΒ β the answer for the problem.
|
[
"1 2 1 2\n",
"1 12 1 12\n",
"50 100 3 30\n"
] |
[
"2\n",
"4\n",
"0\n"
] |
In the first example there are two suitable good pairs of integers (*a*,β*b*): (1,β2) and (2,β1).
In the second example there are four suitable good pairs of integers (*a*,β*b*): (1,β12), (12,β1), (3,β4) and (4,β3).
In the third example there are good pairs of integers, for example, (3,β30), but none of them fits the condition *l*ββ€β*a*,β*b*ββ€β*r*.
| 1,000
|
[
{
"input": "1 2 1 2",
"output": "2"
},
{
"input": "1 12 1 12",
"output": "4"
},
{
"input": "50 100 3 30",
"output": "0"
},
{
"input": "1 1000000000 1 1000000000",
"output": "4"
},
{
"input": "1 1000000000 158260522 200224287",
"output": "0"
},
{
"input": "1 1000000000 2 755829150",
"output": "8"
},
{
"input": "1 1000000000 158260522 158260522",
"output": "1"
},
{
"input": "1 1000000000 877914575 877914575",
"output": "1"
},
{
"input": "232 380232688 116 760465376",
"output": "30"
},
{
"input": "47259 3393570 267 600661890",
"output": "30"
},
{
"input": "1 1000000000 1 672672000",
"output": "64"
},
{
"input": "1000000000 1000000000 1000000000 1000000000",
"output": "1"
},
{
"input": "1 1000000000 1 649209600",
"output": "32"
},
{
"input": "1 1000000000 1 682290000",
"output": "32"
},
{
"input": "1 1000000000 1 228614400",
"output": "16"
},
{
"input": "1 1000000000 1 800280000",
"output": "32"
},
{
"input": "1 1000000000 1 919987200",
"output": "16"
},
{
"input": "1 1000000000 1 456537870",
"output": "64"
},
{
"input": "1 1000000000 1 7198102",
"output": "8"
},
{
"input": "1 1000000000 1 58986263",
"output": "16"
},
{
"input": "1 1000000000 1 316465536",
"output": "16"
},
{
"input": "1 1000000000 1 9558312",
"output": "16"
},
{
"input": "1 1000000000 1 5461344",
"output": "16"
},
{
"input": "58 308939059 29 617878118",
"output": "62"
},
{
"input": "837 16262937 27 504151047",
"output": "28"
},
{
"input": "47275 402550 25 761222050",
"output": "12"
},
{
"input": "22 944623394 22 944623394",
"output": "32"
},
{
"input": "1032 8756124 12 753026664",
"output": "18"
},
{
"input": "7238 939389 11 618117962",
"output": "10"
},
{
"input": "58351 322621 23 818489477",
"output": "6"
},
{
"input": "3450 7068875 25 975504750",
"output": "86"
},
{
"input": "13266 1606792 22 968895576",
"output": "14"
},
{
"input": "21930 632925 15 925336350",
"output": "42"
},
{
"input": "2193 4224517 17 544962693",
"output": "42"
},
{
"input": "526792 39807152 22904 915564496",
"output": "8"
},
{
"input": "67728 122875524 16932 491502096",
"output": "12"
},
{
"input": "319813 63298373 24601 822878849",
"output": "6"
},
{
"input": "572464 23409136 15472 866138032",
"output": "4"
},
{
"input": "39443 809059020 19716 777638472",
"output": "12"
},
{
"input": "2544768 8906688 27072 837228672",
"output": "0"
},
{
"input": "413592 46975344 21768 892531536",
"output": "10"
},
{
"input": "11349 816231429 11349 816231429",
"output": "8"
},
{
"input": "16578 939956022 16578 939956022",
"output": "4"
},
{
"input": "2783175 6882425 21575 887832825",
"output": "2"
},
{
"input": "2862252 7077972 22188 913058388",
"output": "2"
},
{
"input": "1856828 13124976 25436 958123248",
"output": "6"
},
{
"input": "100 1000000000 158260522 158260522",
"output": "1"
},
{
"input": "100 1000000000 877914575 877914575",
"output": "1"
},
{
"input": "100 1000000000 602436426 602436426",
"output": "1"
},
{
"input": "100 1000000000 24979445 24979445",
"output": "1"
},
{
"input": "1 1000000000 18470 112519240",
"output": "4"
},
{
"input": "1 1000000000 22692 2201124",
"output": "2"
},
{
"input": "1 1000000000 24190 400949250",
"output": "16"
},
{
"input": "1 1000000000 33409 694005157",
"output": "2"
},
{
"input": "1 1000000000 24967 470827686",
"output": "16"
},
{
"input": "1 1000000000 35461 152517761",
"output": "8"
},
{
"input": "2 1000000000 158260522 200224287",
"output": "0"
},
{
"input": "2 1000000000 602436426 611751520",
"output": "0"
},
{
"input": "2 1000000000 861648772 942726551",
"output": "0"
},
{
"input": "2 1000000000 433933447 485982495",
"output": "0"
},
{
"input": "2 1000000000 262703497 480832794",
"output": "0"
},
{
"input": "2672374 422235092 1336187 844470184",
"output": "2"
},
{
"input": "1321815 935845020 1321815 935845020",
"output": "8"
},
{
"input": "29259607 69772909 2250739 907047817",
"output": "2"
},
{
"input": "11678540 172842392 2335708 864211960",
"output": "4"
},
{
"input": "297 173688298 2876112 851329152",
"output": "2"
},
{
