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3.99
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|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
343
|
A
|
Rational Resistance
|
PROGRAMMING
| 1,600
|
[
"math",
"number theory"
] | null | null |
Mad scientist Mike is building a time machine in his spare time. To finish the work, he needs a resistor with a certain resistance value.
However, all Mike has is lots of identical resistors with unit resistance *R*0<==<=1. Elements with other resistance can be constructed from these resistors. In this problem, we will consider the following as elements:
1. one resistor; 1. an element and one resistor plugged in sequence; 1. an element and one resistor plugged in parallel.
With the consecutive connection the resistance of the new element equals *R*<==<=*R**e*<=+<=*R*0. With the parallel connection the resistance of the new element equals . In this case *R**e* equals the resistance of the element being connected.
Mike needs to assemble an element with a resistance equal to the fraction . Determine the smallest possible number of resistors he needs to make such an element.
|
The single input line contains two space-separated integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=1018). It is guaranteed that the fraction is irreducible. It is guaranteed that a solution always exists.
|
Print a single number — the answer to the problem.
Please do not use the %lld specifier to read or write 64-bit integers in С++. It is recommended to use the cin, cout streams or the %I64d specifier.
|
[
"1 1\n",
"3 2\n",
"199 200\n"
] |
[
"1\n",
"3\n",
"200\n"
] |
In the first sample, one resistor is enough.
In the second sample one can connect the resistors in parallel, take the resulting element and connect it to a third resistor consecutively. Then, we get an element with resistance <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/5305da389756aab6423d918a08ced468f05604df.png" style="max-width: 100.0%;max-height: 100.0%;"/>. We cannot make this element using two resistors.
| 500
|
[
{
"input": "1 1",
"output": "1"
},
{
"input": "3 2",
"output": "3"
},
{
"input": "199 200",
"output": "200"
},
{
"input": "1 1000000000000000000",
"output": "1000000000000000000"
},
{
"input": "3 1",
"output": "3"
},
{
"input": "21 8",
"output": "7"
},
{
"input": "18 55",
"output": "21"
},
{
"input": "1 2",
"output": "2"
},
{
"input": "2 1",
"output": "2"
},
{
"input": "1 3",
"output": "3"
},
{
"input": "2 3",
"output": "3"
},
{
"input": "1 4",
"output": "4"
},
{
"input": "5 2",
"output": "4"
},
{
"input": "2 5",
"output": "4"
},
{
"input": "4 5",
"output": "5"
},
{
"input": "3 5",
"output": "4"
},
{
"input": "13 4",
"output": "7"
},
{
"input": "21 17",
"output": "9"
},
{
"input": "5 8",
"output": "5"
},
{
"input": "13 21",
"output": "7"
},
{
"input": "74 99",
"output": "28"
},
{
"input": "2377 1055",
"output": "33"
},
{
"input": "645597 134285",
"output": "87"
},
{
"input": "29906716 35911991",
"output": "92"
},
{
"input": "3052460231 856218974",
"output": "82"
},
{
"input": "288565475053 662099878640",
"output": "88"
},
{
"input": "11504415412768 12754036168327",
"output": "163"
},
{
"input": "9958408561221547 4644682781404278",
"output": "196"
},
{
"input": "60236007668635342 110624799949034113",
"output": "179"
},
{
"input": "4 43470202936783249",
"output": "10867550734195816"
},
{
"input": "16 310139055712567491",
"output": "19383690982035476"
},
{
"input": "15 110897893734203629",
"output": "7393192915613582"
},
{
"input": "439910263967866789 38",
"output": "11576585893891241"
},
{
"input": "36 316049483082136289",
"output": "8779152307837131"
},
{
"input": "752278442523506295 52",
"output": "14466893125452056"
},
{
"input": "4052739537881 6557470319842",
"output": "62"
},
{
"input": "44945570212853 72723460248141",
"output": "67"
},
{
"input": "498454011879264 806515533049393",
"output": "72"
},
{
"input": "8944394323791464 5527939700884757",
"output": "77"
},
{
"input": "679891637638612258 420196140727489673",
"output": "86"
},
{
"input": "1 923438",
"output": "923438"
},
{
"input": "3945894354376 1",
"output": "3945894354376"
},
{
"input": "999999999999999999 5",
"output": "200000000000000004"
},
{
"input": "999999999999999999 1000000000000000000",
"output": "1000000000000000000"
},
{
"input": "999999999999999991 1000000000000000000",
"output": "111111111111111120"
},
{
"input": "999999999999999993 999999999999999991",
"output": "499999999999999998"
},
{
"input": "3 1000000000000000000",
"output": "333333333333333336"
},
{
"input": "1000000000000000000 3",
"output": "333333333333333336"
},
{
"input": "10000000000 1000000001",
"output": "100000019"
},
{
"input": "2 999999999999999999",
"output": "500000000000000001"
},
{
"input": "999999999999999999 2",
"output": "500000000000000001"
},
{
"input": "2 1000000001",
"output": "500000002"
},
{
"input": "123 1000000000000000000",
"output": "8130081300813023"
}
] | 1,382,872,114
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 5
| 62
| 0
|
x,y = list(map(int, input().split(' ')))
print(max(x,y))
|
Title: Rational Resistance
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mad scientist Mike is building a time machine in his spare time. To finish the work, he needs a resistor with a certain resistance value.
However, all Mike has is lots of identical resistors with unit resistance *R*0<==<=1. Elements with other resistance can be constructed from these resistors. In this problem, we will consider the following as elements:
1. one resistor; 1. an element and one resistor plugged in sequence; 1. an element and one resistor plugged in parallel.
With the consecutive connection the resistance of the new element equals *R*<==<=*R**e*<=+<=*R*0. With the parallel connection the resistance of the new element equals . In this case *R**e* equals the resistance of the element being connected.
Mike needs to assemble an element with a resistance equal to the fraction . Determine the smallest possible number of resistors he needs to make such an element.
Input Specification:
The single input line contains two space-separated integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=1018). It is guaranteed that the fraction is irreducible. It is guaranteed that a solution always exists.
Output Specification:
Print a single number — the answer to the problem.
Please do not use the %lld specifier to read or write 64-bit integers in С++. It is recommended to use the cin, cout streams or the %I64d specifier.
Demo Input:
['1 1\n', '3 2\n', '199 200\n']
Demo Output:
['1\n', '3\n', '200\n']
Note:
In the first sample, one resistor is enough.
In the second sample one can connect the resistors in parallel, take the resulting element and connect it to a third resistor consecutively. Then, we get an element with resistance <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/5305da389756aab6423d918a08ced468f05604df.png" style="max-width: 100.0%;max-height: 100.0%;"/>. We cannot make this element using two resistors.
|
```python
x,y = list(map(int, input().split(' ')))
print(max(x,y))
```
| 0
|
|
559
|
B
|
Equivalent Strings
|
PROGRAMMING
| 1,700
|
[
"divide and conquer",
"hashing",
"sortings",
"strings"
] | null | null |
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings *a* and *b* of equal length are called equivalent in one of the two cases:
1. They are equal. 1. If we split string *a* into two halves of the same size *a*1 and *a*2, and string *b* into two halves of the same size *b*1 and *b*2, then one of the following is correct: *a*1 is equivalent to *b*1, and *a*2 is equivalent to *b*2 1. *a*1 is equivalent to *b*2, and *a*2 is equivalent to *b*1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn!
|
The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200<=000 and consists of lowercase English letters. The strings have the same length.
|
Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.
|
[
"aaba\nabaa\n",
"aabb\nabab\n"
] |
[
"YES\n",
"NO\n"
] |
In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa".
| 1,000
|
[
{
"input": "aaba\nabaa",
"output": "YES"
},
{
"input": "aabb\nabab",
"output": "NO"
},
{
"input": "a\na",
"output": "YES"
},
{
"input": "a\nb",
"output": "NO"
},
{
"input": "ab\nab",
"output": "YES"
},
{
"input": "ab\nba",
"output": "YES"
},
{
"input": "ab\nbb",
"output": "NO"
},
{
"input": "zzaa\naazz",
"output": "YES"
},
{
"input": "azza\nzaaz",
"output": "YES"
},
{
"input": "abc\nabc",
"output": "YES"
},
{
"input": "abc\nacb",
"output": "NO"
},
{
"input": "azzz\nzzaz",
"output": "YES"
},
{
"input": "abcd\ndcab",
"output": "YES"
},
{
"input": "abcd\ncdab",
"output": "YES"
},
{
"input": "abcd\ndcba",
"output": "YES"
},
{
"input": "abcd\nacbd",
"output": "NO"
},
{
"input": "oloaxgddgujq\noloaxgujqddg",
"output": "YES"
},
{
"input": "uwzwdxfmosmqatyv\ndxfmzwwusomqvyta",
"output": "YES"
},
{
"input": "hagnzomowtledfdotnll\nledfdotnllomowthagnz",
"output": "YES"
},
{
"input": "snyaydaeobufdg\nsnyaydaeobufdg",
"output": "YES"
},
{
"input": "baaaaa\nabaaaa",
"output": "NO"
},
{
"input": "hhiisug\nmzdjwju",
"output": "NO"
},
{
"input": "bbbabbabaaab\naaaabbabbbbb",
"output": "NO"
},
{
"input": "bbaaab\naababb",
"output": "NO"
},
{
"input": "aabbaaaa\naaaaabab",
"output": "NO"
},
{
"input": "aab\naba",
"output": "NO"
},
{
"input": "abcddd\nbacddd",
"output": "NO"
},
{
"input": "qgiufelsfhanx\naaaaaaaaaaaaa",
"output": "NO"
},
{
"input": "aabaababaaba\naababaaababa",
"output": "NO"
},
{
"input": "nocdqzdriyyil\naaaaaaaaaaaaa",
"output": "NO"
},
{
"input": "zdmctxl\nkojqhgw",
"output": "NO"
},
{
"input": "yhwepqwyhwepqwyhwepqweahnqtueahnqtueahnqtuyhwepqwyhwepqwyhwepqwyhwepqweahnqtueahnqtuyhwepqweahnqtueahnqtueahnqtueahnqtueahnqtueahnqtu\neahnqtueahnqtueahnqtuyhwepqweahnqtuyhwepqwyhwepqweahnqtuyhwepqweahnqtuyhwepqweahnqtueahnqtuyhwepqweahnqtueahnqtuyhwepqwyhwepqwyhwepqw",
"output": "NO"
},
{
"input": "abc\nbac",
"output": "NO"
},
{
"input": "ottceez\npcstdvz",
"output": "NO"
}
] | 1,573,508,192
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 4
| 109
| 614,400
|
def eq(stra,strb):
# print('---')
# print(stra,strb)
if (stra == strb):
return True
if (len(stra) == 1):
return False
n = len(stra)//2
# print(n)
stra_1 = stra[:n]
stra_2 = stra[n:]
strb_1 = strb[:n]
strb_2 = strb[n:]
# print(stra_1,stra_2,strb_1,strb_2)
# c = input()
b3 = eq(stra_1,strb_1) and eq(strb_2,stra_2)
b2 = eq(stra_1,strb_2) and eq(strb_1,stra_2)
return b2 or b3
stra = input()
strb = input()
if (eq(stra,strb)):
print("YES")
else:
print("NO")
|
Title: Equivalent Strings
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings *a* and *b* of equal length are called equivalent in one of the two cases:
1. They are equal. 1. If we split string *a* into two halves of the same size *a*1 and *a*2, and string *b* into two halves of the same size *b*1 and *b*2, then one of the following is correct: *a*1 is equivalent to *b*1, and *a*2 is equivalent to *b*2 1. *a*1 is equivalent to *b*2, and *a*2 is equivalent to *b*1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn!
Input Specification:
The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200<=000 and consists of lowercase English letters. The strings have the same length.
Output Specification:
Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.
Demo Input:
['aaba\nabaa\n', 'aabb\nabab\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa".
|
```python
def eq(stra,strb):
# print('---')
# print(stra,strb)
if (stra == strb):
return True
if (len(stra) == 1):
return False
n = len(stra)//2
# print(n)
stra_1 = stra[:n]
stra_2 = stra[n:]
strb_1 = strb[:n]
strb_2 = strb[n:]
# print(stra_1,stra_2,strb_1,strb_2)
# c = input()
b3 = eq(stra_1,strb_1) and eq(strb_2,stra_2)
b2 = eq(stra_1,strb_2) and eq(strb_1,stra_2)
return b2 or b3
stra = input()
strb = input()
if (eq(stra,strb)):
print("YES")
else:
print("NO")
```
| -1
|
|
234
|
A
|
Lefthanders and Righthanders
|
PROGRAMMING
| 1,200
|
[
"implementation"
] | null | null |
One fine October day a mathematics teacher Vasily Petrov went to a class and saw there *n* pupils who sat at the desks, two people at each desk. Vasily quickly realized that number *n* is even. Like all true mathematicians, Vasily has all students numbered from 1 to *n*.
But Vasily Petrov did not like the way the children were seated at the desks. According to him, the students whose numbers differ by 1, can not sit together, as they talk to each other all the time, distract others and misbehave.
On the other hand, if a righthanded student sits at the left end of the desk and a lefthanded student sits at the right end of the desk, they hit elbows all the time and distract each other. In other cases, the students who sit at the same desk, do not interfere with each other.
Vasily knows very well which students are lefthanders and which ones are righthanders, and he asks you to come up with any order that meets these two uncomplicated conditions (students do not talk to each other and do not bump their elbows). It is guaranteed that the input is such that at least one way to seat the students always exists.
|
The first input line contains a single even integer *n* (4<=≤<=*n*<=≤<=100) — the number of students in the class. The second line contains exactly *n* capital English letters "L" and "R". If the *i*-th letter at the second line equals "L", then the student number *i* is a lefthander, otherwise he is a righthander.
|
Print integer pairs, one pair per line. In the *i*-th line print the numbers of students that will sit at the *i*-th desk. The first number in the pair stands for the student who is sitting to the left, and the second number stands for the student who is sitting to the right. Separate the numbers in the pairs by spaces. If there are multiple solutions, print any of them.
|
[
"6\nLLRLLL\n",
"4\nRRLL\n"
] |
[
"1 4\n2 5\n6 3\n",
"3 1\n4 2\n"
] |
none
| 0
|
[
{
"input": "6\nLLRLLL",
"output": "1 4\n2 5\n6 3"
},
{
"input": "4\nRRLL",
"output": "3 1\n4 2"
},
{
"input": "4\nLLRR",
"output": "1 3\n2 4"
},
{
"input": "6\nRLLRRL",
"output": "1 4\n2 5\n3 6"
},
{
"input": "8\nLRLRLLLR",
"output": "1 5\n6 2\n3 7\n4 8"
},
{
"input": "10\nRLLRLRRRLL",
"output": "1 6\n2 7\n3 8\n9 4\n5 10"
},
{
"input": "12\nLRRRRRLRRRRL",
"output": "1 7\n2 8\n3 9\n4 10\n5 11\n12 6"
},
{
"input": "14\nRLLRLLLLRLLLRL",
"output": "8 1\n2 9\n3 10\n11 4\n5 12\n6 13\n7 14"
},
{
"input": "16\nLLLRRRLRRLLRRLLL",
"output": "1 9\n2 10\n3 11\n4 12\n5 13\n14 6\n7 15\n16 8"
},
{
"input": "18\nRRRLLLLRRRLRLRLLRL",
"output": "1 10\n11 2\n3 12\n4 13\n5 14\n6 15\n7 16\n8 17\n18 9"
},
{
"input": "20\nRLRLLRLRRLLRRRRRRLRL",
"output": "11 1\n2 12\n3 13\n4 14\n5 15\n6 16\n7 17\n18 8\n9 19\n10 20"
},
{
"input": "22\nRLLLRLLLRRLRRRLRLLLLLL",
"output": "1 12\n2 13\n3 14\n4 15\n5 16\n6 17\n7 18\n8 19\n20 9\n21 10\n11 22"
},
{
"input": "24\nLRRRLRLLRLRRRRLLLLRRLRLR",
"output": "1 13\n2 14\n15 3\n16 4\n5 17\n18 6\n7 19\n8 20\n21 9\n10 22\n23 11\n12 24"
},
{
"input": "26\nRLRRLLRRLLRLRRLLRLLRRLRLRR",
"output": "1 14\n2 15\n16 3\n4 17\n5 18\n6 19\n7 20\n8 21\n9 22\n10 23\n24 11\n12 25\n13 26"
},
{
"input": "28\nLLLRRRRRLRRLRRRLRLRLRRLRLRRL",
"output": "1 15\n2 16\n3 17\n18 4\n5 19\n20 6\n7 21\n8 22\n9 23\n10 24\n25 11\n12 26\n13 27\n28 14"
},
{
"input": "30\nLRLLRLRRLLRLRLLRRRRRLRLRLRLLLL",
"output": "1 16\n2 17\n3 18\n4 19\n5 20\n6 21\n7 22\n23 8\n9 24\n10 25\n11 26\n12 27\n28 13\n14 29\n15 30"
},
{
"input": "32\nRLRLLRRLLRRLRLLRLRLRLLRLRRRLLRRR",
"output": "17 1\n2 18\n19 3\n4 20\n5 21\n22 6\n7 23\n8 24\n9 25\n10 26\n11 27\n12 28\n29 13\n14 30\n15 31\n16 32"
},
{
"input": "34\nLRRLRLRLLRRRRLLRLRRLRRLRLRRLRRRLLR",
"output": "1 18\n2 19\n20 3\n4 21\n5 22\n6 23\n7 24\n8 25\n9 26\n10 27\n28 11\n12 29\n13 30\n14 31\n15 32\n33 16\n17 34"
},
{
"input": "36\nRRLLLRRRLLLRRLLLRRLLRLLRLRLLRLRLRLLL",
"output": "19 1\n20 2\n3 21\n4 22\n5 23\n6 24\n25 7\n8 26\n9 27\n10 28\n11 29\n30 12\n13 31\n14 32\n15 33\n16 34\n35 17\n36 18"
},
{
"input": "38\nLLRRRLLRRRLRRLRLRRLRRLRLRLLRRRRLLLLRLL",
"output": "1 20\n2 21\n22 3\n4 23\n24 5\n6 25\n7 26\n27 8\n9 28\n10 29\n11 30\n12 31\n32 13\n14 33\n34 15\n16 35\n17 36\n37 18\n19 38"
},
{
"input": "40\nLRRRRRLRLLRRRLLRRLRLLRLRRLRRLLLRRLRRRLLL",
"output": "1 21\n2 22\n23 3\n4 24\n5 25\n26 6\n7 27\n8 28\n9 29\n10 30\n31 11\n12 32\n13 33\n14 34\n15 35\n16 36\n17 37\n18 38\n39 19\n20 40"
},
{
"input": "42\nRLRRLLLLLLLRRRLRLLLRRRLRLLLRLRLRLLLRLRLRRR",
"output": "1 22\n2 23\n3 24\n25 4\n5 26\n6 27\n7 28\n8 29\n9 30\n10 31\n11 32\n33 12\n34 13\n35 14\n15 36\n37 16\n17 38\n18 39\n19 40\n20 41\n21 42"
},
{
"input": "44\nLLLLRRLLRRLLRRLRLLRRRLRLRLLRLRLRRLLRLRRLLLRR",
"output": "1 23\n2 24\n3 25\n4 26\n27 5\n6 28\n7 29\n8 30\n31 9\n10 32\n11 33\n12 34\n35 13\n14 36\n15 37\n16 38\n17 39\n18 40\n41 19\n42 20\n21 43\n22 44"
},
{
"input": "46\nRRRLLLLRRLRLRRRRRLRLLRLRRLRLLLLLLLLRRLRLRLRLLL",
"output": "1 24\n2 25\n26 3\n4 27\n5 28\n6 29\n7 30\n31 8\n32 9\n10 33\n34 11\n12 35\n13 36\n14 37\n38 15\n16 39\n40 17\n18 41\n42 19\n20 43\n21 44\n45 22\n23 46"
},
{
"input": "48\nLLLLRRLRRRRLRRRLRLLLLLRRLLRLLRLLRRLRRLLRLRLRRRRL",
"output": "1 25\n2 26\n3 27\n4 28\n29 5\n6 30\n7 31\n32 8\n9 33\n10 34\n35 11\n12 36\n13 37\n38 14\n39 15\n16 40\n41 17\n18 42\n19 43\n20 44\n21 45\n22 46\n23 47\n48 24"
},
{
"input": "50\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "1 26\n2 27\n3 28\n4 29\n5 30\n6 31\n7 32\n8 33\n9 34\n10 35\n11 36\n12 37\n13 38\n14 39\n15 40\n16 41\n17 42\n18 43\n19 44\n20 45\n21 46\n22 47\n23 48\n24 49\n25 50"
},
{
"input": "52\nLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLL",
"output": "1 27\n2 28\n3 29\n4 30\n5 31\n6 32\n7 33\n8 34\n9 35\n10 36\n11 37\n12 38\n13 39\n14 40\n15 41\n16 42\n17 43\n18 44\n19 45\n20 46\n21 47\n22 48\n23 49\n24 50\n25 51\n26 52"
},
{
"input": "54\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "1 28\n2 29\n3 30\n4 31\n5 32\n6 33\n7 34\n8 35\n9 36\n10 37\n11 38\n12 39\n13 40\n14 41\n15 42\n16 43\n17 44\n18 45\n19 46\n20 47\n21 48\n22 49\n23 50\n24 51\n25 52\n26 53\n27 54"
},
{
"input": "56\nLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLL",
"output": "1 29\n2 30\n3 31\n4 32\n5 33\n6 34\n7 35\n8 36\n9 37\n10 38\n11 39\n12 40\n13 41\n14 42\n15 43\n16 44\n17 45\n18 46\n19 47\n20 48\n21 49\n22 50\n23 51\n24 52\n25 53\n26 54\n27 55\n28 56"
},
{
"input": "58\nRRRLLLRLLLLRRLRRRLLRLLRLRLLRLRRRRLLLLLLRLRRLRLRRRLRLRRLRRL",
"output": "1 30\n2 31\n3 32\n4 33\n5 34\n6 35\n36 7\n8 37\n9 38\n10 39\n11 40\n41 12\n13 42\n14 43\n44 15\n16 45\n46 17\n18 47\n19 48\n20 49\n21 50\n22 51\n52 23\n24 53\n25 54\n26 55\n27 56\n28 57\n29 58"
},
{
"input": "60\nRLLLLRRLLRRRLLLLRRRRRLRRRLRRRLLLRLLLRLRRRLRLLLRLLRRLLRRRRRLL",
"output": "31 1\n2 32\n3 33\n4 34\n5 35\n36 6\n7 37\n8 38\n9 39\n10 40\n11 41\n42 12\n13 43\n14 44\n15 45\n16 46\n17 47\n48 18\n49 19\n20 50\n21 51\n22 52\n53 23\n24 54\n25 55\n26 56\n27 57\n28 58\n59 29\n30 60"
},
{
"input": "62\nLRRLRLRLLLLRRLLLLRRRLRLLLLRRRLLLLLLRRRLLLLRRLRRLRLLLLLLLLRRLRR",
"output": "1 32\n33 2\n34 3\n4 35\n5 36\n6 37\n7 38\n8 39\n9 40\n10 41\n11 42\n12 43\n13 44\n14 45\n15 46\n16 47\n17 48\n18 49\n50 19\n51 20\n21 52\n53 22\n23 54\n24 55\n25 56\n26 57\n27 58\n28 59\n60 29\n30 61\n31 62"
},
{
"input": "64\nRLLLLRRRLRLLRRRRLRLLLRRRLLLRRRLLRLLRLRLRRRLLRRRRLRLRRRLLLLRRLLLL",
"output": "1 33\n2 34\n3 35\n4 36\n5 37\n6 38\n39 7\n8 40\n9 41\n10 42\n11 43\n12 44\n13 45\n14 46\n15 47\n16 48\n17 49\n18 50\n19 51\n20 52\n21 53\n22 54\n55 23\n56 24\n25 57\n26 58\n27 59\n28 60\n61 29\n62 30\n31 63\n32 64"
},
{
"input": "66\nLLRRRLLRLRLLRRRRRRRLLLLRRLLLLLLRLLLRLLLLLLRRRLRRLLRRRRRLRLLRLLLLRR",
"output": "1 34\n2 35\n3 36\n37 4\n38 5\n6 39\n7 40\n41 8\n9 42\n10 43\n11 44\n12 45\n46 13\n14 47\n15 48\n49 16\n50 17\n18 51\n19 52\n20 53\n21 54\n22 55\n23 56\n24 57\n58 25\n26 59\n27 60\n28 61\n29 62\n30 63\n31 64\n32 65\n33 66"
},
{
"input": "68\nRRLRLRLLRLRLRRRRRRLRRRLLLLRLLRLRLRLRRRRLRLRLLRRRRLRRLLRLRRLLRLRRLRRL",
"output": "35 1\n2 36\n3 37\n4 38\n5 39\n40 6\n7 41\n8 42\n9 43\n10 44\n45 11\n12 46\n13 47\n14 48\n15 49\n50 16\n17 51\n18 52\n19 53\n54 20\n21 55\n56 22\n23 57\n24 58\n25 59\n26 60\n27 61\n28 62\n29 63\n30 64\n31 65\n32 66\n33 67\n68 34"
},
{
"input": "70\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "1 36\n2 37\n3 38\n4 39\n5 40\n6 41\n7 42\n8 43\n9 44\n10 45\n11 46\n12 47\n13 48\n14 49\n15 50\n16 51\n17 52\n18 53\n19 54\n20 55\n21 56\n22 57\n23 58\n24 59\n25 60\n26 61\n27 62\n28 63\n29 64\n30 65\n31 66\n32 67\n33 68\n34 69\n35 70"
},
{
"input": "72\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "1 37\n2 38\n3 39\n4 40\n5 41\n6 42\n7 43\n8 44\n9 45\n10 46\n11 47\n12 48\n13 49\n14 50\n15 51\n16 52\n17 53\n18 54\n19 55\n20 56\n21 57\n22 58\n23 59\n24 60\n25 61\n26 62\n27 63\n28 64\n29 65\n30 66\n31 67\n32 68\n33 69\n34 70\n35 71\n36 72"
},
{
"input": "74\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "1 38\n2 39\n3 40\n4 41\n5 42\n6 43\n7 44\n8 45\n9 46\n10 47\n11 48\n12 49\n13 50\n14 51\n15 52\n16 53\n17 54\n18 55\n19 56\n20 57\n21 58\n22 59\n23 60\n24 61\n25 62\n26 63\n27 64\n28 65\n29 66\n30 67\n31 68\n32 69\n33 70\n34 71\n35 72\n36 73\n37 74"
},
{
"input": "76\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "1 39\n2 40\n3 41\n4 42\n5 43\n6 44\n7 45\n8 46\n9 47\n10 48\n11 49\n12 50\n13 51\n14 52\n15 53\n16 54\n17 55\n18 56\n19 57\n20 58\n21 59\n22 60\n23 61\n24 62\n25 63\n26 64\n27 65\n28 66\n29 67\n30 68\n31 69\n32 70\n33 71\n34 72\n35 73\n36 74\n37 75\n38 76"
},
{
"input": "78\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "1 40\n2 41\n3 42\n4 43\n5 44\n6 45\n7 46\n8 47\n9 48\n10 49\n11 50\n12 51\n13 52\n14 53\n15 54\n16 55\n17 56\n18 57\n19 58\n20 59\n21 60\n22 61\n23 62\n24 63\n25 64\n26 65\n27 66\n28 67\n29 68\n30 69\n31 70\n32 71\n33 72\n34 73\n35 74\n36 75\n37 76\n38 77\n39 78"
},
{
"input": "80\nLRLRRRRLRRRRLLLLRLLRLRLLRRLRLLLRRLLLLRLLLRLRLLRRRLRRRLRLRRRRRLRLLRLLRRLLLRLRRRLL",
"output": "1 41\n2 42\n3 43\n4 44\n45 5\n46 6\n7 47\n8 48\n9 49\n50 10\n11 51\n12 52\n13 53\n14 54\n15 55\n16 56\n17 57\n18 58\n19 59\n20 60\n21 61\n62 22\n23 63\n24 64\n65 25\n26 66\n27 67\n68 28\n29 69\n30 70\n31 71\n72 32\n73 33\n34 74\n35 75\n36 76\n37 77\n38 78\n39 79\n40 80"
},
{
"input": "82\nRLRRLLRLRLRLLLRLLLRRLLRRLRRRRLLRLLLLRRRRRLLLRRRLLLLRLRRLRRRLRLLLLRRRLRLRLLLRLLLLLR",
"output": "42 1\n2 43\n44 3\n4 45\n5 46\n6 47\n48 7\n8 49\n50 9\n10 51\n11 52\n12 53\n13 54\n14 55\n56 15\n16 57\n17 58\n18 59\n60 19\n20 61\n21 62\n22 63\n64 23\n65 24\n25 66\n26 67\n27 68\n69 28\n29 70\n30 71\n31 72\n73 32\n33 74\n34 75\n35 76\n36 77\n78 37\n79 38\n80 39\n81 40\n41 82"
},
{
"input": "84\nLRLRRRRRRLLLRLRLLLLLRRLRLRLRRRLLRLLLRLRLLLRRRLRLRRLRLRLLLLLLLLRRRRRRLLLRRLRLRLLLRLRR",
"output": "1 43\n2 44\n3 45\n46 4\n5 47\n48 6\n7 49\n8 50\n51 9\n10 52\n11 53\n12 54\n55 13\n14 56\n57 15\n16 58\n17 59\n18 60\n19 61\n20 62\n21 63\n22 64\n23 65\n24 66\n25 67\n26 68\n27 69\n70 28\n71 29\n30 72\n31 73\n32 74\n33 75\n34 76\n35 77\n36 78\n79 37\n38 80\n39 81\n40 82\n41 83\n42 84"
},
{
"input": "86\nRRRLLLRLLRLLRLRLRLLLRLRLRRLLRLLLRLLLLLLRRRLRLLRLLLRRRLRLLLLRLLRLRRLLRLLLRRRLLRLRLLRLLR",
"output": "1 44\n45 2\n46 3\n4 47\n5 48\n6 49\n50 7\n8 51\n9 52\n10 53\n11 54\n12 55\n56 13\n14 57\n58 15\n16 59\n17 60\n18 61\n19 62\n20 63\n64 21\n22 65\n23 66\n24 67\n68 25\n26 69\n27 70\n28 71\n72 29\n30 73\n31 74\n32 75\n76 33\n34 77\n35 78\n36 79\n37 80\n38 81\n39 82\n40 83\n84 41\n85 42\n43 86"
},
{
"input": "88\nLLRLRLRLLLLRRRRRRLRRLLLLLRRLRRLLLLLRLRLRLLLLLRLRLRRLRLRRLRLLRRLRLLLRLLLLRRLLRRLRLRLRRLRR",
"output": "1 45\n2 46\n47 3\n4 48\n49 5\n6 50\n7 51\n8 52\n9 53\n10 54\n11 55\n12 56\n57 13\n14 58\n59 15\n60 16\n17 61\n18 62\n63 19\n20 64\n21 65\n22 66\n23 67\n24 68\n25 69\n70 26\n71 27\n28 72\n29 73\n30 74\n31 75\n32 76\n33 77\n34 78\n35 79\n36 80\n37 81\n38 82\n39 83\n40 84\n41 85\n42 86\n43 87\n44 88"
},
{
"input": "90\nLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLL",
"output": "1 46\n2 47\n3 48\n4 49\n5 50\n6 51\n7 52\n8 53\n9 54\n10 55\n11 56\n12 57\n13 58\n14 59\n15 60\n16 61\n17 62\n18 63\n19 64\n20 65\n21 66\n22 67\n23 68\n24 69\n25 70\n26 71\n27 72\n28 73\n29 74\n30 75\n31 76\n32 77\n33 78\n34 79\n35 80\n36 81\n37 82\n38 83\n39 84\n40 85\n41 86\n42 87\n43 88\n44 89\n45 90"
},
{
"input": "92\nLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLL",
"output": "1 47\n2 48\n3 49\n4 50\n5 51\n6 52\n7 53\n8 54\n9 55\n10 56\n11 57\n12 58\n13 59\n14 60\n15 61\n16 62\n17 63\n18 64\n19 65\n20 66\n21 67\n22 68\n23 69\n24 70\n25 71\n26 72\n27 73\n28 74\n29 75\n30 76\n31 77\n32 78\n33 79\n34 80\n35 81\n36 82\n37 83\n38 84\n39 85\n40 86\n41 87\n42 88\n43 89\n44 90\n45 91\n46 92"
},
{
"input": "94\nLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLL",
"output": "1 48\n2 49\n3 50\n4 51\n5 52\n6 53\n7 54\n8 55\n9 56\n10 57\n11 58\n12 59\n13 60\n14 61\n15 62\n16 63\n17 64\n18 65\n19 66\n20 67\n21 68\n22 69\n23 70\n24 71\n25 72\n26 73\n27 74\n28 75\n29 76\n30 77\n31 78\n32 79\n33 80\n34 81\n35 82\n36 83\n37 84\n38 85\n39 86\n40 87\n41 88\n42 89\n43 90\n44 91\n45 92\n46 93\n47 94"
},
{
"input": "96\nLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLL",
"output": "1 49\n2 50\n3 51\n4 52\n5 53\n6 54\n7 55\n8 56\n9 57\n10 58\n11 59\n12 60\n13 61\n14 62\n15 63\n16 64\n17 65\n18 66\n19 67\n20 68\n21 69\n22 70\n23 71\n24 72\n25 73\n26 74\n27 75\n28 76\n29 77\n30 78\n31 79\n32 80\n33 81\n34 82\n35 83\n36 84\n37 85\n38 86\n39 87\n40 88\n41 89\n42 90\n43 91\n44 92\n45 93\n46 94\n47 95\n48 96"
},
{
"input": "98\nLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLL",
"output": "1 50\n2 51\n3 52\n4 53\n5 54\n6 55\n7 56\n8 57\n9 58\n10 59\n11 60\n12 61\n13 62\n14 63\n15 64\n16 65\n17 66\n18 67\n19 68\n20 69\n21 70\n22 71\n23 72\n24 73\n25 74\n26 75\n27 76\n28 77\n29 78\n30 79\n31 80\n32 81\n33 82\n34 83\n35 84\n36 85\n37 86\n38 87\n39 88\n40 89\n41 90\n42 91\n43 92\n44 93\n45 94\n46 95\n47 96\n48 97\n49 98"
},
{
"input": "100\nRLRRRRLLLLRRRRLRRRRRRRRLRLRRLLRRRRRRRRLRRRRLLLLRRRRLRRLRLRRRLLRRLRRLLLRLRRLLLLLLRLRLRLRRLRLRLRRRLLLR",
"output": "1 51\n2 52\n3 53\n4 54\n55 5\n6 56\n7 57\n8 58\n9 59\n10 60\n61 11\n62 12\n13 63\n14 64\n15 65\n16 66\n17 67\n68 18\n69 19\n70 20\n21 71\n72 22\n23 73\n24 74\n75 25\n26 76\n77 27\n78 28\n29 79\n30 80\n31 81\n82 32\n33 83\n84 34\n35 85\n86 36\n37 87\n38 88\n39 89\n40 90\n91 41\n42 92\n93 43\n44 94\n45 95\n46 96\n47 97\n98 48\n99 49\n50 100"
},
{
"input": "100\nLRLLLLRLLLLRRRRRLRRRRLRRLRRLRLLRRLRRRRLLRRRLLLRLLLRRRRLLRLRLRRLRLLRRLLRRLRRLRRRRRLRRLRLRLRLLLLLLLLRL",
"output": "1 51\n2 52\n3 53\n4 54\n5 55\n6 56\n7 57\n8 58\n9 59\n10 60\n11 61\n12 62\n63 13\n14 64\n65 15\n66 16\n17 67\n18 68\n69 19\n70 20\n21 71\n22 72\n73 23\n24 74\n25 75\n76 26\n27 77\n28 78\n29 79\n30 80\n31 81\n82 32\n33 83\n34 84\n85 35\n36 86\n87 37\n38 88\n39 89\n40 90\n91 41\n92 42\n93 43\n44 94\n45 95\n46 96\n97 47\n48 98\n49 99\n50 100"
},
{
"input": "100\nLLLRRLLRLRLLLRLLLRLRLLRRRLRRLLLRLRLRRLLRLRRRLLLRRLLRLLRRLLRRRRRLRLRRLRLRRLRLRRLLRLRLLRLLLRLLRLLLLRLL",
"output": "1 51\n2 52\n3 53\n54 4\n5 55\n6 56\n7 57\n58 8\n9 59\n10 60\n11 61\n12 62\n13 63\n64 14\n15 65\n16 66\n17 67\n18 68\n19 69\n20 70\n21 71\n22 72\n23 73\n74 24\n25 75\n26 76\n27 77\n28 78\n29 79\n30 80\n31 81\n82 32\n33 83\n84 34\n35 85\n36 86\n87 37\n38 88\n39 89\n40 90\n41 91\n92 42\n43 93\n94 44\n45 95\n46 96\n47 97\n48 98\n99 49\n50 100"
},
{
"input": "100\nRLLLLRRLLLLRRRRLLRLRRRLLLRLLRLLLLLRRLLLLLLRRLRRRRRLRLLRLRRRLLLRLRLRLLLRRRLLLLLRRRRRLRRLLLLRLLLRRLLLL",
"output": "51 1\n2 52\n3 53\n4 54\n5 55\n56 6\n7 57\n8 58\n9 59\n10 60\n11 61\n62 12\n13 63\n64 14\n15 65\n16 66\n17 67\n68 18\n19 69\n70 20\n21 71\n22 72\n23 73\n24 74\n25 75\n76 26\n27 77\n28 78\n29 79\n30 80\n31 81\n32 82\n33 83\n34 84\n35 85\n36 86\n37 87\n38 88\n39 89\n40 90\n41 91\n42 92\n93 43\n94 44\n45 95\n46 96\n97 47\n98 48\n99 49\n100 50"
},
{
"input": "100\nRLRRLRLRRLRLLRLLRRRLRRLLLLLRLRLRRRRRRRLLRRRLLRLRLLLRRRLLRRRLLRLRLLLLRRLRLLRLLRLLLLRRLRLRRLRLLLLRLRRR",
"output": "51 1\n2 52\n3 53\n4 54\n5 55\n56 6\n7 57\n8 58\n9 59\n10 60\n61 11\n12 62\n13 63\n14 64\n15 65\n16 66\n67 17\n68 18\n19 69\n20 70\n71 21\n22 72\n23 73\n24 74\n25 75\n26 76\n27 77\n28 78\n29 79\n80 30\n31 81\n82 32\n33 83\n34 84\n85 35\n36 86\n87 37\n38 88\n39 89\n40 90\n41 91\n92 42\n93 43\n44 94\n45 95\n46 96\n47 97\n48 98\n49 99\n50 100"
},
{
"input": "100\nLRRLRLRRRRRRLRRLRRLLLLLLRRLLRRLLRLLLLLLRRRLLRLRRRLLRLLRRLRRRLLRLRLLRRLRRRLLLRRRRLLRRRLLLRRRRRLLLLLLR",
"output": "1 51\n2 52\n53 3\n4 54\n5 55\n6 56\n57 7\n8 58\n9 59\n10 60\n61 11\n62 12\n13 63\n64 14\n15 65\n16 66\n67 17\n18 68\n19 69\n20 70\n21 71\n22 72\n23 73\n24 74\n75 25\n76 26\n27 77\n28 78\n29 79\n30 80\n31 81\n32 82\n33 83\n34 84\n35 85\n36 86\n37 87\n38 88\n39 89\n40 90\n41 91\n42 92\n43 93\n44 94\n95 45\n46 96\n97 47\n98 48\n99 49\n50 100"
},
{
"input": "100\nRRLRRLRLRLRRRRLLRRLLRLRRLLRRRLLRLRRLRLRRLLLRRLLRRRRRRLLLRRRLLRRLLLLLLRLLLLLLRLLLRRRLRLLRRRRRLLRLLRRR",
"output": "1 51\n2 52\n3 53\n54 4\n55 5\n6 56\n7 57\n8 58\n9 59\n10 60\n61 11\n12 62\n13 63\n64 14\n15 65\n16 66\n67 17\n68 18\n19 69\n20 70\n71 21\n22 72\n73 23\n74 24\n25 75\n26 76\n27 77\n78 28\n79 29\n30 80\n31 81\n32 82\n33 83\n84 34\n35 85\n36 86\n87 37\n38 88\n39 89\n40 90\n41 91\n42 92\n43 93\n94 44\n45 95\n46 96\n47 97\n48 98\n49 99\n50 100"
},
{
"input": "100\nRRLLLRLRRLRLLRRLRRRLLRRRLRRLLLLLLLLLRRRLLRLRRLRRLRRLRRLRLLLLRLLRRRLLLLRLRRRLLRRRRLRRLLRRRRLRRRLRLLLR",
"output": "1 51\n52 2\n3 53\n4 54\n5 55\n6 56\n7 57\n58 8\n59 9\n10 60\n11 61\n12 62\n13 63\n14 64\n15 65\n16 66\n67 17\n68 18\n69 19\n20 70\n21 71\n72 22\n23 73\n24 74\n25 75\n76 26\n77 27\n28 78\n29 79\n30 80\n31 81\n32 82\n33 83\n34 84\n35 85\n36 86\n37 87\n38 88\n39 89\n40 90\n41 91\n42 92\n43 93\n44 94\n95 45\n46 96\n97 47\n98 48\n49 99\n50 100"
},
{
"input": "100\nLLLLLRRLRRRRRRRLLRRRRRLRRLRLRLLRLRRLLLRRRRLLRRLRLLRLLLRLRLLRRRRRRRRRLRLLLRLRLLLLLRLRRRRLRLLRLRLRLRRL",
"output": "1 51\n2 52\n3 53\n4 54\n5 55\n56 6\n7 57\n8 58\n59 9\n10 60\n11 61\n12 62\n13 63\n14 64\n15 65\n16 66\n17 67\n18 68\n69 19\n20 70\n71 21\n72 22\n23 73\n24 74\n75 25\n26 76\n77 27\n28 78\n79 29\n30 80\n31 81\n32 82\n33 83\n34 84\n35 85\n36 86\n37 87\n38 88\n39 89\n90 40\n91 41\n42 92\n43 93\n44 94\n95 45\n46 96\n47 97\n48 98\n49 99\n50 100"
},
{
"input": "100\nLLRRRLLLRLLLLRLRLRLRRRLLLRRRLRLLRLLLRRRRRLRRLRRLRRRLRRLRRLLLRLRLLRRRRLRLRRRRRLRRLRLLRRRRLLLRRRRRLLLL",
"output": "1 51\n2 52\n3 53\n4 54\n55 5\n6 56\n7 57\n8 58\n59 9\n10 60\n11 61\n12 62\n13 63\n64 14\n15 65\n16 66\n17 67\n18 68\n19 69\n70 20\n21 71\n72 22\n23 73\n24 74\n25 75\n26 76\n27 77\n78 28\n29 79\n30 80\n31 81\n32 82\n83 33\n34 84\n35 85\n36 86\n37 87\n38 88\n89 39\n90 40\n91 41\n42 92\n43 93\n44 94\n45 95\n46 96\n97 47\n48 98\n99 49\n100 50"
}
] | 1,670,352,229
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 9
| 93
| 0
|
def get_ints():
return map(int, input().strip().split())
def get_list():
return list(map(int, input().strip().split()))
def get_string():
return input().strip()
# n = int(input())
# arr = [int(x) for x in input().split()]
import sys
from os import path
if (path.exists('input.txt')):
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
n = int(sys.stdin.readline().strip())
arr = list(sys.stdin.readline().strip())
ptr1, ptr2 = 0, 1
ans = []
# print(arr)
while ptr1+2<n or ptr2+2<n:
if arr[ptr1] == 'R' and arr[ptr1+2] == 'L':
ans.append([ptr1+2, ptr1])
else:
ans.append([ptr1, ptr1+2])
ptr1 += 4
if ptr2+2 >= n:
break
if arr[ptr2] == 'R' and arr[ptr2 + 2] == 'L':
ans.append([ptr2+2, ptr2])
else:
ans.append([ptr2, ptr2+2])
ptr2 += 4
if n%4 != 0:
ans.append([n-2, n-1])
firstPair = ans[0]
lastPair = ans[-1]
if arr[lastPair[0]] == 'L':
firstPair[0], lastPair[0] = lastPair[0], firstPair[0]
else:
firstPair[1], lastPair[1] = lastPair[1], firstPair[1]
if arr[lastPair[0]] == 'R':
lastPair[0], lastPair[1] = lastPair[1], lastPair[0]
for i in ans:
sys.stdout.write(str(i[0] + 1) + " " + str(i[1] + 1) + "\n")
# print(str(i[0] + 1) + " " + str(i[1] + 1) + " : " + str(arr[i[0]]) + " " + str(arr[i[1]]))
# . . . . . . . .
# 1 2 1 2 3 4 3 4
|
Title: Lefthanders and Righthanders
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One fine October day a mathematics teacher Vasily Petrov went to a class and saw there *n* pupils who sat at the desks, two people at each desk. Vasily quickly realized that number *n* is even. Like all true mathematicians, Vasily has all students numbered from 1 to *n*.
But Vasily Petrov did not like the way the children were seated at the desks. According to him, the students whose numbers differ by 1, can not sit together, as they talk to each other all the time, distract others and misbehave.
On the other hand, if a righthanded student sits at the left end of the desk and a lefthanded student sits at the right end of the desk, they hit elbows all the time and distract each other. In other cases, the students who sit at the same desk, do not interfere with each other.
Vasily knows very well which students are lefthanders and which ones are righthanders, and he asks you to come up with any order that meets these two uncomplicated conditions (students do not talk to each other and do not bump their elbows). It is guaranteed that the input is such that at least one way to seat the students always exists.
Input Specification:
The first input line contains a single even integer *n* (4<=≤<=*n*<=≤<=100) — the number of students in the class. The second line contains exactly *n* capital English letters "L" and "R". If the *i*-th letter at the second line equals "L", then the student number *i* is a lefthander, otherwise he is a righthander.
Output Specification:
Print integer pairs, one pair per line. In the *i*-th line print the numbers of students that will sit at the *i*-th desk. The first number in the pair stands for the student who is sitting to the left, and the second number stands for the student who is sitting to the right. Separate the numbers in the pairs by spaces. If there are multiple solutions, print any of them.
Demo Input:
['6\nLLRLLL\n', '4\nRRLL\n']
Demo Output:
['1 4\n2 5\n6 3\n', '3 1\n4 2\n']
Note:
none
|
```python
def get_ints():
return map(int, input().strip().split())
def get_list():
return list(map(int, input().strip().split()))
def get_string():
return input().strip()
# n = int(input())
# arr = [int(x) for x in input().split()]
import sys
from os import path
if (path.exists('input.txt')):
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
n = int(sys.stdin.readline().strip())
arr = list(sys.stdin.readline().strip())
ptr1, ptr2 = 0, 1
ans = []
# print(arr)
while ptr1+2<n or ptr2+2<n:
if arr[ptr1] == 'R' and arr[ptr1+2] == 'L':
ans.append([ptr1+2, ptr1])
else:
ans.append([ptr1, ptr1+2])
ptr1 += 4
if ptr2+2 >= n:
break
if arr[ptr2] == 'R' and arr[ptr2 + 2] == 'L':
ans.append([ptr2+2, ptr2])
else:
ans.append([ptr2, ptr2+2])
ptr2 += 4
if n%4 != 0:
ans.append([n-2, n-1])
firstPair = ans[0]
lastPair = ans[-1]
if arr[lastPair[0]] == 'L':
firstPair[0], lastPair[0] = lastPair[0], firstPair[0]
else:
firstPair[1], lastPair[1] = lastPair[1], firstPair[1]
if arr[lastPair[0]] == 'R':
lastPair[0], lastPair[1] = lastPair[1], lastPair[0]
for i in ans:
sys.stdout.write(str(i[0] + 1) + " " + str(i[1] + 1) + "\n")
# print(str(i[0] + 1) + " " + str(i[1] + 1) + " : " + str(arr[i[0]]) + " " + str(arr[i[1]]))
# . . . . . . . .
# 1 2 1 2 3 4 3 4
```
| 0
|
|
161
|
A
|
Dress'em in Vests!
|
PROGRAMMING
| 1,300
|
[
"binary search",
"brute force",
"greedy",
"two pointers"
] | null | null |
The Two-dimensional kingdom is going through hard times... This morning the Three-Dimensional kingdom declared war on the Two-dimensional one. This (possibly armed) conflict will determine the ultimate owner of the straight line.
The Two-dimensional kingdom has a regular army of *n* people. Each soldier registered himself and indicated the desired size of the bulletproof vest: the *i*-th soldier indicated size *a**i*. The soldiers are known to be unpretentious, so the command staff assumes that the soldiers are comfortable in any vests with sizes from *a**i*<=-<=*x* to *a**i*<=+<=*y*, inclusive (numbers *x*,<=*y*<=≥<=0 are specified).
The Two-dimensional kingdom has *m* vests at its disposal, the *j*-th vest's size equals *b**j*. Help mobilize the Two-dimensional kingdom's army: equip with vests as many soldiers as possible. Each vest can be used only once. The *i*-th soldier can put on the *j*-th vest, if *a**i*<=-<=*x*<=≤<=*b**j*<=≤<=*a**i*<=+<=*y*.
|
The first input line contains four integers *n*, *m*, *x* and *y* (1<=≤<=*n*,<=*m*<=≤<=105, 0<=≤<=*x*,<=*y*<=≤<=109) — the number of soldiers, the number of vests and two numbers that specify the soldiers' unpretentiousness, correspondingly.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) in non-decreasing order, separated by single spaces — the desired sizes of vests.
The third line contains *m* integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**j*<=≤<=109) in non-decreasing order, separated by single spaces — the sizes of the available vests.
|
In the first line print a single integer *k* — the maximum number of soldiers equipped with bulletproof vests.
In the next *k* lines print *k* pairs, one pair per line, as "*u**i* *v**i*" (without the quotes). Pair (*u**i*, *v**i*) means that soldier number *u**i* must wear vest number *v**i*. Soldiers and vests are numbered starting from one in the order in which they are specified in the input. All numbers of soldiers in the pairs should be pairwise different, all numbers of vests in the pairs also should be pairwise different. You can print the pairs in any order.
If there are multiple optimal answers, you are allowed to print any of them.
|
[
"5 3 0 0\n1 2 3 3 4\n1 3 5\n",
"3 3 2 2\n1 5 9\n3 5 7\n"
] |
[
"2\n1 1\n3 2\n",
"3\n1 1\n2 2\n3 3\n"
] |
In the first sample you need the vests' sizes to match perfectly: the first soldier gets the first vest (size 1), the third soldier gets the second vest (size 3). This sample allows another answer, which gives the second vest to the fourth soldier instead of the third one.
In the second sample the vest size can differ from the desired size by at most 2 sizes, so all soldiers can be equipped.
| 1,000
|
[
{
"input": "5 3 0 0\n1 2 3 3 4\n1 3 5",
"output": "2\n1 1\n3 2"
},
{
"input": "3 3 2 2\n1 5 9\n3 5 7",
"output": "3\n1 1\n2 2\n3 3"
},
{
"input": "1 1 0 0\n1\n1",
"output": "1\n1 1"
},
{
"input": "1 1 0 0\n1\n2",
"output": "0"
},
{
"input": "2 3 1 4\n1 5\n1 2 2",
"output": "1\n1 1"
},
{
"input": "20 30 1 4\n1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 4 4 4 5\n1 1 1 1 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 4 4 4 4 4 4 4 4 5 5",
"output": "20\n1 1\n2 2\n3 3\n4 4\n5 5\n6 6\n7 7\n8 8\n9 9\n10 10\n11 11\n12 12\n13 15\n14 16\n15 17\n16 18\n17 19\n18 20\n19 21\n20 22"
},
{
"input": "33 23 17 2\n1 1 2 2 2 3 3 3 3 3 3 4 4 4 4 4 5 5 5 6 6 7 7 7 8 8 8 8 8 9 9 10 10\n1 1 3 3 4 4 4 5 5 6 6 6 7 8 8 8 8 8 8 9 9 10 10",
"output": "23\n1 1\n2 2\n3 3\n4 4\n5 5\n6 6\n7 7\n8 8\n9 9\n12 10\n13 11\n14 12\n17 13\n20 14\n21 15\n22 16\n23 17\n24 18\n25 19\n26 20\n27 21\n28 22\n29 23"
},
{
"input": "2 2 1 4\n1 4\n3 6",
"output": "2\n1 1\n2 2"
},
{
"input": "20 20 1 4\n1 1 1 1 1 2 2 2 2 2 2 2 2 3 3 3 4 4 5 5\n3 3 3 3 3 4 4 4 4 4 4 4 4 5 5 5 6 6 7 7",
"output": "20\n1 1\n2 2\n3 3\n4 4\n5 5\n6 6\n7 7\n8 8\n9 9\n10 10\n11 11\n12 12\n13 13\n14 14\n15 15\n16 16\n17 17\n18 18\n19 19\n20 20"
},
{
"input": "33 23 17 2\n1 1 1 2 3 3 3 3 3 4 4 4 4 5 6 6 6 6 6 6 7 7 7 7 7 8 8 8 8 8 8 10 10\n10 10 10 11 12 12 12 12 12 13 13 13 13 14 15 15 15 15 15 15 16 16 16",
"output": "5\n26 1\n27 2\n28 3\n32 4\n33 5"
},
{
"input": "1 1 1 2\n783266931\n783266932",
"output": "1\n1 1"
},
{
"input": "2 3 1 4\n1 1\n3 3 4",
"output": "2\n1 1\n2 2"
},
{
"input": "20 30 1 4\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2\n3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4",
"output": "20\n1 1\n2 2\n3 3\n4 4\n5 5\n6 6\n7 7\n8 8\n9 9\n10 10\n11 11\n12 12\n13 13\n14 14\n15 15\n16 16\n17 17\n18 18\n19 19\n20 20"
},
{
"input": "33 23 17 2\n1 1 1 1 1 1 1 2 2 2 2 2 2 2 3 3 3 3 3 3 4 4 4 4 4 4 4 5 5 5 5 5 5\n10 10 10 10 10 10 10 11 11 11 11 11 11 11 12 12 12 12 12 12 13 13 13",
"output": "0"
},
{
"input": "10 1 1 1\n1 1 1 5 5 7 7 8 8 8\n3",
"output": "0"
},
{
"input": "10 1 2 1\n1 2 5 8 9 9 9 10 10 10\n7",
"output": "1\n4 1"
},
{
"input": "1 1 2 1\n7\n8",
"output": "1\n1 1"
},
{
"input": "1 1 0 2\n9\n7",
"output": "0"
},
{
"input": "2 2 0 2\n2 8\n3 5",
"output": "1\n1 1"
},
{
"input": "2 2 1 0\n5 5\n6 7",
"output": "0"
},
{
"input": "2 2 3 0\n8 9\n1 5",
"output": "1\n1 2"
}
] | 1,656,760,926
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 1
| 92
| 102,400
|
def popleft_to(elem):
global popped
global b
while b[0] <= elem:
b.popleft()
popped += 1
def elem_to_big_index(elem):
if elem in b:
return b.index(elem)
else:
l = 0
r = len(b) - 1
while r - l > 1:
mid = (r + l) // 2
if b[mid] <= elem:
l = mid
else:
r = mid
return l
from collections import deque
ans = []
n, m, x, y = map(int, input().split())
a = list(map(int, input().split()))
b = deque(map(int, input().split()))
popped = 0
for i in range(n):
idk = elem_to_big_index(a[i] - x)
if b[idk] <= a[i] + y:
ans.append([i + 1, idk + popped + 1])
popleft_to(b[idk])
print(len(ans))
for i in ans:
print(" ".join(list(map(str, i))))
# Sat Jul 02 2022 11:22:06 GMT+0000 (Coordinated Universal Time)
|
Title: Dress'em in Vests!
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Two-dimensional kingdom is going through hard times... This morning the Three-Dimensional kingdom declared war on the Two-dimensional one. This (possibly armed) conflict will determine the ultimate owner of the straight line.
The Two-dimensional kingdom has a regular army of *n* people. Each soldier registered himself and indicated the desired size of the bulletproof vest: the *i*-th soldier indicated size *a**i*. The soldiers are known to be unpretentious, so the command staff assumes that the soldiers are comfortable in any vests with sizes from *a**i*<=-<=*x* to *a**i*<=+<=*y*, inclusive (numbers *x*,<=*y*<=≥<=0 are specified).
The Two-dimensional kingdom has *m* vests at its disposal, the *j*-th vest's size equals *b**j*. Help mobilize the Two-dimensional kingdom's army: equip with vests as many soldiers as possible. Each vest can be used only once. The *i*-th soldier can put on the *j*-th vest, if *a**i*<=-<=*x*<=≤<=*b**j*<=≤<=*a**i*<=+<=*y*.
Input Specification:
The first input line contains four integers *n*, *m*, *x* and *y* (1<=≤<=*n*,<=*m*<=≤<=105, 0<=≤<=*x*,<=*y*<=≤<=109) — the number of soldiers, the number of vests and two numbers that specify the soldiers' unpretentiousness, correspondingly.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) in non-decreasing order, separated by single spaces — the desired sizes of vests.
The third line contains *m* integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**j*<=≤<=109) in non-decreasing order, separated by single spaces — the sizes of the available vests.
Output Specification:
In the first line print a single integer *k* — the maximum number of soldiers equipped with bulletproof vests.
In the next *k* lines print *k* pairs, one pair per line, as "*u**i* *v**i*" (without the quotes). Pair (*u**i*, *v**i*) means that soldier number *u**i* must wear vest number *v**i*. Soldiers and vests are numbered starting from one in the order in which they are specified in the input. All numbers of soldiers in the pairs should be pairwise different, all numbers of vests in the pairs also should be pairwise different. You can print the pairs in any order.
If there are multiple optimal answers, you are allowed to print any of them.
Demo Input:
['5 3 0 0\n1 2 3 3 4\n1 3 5\n', '3 3 2 2\n1 5 9\n3 5 7\n']
Demo Output:
['2\n1 1\n3 2\n', '3\n1 1\n2 2\n3 3\n']
Note:
In the first sample you need the vests' sizes to match perfectly: the first soldier gets the first vest (size 1), the third soldier gets the second vest (size 3). This sample allows another answer, which gives the second vest to the fourth soldier instead of the third one.
In the second sample the vest size can differ from the desired size by at most 2 sizes, so all soldiers can be equipped.
|
```python
def popleft_to(elem):
global popped
global b
while b[0] <= elem:
b.popleft()
popped += 1
def elem_to_big_index(elem):
if elem in b:
return b.index(elem)
else:
l = 0
r = len(b) - 1
while r - l > 1:
mid = (r + l) // 2
if b[mid] <= elem:
l = mid
else:
r = mid
return l
from collections import deque
ans = []
n, m, x, y = map(int, input().split())
a = list(map(int, input().split()))
b = deque(map(int, input().split()))
popped = 0
for i in range(n):
idk = elem_to_big_index(a[i] - x)
if b[idk] <= a[i] + y:
ans.append([i + 1, idk + popped + 1])
popleft_to(b[idk])
print(len(ans))
for i in ans:
print(" ".join(list(map(str, i))))
# Sat Jul 02 2022 11:22:06 GMT+0000 (Coordinated Universal Time)
```
| -1
|
|
119
|
A
|
Epic Game
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Simon and Antisimon play a game. Initially each player receives one fixed positive integer that doesn't change throughout the game. Simon receives number *a* and Antisimon receives number *b*. They also have a heap of *n* stones. The players take turns to make a move and Simon starts. During a move a player should take from the heap the number of stones equal to the greatest common divisor of the fixed number he has received and the number of stones left in the heap. A player loses when he cannot take the required number of stones (i. e. the heap has strictly less stones left than one needs to take).
Your task is to determine by the given *a*, *b* and *n* who wins the game.
|
The only string contains space-separated integers *a*, *b* and *n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=100) — the fixed numbers Simon and Antisimon have received correspondingly and the initial number of stones in the pile.
|
If Simon wins, print "0" (without the quotes), otherwise print "1" (without the quotes).
|
[
"3 5 9\n",
"1 1 100\n"
] |
[
"0",
"1"
] |
The greatest common divisor of two non-negative integers *a* and *b* is such maximum positive integer *k*, that *a* is divisible by *k* without remainder and similarly, *b* is divisible by *k* without remainder. Let *gcd*(*a*, *b*) represent the operation of calculating the greatest common divisor of numbers *a* and *b*. Specifically, *gcd*(*x*, 0) = *gcd*(0, *x*) = *x*.
In the first sample the game will go like that:
- Simon should take *gcd*(3, 9) = 3 stones from the heap. After his move the heap has 6 stones left.- Antisimon should take *gcd*(5, 6) = 1 stone from the heap. After his move the heap has 5 stones left.- Simon should take *gcd*(3, 5) = 1 stone from the heap. After his move the heap has 4 stones left.- Antisimon should take *gcd*(5, 4) = 1 stone from the heap. After his move the heap has 3 stones left.- Simon should take *gcd*(3, 3) = 3 stones from the heap. After his move the heap has 0 stones left.- Antisimon should take *gcd*(5, 0) = 5 stones from the heap. As 0 < 5, it is impossible and Antisimon loses.
In the second sample each player during each move takes one stone from the heap. As *n* is even, Antisimon takes the last stone and Simon can't make a move after that.
| 500
|
[
{
"input": "3 5 9",
"output": "0"
},
{
"input": "1 1 100",
"output": "1"
},
{
"input": "23 12 16",
"output": "1"
},
{
"input": "95 26 29",
"output": "1"
},
{
"input": "73 32 99",
"output": "1"
},
{
"input": "1 1 1",
"output": "0"
},
{
"input": "41 12 65",
"output": "1"
},
{
"input": "13 61 100",
"output": "1"
},
{
"input": "100 100 10",
"output": "0"
},
{
"input": "12 24 26",
"output": "1"
},
{
"input": "73 21 96",
"output": "1"
},
{
"input": "17 22 81",
"output": "1"
},
{
"input": "14 88 97",
"output": "1"
},
{
"input": "42 81 17",
"output": "0"
},
{
"input": "34 36 19",
"output": "1"
},
{
"input": "21 87 12",
"output": "1"
},
{
"input": "97 92 98",
"output": "1"
},
{
"input": "49 25 62",
"output": "1"
},
{
"input": "29 17 44",
"output": "1"
},
{
"input": "10 1 10",
"output": "0"
},
{
"input": "99 98 100",
"output": "1"
},
{
"input": "100 100 1",
"output": "0"
},
{
"input": "11 99 55",
"output": "0"
},
{
"input": "8 13 21",
"output": "0"
},
{
"input": "34 55 89",
"output": "0"
},
{
"input": "21 34 55",
"output": "1"
},
{
"input": "13 21 34",
"output": "1"
},
{
"input": "34 21 55",
"output": "0"
},
{
"input": "10 20 84",
"output": "1"
},
{
"input": "4 2 5",
"output": "0"
},
{
"input": "3 2 3",
"output": "0"
},
{
"input": "31 23 1",
"output": "0"
},
{
"input": "100 100 100",
"output": "0"
},
{
"input": "100 1 100",
"output": "0"
},
{
"input": "1 100 100",
"output": "1"
},
{
"input": "1 1 100",
"output": "1"
},
{
"input": "17 23 100",
"output": "1"
},
{
"input": "99 98 100",
"output": "1"
},
{
"input": "1 1 2",
"output": "1"
},
{
"input": "2 1 1",
"output": "0"
},
{
"input": "1 2 1",
"output": "0"
},
{
"input": "2 1 2",
"output": "0"
},
{
"input": "10 10 3",
"output": "1"
},
{
"input": "2 2 1",
"output": "0"
},
{
"input": "9 5 1",
"output": "0"
}
] | 1,613,979,378
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
import sys
a, b, n = map(int, sys.stdin.readline().split())
gcd(x, y):
for i in reversed(1, range(max(x, y)+1)):
if x%i == 0 and y%i == 0:
return i
cnt = 0
while n > 0:
if cnt == 0:
n -= gcd(a, n)
cnt = 1
elif cnt == 1:
n -= gcd(b, n)
cnt = 0
if cnt == 0:
print(0)
else:
print(1)
|
Title: Epic Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Simon and Antisimon play a game. Initially each player receives one fixed positive integer that doesn't change throughout the game. Simon receives number *a* and Antisimon receives number *b*. They also have a heap of *n* stones. The players take turns to make a move and Simon starts. During a move a player should take from the heap the number of stones equal to the greatest common divisor of the fixed number he has received and the number of stones left in the heap. A player loses when he cannot take the required number of stones (i. e. the heap has strictly less stones left than one needs to take).
Your task is to determine by the given *a*, *b* and *n* who wins the game.
Input Specification:
The only string contains space-separated integers *a*, *b* and *n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=100) — the fixed numbers Simon and Antisimon have received correspondingly and the initial number of stones in the pile.
Output Specification:
If Simon wins, print "0" (without the quotes), otherwise print "1" (without the quotes).
Demo Input:
['3 5 9\n', '1 1 100\n']
Demo Output:
['0', '1']
Note:
The greatest common divisor of two non-negative integers *a* and *b* is such maximum positive integer *k*, that *a* is divisible by *k* without remainder and similarly, *b* is divisible by *k* without remainder. Let *gcd*(*a*, *b*) represent the operation of calculating the greatest common divisor of numbers *a* and *b*. Specifically, *gcd*(*x*, 0) = *gcd*(0, *x*) = *x*.
In the first sample the game will go like that:
- Simon should take *gcd*(3, 9) = 3 stones from the heap. After his move the heap has 6 stones left.- Antisimon should take *gcd*(5, 6) = 1 stone from the heap. After his move the heap has 5 stones left.- Simon should take *gcd*(3, 5) = 1 stone from the heap. After his move the heap has 4 stones left.- Antisimon should take *gcd*(5, 4) = 1 stone from the heap. After his move the heap has 3 stones left.- Simon should take *gcd*(3, 3) = 3 stones from the heap. After his move the heap has 0 stones left.- Antisimon should take *gcd*(5, 0) = 5 stones from the heap. As 0 < 5, it is impossible and Antisimon loses.
In the second sample each player during each move takes one stone from the heap. As *n* is even, Antisimon takes the last stone and Simon can't make a move after that.
|
```python
import sys
a, b, n = map(int, sys.stdin.readline().split())
gcd(x, y):
for i in reversed(1, range(max(x, y)+1)):
if x%i == 0 and y%i == 0:
return i
cnt = 0
while n > 0:
if cnt == 0:
n -= gcd(a, n)
cnt = 1
elif cnt == 1:
n -= gcd(b, n)
cnt = 0
if cnt == 0:
print(0)
else:
print(1)
```
| -1
|
|
474
|
A
|
Keyboard
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Our good friend Mole is trying to code a big message. He is typing on an unusual keyboard with characters arranged in following way:
Unfortunately Mole is blind, so sometimes it is problem for him to put his hands accurately. He accidentally moved both his hands with one position to the left or to the right. That means that now he presses not a button he wants, but one neighboring button (left or right, as specified in input).
We have a sequence of characters he has typed and we want to find the original message.
|
First line of the input contains one letter describing direction of shifting ('L' or 'R' respectively for left or right).
Second line contains a sequence of characters written by Mole. The size of this sequence will be no more than 100. Sequence contains only symbols that appear on Mole's keyboard. It doesn't contain spaces as there is no space on Mole's keyboard.
It is guaranteed that even though Mole hands are moved, he is still pressing buttons on keyboard and not hitting outside it.
|
Print a line that contains the original message.
|
[
"R\ns;;upimrrfod;pbr\n"
] |
[
"allyouneedislove\n"
] |
none
| 500
|
[
{
"input": "R\ns;;upimrrfod;pbr",
"output": "allyouneedislove"
},
{
"input": "R\nwertyuiop;lkjhgfdsxcvbnm,.",
"output": "qwertyuiolkjhgfdsazxcvbnm,"
},
{
"input": "L\nzxcvbnm,kjhgfdsaqwertyuio",
"output": "xcvbnm,.lkjhgfdswertyuiop"
},
{
"input": "R\nbubbuduppudup",
"output": "vyvvysyooysyo"
},
{
"input": "L\ngggggggggggggggggggggggggggggggggggggggggg",
"output": "hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh"
},
{
"input": "R\ngggggggggggggggggggggggggggggggggggggggggg",
"output": "ffffffffffffffffffffffffffffffffffffffffff"
},
{
"input": "L\nggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggg",
"output": "hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh"
},
{
"input": "R\nggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggg",
"output": "fffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff"
},
{
"input": "L\nxgwurenkxkiau,c,vonei.zltazmnkhqtwuogkgvgckvja,z.rhanuy.ybebmzcfwozkwvuuiolaqlgvvvewnbuinrncgjwjdsfw",
"output": "cheitrmlclosi.v.bpmro/x;ysx,mljwyeiphlhbhvlbks.x/tjsmiu/unrn,xvgepxlebiiop;sw;hbbbremniomtmvhkekfdge"
},
{
"input": "L\nuoz.vmks,wxrb,nwcvdzh.m,hwsios.lvu,ktes,,ythddhm.sh,d,c,cfj.wqam,bowofbyx,jathqayhreqvixvbmgdokofmym",
"output": "ipx/b,ld.ectn.mevbfxj/,.jedopd/;bi.lyrd..uyjffj,/dj.f.v.vgk/ews,.npepgnuc.ksyjwsujtrwbocbn,hfplpg,u,"
},
{
"input": "R\noedjyrvuw/rn.v.hdwndbiposiewgsn.pnyf;/tsdohp,hrtd/mx,;coj./billd..mwbneohcikrdes/ucjr,wspthleyp,..f,",
"output": "iwshtecyq.eb,c,gsqbsvuoiauwqfab,obtdl.rasigomgers.nzmlxih,.vukks,,nqvbwigxujeswa.yxhemqaorgkwtom,,dm"
},
{
"input": "R\nvgj;o;ijrtfyck,dthccioltcx,crub;oceooognsuvfx/kgo.fbsudv,yod.erdrxhbeiyltxhnrobbb;ydrgroefcr/f;uvdjd",
"output": "cfhliluherdtxjmsrgxxuikrxzmxeyvlixwiiifbaycdz.jfi,dvayscmtis,wesezgvwutkrzgbeivvvltsefeiwdxe.dlycshs"
},
{
"input": "L\nqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq",
"output": "wwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww"
},
{
"input": "L\noooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo",
"output": "pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp"
},
{
"input": "L\n,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,",
"output": "...................................................................................................."
},
{
"input": "L\nzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz",
"output": "xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx"
},
{
"input": "R\noooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo",
"output": "iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii"
},
{
"input": "R\nwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww",
"output": "qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq"
},
{
"input": "R\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx",
"output": "zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz"
},
{
"input": "L\nq",
"output": "w"
},
{
"input": "L\no",
"output": "p"
},
{
"input": "L\n,",
"output": "."
},
{
"input": "L\nz",
"output": "x"
},
{
"input": "R\n.",
"output": ","
},
{
"input": "R\no",
"output": "i"
},
{
"input": "R\nw",
"output": "q"
},
{
"input": "R\nx",
"output": "z"
},
{
"input": "R\n,./",
"output": "m,."
},
{
"input": "R\nwertuk;;/",
"output": "qweryjll."
},
{
"input": "L\n..",
"output": "//"
}
] | 1,672,744,225
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 31
| 46
| 0
|
def mover(ele,line,shift):
if shift=='R':
idx=line.index(ele)
ele = line[idx-1]
return ele
else:
idx=line.index(ele)
ele = line[idx+1]
return ele
shift=input()
st=input()
one= 'qwertyuiop'
two= 'asdfghjkl;'
three='zxcvbnm,./'
s=''
for ele in st:
if ele in one:
s+=mover(ele,one,shift)
elif ele in two:
s+=mover(ele,two,shift)
else:
s+=mover(ele,three,shift)
print(s)
|
Title: Keyboard
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Our good friend Mole is trying to code a big message. He is typing on an unusual keyboard with characters arranged in following way:
Unfortunately Mole is blind, so sometimes it is problem for him to put his hands accurately. He accidentally moved both his hands with one position to the left or to the right. That means that now he presses not a button he wants, but one neighboring button (left or right, as specified in input).
We have a sequence of characters he has typed and we want to find the original message.
Input Specification:
First line of the input contains one letter describing direction of shifting ('L' or 'R' respectively for left or right).
Second line contains a sequence of characters written by Mole. The size of this sequence will be no more than 100. Sequence contains only symbols that appear on Mole's keyboard. It doesn't contain spaces as there is no space on Mole's keyboard.
It is guaranteed that even though Mole hands are moved, he is still pressing buttons on keyboard and not hitting outside it.
Output Specification:
Print a line that contains the original message.
Demo Input:
['R\ns;;upimrrfod;pbr\n']
Demo Output:
['allyouneedislove\n']
Note:
none
|
```python
def mover(ele,line,shift):
if shift=='R':
idx=line.index(ele)
ele = line[idx-1]
return ele
else:
idx=line.index(ele)
ele = line[idx+1]
return ele
shift=input()
st=input()
one= 'qwertyuiop'
two= 'asdfghjkl;'
three='zxcvbnm,./'
s=''
for ele in st:
if ele in one:
s+=mover(ele,one,shift)
elif ele in two:
s+=mover(ele,two,shift)
else:
s+=mover(ele,three,shift)
print(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 — 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* — 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"
}
] | 1,610,306,166
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 30
| 0
|
n=input()
a,b=0,0
for i in range(len(n)):
if n[i].islower()==True:
a+=1
else:
b+=1
if x>=y:
print(n.lower())
else:
print(n.upper())
|
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 — 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* — 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
n=input()
a,b=0,0
for i in range(len(n)):
if n[i].islower()==True:
a+=1
else:
b+=1
if x>=y:
print(n.lower())
else:
print(n.upper())
```
| -1
|
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<=≤<=*n*<=≤<=100) — the number of elements in the array. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000) — 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"
}
] | 1,449,339,710
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 0
| 156
| 0
|
n = int(input())
a = set()
d = dict()
v = list(map(int, input().split()))
mx, sum = 0, 0
for i in range(n):
if v[i] not in a:
a.add(v[i])
d[v[i]] = 1
else:
d[v[i]] += 1
for x in a:
if d[x] > mx:
sum += mx
mx = d[x]
else:
sum += d[x]
print(sum, mx)
if (sum + 1) >= mx:
print("YES")
else:
print("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<=≤<=*n*<=≤<=100) — the number of elements in the array. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000) — 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())
a = set()
d = dict()
v = list(map(int, input().split()))
mx, sum = 0, 0
for i in range(n):
if v[i] not in a:
a.add(v[i])
d[v[i]] = 1
else:
d[v[i]] += 1
for x in a:
if d[x] > mx:
sum += mx
mx = d[x]
else:
sum += d[x]
print(sum, mx)
if (sum + 1) >= mx:
print("YES")
else:
print("NO")
```
| 0
|
|
711
|
B
|
Chris and Magic Square
|
PROGRAMMING
| 1,400
|
[
"constructive algorithms",
"implementation"
] | null | null |
ZS the Coder and Chris the Baboon arrived at the entrance of Udayland. There is a *n*<=×<=*n* magic grid on the entrance which is filled with integers. Chris noticed that exactly one of the cells in the grid is empty, and to enter Udayland, they need to fill a positive integer into the empty cell.
Chris tried filling in random numbers but it didn't work. ZS the Coder realizes that they need to fill in a positive integer such that the numbers in the grid form a magic square. This means that he has to fill in a positive integer so that the sum of the numbers in each row of the grid (), each column of the grid (), and the two long diagonals of the grid (the main diagonal — and the secondary diagonal — ) are equal.
Chris doesn't know what number to fill in. Can you help Chris find the correct positive integer to fill in or determine that it is impossible?
|
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=500) — the number of rows and columns of the magic grid.
*n* lines follow, each of them contains *n* integers. The *j*-th number in the *i*-th of them denotes *a**i*,<=*j* (1<=≤<=*a**i*,<=*j*<=≤<=109 or *a**i*,<=*j*<==<=0), the number in the *i*-th row and *j*-th column of the magic grid. If the corresponding cell is empty, *a**i*,<=*j* will be equal to 0. Otherwise, *a**i*,<=*j* is positive.
It is guaranteed that there is exactly one pair of integers *i*,<=*j* (1<=≤<=*i*,<=*j*<=≤<=*n*) such that *a**i*,<=*j*<==<=0.
|
Output a single integer, the positive integer *x* (1<=≤<=*x*<=≤<=1018) that should be filled in the empty cell so that the whole grid becomes a magic square. If such positive integer *x* does not exist, output <=-<=1 instead.
If there are multiple solutions, you may print any of them.
|
[
"3\n4 0 2\n3 5 7\n8 1 6\n",
"4\n1 1 1 1\n1 1 0 1\n1 1 1 1\n1 1 1 1\n",
"4\n1 1 1 1\n1 1 0 1\n1 1 2 1\n1 1 1 1\n"
] |
[
"9\n",
"1\n",
"-1\n"
] |
In the first sample case, we can fill in 9 into the empty cell to make the resulting grid a magic square. Indeed,
The sum of numbers in each row is:
4 + 9 + 2 = 3 + 5 + 7 = 8 + 1 + 6 = 15.
The sum of numbers in each column is:
4 + 3 + 8 = 9 + 5 + 1 = 2 + 7 + 6 = 15.
The sum of numbers in the two diagonals is:
4 + 5 + 6 = 2 + 5 + 8 = 15.
In the third sample case, it is impossible to fill a number in the empty square such that the resulting grid is a magic square.
| 1,000
|
[
{
"input": "3\n4 0 2\n3 5 7\n8 1 6",
"output": "9"
},
{
"input": "4\n1 1 1 1\n1 1 0 1\n1 1 1 1\n1 1 1 1",
"output": "1"
},
{
"input": "4\n1 1 1 1\n1 1 0 1\n1 1 2 1\n1 1 1 1",
"output": "-1"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "10\n92 67 99 74 1 51 8 58 15 40\n17 42 24 49 0 26 83 33 90 65\n98 73 80 55 7 57 14 64 16 41\n23 48 5 30 82 32 89 39 91 66\n4 54 81 56 88 63 20 70 22 47\n79 29 6 31 13 38 95 45 97 72\n85 60 87 62 19 69 21 71 3 28\n10 35 12 37 94 44 96 46 78 53\n86 61 93 68 25 75 2 52 9 34\n11 36 18 43 100 50 77 27 84 59",
"output": "76"
},
{
"input": "4\n1000000000 1000000000 1000000000 1000000000\n1000000000 1000000000 1000000000 1000000000\n1000000000 1000000000 0 1000000000\n1000000000 1000000000 1000000000 1000000000",
"output": "1000000000"
},
{
"input": "3\n3 8 1\n2 4 6\n7 0 5",
"output": "-1"
},
{
"input": "3\n1 2 2\n2 2 1\n0 1 2",
"output": "-1"
},
{
"input": "3\n1 6 10\n5 6 16\n0 5 1",
"output": "-1"
},
{
"input": "3\n2 2 1\n1 2 2\n0 1 2",
"output": "-1"
},
{
"input": "3\n1 2 2\n2 2 1\n2 1 0",
"output": "-1"
},
{
"input": "3\n2016 2016 2016\n2016 0 2016\n2016 2016 2016",
"output": "2016"
},
{
"input": "10\n92 67 99 74 1 51 8 58 15 40\n17 42 24 49 76 26 83 33 90 65\n98 73 80 55 7 57 14 64 16 41\n23 48 5 30 82 32 89 39 91 66\n4 54 81 56 88 63 20 70 22 47\n79 29 6 31 13 38 95 45 97 72\n85 60 87 62 19 69 21 71 3 28\n10 35 12 37 94 44 96 46 78 53\n86 61 93 68 25 75 2 52 0 34\n11 36 18 43 100 50 77 27 84 59",
"output": "9"
},
{
"input": "10\n92 67 99 74 1 51 8 58 15 40\n17 42 24 49 76 26 83 33 90 65\n98 73 80 55 7 57 14 64 16 41\n23 48 5 30 82 32 89 39 91 66\n4 54 81 56 0 63 20 70 22 47\n79 29 6 31 13 38 95 45 97 72\n85 60 87 62 19 69 21 71 3 28\n10 35 12 37 94 44 96 46 78 53\n86 61 93 68 25 75 2 52 9 34\n11 36 18 43 100 50 77 27 84 59",
"output": "88"
},
{
"input": "3\n2 2 1\n1 2 2\n2 1 0",
"output": "-1"
},
{
"input": "10\n92 67 99 74 1 51 8 58 15 0\n17 42 24 49 76 26 83 33 90 65\n98 73 80 55 7 57 14 64 16 41\n23 48 5 30 82 32 89 39 91 66\n4 54 81 56 88 63 20 70 22 47\n79 29 6 31 13 38 95 45 97 72\n85 60 87 62 19 69 21 71 3 28\n10 35 12 37 94 44 96 46 78 53\n86 61 93 68 25 75 2 52 9 34\n11 36 18 43 100 50 77 27 84 59",
"output": "40"
},
{
"input": "4\n2 2 2 2\n2 0 2 2\n3 2 2 1\n2 2 2 2",
"output": "-1"
},
{
"input": "3\n1 15 5\n11 7 3\n9 0 13",
"output": "-1"
},
{
"input": "3\n61 0 41\n11 31 51\n21 71 1",
"output": "-1"
},
{
"input": "3\n3 0 3\n2 3 2\n2 3 2",
"output": "-1"
},
{
"input": "3\n0 2 2\n3 1 1\n1 2 2",
"output": "-1"
},
{
"input": "3\n1 0 1\n1 1 2\n1 1 1",
"output": "-1"
},
{
"input": "3\n1 0 1\n2 1 2\n2 1 2",
"output": "-1"
},
{
"input": "3\n1 0 1\n4 1 4\n1 1 1",
"output": "-1"
},
{
"input": "3\n1 1 1\n1 1 0\n1 2 1",
"output": "-1"
},
{
"input": "3\n2 0 1\n1 2 1\n1 1 2",
"output": "-1"
},
{
"input": "3\n1 2 2\n3 1 1\n0 2 2",
"output": "-1"
},
{
"input": "4\n0 1 1 1\n1 1 1 1\n1 1 1 2\n1 1 2 1",
"output": "-1"
},
{
"input": "4\n1 1 0 1\n1 1 1 1\n1 1 1 1\n1 2 1 1",
"output": "-1"
},
{
"input": "5\n1 1 1000000000 1000000000 1000000000\n1 1000000000 1 1000000000 1000000000\n0 1 1 1 1\n1 1000000000 1000000000 1000000000 1\n1 1000000000 1000000000 1 1000000000",
"output": "2999999998"
},
{
"input": "3\n5 5 5\n6 5 0\n5 5 5",
"output": "-1"
},
{
"input": "3\n1 0 1\n50 1 500\n2 1 2",
"output": "-1"
},
{
"input": "9\n1 1000000000 1 1000000000 1 1000000000 1 1000000000 1\n1000000000 1 1000000000 1 1 1 1000000000 1 1000000000\n1 1000000000 1 1000000000 1 1000000000 1 1000000000 1\n1000000000 1 1000000000 1 1 1 1000000000 1 1000000000\n1 1 1 1 0 1 1 1 1\n1000000000 1 1000000000 1 1 1 1000000000 1 1000000000\n1 1000000000 1 1000000000 1 1000000000 1 1000000000 1\n1000000000 1 1000000000 1 1 1 1000000000 1 1000000000\n1 1000000000 1 1000000000 1 1000000000 1 1000000000 1",
"output": "3999999997"
},
{
"input": "3\n7 22 1\n4 10 16\n19 0 13",
"output": "-1"
},
{
"input": "5\n1 1 1 1 1\n1 1 1 1 0\n1 2 1 1 1\n1 1 1 1 1\n1 1 1 1 1",
"output": "-1"
},
{
"input": "4\n3 6 0 2\n5 5 7 1\n1 7 4 6\n2 9 1 6",
"output": "-1"
},
{
"input": "5\n1 2 1 1 1\n1 1 2 1 1\n2 1 1 0 1\n1 1 1 1 2\n1 1 1 2 1",
"output": "-1"
},
{
"input": "11\n5 5 5 5 5 5 5 5 5 5 5\n5 5 5 5 5 5 5 5 5 5 5\n5 5 5 5 5 5 5 5 5 5 5\n5 5 5 5 5 5 5 5 5 5 5\n5 5 5 5 5 5 5 5 5 5 5\n5 5 5 5 5 5 5 5 5 5 5\n5 5 5 5 5 5 5 5 5 5 5\n5 5 5 5 5 5 5 5 5 5 5\n5 5 5 5 13 1 1 5 5 5 5\n5 5 5 5 5 9 1 5 5 5 5\n5 5 5 5 0 5 13 5 5 5 5",
"output": "-1"
},
{
"input": "2\n5 5\n5 0",
"output": "5"
},
{
"input": "5\n10 10 1 10 10\n1 1 0 1 1\n10 10 1 10 10\n10 10 1 10 10\n10 10 1 10 10",
"output": "-1"
},
{
"input": "5\n1 1 1 2 1\n1 1 1 1 1\n1 1 0 1 1\n1 1 1 1 1\n1 1 1 1 1",
"output": "-1"
},
{
"input": "3\n1000000000 1000000000 1000000000\n1000000000 1000000000 1000000000\n1000000000 0 1000000000",
"output": "1000000000"
},
{
"input": "3\n3 3 3\n0 2 5\n1 1 1",
"output": "-1"
},
{
"input": "4\n2 2 3 1\n1 0 3 3\n4 3 4 1\n1 2 3 1",
"output": "-1"
},
{
"input": "3\n1 1 2\n2 1 0\n1 2 1",
"output": "-1"
},
{
"input": "2\n1 2\n1 0",
"output": "-1"
},
{
"input": "2\n0 535\n535 535",
"output": "535"
},
{
"input": "6\n0 1 1 1 1 1\n1 1 1000000000 1000000000 1000000000 1000000000\n1 1000000000 1 1000000000 1000000000 1000000000\n1 1000000000 1000000000 1 1000000000 1000000000\n1 1000000000 1000000000 1000000000 1 1000000000\n1 1000000000 1000000000 1000000000 1000000000 1",
"output": "3999999997"
},
{
"input": "4\n2 6 0 3\n5 5 7 1\n5 1 3 9\n6 6 1 5",
"output": "-1"
},
{
"input": "5\n2 1 2 1 2\n2 2 2 2 2\n2 2 0 2 2\n2 2 2 2 2\n2 2 2 2 2",
"output": "-1"
},
{
"input": "3\n1 2 3\n1 0 3\n1 2 3",
"output": "-1"
},
{
"input": "3\n0 1 2\n1 2 1\n2 1 1",
"output": "-1"
},
{
"input": "4\n2 3 2 3\n3 2 3 0\n2 4 2 2\n3 1 3 3",
"output": "-1"
},
{
"input": "3\n1 1 1\n1 0 1\n1 2 1",
"output": "-1"
},
{
"input": "3\n1 1 1\n1 4 1\n1 1 0",
"output": "-1"
},
{
"input": "5\n1 1 2 1 1\n1 1 1 1 1\n1 1 1 0 1\n1 1 1 1 1\n1 1 1 1 1",
"output": "-1"
},
{
"input": "3\n0 1 1\n1 1 1\n1 1 2",
"output": "-1"
},
{
"input": "3\n1 2 1\n1 0 1\n1 2 1",
"output": "-1"
},
{
"input": "3\n6 7 2\n1 0 9\n8 3 4",
"output": "5"
},
{
"input": "3\n1 1 1\n1 1 1\n1 0 1",
"output": "1"
},
{
"input": "3\n3 6 0\n3 3 5\n5 2 4",
"output": "-1"
},
{
"input": "5\n1 2 2 2 1\n1 1 1 1 0\n2 2 1 2 1\n2 1 2 1 1\n1 2 2 2 1",
"output": "-1"
},
{
"input": "4\n1 1 1 1\n1 1 1 0\n1 1 2 1\n1 1 1 1",
"output": "-1"
},
{
"input": "3\n13 0 19\n16 10 4\n1 22 7",
"output": "-1"
},
{
"input": "4\n1 2 2 1\n2 1 0 2\n2 1 1 2\n1 2 2 1",
"output": "-1"
}
] | 1,614,256,414
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
#include <iostream>
using namespace std;
typedef long long ll;
ll a[1001][1001];
ll r[1001];
ll c[1001];
int n;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cin >> n;
int x, y;
ll pdsum = 0;
ll sdsum = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
cin >> a[i][j];
if (a[i][j] == 0) {
x = i;
y = j;
} else {
r[i] += a[i][j];
c[j] += a[i][j];
if (i == j) {
pdsum += a[i][j];
}
if (i + j == n - 1) {
sdsum += a[i][j];
}
}
}
}
if (n == 1) {
cout << 233 << '\n';
return 0;
}
ll commonsum = r[0];
if (x == 0) {
commonsum = r[1];
}
ll rowsum = -1;
ll colsum = -1;
ll d1sum = -1;
ll d2sum = -1;
for (int i = 0; i < n; i++) {
if (i != x) {
if (r[i] != commonsum) {
cout << -1 << endl;
return 0;
}
} else {
rowsum = r[i];
}
}
for (int i = 0; i < n; i++) {
if (i != y) {
if (c[i] != commonsum) {
cout << -1 << endl;
return 0;
}
} else {
colsum = c[i];
}
}
bool isd1 = false;
bool isd2 = false;
if (x == y)
isd1 = true;
if (x + y == n - 1)
isd2 = true;
if (!isd1) {
if (pdsum != commonsum) {
cout << -1 << endl;
return 0;
}
} else {
d1sum = pdsum;
}
if (!isd2) {
if (sdsum != commonsum) {
cout << -1 << endl;
return 0;
}
} else {
d2sum = sdsum;
}
if (rowsum == colsum) {
if (isd1 && d1sum != rowsum) {
cout << -1 << endl;
return 0;
}
if (isd2 && d2sum != rowsum) {
cout << -1 << endl;
return 0;
}
ll value = commonsum - rowsum;
if (value > 0) {
cout << value << '\n';
return 0;
} else {
cout << -1 << endl;
return 0;
}
} else {
cout << -1 << endl;
return 0;
}
}
|
Title: Chris and Magic Square
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
ZS the Coder and Chris the Baboon arrived at the entrance of Udayland. There is a *n*<=×<=*n* magic grid on the entrance which is filled with integers. Chris noticed that exactly one of the cells in the grid is empty, and to enter Udayland, they need to fill a positive integer into the empty cell.
Chris tried filling in random numbers but it didn't work. ZS the Coder realizes that they need to fill in a positive integer such that the numbers in the grid form a magic square. This means that he has to fill in a positive integer so that the sum of the numbers in each row of the grid (), each column of the grid (), and the two long diagonals of the grid (the main diagonal — and the secondary diagonal — ) are equal.
Chris doesn't know what number to fill in. Can you help Chris find the correct positive integer to fill in or determine that it is impossible?
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=500) — the number of rows and columns of the magic grid.
*n* lines follow, each of them contains *n* integers. The *j*-th number in the *i*-th of them denotes *a**i*,<=*j* (1<=≤<=*a**i*,<=*j*<=≤<=109 or *a**i*,<=*j*<==<=0), the number in the *i*-th row and *j*-th column of the magic grid. If the corresponding cell is empty, *a**i*,<=*j* will be equal to 0. Otherwise, *a**i*,<=*j* is positive.
It is guaranteed that there is exactly one pair of integers *i*,<=*j* (1<=≤<=*i*,<=*j*<=≤<=*n*) such that *a**i*,<=*j*<==<=0.
Output Specification:
Output a single integer, the positive integer *x* (1<=≤<=*x*<=≤<=1018) that should be filled in the empty cell so that the whole grid becomes a magic square. If such positive integer *x* does not exist, output <=-<=1 instead.
If there are multiple solutions, you may print any of them.
Demo Input:
['3\n4 0 2\n3 5 7\n8 1 6\n', '4\n1 1 1 1\n1 1 0 1\n1 1 1 1\n1 1 1 1\n', '4\n1 1 1 1\n1 1 0 1\n1 1 2 1\n1 1 1 1\n']
Demo Output:
['9\n', '1\n', '-1\n']
Note:
In the first sample case, we can fill in 9 into the empty cell to make the resulting grid a magic square. Indeed,
The sum of numbers in each row is:
4 + 9 + 2 = 3 + 5 + 7 = 8 + 1 + 6 = 15.