"input": "7249 55497026 659 610467286",
"output": "28"
},
{
"input": "398520 1481490 810 728893080",
"output": "4"
},
{
"input": "2354 369467362 1177 738934724",
"output": "14"
},
{
"input": "407264 2497352 1144 889057312",
"output": "2"
},
{
"input": "321399 1651014 603 879990462",
"output": "4"
},
{
"input": "475640 486640 440 526057840",
"output": "2"
},
{
"input": "631714 179724831 1136 717625968",
"output": "0"
},
{
"input": "280476 1595832 588 761211864",
"output": "8"
},
{
"input": "10455 39598005 615 673166085",
"output": "6"
},
{
"input": "24725 19759875 575 849674625",
"output": "22"
},
{
"input": "22 158 2 1738",
"output": "2"
},
{
"input": "1 2623 1 2623",
"output": "4"
},
{
"input": "7 163677675 3 18",
"output": "0"
},
{
"input": "159 20749927 1 158",
"output": "0"
},
{
"input": "5252 477594071 1 5251",
"output": "0"
},
{
"input": "2202 449433679 3 6603",
"output": "0"
},
{
"input": "6 111 3 222",
"output": "2"
},
{
"input": "26 46 2 598",
"output": "2"
},
{
"input": "26 82 2 1066",
"output": "2"
},
{
"input": "1 2993 1 2993",
"output": "4"
},
{
"input": "17 17 1 289",
"output": "0"
},
{
"input": "177 267 3 15753",
"output": "2"
},
{
"input": "7388 22705183 1 7387",
"output": "0"
},
{
"input": "1 100 3 100",
"output": "0"
},
{
"input": "1 1000 6 1024",
"output": "0"
},
{
"input": "1 100 2 4",
"output": "2"
},
{
"input": "1 10000 2 455",
"output": "0"
},
{
"input": "1 1000000000 250000000 1000000000",
"output": "2"
},
{
"input": "3 3 1 1",
"output": "0"
},
{
"input": "1 1000000000 100000000 1000000000",
"output": "4"
},
{
"input": "5 10 3 3",
"output": "0"
},
{
"input": "1 1000 5 13",
"output": "0"
},
{
"input": "2 2 3 3",
"output": "0"
},
{
"input": "1 1000000000 499999993 999999986",
"output": "2"
},
{
"input": "1 1 1 10",
"output": "0"
},
{
"input": "1 10 10 100",
"output": "0"
},
{
"input": "1 1000 4 36",
"output": "2"
},
{
"input": "1 1000000000 10000000 20000000",
"output": "2"
},
{
"input": "100 100 5 5",
"output": "0"
},
{
"input": "3 3 3 9",
"output": "0"
},
{
"input": "36 200 24 144",
"output": "2"
},
{
"input": "1 100 3 10",
"output": "0"
}
] | 1,569,652,152
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 108
| 124
| 0
|
import math
l,r,x,y=map(int,input().split())
def gc(a,b):
if(a==0):
return b
elif(b==0):
return a
elif(a==b):
return b
else:
return gc(b%a,a)
if(y%x!=0):
print(0)
else:
n=y/x
ans=0
for i in range(1,int(math.sqrt(n))+1):
if(n%i==0):
c=n/i
if(c*x>=l and c*x<=r and i*x<=r and i*x>=l and gc(c,i)==1):
if(i*i==n):
ans+=1
else:
ans+=2
print(ans)
|
Title: Nastya Studies Informatics
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today on Informatics class Nastya learned about GCD and LCM (see links below). Nastya is very intelligent, so she solved all the tasks momentarily and now suggests you to solve one of them as well.
We define a pair of integers (*a*,<=*b*) good, if *GCD*(*a*,<=*b*)<==<=*x* and *LCM*(*a*,<=*b*)<==<=*y*, where *GCD*(*a*,<=*b*) denotes the [greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor) of *a* and *b*, and *LCM*(*a*,<=*b*) denotes the [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) of *a* and *b*.
You are given two integers *x* and *y*. You are to find the number of good pairs of integers (*a*,<=*b*) such that *l*<=β€<=*a*,<=*b*<=β€<=*r*. Note that pairs (*a*,<=*b*) and (*b*,<=*a*) are considered different if *a*<=β <=*b*.
Input Specification:
The only line contains four integers *l*,<=*r*,<=*x*,<=*y* (1<=β€<=*l*<=β€<=*r*<=β€<=109, 1<=β€<=*x*<=β€<=*y*<=β€<=109).
Output Specification:
In the only line print the only integerΒ β the answer for the problem.
Demo Input:
['1 2 1 2\n', '1 12 1 12\n', '50 100 3 30\n']
Demo Output:
['2\n', '4\n', '0\n']
Note:
In the first example there are two suitable good pairs of integers (*a*,β*b*): (1,β2) and (2,β1).
In the second example there are four suitable good pairs of integers (*a*,β*b*): (1,β12), (12,β1), (3,β4) and (4,β3).
In the third example there are good pairs of integers, for example, (3,β30), but none of them fits the condition *l*ββ€β*a*,β*b*ββ€β*r*.
|
```python
import math
l,r,x,y=map(int,input().split())
def gc(a,b):
if(a==0):
return b
elif(b==0):
return a
elif(a==b):
return b
else:
return gc(b%a,a)
if(y%x!=0):
print(0)
else:
n=y/x
ans=0
for i in range(1,int(math.sqrt(n))+1):
if(n%i==0):
c=n/i
if(c*x>=l and c*x<=r and i*x<=r and i*x>=l and gc(c,i)==1):
if(i*i==n):
ans+=1
else:
ans+=2
print(ans)
```
| 3
|
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