The sum of numbers in each column is:
4 + 3 + 8 = 9 + 5 + 1 = 2 + 7 + 6 = 15.
The sum of numbers in the two diagonals is:
4 + 5 + 6 = 2 + 5 + 8 = 15.
In the third sample case, it is impossible to fill a number in the empty square such that the resulting grid is a magic square.
|
```python
#include <iostream>
using namespace std;
typedef long long ll;
ll a[1001][1001];
ll r[1001];
ll c[1001];
int n;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cin >> n;
int x, y;
ll pdsum = 0;
ll sdsum = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
cin >> a[i][j];
if (a[i][j] == 0) {
x = i;
y = j;
} else {
r[i] += a[i][j];
c[j] += a[i][j];
if (i == j) {
pdsum += a[i][j];
}
if (i + j == n - 1) {
sdsum += a[i][j];
}
}
}
}
if (n == 1) {
cout << 233 << '\n';
return 0;
}
ll commonsum = r[0];
if (x == 0) {
commonsum = r[1];
}
ll rowsum = -1;
ll colsum = -1;
ll d1sum = -1;
ll d2sum = -1;
for (int i = 0; i < n; i++) {
if (i != x) {
if (r[i] != commonsum) {
cout << -1 << endl;
return 0;
}
} else {
rowsum = r[i];
}
}
for (int i = 0; i < n; i++) {
if (i != y) {
if (c[i] != commonsum) {
cout << -1 << endl;
return 0;
}
} else {
colsum = c[i];
}
}
bool isd1 = false;
bool isd2 = false;
if (x == y)
isd1 = true;
if (x + y == n - 1)
isd2 = true;
if (!isd1) {
if (pdsum != commonsum) {
cout << -1 << endl;
return 0;
}
} else {
d1sum = pdsum;
}
if (!isd2) {
if (sdsum != commonsum) {
cout << -1 << endl;
return 0;
}
} else {
d2sum = sdsum;
}
if (rowsum == colsum) {
if (isd1 && d1sum != rowsum) {
cout << -1 << endl;
return 0;
}
if (isd2 && d2sum != rowsum) {
cout << -1 << endl;
return 0;
}
ll value = commonsum - rowsum;
if (value > 0) {
cout << value << '\n';
return 0;
} else {
cout << -1 << endl;
return 0;
}
} else {
cout << -1 << endl;
return 0;
}
}
```
| -1
|
|
518
|
C
|
Anya and Smartphone
|
PROGRAMMING
| 1,600
|
[
"constructive algorithms",
"data structures",
"implementation"
] | null | null |
Anya has bought a new smartphone that uses Berdroid operating system. The smartphone menu has exactly *n* applications, each application has its own icon. The icons are located on different screens, one screen contains *k* icons. The icons from the first to the *k*-th one are located on the first screen, from the (*k*<=+<=1)-th to the 2*k*-th ones are on the second screen and so on (the last screen may be partially empty).
Initially the smartphone menu is showing the screen number 1. To launch the application with the icon located on the screen *t*, Anya needs to make the following gestures: first she scrolls to the required screen number *t*, by making *t*<=-<=1 gestures (if the icon is on the screen *t*), and then make another gesture — press the icon of the required application exactly once to launch it.
After the application is launched, the menu returns to the first screen. That is, to launch the next application you need to scroll through the menu again starting from the screen number 1.
All applications are numbered from 1 to *n*. We know a certain order in which the icons of the applications are located in the menu at the beginning, but it changes as long as you use the operating system. Berdroid is intelligent system, so it changes the order of the icons by moving the more frequently used icons to the beginning of the list. Formally, right after an application is launched, Berdroid swaps the application icon and the icon of a preceding application (that is, the icon of an application on the position that is smaller by one in the order of menu). The preceding icon may possibly be located on the adjacent screen. The only exception is when the icon of the launched application already occupies the first place, in this case the icon arrangement doesn't change.
Anya has planned the order in which she will launch applications. How many gestures should Anya make to launch the applications in the planned order?
Note that one application may be launched multiple times.
|
The first line of the input contains three numbers *n*,<=*m*,<=*k* (1<=≤<=*n*,<=*m*,<=*k*<=≤<=105) — the number of applications that Anya has on her smartphone, the number of applications that will be launched and the number of icons that are located on the same screen.
The next line contains *n* integers, permutation *a*1,<=*a*2,<=...,<=*a**n* — the initial order of icons from left to right in the menu (from the first to the last one), *a**i* — is the id of the application, whose icon goes *i*-th in the menu. Each integer from 1 to *n* occurs exactly once among *a**i*.
The third line contains *m* integers *b*1,<=*b*2,<=...,<=*b**m*(1<=≤<=*b**i*<=≤<=*n*) — the ids of the launched applications in the planned order. One application may be launched multiple times.
|
Print a single number — the number of gestures that Anya needs to make to launch all the applications in the desired order.
|
[
"8 3 3\n1 2 3 4 5 6 7 8\n7 8 1\n",
"5 4 2\n3 1 5 2 4\n4 4 4 4\n"
] |
[
"7\n",
"8\n"
] |
In the first test the initial configuration looks like (123)(456)(78), that is, the first screen contains icons of applications 1, 2, 3, the second screen contains icons 4, 5, 6, the third screen contains icons 7, 8.
After application 7 is launched, we get the new arrangement of the icons — (123)(457)(68). To launch it Anya makes 3 gestures.
After application 8 is launched, we get configuration (123)(457)(86). To launch it Anya makes 3 gestures.
After application 1 is launched, the arrangement of icons in the menu doesn't change. To launch it Anya makes 1 gesture.
In total, Anya makes 7 gestures.
| 1,500
|
[
{
"input": "8 3 3\n1 2 3 4 5 6 7 8\n7 8 1",
"output": "7"
},
{
"input": "5 4 2\n3 1 5 2 4\n4 4 4 4",
"output": "8"
},
{
"input": "10 10 3\n1 2 3 4 5 6 7 8 9 10\n2 3 4 5 6 7 8 9 10 1",
"output": "25"
},
{
"input": "10 12 3\n6 1 2 9 3 10 8 5 7 4\n3 9 9 4 8 2 3 8 10 8 3 4",
"output": "29"
},
{
"input": "16 12 8\n6 11 12 5 1 10 16 3 13 15 9 4 8 2 14 7\n6 13 8 12 8 15 12 16 8 11 3 16",
"output": "18"
},
{
"input": "4 3 5\n3 1 4 2\n4 2 4",
"output": "3"
},
{
"input": "4 2 15\n2 4 3 1\n1 2",
"output": "2"
},
{
"input": "71 96 75634\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\n27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26",
"output": "96"
},
{
"input": "71 81 2149\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 29 28 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\n29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29",
"output": "81"
}
] | 1,583,939,744
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 43
| 327
| 15,769,600
|
from sys import stdin
from collections import *
def arr_inp(n):
if n == 1:
return [int(x) for x in stdin.readline().split()]
elif n == 2:
return [float(x) for x in stdin.readline().split()]
else:
return list(stdin.readline()[:-1])
n, m, k = arr_inp(1)
a, b, ans = arr_inp(1), arr_inp(1), 0
dic = defaultdict(int, {a[i]: i for i in range(n)})
for i in range(m):
ans += (dic[b[i]] // k) + 1
if dic[b[i]]:
a[dic[b[i]]], a[dic[b[i]] - 1] = a[dic[b[i]] - 1], a[dic[b[i]]]
dic[a[dic[b[i]]]] += 1
dic[b[i]] -= 1
# print(ans)
print(ans)
|
Title: Anya and Smartphone
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Anya has bought a new smartphone that uses Berdroid operating system. The smartphone menu has exactly *n* applications, each application has its own icon. The icons are located on different screens, one screen contains *k* icons. The icons from the first to the *k*-th one are located on the first screen, from the (*k*<=+<=1)-th to the 2*k*-th ones are on the second screen and so on (the last screen may be partially empty).
Initially the smartphone menu is showing the screen number 1. To launch the application with the icon located on the screen *t*, Anya needs to make the following gestures: first she scrolls to the required screen number *t*, by making *t*<=-<=1 gestures (if the icon is on the screen *t*), and then make another gesture — press the icon of the required application exactly once to launch it.
After the application is launched, the menu returns to the first screen. That is, to launch the next application you need to scroll through the menu again starting from the screen number 1.
All applications are numbered from 1 to *n*. We know a certain order in which the icons of the applications are located in the menu at the beginning, but it changes as long as you use the operating system. Berdroid is intelligent system, so it changes the order of the icons by moving the more frequently used icons to the beginning of the list. Formally, right after an application is launched, Berdroid swaps the application icon and the icon of a preceding application (that is, the icon of an application on the position that is smaller by one in the order of menu). The preceding icon may possibly be located on the adjacent screen. The only exception is when the icon of the launched application already occupies the first place, in this case the icon arrangement doesn't change.
Anya has planned the order in which she will launch applications. How many gestures should Anya make to launch the applications in the planned order?
Note that one application may be launched multiple times.
Input Specification:
The first line of the input contains three numbers *n*,<=*m*,<=*k* (1<=≤<=*n*,<=*m*,<=*k*<=≤<=105) — the number of applications that Anya has on her smartphone, the number of applications that will be launched and the number of icons that are located on the same screen.
The next line contains *n* integers, permutation *a*1,<=*a*2,<=...,<=*a**n* — the initial order of icons from left to right in the menu (from the first to the last one), *a**i* — is the id of the application, whose icon goes *i*-th in the menu. Each integer from 1 to *n* occurs exactly once among *a**i*.
The third line contains *m* integers *b*1,<=*b*2,<=...,<=*b**m*(1<=≤<=*b**i*<=≤<=*n*) — the ids of the launched applications in the planned order. One application may be launched multiple times.
Output Specification:
Print a single number — the number of gestures that Anya needs to make to launch all the applications in the desired order.
Demo Input:
['8 3 3\n1 2 3 4 5 6 7 8\n7 8 1\n', '5 4 2\n3 1 5 2 4\n4 4 4 4\n']
Demo Output:
['7\n', '8\n']
Note:
In the first test the initial configuration looks like (123)(456)(78), that is, the first screen contains icons of applications 1, 2, 3, the second screen contains icons 4, 5, 6, the third screen contains icons 7, 8.
After application 7 is launched, we get the new arrangement of the icons — (123)(457)(68). To launch it Anya makes 3 gestures.
After application 8 is launched, we get configuration (123)(457)(86). To launch it Anya makes 3 gestures.
After application 1 is launched, the arrangement of icons in the menu doesn't change. To launch it Anya makes 1 gesture.
In total, Anya makes 7 gestures.
|
```python
from sys import stdin
from collections import *
def arr_inp(n):
if n == 1:
return [int(x) for x in stdin.readline().split()]
elif n == 2:
return [float(x) for x in stdin.readline().split()]
else:
return list(stdin.readline()[:-1])
n, m, k = arr_inp(1)
a, b, ans = arr_inp(1), arr_inp(1), 0
dic = defaultdict(int, {a[i]: i for i in range(n)})
for i in range(m):
ans += (dic[b[i]] // k) + 1
if dic[b[i]]:
a[dic[b[i]]], a[dic[b[i]] - 1] = a[dic[b[i]] - 1], a[dic[b[i]]]
dic[a[dic[b[i]]]] += 1
dic[b[i]] -= 1
# print(ans)
print(ans)
```
| 3
|
|
9
|
A
|
Die Roll
|
PROGRAMMING
| 800
|
[
"math",
"probabilities"
] |
A. Die Roll
|
1
|
64
|
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
|
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
|
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
|
[
"4 2\n"
] |
[
"1/2\n"
] |
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points.
| 0
|
[
{
"input": "4 2",
"output": "1/2"
},
{
"input": "1 1",
"output": "1/1"
},
{
"input": "1 2",
"output": "5/6"
},
{
"input": "1 3",
"output": "2/3"
},
{
"input": "1 4",
"output": "1/2"
},
{
"input": "1 5",
"output": "1/3"
},
{
"input": "1 6",
"output": "1/6"
},
{
"input": "2 1",
"output": "5/6"
},
{
"input": "2 2",
"output": "5/6"
},
{
"input": "2 3",
"output": "2/3"
},
{
"input": "2 4",
"output": "1/2"
},
{
"input": "2 5",
"output": "1/3"
},
{
"input": "2 6",
"output": "1/6"
},
{
"input": "3 1",
"output": "2/3"
},
{
"input": "3 2",
"output": "2/3"
},
{
"input": "3 3",
"output": "2/3"
},
{
"input": "3 4",
"output": "1/2"
},
{
"input": "3 5",
"output": "1/3"
},
{
"input": "3 6",
"output": "1/6"
},
{
"input": "4 1",
"output": "1/2"
},
{
"input": "4 3",
"output": "1/2"
},
{
"input": "4 4",
"output": "1/2"
},
{
"input": "4 5",
"output": "1/3"
},
{
"input": "4 6",
"output": "1/6"
},
{
"input": "5 1",
"output": "1/3"
},
{
"input": "5 2",
"output": "1/3"
},
{
"input": "5 3",
"output": "1/3"
},
{
"input": "5 4",
"output": "1/3"
},
{
"input": "5 5",
"output": "1/3"
},
{
"input": "5 6",
"output": "1/6"
},
{
"input": "6 1",
"output": "1/6"
},
{
"input": "6 2",
"output": "1/6"
},
{
"input": "6 3",
"output": "1/6"
},
{
"input": "6 4",
"output": "1/6"
},
{
"input": "6 5",
"output": "1/6"
},
{
"input": "6 6",
"output": "1/6"
}
] | 1,642,323,039
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 36
| 92
| 0
|
a, b = input().split()
a, b = int(a), int(b)
d = 6
if a > b:
c = 6 - a + 1
else:
c = 6 - b + 1
for i in range(2, 4):
if d % i == 0 and c % i == 0:
d = int(d / i)
c = int(c / i)
print(f"{c}/{d}")
|
Title: Die Roll
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
Input Specification:
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
Output Specification:
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
Demo Input:
['4 2\n']
Demo Output:
['1/2\n']
Note:
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points.
|
```python
a, b = input().split()
a, b = int(a), int(b)
d = 6
if a > b:
c = 6 - a + 1
else:
c = 6 - b + 1
for i in range(2, 4):
if d % i == 0 and c % i == 0:
d = int(d / i)
c = int(c / i)
print(f"{c}/{d}")
```
| 3.954
|
9
|
A
|
Die Roll
|
PROGRAMMING
| 800
|
[
"math",
"probabilities"
] |
A. Die Roll
|
1
|
64
|
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
|
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
|
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
|
[
"4 2\n"
] |
[
"1/2\n"
] |
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points.
| 0
|
[
{
"input": "4 2",
"output": "1/2"
},
{
"input": "1 1",
"output": "1/1"
},
{
"input": "1 2",
"output": "5/6"
},
{
"input": "1 3",
"output": "2/3"
},
{
"input": "1 4",
"output": "1/2"
},
{
"input": "1 5",
"output": "1/3"
},
{
"input": "1 6",
"output": "1/6"
},
{
"input": "2 1",
"output": "5/6"
},
{
"input": "2 2",
"output": "5/6"
},
{
"input": "2 3",
"output": "2/3"
},
{
"input": "2 4",
"output": "1/2"
},
{
"input": "2 5",
"output": "1/3"
},
{
"input": "2 6",
"output": "1/6"
},
{
"input": "3 1",
"output": "2/3"
},
{
"input": "3 2",
"output": "2/3"
},
{
"input": "3 3",
"output": "2/3"
},
{
"input": "3 4",
"output": "1/2"
},
{
"input": "3 5",
"output": "1/3"
},
{
"input": "3 6",
"output": "1/6"
},
{
"input": "4 1",
"output": "1/2"
},
{
"input": "4 3",
"output": "1/2"
},
{
"input": "4 4",
"output": "1/2"
},
{
"input": "4 5",
"output": "1/3"
},
{
"input": "4 6",
"output": "1/6"
},
{
"input": "5 1",
"output": "1/3"
},
{
"input": "5 2",
"output": "1/3"
},
{
"input": "5 3",
"output": "1/3"
},
{
"input": "5 4",
"output": "1/3"
},
{
"input": "5 5",
"output": "1/3"
},
{
"input": "5 6",
"output": "1/6"
},
{
"input": "6 1",
"output": "1/6"
},
{
"input": "6 2",
"output": "1/6"
},
{
"input": "6 3",
"output": "1/6"
},
{
"input": "6 4",
"output": "1/6"
},
{
"input": "6 5",
"output": "1/6"
},
{
"input": "6 6",
"output": "1/6"
}
] | 1,659,544,815
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 62
| 0
|
y,w = map(int,input().split())
if y > 5 or w > 5:
print('0/1')
elif y == w:
print('1/1')
else:
print('1/2')
|
Title: Die Roll
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
Input Specification:
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
Output Specification:
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
Demo Input:
['4 2\n']
Demo Output:
['1/2\n']
Note:
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points.
|
```python
y,w = map(int,input().split())
if y > 5 or w > 5:
print('0/1')
elif y == w:
print('1/1')
else:
print('1/2')
```
| 0
|
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<=≤<=*n*<=≤<=105) — the number of sock pairs.
The second line contains 2*n* integers *x*1,<=*x*2,<=...,<=*x*2*n* (1<=≤<=*x**i*<=≤<=*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 integer — 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, 2) 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, 3) 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"
}
] | 1,488,720,382
| 1,282
|
Python 3
|
OK
|
TESTS
| 56
| 187
| 18,227,200
|
#yl1
n=int(input())
li=list(map(int,input().split()))
temp=set()
mat=0
for i in li:
if i not in temp:
temp.add(i)
if len(temp)>mat:mat=len(temp)
else:temp.remove(i)
print(mat)
|
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<=≤<=*n*<=≤<=105) — the number of sock pairs.
The second line contains 2*n* integers *x*1,<=*x*2,<=...,<=*x*2*n* (1<=≤<=*x**i*<=≤<=*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 integer — 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, 2) 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, 3) 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
#yl1
n=int(input())
li=list(map(int,input().split()))
temp=set()
mat=0
for i in li:
if i not in temp:
temp.add(i)
if len(temp)>mat:mat=len(temp)
else:temp.remove(i)
print(mat)
```
| 3
|
|
271
|
A
|
Beautiful Year
|
PROGRAMMING
| 800
|
[
"brute force"
] | null | null |
It seems like the year of 2013 came only yesterday. Do you know a curious fact? The year of 2013 is the first year after the old 1987 with only distinct digits.
Now you are suggested to solve the following problem: given a year number, find the minimum year number which is strictly larger than the given one and has only distinct digits.
|
The single line contains integer *y* (1000<=≤<=*y*<=≤<=9000) — the year number.
|
Print a single integer — the minimum year number that is strictly larger than *y* and all it's digits are distinct. It is guaranteed that the answer exists.
|
[
"1987\n",
"2013\n"
] |
[
"2013\n",
"2014\n"
] |
none
| 500
|
[
{
"input": "1987",
"output": "2013"
},
{
"input": "2013",
"output": "2014"
},
{
"input": "1000",
"output": "1023"
},
{
"input": "1001",
"output": "1023"
},
{
"input": "1234",
"output": "1235"
},
{
"input": "5555",
"output": "5601"
},
{
"input": "9000",
"output": "9012"
},
{
"input": "1111",
"output": "1203"
},
{
"input": "8999",
"output": "9012"
},
{
"input": "4572",
"output": "4573"
},
{
"input": "6666",
"output": "6701"
},
{
"input": "2001",
"output": "2013"
},
{
"input": "3000",
"output": "3012"
},
{
"input": "7712",
"output": "7801"
},
{
"input": "8088",
"output": "8091"
},
{
"input": "1594",
"output": "1596"
},
{
"input": "6016",
"output": "6017"
},
{
"input": "8800",
"output": "8901"
},
{
"input": "3331",
"output": "3401"
},
{
"input": "5090",
"output": "5091"
},
{
"input": "2342",
"output": "2345"
},
{
"input": "2334",
"output": "2340"
},
{
"input": "1123",
"output": "1203"
},
{
"input": "8989",
"output": "9012"
},
{
"input": "8977",
"output": "9012"
},
{
"input": "6869",
"output": "6870"
},
{
"input": "8999",
"output": "9012"
}
] | 1,694,519,735
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 27
| 92
| 0
|
year = input()
def judge(y):
ans = False
y = set(str(y))
if len(y) == 4:
ans = True
return ans
while True:
year = int(year)
year = year + 1
if judge(year) == True:
break
print(year)
|
Title: Beautiful Year
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It seems like the year of 2013 came only yesterday. Do you know a curious fact? The year of 2013 is the first year after the old 1987 with only distinct digits.
Now you are suggested to solve the following problem: given a year number, find the minimum year number which is strictly larger than the given one and has only distinct digits.
Input Specification:
The single line contains integer *y* (1000<=≤<=*y*<=≤<=9000) — the year number.
Output Specification:
Print a single integer — the minimum year number that is strictly larger than *y* and all it's digits are distinct. It is guaranteed that the answer exists.
Demo Input:
['1987\n', '2013\n']
Demo Output:
['2013\n', '2014\n']
Note:
none
|
```python
year = input()
def judge(y):
ans = False
y = set(str(y))
if len(y) == 4:
ans = True
return ans
while True:
year = int(year)
year = year + 1
if judge(year) == True:
break
print(year)
```
| 3
|
|
20
|
B
|
Equation
|
PROGRAMMING
| 2,000
|
[
"math"
] |
B. Equation
|
1
|
256
|
You are given an equation:
Your task is to find the number of distinct roots of the equation and print all of them in ascending order.
|
The first line contains three integer numbers *A*,<=*B* and *C* (<=-<=105<=≤<=*A*,<=*B*,<=*C*<=≤<=105). Any coefficient may be equal to 0.
|
In case of infinite root count print the only integer -1. In case of no roots print the only integer 0. In other cases print the number of root on the first line and the roots on the following lines in the ascending order. Print roots with at least 5 digits after the decimal point.
|
[
"1 -5 6\n"
] |
[
"2\n2.0000000000\n3.0000000000"
] |
none
| 1,000
|
[
{
"input": "1 -5 6",
"output": "2\n2.0000000000\n3.0000000000"
},
{
"input": "1 1 1",
"output": "0"
},
{
"input": "1 2 1",
"output": "1\n-1.0000000000"
},
{
"input": "0 0 0",
"output": "-1"
},
{
"input": "0 -2 1",
"output": "1\n0.5000000000"
},
{
"input": "0 -2 0",
"output": "1\n0.0000000000"
},
{
"input": "0 0 1",
"output": "0"
},
{
"input": "0 0 -100000",
"output": "0"
},
{
"input": "0 10000 -100000",
"output": "1\n10.0000000000"
},
{
"input": "1 100000 -100000",
"output": "2\n-100000.9999900002\n0.9999900002"
},
{
"input": "0 3431 43123",
"output": "1\n-12.5686388808"
},
{
"input": "100 200 100",
"output": "1\n-1.0000000000"
},
{
"input": "50000 100000 50000",
"output": "1\n-1.0000000000"
},
{
"input": "-1 10 20",
"output": "2\n-1.7082039325\n11.7082039325"
},
{
"input": "-50000 100000 -50000",
"output": "1\n1.0000000000"
},
{
"input": "1 -2 1",
"output": "1\n1.0000000000"
},
{
"input": "1000 -5000 6000",
"output": "2\n2.0000000000\n3.0000000000"
},
{
"input": "0 -100000 0",
"output": "1\n0.0000000000"
},
{
"input": "1 -100000 0",
"output": "2\n0.0000000000\n100000.0000000000"
},
{
"input": "1223 -23532 1232",
"output": "2\n0.0524974745\n19.1887126645"
},
{
"input": "-1 -2 -1",
"output": "1\n-1.0000000000"
},
{
"input": "1 0 0",
"output": "1\n0.0000000000"
},
{
"input": "0 1 0",
"output": "1\n-0.0000000000"
},
{
"input": "0 0 1",
"output": "0"
},
{
"input": "0 1 -1",
"output": "1\n1.0000000000"
},
{
"input": "5 0 5",
"output": "0"
},
{
"input": "-2 -5 0",
"output": "2\n-2.5000000000\n-0.0000000000"
},
{
"input": "-2 -4 0",
"output": "2\n-2.0000000000\n-0.0000000000"
},
{
"input": "-2 0 0",
"output": "1\n-0.0000000000"
},
{
"input": "0 -4 -4",
"output": "1\n-1.0000000000"
},
{
"input": "1 1 0",
"output": "2\n-1.0000000000\n0.0000000000"
},
{
"input": "1 0 1",
"output": "0"
},
{
"input": "1 1 1",
"output": "0"
},
{
"input": "0 0 0",
"output": "-1"
}
] | 1,580,888,512
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 34
| 140
| 0
|
import math
a, b, c = map(int, input().split())
d = 0
if a == 0 and b == 0 and c == 0:
print(-1)
exit()
if a == 0 and b == 0:
print(0)
exit()
if a == 0:
d -= c
d /= b
print(1)
if d == -0:
d = 0
print(d)
exit()
d = b * b - (4 * a * c)
if d < 0:
print(0)
exit()
if d == 0:
print(1)
d -= b
d /= 2 * a
print(d)
exit()
if d > 0:
print(2)
q = -b + math.sqrt(d)
q /= 2 * a
w = -b - math.sqrt(d)
w /= 2 * a
if q > w:
q, w = w, q
print(q)
print(w)
exit()
|
Title: Equation
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given an equation:
Your task is to find the number of distinct roots of the equation and print all of them in ascending order.
Input Specification:
The first line contains three integer numbers *A*,<=*B* and *C* (<=-<=105<=≤<=*A*,<=*B*,<=*C*<=≤<=105). Any coefficient may be equal to 0.
Output Specification:
In case of infinite root count print the only integer -1. In case of no roots print the only integer 0. In other cases print the number of root on the first line and the roots on the following lines in the ascending order. Print roots with at least 5 digits after the decimal point.
Demo Input:
['1 -5 6\n']
Demo Output:
['2\n2.0000000000\n3.0000000000']
Note:
none
|
```python
import math
a, b, c = map(int, input().split())
d = 0
if a == 0 and b == 0 and c == 0:
print(-1)
exit()
if a == 0 and b == 0:
print(0)
exit()
if a == 0:
d -= c
d /= b
print(1)
if d == -0:
d = 0
print(d)
exit()
d = b * b - (4 * a * c)
if d < 0:
print(0)
exit()
if d == 0:
print(1)
d -= b
d /= 2 * a
print(d)
exit()
if d > 0:
print(2)
q = -b + math.sqrt(d)
q /= 2 * a
w = -b - math.sqrt(d)
w /= 2 * a
if q > w:
q, w = w, q
print(q)
print(w)
exit()
```
| 3.93
|
48
|
A
|
Rock-paper-scissors
|
PROGRAMMING
| 900
|
[
"implementation",
"schedules"
] |
A. Rock-paper-scissors
|
2
|
256
|
Uncle Fyodor, Matroskin the Cat and Sharic the Dog live their simple but happy lives in Prostokvashino. Sometimes they receive parcels from Uncle Fyodor’s parents and sometimes from anonymous benefactors, in which case it is hard to determine to which one of them the package has been sent. A photographic rifle is obviously for Sharic who loves hunting and fish is for Matroskin, but for whom was a new video game console meant? Every one of the three friends claimed that the present is for him and nearly quarreled. Uncle Fyodor had an idea how to solve the problem justly: they should suppose that the console was sent to all three of them and play it in turns. Everybody got relieved but then yet another burning problem popped up — who will play first? This time Matroskin came up with a brilliant solution, suggesting the most fair way to find it out: play rock-paper-scissors together. The rules of the game are very simple. On the count of three every player shows a combination with his hand (or paw). The combination corresponds to one of three things: a rock, scissors or paper. Some of the gestures win over some other ones according to well-known rules: the rock breaks the scissors, the scissors cut the paper, and the paper gets wrapped over the stone. Usually there are two players. Yet there are three friends, that’s why they decided to choose the winner like that: If someone shows the gesture that wins over the other two players, then that player wins. Otherwise, another game round is required. Write a program that will determine the winner by the gestures they have shown.
|
The first input line contains the name of the gesture that Uncle Fyodor showed, the second line shows which gesture Matroskin showed and the third line shows Sharic’s gesture.
|
Print "F" (without quotes) if Uncle Fyodor wins. Print "M" if Matroskin wins and "S" if Sharic wins. If it is impossible to find the winner, print "?".
|
[
"rock\nrock\nrock\n",
"paper\nrock\nrock\n",
"scissors\nrock\nrock\n",
"scissors\npaper\nrock\n"
] |
[
"?\n",
"F\n",
"?\n",
"?\n"
] |
none
| 0
|
[
{
"input": "rock\nrock\nrock",
"output": "?"
},
{
"input": "paper\nrock\nrock",
"output": "F"
},
{
"input": "scissors\nrock\nrock",
"output": "?"
},
{
"input": "scissors\npaper\nrock",
"output": "?"
},
{
"input": "paper\npaper\nrock",
"output": "?"
},
{
"input": "rock\npaper\nrock",
"output": "M"
},
{
"input": "rock\nscissors\nrock",
"output": "?"
},
{
"input": "paper\nscissors\nrock",
"output": "?"
},
{
"input": "scissors\nscissors\nrock",
"output": "S"
},
{
"input": "rock\nrock\npaper",
"output": "S"
},
{
"input": "paper\nrock\npaper",
"output": "?"
},
{
"input": "scissors\nrock\npaper",
"output": "?"
},
{
"input": "rock\npaper\npaper",
"output": "?"
},
{
"input": "paper\npaper\npaper",
"output": "?"
},
{
"input": "scissors\npaper\npaper",
"output": "F"
},
{
"input": "rock\nscissors\npaper",
"output": "?"
},
{
"input": "paper\nscissors\npaper",
"output": "M"
},
{
"input": "scissors\nscissors\npaper",
"output": "?"
},
{
"input": "rock\nrock\nscissors",
"output": "?"
},
{
"input": "paper\nrock\nscissors",
"output": "?"
},
{
"input": "scissors\nrock\nscissors",
"output": "M"
},
{
"input": "rock\npaper\nscissors",
"output": "?"
},
{
"input": "paper\npaper\nscissors",
"output": "S"
},
{
"input": "scissors\npaper\nscissors",
"output": "?"
},
{
"input": "rock\nscissors\nscissors",
"output": "F"
},
{
"input": "paper\nscissors\nscissors",
"output": "?"
},
{
"input": "scissors\nscissors\nscissors",
"output": "?"
}
] | 1,555,753,313
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 27
| 218
| 0
|
def main():
a = [input(), input(), input()]
winner = -1
if a.count('rock') == 1 and a.count('scissors') == 2:
winner = a.index('rock')
elif a.count('scissors') == 1 and a.count('paper') == 2:
winner = a.index('scissors')
elif a.count('paper') == 1 and a.count('rock') == 2:
winner = a.index('paper')
print('?' if winner == -1 else
'F' if winner == 0 else
'M' if winner == 1 else
'S')
if __name__ == '__main__':
main()
|
Title: Rock-paper-scissors
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Uncle Fyodor, Matroskin the Cat and Sharic the Dog live their simple but happy lives in Prostokvashino. Sometimes they receive parcels from Uncle Fyodor’s parents and sometimes from anonymous benefactors, in which case it is hard to determine to which one of them the package has been sent. A photographic rifle is obviously for Sharic who loves hunting and fish is for Matroskin, but for whom was a new video game console meant? Every one of the three friends claimed that the present is for him and nearly quarreled. Uncle Fyodor had an idea how to solve the problem justly: they should suppose that the console was sent to all three of them and play it in turns. Everybody got relieved but then yet another burning problem popped up — who will play first? This time Matroskin came up with a brilliant solution, suggesting the most fair way to find it out: play rock-paper-scissors together. The rules of the game are very simple. On the count of three every player shows a combination with his hand (or paw). The combination corresponds to one of three things: a rock, scissors or paper. Some of the gestures win over some other ones according to well-known rules: the rock breaks the scissors, the scissors cut the paper, and the paper gets wrapped over the stone. Usually there are two players. Yet there are three friends, that’s why they decided to choose the winner like that: If someone shows the gesture that wins over the other two players, then that player wins. Otherwise, another game round is required. Write a program that will determine the winner by the gestures they have shown.
Input Specification:
The first input line contains the name of the gesture that Uncle Fyodor showed, the second line shows which gesture Matroskin showed and the third line shows Sharic’s gesture.
Output Specification:
Print "F" (without quotes) if Uncle Fyodor wins. Print "M" if Matroskin wins and "S" if Sharic wins. If it is impossible to find the winner, print "?".
Demo Input:
['rock\nrock\nrock\n', 'paper\nrock\nrock\n', 'scissors\nrock\nrock\n', 'scissors\npaper\nrock\n']
Demo Output:
['?\n', 'F\n', '?\n', '?\n']
Note:
none
|
```python
def main():
a = [input(), input(), input()]
winner = -1
if a.count('rock') == 1 and a.count('scissors') == 2:
winner = a.index('rock')
elif a.count('scissors') == 1 and a.count('paper') == 2:
winner = a.index('scissors')
elif a.count('paper') == 1 and a.count('rock') == 2:
winner = a.index('paper')
print('?' if winner == -1 else
'F' if winner == 0 else
'M' if winner == 1 else
'S')
if __name__ == '__main__':
main()
```
| 3.9455
|
610
|
A
|
Pasha and Stick
|
PROGRAMMING
| 1,000
|
[
"combinatorics",
"math"
] | null | null |
Pasha has a wooden stick of some positive integer length *n*. He wants to perform exactly three cuts to get four parts of the stick. Each part must have some positive integer length and the sum of these lengths will obviously be *n*.
Pasha likes rectangles but hates squares, so he wonders, how many ways are there to split a stick into four parts so that it's possible to form a rectangle using these parts, but is impossible to form a square.
Your task is to help Pasha and count the number of such ways. Two ways to cut the stick are considered distinct if there exists some integer *x*, such that the number of parts of length *x* in the first way differ from the number of parts of length *x* in the second way.
|
The first line of the input contains a positive integer *n* (1<=≤<=*n*<=≤<=2·109) — the length of Pasha's stick.
|
The output should contain a single integer — the number of ways to split Pasha's stick into four parts of positive integer length so that it's possible to make a rectangle by connecting the ends of these parts, but is impossible to form a square.
|
[
"6\n",
"20\n"
] |
[
"1\n",
"4\n"
] |
There is only one way to divide the stick in the first sample {1, 1, 2, 2}.
Four ways to divide the stick in the second sample are {1, 1, 9, 9}, {2, 2, 8, 8}, {3, 3, 7, 7} and {4, 4, 6, 6}. Note that {5, 5, 5, 5} doesn't work.
| 500
|
[
{
"input": "6",
"output": "1"
},
{
"input": "20",
"output": "4"
},
{
"input": "1",
"output": "0"
},
{
"input": "2",
"output": "0"
},
{
"input": "3",
"output": "0"
},
{
"input": "4",
"output": "0"
},
{
"input": "2000000000",
"output": "499999999"
},
{
"input": "1924704072",
"output": "481176017"
},
{
"input": "73740586",
"output": "18435146"
},
{
"input": "1925088820",
"output": "481272204"
},
{
"input": "593070992",
"output": "148267747"
},
{
"input": "1925473570",
"output": "481368392"
},
{
"input": "629490186",
"output": "157372546"
},
{
"input": "1980649112",
"output": "495162277"
},
{
"input": "36661322",
"output": "9165330"
},
{
"input": "1943590793",
"output": "0"
},
{
"input": "71207034",
"output": "17801758"
},
{
"input": "1757577394",
"output": "439394348"
},
{
"input": "168305294",
"output": "42076323"
},
{
"input": "1934896224",
"output": "483724055"
},
{
"input": "297149088",
"output": "74287271"
},
{
"input": "1898001634",
"output": "474500408"
},
{
"input": "176409698",
"output": "44102424"
},
{
"input": "1873025522",
"output": "468256380"
},
{
"input": "5714762",
"output": "1428690"
},
{
"input": "1829551192",
"output": "457387797"
},
{
"input": "16269438",
"output": "4067359"
},
{
"input": "1663283390",
"output": "415820847"
},
{
"input": "42549941",
"output": "0"
},
{
"input": "1967345604",
"output": "491836400"
},
{
"input": "854000",
"output": "213499"
},
{
"input": "1995886626",
"output": "498971656"
},
{
"input": "10330019",
"output": "0"
},
{
"input": "1996193634",
"output": "499048408"
},
{
"input": "9605180",
"output": "2401294"
},
{
"input": "1996459740",
"output": "499114934"
},
{
"input": "32691948",
"output": "8172986"
},
{
"input": "1975903308",
"output": "493975826"
},
{
"input": "1976637136",
"output": "494159283"
},
{
"input": "29803038",
"output": "7450759"
},
{
"input": "1977979692",
"output": "494494922"
},
{
"input": "1978595336",
"output": "494648833"
},
{
"input": "27379344",
"output": "6844835"
},
{
"input": "1979729912",
"output": "494932477"
},
{
"input": "1980253780",
"output": "495063444"
},
{
"input": "1980751584",
"output": "495187895"
},
{
"input": "53224878",
"output": "13306219"
},
{
"input": "5",
"output": "0"
},
{
"input": "7",
"output": "0"
},
{
"input": "8",
"output": "1"
},
{
"input": "9",
"output": "0"
},
{
"input": "10",
"output": "2"
},
{
"input": "11",
"output": "0"
},
{
"input": "12",
"output": "2"
},
{
"input": "13",
"output": "0"
},
{
"input": "14",
"output": "3"
},
{
"input": "15",
"output": "0"
},
{
"input": "16",
"output": "3"
},
{
"input": "17",
"output": "0"
},
{
"input": "18",
"output": "4"
},
{
"input": "19",
"output": "0"
},
{
"input": "21",
"output": "0"
},
{
"input": "22",
"output": "5"
},
{
"input": "23",
"output": "0"
},
{
"input": "24",
"output": "5"
},
{
"input": "25",
"output": "0"
},
{
"input": "26",
"output": "6"
},
{
"input": "27",
"output": "0"
},
{
"input": "28",
"output": "6"
},
{
"input": "29",
"output": "0"
},
{
"input": "30",
"output": "7"
},
{
"input": "111",
"output": "0"
},
{
"input": "55",
"output": "0"
},
{
"input": "105",
"output": "0"
},
{
"input": "199",
"output": "0"
},
{
"input": "151",
"output": "0"
}
] | 1,699,030,750
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 15
| 61
| 0
|
import math
n = int(input())
ans = math.ceil(n / 4)
print(ans - 1)
|
Title: Pasha and Stick
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pasha has a wooden stick of some positive integer length *n*. He wants to perform exactly three cuts to get four parts of the stick. Each part must have some positive integer length and the sum of these lengths will obviously be *n*.
Pasha likes rectangles but hates squares, so he wonders, how many ways are there to split a stick into four parts so that it's possible to form a rectangle using these parts, but is impossible to form a square.
Your task is to help Pasha and count the number of such ways. Two ways to cut the stick are considered distinct if there exists some integer *x*, such that the number of parts of length *x* in the first way differ from the number of parts of length *x* in the second way.
Input Specification:
The first line of the input contains a positive integer *n* (1<=≤<=*n*<=≤<=2·109) — the length of Pasha's stick.
Output Specification:
The output should contain a single integer — the number of ways to split Pasha's stick into four parts of positive integer length so that it's possible to make a rectangle by connecting the ends of these parts, but is impossible to form a square.
Demo Input:
['6\n', '20\n']
Demo Output:
['1\n', '4\n']
Note:
There is only one way to divide the stick in the first sample {1, 1, 2, 2}.
Four ways to divide the stick in the second sample are {1, 1, 9, 9}, {2, 2, 8, 8}, {3, 3, 7, 7} and {4, 4, 6, 6}. Note that {5, 5, 5, 5} doesn't work.
|
```python
import math
n = int(input())
ans = math.ceil(n / 4)
print(ans - 1)
```
| 0
|
|
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 — 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 — 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 1 × 1 × 1, in the second one — 2 × 2 × 3.
| 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"
}
] | 1,672,180,180
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
areas = [int(i) for i in input().split()]
vals = []
for i in range(1, min(areas) + 1):
if areas[0]%i == 0 and areas[1]%i == 0 and (areas[0]/i)*(areas[1]/i) == areas[2]:
vals = [i, areas[0]/i, areas[1]/i]
break
if areas[1]%i == 0 and areas[2]%i == 0 and (areas[1]/i)*(areas[2]/i) == areas[0]:
vals = [i, areas[1]/i, areas[2]/i]
break
if areas[0]%i == 0 and areas[2]%i == 0 and (areas[0]/i)*(areas[2]/i) == areas[1]:
vals = [i, areas[0]/i, areas[2]/i]
print(4*sum(vals))
|
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 — 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 — 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 1 × 1 × 1, in the second one — 2 × 2 × 3.
|
```python
areas = [int(i) for i in input().split()]
vals = []
for i in range(1, min(areas) + 1):
if areas[0]%i == 0 and areas[1]%i == 0 and (areas[0]/i)*(areas[1]/i) == areas[2]:
vals = [i, areas[0]/i, areas[1]/i]
break
if areas[1]%i == 0 and areas[2]%i == 0 and (areas[1]/i)*(areas[2]/i) == areas[0]:
vals = [i, areas[1]/i, areas[2]/i]
break
if areas[0]%i == 0 and areas[2]%i == 0 and (areas[0]/i)*(areas[2]/i) == areas[1]:
vals = [i, areas[0]/i, areas[2]/i]
print(4*sum(vals))
```
| 0
|
|
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" — 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 "
}
] | 1,684,068,860
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 92
| 0
|
a=input()
b=a.replace('WUB', '')
print(b)
|
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" — between them Vasya added two "WUB".
|
```python
a=input()
b=a.replace('WUB', '')
print(b)
```
| 0
|
|
844
|
B
|
Rectangles
|
PROGRAMMING
| 1,300
|
[
"combinatorics",
"math"
] | null | null |
You are given *n*<=×<=*m* table. Each cell of the table is colored white or black. Find the number of non-empty sets of cells such that:
1. All cells in a set have the same color. 1. Every two cells in a set share row or column.
|
The first line of input contains integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=50) — the number of rows and the number of columns correspondingly.
The next *n* lines of input contain descriptions of rows. There are *m* integers, separated by spaces, in each line. The number equals 0 if the corresponding cell is colored white and equals 1 if the corresponding cell is colored black.
|
Output single integer — the number of non-empty sets from the problem description.
|
[
"1 1\n0\n",
"2 3\n1 0 1\n0 1 0\n"
] |
[
"1\n",
"8\n"
] |
In the second example, there are six one-element sets. Additionally, there are two two-element sets, the first one consists of the first and the third cells of the first row, the second one consists of the first and the third cells of the second row. To sum up, there are 8 sets.
| 1,000
|
[
{
"input": "1 1\n0",
"output": "1"
},
{
"input": "2 3\n1 0 1\n0 1 0",
"output": "8"
},
{
"input": "2 2\n1 1\n1 1",
"output": "8"
},
{
"input": "1 10\n0 0 0 0 0 0 0 0 0 0",
"output": "1023"
},
{
"input": "11 1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1",
"output": "2047"
},
{
"input": "10 11\n1 1 0 1 1 0 0 0 1 0 0\n1 0 0 1 1 1 0 0 1 1 0\n0 0 1 0 1 1 0 1 0 1 1\n0 1 1 1 0 1 0 1 0 0 0\n1 1 1 1 1 1 1 0 1 0 0\n1 1 0 1 1 1 1 0 0 1 1\n1 0 1 0 1 0 0 1 1 1 0\n1 1 0 0 0 0 0 1 0 1 1\n1 1 0 1 1 1 0 0 1 1 0\n1 0 1 1 0 0 1 0 0 1 1",
"output": "2444"
},
{
"input": "50 1\n0\n1\n0\n1\n0\n1\n0\n1\n1\n1\n0\n0\n1\n0\n0\n1\n1\n1\n1\n0\n1\n1\n0\n1\n1\n1\n0\n1\n0\n0\n0\n1\n1\n0\n1\n1\n0\n1\n0\n1\n0\n0\n1\n0\n0\n0\n1\n1\n0\n1",
"output": "142606334"
},
{
"input": "1 50\n0 1 0 1 0 1 0 1 1 1 0 0 1 0 0 1 1 1 1 0 1 1 0 1 1 1 0 1 0 0 0 1 1 0 1 1 0 1 0 1 0 0 1 0 0 0 1 1 0 1",
"output": "142606334"
},
{
"input": "2 20\n0 1 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0",
"output": "589853"
},
{
"input": "5 5\n0 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": "285"
},
{
"input": "6 6\n1 1 1 1 1 1\n1 1 1 1 1 1\n1 1 1 1 1 1\n1 1 1 1 1 1\n1 1 1 1 1 1\n1 1 1 1 1 1",
"output": "720"
},
{
"input": "21 2\n0 1\n1 1\n0 1\n0 0\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1",
"output": "1310745"
},
{
"input": "3 15\n1 0 0 0 0 0 0 0 0 0 0 0 0 0 0\n0 0 0 0 0 1 0 1 0 0 0 0 0 1 0\n1 0 0 1 0 0 0 0 0 0 0 0 1 0 1",
"output": "22587"
},
{
"input": "10 11\n0 1 0 0 0 0 0 0 0 0 0\n0 1 0 1 0 0 1 0 0 0 0\n0 0 0 0 0 0 1 1 1 0 0\n0 0 0 0 0 0 0 0 0 0 0\n0 0 0 0 1 0 0 0 0 1 0\n0 0 0 0 0 0 1 0 0 0 0\n0 0 0 0 0 0 0 0 0 1 0\n0 0 1 0 0 0 1 1 0 0 0\n0 0 0 0 0 0 0 0 1 0 0\n0 0 1 0 1 0 0 0 0 1 1",
"output": "12047"
},
{
"input": "14 15\n0 1 0 0 0 0 0 0 1 0 0 0 1 0 1\n0 0 0 1 1 1 1 0 1 0 0 1 1 0 0\n1 0 0 0 0 1 1 0 0 0 0 0 0 0 0\n0 1 0 0 0 1 0 1 1 0 0 1 0 0 0\n0 0 1 1 0 1 0 1 0 1 1 0 1 0 0\n0 0 0 1 1 0 0 0 0 0 1 1 0 1 0\n0 0 1 0 0 0 0 0 0 1 0 0 1 1 0\n1 1 0 0 0 1 0 0 0 0 0 0 1 1 0\n0 0 0 0 1 0 1 1 1 0 0 0 1 0 1\n1 0 1 1 0 1 0 0 1 0 0 1 1 1 0\n1 0 0 0 0 1 0 0 0 0 0 1 0 0 0\n0 0 0 1 0 1 0 0 0 0 1 0 0 0 1\n0 0 1 0 1 0 0 0 1 1 1 1 0 0 0\n0 0 0 0 0 0 0 0 0 0 0 1 0 0 0",
"output": "53166"
},
{
"input": "1 50\n0 0 0 0 0 0 0 1 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 1 0 0 0 1 0 0",
"output": "1099511628798"
},
{
"input": "50 1\n0\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n0\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n0\n1\n1\n1\n1\n1\n1\n1\n0\n1\n1\n1\n1\n0\n1\n1\n1\n1\n1",
"output": "35184372088862"
},
{
"input": "1 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",
"output": "1125899906842623"
},
{
"input": "5 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\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\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\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\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",
"output": "5629499534214415"
},
{
"input": "32 2\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1",
"output": "8589934622"
},
{
"input": "1 50\n0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "562949953421312"
},
{
"input": "50 1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1",
"output": "1125899906842623"
},
{
"input": "1 49\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",
"output": "562949953421311"
},
{
"input": "2 50\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\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",
"output": "2251799813685296"
},
{
"input": "3 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\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\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",
"output": "3377699720528069"
},
{
"input": "1 50\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",
"output": "1125899906842623"
},
{
"input": "1 40\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",
"output": "1099511627775"
},
{
"input": "1 33\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",
"output": "8589934591"
},
{
"input": "2 40\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\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",
"output": "2199023255590"
},
{
"input": "1 35\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",
"output": "34359738367"
},
{
"input": "50 1\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0",
"output": "1125899906842623"
}
] | 1,503,593,661
| 1,161
|
Python 3
|
OK
|
TESTS
| 55
| 77
| 0
|
n,m=map(int,input().split())
l=[]
for i in range(n):
l.append(list(map(int,input().split())))
ans=0
for i in range(n):
r=l[i]
c1=r.count(1)
c2=r.count(0)
ans+=(((2**c1)-1)+((2**c2)-1))
for i in range(m):
ls=[]
for j in range(n):
ls.append(l[j][i])
c1=ls.count(1)
c2=ls.count(0)
ans+=(((2**c1)-1)+((2**c2)-1)-c1-c2)
print(ans)
|
Title: Rectangles
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given *n*<=×<=*m* table. Each cell of the table is colored white or black. Find the number of non-empty sets of cells such that:
1. All cells in a set have the same color. 1. Every two cells in a set share row or column.
Input Specification:
The first line of input contains integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=50) — the number of rows and the number of columns correspondingly.
The next *n* lines of input contain descriptions of rows. There are *m* integers, separated by spaces, in each line. The number equals 0 if the corresponding cell is colored white and equals 1 if the corresponding cell is colored black.
Output Specification:
Output single integer — the number of non-empty sets from the problem description.
Demo Input:
['1 1\n0\n', '2 3\n1 0 1\n0 1 0\n']
Demo Output:
['1\n', '8\n']
Note:
In the second example, there are six one-element sets. Additionally, there are two two-element sets, the first one consists of the first and the third cells of the first row, the second one consists of the first and the third cells of the second row. To sum up, there are 8 sets.
|
```python
n,m=map(int,input().split())
l=[]
for i in range(n):
l.append(list(map(int,input().split())))
ans=0
for i in range(n):
r=l[i]
c1=r.count(1)
c2=r.count(0)
ans+=(((2**c1)-1)+((2**c2)-1))
for i in range(m):
ls=[]
for j in range(n):
ls.append(l[j][i])
c1=ls.count(1)
c2=ls.count(0)
ans+=(((2**c1)-1)+((2**c2)-1)-c1-c2)
print(ans)
```
| 3
|
|
339
|
A
|
Helpful Maths
|
PROGRAMMING
| 800
|
[
"greedy",
"implementation",
"sortings",
"strings"
] | null | null |
Xenia the beginner mathematician is a third year student at elementary school. She is now learning the addition operation.
The teacher has written down the sum of multiple numbers. Pupils should calculate the sum. To make the calculation easier, the sum only contains numbers 1, 2 and 3. Still, that isn't enough for Xenia. She is only beginning to count, so she can calculate a sum only if the summands follow in non-decreasing order. For example, she can't calculate sum 1+3+2+1 but she can calculate sums 1+1+2 and 3+3.
You've got the sum that was written on the board. Rearrange the summans and print the sum in such a way that Xenia can calculate the sum.
|
The first line contains a non-empty string *s* — the sum Xenia needs to count. String *s* contains no spaces. It only contains digits and characters "+". Besides, string *s* is a correct sum of numbers 1, 2 and 3. String *s* is at most 100 characters long.
|
Print the new sum that Xenia can count.
|
[
"3+2+1\n",
"1+1+3+1+3\n",
"2\n"
] |
[
"1+2+3\n",
"1+1+1+3+3\n",
"2\n"
] |
none
| 500
|
[
{
"input": "3+2+1",
"output": "1+2+3"
},
{
"input": "1+1+3+1+3",
"output": "1+1+1+3+3"
},
{
"input": "2",
"output": "2"
},
{
"input": "2+2+1+1+3",
"output": "1+1+2+2+3"
},
{
"input": "2+1+2+2+2+3+1+3+1+2",
"output": "1+1+1+2+2+2+2+2+3+3"
},
{
"input": "1+2+1+2+2+2+2+1+3+3",
"output": "1+1+1+2+2+2+2+2+3+3"
},
{
"input": "2+3+3+1+2+2+2+1+1+2+1+3+2+2+3+3+2+2+3+3+3+1+1+1+3+3+3+2+1+3+2+3+2+1+1+3+3+3+1+2+2+1+2+2+1+2+1+3+1+1",
"output": "1+1+1+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+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3"
},
{
"input": "1",
"output": "1"
},
{
"input": "2+1+2+2+1+3+2+3+1+1+2+1+2+2+3+1+1+3+3+3+2+2+3+2+2+2+1+2+1+2+3+2+2+2+1+3+1+3+3+3+1+2+1+2+2+2+2+3+1+1",
"output": "1+1+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+3+3+3+3+3+3+3+3+3+3+3+3+3"
},
{
"input": "2+2+1+1+1+3+1+1+3+3+2+3+1+3+1+1+3+1+1+2+2+2+2+1+2+1+2+1+1+1+3+1+3+2+3+2+3+3+1+1+1+2+3+2+1+3+1+3+2+2",
"output": "1+1+1+1+1+1+1+1+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+3+3+3+3+3+3+3+3+3+3+3+3+3+3"
},
{
"input": "3+2+3+3+2+2+1+2+1+2+3+1+2+3+2+3+2+1+2+2+1+1+2+2+3+2+1+3+1+1+3+2+2+2+2+3+3+2+2+3+3+1+1+2+3+3+2+3+3+3",
"output": "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+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3"
},
{
"input": "3",
"output": "3"
},
{
"input": "1+1",
"output": "1+1"
},
{
"input": "1+2",
"output": "1+2"
},
{
"input": "1+3",
"output": "1+3"
},
{
"input": "2+1",
"output": "1+2"
},
{
"input": "2+2",
"output": "2+2"
},
{
"input": "2+3",
"output": "2+3"
},
{
"input": "3+1",
"output": "1+3"
},
{
"input": "3+2",
"output": "2+3"
},
{
"input": "3+3",
"output": "3+3"
}
] | 1,695,392,220
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 21
| 92
| 0
|
arr = input().split("+")
arr.sort()
print('+'.join(arr))
|
Title: Helpful Maths
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Xenia the beginner mathematician is a third year student at elementary school. She is now learning the addition operation.
The teacher has written down the sum of multiple numbers. Pupils should calculate the sum. To make the calculation easier, the sum only contains numbers 1, 2 and 3. Still, that isn't enough for Xenia. She is only beginning to count, so she can calculate a sum only if the summands follow in non-decreasing order. For example, she can't calculate sum 1+3+2+1 but she can calculate sums 1+1+2 and 3+3.
You've got the sum that was written on the board. Rearrange the summans and print the sum in such a way that Xenia can calculate the sum.
Input Specification:
The first line contains a non-empty string *s* — the sum Xenia needs to count. String *s* contains no spaces. It only contains digits and characters "+". Besides, string *s* is a correct sum of numbers 1, 2 and 3. String *s* is at most 100 characters long.
Output Specification:
Print the new sum that Xenia can count.
Demo Input:
['3+2+1\n', '1+1+3+1+3\n', '2\n']
Demo Output:
['1+2+3\n', '1+1+1+3+3\n', '2\n']
Note:
none
|
```python
arr = input().split("+")
arr.sort()
print('+'.join(arr))
```
| 3
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | 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<=≤<=*n*<=≤<=3·105). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=106) — the initial group that is given to Toastman.
|
Print a single integer — 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.
| 0
|
[
{
"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"
}
] | 1,409,067,417
| 5,817
|
Python 3
|
WRONG_ANSWER
|
PRETESTS
| 2
| 62
| 0
|
def yabl(b):
#print(b)
if len(b) == 1:
return 1
else:
x = len(b) // 2
tost(b[:x])
tost(b[x:])
def tost(a):
global cnt
cnt += sum(a)
# print('sum = ', sum(a))
yabl(a)
cnt = 0
n = int(input())
arr = list(map(int, input().split()))
arr.sort()
tost(arr)
#x = len(arr) // 2
#print(arr[x + 1:])
print(cnt)
|
Title: none
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<=≤<=*n*<=≤<=3·105). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=106) — the initial group that is given to Toastman.
Output Specification:
Print a single integer — 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
def yabl(b):
#print(b)
if len(b) == 1:
return 1
else:
x = len(b) // 2
tost(b[:x])
tost(b[x:])
def tost(a):
global cnt
cnt += sum(a)
# print('sum = ', sum(a))
yabl(a)
cnt = 0
n = int(input())
arr = list(map(int, input().split()))
arr.sort()
tost(arr)
#x = len(arr) // 2
#print(arr[x + 1:])
print(cnt)
```
| 0
|
|
664
|
A
|
Complicated GCD
|
PROGRAMMING
| 800
|
[
"math",
"number theory"
] | null | null |
Greatest common divisor *GCD*(*a*,<=*b*) of two positive integers *a* and *b* is equal to the biggest integer *d* such that both integers *a* and *b* are divisible by *d*. There are many efficient algorithms to find greatest common divisor *GCD*(*a*,<=*b*), for example, Euclid algorithm.
Formally, find the biggest integer *d*, such that all integers *a*,<=*a*<=+<=1,<=*a*<=+<=2,<=...,<=*b* are divisible by *d*. To make the problem even more complicated we allow *a* and *b* to be up to googol, 10100 — such number do not fit even in 64-bit integer type!
|
The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=10100).
|
Output one integer — greatest common divisor of all integers from *a* to *b* inclusive.
|
[
"1 2\n",
"61803398874989484820458683436563811772030917980576 61803398874989484820458683436563811772030917980576\n"
] |
[
"1\n",
"61803398874989484820458683436563811772030917980576\n"
] |
none
| 500
|
[
{
"input": "1 2",
"output": "1"
},
{
"input": "61803398874989484820458683436563811772030917980576 61803398874989484820458683436563811772030917980576",
"output": "61803398874989484820458683436563811772030917980576"
},
{
"input": "1 100",
"output": "1"
},
{
"input": "100 100000",
"output": "1"
},
{
"input": "12345 67890123456789123457",
"output": "1"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "8392739158839273915883927391588392739158839273915883927391588392739158839273915883927391588392739158 8392739158839273915883927391588392739158839273915883927391588392739158839273915883927391588392739158",
"output": "8392739158839273915883927391588392739158839273915883927391588392739158839273915883927391588392739158"
},
{
"input": "1 10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "1"
},
{
"input": "8328748239473982794239847237438782379810988324751 9328748239473982794239847237438782379810988324751",
"output": "1"
},
{
"input": "1029398958432734901284327523909481928483573793 1029398958432734901284327523909481928483573794",
"output": "1"
},
{
"input": "10000 1000000000",
"output": "1"
},
{
"input": "10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "11210171722243 65715435710585778347",
"output": "1"
},
{
"input": "2921881079263974825226940825843 767693191032295360887755303860323261471",
"output": "1"
},
{
"input": "8025352957265704896940312528736939363590612908210603 96027920417708260814607687034511406492969694925539085",
"output": "1"
},
{
"input": "23510978780782786207241069904470895053213996267165977112058175452757132930 210352653280909370107314249722987050753257161175393375412301228883856435481424",
"output": "1"
},
{
"input": "8150070767079366215626260746398623663859344142817267779361251788637547414925170226504788118262 49924902262298336032630839998470954964895251605110946547855439236151401194070172107435992986913614",
"output": "1"
},
{
"input": "15943150466658398903 15943150466658398903",
"output": "15943150466658398903"
},
{
"input": "410470228200245407491525399055972 410470228200245407491525399055972",
"output": "410470228200245407491525399055972"
},
{
"input": "51894705655711504622197349350106792045098781545973899451307 51894705655711504622197349350106792045098781545973899451307",
"output": "51894705655711504622197349350106792045098781545973899451307"
},
{
"input": "60353594589897438036015726222485085035927634677598681595162804007836722215668410 60353594589897438036015726222485085035927634677598681595162804007836722215668410",
"output": "60353594589897438036015726222485085035927634677598681595162804007836722215668410"
},
{
"input": "761733780145118977868180796896376577405349682060892737466239031663476251177476275459280340045369535 761733780145118977868180796896376577405349682060892737466239031663476251177476275459280340045369535",
"output": "761733780145118977868180796896376577405349682060892737466239031663476251177476275459280340045369535"
},
{
"input": "1 2000000000",
"output": "1"
},
{
"input": "13 1928834874",
"output": "1"
},
{
"input": "87 2938984237482934238",
"output": "1"
},
{
"input": "213 413",
"output": "1"
},
{
"input": "3 4",
"output": "1"
}
] | 1,560,021,743
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 78
| 0
|
def gcd(a, b):
if a==0:
return b
else:
return gcd(b%a, a)
a = int(input())
b = int(input())
print(gcd(a, b))
|
Title: Complicated GCD
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Greatest common divisor *GCD*(*a*,<=*b*) of two positive integers *a* and *b* is equal to the biggest integer *d* such that both integers *a* and *b* are divisible by *d*. There are many efficient algorithms to find greatest common divisor *GCD*(*a*,<=*b*), for example, Euclid algorithm.
Formally, find the biggest integer *d*, such that all integers *a*,<=*a*<=+<=1,<=*a*<=+<=2,<=...,<=*b* are divisible by *d*. To make the problem even more complicated we allow *a* and *b* to be up to googol, 10100 — such number do not fit even in 64-bit integer type!
Input Specification:
The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=10100).
Output Specification:
Output one integer — greatest common divisor of all integers from *a* to *b* inclusive.
Demo Input:
['1 2\n', '61803398874989484820458683436563811772030917980576 61803398874989484820458683436563811772030917980576\n']
Demo Output:
['1\n', '61803398874989484820458683436563811772030917980576\n']
Note:
none
|
```python
def gcd(a, b):
if a==0:
return b
else:
return gcd(b%a, a)
a = int(input())
b = int(input())
print(gcd(a, b))
```
| -1
|
|
263
|
A
|
Beautiful Matrix
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
You've got a 5<=×<=5 matrix, consisting of 24 zeroes and a single number one. Let's index the matrix rows by numbers from 1 to 5 from top to bottom, let's index the matrix columns by numbers from 1 to 5 from left to right. In one move, you are allowed to apply one of the two following transformations to the matrix:
1. Swap two neighboring matrix rows, that is, rows with indexes *i* and *i*<=+<=1 for some integer *i* (1<=≤<=*i*<=<<=5). 1. Swap two neighboring matrix columns, that is, columns with indexes *j* and *j*<=+<=1 for some integer *j* (1<=≤<=*j*<=<<=5).
You think that a matrix looks beautiful, if the single number one of the matrix is located in its middle (in the cell that is on the intersection of the third row and the third column). Count the minimum number of moves needed to make the matrix beautiful.
|
The input consists of five lines, each line contains five integers: the *j*-th integer in the *i*-th line of the input represents the element of the matrix that is located on the intersection of the *i*-th row and the *j*-th column. It is guaranteed that the matrix consists of 24 zeroes and a single number one.
|
Print a single integer — the minimum number of moves needed to make the matrix beautiful.
|
[
"0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n",
"0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n"
] |
[
"3\n",
"1\n"
] |
none
| 500
|
[
{
"input": "0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "0"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1",
"output": "4"
},
{
"input": "0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "4"
},
{
"input": "0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "4"
},
{
"input": "0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0",
"output": "4"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0",
"output": "3"
}
] | 1,696,248,716
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 60
| 0
|
for i in range (1,6):
for j in range (1,6):
n=int(input())
if n==1:
a=abs(i-3)+abs(j-3)
print(a)
|
Title: Beautiful Matrix
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got a 5<=×<=5 matrix, consisting of 24 zeroes and a single number one. Let's index the matrix rows by numbers from 1 to 5 from top to bottom, let's index the matrix columns by numbers from 1 to 5 from left to right. In one move, you are allowed to apply one of the two following transformations to the matrix:
1. Swap two neighboring matrix rows, that is, rows with indexes *i* and *i*<=+<=1 for some integer *i* (1<=≤<=*i*<=<<=5). 1. Swap two neighboring matrix columns, that is, columns with indexes *j* and *j*<=+<=1 for some integer *j* (1<=≤<=*j*<=<<=5).
You think that a matrix looks beautiful, if the single number one of the matrix is located in its middle (in the cell that is on the intersection of the third row and the third column). Count the minimum number of moves needed to make the matrix beautiful.
Input Specification:
The input consists of five lines, each line contains five integers: the *j*-th integer in the *i*-th line of the input represents the element of the matrix that is located on the intersection of the *i*-th row and the *j*-th column. It is guaranteed that the matrix consists of 24 zeroes and a single number one.
Output Specification:
Print a single integer — the minimum number of moves needed to make the matrix beautiful.
Demo Input:
['0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n', '0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n']
Demo Output:
['3\n', '1\n']
Note:
none
|
```python
for i in range (1,6):
for j in range (1,6):
n=int(input())
if n==1:
a=abs(i-3)+abs(j-3)
print(a)
```
| -1
|
|
918
|
B
|
Radio Station
|
PROGRAMMING
| 900
|
[
"implementation",
"strings"
] | null | null |
As the guys fried the radio station facilities, the school principal gave them tasks as a punishment. Dustin's task was to add comments to nginx configuration for school's website. The school has *n* servers. Each server has a name and an ip (names aren't necessarily unique, but ips are). Dustin knows the ip and name of each server. For simplicity, we'll assume that an nginx command is of form "command ip;" where command is a string consisting of English lowercase letter only, and ip is the ip of one of school servers.
Each ip is of form "a.b.c.d" where *a*, *b*, *c* and *d* are non-negative integers less than or equal to 255 (with no leading zeros). The nginx configuration file Dustin has to add comments to has *m* commands. Nobody ever memorizes the ips of servers, so to understand the configuration better, Dustin has to comment the name of server that the ip belongs to at the end of each line (after each command). More formally, if a line is "command ip;" Dustin has to replace it with "command ip; #name" where name is the name of the server with ip equal to ip.
Dustin doesn't know anything about nginx, so he panicked again and his friends asked you to do his task for him.
|
The first line of input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000).
The next *n* lines contain the names and ips of the servers. Each line contains a string name, name of the server and a string ip, ip of the server, separated by space (1<=≤<=|*name*|<=≤<=10, *name* only consists of English lowercase letters). It is guaranteed that all ip are distinct.
The next *m* lines contain the commands in the configuration file. Each line is of form "command ip;" (1<=≤<=|*command*|<=≤<=10, command only consists of English lowercase letters). It is guaranteed that ip belongs to one of the *n* school servers.
|
Print *m* lines, the commands in the configuration file after Dustin did his task.
|
[
"2 2\nmain 192.168.0.2\nreplica 192.168.0.1\nblock 192.168.0.1;\nproxy 192.168.0.2;\n",
"3 5\ngoogle 8.8.8.8\ncodeforces 212.193.33.27\nserver 138.197.64.57\nredirect 138.197.64.57;\nblock 8.8.8.8;\ncf 212.193.33.27;\nunblock 8.8.8.8;\ncheck 138.197.64.57;\n"
] |
[
"block 192.168.0.1; #replica\nproxy 192.168.0.2; #main\n",
"redirect 138.197.64.57; #server\nblock 8.8.8.8; #google\ncf 212.193.33.27; #codeforces\nunblock 8.8.8.8; #google\ncheck 138.197.64.57; #server\n"
] |
none
| 1,000
|
[
{
"input": "2 2\nmain 192.168.0.2\nreplica 192.168.0.1\nblock 192.168.0.1;\nproxy 192.168.0.2;",
"output": "block 192.168.0.1; #replica\nproxy 192.168.0.2; #main"
},
{
"input": "3 5\ngoogle 8.8.8.8\ncodeforces 212.193.33.27\nserver 138.197.64.57\nredirect 138.197.64.57;\nblock 8.8.8.8;\ncf 212.193.33.27;\nunblock 8.8.8.8;\ncheck 138.197.64.57;",
"output": "redirect 138.197.64.57; #server\nblock 8.8.8.8; #google\ncf 212.193.33.27; #codeforces\nunblock 8.8.8.8; #google\ncheck 138.197.64.57; #server"
},
{
"input": "10 10\nittmcs 112.147.123.173\njkt 228.40.73.178\nfwckqtz 88.28.31.198\nkal 224.226.34.213\nnacuyokm 49.57.13.44\nfouynv 243.18.250.17\ns 45.248.83.247\ne 75.69.23.169\nauwoqlch 100.44.219.187\nlkldjq 46.123.169.140\ngjcylatwzi 46.123.169.140;\ndxfi 88.28.31.198;\ngv 46.123.169.140;\nety 88.28.31.198;\notbmgcrn 46.123.169.140;\nw 112.147.123.173;\np 75.69.23.169;\nvdsnigk 46.123.169.140;\nmmc 46.123.169.140;\ngtc 49.57.13.44;",
"output": "gjcylatwzi 46.123.169.140; #lkldjq\ndxfi 88.28.31.198; #fwckqtz\ngv 46.123.169.140; #lkldjq\nety 88.28.31.198; #fwckqtz\notbmgcrn 46.123.169.140; #lkldjq\nw 112.147.123.173; #ittmcs\np 75.69.23.169; #e\nvdsnigk 46.123.169.140; #lkldjq\nmmc 46.123.169.140; #lkldjq\ngtc 49.57.13.44; #nacuyokm"
},
{
"input": "1 1\nervbfot 185.32.99.2\nzygoumbmx 185.32.99.2;",
"output": "zygoumbmx 185.32.99.2; #ervbfot"
},
{
"input": "1 2\ny 245.182.246.189\nlllq 245.182.246.189;\nxds 245.182.246.189;",
"output": "lllq 245.182.246.189; #y\nxds 245.182.246.189; #y"
},
{
"input": "2 1\ntdwmshz 203.115.124.110\neksckjya 201.80.191.212\nzbtjzzue 203.115.124.110;",
"output": "zbtjzzue 203.115.124.110; #tdwmshz"
},
{
"input": "8 5\nfhgkq 5.19.189.178\nphftablcr 75.18.177.178\nxnpcg 158.231.167.176\ncfahrkq 26.165.124.191\nfkgtnqtfoh 230.13.13.129\nt 101.24.94.85\nvjoirslx 59.6.179.72\ntwktmskb 38.194.117.184\nrvzzlygosc 26.165.124.191;\ndcsgxrkgv 101.24.94.85;\nyvmyppn 59.6.179.72;\ngpdjjuq 75.18.177.178;\nvdviz 101.24.94.85;",
"output": "rvzzlygosc 26.165.124.191; #cfahrkq\ndcsgxrkgv 101.24.94.85; #t\nyvmyppn 59.6.179.72; #vjoirslx\ngpdjjuq 75.18.177.178; #phftablcr\nvdviz 101.24.94.85; #t"
},
{
"input": "2 1\nmain 1.1.1.1\nget 11.1.1.1\ncommand 11.1.1.1;",
"output": "command 11.1.1.1; #get"
},
{
"input": "2 2\nmain 0.0.0.255\nmainn 0.0.2.55\nblock 0.0.0.255;\nblock 0.0.2.55;",
"output": "block 0.0.0.255; #main\nblock 0.0.2.55; #mainn"
},
{
"input": "2 2\nmain 0.0.63.0\nreplica 0.38.0.56\nblock 0.0.63.0;\nproxy 0.38.0.56;",
"output": "block 0.0.63.0; #main\nproxy 0.38.0.56; #replica"
},
{
"input": "2 2\nmain 192.168.10.12\nreplica 192.167.10.12\nblock 192.168.10.12;\nproxy 192.167.10.12;",
"output": "block 192.168.10.12; #main\nproxy 192.167.10.12; #replica"
},
{
"input": "2 1\nneserver 185.218.47.91\nserver 255.255.255.255\nblock 255.255.255.255;",
"output": "block 255.255.255.255; #server"
},
{
"input": "2 1\nalp 22.222.30.10\nbet 222.22.30.10\nblock 22.222.30.10;",
"output": "block 22.222.30.10; #alp"
}
] | 1,642,586,847
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 28
| 46
| 0
|
n,o=map(int,input().split())
fre={}
for i in range(n):
x,y=map(str,input().split())
fre[y]=x
for j in range(o):
q,r=map(str,input().split())
print(q," ",r," ","#",fre[r[:len(r)-1]],sep="")
|
Title: Radio Station
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As the guys fried the radio station facilities, the school principal gave them tasks as a punishment. Dustin's task was to add comments to nginx configuration for school's website. The school has *n* servers. Each server has a name and an ip (names aren't necessarily unique, but ips are). Dustin knows the ip and name of each server. For simplicity, we'll assume that an nginx command is of form "command ip;" where command is a string consisting of English lowercase letter only, and ip is the ip of one of school servers.
Each ip is of form "a.b.c.d" where *a*, *b*, *c* and *d* are non-negative integers less than or equal to 255 (with no leading zeros). The nginx configuration file Dustin has to add comments to has *m* commands. Nobody ever memorizes the ips of servers, so to understand the configuration better, Dustin has to comment the name of server that the ip belongs to at the end of each line (after each command). More formally, if a line is "command ip;" Dustin has to replace it with "command ip; #name" where name is the name of the server with ip equal to ip.
Dustin doesn't know anything about nginx, so he panicked again and his friends asked you to do his task for him.
Input Specification:
The first line of input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000).
The next *n* lines contain the names and ips of the servers. Each line contains a string name, name of the server and a string ip, ip of the server, separated by space (1<=≤<=|*name*|<=≤<=10, *name* only consists of English lowercase letters). It is guaranteed that all ip are distinct.
The next *m* lines contain the commands in the configuration file. Each line is of form "command ip;" (1<=≤<=|*command*|<=≤<=10, command only consists of English lowercase letters). It is guaranteed that ip belongs to one of the *n* school servers.
Output Specification:
Print *m* lines, the commands in the configuration file after Dustin did his task.
Demo Input:
['2 2\nmain 192.168.0.2\nreplica 192.168.0.1\nblock 192.168.0.1;\nproxy 192.168.0.2;\n', '3 5\ngoogle 8.8.8.8\ncodeforces 212.193.33.27\nserver 138.197.64.57\nredirect 138.197.64.57;\nblock 8.8.8.8;\ncf 212.193.33.27;\nunblock 8.8.8.8;\ncheck 138.197.64.57;\n']
Demo Output:
['block 192.168.0.1; #replica\nproxy 192.168.0.2; #main\n', 'redirect 138.197.64.57; #server\nblock 8.8.8.8; #google\ncf 212.193.33.27; #codeforces\nunblock 8.8.8.8; #google\ncheck 138.197.64.57; #server\n']
Note:
none
|
```python
n,o=map(int,input().split())
fre={}
for i in range(n):
x,y=map(str,input().split())
fre[y]=x
for j in range(o):
q,r=map(str,input().split())
print(q," ",r," ","#",fre[r[:len(r)-1]],sep="")
```
| 3
|
|
957
|
A
|
Tritonic Iridescence
|
PROGRAMMING
| 1,300
|
[
"implementation"
] | null | null |
Overlooking the captivating blend of myriads of vernal hues, Arkady the painter lays out a long, long canvas.
Arkady has a sufficiently large amount of paint of three colours: cyan, magenta, and yellow. On the one-dimensional canvas split into *n* consecutive segments, each segment needs to be painted in one of the colours.
Arkady has already painted some (possibly none or all) segments and passes the paintbrush to you. You are to determine whether there are at least two ways of colouring all the unpainted segments so that no two adjacent segments are of the same colour. Two ways are considered different if and only if a segment is painted in different colours in them.
|
The first line contains a single positive integer *n* (1<=≤<=*n*<=≤<=100) — the length of the canvas.
The second line contains a string *s* of *n* characters, the *i*-th of which is either 'C' (denoting a segment painted in cyan), 'M' (denoting one painted in magenta), 'Y' (one painted in yellow), or '?' (an unpainted one).
|
If there are at least two different ways of painting, output "Yes"; otherwise output "No" (both without quotes).
You can print each character in any case (upper or lower).
|
[
"5\nCY??Y\n",
"5\nC?C?Y\n",
"5\n?CYC?\n",
"5\nC??MM\n",
"3\nMMY\n"
] |
[
"Yes\n",
"Yes\n",
"Yes\n",
"No\n",
"No\n"
] |
For the first example, there are exactly two different ways of colouring: CYCMY and CYMCY.
For the second example, there are also exactly two different ways of colouring: CMCMY and CYCMY.
For the third example, there are four ways of colouring: MCYCM, MCYCY, YCYCM, and YCYCY.
For the fourth example, no matter how the unpainted segments are coloured, the existing magenta segments will prevent the painting from satisfying the requirements. The similar is true for the fifth example.
| 500
|
[
{
"input": "5\nCY??Y",
"output": "Yes"
},
{
"input": "5\nC?C?Y",
"output": "Yes"
},
{
"input": "5\n?CYC?",
"output": "Yes"
},
{
"input": "5\nC??MM",
"output": "No"
},
{
"input": "3\nMMY",
"output": "No"
},
{
"input": "15\n??YYYYYY??YYYY?",
"output": "No"
},
{
"input": "100\nYCY?CMCMCYMYMYC?YMYMYMY?CMC?MCMYCMYMYCM?CMCM?CMYMYCYCMCMCMCMCMYM?CYCYCMCM?CY?MYCYCMYM?CYCYCYMY?CYCYC",
"output": "No"
},
{
"input": "1\nC",
"output": "No"
},
{
"input": "1\n?",
"output": "Yes"
},
{
"input": "2\nMY",
"output": "No"
},
{
"input": "2\n?M",
"output": "Yes"
},
{
"input": "2\nY?",
"output": "Yes"
},
{
"input": "2\n??",
"output": "Yes"
},
{
"input": "3\n??C",
"output": "Yes"
},
{
"input": "3\nM??",
"output": "Yes"
},
{
"input": "3\nYCM",
"output": "No"
},
{
"input": "3\n?C?",
"output": "Yes"
},
{
"input": "3\nMC?",
"output": "Yes"
},
{
"input": "4\nCYCM",
"output": "No"
},
{
"input": "4\nM?CM",
"output": "No"
},
{
"input": "4\n??YM",
"output": "Yes"
},
{
"input": "4\nC???",
"output": "Yes"
},
{
"input": "10\nMCYM?MYM?C",
"output": "Yes"
},
{
"input": "50\nCMCMCYM?MY?C?MC??YM?CY?YM??M?MCMCYCYMCYCMCM?MCM?MC",
"output": "Yes"
},
{
"input": "97\nMCM?YCMYM?YMY?MY?MYCY?CMCMCYC?YMY?MYCMC?M?YCMC?YM?C?MCMCMYMCMY?MCM?YC?YMYMY?MYCYCM?YC?YCY?MYMYMYC",
"output": "No"
},
{
"input": "100\nC?M?M?M?YM??YMYC?MCYMYM??Y??YC?CYC???YM?YM??MYMY?CYCYMYC?YC?C?CYCMY??CMC?YMCMYCYCYMYM?CYM?M?MCMCMY?Y",
"output": "Yes"
},
{
"input": "100\n?YYYYYYYYYYYYYYYYYYYYYYYYYYYYY??YYY?YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY?",
"output": "No"
},
{
"input": "100\n????????????????????????????????????????????????????????????????????????????????????????????????????",
"output": "Yes"
},
{
"input": "100\nY?CYMYMYMYCYMY?CMCYMYMC?MYCY?CMCYCYMCM?YCMCYMYCYCY?MYMYCYCYCM?CYMYMY?CY?MYCYCY?M?YMYCY?CMCMCY?CY?M?C",
"output": "No"
},
{
"input": "4\nM??C",
"output": "Yes"
},
{
"input": "6\nY?C??C",
"output": "Yes"
},
{
"input": "5\nC???Y",
"output": "Yes"
},
{
"input": "5\nC??MY",
"output": "Yes"
},
{
"input": "5\nCY??M",
"output": "Yes"
},
{
"input": "4\nC??Y",
"output": "Yes"
},
{
"input": "52\n??????????????????????????????????????????????????YY",
"output": "No"
},
{
"input": "3\nYY?",
"output": "No"
},
{
"input": "5\nCC??Y",
"output": "No"
},
{
"input": "8\nCMC??MCM",
"output": "Yes"
},
{
"input": "7\nM?YCM??",
"output": "Yes"
},
{
"input": "6\n?CC???",
"output": "No"
},
{
"input": "100\n??????????????????????????????????????????????????????????????????????????????????????????????????MM",
"output": "No"
},
{
"input": "4\nC??M",
"output": "Yes"
},
{
"input": "4\n?C?M",
"output": "Yes"
},
{
"input": "6\nMC??MC",
"output": "Yes"
}
] | 1,521,982,332
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 6
| 78
| 7,065,600
|
def checker(paints):
for i in range(1, len(paints) - 1, 1):
if paints[i] == '?' and paints[i - 1] != '?' and paints[i + 1] != '?' and paints[i - 1] != paints[i + 1]:
paints.replace(paints[i], "")
for i in range(0, len(paints) - 1, 1):
if paints[i] == paints[i + 1] and paints[i] != '?' and paints[i + 1] != '?':
return False
return paints.__contains__('?')
lengthOfTheCanvas = int(input())
paints = input()
if checker(paints):
print("Yes")
else:
print("No")
|
Title: Tritonic Iridescence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Overlooking the captivating blend of myriads of vernal hues, Arkady the painter lays out a long, long canvas.
Arkady has a sufficiently large amount of paint of three colours: cyan, magenta, and yellow. On the one-dimensional canvas split into *n* consecutive segments, each segment needs to be painted in one of the colours.
Arkady has already painted some (possibly none or all) segments and passes the paintbrush to you. You are to determine whether there are at least two ways of colouring all the unpainted segments so that no two adjacent segments are of the same colour. Two ways are considered different if and only if a segment is painted in different colours in them.
Input Specification:
The first line contains a single positive integer *n* (1<=≤<=*n*<=≤<=100) — the length of the canvas.
The second line contains a string *s* of *n* characters, the *i*-th of which is either 'C' (denoting a segment painted in cyan), 'M' (denoting one painted in magenta), 'Y' (one painted in yellow), or '?' (an unpainted one).
Output Specification:
If there are at least two different ways of painting, output "Yes"; otherwise output "No" (both without quotes).
You can print each character in any case (upper or lower).
Demo Input:
['5\nCY??Y\n', '5\nC?C?Y\n', '5\n?CYC?\n', '5\nC??MM\n', '3\nMMY\n']
Demo Output:
['Yes\n', 'Yes\n', 'Yes\n', 'No\n', 'No\n']
Note:
For the first example, there are exactly two different ways of colouring: CYCMY and CYMCY.
For the second example, there are also exactly two different ways of colouring: CMCMY and CYCMY.
For the third example, there are four ways of colouring: MCYCM, MCYCY, YCYCM, and YCYCY.
For the fourth example, no matter how the unpainted segments are coloured, the existing magenta segments will prevent the painting from satisfying the requirements. The similar is true for the fifth example.
|
```python
def checker(paints):
for i in range(1, len(paints) - 1, 1):
if paints[i] == '?' and paints[i - 1] != '?' and paints[i + 1] != '?' and paints[i - 1] != paints[i + 1]:
paints.replace(paints[i], "")
for i in range(0, len(paints) - 1, 1):
if paints[i] == paints[i + 1] and paints[i] != '?' and paints[i + 1] != '?':
return False
return paints.__contains__('?')
lengthOfTheCanvas = int(input())
paints = input()
if checker(paints):
print("Yes")
else:
print("No")
```
| 0
|
|
688
|
A
|
Opponents
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Arya has *n* opponents in the school. Each day he will fight with all opponents who are present this day. His opponents have some fighting plan that guarantees they will win, but implementing this plan requires presence of them all. That means if one day at least one of Arya's opponents is absent at the school, then Arya will beat all present opponents. Otherwise, if all opponents are present, then they will beat Arya.
For each opponent Arya knows his schedule — whether or not he is going to present on each particular day. Tell him the maximum number of consecutive days that he will beat all present opponents.
Note, that if some day there are no opponents present, Arya still considers he beats all the present opponents.
|
The first line of the input contains two integers *n* and *d* (1<=≤<=*n*,<=*d*<=≤<=100) — the number of opponents and the number of days, respectively.
The *i*-th of the following *d* lines contains a string of length *n* consisting of characters '0' and '1'. The *j*-th character of this string is '0' if the *j*-th opponent is going to be absent on the *i*-th day.
|
Print the only integer — the maximum number of consecutive days that Arya will beat all present opponents.
|
[
"2 2\n10\n00\n",
"4 1\n0100\n",
"4 5\n1101\n1111\n0110\n1011\n1111\n"
] |
[
"2\n",
"1\n",
"2\n"
] |
In the first and the second samples, Arya will beat all present opponents each of the *d* days.
In the third sample, Arya will beat his opponents on days 1, 3 and 4 and his opponents will beat him on days 2 and 5. Thus, the maximum number of consecutive winning days is 2, which happens on days 3 and 4.
| 500
|
[
{
"input": "2 2\n10\n00",
"output": "2"
},
{
"input": "4 1\n0100",
"output": "1"
},
{
"input": "4 5\n1101\n1111\n0110\n1011\n1111",
"output": "2"
},
{
"input": "3 2\n110\n110",
"output": "2"
},
{
"input": "10 6\n1111111111\n0100110101\n1111111111\n0000011010\n1111111111\n1111111111",
"output": "1"
},
{
"input": "10 10\n1111111111\n0001001000\n1111111111\n1111111111\n1111111111\n1000000100\n1111111111\n0000011100\n1111111111\n1111111111",
"output": "1"
},
{
"input": "10 10\n0000100011\n0100001111\n1111111111\n1100011111\n1111111111\n1000111000\n1111000010\n0111001001\n1101010110\n1111111111",
"output": "4"
},
{
"input": "10 10\n1100110010\n0000000001\n1011100111\n1111111111\n1111111111\n1111111111\n1100010110\n1111111111\n1001001010\n1111111111",
"output": "3"
},
{
"input": "10 7\n0000111001\n1111111111\n0110110001\n1111111111\n1111111111\n1000111100\n0110000111",
"output": "2"
},
{
"input": "5 10\n00110\n11000\n10010\n00010\n11110\n01101\n11111\n10001\n11111\n01001",
"output": "6"
},
{
"input": "5 9\n11111\n11101\n11111\n11111\n01010\n01010\n00000\n11111\n00111",
"output": "3"
},
{
"input": "5 10\n11111\n00010\n11010\n11111\n11111\n00100\n11111\n11111\n01000\n11111",
"output": "2"
},
{
"input": "5 9\n11111\n11111\n11111\n11111\n11100\n11111\n11111\n11111\n00000",
"output": "1"
},
{
"input": "5 8\n11111\n10110\n01001\n11111\n01100\n10010\n11111\n11111",
"output": "2"
},
{
"input": "1 1\n1",
"output": "0"
},
{
"input": "100 1\n0011001100100010000011001100000001011101110110010001110001101100110011111101001011011001000010001111",
"output": "1"
},
{
"input": "100 1\n1011011100000101000111110000110111010101110010010011110010001110100011001110110101111100100110000000",
"output": "1"
},
{
"input": "100 1\n1110000011110101010111111100011001100000101101010110100101110000011100110110110101011100110110010011",
"output": "1"
},
{
"input": "100 1\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111",
"output": "0"
},
{
"input": "1 100\n1\n0\n0\n0\n1\n1\n0\n0\n0\n0\n1\n1\n0\n1\n1\n0\n0\n1\n1\n1\n0\n0\n1\n1\n1\n1\n1\n0\n1\n0\n0\n0\n1\n1\n0\n1\n0\n1\n0\n0\n0\n1\n0\n1\n0\n0\n0\n1\n1\n1\n0\n1\n1\n1\n0\n1\n0\n1\n1\n1\n1\n0\n0\n0\n0\n0\n0\n1\n1\n0\n1\n1\n1\n1\n1\n0\n1\n1\n1\n1\n1\n0\n1\n0\n0\n1\n0\n0\n1\n0\n0\n1\n0\n1\n1\n1\n0\n1\n0\n0",
"output": "6"
},
{
"input": "1 100\n0\n0\n0\n0\n1\n0\n0\n0\n0\n1\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n1\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0",
"output": "49"
},
{
"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": "0"
},
{
"input": "1 100\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0",
"output": "100"
},
{
"input": "2 2\n11\n10",
"output": "1"
},
{
"input": "1 1\n0",
"output": "1"
}
] | 1,592,070,404
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 56
| 109
| 0
|
n,d=map(int,input().split())
brr=[]
count=0
max=0
while d!=0:
s=input()
li=list(s)
if li.count('0')>=1:
count+=1
else:
count=0
if count!=0:
if max<count:
max=count
d=d-1
print(max)
|
Title: Opponents
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Arya has *n* opponents in the school. Each day he will fight with all opponents who are present this day. His opponents have some fighting plan that guarantees they will win, but implementing this plan requires presence of them all. That means if one day at least one of Arya's opponents is absent at the school, then Arya will beat all present opponents. Otherwise, if all opponents are present, then they will beat Arya.
For each opponent Arya knows his schedule — whether or not he is going to present on each particular day. Tell him the maximum number of consecutive days that he will beat all present opponents.
Note, that if some day there are no opponents present, Arya still considers he beats all the present opponents.
Input Specification:
The first line of the input contains two integers *n* and *d* (1<=≤<=*n*,<=*d*<=≤<=100) — the number of opponents and the number of days, respectively.
The *i*-th of the following *d* lines contains a string of length *n* consisting of characters '0' and '1'. The *j*-th character of this string is '0' if the *j*-th opponent is going to be absent on the *i*-th day.
Output Specification:
Print the only integer — the maximum number of consecutive days that Arya will beat all present opponents.
Demo Input:
['2 2\n10\n00\n', '4 1\n0100\n', '4 5\n1101\n1111\n0110\n1011\n1111\n']
Demo Output:
['2\n', '1\n', '2\n']
Note:
In the first and the second samples, Arya will beat all present opponents each of the *d* days.
In the third sample, Arya will beat his opponents on days 1, 3 and 4 and his opponents will beat him on days 2 and 5. Thus, the maximum number of consecutive winning days is 2, which happens on days 3 and 4.
|
```python
n,d=map(int,input().split())
brr=[]
count=0
max=0
while d!=0:
s=input()
li=list(s)
if li.count('0')>=1:
count+=1
else:
count=0
if count!=0:
if max<count:
max=count
d=d-1
print(max)
```
| 3
|
|
862
|
A
|
Mahmoud and Ehab and the MEX
|
PROGRAMMING
| 1,000
|
[
"greedy",
"implementation"
] | null | null |
Dr. Evil kidnapped Mahmoud and Ehab in the evil land because of their performance in the Evil Olympiad in Informatics (EOI). He decided to give them some problems to let them go.
Dr. Evil is interested in sets, He has a set of *n* integers. Dr. Evil calls a set of integers evil if the MEX of it is exactly *x*. the MEX of a set of integers is the minimum non-negative integer that doesn't exist in it. For example, the MEX of the set {0,<=2,<=4} is 1 and the MEX of the set {1,<=2,<=3} is 0 .
Dr. Evil is going to make his set evil. To do this he can perform some operations. During each operation he can add some non-negative integer to his set or erase some element from it. What is the minimal number of operations Dr. Evil has to perform to make his set evil?
|
The first line contains two integers *n* and *x* (1<=≤<=*n*<=≤<=100, 0<=≤<=*x*<=≤<=100) — the size of the set Dr. Evil owns, and the desired MEX.
The second line contains *n* distinct non-negative integers not exceeding 100 that represent the set.
|
The only line should contain one integer — the minimal number of operations Dr. Evil should perform.
|
[
"5 3\n0 4 5 6 7\n",
"1 0\n0\n",
"5 0\n1 2 3 4 5\n"
] |
[
"2\n",
"1\n",
"0\n"
] |
For the first test case Dr. Evil should add 1 and 2 to the set performing 2 operations.
For the second test case Dr. Evil should erase 0 from the set. After that, the set becomes empty, so the MEX of it is 0.
In the third test case the set is already evil.
| 500
|
[
{
"input": "5 3\n0 4 5 6 7",
"output": "2"
},
{
"input": "1 0\n0",
"output": "1"
},
{
"input": "5 0\n1 2 3 4 5",
"output": "0"
},
{
"input": "10 5\n57 1 47 9 93 37 76 70 78 15",
"output": "4"
},
{
"input": "10 5\n99 98 93 97 95 100 92 94 91 96",
"output": "5"
},
{
"input": "10 5\n1 2 3 4 59 45 0 58 51 91",
"output": "0"
},
{
"input": "100 100\n79 13 21 11 3 87 28 40 29 4 96 34 8 78 61 46 33 45 99 30 92 67 22 97 39 86 73 31 74 44 62 55 57 2 54 63 80 69 25 48 77 98 17 93 15 16 89 12 43 23 37 95 14 38 83 90 49 56 72 10 20 0 50 71 70 88 19 1 76 81 52 41 82 68 85 47 6 7 35 60 18 64 75 84 27 9 65 91 94 42 53 24 66 26 59 36 51 32 5 58",
"output": "0"
},
{
"input": "100 50\n95 78 46 92 80 18 79 58 30 72 19 89 39 29 44 65 15 100 59 8 96 9 62 67 41 42 82 14 57 32 71 77 40 5 7 51 28 53 85 23 16 35 3 91 6 11 75 61 17 66 13 47 36 56 10 22 83 60 48 24 26 97 4 33 76 86 70 0 34 64 52 43 21 49 55 74 1 73 81 25 54 63 94 84 20 68 87 12 31 88 38 93 37 90 98 69 99 45 27 2",
"output": "0"
},
{
"input": "100 33\n28 11 79 92 88 62 77 72 7 41 96 97 67 84 44 8 81 35 38 1 64 68 46 17 98 83 31 12 74 21 2 22 47 6 36 75 65 61 37 26 25 45 59 48 100 51 93 76 78 49 3 57 16 4 87 29 55 82 70 39 53 0 60 15 24 71 58 20 66 89 95 42 13 43 63 90 85 52 50 30 54 40 56 23 27 34 32 18 10 19 69 9 99 73 91 14 5 80 94 86",
"output": "0"
},
{
"input": "99 33\n25 76 41 95 55 20 47 59 58 84 87 92 16 27 35 65 72 63 93 54 36 96 15 86 5 69 24 46 67 73 48 60 40 6 61 74 97 10 100 8 52 26 77 18 7 62 37 2 14 66 11 56 68 91 0 64 75 99 30 21 53 1 89 81 3 98 12 88 39 38 29 83 22 90 9 28 45 43 78 44 32 57 4 50 70 17 13 51 80 85 71 94 82 19 34 42 23 79 49",
"output": "1"
},
{
"input": "100 100\n65 56 84 46 44 33 99 74 62 72 93 67 43 92 75 88 38 34 66 12 55 76 58 90 78 8 14 45 97 59 48 32 64 18 39 89 31 51 54 81 29 36 70 77 40 22 49 27 3 1 73 13 98 42 87 37 2 57 4 6 50 25 23 79 28 86 68 61 80 17 19 10 15 63 52 11 35 60 21 16 24 85 30 91 7 5 69 20 71 82 53 94 41 95 96 9 26 83 0 47",
"output": "0"
},
{
"input": "100 100\n58 88 12 71 22 1 40 19 73 20 67 48 57 17 69 36 100 35 33 37 72 55 52 8 89 85 47 42 78 70 81 86 11 9 68 99 6 16 21 61 53 98 23 62 32 59 51 0 87 24 50 30 65 10 80 95 7 92 25 74 60 79 91 5 13 31 75 38 90 94 46 66 93 34 14 41 28 2 76 84 43 96 3 56 49 82 27 77 64 63 4 45 18 29 54 39 15 26 83 44",
"output": "2"
},
{
"input": "89 100\n58 96 17 41 86 34 28 84 18 40 8 77 87 89 68 79 33 35 53 49 0 6 22 12 72 90 48 55 21 50 56 62 75 2 37 95 69 74 14 20 44 46 27 32 31 59 63 60 10 85 71 70 38 52 94 30 61 51 80 26 36 23 39 47 76 45 100 57 15 78 97 66 54 13 99 16 93 73 24 4 83 5 98 81 92 25 29 88 65",
"output": "13"
},
{
"input": "100 50\n7 95 24 76 81 78 60 69 83 84 100 1 65 31 48 92 73 39 18 89 38 97 10 42 8 55 98 51 21 90 62 77 16 91 0 94 4 37 19 17 67 35 45 41 56 20 15 85 75 28 59 27 12 54 61 68 36 5 79 93 66 11 70 49 50 34 30 25 96 46 64 14 32 22 47 40 58 23 43 9 87 82 26 53 80 52 3 86 13 99 33 71 6 88 57 74 2 44 72 63",
"output": "2"
},
{
"input": "77 0\n27 8 20 92 21 41 53 98 17 65 67 35 81 11 55 49 61 44 2 66 51 89 40 28 52 62 86 91 64 24 18 5 94 82 96 99 71 6 39 83 26 29 16 30 45 97 80 90 69 12 13 33 76 73 46 19 78 56 88 38 42 34 57 77 47 4 59 58 7 100 95 72 9 74 15 43 54",
"output": "0"
},
{
"input": "100 50\n55 36 0 32 81 6 17 43 24 13 30 19 8 59 71 45 15 74 3 41 99 42 86 47 2 94 35 1 66 95 38 49 4 27 96 89 34 44 92 25 51 39 54 28 80 77 20 14 48 40 68 56 31 63 33 78 69 37 18 26 83 70 23 82 91 65 67 52 61 53 7 22 60 21 12 73 72 87 75 100 90 29 64 79 98 85 5 62 93 84 50 46 97 58 57 16 9 10 76 11",
"output": "1"
},
{
"input": "77 0\n12 8 19 87 9 54 55 86 97 7 27 85 25 48 94 73 26 1 13 57 72 69 76 39 38 91 75 40 42 28 93 21 70 84 65 11 60 90 20 95 66 89 59 47 34 99 6 61 52 100 50 3 77 81 82 53 15 24 0 45 44 14 68 96 58 5 18 35 10 98 29 74 92 49 83 71 17",
"output": "1"
},
{
"input": "100 70\n25 94 66 65 10 99 89 6 70 31 7 40 20 92 64 27 21 72 77 98 17 43 47 44 48 81 38 56 100 39 90 22 88 76 3 83 86 29 33 55 82 79 49 11 2 16 12 78 85 69 32 97 26 15 53 24 23 91 51 67 34 35 52 5 62 50 95 18 71 13 75 8 30 42 93 36 45 60 63 46 57 41 87 0 84 54 74 37 4 58 28 19 96 61 80 9 1 14 73 68",
"output": "2"
},
{
"input": "89 19\n14 77 85 81 79 38 91 45 55 51 50 11 62 67 73 76 2 27 16 23 3 29 65 98 78 17 4 58 22 20 34 66 64 31 72 5 32 44 12 75 80 47 18 25 99 0 61 56 71 84 48 88 10 7 86 8 49 24 43 21 37 28 33 54 46 57 40 89 36 97 6 96 39 95 26 74 1 69 9 100 52 30 83 87 68 60 92 90 35",
"output": "2"
},
{
"input": "89 100\n69 61 56 45 11 41 42 32 28 29 0 76 7 65 13 35 36 82 10 39 26 34 38 40 92 12 17 54 24 46 88 70 66 27 100 52 85 62 22 48 86 68 21 49 53 94 67 20 1 90 77 84 31 87 58 47 95 33 4 72 93 83 8 51 91 80 99 43 71 19 44 59 98 97 64 9 81 16 79 63 25 37 3 75 2 55 50 6 18",
"output": "13"
},
{
"input": "77 0\n38 76 24 74 42 88 29 75 96 46 90 32 59 97 98 60 41 57 80 37 100 49 25 63 95 31 61 68 53 78 27 66 84 48 94 83 30 26 36 99 71 62 45 47 70 28 35 54 34 85 79 43 91 72 86 33 67 92 77 65 69 52 82 55 87 64 56 40 50 44 51 73 89 81 58 93 39",
"output": "0"
},
{
"input": "89 100\n38 90 80 64 35 44 56 11 15 89 23 12 49 70 72 60 63 85 92 10 45 83 8 88 41 33 16 6 61 76 62 71 87 13 25 77 74 0 1 37 96 93 7 94 21 82 34 78 4 73 65 20 81 95 50 32 48 17 69 55 68 5 51 27 53 43 91 67 59 46 86 84 99 24 22 3 97 98 40 36 26 58 57 9 42 30 52 2 47",
"output": "11"
},
{
"input": "77 0\n55 71 78 86 68 35 53 10 59 32 81 19 74 97 62 61 93 87 96 44 25 18 43 82 84 16 34 48 92 39 64 36 49 91 45 76 95 31 57 29 75 79 13 2 14 24 52 23 33 20 47 99 63 15 5 80 58 67 12 3 85 6 1 27 73 90 4 42 37 70 8 11 89 77 9 22 94",
"output": "0"
},
{
"input": "77 0\n12 75 31 71 44 8 3 82 21 77 50 29 57 74 40 10 15 42 84 2 100 9 28 72 92 0 49 11 90 55 17 36 19 54 68 52 4 69 97 91 5 39 59 45 89 62 53 83 16 94 76 60 95 47 30 51 7 48 20 70 67 32 58 78 63 34 56 93 99 88 24 1 66 22 25 14 13",
"output": "1"
},
{
"input": "100 70\n91 82 8 85 26 25 95 97 40 87 81 93 7 73 38 94 64 96 74 18 90 19 65 68 72 61 23 43 36 41 60 88 30 33 71 24 52 39 15 3 16 89 86 79 55 4 9 58 67 44 46 29 6 48 84 69 27 21 78 54 51 57 80 53 76 50 47 77 45 12 34 10 100 0 17 31 56 99 98 11 92 5 2 42 32 59 66 62 37 63 28 75 35 1 22 13 83 49 20 14",
"output": "0"
},
{
"input": "77 0\n51 5 81 62 30 22 11 0 83 16 79 85 52 70 69 10 8 47 58 3 24 34 44 14 82 66 99 17 28 31 64 67 23 49 94 45 4 12 27 15 21 6 43 72 87 2 63 92 35 39 59 9 90 78 93 20 65 36 60 89 50 41 61 84 77 86 76 100 38 68 53 97 96 95 7 19 88",
"output": "1"
},
{
"input": "1 100\n0",
"output": "99"
},
{
"input": "1 0\n100",
"output": "0"
},
{
"input": "1 100\n100",
"output": "101"
},
{
"input": "2 100\n0 100",
"output": "100"
},
{
"input": "5 5\n1 2 3 4 5",
"output": "2"
},
{
"input": "5 3\n0 3 4 5 6",
"output": "3"
},
{
"input": "7 10\n0 1 2 3 4 5 10",
"output": "5"
},
{
"input": "2 2\n0 2",
"output": "2"
},
{
"input": "2 1\n1 2",
"output": "2"
},
{
"input": "5 1\n1 2 3 4 5",
"output": "2"
},
{
"input": "5 2\n1 2 3 4 5",
"output": "2"
},
{
"input": "5 6\n0 1 2 3 4",
"output": "1"
},
{
"input": "3 2\n3 4 5",
"output": "2"
}
] | 1,507,375,947
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 39
| 62
| 0
|
g,x = [int(i) for i in input().split()]
n = [int(i) for i in input().split()]
h = [False] * 101
for i in n:
h[i] = True
p = 0
for i in range(x):
if h[i] == False:
p+=1
if h[x] == True:
p+=1
print(p)
|
Title: Mahmoud and Ehab and the MEX
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dr. Evil kidnapped Mahmoud and Ehab in the evil land because of their performance in the Evil Olympiad in Informatics (EOI). He decided to give them some problems to let them go.
Dr. Evil is interested in sets, He has a set of *n* integers. Dr. Evil calls a set of integers evil if the MEX of it is exactly *x*. the MEX of a set of integers is the minimum non-negative integer that doesn't exist in it. For example, the MEX of the set {0,<=2,<=4} is 1 and the MEX of the set {1,<=2,<=3} is 0 .
Dr. Evil is going to make his set evil. To do this he can perform some operations. During each operation he can add some non-negative integer to his set or erase some element from it. What is the minimal number of operations Dr. Evil has to perform to make his set evil?
Input Specification:
The first line contains two integers *n* and *x* (1<=≤<=*n*<=≤<=100, 0<=≤<=*x*<=≤<=100) — the size of the set Dr. Evil owns, and the desired MEX.
The second line contains *n* distinct non-negative integers not exceeding 100 that represent the set.
Output Specification:
The only line should contain one integer — the minimal number of operations Dr. Evil should perform.
Demo Input:
['5 3\n0 4 5 6 7\n', '1 0\n0\n', '5 0\n1 2 3 4 5\n']
Demo Output:
['2\n', '1\n', '0\n']
Note:
For the first test case Dr. Evil should add 1 and 2 to the set performing 2 operations.
For the second test case Dr. Evil should erase 0 from the set. After that, the set becomes empty, so the MEX of it is 0.
In the third test case the set is already evil.
|
```python
g,x = [int(i) for i in input().split()]
n = [int(i) for i in input().split()]
h = [False] * 101
for i in n:
h[i] = True
p = 0
for i in range(x):
if h[i] == False:
p+=1
if h[x] == True:
p+=1
print(p)
```
| 3
|
|
49
|
A
|
Sleuth
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Sleuth
|
2
|
256
|
Vasya plays the sleuth with his friends. The rules of the game are as follows: those who play for the first time, that is Vasya is the sleuth, he should investigate a "crime" and find out what is happening. He can ask any questions whatsoever that can be answered with "Yes" or "No". All the rest agree beforehand to answer the questions like that: if the question’s last letter is a vowel, they answer "Yes" and if the last letter is a consonant, they answer "No". Of course, the sleuth knows nothing about it and his task is to understand that.
Unfortunately, Vasya is not very smart. After 5 hours of endless stupid questions everybody except Vasya got bored. That’s why Vasya’s friends ask you to write a program that would give answers instead of them.
The English alphabet vowels are: A, E, I, O, U, Y
The English alphabet consonants are: B, C, D, F, G, H, J, K, L, M, N, P, Q, R, S, T, V, W, X, Z
|
The single line contains a question represented by a non-empty line consisting of large and small Latin letters, spaces and a question mark. The line length does not exceed 100. It is guaranteed that the question mark occurs exactly once in the line — as the last symbol and that the line contains at least one letter.
|
Print answer for the question in a single line: YES if the answer is "Yes", NO if the answer is "No".
Remember that in the reply to the question the last letter, not the last character counts. I. e. the spaces and the question mark do not count as letters.
|
[
"Is it a melon?\n",
"Is it an apple?\n",
"Is it a banana ?\n",
"Is it an apple and a banana simultaneouSLY?\n"
] |
[
"NO\n",
"YES\n",
"YES\n",
"YES\n"
] |
none
| 500
|
[
{
"input": "Is it a melon?",
"output": "NO"
},
{
"input": "Is it an apple?",
"output": "YES"
},
{
"input": " Is it a banana ?",
"output": "YES"
},
{
"input": "Is it an apple and a banana simultaneouSLY?",
"output": "YES"
},
{
"input": "oHtSbDwzHb?",
"output": "NO"
},
{
"input": "sZecYdUvZHrXx?",
"output": "NO"
},
{
"input": "uMtXK?",
"output": "NO"
},
{
"input": "U?",
"output": "YES"
},
{
"input": "aqFDkCUKeHMyvZFcAyWlMUSQTFomtaWjoKLVyxLCw vcufPBFbaljOuHWiDCROYTcmbgzbaqHXKPOYEbuEtRqqoxBbOETCsQzhw?",
"output": "NO"
},
{
"input": "dJcNqQiFXzcbsj fItCpBLyXOnrSBPebwyFHlxUJHqCUzzCmcAvMiKL NunwOXnKeIxUZmBVwiCUfPkjRAkTPbkYCmwRRnDSLaz?",
"output": "NO"
},
{
"input": "gxzXbdcAQMuFKuuiPohtMgeypr wpDIoDSyOYTdvylcg SoEBZjnMHHYZGEqKgCgBeTbyTwyGuPZxkxsnSczotBdYyfcQsOVDVC?",
"output": "NO"
},
{
"input": "FQXBisXaJFMiHFQlXjixBDMaQuIbyqSBKGsBfTmBKCjszlGVZxEOqYYqRTUkGpSDDAoOXyXcQbHcPaegeOUBNeSD JiKOdECPOF?",
"output": "NO"
},
{
"input": "YhCuZnrWUBEed?",
"output": "NO"
},
{
"input": "hh?",
"output": "NO"
},
{
"input": "whU?",
"output": "YES"
},
{
"input": "fgwg?",
"output": "NO"
},
{
"input": "GlEmEPKrYcOnBNJUIFjszWUyVdvWw DGDjoCMtRJUburkPToCyDrOtMr?",
"output": "NO"
},
{
"input": "n?",
"output": "NO"
},
{
"input": "BueDOlxgzeNlxrzRrMbKiQdmGujEKmGxclvaPpTuHmTqBp?",
"output": "NO"
},
{
"input": "iehvZNQXDGCuVmJPOEysLyUryTdfaIxIuTzTadDbqRQGoCLXkxnyfWSGoLXebNnQQNTqAQJebbyYvHOfpUnXeWdjx?",
"output": "NO"
},
{
"input": " J ?",
"output": "NO"
},
{
"input": " j ?",
"output": "NO"
},
{
"input": " o ?",
"output": "YES"
},
{
"input": " T ?",
"output": "NO"
},
{
"input": " q ?",
"output": "NO"
},
{
"input": " j ?",
"output": "NO"
},
{
"input": " c ?",
"output": "NO"
},
{
"input": " B ?",
"output": "NO"
},
{
"input": "LuhxDHVwMPTtUIUMIQTuQETgXCOQPsfdFlyHvpfOVedjUTpGLAZGOHloIjJJtOLAlHPivzA?",
"output": "YES"
},
{
"input": "wmztmzFfwbGyOmNHENUFMTsFEMWYA?",
"output": "YES"
},
{
"input": "wGsfZCSwN PEUhNUrLfABrxA?",
"output": "YES"
},
{
"input": "mCDHENXjYbgMdBimAdPnewaHfpGWowjWrVAdvWczjw iDcUbyzMsmsnwbviiKiAyGVA?",
"output": "YES"
},
{
"input": "ARIWnwqFqxsQXsXXzHqvFjxOCttAGPUzDtWzsenPYdNXuFOIUGYZsLLK IaoxiyjBBRThoelwdPTkuCQfcBLUEJpCPIrVZlvUWA?",
"output": "YES"
},
{
"input": " PslvVpgpN BXkMFBEVXsyZFIQbBEFxGkYTeXKrOdcmhbiTUatYRUoYAayrchqbksswIlfIjerZPqptvCGnMUhyrQSvwltRhFzA?",
"output": "YES"
},
{
"input": "HpBkttwSjBXDmyleGiRWNUMPaAIE uzTrp KJDzaUiCdsMYOoWKHoUhWUoecCPmACymMUUbGav UMRpCytPETwNFAObZJA?",
"output": "YES"
}
] | 1,575,446,551
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 35
| 218
| 0
|
n = input()
n = n[:len(n)-1]
v = ['A','E','I','O','U','Y']
n = n.upper().strip()
if n[len(n)-1] in v:
print("YES")
else:
print("NO")
|
Title: Sleuth
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya plays the sleuth with his friends. The rules of the game are as follows: those who play for the first time, that is Vasya is the sleuth, he should investigate a "crime" and find out what is happening. He can ask any questions whatsoever that can be answered with "Yes" or "No". All the rest agree beforehand to answer the questions like that: if the question’s last letter is a vowel, they answer "Yes" and if the last letter is a consonant, they answer "No". Of course, the sleuth knows nothing about it and his task is to understand that.
Unfortunately, Vasya is not very smart. After 5 hours of endless stupid questions everybody except Vasya got bored. That’s why Vasya’s friends ask you to write a program that would give answers instead of them.
The English alphabet vowels are: A, E, I, O, U, Y
The English alphabet consonants are: B, C, D, F, G, H, J, K, L, M, N, P, Q, R, S, T, V, W, X, Z
Input Specification:
The single line contains a question represented by a non-empty line consisting of large and small Latin letters, spaces and a question mark. The line length does not exceed 100. It is guaranteed that the question mark occurs exactly once in the line — as the last symbol and that the line contains at least one letter.
Output Specification:
Print answer for the question in a single line: YES if the answer is "Yes", NO if the answer is "No".
Remember that in the reply to the question the last letter, not the last character counts. I. e. the spaces and the question mark do not count as letters.
Demo Input:
['Is it a melon?\n', 'Is it an apple?\n', 'Is it a banana ?\n', 'Is it an apple and a banana simultaneouSLY?\n']
Demo Output:
['NO\n', 'YES\n', 'YES\n', 'YES\n']
Note:
none
|
```python
n = input()
n = n[:len(n)-1]
v = ['A','E','I','O','U','Y']
n = n.upper().strip()
if n[len(n)-1] in v:
print("YES")
else:
print("NO")
```
| 3.9455
|
407
|
A
|
Triangle
|
PROGRAMMING
| 1,600
|
[
"brute force",
"geometry",
"implementation",
"math"
] | null | null |
There is a right triangle with legs of length *a* and *b*. Your task is to determine whether it is possible to locate the triangle on the plane in such a way that none of its sides is parallel to the coordinate axes. All the vertices must have integer coordinates. If there exists such a location, you have to output the appropriate coordinates of vertices.
|
The first line contains two integers *a*,<=*b* (1<=≤<=*a*,<=*b*<=≤<=1000), separated by a single space.
|
In the first line print either "YES" or "NO" (without the quotes) depending on whether the required location exists. If it does, print in the next three lines three pairs of integers — the coordinates of the triangle vertices, one pair per line. The coordinates must be integers, not exceeding 109 in their absolute value.
|
[
"1 1\n",
"5 5\n",
"5 10\n"
] |
[
"NO\n",
"YES\n2 1\n5 5\n-2 4\n",
"YES\n-10 4\n-2 -2\n1 2\n"
] |
none
| 500
|
[
{
"input": "1 1",
"output": "NO"
},
{
"input": "5 5",
"output": "YES\n2 1\n5 5\n-2 4"
},
{
"input": "5 10",
"output": "YES\n-10 4\n-2 -2\n1 2"
},
{
"input": "2 2",
"output": "NO"
},
{
"input": "5 6",
"output": "NO"
},
{
"input": "5 11",
"output": "NO"
},
{
"input": "10 15",
"output": "YES\n0 0\n6 8\n-12 9"
},
{
"input": "935 938",
"output": "NO"
},
{
"input": "999 1000",
"output": "NO"
},
{
"input": "1000 1000",
"output": "YES\n0 0\n280 960\n-960 280"
},
{
"input": "15 20",
"output": "YES\n0 0\n12 9\n-12 16"
},
{
"input": "20 15",
"output": "YES\n0 0\n12 16\n-12 9"
},
{
"input": "629 865",
"output": "NO"
},
{
"input": "45 872",
"output": "NO"
},
{
"input": "757 582",
"output": "NO"
},
{
"input": "173 588",
"output": "NO"
},
{
"input": "533 298",
"output": "NO"
},
{
"input": "949 360",
"output": "NO"
},
{
"input": "661 175",
"output": "NO"
},
{
"input": "728 299",
"output": "YES\n0 0\n280 672\n-276 115"
},
{
"input": "575 85",
"output": "YES\n0 0\n345 460\n-68 51"
},
{
"input": "385 505",
"output": "YES\n0 0\n231 308\n-404 303"
},
{
"input": "755 865",
"output": "YES\n0 0\n453 604\n-692 519"
},
{
"input": "395 55",
"output": "YES\n0 0\n237 316\n-44 33"
},
{
"input": "600 175",
"output": "YES\n0 0\n168 576\n-168 49"
},
{
"input": "280 210",
"output": "YES\n0 0\n168 224\n-168 126"
},
{
"input": "180 135",
"output": "YES\n0 0\n108 144\n-108 81"
},
{
"input": "140 105",
"output": "YES\n0 0\n84 112\n-84 63"
},
{
"input": "440 330",
"output": "YES\n0 0\n264 352\n-264 198"
},
{
"input": "130 312",
"output": "YES\n0 0\n120 50\n-120 288"
},
{
"input": "65 156",
"output": "YES\n0 0\n60 25\n-60 144"
},
{
"input": "105 140",
"output": "YES\n0 0\n84 63\n-84 112"
},
{
"input": "408 765",
"output": "YES\n0 0\n360 192\n-360 675"
},
{
"input": "195 468",
"output": "YES\n0 0\n180 75\n-180 432"
},
{
"input": "305 949",
"output": "NO"
},
{
"input": "80 60",
"output": "YES\n0 0\n48 64\n-48 36"
},
{
"input": "15 15",
"output": "YES\n0 0\n9 12\n-12 9"
},
{
"input": "120 90",
"output": "YES\n0 0\n72 96\n-72 54"
},
{
"input": "60 80",
"output": "YES\n0 0\n48 36\n-48 64"
},
{
"input": "5 25",
"output": "YES\n0 0\n3 4\n-20 15"
},
{
"input": "5 1000",
"output": "YES\n0 0\n3 4\n-800 600"
},
{
"input": "25 35",
"output": "YES\n0 0\n15 20\n-28 21"
},
{
"input": "25 5",
"output": "YES\n0 0\n15 20\n-4 3"
},
{
"input": "30 40",
"output": "YES\n0 0\n24 18\n-24 32"
},
{
"input": "65 156",
"output": "YES\n0 0\n60 25\n-60 144"
},
{
"input": "455 470",
"output": "YES\n0 0\n273 364\n-376 282"
},
{
"input": "90 120",
"output": "YES\n0 0\n72 54\n-72 96"
},
{
"input": "5 205",
"output": "YES\n0 0\n3 4\n-164 123"
},
{
"input": "111 111",
"output": "YES\n0 0\n36 105\n-105 36"
},
{
"input": "17 17",
"output": "YES\n0 0\n8 15\n-15 8"
},
{
"input": "1000 5",
"output": "YES\n0 0\n600 800\n-4 3"
}
] | 1,639,711,675
| 2,147,483,647
|
PyPy 3-64
|
MEMORY_LIMIT_EXCEEDED
|
TESTS
| 7
| 982
| 268,390,400
|
def solv(a,b):
n = max(a,b)
S = [i*i for i in range(1,n)]
D = {i:[] for i in range(1,4*n*n+1)}
for i in range(1,n):
for j in range(i,n):
D[i*i+j*j].append([i,j])
L1 = D[a*a]
L2 = D[b*b]
if not L1: return ["NO"]
if not L2: return ["NO"]
for a1,a2 in L1:
for b1,b2 in L2:
if a1*b2 == a2*b1:
return ["YES","0 0","%s %s"%(a1,a2),"%s %s"%(-b2,b1)]
return ["NO"]
a,b = map(int,input().split())
[print(s) for s in solv(a,b)]
|
Title: Triangle
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There is a right triangle with legs of length *a* and *b*. Your task is to determine whether it is possible to locate the triangle on the plane in such a way that none of its sides is parallel to the coordinate axes. All the vertices must have integer coordinates. If there exists such a location, you have to output the appropriate coordinates of vertices.
Input Specification:
The first line contains two integers *a*,<=*b* (1<=≤<=*a*,<=*b*<=≤<=1000), separated by a single space.
Output Specification:
In the first line print either "YES" or "NO" (without the quotes) depending on whether the required location exists. If it does, print in the next three lines three pairs of integers — the coordinates of the triangle vertices, one pair per line. The coordinates must be integers, not exceeding 109 in their absolute value.
Demo Input:
['1 1\n', '5 5\n', '5 10\n']
Demo Output:
['NO\n', 'YES\n2 1\n5 5\n-2 4\n', 'YES\n-10 4\n-2 -2\n1 2\n']
Note:
none
|
```python
def solv(a,b):
n = max(a,b)
S = [i*i for i in range(1,n)]
D = {i:[] for i in range(1,4*n*n+1)}
for i in range(1,n):
for j in range(i,n):
D[i*i+j*j].append([i,j])
L1 = D[a*a]
L2 = D[b*b]
if not L1: return ["NO"]
if not L2: return ["NO"]
for a1,a2 in L1:
for b1,b2 in L2:
if a1*b2 == a2*b1:
return ["YES","0 0","%s %s"%(a1,a2),"%s %s"%(-b2,b1)]
return ["NO"]
a,b = map(int,input().split())
[print(s) for s in solv(a,b)]
```
| 0
|
|
785
|
A
|
Anton and Polyhedrons
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] | null | null |
Anton's favourite geometric figures are regular polyhedrons. Note that there are five kinds of regular polyhedrons:
- Tetrahedron. Tetrahedron has 4 triangular faces. - Cube. Cube has 6 square faces. - Octahedron. Octahedron has 8 triangular faces. - Dodecahedron. Dodecahedron has 12 pentagonal faces. - Icosahedron. Icosahedron has 20 triangular faces.
All five kinds of polyhedrons are shown on the picture below:
Anton has a collection of *n* polyhedrons. One day he decided to know, how many faces his polyhedrons have in total. Help Anton and find this number!
|
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of polyhedrons in Anton's collection.
Each of the following *n* lines of the input contains a string *s**i* — the name of the *i*-th polyhedron in Anton's collection. The string can look like this:
- "Tetrahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a tetrahedron. - "Cube" (without quotes), if the *i*-th polyhedron in Anton's collection is a cube. - "Octahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an octahedron. - "Dodecahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a dodecahedron. - "Icosahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an icosahedron.
|
Output one number — the total number of faces in all the polyhedrons in Anton's collection.
|
[
"4\nIcosahedron\nCube\nTetrahedron\nDodecahedron\n",
"3\nDodecahedron\nOctahedron\nOctahedron\n"
] |
[
"42\n",
"28\n"
] |
In the first sample Anton has one icosahedron, one cube, one tetrahedron and one dodecahedron. Icosahedron has 20 faces, cube has 6 faces, tetrahedron has 4 faces and dodecahedron has 12 faces. In total, they have 20 + 6 + 4 + 12 = 42 faces.
| 500
|
[
{
"input": "4\nIcosahedron\nCube\nTetrahedron\nDodecahedron",
"output": "42"
},
{
"input": "3\nDodecahedron\nOctahedron\nOctahedron",
"output": "28"
},
{
"input": "25\nIcosahedron\nOctahedron\nTetrahedron\nDodecahedron\nCube\nIcosahedron\nOctahedron\nCube\nTetrahedron\nIcosahedron\nIcosahedron\nTetrahedron\nOctahedron\nDodecahedron\nIcosahedron\nOctahedron\nIcosahedron\nTetrahedron\nDodecahedron\nTetrahedron\nOctahedron\nCube\nCube\nDodecahedron\nTetrahedron",
"output": "256"
},
{
"input": "1\nTetrahedron",
"output": "4"
},
{
"input": "1\nCube",
"output": "6"
},
{
"input": "1\nOctahedron",
"output": "8"
},
{
"input": "1\nDodecahedron",
"output": "12"
},
{
"input": "1\nIcosahedron",
"output": "20"
},
{
"input": "28\nOctahedron\nDodecahedron\nOctahedron\nOctahedron\nDodecahedron\nIcosahedron\nIcosahedron\nDodecahedron\nDodecahedron\nDodecahedron\nCube\nDodecahedron\nCube\nTetrahedron\nCube\nCube\nTetrahedron\nDodecahedron\nDodecahedron\nDodecahedron\nIcosahedron\nIcosahedron\nDodecahedron\nIcosahedron\nDodecahedron\nDodecahedron\nIcosahedron\nIcosahedron",
"output": "340"
}
] | 1,681,062,813
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 30
| 1,045
| 9,216,000
|
Tetrahedron = 4
Cube = 6
Octahedron = 8
Dodecahedron = 12
Icosahedron = 20
a = 0
for i in range(int(input())):
b = input()
if b=="Tetrahedron":
a = a + 4
elif b=="Cube":
a = a + 6
elif b=="Octahedron":
a = a + 8
elif b=="Dodecahedron":
a = a + 12
elif b=="Icosahedron":
a = a + 20
print(a)
|
Title: Anton and Polyhedrons
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Anton's favourite geometric figures are regular polyhedrons. Note that there are five kinds of regular polyhedrons:
- Tetrahedron. Tetrahedron has 4 triangular faces. - Cube. Cube has 6 square faces. - Octahedron. Octahedron has 8 triangular faces. - Dodecahedron. Dodecahedron has 12 pentagonal faces. - Icosahedron. Icosahedron has 20 triangular faces.
All five kinds of polyhedrons are shown on the picture below:
Anton has a collection of *n* polyhedrons. One day he decided to know, how many faces his polyhedrons have in total. Help Anton and find this number!
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of polyhedrons in Anton's collection.
Each of the following *n* lines of the input contains a string *s**i* — the name of the *i*-th polyhedron in Anton's collection. The string can look like this:
- "Tetrahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a tetrahedron. - "Cube" (without quotes), if the *i*-th polyhedron in Anton's collection is a cube. - "Octahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an octahedron. - "Dodecahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a dodecahedron. - "Icosahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an icosahedron.
Output Specification:
Output one number — the total number of faces in all the polyhedrons in Anton's collection.
Demo Input:
['4\nIcosahedron\nCube\nTetrahedron\nDodecahedron\n', '3\nDodecahedron\nOctahedron\nOctahedron\n']
Demo Output:
['42\n', '28\n']
Note:
In the first sample Anton has one icosahedron, one cube, one tetrahedron and one dodecahedron. Icosahedron has 20 faces, cube has 6 faces, tetrahedron has 4 faces and dodecahedron has 12 faces. In total, they have 20 + 6 + 4 + 12 = 42 faces.
|
```python
Tetrahedron = 4
Cube = 6
Octahedron = 8
Dodecahedron = 12
Icosahedron = 20
a = 0
for i in range(int(input())):
b = input()
if b=="Tetrahedron":
a = a + 4
elif b=="Cube":
a = a + 6
elif b=="Octahedron":
a = a + 8
elif b=="Dodecahedron":
a = a + 12
elif b=="Icosahedron":
a = a + 20
print(a)
```
| 3
|
|
883
|
A
|
Automatic Door
|
PROGRAMMING
| 2,200
|
[
"implementation"
] | null | null |
There is an automatic door at the entrance of a factory. The door works in the following way:
- when one or several people come to the door and it is closed, the door immediately opens automatically and all people immediately come inside, - when one or several people come to the door and it is open, all people immediately come inside, - opened door immediately closes in *d* seconds after its opening, - if the door is closing and one or several people are coming to the door at the same moment, then all of them will have enough time to enter and only after that the door will close.
For example, if *d*<==<=3 and four people are coming at four different moments of time *t*1<==<=4, *t*2<==<=7, *t*3<==<=9 and *t*4<==<=13 then the door will open three times: at moments 4, 9 and 13. It will close at moments 7 and 12.
It is known that *n* employees will enter at moments *a*,<=2·*a*,<=3·*a*,<=...,<=*n*·*a* (the value *a* is positive integer). Also *m* clients will enter at moments *t*1,<=*t*2,<=...,<=*t**m*.
Write program to find the number of times the automatic door will open. Assume that the door is initially closed.
|
The first line contains four integers *n*, *m*, *a* and *d* (1<=≤<=*n*,<=*a*<=≤<=109, 1<=≤<=*m*<=≤<=105, 1<=≤<=*d*<=≤<=1018) — the number of the employees, the number of the clients, the moment of time when the first employee will come and the period of time in which the door closes.
The second line contains integer sequence *t*1,<=*t*2,<=...,<=*t**m* (1<=≤<=*t**i*<=≤<=1018) — moments of time when clients will come. The values *t**i* are given in non-decreasing order.
|
Print the number of times the door will open.
|
[
"1 1 3 4\n7\n",
"4 3 4 2\n7 9 11\n"
] |
[
"1\n",
"4\n"
] |
In the first example the only employee will come at moment 3. At this moment the door will open and will stay open until the moment 7. At the same moment of time the client will come, so at first he will enter and only after it the door will close. Thus the door will open one time.
| 0
|
[
{
"input": "1 1 3 4\n7",
"output": "1"
},
{
"input": "4 3 4 2\n7 9 11",
"output": "4"
},
{
"input": "10 10 51 69\n154 170 170 183 251 337 412 426 445 452",
"output": "6"
},
{
"input": "70 10 26 17\n361 371 579 585 629 872 944 1017 1048 1541",
"output": "70"
},
{
"input": "100 20 49 52\n224 380 690 1585 1830 1973 2490 2592 3240 3341 3406 3429 3549 3560 3895 3944 4344 4390 4649 4800",
"output": "55"
},
{
"input": "100 30 36 47\n44 155 275 390 464 532 1186 1205 1345 1349 1432 1469 1482 1775 1832 1856 1869 2049 2079 2095 2374 2427 2577 2655 2792 2976 3020 3317 3482 3582",
"output": "51"
},
{
"input": "97 60 1 1\n5 6 6 7 9 10 10 11 11 11 12 13 13 13 13 14 14 15 16 18 20 23 23 24 25 26 29 31 32 35 38 41 43 43 46 47 48 48 49 52 53 54 55 56 58 59 68 70 72 74 78 81 81 82 91 92 96 96 97 98",
"output": "49"
},
{
"input": "1000000000 1 157 468\n57575875712",
"output": "333333334"
},
{
"input": "1000000000 1 1000000000 1000000000000000000\n1000000000000000000",
"output": "1"
}
] | 1,650,615,065
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 46
| 0
|
import sys
import math
from bisect import *
def read_input(input_path=None):
if input_path is None:
f = sys.stdin
else:
f = open(input_path, 'r')
n, m, a, d = map(int, f.readline().split())
t = list(map(int, f.readline().split()))
return n, m, a, d, t
def sol(n, m, a, d, t):
insort(t, a*n)
pred = 0
k = 0
n = 0
step = d // a + 1
answer = 0
fl = 0
for i in t:
if i > pred:
if fl == 0:
n = (i - pred + (pred % a)) // a
if n != 0:
k += (n // step) * step - step * (n % step == 0) + 1
if k > n:
k = n
fl = 1
if (k * a + d >= i) and (n != 0):
pred = k * a + d
else:
pred = i + d
k = math.floor(pred // a)
answer += 1
k = min(math.floor(pred // a), n)
answer += n // step + (n % step != 0)
else:
answer += 1
pred = i + d
if i == a * n:
fl = 1
return [f"{answer}"]
def solve(input_path=None):
return sol(*read_input(input_path))
def main():
for line in sol(*read_input()):
print(f"{line}")
if __name__ == '__main__':
main()
|
Title: Automatic Door
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There is an automatic door at the entrance of a factory. The door works in the following way:
- when one or several people come to the door and it is closed, the door immediately opens automatically and all people immediately come inside, - when one or several people come to the door and it is open, all people immediately come inside, - opened door immediately closes in *d* seconds after its opening, - if the door is closing and one or several people are coming to the door at the same moment, then all of them will have enough time to enter and only after that the door will close.
For example, if *d*<==<=3 and four people are coming at four different moments of time *t*1<==<=4, *t*2<==<=7, *t*3<==<=9 and *t*4<==<=13 then the door will open three times: at moments 4, 9 and 13. It will close at moments 7 and 12.
It is known that *n* employees will enter at moments *a*,<=2·*a*,<=3·*a*,<=...,<=*n*·*a* (the value *a* is positive integer). Also *m* clients will enter at moments *t*1,<=*t*2,<=...,<=*t**m*.
Write program to find the number of times the automatic door will open. Assume that the door is initially closed.
Input Specification:
The first line contains four integers *n*, *m*, *a* and *d* (1<=≤<=*n*,<=*a*<=≤<=109, 1<=≤<=*m*<=≤<=105, 1<=≤<=*d*<=≤<=1018) — the number of the employees, the number of the clients, the moment of time when the first employee will come and the period of time in which the door closes.
The second line contains integer sequence *t*1,<=*t*2,<=...,<=*t**m* (1<=≤<=*t**i*<=≤<=1018) — moments of time when clients will come. The values *t**i* are given in non-decreasing order.
Output Specification:
Print the number of times the door will open.
Demo Input:
['1 1 3 4\n7\n', '4 3 4 2\n7 9 11\n']
Demo Output:
['1\n', '4\n']
Note:
In the first example the only employee will come at moment 3. At this moment the door will open and will stay open until the moment 7. At the same moment of time the client will come, so at first he will enter and only after it the door will close. Thus the door will open one time.
|
```python
import sys
import math
from bisect import *
def read_input(input_path=None):
if input_path is None:
f = sys.stdin
else:
f = open(input_path, 'r')
n, m, a, d = map(int, f.readline().split())
t = list(map(int, f.readline().split()))
return n, m, a, d, t
def sol(n, m, a, d, t):
insort(t, a*n)
pred = 0
k = 0
n = 0
step = d // a + 1
answer = 0
fl = 0
for i in t:
if i > pred:
if fl == 0:
n = (i - pred + (pred % a)) // a
if n != 0:
k += (n // step) * step - step * (n % step == 0) + 1
if k > n:
k = n
fl = 1
if (k * a + d >= i) and (n != 0):
pred = k * a + d
else:
pred = i + d
k = math.floor(pred // a)
answer += 1
k = min(math.floor(pred // a), n)
answer += n // step + (n % step != 0)
else:
answer += 1
pred = i + d
if i == a * n:
fl = 1
return [f"{answer}"]
def solve(input_path=None):
return sol(*read_input(input_path))
def main():
for line in sol(*read_input()):
print(f"{line}")
if __name__ == '__main__':
main()
```
| 0
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
PMP is getting a warrior. He is practicing a lot, but the results are not acceptable yet. This time instead of programming contests, he decided to compete in a car racing to increase the spirit of victory. He decides to choose a competition that also exhibits algorithmic features.
AlgoRace is a special league of car racing where different teams compete in a country of *n* cities. Cities are numbered 1 through *n*. Every two distinct cities in the country are connected with one bidirectional road. Each competing team should introduce one driver and a set of cars.
The competition is held in *r* rounds. In *i*-th round, drivers will start at city *s**i* and finish at city *t**i*. Drivers are allowed to change their cars at most *k**i* times. Changing cars can take place in any city in no time. One car can be used multiple times in one round, but total number of changes should not exceed *k**i*. Drivers can freely choose their path to destination.
PMP has prepared *m* type of purpose-built cars. Beside for PMP’s driving skills, depending on properties of the car and the road, a car traverses each road in each direction in different times.
PMP Warriors wants to devise best strategies of choosing car and roads in each round to maximize the chance of winning the cup. For each round they want to find the minimum time required to finish it.
|
The first line contains three space-separated integers *n*,<=*m*,<=*r* (2<=≤<=*n*<=≤<=60,<=1<=≤<=*m*<=≤<=60,<=1<=≤<=*r*<=≤<=105) — the number of cities, the number of different types of cars and the number of rounds in the competition, correspondingly.
Next *m* sets of *n*<=×<=*n* matrices of integers between 0 to 106 (inclusive) will follow — describing the time one car requires to traverse different roads. The *k*-th integer in *j*-th line of the *i*-th set is the time that *i*-th car requires to traverse the road from *j*-th city to *k*-th city. These matrices are not necessarily symmetric, but their diagonal is always zero.
Next *r* lines contain description of the rounds. The *i*-th of these lines contains space-separated integers *s**i*,<=*t**i*,<=*k**i* (1<=≤<=*s**i*,<=*t**i*<=≤<=*n*,<=*s**i*<=≠<=*t**i*,<=0<=≤<=*k**i*<=≤<=1000) — the number of starting city, finishing city and the number of possible car changes in *i*-th round, correspondingly.
|
For each round you should print the minimum required time to complete the round in a single line.
|
[
"4 2 3\n0 1 5 6\n2 0 3 6\n1 3 0 1\n6 6 7 0\n0 3 5 6\n2 0 1 6\n1 3 0 2\n6 6 7 0\n1 4 2\n1 4 1\n1 4 3\n",
"4 2 3\n0 7 3 3\n8 0 10 5\n1 1 0 4\n8 9 2 0\n0 3 3 9\n7 0 4 9\n3 8 0 4\n4 8 9 0\n2 3 3\n2 1 3\n1 2 2\n"
] |
[
"3\n4\n3\n",
"4\n5\n3\n"
] |
In the first sample, in all rounds PMP goes from city #1 to city #2, then city #3 and finally city #4. But the sequences of types of the cars he uses are (1, 2, 1) in the first round and (1, 2, 2) in the second round. In the third round, although he can change his car three times, he uses the same strategy as the first round which only needs two car changes.
| 0
|
[] | 1,689,194,960
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 62
| 0
|
print("_RANDOM_GUESS_1689194960.401511")# 1689194960.401546
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
PMP is getting a warrior. He is practicing a lot, but the results are not acceptable yet. This time instead of programming contests, he decided to compete in a car racing to increase the spirit of victory. He decides to choose a competition that also exhibits algorithmic features.
AlgoRace is a special league of car racing where different teams compete in a country of *n* cities. Cities are numbered 1 through *n*. Every two distinct cities in the country are connected with one bidirectional road. Each competing team should introduce one driver and a set of cars.
The competition is held in *r* rounds. In *i*-th round, drivers will start at city *s**i* and finish at city *t**i*. Drivers are allowed to change their cars at most *k**i* times. Changing cars can take place in any city in no time. One car can be used multiple times in one round, but total number of changes should not exceed *k**i*. Drivers can freely choose their path to destination.
PMP has prepared *m* type of purpose-built cars. Beside for PMP’s driving skills, depending on properties of the car and the road, a car traverses each road in each direction in different times.
PMP Warriors wants to devise best strategies of choosing car and roads in each round to maximize the chance of winning the cup. For each round they want to find the minimum time required to finish it.
Input Specification:
The first line contains three space-separated integers *n*,<=*m*,<=*r* (2<=≤<=*n*<=≤<=60,<=1<=≤<=*m*<=≤<=60,<=1<=≤<=*r*<=≤<=105) — the number of cities, the number of different types of cars and the number of rounds in the competition, correspondingly.
Next *m* sets of *n*<=×<=*n* matrices of integers between 0 to 106 (inclusive) will follow — describing the time one car requires to traverse different roads. The *k*-th integer in *j*-th line of the *i*-th set is the time that *i*-th car requires to traverse the road from *j*-th city to *k*-th city. These matrices are not necessarily symmetric, but their diagonal is always zero.
Next *r* lines contain description of the rounds. The *i*-th of these lines contains space-separated integers *s**i*,<=*t**i*,<=*k**i* (1<=≤<=*s**i*,<=*t**i*<=≤<=*n*,<=*s**i*<=≠<=*t**i*,<=0<=≤<=*k**i*<=≤<=1000) — the number of starting city, finishing city and the number of possible car changes in *i*-th round, correspondingly.
Output Specification:
For each round you should print the minimum required time to complete the round in a single line.
Demo Input:
['4 2 3\n0 1 5 6\n2 0 3 6\n1 3 0 1\n6 6 7 0\n0 3 5 6\n2 0 1 6\n1 3 0 2\n6 6 7 0\n1 4 2\n1 4 1\n1 4 3\n', '4 2 3\n0 7 3 3\n8 0 10 5\n1 1 0 4\n8 9 2 0\n0 3 3 9\n7 0 4 9\n3 8 0 4\n4 8 9 0\n2 3 3\n2 1 3\n1 2 2\n']
Demo Output:
['3\n4\n3\n', '4\n5\n3\n']
Note:
In the first sample, in all rounds PMP goes from city #1 to city #2, then city #3 and finally city #4. But the sequences of types of the cars he uses are (1, 2, 1) in the first round and (1, 2, 2) in the second round. In the third round, although he can change his car three times, he uses the same strategy as the first round which only needs two car changes.
|
```python
print("_RANDOM_GUESS_1689194960.401511")# 1689194960.401546
```
| 0
|
|
23
|
B
|
Party
|
PROGRAMMING
| 1,600
|
[
"constructive algorithms",
"graphs",
"math"
] |
B. Party
|
2
|
256
|
*n* people came to a party. Then those, who had no friends among people at the party, left. Then those, who had exactly 1 friend among those who stayed, left as well. Then those, who had exactly 2,<=3,<=...,<=*n*<=-<=1 friends among those who stayed by the moment of their leaving, did the same.
What is the maximum amount of people that could stay at the party in the end?
|
The first input line contains one number *t* — amount of tests (1<=≤<=*t*<=≤<=105). Each of the following *t* lines contains one integer number *n* (1<=≤<=*n*<=≤<=105).
|
For each test output in a separate line one number — the maximum amount of people that could stay in the end.
|
[
"1\n3\n"
] |
[
"1\n"
] |
none
| 0
|
[
{
"input": "1\n3",
"output": "1"
}
] | 1,452,084,615
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 1
| 2,000
| 0
|
n = int(input())
for i in range(n):
t = int(input())
print(max(0, t - 2))
|
Title: Party
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
*n* people came to a party. Then those, who had no friends among people at the party, left. Then those, who had exactly 1 friend among those who stayed, left as well. Then those, who had exactly 2,<=3,<=...,<=*n*<=-<=1 friends among those who stayed by the moment of their leaving, did the same.
What is the maximum amount of people that could stay at the party in the end?
Input Specification:
The first input line contains one number *t* — amount of tests (1<=≤<=*t*<=≤<=105). Each of the following *t* lines contains one integer number *n* (1<=≤<=*n*<=≤<=105).
Output Specification:
For each test output in a separate line one number — the maximum amount of people that could stay in the end.
Demo Input:
['1\n3\n']
Demo Output:
['1\n']
Note:
none
|
```python
n = int(input())
for i in range(n):
t = int(input())
print(max(0, t - 2))
```
| 0
|
467
|
A
|
George and Accommodation
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
George has recently entered the BSUCP (Berland State University for Cool Programmers). George has a friend Alex who has also entered the university. Now they are moving into a dormitory.
George and Alex want to live in the same room. The dormitory has *n* rooms in total. At the moment the *i*-th room has *p**i* people living in it and the room can accommodate *q**i* people in total (*p**i*<=≤<=*q**i*). Your task is to count how many rooms has free place for both George and Alex.
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of rooms.
The *i*-th of the next *n* lines contains two integers *p**i* and *q**i* (0<=≤<=*p**i*<=≤<=*q**i*<=≤<=100) — the number of people who already live in the *i*-th room and the room's capacity.
|
Print a single integer — the number of rooms where George and Alex can move in.
|
[
"3\n1 1\n2 2\n3 3\n",
"3\n1 10\n0 10\n10 10\n"
] |
[
"0\n",
"2\n"
] |
none
| 500
|
[
{
"input": "3\n1 1\n2 2\n3 3",
"output": "0"
},
{
"input": "3\n1 10\n0 10\n10 10",
"output": "2"
},
{
"input": "2\n36 67\n61 69",
"output": "2"
},
{
"input": "3\n21 71\n10 88\n43 62",
"output": "3"
},
{
"input": "3\n1 2\n2 3\n3 4",
"output": "0"
},
{
"input": "10\n0 10\n0 20\n0 30\n0 40\n0 50\n0 60\n0 70\n0 80\n0 90\n0 100",
"output": "10"
},
{
"input": "13\n14 16\n30 31\n45 46\n19 20\n15 17\n66 67\n75 76\n95 97\n29 30\n37 38\n0 2\n36 37\n8 9",
"output": "4"
},
{
"input": "19\n66 67\n97 98\n89 91\n67 69\n67 68\n18 20\n72 74\n28 30\n91 92\n27 28\n75 77\n17 18\n74 75\n28 30\n16 18\n90 92\n9 11\n22 24\n52 54",
"output": "12"
},
{
"input": "15\n55 57\n95 97\n57 59\n34 36\n50 52\n96 98\n39 40\n13 15\n13 14\n74 76\n47 48\n56 58\n24 25\n11 13\n67 68",
"output": "10"
},
{
"input": "17\n68 69\n47 48\n30 31\n52 54\n41 43\n33 35\n38 40\n56 58\n45 46\n92 93\n73 74\n61 63\n65 66\n37 39\n67 68\n77 78\n28 30",
"output": "8"
},
{
"input": "14\n64 66\n43 44\n10 12\n76 77\n11 12\n25 27\n87 88\n62 64\n39 41\n58 60\n10 11\n28 29\n57 58\n12 14",
"output": "7"
},
{
"input": "38\n74 76\n52 54\n78 80\n48 49\n40 41\n64 65\n28 30\n6 8\n49 51\n68 70\n44 45\n57 59\n24 25\n46 48\n49 51\n4 6\n63 64\n76 78\n57 59\n18 20\n63 64\n71 73\n88 90\n21 22\n89 90\n65 66\n89 91\n96 98\n42 44\n1 1\n74 76\n72 74\n39 40\n75 76\n29 30\n48 49\n87 89\n27 28",
"output": "22"
},
{
"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": "26\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2",
"output": "0"
},
{
"input": "68\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2",
"output": "68"
},
{
"input": "7\n0 1\n1 5\n2 4\n3 5\n4 6\n5 6\n6 8",
"output": "5"
},
{
"input": "1\n0 0",
"output": "0"
},
{
"input": "1\n100 100",
"output": "0"
},
{
"input": "44\n0 8\n1 11\n2 19\n3 5\n4 29\n5 45\n6 6\n7 40\n8 19\n9 22\n10 18\n11 26\n12 46\n13 13\n14 27\n15 48\n16 25\n17 20\n18 29\n19 27\n20 45\n21 39\n22 29\n23 39\n24 42\n25 37\n26 52\n27 36\n28 43\n29 35\n30 38\n31 70\n32 47\n33 38\n34 61\n35 71\n36 51\n37 71\n38 59\n39 77\n40 70\n41 80\n42 77\n43 73",
"output": "42"
},
{
"input": "3\n1 3\n2 7\n8 9",
"output": "2"
},
{
"input": "53\n0 1\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 19\n19 20\n20 21\n21 22\n22 23\n23 24\n24 25\n25 26\n26 27\n27 28\n28 29\n29 30\n30 31\n31 32\n32 33\n33 34\n34 35\n35 36\n36 37\n37 38\n38 39\n39 40\n40 41\n41 42\n42 43\n43 44\n44 45\n45 46\n46 47\n47 48\n48 49\n49 50\n50 51\n51 52\n52 53",
"output": "0"
},
{
"input": "55\n0 0\n1 1\n2 2\n3 3\n4 4\n5 5\n6 6\n7 7\n8 8\n9 9\n10 10\n11 11\n12 12\n13 13\n14 14\n15 15\n16 16\n17 17\n18 18\n19 19\n20 20\n21 21\n22 22\n23 23\n24 24\n25 25\n26 26\n27 27\n28 28\n29 29\n30 30\n31 31\n32 32\n33 33\n34 34\n35 35\n36 36\n37 37\n38 38\n39 39\n40 40\n41 41\n42 42\n43 43\n44 44\n45 45\n46 46\n47 47\n48 48\n49 49\n50 50\n51 51\n52 52\n53 53\n54 54",
"output": "0"
},
{
"input": "51\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 62\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 73\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 82\n55 68\n55 70\n55 63\n55 55\n55 55\n55 55\n55 75\n55 75\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 73\n55 55\n55 82\n55 99\n55 60",
"output": "12"
},
{
"input": "14\n1 1\n1 1\n1 55\n1 16\n1 1\n1 1\n1 55\n1 62\n1 53\n1 26\n1 1\n1 36\n1 2\n1 3",
"output": "8"
}
] | 1,693,139,241
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 31
| 0
|
n = int(input())
res = 0
for i in range(n):
curr, cap = [int(x) for x in input().split()]
if(curr < cap): res += 1
print(res)
|
Title: George and Accommodation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
George has recently entered the BSUCP (Berland State University for Cool Programmers). George has a friend Alex who has also entered the university. Now they are moving into a dormitory.
George and Alex want to live in the same room. The dormitory has *n* rooms in total. At the moment the *i*-th room has *p**i* people living in it and the room can accommodate *q**i* people in total (*p**i*<=≤<=*q**i*). Your task is to count how many rooms has free place for both George and Alex.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of rooms.
The *i*-th of the next *n* lines contains two integers *p**i* and *q**i* (0<=≤<=*p**i*<=≤<=*q**i*<=≤<=100) — the number of people who already live in the *i*-th room and the room's capacity.
Output Specification:
Print a single integer — the number of rooms where George and Alex can move in.
Demo Input:
['3\n1 1\n2 2\n3 3\n', '3\n1 10\n0 10\n10 10\n']
Demo Output:
['0\n', '2\n']
Note:
none
|
```python
n = int(input())
res = 0
for i in range(n):
curr, cap = [int(x) for x in input().split()]
if(curr < cap): res += 1
print(res)
```
| 0
|
|
946
|
B
|
Weird Subtraction Process
|
PROGRAMMING
| 1,100
|
[
"math",
"number theory"
] | null | null |
You have two variables *a* and *b*. Consider the following sequence of actions performed with these variables:
1. If *a*<==<=0 or *b*<==<=0, end the process. Otherwise, go to step 2;1. If *a*<=≥<=2·*b*, then set the value of *a* to *a*<=-<=2·*b*, and repeat step 1. Otherwise, go to step 3;1. If *b*<=≥<=2·*a*, then set the value of *b* to *b*<=-<=2·*a*, and repeat step 1. Otherwise, end the process.
Initially the values of *a* and *b* are positive integers, and so the process will be finite.
You have to determine the values of *a* and *b* after the process ends.
|
The only line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1018). *n* is the initial value of variable *a*, and *m* is the initial value of variable *b*.
|
Print two integers — the values of *a* and *b* after the end of the process.
|
[
"12 5\n",
"31 12\n"
] |
[
"0 1\n",
"7 12\n"
] |
Explanations to the samples:
1. *a* = 12, *b* = 5 <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> *a* = 2, *b* = 5 <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> *a* = 2, *b* = 1 <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> *a* = 0, *b* = 1;1. *a* = 31, *b* = 12 <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> *a* = 7, *b* = 12.
| 0
|
[
{
"input": "12 5",
"output": "0 1"
},
{
"input": "31 12",
"output": "7 12"
},
{
"input": "1000000000000000000 7",
"output": "8 7"
},
{
"input": "31960284556200 8515664064180",
"output": "14928956427840 8515664064180"
},
{
"input": "1000000000000000000 1000000000000000000",
"output": "1000000000000000000 1000000000000000000"
},
{
"input": "1 1000",
"output": "1 0"
},
{
"input": "1 1000000",
"output": "1 0"
},
{
"input": "1 1000000000000000",
"output": "1 0"
},
{
"input": "1 99999999999999999",
"output": "1 1"
},
{
"input": "1 4",
"output": "1 0"
},
{
"input": "1000000000000001 500000000000000",
"output": "1 0"
},
{
"input": "1 1000000000000000000",
"output": "1 0"
},
{
"input": "2 4",
"output": "2 0"
},
{
"input": "2 1",
"output": "0 1"
},
{
"input": "6 19",
"output": "6 7"
},
{
"input": "22 5",
"output": "0 1"
},
{
"input": "10000000000000000 100000000000000001",
"output": "0 1"
},
{
"input": "1 1000000000000",
"output": "1 0"
},
{
"input": "2 1000000000000000",
"output": "2 0"
},
{
"input": "2 10",
"output": "2 2"
},
{
"input": "51 100",
"output": "51 100"
},
{
"input": "3 1000000000000000000",
"output": "3 4"
},
{
"input": "1000000000000000000 3",
"output": "4 3"
},
{
"input": "1 10000000000000000",
"output": "1 0"
},
{
"input": "8796203 7556",
"output": "1019 1442"
},
{
"input": "5 22",
"output": "1 0"
},
{
"input": "1000000000000000000 1",
"output": "0 1"
},
{
"input": "1 100000000000",
"output": "1 0"
},
{
"input": "2 1000000000000",
"output": "2 0"
},
{
"input": "5 4567865432345678",
"output": "5 8"
},
{
"input": "576460752303423487 288230376151711743",
"output": "1 1"
},
{
"input": "499999999999999999 1000000000000000000",
"output": "3 2"
},
{
"input": "1 9999999999999",
"output": "1 1"
},
{
"input": "103 1000000000000000000",
"output": "103 196"
},
{
"input": "7 1",
"output": "1 1"
},
{
"input": "100000000000000001 10000000000000000",
"output": "1 0"
},
{
"input": "5 10",
"output": "5 0"
},
{
"input": "7 11",
"output": "7 11"
},
{
"input": "1 123456789123456",
"output": "1 0"
},
{
"input": "5000000000000 100000000000001",
"output": "0 1"
},
{
"input": "1000000000000000 1",
"output": "0 1"
},
{
"input": "1000000000000000000 499999999999999999",
"output": "2 3"
},
{
"input": "10 5",
"output": "0 5"
},
{
"input": "9 18917827189272",
"output": "9 0"
},
{
"input": "179 100000000000497000",
"output": "179 270"
},
{
"input": "5 100000000000001",
"output": "1 1"
},
{
"input": "5 20",
"output": "5 0"
},
{
"input": "100000001 50000000",
"output": "1 0"
},
{
"input": "345869461223138161 835002744095575440",
"output": "1 0"
},
{
"input": "8589934592 4294967296",
"output": "0 4294967296"
},
{
"input": "4 8",
"output": "4 0"
},
{
"input": "1 100000000000000000",
"output": "1 0"
},
{
"input": "1000000000000000000 333333333333333",
"output": "1000 1333"
},
{
"input": "25 12",
"output": "1 0"
},
{
"input": "24 54",
"output": "0 6"
},
{
"input": "6 12",
"output": "6 0"
},
{
"input": "129200000000305 547300000001292",
"output": "1 0"
},
{
"input": "1000000000000000000 49999999999999999",
"output": "20 39"
},
{
"input": "1 2",
"output": "1 0"
},
{
"input": "1 123456789876",
"output": "1 0"
},
{
"input": "2 3",
"output": "2 3"
},
{
"input": "1 3",
"output": "1 1"
},
{
"input": "1 1",
"output": "1 1"
},
{
"input": "19 46",
"output": "3 2"
},
{
"input": "3 6",
"output": "3 0"
},
{
"input": "129 1000000000000000000",
"output": "1 0"
},
{
"input": "12 29",
"output": "0 1"
},
{
"input": "8589934592 2147483648",
"output": "0 2147483648"
},
{
"input": "2147483648 8589934592",
"output": "2147483648 0"
},
{
"input": "5 6",
"output": "5 6"
},
{
"input": "1000000000000000000 2",
"output": "0 2"
},
{
"input": "2 7",
"output": "2 3"
},
{
"input": "17174219820754872 61797504734333370",
"output": "17174219820754872 27449065092823626"
},
{
"input": "49 100",
"output": "1 0"
},
{
"input": "7 17",
"output": "1 1"
},
{
"input": "1000000000000000000 10000001",
"output": "0 1"
},
{
"input": "49999999999999999 2",
"output": "3 2"
},
{
"input": "49999999999999999 1",
"output": "1 1"
},
{
"input": "576460752303423487 2",
"output": "3 2"
},
{
"input": "19395 19395",
"output": "19395 19395"
},
{
"input": "19394 19394",
"output": "19394 19394"
}
] | 1,548,421,767
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 77
| 0
|
a,b=map(int,input().split())
for i in range(0,10000):
if (a == 0 or b == 0):
break
elif (a >= 2 * b):
a = a - 2 * b
elif (b >= 2 * a):
b = b - 2 * a
else:
break
print(a,b)
|
Title: Weird Subtraction Process
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have two variables *a* and *b*. Consider the following sequence of actions performed with these variables:
1. If *a*<==<=0 or *b*<==<=0, end the process. Otherwise, go to step 2;1. If *a*<=≥<=2·*b*, then set the value of *a* to *a*<=-<=2·*b*, and repeat step 1. Otherwise, go to step 3;1. If *b*<=≥<=2·*a*, then set the value of *b* to *b*<=-<=2·*a*, and repeat step 1. Otherwise, end the process.
Initially the values of *a* and *b* are positive integers, and so the process will be finite.
You have to determine the values of *a* and *b* after the process ends.
Input Specification:
The only line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1018). *n* is the initial value of variable *a*, and *m* is the initial value of variable *b*.
Output Specification:
Print two integers — the values of *a* and *b* after the end of the process.
Demo Input:
['12 5\n', '31 12\n']
Demo Output:
['0 1\n', '7 12\n']
Note:
Explanations to the samples:
1. *a* = 12, *b* = 5 <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> *a* = 2, *b* = 5 <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> *a* = 2, *b* = 1 <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> *a* = 0, *b* = 1;1. *a* = 31, *b* = 12 <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> *a* = 7, *b* = 12.
|
```python
a,b=map(int,input().split())
for i in range(0,10000):
if (a == 0 or b == 0):
break
elif (a >= 2 * b):
a = a - 2 * b
elif (b >= 2 * a):
b = b - 2 * a
else:
break
print(a,b)
```
| 0
|
|
771
|
E
|
Bear and Rectangle Strips
|
PROGRAMMING
| 3,000
|
[
"dp",
"greedy"
] | null | null |
Limak has a grid that consists of 2 rows and *n* columns. The *j*-th cell in the *i*-th row contains an integer *t**i*,<=*j* which can be positive, negative or zero.
A non-empty rectangle of cells is called nice if and only if the sum of numbers in its cells is equal to 0.
Limak wants to choose some nice rectangles and give them to his friends, as gifts. No two chosen rectangles should share a cell. What is the maximum possible number of nice rectangles Limak can choose?
|
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=300<=000) — the number of columns in the grid.
The next two lines contain numbers in the grid. The *i*-th of those two lines contains *n* integers *t**i*,<=1,<=*t**i*,<=2,<=...,<=*t**i*,<=*n* (<=-<=109<=≤<=*t**i*,<=*j*<=≤<=109).
|
Print one integer, denoting the maximum possible number of cell-disjoint nice rectangles.
|
[
"6\n70 70 70 70 70 -15\n90 -60 -30 30 -30 15\n",
"4\n0 -1 0 0\n0 0 1 0\n",
"3\n1000000000 999999999 -1000000000\n999999999 -1000000000 -999999998\n"
] |
[
"3\n",
"6\n",
"1\n"
] |
In the first sample, there are four nice rectangles:
Limak can't choose all of them because they are not disjoint. He should take three nice rectangles: those denoted as blue frames on the drawings.
In the second sample, it's optimal to choose six nice rectangles, each consisting of one cell with a number 0.
In the third sample, the only nice rectangle is the whole grid — the sum of all numbers is 0. Clearly, Limak can choose at most one nice rectangle, so the answer is 1.
| 2,250
|
[] | 1,492,790,315
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
#include<bits/stdc++.h>
#define ll long long int
#define mp make_pair
#define ff first
#define ss second
#define pb push_back
using namespace std ;
ll n,m,i,j,k,ans,x,y,z ;
ll dp[1000][1000],a[1000][1000],b[1000][1000],c[1000][1000] ;
vector<ll> v;
ll inf = 2334233442423;
map<ll,ll>m1,m2,m3 ;
int main(){
cin >> n ;
for(i=0;i<2;i++){
for(j=0;j<n;j++){
cin >> a[i][j];
}
}
for(i=0;i<3;i++){
b[2][0]=a[1][0]+a[0][0] ;
b[1][0]=a[1][0] ;
b[0][0]=a[0][0] ;
for(j=1;j<n;j++){
if(i==0||i==1){ b[i][j] = b[i][j-1]+a[i][j] ; }
else{b[i][j]=b[i][j-1]+a[i-1][j]+a[i-2][j];}
}
}
for(i=0;i<3;i++){for(j=0;j<n;j++){ c[i][j]= -1*inf ;} }
m1[0]=-1;m2[0]=-1;m3[0]=-1;
for(j=0;j<n;j++){
if(m1[b[0][j]]!=0){c[0][j]=m1[b[0][j]];}
if(m2[b[1][j]]!=0){c[1][j]=m2[b[1][j]];}
if(m3[b[2][j]]!=0){c[2][j]=m3[b[2][j]];}
m1[b[0][j]]=j ;
m2[b[1][j]]=j;
m3[b[2][j]]=j;
}
//for(i=0;i<3;i++){for(j=0;j<n;j++){cout << c[i][j] << " " ; }cout << endl ;}
for(i=1;i<=n;i++){
for(j=1;j<=n;j++){
if(i==1&&j==1&&(a[0][0]==0||a[1][0]==0)){dp[1][1]=1;}
if(i>1){
if(a[0][i-1]==0){x=1;}
else{x=0;}
dp[i][j]=max(dp[i][j],(dp[i-1][j]+x)) ;}
if(j>1){
if(a[1][j-1]==0){x=1;}
else{x=0;}
dp[i][j]=max(dp[i][j],(dp[i][j-1]+x)) ;}
if(i>=1&&j>=1&&a[0][i-1]==0&&a[1][j-1]==0){
dp[i][j]=max(dp[i][j],dp[i-1][j-1]+2) ;
}
x = c[0][i-1]+1 ; y= c[1][j-1]+1 ; z= c[2][i-1]+1 ;
if(x>=0){dp[i][j]=max(dp[i][j],(1+dp[x][j]));}
if(y>=0){dp[i][j]=max(dp[i][j],(1+dp[i][y]));}
if(i==j&&z>=0){dp[i][j]=max(dp[i][j],(1+dp[z][z]));}
}
}
for(i=0;i<n;i++){for(j=0;j<n;j++){cout << dp[i][j] << " " ;} cout << endl ;}
cout << dp[n][n] ;
}
|
Title: Bear and Rectangle Strips
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Limak has a grid that consists of 2 rows and *n* columns. The *j*-th cell in the *i*-th row contains an integer *t**i*,<=*j* which can be positive, negative or zero.
A non-empty rectangle of cells is called nice if and only if the sum of numbers in its cells is equal to 0.
Limak wants to choose some nice rectangles and give them to his friends, as gifts. No two chosen rectangles should share a cell. What is the maximum possible number of nice rectangles Limak can choose?
Input Specification:
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=300<=000) — the number of columns in the grid.
The next two lines contain numbers in the grid. The *i*-th of those two lines contains *n* integers *t**i*,<=1,<=*t**i*,<=2,<=...,<=*t**i*,<=*n* (<=-<=109<=≤<=*t**i*,<=*j*<=≤<=109).
Output Specification:
Print one integer, denoting the maximum possible number of cell-disjoint nice rectangles.
Demo Input:
['6\n70 70 70 70 70 -15\n90 -60 -30 30 -30 15\n', '4\n0 -1 0 0\n0 0 1 0\n', '3\n1000000000 999999999 -1000000000\n999999999 -1000000000 -999999998\n']
Demo Output:
['3\n', '6\n', '1\n']
Note:
In the first sample, there are four nice rectangles:
Limak can't choose all of them because they are not disjoint. He should take three nice rectangles: those denoted as blue frames on the drawings.
In the second sample, it's optimal to choose six nice rectangles, each consisting of one cell with a number 0.
In the third sample, the only nice rectangle is the whole grid — the sum of all numbers is 0. Clearly, Limak can choose at most one nice rectangle, so the answer is 1.
|
```python
#include<bits/stdc++.h>
#define ll long long int
#define mp make_pair
#define ff first
#define ss second
#define pb push_back
using namespace std ;
ll n,m,i,j,k,ans,x,y,z ;
ll dp[1000][1000],a[1000][1000],b[1000][1000],c[1000][1000] ;
vector<ll> v;
ll inf = 2334233442423;
map<ll,ll>m1,m2,m3 ;
int main(){
cin >> n ;
for(i=0;i<2;i++){
for(j=0;j<n;j++){
cin >> a[i][j];
}
}
for(i=0;i<3;i++){
b[2][0]=a[1][0]+a[0][0] ;
b[1][0]=a[1][0] ;
b[0][0]=a[0][0] ;
for(j=1;j<n;j++){
if(i==0||i==1){ b[i][j] = b[i][j-1]+a[i][j] ; }
else{b[i][j]=b[i][j-1]+a[i-1][j]+a[i-2][j];}
}
}
for(i=0;i<3;i++){for(j=0;j<n;j++){ c[i][j]= -1*inf ;} }
m1[0]=-1;m2[0]=-1;m3[0]=-1;
for(j=0;j<n;j++){
if(m1[b[0][j]]!=0){c[0][j]=m1[b[0][j]];}
if(m2[b[1][j]]!=0){c[1][j]=m2[b[1][j]];}
if(m3[b[2][j]]!=0){c[2][j]=m3[b[2][j]];}
m1[b[0][j]]=j ;
m2[b[1][j]]=j;
m3[b[2][j]]=j;
}
//for(i=0;i<3;i++){for(j=0;j<n;j++){cout << c[i][j] << " " ; }cout << endl ;}
for(i=1;i<=n;i++){
for(j=1;j<=n;j++){
if(i==1&&j==1&&(a[0][0]==0||a[1][0]==0)){dp[1][1]=1;}
if(i>1){
if(a[0][i-1]==0){x=1;}
else{x=0;}
dp[i][j]=max(dp[i][j],(dp[i-1][j]+x)) ;}
if(j>1){
if(a[1][j-1]==0){x=1;}
else{x=0;}
dp[i][j]=max(dp[i][j],(dp[i][j-1]+x)) ;}
if(i>=1&&j>=1&&a[0][i-1]==0&&a[1][j-1]==0){
dp[i][j]=max(dp[i][j],dp[i-1][j-1]+2) ;
}
x = c[0][i-1]+1 ; y= c[1][j-1]+1 ; z= c[2][i-1]+1 ;
if(x>=0){dp[i][j]=max(dp[i][j],(1+dp[x][j]));}
if(y>=0){dp[i][j]=max(dp[i][j],(1+dp[i][y]));}
if(i==j&&z>=0){dp[i][j]=max(dp[i][j],(1+dp[z][z]));}
}
}
for(i=0;i<n;i++){for(j=0;j<n;j++){cout << dp[i][j] << " " ;} cout << endl ;}
cout << dp[n][n] ;
}
```
| -1
|
|
978
|
E
|
Bus Video System
|
PROGRAMMING
| 1,400
|
[
"combinatorics",
"math"
] | null | null |
The busses in Berland are equipped with a video surveillance system. The system records information about changes in the number of passengers in a bus after stops.
If $x$ is the number of passengers in a bus just before the current bus stop and $y$ is the number of passengers in the bus just after current bus stop, the system records the number $y-x$. So the system records show how number of passengers changed.
The test run was made for single bus and $n$ bus stops. Thus, the system recorded the sequence of integers $a_1, a_2, \dots, a_n$ (exactly one number for each bus stop), where $a_i$ is the record for the bus stop $i$. The bus stops are numbered from $1$ to $n$ in chronological order.
Determine the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to $w$ (that is, at any time in the bus there should be from $0$ to $w$ passengers inclusive).
|
The first line contains two integers $n$ and $w$ $(1 \le n \le 1\,000, 1 \le w \le 10^{9})$ — the number of bus stops and the capacity of the bus.
The second line contains a sequence $a_1, a_2, \dots, a_n$ $(-10^{6} \le a_i \le 10^{6})$, where $a_i$ equals to the number, which has been recorded by the video system after the $i$-th bus stop.
|
Print the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to $w$. If the situation is contradictory (i.e. for any initial number of passengers there will be a contradiction), print 0.
|
[
"3 5\n2 1 -3\n",
"2 4\n-1 1\n",
"4 10\n2 4 1 2\n"
] |
[
"3\n",
"4\n",
"2\n"
] |
In the first example initially in the bus could be $0$, $1$ or $2$ passengers.
In the second example initially in the bus could be $1$, $2$, $3$ or $4$ passengers.
In the third example initially in the bus could be $0$ or $1$ passenger.
| 0
|
[
{
"input": "3 5\n2 1 -3",
"output": "3"
},
{
"input": "2 4\n-1 1",
"output": "4"
},
{
"input": "4 10\n2 4 1 2",
"output": "2"
},
{
"input": "2 10\n-1 2",
"output": "9"
},
{
"input": "3 4\n-3 -4 4",
"output": "0"
},
{
"input": "10 1\n-1 -1 3 -4 2 3 0 -3 3 2",
"output": "0"
},
{
"input": "5 21\n-3 2 -4 -1 -5",
"output": "11"
},
{
"input": "5 9\n-2 -1 2 -1 -2",
"output": "6"
},
{
"input": "8 7\n-5 0 -3 1 -1 5 0 -6",
"output": "0"
},
{
"input": "3 4\n-2 -1 0",
"output": "2"
},
{
"input": "1 1000000000\n0",
"output": "1000000001"
},
{
"input": "2 1000000000\n-1000000 -1000000",
"output": "998000001"
},
{
"input": "2 1000000000\n1000000 -1000000",
"output": "999000001"
},
{
"input": "2 1000000000\n-1000000 1000000",
"output": "999000001"
},
{
"input": "2 1000000000\n1000000 1000000",
"output": "998000001"
},
{
"input": "102 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 1 1",
"output": "0"
},
{
"input": "1 1234564\n1",
"output": "1234564"
},
{
"input": "3 4\n1 2 3",
"output": "0"
}
] | 1,677,644,341
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 42
| 46
| 0
|
n, w = map(int, input().split())
a = list(map(int, input().split()))
# Минимальное и максимальное количество пассажиров в автобусе до первой остановки
min_p = max_p = 0
cur_p = 0 # Текущее количество пассажиров в автобусе
for i in range(n):
cur_p += a[i]
min_p = min(min_p, cur_p)
max_p = max(max_p, cur_p)
# Количество возможных способов сколько человек могло быть в автобусе до первой остановки
ans = max(0, w - max_p + min_p + 1)
print(ans)
|
Title: Bus Video System
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The busses in Berland are equipped with a video surveillance system. The system records information about changes in the number of passengers in a bus after stops.
If $x$ is the number of passengers in a bus just before the current bus stop and $y$ is the number of passengers in the bus just after current bus stop, the system records the number $y-x$. So the system records show how number of passengers changed.
The test run was made for single bus and $n$ bus stops. Thus, the system recorded the sequence of integers $a_1, a_2, \dots, a_n$ (exactly one number for each bus stop), where $a_i$ is the record for the bus stop $i$. The bus stops are numbered from $1$ to $n$ in chronological order.
Determine the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to $w$ (that is, at any time in the bus there should be from $0$ to $w$ passengers inclusive).
Input Specification:
The first line contains two integers $n$ and $w$ $(1 \le n \le 1\,000, 1 \le w \le 10^{9})$ — the number of bus stops and the capacity of the bus.
The second line contains a sequence $a_1, a_2, \dots, a_n$ $(-10^{6} \le a_i \le 10^{6})$, where $a_i$ equals to the number, which has been recorded by the video system after the $i$-th bus stop.
Output Specification:
Print the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to $w$. If the situation is contradictory (i.e. for any initial number of passengers there will be a contradiction), print 0.
Demo Input:
['3 5\n2 1 -3\n', '2 4\n-1 1\n', '4 10\n2 4 1 2\n']
Demo Output:
['3\n', '4\n', '2\n']
Note:
In the first example initially in the bus could be $0$, $1$ or $2$ passengers.
In the second example initially in the bus could be $1$, $2$, $3$ or $4$ passengers.
In the third example initially in the bus could be $0$ or $1$ passenger.
|
```python
n, w = map(int, input().split())
a = list(map(int, input().split()))
# Минимальное и максимальное количество пассажиров в автобусе до первой остановки
min_p = max_p = 0
cur_p = 0 # Текущее количество пассажиров в автобусе
for i in range(n):
cur_p += a[i]
min_p = min(min_p, cur_p)
max_p = max(max_p, cur_p)
# Количество возможных способов сколько человек могло быть в автобусе до первой остановки
ans = max(0, w - max_p + min_p + 1)
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"
}
] | 1,690,878,010
| 2,147,483,647
|
PyPy 3-64
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 5
| 1,000
| 20,889,600
|
a = list(input())
b = list(input())
ans = 0
while a!= b:
if len(a) == len(b):
a.pop(0)
b.pop(0)
ans += 2
elif len(a) > len(b):
a.pop(0)
ans += 1
else:
b.pop(0)
ans += 1
print(ans)
|
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 = list(input())
b = list(input())
ans = 0
while a!= b:
if len(a) == len(b):
a.pop(0)
b.pop(0)
ans += 2
elif len(a) > len(b):
a.pop(0)
ans += 1
else:
b.pop(0)
ans += 1
print(ans)
```
| 0
|
|
435
|
A
|
Queue on Bus Stop
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
It's that time of the year when the Russians flood their countryside summer cottages (dachas) and the bus stop has a lot of people. People rarely go to the dacha on their own, it's usually a group, so the people stand in queue by groups.
The bus stop queue has *n* groups of people. The *i*-th group from the beginning has *a**i* people. Every 30 minutes an empty bus arrives at the bus stop, it can carry at most *m* people. Naturally, the people from the first group enter the bus first. Then go the people from the second group and so on. Note that the order of groups in the queue never changes. Moreover, if some group cannot fit all of its members into the current bus, it waits for the next bus together with other groups standing after it in the queue.
Your task is to determine how many buses is needed to transport all *n* groups to the dacha countryside.
|
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100). The next line contains *n* integers: *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*m*).
|
Print a single integer — the number of buses that is needed to transport all *n* groups to the dacha countryside.
|
[
"4 3\n2 3 2 1\n",
"3 4\n1 2 1\n"
] |
[
"3\n",
"1\n"
] |
none
| 500
|
[
{
"input": "4 3\n2 3 2 1",
"output": "3"
},
{
"input": "3 4\n1 2 1",
"output": "1"
},
{
"input": "1 5\n4",
"output": "1"
},
{
"input": "5 1\n1 1 1 1 1",
"output": "5"
},
{
"input": "6 4\n1 3 2 3 4 1",
"output": "5"
},
{
"input": "6 8\n6 1 1 1 4 5",
"output": "3"
},
{
"input": "10 10\n1 10 1 10 1 1 7 8 6 7",
"output": "8"
},
{
"input": "100 100\n85 50 17 89 65 89 5 20 86 26 16 21 85 14 44 31 87 31 6 2 48 67 8 80 79 1 48 36 97 1 5 30 79 50 78 12 2 55 76 100 54 40 26 81 97 96 68 56 87 14 51 17 54 37 52 33 69 62 38 63 74 15 62 78 9 19 67 2 60 58 93 60 18 96 55 48 34 7 79 82 32 58 90 67 20 50 27 15 7 89 98 10 11 15 99 49 4 51 77 52",
"output": "63"
},
{
"input": "10 1\n1 1 1 1 1 1 1 1 1 1",
"output": "10"
},
{
"input": "10 2\n2 2 1 1 1 1 1 2 1 2",
"output": "8"
},
{
"input": "10 3\n1 3 1 1 3 2 2 2 3 3",
"output": "9"
},
{
"input": "10 4\n2 1 1 1 3 4 4 4 1 2",
"output": "6"
},
{
"input": "10 5\n2 2 3 4 4 1 5 3 1 2",
"output": "7"
},
{
"input": "100 3\n1 2 3 2 1 2 2 3 1 3 3 2 2 1 1 2 2 1 1 1 1 2 3 3 2 1 1 2 2 2 3 3 3 2 1 3 1 3 3 2 3 1 2 2 2 3 2 1 1 3 3 3 3 2 1 1 2 3 2 2 3 2 3 2 2 3 2 2 2 2 3 3 3 1 3 3 1 1 2 3 2 2 2 2 3 3 3 2 1 2 3 1 1 2 3 3 1 3 3 2",
"output": "83"
},
{
"input": "100 7\n4 7 4 7 7 4 7 3 5 6 3 5 4 3 7 2 7 2 4 1 6 3 3 7 4 4 5 4 3 6 4 3 2 2 1 4 4 1 7 3 7 7 1 3 1 5 4 1 5 3 5 2 2 1 5 5 1 5 2 7 5 5 1 5 5 4 6 5 1 3 5 6 7 4 1 3 3 4 3 2 7 6 5 7 2 7 1 1 2 2 3 1 3 7 1 3 2 1 1 7",
"output": "71"
},
{
"input": "100 10\n3 4 8 10 8 6 4 3 7 7 6 2 3 1 3 10 1 7 9 3 5 5 2 6 2 9 1 7 4 2 4 1 6 1 7 10 2 5 3 7 6 4 6 2 8 8 8 6 6 10 3 7 4 3 4 1 7 9 3 6 3 6 1 4 9 3 8 1 10 1 4 10 7 7 9 5 3 8 10 2 1 10 8 7 10 8 5 3 1 2 1 10 6 1 5 3 3 5 7 2",
"output": "64"
},
{
"input": "100 15\n3 12 8 3 11 14 12 14 1 11 13 3 5 13 4 14 2 11 7 8 12 9 15 7 15 1 4 11 6 12 1 3 8 13 1 8 14 4 3 14 1 3 1 6 10 15 13 11 12 1 14 13 11 14 11 3 12 7 3 15 14 4 5 6 5 14 7 14 6 2 6 12 6 13 13 1 9 13 15 11 6 3 15 11 9 4 15 8 15 12 1 15 10 10 4 1 15 1 4 1",
"output": "71"
},
{
"input": "100 30\n7 14 22 16 11 13 7 29 20 19 22 6 12 16 1 8 27 21 22 3 15 27 20 12 4 19 1 26 26 22 25 17 29 25 16 29 29 28 16 26 25 14 16 20 5 21 5 15 19 13 17 21 17 19 23 13 1 25 6 30 16 19 12 10 28 8 15 13 14 24 19 30 12 19 22 1 3 14 16 3 20 26 15 19 9 10 19 27 2 16 10 22 15 13 19 3 24 9 8 13",
"output": "71"
},
{
"input": "100 40\n39 19 13 36 11 21 32 12 1 2 39 26 32 39 24 1 4 19 10 4 16 39 32 34 13 24 30 35 3 10 8 18 13 12 39 27 31 40 37 20 17 17 37 5 10 12 22 17 7 1 31 13 11 10 2 6 22 16 2 4 9 27 6 35 22 16 22 30 33 2 26 20 35 19 40 37 19 17 21 28 37 28 40 4 5 4 35 19 26 36 19 12 21 20 21 30 9 16 9 32",
"output": "65"
},
{
"input": "100 50\n2 46 4 6 38 19 15 34 10 35 37 30 3 25 5 45 40 45 33 31 6 20 10 44 11 9 2 14 35 5 9 23 20 2 48 22 25 35 38 31 24 33 35 16 4 30 27 10 12 22 6 24 12 30 23 21 14 12 32 21 7 12 25 43 18 34 34 28 47 13 28 43 18 39 44 42 35 26 35 14 8 29 32 20 29 3 20 6 20 9 9 27 8 42 10 37 42 27 8 1",
"output": "60"
},
{
"input": "100 60\n34 21 39 17 48 46 23 56 46 52 50 39 55 48 54 38 32 38 24 26 44 12 28 9 25 26 10 52 42 60 41 3 16 60 44 29 27 55 19 19 19 57 45 59 29 35 5 14 50 47 57 48 16 7 12 36 58 31 37 58 30 50 19 11 10 41 59 57 49 41 33 9 12 11 53 50 60 51 21 9 44 23 1 16 4 15 17 57 15 17 46 50 18 52 43 24 47 50 19 18",
"output": "74"
},
{
"input": "100 90\n74 65 49 41 3 79 61 83 50 40 13 57 90 14 62 77 36 10 3 5 5 40 50 75 32 26 3 71 79 54 88 50 46 20 42 59 30 36 83 86 60 62 82 68 62 80 18 65 28 28 81 74 62 33 61 35 33 83 90 72 6 6 51 4 22 20 29 10 8 3 84 69 12 17 24 16 12 64 80 74 68 59 1 59 15 59 37 58 79 83 51 56 81 14 37 45 19 31 61 90",
"output": "67"
},
{
"input": "100 99\n69 46 76 47 71 9 66 46 78 17 96 83 56 96 29 3 43 48 79 23 93 61 19 9 29 72 15 84 93 46 71 87 11 43 96 44 54 75 3 66 2 95 46 32 69 52 79 38 57 53 37 60 71 82 28 31 84 58 89 40 62 74 22 50 45 38 99 67 24 28 28 12 69 88 33 10 31 71 46 7 42 81 54 81 96 44 8 1 20 24 28 19 54 35 69 32 71 13 66 15",
"output": "68"
},
{
"input": "90 100\n25 52 88 89 36 17 57 64 66 11 89 61 54 92 48 51 18 42 44 92 6 14 67 100 16 21 17 88 85 73 33 11 94 84 56 72 4 80 90 78 96 5 62 70 54 70 94 80 10 91 100 89 98 87 69 74 88 63 53 79 38 94 89 52 21 82 67 79 100 81 2 40 30 69 34 15 12 33 87 52 95 18 51 30 15 39 30 99 46 84",
"output": "67"
},
{
"input": "5 100\n14 67 15 28 21",
"output": "2"
},
{
"input": "10 100\n2 17 53 94 95 57 36 47 68 48",
"output": "7"
},
{
"input": "1 100\n18",
"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": "100"
},
{
"input": "30 100\n56 7 99 83 2 65 35 53 99 36 42 57 13 37 68 52 87 11 50 23 86 24 32 39 97 6 64 1 18 86",
"output": "18"
},
{
"input": "60 100\n18 75 43 88 45 43 20 59 59 79 62 39 53 21 28 46 54 53 97 81 18 15 2 95 84 9 36 70 30 76 17 19 83 40 45 32 31 70 23 14 44 35 79 84 97 96 99 60 3 73 64 83 6 12 67 86 70 89 18 61",
"output": "40"
},
{
"input": "1 73\n24",
"output": "1"
},
{
"input": "4 4\n1 4 4 4",
"output": "4"
},
{
"input": "2 6\n1 4",
"output": "1"
},
{
"input": "2 5\n5 5",
"output": "2"
}
] | 1,546,448,780
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 34
| 140
| 102,400
|
import itertools
import math
from collections import defaultdict
def input_ints():
return list(map(int, input().split()))
def solve():
n, m = input_ints()
ans = 0
s = 0
for x in input_ints():
if s + x > m:
ans += 1
s = 0
s += x
ans += 1
print(ans)
if __name__ == '__main__':
solve()
|
Title: Queue on Bus Stop
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It's that time of the year when the Russians flood their countryside summer cottages (dachas) and the bus stop has a lot of people. People rarely go to the dacha on their own, it's usually a group, so the people stand in queue by groups.
The bus stop queue has *n* groups of people. The *i*-th group from the beginning has *a**i* people. Every 30 minutes an empty bus arrives at the bus stop, it can carry at most *m* people. Naturally, the people from the first group enter the bus first. Then go the people from the second group and so on. Note that the order of groups in the queue never changes. Moreover, if some group cannot fit all of its members into the current bus, it waits for the next bus together with other groups standing after it in the queue.
Your task is to determine how many buses is needed to transport all *n* groups to the dacha countryside.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100). The next line contains *n* integers: *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*m*).
Output Specification:
Print a single integer — the number of buses that is needed to transport all *n* groups to the dacha countryside.
Demo Input:
['4 3\n2 3 2 1\n', '3 4\n1 2 1\n']
Demo Output:
['3\n', '1\n']
Note:
none
|
```python
import itertools
import math
from collections import defaultdict
def input_ints():
return list(map(int, input().split()))
def solve():
n, m = input_ints()
ans = 0
s = 0
for x in input_ints():
if s + x > m:
ans += 1
s = 0
s += x
ans += 1
print(ans)
if __name__ == '__main__':
solve()
```
| 3
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
An atom of element X can exist in *n* distinct states with energies *E*1<=<<=*E*2<=<<=...<=<<=*E**n*. Arkady wants to build a laser on this element, using a three-level scheme. Here is a simplified description of the scheme.
Three distinct states *i*, *j* and *k* are selected, where *i*<=<<=*j*<=<<=*k*. After that the following process happens:
1. initially the atom is in the state *i*,1. we spend *E**k*<=-<=*E**i* energy to put the atom in the state *k*,1. the atom emits a photon with useful energy *E**k*<=-<=*E**j* and changes its state to the state *j*,1. the atom spontaneously changes its state to the state *i*, losing energy *E**j*<=-<=*E**i*,1. the process repeats from step 1.
Let's define the energy conversion efficiency as , i. e. the ration between the useful energy of the photon and spent energy.
Due to some limitations, Arkady can only choose such three states that *E**k*<=-<=*E**i*<=≤<=*U*.
Help Arkady to find such the maximum possible energy conversion efficiency within the above constraints.
|
The first line contains two integers *n* and *U* (3<=≤<=*n*<=≤<=105, 1<=≤<=*U*<=≤<=109) — the number of states and the maximum possible difference between *E**k* and *E**i*.
The second line contains a sequence of integers *E*1,<=*E*2,<=...,<=*E**n* (1<=≤<=*E*1<=<<=*E*2...<=<<=*E**n*<=≤<=109). It is guaranteed that all *E**i* are given in increasing order.
|
If it is not possible to choose three states that satisfy all constraints, print -1.
Otherwise, print one real number η — the maximum possible energy conversion efficiency. Your answer is considered correct its absolute or relative error does not exceed 10<=-<=9.
Formally, let your answer be *a*, and the jury's answer be *b*. Your answer is considered correct if .
|
[
"4 4\n1 3 5 7\n",
"10 8\n10 13 15 16 17 19 20 22 24 25\n",
"3 1\n2 5 10\n"
] |
[
"0.5\n",
"0.875\n",
"-1\n"
] |
In the first example choose states 1, 2 and 3, so that the energy conversion efficiency becomes equal to <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/147ae7a830722917b0aa37d064df8eb74cfefb97.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second example choose states 4, 5 and 9, so that the energy conversion efficiency becomes equal to <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/f68f268de4eb2242167e6ec64e6b8aa60a5703ae.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
| 0
|
[
{
"input": "4 4\n1 3 5 7",
"output": "0.5"
},
{
"input": "10 8\n10 13 15 16 17 19 20 22 24 25",
"output": "0.875"
},
{
"input": "3 1\n2 5 10",
"output": "-1"
},
{
"input": "5 3\n4 6 8 9 10",
"output": "0.5"
},
{
"input": "10 128\n110 121 140 158 174 188 251 271 272 277",
"output": "0.86554621848739499157"
},
{
"input": "20 17\n104 107 121 131 138 140 143 144 178 192 193 198 201 206 238 242 245 248 255 265",
"output": "0.92857142857142860315"
},
{
"input": "30 23\n102 104 105 107 108 109 110 111 116 118 119 122 127 139 140 142 145 157 166 171 173 174 175 181 187 190 191 193 195 196",
"output": "0.95652173913043481157"
},
{
"input": "50 64\n257 258 350 375 1014 1017 1051 1097 1169 1177 1223 1836 1942 1983 2111 2131 2341 2418 2593 2902 2948 3157 3243 3523 3566 4079 4499 4754 5060 5624 6279 6976 7011 7071 7278 7366 7408 7466 7526 7837 7934 8532 8577 8680 9221 9271 9327 9411 9590 9794",
"output": "0.91891891891891896993"
},
{
"input": "5 2\n4 6 8 9 10",
"output": "0.5"
},
{
"input": "10 2\n110 121 140 158 174 188 251 271 272 277",
"output": "-1"
},
{
"input": "30 5\n102 104 105 107 108 109 110 111 116 118 119 122 127 139 140 142 145 157 166 171 173 174 175 181 187 190 191 193 195 196",
"output": "0.80000000000000004441"
},
{
"input": "10 6\n110 121 140 158 174 188 251 271 272 277",
"output": "0.83333333333333337034"
},
{
"input": "20 4\n104 107 121 131 138 140 143 144 178 192 193 198 201 206 238 242 245 248 255 265",
"output": "0.25"
},
{
"input": "3 1000000000\n1 2 1000000000",
"output": "0.99999999900000002828"
},
{
"input": "3 1\n1 2 3",
"output": "-1"
},
{
"input": "5 1000000000\n1 2 3 999999999 1000000000",
"output": "0.99999999900000002828"
},
{
"input": "10 199\n1 3 190 191 193 195 196 197 199 200",
"output": "0.98994974874371854945"
},
{
"input": "10 300\n80 100 103 140 146 159 392 393 396 398",
"output": "0.98993288590604022747"
},
{
"input": "10 92\n44 119 252 281 303 323 351 363 377 392",
"output": "0.77528089887640450062"
},
{
"input": "4 2\n1 3 5 7",
"output": "-1"
},
{
"input": "8 2\n1 3 7 9 15 17 23 25",
"output": "-1"
},
{
"input": "3 5\n1 2 10",
"output": "-1"
},
{
"input": "4 7\n1 5 8 9",
"output": "0.42857142857142854764"
}
] | 1,669,541,501
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 3
| 46
| 0
|
n, U = map(int, input().split())
row = list(map(int, input().split()))
s = False
for i in range(1, U):
for x in row:
if (x + i) in row and (x + U) in row:
print((U-i)/(U))
s = True
break
if s: break
else:
print(-1)
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
An atom of element X can exist in *n* distinct states with energies *E*1<=<<=*E*2<=<<=...<=<<=*E**n*. Arkady wants to build a laser on this element, using a three-level scheme. Here is a simplified description of the scheme.
Three distinct states *i*, *j* and *k* are selected, where *i*<=<<=*j*<=<<=*k*. After that the following process happens:
1. initially the atom is in the state *i*,1. we spend *E**k*<=-<=*E**i* energy to put the atom in the state *k*,1. the atom emits a photon with useful energy *E**k*<=-<=*E**j* and changes its state to the state *j*,1. the atom spontaneously changes its state to the state *i*, losing energy *E**j*<=-<=*E**i*,1. the process repeats from step 1.
Let's define the energy conversion efficiency as , i. e. the ration between the useful energy of the photon and spent energy.
Due to some limitations, Arkady can only choose such three states that *E**k*<=-<=*E**i*<=≤<=*U*.
Help Arkady to find such the maximum possible energy conversion efficiency within the above constraints.
Input Specification:
The first line contains two integers *n* and *U* (3<=≤<=*n*<=≤<=105, 1<=≤<=*U*<=≤<=109) — the number of states and the maximum possible difference between *E**k* and *E**i*.
The second line contains a sequence of integers *E*1,<=*E*2,<=...,<=*E**n* (1<=≤<=*E*1<=<<=*E*2...<=<<=*E**n*<=≤<=109). It is guaranteed that all *E**i* are given in increasing order.
Output Specification:
If it is not possible to choose three states that satisfy all constraints, print -1.
Otherwise, print one real number η — the maximum possible energy conversion efficiency. Your answer is considered correct its absolute or relative error does not exceed 10<=-<=9.
Formally, let your answer be *a*, and the jury's answer be *b*. Your answer is considered correct if .
Demo Input:
['4 4\n1 3 5 7\n', '10 8\n10 13 15 16 17 19 20 22 24 25\n', '3 1\n2 5 10\n']
Demo Output:
['0.5\n', '0.875\n', '-1\n']
Note:
In the first example choose states 1, 2 and 3, so that the energy conversion efficiency becomes equal to <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/147ae7a830722917b0aa37d064df8eb74cfefb97.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second example choose states 4, 5 and 9, so that the energy conversion efficiency becomes equal to <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/f68f268de4eb2242167e6ec64e6b8aa60a5703ae.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
|
```python
n, U = map(int, input().split())
row = list(map(int, input().split()))
s = False
for i in range(1, U):
for x in row:
if (x + i) in row and (x + U) in row:
print((U-i)/(U))
s = True
break
if s: break
else:
print(-1)
```
| 0
|
|
76
|
D
|
Plus and xor
|
PROGRAMMING
| 1,700
|
[
"dp",
"greedy",
"math"
] |
D. Plus and xor
|
0
|
256
|
Bitwise exclusive OR (or bitwise addition modulo two) is a binary operation which is equivalent to applying logical exclusive OR to every pair of bits located on the same positions in binary notation of operands. In other words, a binary digit of the result is equal to 1 if and only if bits on the respective positions in the operands are different.
For example, if *X*<==<=10910<==<=11011012, *Y*<==<=4110<==<=1010012, then:
Write a program, which takes two non-negative integers *A* and *B* as an input and finds two non-negative integers *X* and *Y*, which satisfy the following conditions:
- *A*<==<=*X*<=+<=*Y* - *B* <==<= *X* xor *Y*, where xor is bitwise exclusive or. - *X* is the smallest number among all numbers for which the first two conditions are true.
|
The first line contains integer number *A* and the second line contains integer number *B* (0<=≤<=*A*,<=*B*<=≤<=264<=-<=1).
|
The only output line should contain two integer non-negative numbers *X* and *Y*. Print the only number -1 if there is no answer.
|
[
"142\n76\n"
] |
[
"33 109\n"
] |
none
| 0
|
[
{
"input": "142\n76",
"output": "33 109"
},
{
"input": "638\n126",
"output": "256 382"
},
{
"input": "1639\n1176",
"output": "-1"
},
{
"input": "12608\n0",
"output": "6304 6304"
},
{
"input": "104066\n104066",
"output": "0 104066"
},
{
"input": "1024996\n990106",
"output": "17445 1007551"
},
{
"input": "1215996781\n108302929",
"output": "553846926 662149855"
},
{
"input": "1870807699\n259801747",
"output": "805502976 1065304723"
},
{
"input": "339671107814\n208405902980",
"output": "65632602417 274038505397"
},
{
"input": "1367480970723947\n584615739735395",
"output": "391432615494276 976048355229671"
},
{
"input": "9992164445234764941\n8162963574901971597",
"output": "914600435166396672 9077564010068368269"
}
] | 1,686,930,260
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 11
| 92
| 0
|
a = int(input())
b = int(input())
if a < b or ((a & 1) != (b & 1)):
print(-1)
else:
print((a - b) // 2, (a + b) // 2)
|
Title: Plus and xor
Time Limit: 0 seconds
Memory Limit: 256 megabytes
Problem Description:
Bitwise exclusive OR (or bitwise addition modulo two) is a binary operation which is equivalent to applying logical exclusive OR to every pair of bits located on the same positions in binary notation of operands. In other words, a binary digit of the result is equal to 1 if and only if bits on the respective positions in the operands are different.
For example, if *X*<==<=10910<==<=11011012, *Y*<==<=4110<==<=1010012, then:
Write a program, which takes two non-negative integers *A* and *B* as an input and finds two non-negative integers *X* and *Y*, which satisfy the following conditions:
- *A*<==<=*X*<=+<=*Y* - *B* <==<= *X* xor *Y*, where xor is bitwise exclusive or. - *X* is the smallest number among all numbers for which the first two conditions are true.
Input Specification:
The first line contains integer number *A* and the second line contains integer number *B* (0<=≤<=*A*,<=*B*<=≤<=264<=-<=1).
Output Specification:
The only output line should contain two integer non-negative numbers *X* and *Y*. Print the only number -1 if there is no answer.
Demo Input:
['142\n76\n']
Demo Output:
['33 109\n']
Note:
none
|
```python
a = int(input())
b = int(input())
if a < b or ((a & 1) != (b & 1)):
print(-1)
else:
print((a - b) // 2, (a + b) // 2)
```
| 3
|
199
|
A
|
Hexadecimal's theorem
|
PROGRAMMING
| 900
|
[
"brute force",
"constructive algorithms",
"implementation",
"number theory"
] | null | null |
Recently, a chaotic virus Hexadecimal advanced a new theorem which will shake the Universe. She thinks that each Fibonacci number can be represented as sum of three not necessary different Fibonacci numbers.
Let's remember how Fibonacci numbers can be calculated. *F*0<==<=0, *F*1<==<=1, and all the next numbers are *F**i*<==<=*F**i*<=-<=2<=+<=*F**i*<=-<=1.
So, Fibonacci numbers make a sequence of numbers: 0, 1, 1, 2, 3, 5, 8, 13, ...
If you haven't run away from the PC in fear, you have to help the virus. Your task is to divide given Fibonacci number *n* by three not necessary different Fibonacci numbers or say that it is impossible.
|
The input contains of a single integer *n* (0<=≤<=*n*<=<<=109) — the number that should be represented by the rules described above. It is guaranteed that *n* is a Fibonacci number.
|
Output three required numbers: *a*, *b* and *c*. If there is no answer for the test you have to print "I'm too stupid to solve this problem" without the quotes.
If there are multiple answers, print any of them.
|
[
"3\n",
"13\n"
] |
[
"1 1 1\n",
"2 3 8\n"
] |
none
| 500
|
[
{
"input": "3",
"output": "1 1 1"
},
{
"input": "13",
"output": "2 3 8"
},
{
"input": "0",
"output": "0 0 0"
},
{
"input": "1",
"output": "1 0 0"
},
{
"input": "2",
"output": "1 1 0"
},
{
"input": "1597",
"output": "233 377 987"
},
{
"input": "0",
"output": "0 0 0"
},
{
"input": "1",
"output": "1 0 0"
},
{
"input": "1",
"output": "1 0 0"
},
{
"input": "2",
"output": "1 1 0"
},
{
"input": "3",
"output": "1 1 1"
},
{
"input": "5",
"output": "1 1 3"
},
{
"input": "8",
"output": "1 2 5"
},
{
"input": "13",
"output": "2 3 8"
},
{
"input": "21",
"output": "3 5 13"
},
{
"input": "34",
"output": "5 8 21"
},
{
"input": "55",
"output": "8 13 34"
},
{
"input": "89",
"output": "13 21 55"
},
{
"input": "144",
"output": "21 34 89"
},
{
"input": "233",
"output": "34 55 144"
},
{
"input": "377",
"output": "55 89 233"
},
{
"input": "610",
"output": "89 144 377"
},
{
"input": "987",
"output": "144 233 610"
},
{
"input": "1597",
"output": "233 377 987"
},
{
"input": "2584",
"output": "377 610 1597"
},
{
"input": "4181",
"output": "610 987 2584"
},
{
"input": "6765",
"output": "987 1597 4181"
},
{
"input": "10946",
"output": "1597 2584 6765"
},
{
"input": "17711",
"output": "2584 4181 10946"
},
{
"input": "28657",
"output": "4181 6765 17711"
},
{
"input": "46368",
"output": "6765 10946 28657"
},
{
"input": "75025",
"output": "10946 17711 46368"
},
{
"input": "121393",
"output": "17711 28657 75025"
},
{
"input": "196418",
"output": "28657 46368 121393"
},
{
"input": "317811",
"output": "46368 75025 196418"
},
{
"input": "514229",
"output": "75025 121393 317811"
},
{
"input": "832040",
"output": "121393 196418 514229"
},
{
"input": "1346269",
"output": "196418 317811 832040"
},
{
"input": "2178309",
"output": "317811 514229 1346269"
},
{
"input": "3524578",
"output": "514229 832040 2178309"
},
{
"input": "5702887",
"output": "832040 1346269 3524578"
},
{
"input": "9227465",
"output": "1346269 2178309 5702887"
},
{
"input": "14930352",
"output": "2178309 3524578 9227465"
},
{
"input": "24157817",
"output": "3524578 5702887 14930352"
},
{
"input": "39088169",
"output": "5702887 9227465 24157817"
},
{
"input": "63245986",
"output": "9227465 14930352 39088169"
},
{
"input": "102334155",
"output": "14930352 24157817 63245986"
},
{
"input": "165580141",
"output": "24157817 39088169 102334155"
},
{
"input": "267914296",
"output": "39088169 63245986 165580141"
},
{
"input": "433494437",
"output": "63245986 102334155 267914296"
},
{
"input": "701408733",
"output": "102334155 165580141 433494437"
},
{
"input": "701408733",
"output": "102334155 165580141 433494437"
},
{
"input": "102334155",
"output": "14930352 24157817 63245986"
},
{
"input": "63245986",
"output": "9227465 14930352 39088169"
}
] | 1,552,580,730
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 54
| 216
| 0
|
n=int(input())
def fn(n):
if n==0:
print('0 0 0')
return
if n==1:
print('1 0 0')
return
f1,f2=0,1
f=0
while f<n:
f=f1+f2
f1=f2
f2=f
print(f1,f-f1,0)
fn(n)
|
Title: Hexadecimal's theorem
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently, a chaotic virus Hexadecimal advanced a new theorem which will shake the Universe. She thinks that each Fibonacci number can be represented as sum of three not necessary different Fibonacci numbers.
Let's remember how Fibonacci numbers can be calculated. *F*0<==<=0, *F*1<==<=1, and all the next numbers are *F**i*<==<=*F**i*<=-<=2<=+<=*F**i*<=-<=1.
So, Fibonacci numbers make a sequence of numbers: 0, 1, 1, 2, 3, 5, 8, 13, ...
If you haven't run away from the PC in fear, you have to help the virus. Your task is to divide given Fibonacci number *n* by three not necessary different Fibonacci numbers or say that it is impossible.
Input Specification:
The input contains of a single integer *n* (0<=≤<=*n*<=<<=109) — the number that should be represented by the rules described above. It is guaranteed that *n* is a Fibonacci number.
Output Specification:
Output three required numbers: *a*, *b* and *c*. If there is no answer for the test you have to print "I'm too stupid to solve this problem" without the quotes.
If there are multiple answers, print any of them.
Demo Input:
['3\n', '13\n']
Demo Output:
['1 1 1\n', '2 3 8\n']
Note:
none
|
```python
n=int(input())
def fn(n):
if n==0:
print('0 0 0')
return
if n==1:
print('1 0 0')
return
f1,f2=0,1
f=0
while f<n:
f=f1+f2
f1=f2
f2=f
print(f1,f-f1,0)
fn(n)
```
| 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*<=×<=*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*<=×<=*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<=≤<=<=*n*,<=*m*,<=*a*<=≤<=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"
}
] | 1,657,007,700
| 2,147,483,647
|
PyPy 3-64
|
RUNTIME_ERROR
|
TESTS
| 0
| 186
| 3,788,800
|
import numpy as np
j=np.int32()
for i in range(100000000):
j+=1
n,m,a=map(int,input().split())
print(-n//a*(-m//a))
|
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*<=×<=*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*<=×<=*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<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109).
Output Specification:
Write the needed number of flagstones.
Demo Input:
['6 6 4\n']
Demo Output:
['4\n']
Note:
none
|
```python
import numpy as np
j=np.int32()
for i in range(100000000):
j+=1
n,m,a=map(int,input().split())
print(-n//a*(-m//a))
```
| -1
|
445
|
A
|
DZY Loves Chessboard
|
PROGRAMMING
| 1,200
|
[
"dfs and similar",
"implementation"
] | null | null |
DZY loves chessboard, and he enjoys playing with it.
He has a chessboard of *n* rows and *m* columns. Some cells of the chessboard are bad, others are good. For every good cell, DZY wants to put a chessman on it. Each chessman is either white or black. After putting all chessmen, DZY wants that no two chessmen with the same color are on two adjacent cells. Two cells are adjacent if and only if they share a common edge.
You task is to find any suitable placement of chessmen on the given chessboard.
|
The first line contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
Each of the next *n* lines contains a string of *m* characters: the *j*-th character of the *i*-th string is either "." or "-". A "." means that the corresponding cell (in the *i*-th row and the *j*-th column) is good, while a "-" means it is bad.
|
Output must contain *n* lines, each line must contain a string of *m* characters. The *j*-th character of the *i*-th string should be either "W", "B" or "-". Character "W" means the chessman on the cell is white, "B" means it is black, "-" means the cell is a bad cell.
If multiple answers exist, print any of them. It is guaranteed that at least one answer exists.
|
[
"1 1\n.\n",
"2 2\n..\n..\n",
"3 3\n.-.\n---\n--."
] |
[
"B\n",
"BW\nWB\n",
"B-B\n---\n--B"
] |
In the first sample, DZY puts a single black chessman. Of course putting a white one is also OK.
In the second sample, all 4 cells are good. No two same chessmen share an edge in the sample output.
In the third sample, no good cells are adjacent. So you can just put 3 chessmen, no matter what their colors are.
| 500
|
[
{
"input": "1 1\n.",
"output": "B"
},
{
"input": "2 2\n..\n..",
"output": "BW\nWB"
},
{
"input": "3 3\n.-.\n---\n--.",
"output": "B-B\n---\n--B"
},
{
"input": "3 1\n-\n.\n.",
"output": "-\nW\nB"
},
{
"input": "11 11\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------",
"output": "-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------"
},
{
"input": "1 1\n-",
"output": "-"
},
{
"input": "4 2\n..\n..\n..\n..",
"output": "BW\nWB\nBW\nWB"
},
{
"input": "5 1\n.\n.\n.\n.\n.",
"output": "B\nW\nB\nW\nB"
}
] | 1,634,572,357
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 1
| 62
| 0
|
n, m = map(int, input().split())
chess = []
pattern1 = []
pattern2 = []
flag = 0
for i in range(n):
temp = input()
chess.append(temp)
for i in range(m):
if flag == 0:
flag = 1
pattern1.append("W")
pattern2.append("B")
else:
flag = 0
pattern1.append("B")
pattern2.append("W")
# print(chess)
flag = 0
for i in chess:
for j in range(m):
if i[j] == ".":
if flag == 0:
print(pattern1[j], end=" ")
else:
print(pattern2[j], end=" ")
else:
print("-", end=" ")
if flag == 0:
flag = 1
else:
flag = 0
print()
|
Title: DZY Loves Chessboard
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
DZY loves chessboard, and he enjoys playing with it.
He has a chessboard of *n* rows and *m* columns. Some cells of the chessboard are bad, others are good. For every good cell, DZY wants to put a chessman on it. Each chessman is either white or black. After putting all chessmen, DZY wants that no two chessmen with the same color are on two adjacent cells. Two cells are adjacent if and only if they share a common edge.
You task is to find any suitable placement of chessmen on the given chessboard.
Input Specification:
The first line contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
Each of the next *n* lines contains a string of *m* characters: the *j*-th character of the *i*-th string is either "." or "-". A "." means that the corresponding cell (in the *i*-th row and the *j*-th column) is good, while a "-" means it is bad.
Output Specification:
Output must contain *n* lines, each line must contain a string of *m* characters. The *j*-th character of the *i*-th string should be either "W", "B" or "-". Character "W" means the chessman on the cell is white, "B" means it is black, "-" means the cell is a bad cell.
If multiple answers exist, print any of them. It is guaranteed that at least one answer exists.
Demo Input:
['1 1\n.\n', '2 2\n..\n..\n', '3 3\n.-.\n---\n--.']
Demo Output:
['B\n', 'BW\nWB\n', 'B-B\n---\n--B']
Note:
In the first sample, DZY puts a single black chessman. Of course putting a white one is also OK.
In the second sample, all 4 cells are good. No two same chessmen share an edge in the sample output.
In the third sample, no good cells are adjacent. So you can just put 3 chessmen, no matter what their colors are.
|
```python
n, m = map(int, input().split())
chess = []
pattern1 = []
pattern2 = []
flag = 0
for i in range(n):
temp = input()
chess.append(temp)
for i in range(m):
if flag == 0:
flag = 1
pattern1.append("W")
pattern2.append("B")
else:
flag = 0
pattern1.append("B")
pattern2.append("W")
# print(chess)
flag = 0
for i in chess:
for j in range(m):
if i[j] == ".":
if flag == 0:
print(pattern1[j], end=" ")
else:
print(pattern2[j], end=" ")
else:
print("-", end=" ")
if flag == 0:
flag = 1
else:
flag = 0
print()
```
| 0
|
|
431
|
C
|
k-Tree
|
PROGRAMMING
| 1,600
|
[
"dp",
"implementation",
"trees"
] | null | null |
Quite recently a creative student Lesha had a lecture on trees. After the lecture Lesha was inspired and came up with the tree of his own which he called a *k*-tree.
A *k*-tree is an infinite rooted tree where:
- each vertex has exactly *k* children; - each edge has some weight; - if we look at the edges that goes from some vertex to its children (exactly *k* edges), then their weights will equal 1,<=2,<=3,<=...,<=*k*.
The picture below shows a part of a 3-tree.
Help Dima find an answer to his question. As the number of ways can be rather large, print it modulo 1000000007 (109<=+<=7).
|
A single line contains three space-separated integers: *n*, *k* and *d* (1<=≤<=*n*,<=*k*<=≤<=100; 1<=≤<=*d*<=≤<=*k*).
|
Print a single integer — the answer to the problem modulo 1000000007 (109<=+<=7).
|
[
"3 3 2\n",
"3 3 3\n",
"4 3 2\n",
"4 5 2\n"
] |
[
"3\n",
"1\n",
"6\n",
"7\n"
] |
none
| 1,500
|
[
{
"input": "3 3 2",
"output": "3"
},
{
"input": "3 3 3",
"output": "1"
},
{
"input": "4 3 2",
"output": "6"
},
{
"input": "4 5 2",
"output": "7"
},
{
"input": "28 6 3",
"output": "110682188"
},
{
"input": "5 100 1",
"output": "16"
},
{
"input": "50 6 3",
"output": "295630102"
},
{
"input": "10 13 6",
"output": "48"
},
{
"input": "20 16 14",
"output": "236"
},
{
"input": "1 10 1",
"output": "1"
},
{
"input": "8 11 4",
"output": "47"
},
{
"input": "16 5 4",
"output": "16175"
},
{
"input": "5 26 17",
"output": "0"
},
{
"input": "35 47 42",
"output": "0"
},
{
"input": "11 6 2",
"output": "975"
},
{
"input": "54 60 16",
"output": "931055544"
},
{
"input": "47 5 1",
"output": "164058640"
},
{
"input": "70 6 1",
"output": "592826579"
},
{
"input": "40 77 77",
"output": "0"
},
{
"input": "96 9 6",
"output": "362487247"
},
{
"input": "52 46 4",
"output": "27907693"
},
{
"input": "74 41 28",
"output": "806604424"
},
{
"input": "100 100 100",
"output": "1"
},
{
"input": "99 99 5",
"output": "700732369"
},
{
"input": "100 100 1",
"output": "988185646"
},
{
"input": "98 98 64",
"output": "237643149"
},
{
"input": "28 74 2",
"output": "134217727"
},
{
"input": "86 69 62",
"output": "217513984"
},
{
"input": "9 17 14",
"output": "0"
},
{
"input": "73 72 12",
"output": "426374014"
},
{
"input": "41 98 76",
"output": "0"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "1 100 100",
"output": "0"
},
{
"input": "1 100 1",
"output": "1"
},
{
"input": "1 100 2",
"output": "0"
},
{
"input": "2 100 2",
"output": "1"
},
{
"input": "2 100 1",
"output": "2"
},
{
"input": "50 50 1",
"output": "949480669"
},
{
"input": "100 50 50",
"output": "661237556"
},
{
"input": "3 2 2",
"output": "2"
},
{
"input": "100 50 3",
"output": "494224664"
},
{
"input": "90 97 24",
"output": "413496885"
},
{
"input": "31 8 8",
"output": "52532592"
},
{
"input": "78 90 38",
"output": "744021655"
},
{
"input": "100 13 11",
"output": "883875774"
},
{
"input": "100 45 8",
"output": "367847193"
},
{
"input": "31 8 4",
"output": "924947104"
},
{
"input": "35 95 9",
"output": "927164672"
},
{
"input": "45 5 3",
"output": "252804490"
},
{
"input": "1 5 5",
"output": "0"
},
{
"input": "89 75 59",
"output": "179807625"
},
{
"input": "90 100 30",
"output": "697322870"
},
{
"input": "89 73 32",
"output": "152673180"
},
{
"input": "100 90 80",
"output": "11531520"
}
] | 1,677,932,299
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 4
| 1,000
| 0
|
def f(n,k,s):
if s>n:
return 0
if s==n:
return 1
ans=0
for j in range(1,k+1):
ans+=f(n,k,s+j)
return ans
n,k,d=map(int,input().split())
print(f(n,k,0)-f(n,d-1,0))
|
Title: k-Tree
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Quite recently a creative student Lesha had a lecture on trees. After the lecture Lesha was inspired and came up with the tree of his own which he called a *k*-tree.
A *k*-tree is an infinite rooted tree where:
- each vertex has exactly *k* children; - each edge has some weight; - if we look at the edges that goes from some vertex to its children (exactly *k* edges), then their weights will equal 1,<=2,<=3,<=...,<=*k*.
The picture below shows a part of a 3-tree.
Help Dima find an answer to his question. As the number of ways can be rather large, print it modulo 1000000007 (109<=+<=7).
Input Specification:
A single line contains three space-separated integers: *n*, *k* and *d* (1<=≤<=*n*,<=*k*<=≤<=100; 1<=≤<=*d*<=≤<=*k*).
Output Specification:
Print a single integer — the answer to the problem modulo 1000000007 (109<=+<=7).
Demo Input:
['3 3 2\n', '3 3 3\n', '4 3 2\n', '4 5 2\n']
Demo Output:
['3\n', '1\n', '6\n', '7\n']
Note:
none
|
```python
def f(n,k,s):
if s>n:
return 0
if s==n:
return 1
ans=0
for j in range(1,k+1):
ans+=f(n,k,s+j)
return ans
n,k,d=map(int,input().split())
print(f(n,k,0)-f(n,d-1,0))
```
| 0
|
|
796
|
A
|
Buying A House
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation"
] | null | null |
Zane the wizard had never loved anyone before, until he fell in love with a girl, whose name remains unknown to us.
The girl lives in house *m* of a village. There are *n* houses in that village, lining in a straight line from left to right: house 1, house 2, ..., house *n*. The village is also well-structured: house *i* and house *i*<=+<=1 (1<=≤<=*i*<=<<=*n*) are exactly 10 meters away. In this village, some houses are occupied, and some are not. Indeed, unoccupied houses can be purchased.
You will be given *n* integers *a*1,<=*a*2,<=...,<=*a**n* that denote the availability and the prices of the houses. If house *i* is occupied, and therefore cannot be bought, then *a**i* equals 0. Otherwise, house *i* can be bought, and *a**i* represents the money required to buy it, in dollars.
As Zane has only *k* dollars to spare, it becomes a challenge for him to choose the house to purchase, so that he could live as near as possible to his crush. Help Zane determine the minimum distance from his crush's house to some house he can afford, to help him succeed in his love.
|
The first line contains three integers *n*, *m*, and *k* (2<=≤<=*n*<=≤<=100, 1<=≤<=*m*<=≤<=*n*, 1<=≤<=*k*<=≤<=100) — the number of houses in the village, the house where the girl lives, and the amount of money Zane has (in dollars), respectively.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100) — denoting the availability and the prices of the houses.
It is guaranteed that *a**m*<==<=0 and that it is possible to purchase some house with no more than *k* dollars.
|
Print one integer — the minimum distance, in meters, from the house where the girl Zane likes lives to the house Zane can buy.
|
[
"5 1 20\n0 27 32 21 19\n",
"7 3 50\n62 0 0 0 99 33 22\n",
"10 5 100\n1 0 1 0 0 0 0 0 1 1\n"
] |
[
"40",
"30",
"20"
] |
In the first sample, with *k* = 20 dollars, Zane can buy only house 5. The distance from house *m* = 1 to house 5 is 10 + 10 + 10 + 10 = 40 meters.
In the second sample, Zane can buy houses 6 and 7. It is better to buy house 6 than house 7, since house *m* = 3 and house 6 are only 30 meters away, while house *m* = 3 and house 7 are 40 meters away.
| 500
|
[
{
"input": "5 1 20\n0 27 32 21 19",
"output": "40"
},
{
"input": "7 3 50\n62 0 0 0 99 33 22",
"output": "30"
},
{
"input": "10 5 100\n1 0 1 0 0 0 0 0 1 1",
"output": "20"
},
{
"input": "5 3 1\n1 1 0 0 1",
"output": "10"
},
{
"input": "5 5 5\n1 0 5 6 0",
"output": "20"
},
{
"input": "15 10 50\n20 0 49 50 50 50 50 50 50 0 50 50 49 0 20",
"output": "10"
},
{
"input": "7 5 1\n0 100 2 2 0 2 1",
"output": "20"
},
{
"input": "100 50 100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 0 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100",
"output": "10"
},
{
"input": "100 50 1\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 0 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100",
"output": "490"
},
{
"input": "100 77 50\n50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 0 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0",
"output": "10"
},
{
"input": "100 1 1\n0 98 97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 80 79 78 77 76 75 74 73 72 71 70 69 68 67 66 65 64 63 62 61 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0",
"output": "980"
},
{
"input": "100 1 100\n0 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "10"
},
{
"input": "100 10 99\n0 0 0 0 0 0 0 0 0 0 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 99 98",
"output": "890"
},
{
"input": "7 4 5\n1 0 6 0 5 6 0",
"output": "10"
},
{
"input": "7 4 5\n1 6 5 0 0 6 0",
"output": "10"
},
{
"input": "100 42 59\n50 50 50 50 50 50 50 50 50 50 59 59 59 59 59 59 59 59 59 59 59 59 59 59 59 59 59 59 59 59 59 59 59 60 60 60 60 60 60 60 60 0 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 0",
"output": "90"
},
{
"input": "2 1 100\n0 1",
"output": "10"
},
{
"input": "2 2 100\n1 0",
"output": "10"
},
{
"input": "10 1 88\n0 95 0 0 0 0 0 94 0 85",
"output": "90"
},
{
"input": "10 2 14\n2 0 1 26 77 39 41 100 13 32",
"output": "10"
},
{
"input": "10 3 11\n0 0 0 0 0 62 0 52 1 35",
"output": "60"
},
{
"input": "20 12 44\n27 40 58 69 53 38 31 39 75 95 8 0 28 81 77 90 38 61 21 88",
"output": "10"
},
{
"input": "30 29 10\n59 79 34 12 100 6 1 58 18 73 54 11 37 46 89 90 80 85 73 45 64 5 31 0 89 19 0 74 0 82",
"output": "70"
},
{
"input": "40 22 1\n7 95 44 53 0 0 19 93 0 68 65 0 24 91 10 58 17 0 71 0 100 0 94 90 79 73 0 73 4 61 54 81 7 13 21 84 5 41 0 1",
"output": "180"
},
{
"input": "40 22 99\n60 0 100 0 0 100 100 0 0 0 0 100 100 0 0 100 100 0 100 100 100 0 100 100 100 0 100 100 0 0 100 100 100 0 0 100 0 100 0 0",
"output": "210"
},
{
"input": "50 10 82\n56 54 0 0 0 0 88 93 0 0 83 93 0 0 91 89 0 30 62 52 24 84 80 8 38 13 92 78 16 87 23 30 71 55 16 63 15 99 4 93 24 6 3 35 4 42 73 27 86 37",
"output": "80"
},
{
"input": "63 49 22\n18 3 97 52 75 2 12 24 58 75 80 97 22 10 79 51 30 60 68 99 75 2 35 3 97 88 9 7 18 5 0 0 0 91 0 91 56 36 76 0 0 0 52 27 35 0 51 72 0 96 57 0 0 0 0 92 55 28 0 30 0 78 77",
"output": "190"
},
{
"input": "74 38 51\n53 36 55 42 64 5 87 9 0 16 86 78 9 22 19 1 25 72 1 0 0 0 79 0 0 0 77 58 70 0 0 100 64 0 99 59 0 0 0 0 65 74 0 96 0 58 89 93 61 88 0 0 82 89 0 0 49 24 7 77 89 87 94 61 100 31 93 70 39 49 39 14 20 84",
"output": "190"
},
{
"input": "89 22 11\n36 0 68 89 0 85 72 0 38 56 0 44 0 94 0 28 71 0 0 18 0 0 0 89 0 0 0 75 0 0 0 32 66 0 0 0 0 0 0 48 63 0 64 58 0 23 48 0 0 52 93 61 57 0 18 0 0 34 62 17 0 41 0 0 53 59 44 0 0 51 40 0 0 100 100 54 0 88 0 5 45 56 57 67 24 16 88 86 15",
"output": "580"
},
{
"input": "97 44 100\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 51 19",
"output": "520"
},
{
"input": "100 1 1\n0 0 0 0 10 54 84 6 17 94 65 82 34 0 61 46 42 0 2 16 56 0 100 0 82 0 0 0 89 78 96 56 0 0 0 0 0 0 0 0 77 70 0 96 67 0 0 32 44 1 72 50 14 11 24 61 100 64 19 5 67 69 44 82 93 22 67 93 22 61 53 64 79 41 84 48 43 97 7 24 8 49 23 16 72 52 97 29 69 47 29 49 64 91 4 73 17 18 51 67",
"output": "490"
},
{
"input": "100 1 50\n0 0 0 60 0 0 54 0 80 0 0 0 97 0 68 97 84 0 0 93 0 0 0 0 68 0 0 62 0 0 55 68 65 87 0 69 0 0 0 0 0 52 61 100 0 71 0 82 88 78 0 81 0 95 0 57 0 67 0 0 0 55 86 0 60 72 0 0 73 0 83 0 0 60 64 0 56 0 0 77 84 0 58 63 84 0 0 67 0 16 3 88 0 98 31 52 40 35 85 23",
"output": "890"
},
{
"input": "100 1 100\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 91 70 14",
"output": "970"
},
{
"input": "100 1 29\n0 0 0 0 64 0 89 97 0 0 0 59 0 67 62 0 59 0 0 80 0 0 0 0 0 97 0 57 0 64 32 0 44 0 0 48 0 47 38 0 42 0 0 0 0 0 0 46 74 0 86 33 33 0 44 0 79 0 0 0 0 91 59 0 59 65 55 0 0 58 33 95 0 97 76 0 81 0 41 0 38 81 80 0 85 0 31 0 0 92 0 0 45 96 0 85 91 87 0 10",
"output": "990"
},
{
"input": "100 50 20\n3 0 32 0 48 32 64 0 54 26 0 0 0 0 0 28 0 0 54 0 0 45 49 0 38 74 0 0 39 42 62 48 75 96 89 42 0 44 0 0 30 21 76 0 50 0 79 0 0 0 0 99 0 84 62 0 0 0 0 53 80 0 28 0 0 53 0 0 38 0 62 0 0 62 0 0 88 0 44 32 0 81 35 45 49 0 69 73 38 27 72 0 96 72 69 0 0 22 76 10",
"output": "490"
},
{
"input": "100 50 20\n49 0 56 0 87 25 40 0 50 0 0 97 0 0 36 29 0 0 0 0 0 73 29 71 44 0 0 0 91 92 69 0 0 60 81 49 48 38 0 87 0 82 0 32 0 82 46 39 0 0 29 0 0 29 0 79 47 0 0 0 0 0 49 0 24 33 70 0 63 45 97 90 0 0 29 53 55 0 84 0 0 100 26 0 88 0 0 0 0 81 70 0 30 80 0 75 59 98 0 2",
"output": "500"
},
{
"input": "100 2 2\n0 0 43 90 47 5 2 97 52 69 21 48 64 10 34 97 97 74 8 19 68 56 55 24 47 38 43 73 72 72 60 60 51 36 33 44 100 45 13 54 72 52 0 15 3 6 50 8 88 4 78 26 40 27 30 63 67 83 61 91 33 97 54 20 92 27 89 35 10 7 84 50 11 95 74 88 24 44 74 100 18 56 34 91 41 34 51 51 11 91 89 54 19 100 83 89 10 17 76 20",
"output": "50"
},
{
"input": "100 100 34\n5 73 0 0 44 0 0 0 79 55 0 0 0 0 0 0 0 0 83 67 75 0 0 0 0 59 0 74 0 0 47 98 0 0 72 41 0 55 87 0 0 78 84 0 0 39 0 79 72 95 0 0 0 0 0 85 53 84 0 0 0 0 37 75 0 66 0 0 0 0 61 0 70 0 37 60 42 78 92 52 0 0 0 55 77 57 0 63 37 0 0 0 96 70 0 94 97 0 0 0",
"output": "990"
},
{
"input": "100 100 100\n43 79 21 87 84 14 28 69 92 16 3 71 79 37 48 37 72 58 12 72 62 49 37 17 60 54 41 99 15 72 40 89 76 1 99 87 14 56 63 48 69 37 96 64 7 14 1 73 85 33 98 70 97 71 96 28 49 71 56 2 67 22 100 2 98 100 62 77 92 76 98 98 47 26 22 47 50 56 9 16 72 47 5 62 29 78 81 1 0 63 32 65 87 3 40 53 8 80 93 0",
"output": "10"
},
{
"input": "100 38 1\n3 59 12 81 33 95 0 41 36 17 63 76 42 77 85 56 3 96 55 41 24 87 18 9 0 37 0 61 69 0 0 0 67 0 0 0 0 0 0 18 0 0 47 56 74 0 0 80 0 42 0 1 60 59 62 9 19 87 92 48 58 30 98 51 99 10 42 94 51 53 50 89 24 5 52 82 50 39 98 8 95 4 57 21 10 0 44 32 19 14 64 34 79 76 17 3 15 22 71 51",
"output": "140"
},
{
"input": "100 72 1\n56 98 8 27 9 23 16 76 56 1 34 43 96 73 75 49 62 20 18 23 51 55 30 84 4 20 89 40 75 16 69 35 1 0 16 0 80 0 41 17 0 0 76 23 0 92 0 34 0 91 82 54 0 0 0 63 85 59 98 24 29 0 8 77 26 0 34 95 39 0 0 0 74 0 0 0 0 12 0 92 0 0 55 95 66 30 0 0 29 98 0 0 0 47 0 0 80 0 0 4",
"output": "390"
},
{
"input": "100 66 1\n38 50 64 91 37 44 74 21 14 41 80 90 26 51 78 85 80 86 44 14 49 75 93 48 78 89 23 72 35 22 14 48 100 71 62 22 7 95 80 66 32 20 17 47 79 30 41 52 15 62 67 71 1 6 0 9 0 0 0 11 0 0 24 0 31 0 77 0 51 0 0 0 0 0 0 77 0 36 44 19 90 45 6 25 100 87 93 30 4 97 36 88 33 50 26 71 97 71 51 68",
"output": "130"
},
{
"input": "100 55 1\n0 33 45 83 56 96 58 24 45 30 38 60 39 69 21 87 59 21 72 73 27 46 61 61 11 97 77 5 39 3 3 35 76 37 53 84 24 75 9 48 31 90 100 84 74 81 83 83 42 23 29 94 18 1 0 53 52 99 86 37 94 54 28 75 28 80 17 14 98 68 76 20 32 23 42 31 57 79 60 14 18 27 1 98 32 3 96 25 15 38 2 6 3 28 59 54 63 2 43 59",
"output": "10"
},
{
"input": "100 55 1\n24 52 41 6 55 11 58 25 63 12 70 39 23 28 72 17 96 85 7 84 21 13 34 37 97 43 36 32 15 30 58 5 14 71 40 70 9 92 44 73 31 58 96 90 19 35 29 91 25 36 48 95 61 78 0 1 99 61 81 88 42 53 61 57 42 55 74 45 41 92 99 30 20 25 89 50 37 4 17 24 6 65 15 44 40 2 38 43 7 90 38 59 75 87 96 28 12 67 24 32",
"output": "10"
},
{
"input": "100 21 1\n62 5 97 80 81 28 83 0 26 0 0 0 0 23 0 0 90 0 0 0 0 0 0 0 0 54 71 8 0 0 42 0 73 0 17 0 1 31 71 78 58 72 84 39 54 59 13 29 16 41 71 35 88 55 70 50 33 100 100 60 52 90 7 66 44 55 51 42 90 17 86 44 46 8 52 74 8 22 2 92 34 37 58 98 70 74 19 91 74 25 4 38 71 68 50 68 63 14 60 98",
"output": "160"
},
{
"input": "5 2 20\n27 0 32 21 19",
"output": "30"
},
{
"input": "6 4 10\n10 0 0 0 0 10",
"output": "20"
},
{
"input": "8 7 100\n1 0 0 0 0 0 0 1",
"output": "10"
},
{
"input": "5 3 20\n1 21 0 0 1",
"output": "20"
},
{
"input": "4 3 1\n0 0 0 1",
"output": "10"
},
{
"input": "5 2 3\n4 0 5 6 1",
"output": "30"
},
{
"input": "5 3 87\n88 89 0 1 90",
"output": "10"
},
{
"input": "5 3 20\n15 30 0 15 35",
"output": "10"
},
{
"input": "6 3 50\n0 0 0 1 2 0",
"output": "10"
},
{
"input": "6 4 9\n100 9 10 0 0 9",
"output": "20"
},
{
"input": "5 4 20\n0 20 0 0 20",
"output": "10"
},
{
"input": "6 3 3\n1 5 0 2 2 0",
"output": "10"
},
{
"input": "5 4 100\n0 1 0 0 1",
"output": "10"
}
] | 1,551,288,202
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 3
| 93
| 0
|
n,m,k=map(int,input().split())
a=list(map(int,input().split()))
i=m
l=[]
while i<n:
if a[i]!=0 and a[i]<k:
r1=i
break
i=i+1
i=m
while i>0 and i!=1:
if a[i]!=0 and a[i]<k:
l.append(i)
break
i=i-1
if len(l)==0:
print((r1-m)*10+10)
else:
x1=(r1-m)*10+10
x2=(m-l[0])*10-10
print(min(x1,x2))
|
Title: Buying A House
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Zane the wizard had never loved anyone before, until he fell in love with a girl, whose name remains unknown to us.
The girl lives in house *m* of a village. There are *n* houses in that village, lining in a straight line from left to right: house 1, house 2, ..., house *n*. The village is also well-structured: house *i* and house *i*<=+<=1 (1<=≤<=*i*<=<<=*n*) are exactly 10 meters away. In this village, some houses are occupied, and some are not. Indeed, unoccupied houses can be purchased.
You will be given *n* integers *a*1,<=*a*2,<=...,<=*a**n* that denote the availability and the prices of the houses. If house *i* is occupied, and therefore cannot be bought, then *a**i* equals 0. Otherwise, house *i* can be bought, and *a**i* represents the money required to buy it, in dollars.
As Zane has only *k* dollars to spare, it becomes a challenge for him to choose the house to purchase, so that he could live as near as possible to his crush. Help Zane determine the minimum distance from his crush's house to some house he can afford, to help him succeed in his love.
Input Specification:
The first line contains three integers *n*, *m*, and *k* (2<=≤<=*n*<=≤<=100, 1<=≤<=*m*<=≤<=*n*, 1<=≤<=*k*<=≤<=100) — the number of houses in the village, the house where the girl lives, and the amount of money Zane has (in dollars), respectively.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100) — denoting the availability and the prices of the houses.
It is guaranteed that *a**m*<==<=0 and that it is possible to purchase some house with no more than *k* dollars.
Output Specification:
Print one integer — the minimum distance, in meters, from the house where the girl Zane likes lives to the house Zane can buy.
Demo Input:
['5 1 20\n0 27 32 21 19\n', '7 3 50\n62 0 0 0 99 33 22\n', '10 5 100\n1 0 1 0 0 0 0 0 1 1\n']
Demo Output:
['40', '30', '20']
Note:
In the first sample, with *k* = 20 dollars, Zane can buy only house 5. The distance from house *m* = 1 to house 5 is 10 + 10 + 10 + 10 = 40 meters.
In the second sample, Zane can buy houses 6 and 7. It is better to buy house 6 than house 7, since house *m* = 3 and house 6 are only 30 meters away, while house *m* = 3 and house 7 are 40 meters away.
|
```python
n,m,k=map(int,input().split())
a=list(map(int,input().split()))
i=m
l=[]
while i<n:
if a[i]!=0 and a[i]<k:
r1=i
break
i=i+1
i=m
while i>0 and i!=1:
if a[i]!=0 and a[i]<k:
l.append(i)
break
i=i-1
if len(l)==0:
print((r1-m)*10+10)
else:
x1=(r1-m)*10+10
x2=(m-l[0])*10-10
print(min(x1,x2))
```
| -1
|
|
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" — 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 "
}
] | 1,697,262,166
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 71
| 92
| 0
|
print(*[i for i in input().split('WUB') if i != ''])
|
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" — between them Vasya added two "WUB".
|
```python
print(*[i for i in input().split('WUB') if i != ''])
```
| 3
|
|
440
|
B
|
Balancer
|
PROGRAMMING
| 1,600
|
[
"greedy",
"implementation"
] | null | null |
Petya has *k* matches, placed in *n* matchboxes lying in a line from left to right. We know that *k* is divisible by *n*. Petya wants all boxes to have the same number of matches inside. For that, he can move a match from its box to the adjacent one in one move. How many such moves does he need to achieve the desired configuration?
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=50000). The second line contains *n* non-negative numbers that do not exceed 109, the *i*-th written number is the number of matches in the *i*-th matchbox. It is guaranteed that the total number of matches is divisible by *n*.
|
Print the total minimum number of moves.
|
[
"6\n1 6 2 5 3 7\n"
] |
[
"12\n"
] |
none
| 1,000
|
[
{
"input": "6\n1 6 2 5 3 7",
"output": "12"
},
{
"input": "6\n6 6 6 0 0 0",
"output": "27"
},
{
"input": "6\n0 0 0 6 6 6",
"output": "27"
},
{
"input": "6\n6 6 0 0 6 6",
"output": "12"
},
{
"input": "5\n0 0 0 0 0",
"output": "0"
},
{
"input": "10\n0 100 0 100 0 100 0 100 0 100",
"output": "250"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "2\n0 0",
"output": "0"
},
{
"input": "3\n0 0 0",
"output": "0"
},
{
"input": "4\n0 0 0 0",
"output": "0"
},
{
"input": "2\n921 29111",
"output": "14095"
},
{
"input": "2\n0 1000000000",
"output": "500000000"
},
{
"input": "2\n291911 1",
"output": "145955"
},
{
"input": "2\n20180000 0",
"output": "10090000"
},
{
"input": "10\n10 9 7 13 7 5 13 15 10 11",
"output": "27"
},
{
"input": "100\n6 3 4 5 3 4 2 4 1 2 4 1 8 5 2 2 4 4 6 8 4 10 4 4 6 8 6 5 5 4 8 4 3 3 6 5 7 2 9 7 6 5 6 3 2 6 8 10 3 6 8 7 2 3 5 4 8 6 5 6 6 8 4 1 5 6 1 8 12 5 3 3 8 2 4 2 4 5 6 6 9 5 1 2 8 8 3 7 5 3 4 5 7 6 3 9 4 6 3 6",
"output": "867"
},
{
"input": "10\n1 1 1 1 1 999999999 999999999 999999999 999999999 999999999",
"output": "12499999975"
},
{
"input": "10\n1 1 1 1 1 1 1 1 2 1000000000",
"output": "4499999999"
},
{
"input": "14\n0 0 0 0 0 0 0 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000",
"output": "24500000000"
},
{
"input": "20\n1 1 1 1 1 1 1 1 1 1 999999999 999999999 999999999 999999999 999999999 999999999 999999999 999999999 999999999 999999999",
"output": "49999999900"
},
{
"input": "20\n1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 2 2 2 2 2 2 2 2 2 2",
"output": "49999999900"
}
] | 1,402,602,790
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 19
| 500
| 2,252,800
|
n = int(input())
xs = list(map(int, input().split()))
u = sum(xs)//n
moves = 0
for i in range(n-1):
moves += abs(u-xs[i])
xs[i+1] += xs[i]-u
print(moves)
|
Title: Balancer
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya has *k* matches, placed in *n* matchboxes lying in a line from left to right. We know that *k* is divisible by *n*. Petya wants all boxes to have the same number of matches inside. For that, he can move a match from its box to the adjacent one in one move. How many such moves does he need to achieve the desired configuration?
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=50000). The second line contains *n* non-negative numbers that do not exceed 109, the *i*-th written number is the number of matches in the *i*-th matchbox. It is guaranteed that the total number of matches is divisible by *n*.
Output Specification:
Print the total minimum number of moves.
Demo Input:
['6\n1 6 2 5 3 7\n']
Demo Output:
['12\n']
Note:
none
|
```python
n = int(input())
xs = list(map(int, input().split()))
u = sum(xs)//n
moves = 0
for i in range(n-1):
moves += abs(u-xs[i])
xs[i+1] += xs[i]-u
print(moves)
```
| 0
|
|
620
|
A
|
Professor GukiZ's Robot
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
Professor GukiZ makes a new robot. The robot are in the point with coordinates (*x*1,<=*y*1) and should go to the point (*x*2,<=*y*2). In a single step the robot can change any of its coordinates (maybe both of them) by one (decrease or increase). So the robot can move in one of the 8 directions. Find the minimal number of steps the robot should make to get the finish position.
|
The first line contains two integers *x*1,<=*y*1 (<=-<=109<=≤<=*x*1,<=*y*1<=≤<=109) — the start position of the robot.
The second line contains two integers *x*2,<=*y*2 (<=-<=109<=≤<=*x*2,<=*y*2<=≤<=109) — the finish position of the robot.
|
Print the only integer *d* — the minimal number of steps to get the finish position.
|
[
"0 0\n4 5\n",
"3 4\n6 1\n"
] |
[
"5\n",
"3\n"
] |
In the first example robot should increase both of its coordinates by one four times, so it will be in position (4, 4). After that robot should simply increase its *y* coordinate and get the finish position.
In the second example robot should simultaneously increase *x* coordinate and decrease *y* coordinate by one three times.
| 0
|
[
{
"input": "0 0\n4 5",
"output": "5"
},
{
"input": "3 4\n6 1",
"output": "3"
},
{
"input": "0 0\n4 6",
"output": "6"
},
{
"input": "1 1\n-3 -5",
"output": "6"
},
{
"input": "-1 -1\n-10 100",
"output": "101"
},
{
"input": "1 -1\n100 -100",
"output": "99"
},
{
"input": "-1000000000 -1000000000\n1000000000 1000000000",
"output": "2000000000"
},
{
"input": "-1000000000 -1000000000\n0 999999999",
"output": "1999999999"
},
{
"input": "0 0\n2 1",
"output": "2"
},
{
"input": "10 0\n100 0",
"output": "90"
},
{
"input": "1 5\n6 4",
"output": "5"
},
{
"input": "0 0\n5 4",
"output": "5"
},
{
"input": "10 1\n20 1",
"output": "10"
},
{
"input": "1 1\n-3 4",
"output": "4"
},
{
"input": "-863407280 504312726\n786535210 -661703810",
"output": "1649942490"
},
{
"input": "-588306085 -741137832\n341385643 152943311",
"output": "929691728"
},
{
"input": "0 0\n4 0",
"output": "4"
},
{
"input": "93097194 -48405232\n-716984003 -428596062",
"output": "810081197"
},
{
"input": "9 1\n1 1",
"output": "8"
},
{
"input": "4 6\n0 4",
"output": "4"
},
{
"input": "2 4\n5 2",
"output": "3"
},
{
"input": "-100000000 -100000000\n100000000 100000123",
"output": "200000123"
},
{
"input": "5 6\n5 7",
"output": "1"
},
{
"input": "12 16\n12 1",
"output": "15"
},
{
"input": "0 0\n5 1",
"output": "5"
},
{
"input": "0 1\n1 1",
"output": "1"
},
{
"input": "-44602634 913365223\n-572368780 933284951",
"output": "527766146"
},
{
"input": "-2 0\n2 -2",
"output": "4"
},
{
"input": "0 0\n3 1",
"output": "3"
},
{
"input": "-458 2\n1255 4548",
"output": "4546"
},
{
"input": "-5 -4\n-3 -3",
"output": "2"
},
{
"input": "4 5\n7 3",
"output": "3"
},
{
"input": "-1000000000 -999999999\n1000000000 999999998",
"output": "2000000000"
},
{
"input": "-1000000000 -1000000000\n1000000000 -1000000000",
"output": "2000000000"
},
{
"input": "-464122675 -898521847\n656107323 -625340409",
"output": "1120229998"
},
{
"input": "-463154699 -654742385\n-699179052 -789004997",
"output": "236024353"
},
{
"input": "982747270 -593488945\n342286841 -593604186",
"output": "640460429"
},
{
"input": "-80625246 708958515\n468950878 574646184",
"output": "549576124"
},
{
"input": "0 0\n1 0",
"output": "1"
},
{
"input": "109810 1\n2 3",
"output": "109808"
},
{
"input": "-9 0\n9 9",
"output": "18"
},
{
"input": "9 9\n9 9",
"output": "0"
},
{
"input": "1 1\n4 3",
"output": "3"
},
{
"input": "1 2\n45 1",
"output": "44"
},
{
"input": "207558188 -313753260\n-211535387 -721675423",
"output": "419093575"
},
{
"input": "-11 0\n0 0",
"output": "11"
},
{
"input": "-1000000000 1000000000\n1000000000 -1000000000",
"output": "2000000000"
},
{
"input": "0 0\n1 1",
"output": "1"
},
{
"input": "0 0\n0 1",
"output": "1"
},
{
"input": "0 0\n-1 1",
"output": "1"
},
{
"input": "0 0\n-1 0",
"output": "1"
},
{
"input": "0 0\n-1 -1",
"output": "1"
},
{
"input": "0 0\n0 -1",
"output": "1"
},
{
"input": "0 0\n1 -1",
"output": "1"
},
{
"input": "10 90\n90 10",
"output": "80"
},
{
"input": "851016864 573579544\n-761410925 -380746263",
"output": "1612427789"
},
{
"input": "1 9\n9 9",
"output": "8"
},
{
"input": "1000 1000\n1000 1000",
"output": "0"
},
{
"input": "1 9\n9 1",
"output": "8"
},
{
"input": "1 90\n90 90",
"output": "89"
},
{
"input": "100 100\n1000 1000",
"output": "900"
},
{
"input": "-1 0\n0 0",
"output": "1"
},
{
"input": "-750595959 -2984043\n649569876 -749608783",
"output": "1400165835"
},
{
"input": "958048496 712083589\n423286949 810566863",
"output": "534761547"
},
{
"input": "146316710 53945094\n-523054748 147499505",
"output": "669371458"
},
{
"input": "50383856 -596516251\n-802950224 -557916272",
"output": "853334080"
},
{
"input": "-637204864 -280290367\n-119020929 153679771",
"output": "518183935"
},
{
"input": "-100 -100\n-60 -91",
"output": "40"
},
{
"input": "337537326 74909428\n-765558776 167951547",
"output": "1103096102"
},
{
"input": "0 81\n18 90",
"output": "18"
},
{
"input": "283722202 -902633305\n-831696497 -160868946",
"output": "1115418699"
},
{
"input": "1000 1000\n-1000 1000",
"output": "2000"
},
{
"input": "5 6\n4 8",
"output": "2"
},
{
"input": "40572000 597493595\n-935051731 368493185",
"output": "975623731"
},
{
"input": "-5 5\n5 5",
"output": "10"
}
] | 1,626,985,251
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 75
| 77
| 6,758,400
|
[x1, y1] = map(int, input().split())
[x2, y2] = map(int, input().split())
a = x1-x2
b = y1-y2
a = abs(a)
b = abs(b)
if a>b:
print(a)
else:
print(b)
|
Title: Professor GukiZ's Robot
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Professor GukiZ makes a new robot. The robot are in the point with coordinates (*x*1,<=*y*1) and should go to the point (*x*2,<=*y*2). In a single step the robot can change any of its coordinates (maybe both of them) by one (decrease or increase). So the robot can move in one of the 8 directions. Find the minimal number of steps the robot should make to get the finish position.
Input Specification:
The first line contains two integers *x*1,<=*y*1 (<=-<=109<=≤<=*x*1,<=*y*1<=≤<=109) — the start position of the robot.
The second line contains two integers *x*2,<=*y*2 (<=-<=109<=≤<=*x*2,<=*y*2<=≤<=109) — the finish position of the robot.
Output Specification:
Print the only integer *d* — the minimal number of steps to get the finish position.
Demo Input:
['0 0\n4 5\n', '3 4\n6 1\n']
Demo Output:
['5\n', '3\n']
Note:
In the first example robot should increase both of its coordinates by one four times, so it will be in position (4, 4). After that robot should simply increase its *y* coordinate and get the finish position.
In the second example robot should simultaneously increase *x* coordinate and decrease *y* coordinate by one three times.
|
```python
[x1, y1] = map(int, input().split())
[x2, y2] = map(int, input().split())
a = x1-x2
b = y1-y2
a = abs(a)
b = abs(b)
if a>b:
print(a)
else:
print(b)
```
| 3
|
|
802
|
G
|
Fake News (easy)
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] | null | null |
As it's the first of April, Heidi is suspecting that the news she reads today are fake, and she does not want to look silly in front of all the contestants. She knows that a newspiece is fake if it contains heidi as a subsequence. Help Heidi assess whether the given piece is true, but please be discreet about it...
|
The first and only line of input contains a single nonempty string *s* of length at most 1000 composed of lowercase letters (a-z).
|
Output YES if the string *s* contains heidi as a subsequence and NO otherwise.
|
[
"abcheaibcdi\n",
"hiedi\n"
] |
[
"YES",
"NO"
] |
A string *s* contains another string *p* as a subsequence if it is possible to delete some characters from *s* and obtain *p*.
| 0
|
[
{
"input": "abcheaibcdi",
"output": "YES"
},
{
"input": "hiedi",
"output": "NO"
},
{
"input": "ihied",
"output": "NO"
},
{
"input": "diehi",
"output": "NO"
},
{
"input": "deiih",
"output": "NO"
},
{
"input": "iheid",
"output": "NO"
},
{
"input": "eihdi",
"output": "NO"
},
{
"input": "ehdii",
"output": "NO"
},
{
"input": "edhii",
"output": "NO"
},
{
"input": "deiih",
"output": "NO"
},
{
"input": "ehdii",
"output": "NO"
},
{
"input": "eufyajkssayhjhqcwxmctecaeepjwmfoscqprpcxsqfwnlgzsmmuwuoruantipholrauvxydfvftwfzhnckxswussvlidcojiciflpvkcxkkcmmvtfvxrkwcpeelwsuzqgamamdtdgzscmikvojfvqehblmjczkvtdeymgertgkwfwfukafqlfdhtedcctixhyetdypswgagrpyto",
"output": "YES"
},
{
"input": "arfbvxgdvqzuloojjrwoyqqbxamxybaqltfimofulusfebodjkwwrgwcppkwiodtpjaraglyplgerrpqjkpoggjmfxhwtqrijpijrcyxnoodvwpyjfpvqaoazllbrpzananbrvvybboedidtuvqquklkpeflfaltukjhzjgiofombhbmqbihgtapswykfvlgdoapjqntvqsaohmbvnphvyyhvhavslamczuqifxnwknkaenqmlvetrqogqxmlptgrmqvxzdxdmwobjesmgxckpmawtioavwdngyiwkzypfnxcovwzdohshwlavwsthdssiadhiwmhpvgkrbezm",
"output": "YES"
},
{
"input": "zcectngbqnejjjtsfrluummmqabzqbyccshjqbrjthzhlbmzjfxugvjouwhumsgrnopiyakfadjnbsesamhynsbfbfunupwbxvohfmpwlcpxhovwpfpciclatgmiufwdvtsqrsdcymvkldpnhfeisrzhyhhlkwdzthgprvkpyldeysvbmcibqkpudyrraqdlxpjecvwcvuiklcrsbgvqasmxmtxqzmawcjtozioqlfflinnxpeexbzloaeqjvglbdeufultpjqexvjjjkzemtzuzmxvawilcqdrcjzpqyhtwfphuonzwkotthsaxrmwtnlmcdylxqcfffyndqeouztluqwlhnkkvzwcfiscikv",
"output": "YES"
},
{
"input": "plqaykgovxkvsiahdbglktdlhcqwelxxmtlyymrsyubxdskvyjkrowvcbpdofpjqspsrgpakdczletxujzlsegepzleipiyycpinzxgwjsgslnxsotouddgfcybozfpjhhocpybfjbaywsehbcfrayvancbrumdfngqytnhihyxnlvilrqyhnxeckprqafofelospffhtwguzjbbjlzbqrtiielbvzutzgpqxosiaqznndgobcluuqlhmffiowkjdlkokehtjdyjvmxsiyxureflmdomerfekxdvtitvwzmdsdzplkpbtafxqfpudnhfqpoiwvjnylanunmagoweobdvfjgepbsymfutrjarlxclhgavpytiiqwvojrptofuvlohzeguxdsrihsbucelhhuedltnnjgzxwyblbqvnoliiydfinzlogbvucwykryzcyibnniggbkdkdcdgcsbvvnavtyhtkanrblpvomvjs",
"output": "YES"
},
{
"input": "fbldqzggeunkpwcfirxanmntbfrudijltoertsdvcvcmbwodbibsrxendzebvxwydpasaqnisrijctsuatihxxygbeovhxjdptdcppkvfytdpjspvrannxavmkmisqtygntxkdlousdypyfkrpzapysfpdbyprufwzhunlsfugojddkmxzinatiwfxdqmgyrnjnxvrclhxyuwxtshoqdjptmeecvgmrlvuwqtmnfnfeeiwcavwnqmyustawbjodzwsqmnjxhpqmgpysierlwbbdzcwprpsexyvreewcmlbvaiytjlxdqdaqftefdlmtmmjcwvfejshymhnouoshdzqcwzxpzupkbcievodzqkqvyjuuxxwepxjalvkzufnveji",
"output": "YES"
},
{
"input": "htsyljgoelbbuipivuzrhmfpkgderqpoprlxdpasxhpmxvaztccldtmujjzjmcpdvsdghzpretlsyyiljhjznseaacruriufswuvizwwuvdioazophhyytvbiogttnnouauxllbdn",
"output": "YES"
},
{
"input": "ikmxzqdzxqlvgeojsnhqzciujslwjyzzexnregabdqztpplosdakimjxmuqccbnwvzbajoiqgdobccwnrwmixohrbdarhoeeelzbpigiybtesybwefpcfx",
"output": "YES"
},
{
"input": "bpvbpjvbdfiodsmahxpcubjxdykesubnypalhypantshkjffmxjmelblqnjdmtaltneuyudyevkgedkqrdmrfeemgpghwrifcwincfixokfgurhqbcfzeajrgkgpwqwsepudxulywowwxzdxkumsicsvnzfxspmjpaixgejeaoyoibegosqoyoydmphfpbutrrewyjecowjckvpcceoamtfbitdneuwqfvnagswlskmsmkhmxyfsrpqwhxzocyffiumcy",
"output": "YES"
},
{
"input": "vllsexwrazvlfvhvrtqeohvzzresjdiuhomfpgqcxpqdevplecuaepixhlijatxzegciizpvyvxuembiplwklahlqibykfideysjygagjbgqkbhdhkatddcwlxboinfuomnpc",
"output": "YES"
},
{
"input": "pnjdwpxmvfoqkjtbhquqcuredrkwqzzfjmdvpnbqtypzdovemhhclkvigjvtprrpzbrbcbatkucaqteuciuozytsptvsskkeplaxdaqmjkmef",
"output": "NO"
},
{
"input": "jpwfhvlxvsdhtuozvlmnfiotrgapgjxtcsgcjnodcztupysvvvmjpzqkpommadppdrykuqkcpzojcwvlogvkddedwbggkrhuvtsvdiokehlkdlnukcufjvqxnikcdawvexxwffxtriqbdmkahxdtygodzohwtdmmuvmatdkvweqvaehaxiefpevkvqpyxsrhtmgjsdfcwzqobibeduooldrmglbinrepmunizheqzvgqvpdskhxfidxfnbisyizhepwyrcykcmjxnkyfjgrqlkixcvysa",
"output": "YES"
},
{
"input": "aftcrvuumeqbfvaqlltscnuhkpcifrrhnutjinxdhhdbzvizlrapzjdatuaynoplgjketupgaejciosofuhcgcjdcucarfvtsofgubtphijciswsvidnvpztlaarydkeqxzwdhfbmullkimerukusbrdnnujviydldrwhdfllsjtziwfeaiqotbiprespmxjulnyunkdtcghrzvhtcychkwatqqmladxpvmvlkzscthylbzkpgwlzfjqwarqvdeyngekqvrhrftpxnkfcibbowvnqdkulcdydspcubwlgoyinpnzgidbgunparnueddzwtzdiavbprbbg",
"output": "YES"
},
{
"input": "oagjghsidigeh",
"output": "NO"
},
{
"input": "chdhzpfzabupskiusjoefrwmjmqkbmdgboicnszkhdrlegeqjsldurmbshijadlwsycselhlnudndpdhcnhruhhvsgbthpruiqfirxkhpqhzhqdfpyozolbionodypfcqfeqbkcgmqkizgeyyelzeoothexcoaahedgrvoemqcwccbvoeqawqeuusyjxmgjkpfwcdttfmwunzuwvsihliexlzygqcgpbdiawfvqukikhbjerjkyhpcknlndaystrgsinghlmekbvhntcpypmchcwoglsmwwdulqneuabuuuvtyrnjxfcgoothalwkzzfxakneusezgnnepkpipzromqubraiggqndliz",
"output": "YES"
},
{
"input": "lgirxqkrkgjcutpqitmffvbujcljkqardlalyigxorscczuzikoylcxenryhskoavymexysvmhbsvhtycjlmzhijpuvcjshyfeycvvcfyzytzoyvxajpqdjtfiatnvxnyeqtfcagfftafllhhjhplbdsrfpctkqpinpdfrtlzyjllfbeffputywcckupyslkbbzpgcnxgbmhtqeqqehpdaokkjtatrhyiuusjhwgiiiikxpzdueasemosmmccoakafgvxduwiuflovhhfhffgnnjhoperhhjtvocpqytjxkmrknnknqeglffhfuplopmktykxuvcmbwpoeisrlyyhdpxfvzseucofyhziuiikihpqheqdyzwigeaqzhxzvporgisxgvhyicqyejovqloibhbunsvsunpvmdckkbuokitdzleilfwutcvuuytpupizinfjrzhxudsmjcjyfcpfgthujjowdwtgbvi",
"output": "YES"
},
{
"input": "uuehrvufgerqbzyzksmqnewacotuimawhlbycdbsmhshrsbqwybbkwjwsrkwptvlbbwjiivqugzrxxwgidrcrhrwsmwgeoleptfamzefgaeyxouxocrpvomjrazmxrnffdwrrmblgdiabdncvfougtmjgvvazasnygdrigbsrieoonirlivfyodvulouslxosswgpdexuldmkdbpdlgutiotvxjyecbrsvbmqxrlcpcipjjncduyqtohlzybvlemmfdeubihwlwqglkgjvnwrbgydcpwklmjeewqklmqdbajqgrpnynaxfvxjzgibqerxyhnxenrmcdqaaeksbzyrcaepozqpetaurlhjuxxhwppuhgoihxdxbmxeiahyaqkbknktlzkheaarjoqqrsyeducvoygwalgarldcdlqogfvsncejssmx",
"output": "YES"
},
{
"input": "iiopulfjxoitgiusqrhgbkiyzinphjtclodbkkydetylvuimkhdkklmyoacmekdvjpuxcrvqnjhqhhbfenlpzpwijtykqziocilvtpqhxuyrphdlamawjuzgjwiebkqyrzyqgtkcrhntjxqmcgkrqcslogjegfrivzidfedeegwbbsopvxvdoididlpypjogxaodtueebbwuwilacunqpozczcgrpaxxrtogpxgfkudtxchoravrrdtimhenwmnafxaruuojduxxglefejycfcyxllfimkszmbrhcwpnwjqgwvzysmlaaifdxfjjpgjmcksiigpanappjdloiolojmcqbnpnjjzaufdpjaknylmyvolhwypygwepmqwpiglpcnpypnudhdzpdvgzosyjthzcwtytxq",
"output": "YES"
}
] | 1,615,557,018
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 58
| 62
| 0
|
s = input()
a = 0
pat = "heidi"
for i in s:
if i == pat[a]:
a+=1
if(a == 5):
break
print("YES") if a==5 else print("NO")
|
Title: Fake News (easy)
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As it's the first of April, Heidi is suspecting that the news she reads today are fake, and she does not want to look silly in front of all the contestants. She knows that a newspiece is fake if it contains heidi as a subsequence. Help Heidi assess whether the given piece is true, but please be discreet about it...
Input Specification:
The first and only line of input contains a single nonempty string *s* of length at most 1000 composed of lowercase letters (a-z).
Output Specification:
Output YES if the string *s* contains heidi as a subsequence and NO otherwise.
Demo Input:
['abcheaibcdi\n', 'hiedi\n']
Demo Output:
['YES', 'NO']
Note:
A string *s* contains another string *p* as a subsequence if it is possible to delete some characters from *s* and obtain *p*.
|
```python
s = input()
a = 0
pat = "heidi"
for i in s:
if i == pat[a]:
a+=1
if(a == 5):
break
print("YES") if a==5 else print("NO")
```
| 3
|
|
50
|
A
|
Domino piling
|
PROGRAMMING
| 800
|
[
"greedy",
"math"
] |
A. Domino piling
|
2
|
256
|
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
|
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
|
Output one number — the maximal number of dominoes, which can be placed.
|
[
"2 4\n",
"3 3\n"
] |
[
"4\n",
"4\n"
] |
none
| 500
|
[
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "2 6",
"output": "6"
},
{
"input": "2 7",
"output": "7"
},
{
"input": "2 14",
"output": "14"
},
{
"input": "2 15",
"output": "15"
},
{
"input": "1 4",
"output": "2"
},
{
"input": "2 16",
"output": "16"
},
{
"input": "3 5",
"output": "7"
},
{
"input": "3 6",
"output": "9"
},
{
"input": "3 10",
"output": "15"
},
{
"input": "3 14",
"output": "21"
},
{
"input": "3 15",
"output": "22"
},
{
"input": "3 16",
"output": "24"
},
{
"input": "5 7",
"output": "17"
},
{
"input": "16 16",
"output": "128"
},
{
"input": "15 16",
"output": "120"
},
{
"input": "2 3",
"output": "3"
},
{
"input": "15 15",
"output": "112"
},
{
"input": "14 16",
"output": "112"
},
{
"input": "11 13",
"output": "71"
},
{
"input": "5 16",
"output": "40"
},
{
"input": "8 15",
"output": "60"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "3 4",
"output": "6"
},
{
"input": "4 4",
"output": "8"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "1 3",
"output": "1"
},
{
"input": "14 15",
"output": "105"
}
] | 1,632,316,793
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 62
| 6,758,400
|
a, b = map(int(), input(), split())
c = a * b
print(c//2)
|
Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
Input Specification:
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output Specification:
Output one number — the maximal number of dominoes, which can be placed.
Demo Input:
['2 4\n', '3 3\n']
Demo Output:
['4\n', '4\n']
Note:
none
|
```python
a, b = map(int(), input(), split())
c = a * b
print(c//2)
```
| -1
|
535
|
B
|
Tavas and SaDDas
|
PROGRAMMING
| 1,100
|
[
"bitmasks",
"brute force",
"combinatorics",
"implementation"
] | null | null |
Once again Tavas started eating coffee mix without water! Keione told him that it smells awful, but he didn't stop doing that. That's why Keione told his smart friend, SaDDas to punish him! SaDDas took Tavas' headphones and told him: "If you solve the following problem, I'll return it to you."
The problem is:
You are given a lucky number *n*. Lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
If we sort all lucky numbers in increasing order, what's the 1-based index of *n*?
Tavas is not as smart as SaDDas, so he asked you to do him a favor and solve this problem so he can have his headphones back.
|
The first and only line of input contains a lucky number *n* (1<=≤<=*n*<=≤<=109).
|
Print the index of *n* among all lucky numbers.
|
[
"4\n",
"7\n",
"77\n"
] |
[
"1\n",
"2\n",
"6\n"
] |
none
| 1,000
|
[
{
"input": "4",
"output": "1"
},
{
"input": "7",
"output": "2"
},
{
"input": "77",
"output": "6"
},
{
"input": "4",
"output": "1"
},
{
"input": "474744",
"output": "83"
},
{
"input": "777774",
"output": "125"
},
{
"input": "447",
"output": "8"
},
{
"input": "774",
"output": "13"
},
{
"input": "4",
"output": "1"
},
{
"input": "4447747",
"output": "140"
},
{
"input": "7747474",
"output": "233"
},
{
"input": "4444",
"output": "15"
},
{
"input": "4447",
"output": "16"
},
{
"input": "7",
"output": "2"
},
{
"input": "4",
"output": "1"
},
{
"input": "4447744",
"output": "139"
},
{
"input": "77474",
"output": "57"
},
{
"input": "7747",
"output": "28"
},
{
"input": "444",
"output": "7"
},
{
"input": "7",
"output": "2"
},
{
"input": "7774477",
"output": "242"
},
{
"input": "4477774",
"output": "157"
},
{
"input": "7444",
"output": "23"
},
{
"input": "7474747",
"output": "212"
},
{
"input": "77",
"output": "6"
},
{
"input": "774477",
"output": "114"
},
{
"input": "7",
"output": "2"
},
{
"input": "47",
"output": "4"
},
{
"input": "747777",
"output": "110"
},
{
"input": "444444444",
"output": "511"
},
{
"input": "777777777",
"output": "1022"
},
{
"input": "477477447",
"output": "728"
},
{
"input": "777744747",
"output": "996"
}
] | 1,447,451,388
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 46
| 0
|
n = input()
res = 1
for el in n:
res *= 1 if el == "4" else 2
if len(n) > 1:
res += 2 ** (len(n) - 1)
print(res)
|
Title: Tavas and SaDDas
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Once again Tavas started eating coffee mix without water! Keione told him that it smells awful, but he didn't stop doing that. That's why Keione told his smart friend, SaDDas to punish him! SaDDas took Tavas' headphones and told him: "If you solve the following problem, I'll return it to you."
The problem is:
You are given a lucky number *n*. Lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
If we sort all lucky numbers in increasing order, what's the 1-based index of *n*?
Tavas is not as smart as SaDDas, so he asked you to do him a favor and solve this problem so he can have his headphones back.
Input Specification:
The first and only line of input contains a lucky number *n* (1<=≤<=*n*<=≤<=109).
Output Specification:
Print the index of *n* among all lucky numbers.
Demo Input:
['4\n', '7\n', '77\n']
Demo Output:
['1\n', '2\n', '6\n']
Note:
none
|
```python
n = input()
res = 1
for el in n:
res *= 1 if el == "4" else 2
if len(n) > 1:
res += 2 ** (len(n) - 1)
print(res)
```
| 0
|
|
462
|
B
|
Appleman and Card Game
|
PROGRAMMING
| 1,300
|
[
"greedy"
] | null | null |
Appleman has *n* cards. Each card has an uppercase letter written on it. Toastman must choose *k* cards from Appleman's cards. Then Appleman should give Toastman some coins depending on the chosen cards. Formally, for each Toastman's card *i* you should calculate how much Toastman's cards have the letter equal to letter on *i*th, then sum up all these quantities, such a number of coins Appleman should give to Toastman.
Given the description of Appleman's cards. What is the maximum number of coins Toastman can get?
|
The first line contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105). The next line contains *n* uppercase letters without spaces — the *i*-th letter describes the *i*-th card of the Appleman.
|
Print a single integer – the answer to the problem.
|
[
"15 10\nDZFDFZDFDDDDDDF\n",
"6 4\nYJSNPI\n"
] |
[
"82\n",
"4\n"
] |
In the first test example Toastman can choose nine cards with letter D and one additional card with any letter. For each card with D he will get 9 coins and for the additional card he will get 1 coin.
| 1,000
|
[
{
"input": "15 10\nDZFDFZDFDDDDDDF",
"output": "82"
},
{
"input": "6 4\nYJSNPI",
"output": "4"
},
{
"input": "5 3\nAOWBY",
"output": "3"
},
{
"input": "1 1\nV",
"output": "1"
},
{
"input": "2 1\nWT",
"output": "1"
},
{
"input": "2 2\nBL",
"output": "2"
},
{
"input": "5 1\nFACJT",
"output": "1"
},
{
"input": "5 5\nMJDIJ",
"output": "7"
},
{
"input": "15 5\nAZBIPTOFTJCJJIK",
"output": "13"
},
{
"input": "100 1\nEVEEVEEEGGECFEHEFVFVFHVHEEEEEFCVEEEEEEVFVEEVEEHEEVEFEVVEFEEEFEVECEHGHEEFGEEVCEECCECEFHEVEEEEEEGEEHVH",
"output": "1"
},
{
"input": "100 15\nKKTFFUTFCKUIKKKKFIFFKTUKUUKUKKIKKKTIFKTKUCFFKKKIIKKKKKKTFKFKKIRKKKFKUUKIKUUUFFKKKKTUZKITUIKKIKUKKTIK",
"output": "225"
},
{
"input": "100 50\nYYIYYAAAIEAAYAYAEAIIIAAEAAYEAEYYYIAEYAYAYYAAAIAYAEAAYAYYIYAAYYAAAAAAIYYYAAYAAEAAYAIEIYIYAYAYAYIIAAEY",
"output": "1972"
},
{
"input": "100 90\nFAFAOOAOOAFAOTFAFAFFATAAAOFAAOAFBAAAFBOAOFFFOAOAFAPFOFAOFAAFOAAAAFAAFOFAAOFPPAAOOAAOOFFOFFFOFAOTOFAF",
"output": "2828"
},
{
"input": "100 99\nBFFBBFBFBQFFFFFQBFFBFFBQFBFQFBBFQFFFBFFFBFQFQFBFFBBFYQFBFFFFFFFBQQFQBFBQBQFFFBQQFFFBQFYFBFBFFFBBBQQY",
"output": "3713"
},
{
"input": "100 100\nMQSBDAJABILIBCUEOWGWCEXMUTEYQKAIWGINXVQEOFDUBSVULROQHQRZZAALVQFEFRAAAYUIMGCAFQGIAEFBETRECGSFQJNXHHDN",
"output": "514"
},
{
"input": "100 50\nBMYIXQSJNHGFVFPJBIOBXIKSFNUFPVODCUBQYSIIQNVNXXCWXWRHKFEUPPIIDDGRDBJLZDCBMNJMYRMWFIHOSTDJJHXHPNRKWNFD",
"output": "328"
},
{
"input": "100 50\nENFNEMLJEMDMFMNNGNIMNINALGLLLAEMENEMNLMMIEIJNAINBJEJMFJLLIMINELGFLAIAMJMHMGNLIEFJIEEFEFGLLLDLMEAEIMM",
"output": "748"
}
] | 1,588,436,572
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 77
| 204,800
|
n, k = list(map(int,input().split()))
s = input()
d = {}
l = []
count = 0
for i in range(len(s)):
d[s[i]] = s.count(s[i])
for i in d.values():
l.append(i)
l.sort()
for i in range(len(l), -1, -1):
if k<=0:
break
if k-l[i]>=0:
count += l[i]**2
k-=l[i]
print(count)
|
Title: Appleman and Card Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Appleman has *n* cards. Each card has an uppercase letter written on it. Toastman must choose *k* cards from Appleman's cards. Then Appleman should give Toastman some coins depending on the chosen cards. Formally, for each Toastman's card *i* you should calculate how much Toastman's cards have the letter equal to letter on *i*th, then sum up all these quantities, such a number of coins Appleman should give to Toastman.
Given the description of Appleman's cards. What is the maximum number of coins Toastman can get?
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105). The next line contains *n* uppercase letters without spaces — the *i*-th letter describes the *i*-th card of the Appleman.
Output Specification:
Print a single integer – the answer to the problem.
Demo Input:
['15 10\nDZFDFZDFDDDDDDF\n', '6 4\nYJSNPI\n']
Demo Output:
['82\n', '4\n']
Note:
In the first test example Toastman can choose nine cards with letter D and one additional card with any letter. For each card with D he will get 9 coins and for the additional card he will get 1 coin.
|
```python
n, k = list(map(int,input().split()))
s = input()
d = {}
l = []
count = 0
for i in range(len(s)):
d[s[i]] = s.count(s[i])
for i in d.values():
l.append(i)
l.sort()
for i in range(len(l), -1, -1):
if k<=0:
break
if k-l[i]>=0:
count += l[i]**2
k-=l[i]
print(count)
```
| -1
|
|
567
|
A
|
Lineland Mail
|
PROGRAMMING
| 900
|
[
"greedy",
"implementation"
] | null | null |
All cities of Lineland are located on the *Ox* coordinate axis. Thus, each city is associated with its position *x**i* — a coordinate on the *Ox* axis. No two cities are located at a single point.
Lineland residents love to send letters to each other. A person may send a letter only if the recipient lives in another city (because if they live in the same city, then it is easier to drop in).
Strange but true, the cost of sending the letter is exactly equal to the distance between the sender's city and the recipient's city.
For each city calculate two values *min**i* and *max**i*, where *min**i* is the minimum cost of sending a letter from the *i*-th city to some other city, and *max**i* is the the maximum cost of sending a letter from the *i*-th city to some other city
|
The first line of the input contains integer *n* (2<=≤<=*n*<=≤<=105) — the number of cities in Lineland. The second line contains the sequence of *n* distinct integers *x*1,<=*x*2,<=...,<=*x**n* (<=-<=109<=≤<=*x**i*<=≤<=109), where *x**i* is the *x*-coordinate of the *i*-th city. All the *x**i*'s are distinct and follow in ascending order.
|
Print *n* lines, the *i*-th line must contain two integers *min**i*,<=*max**i*, separated by a space, where *min**i* is the minimum cost of sending a letter from the *i*-th city, and *max**i* is the maximum cost of sending a letter from the *i*-th city.
|
[
"4\n-5 -2 2 7\n",
"2\n-1 1\n"
] |
[
"3 12\n3 9\n4 7\n5 12\n",
"2 2\n2 2\n"
] |
none
| 500
|
[
{
"input": "4\n-5 -2 2 7",
"output": "3 12\n3 9\n4 7\n5 12"
},
{
"input": "2\n-1 1",
"output": "2 2\n2 2"
},
{
"input": "3\n-1 0 1",
"output": "1 2\n1 1\n1 2"
},
{
"input": "4\n-1 0 1 3",
"output": "1 4\n1 3\n1 2\n2 4"
},
{
"input": "3\n-1000000000 0 1000000000",
"output": "1000000000 2000000000\n1000000000 1000000000\n1000000000 2000000000"
},
{
"input": "2\n-1000000000 1000000000",
"output": "2000000000 2000000000\n2000000000 2000000000"
},
{
"input": "10\n1 10 12 15 59 68 130 912 1239 9123",
"output": "9 9122\n2 9113\n2 9111\n3 9108\n9 9064\n9 9055\n62 8993\n327 8211\n327 7884\n7884 9122"
},
{
"input": "5\n-2 -1 0 1 2",
"output": "1 4\n1 3\n1 2\n1 3\n1 4"
},
{
"input": "5\n-2 -1 0 1 3",
"output": "1 5\n1 4\n1 3\n1 3\n2 5"
},
{
"input": "3\n-10000 1 10000",
"output": "10001 20000\n9999 10001\n9999 20000"
},
{
"input": "5\n-1000000000 -999999999 -999999998 -999999997 -999999996",
"output": "1 4\n1 3\n1 2\n1 3\n1 4"
},
{
"input": "10\n-857422304 -529223472 82412729 145077145 188538640 265299215 527377039 588634631 592896147 702473706",
"output": "328198832 1559896010\n328198832 1231697178\n62664416 939835033\n43461495 1002499449\n43461495 1045960944\n76760575 1122721519\n61257592 1384799343\n4261516 1446056935\n4261516 1450318451\n109577559 1559896010"
},
{
"input": "10\n-876779400 -829849659 -781819137 -570920213 18428128 25280705 121178189 219147240 528386329 923854124",
"output": "46929741 1800633524\n46929741 1753703783\n48030522 1705673261\n210898924 1494774337\n6852577 905425996\n6852577 902060105\n95897484 997957589\n97969051 1095926640\n309239089 1405165729\n395467795 1800633524"
},
{
"input": "30\n-15 1 21 25 30 40 59 60 77 81 97 100 103 123 139 141 157 158 173 183 200 215 226 231 244 256 267 279 289 292",
"output": "16 307\n16 291\n4 271\n4 267\n5 262\n10 252\n1 233\n1 232\n4 215\n4 211\n3 195\n3 192\n3 189\n16 169\n2 154\n2 156\n1 172\n1 173\n10 188\n10 198\n15 215\n11 230\n5 241\n5 246\n12 259\n11 271\n11 282\n10 294\n3 304\n3 307"
},
{
"input": "10\n-1000000000 -999999999 -999999997 -999999996 -999999995 -999999994 -999999992 -999999990 -999999988 -999999986",
"output": "1 14\n1 13\n1 11\n1 10\n1 9\n1 8\n2 8\n2 10\n2 12\n2 14"
},
{
"input": "50\n-50000 -49459 -48875 -48456 -48411 -48096 -47901 -47500 -47150 -46808 -46687 -46679 -46337 -45747 -45604 -45194 -44752 -44242 -44231 -44122 -43636 -43274 -42916 -42881 -42386 -42095 -41830 -41618 -41145 -40897 -40534 -40007 -39761 -39389 -39104 -38909 -38630 -38561 -38364 -38255 -38214 -38084 -37959 -37607 -37202 -36890 -36681 -36136 -36123 -35886",
"output": "541 14114\n541 13573\n419 12989\n45 12570\n45 12525\n195 12210\n195 12015\n350 11614\n342 11264\n121 10922\n8 10801\n8 10793\n342 10451\n143 9861\n143 9718\n410 9308\n442 8866\n11 8356\n11 8345\n109 8236\n362 7750\n358 7388\n35 7084\n35 7119\n291 7614\n265 7905\n212 8170\n212 8382\n248 8855\n248 9103\n363 9466\n246 9993\n246 10239\n285 10611\n195 10896\n195 11091\n69 11370\n69 11439\n109 11636\n41 11745\n41 11786\n125 11916\n125 12041\n352 12393\n312 12798\n209 13110\n209 13319\n13 13864\n13 13877\n237 141..."
},
{
"input": "3\n-1000000000 999999999 1000000000",
"output": "1999999999 2000000000\n1 1999999999\n1 2000000000"
}
] | 1,613,726,443
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 54
| 498
| 8,704,000
|
n = int(input())
x = list(map(int, input().split()))
min_x = 0
max_x = 0
for i in range(len(x)):
if 0 < i < len(x) - 1:
min_x = min(x[i]-x[i-1], x[i+1]-x[i])
max_x = max(x[i] - x[0], x[-1] - x[i])
elif i == 0:
min_x = x[i+1] - x[0]
max_x = x[-1] - x[0]
elif i == len(x)-1:
min_x = x[i] - x[i-1]
max_x = x[i] - x[0]
print(str(min_x) + " " + str(max_x))
|
Title: Lineland Mail
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
All cities of Lineland are located on the *Ox* coordinate axis. Thus, each city is associated with its position *x**i* — a coordinate on the *Ox* axis. No two cities are located at a single point.
Lineland residents love to send letters to each other. A person may send a letter only if the recipient lives in another city (because if they live in the same city, then it is easier to drop in).
Strange but true, the cost of sending the letter is exactly equal to the distance between the sender's city and the recipient's city.
For each city calculate two values *min**i* and *max**i*, where *min**i* is the minimum cost of sending a letter from the *i*-th city to some other city, and *max**i* is the the maximum cost of sending a letter from the *i*-th city to some other city
Input Specification:
The first line of the input contains integer *n* (2<=≤<=*n*<=≤<=105) — the number of cities in Lineland. The second line contains the sequence of *n* distinct integers *x*1,<=*x*2,<=...,<=*x**n* (<=-<=109<=≤<=*x**i*<=≤<=109), where *x**i* is the *x*-coordinate of the *i*-th city. All the *x**i*'s are distinct and follow in ascending order.
Output Specification:
Print *n* lines, the *i*-th line must contain two integers *min**i*,<=*max**i*, separated by a space, where *min**i* is the minimum cost of sending a letter from the *i*-th city, and *max**i* is the maximum cost of sending a letter from the *i*-th city.
Demo Input:
['4\n-5 -2 2 7\n', '2\n-1 1\n']
Demo Output:
['3 12\n3 9\n4 7\n5 12\n', '2 2\n2 2\n']
Note:
none
|
```python
n = int(input())
x = list(map(int, input().split()))
min_x = 0
max_x = 0
for i in range(len(x)):
if 0 < i < len(x) - 1:
min_x = min(x[i]-x[i-1], x[i+1]-x[i])
max_x = max(x[i] - x[0], x[-1] - x[i])
elif i == 0:
min_x = x[i+1] - x[0]
max_x = x[-1] - x[0]
elif i == len(x)-1:
min_x = x[i] - x[i-1]
max_x = x[i] - x[0]
print(str(min_x) + " " + str(max_x))
```
| 3
|
|
854
|
B
|
Maxim Buys an Apartment
|
PROGRAMMING
| 1,200
|
[
"constructive algorithms",
"math"
] | null | null |
Maxim wants to buy an apartment in a new house at Line Avenue of Metropolis. The house has *n* apartments that are numbered from 1 to *n* and are arranged in a row. Two apartments are adjacent if their indices differ by 1. Some of the apartments can already be inhabited, others are available for sale.
Maxim often visits his neighbors, so apartment is good for him if it is available for sale and there is at least one already inhabited apartment adjacent to it. Maxim knows that there are exactly *k* already inhabited apartments, but he doesn't know their indices yet.
Find out what could be the minimum possible and the maximum possible number of apartments that are good for Maxim.
|
The only line of the input contains two integers: *n* and *k* (1<=≤<=*n*<=≤<=109, 0<=≤<=*k*<=≤<=*n*).
|
Print the minimum possible and the maximum possible number of apartments good for Maxim.
|
[
"6 3\n"
] |
[
"1 3\n"
] |
In the sample test, the number of good apartments could be minimum possible if, for example, apartments with indices 1, 2 and 3 were inhabited. In this case only apartment 4 is good. The maximum possible number could be, for example, if apartments with indices 1, 3 and 5 were inhabited. In this case all other apartments: 2, 4 and 6 are good.
| 1,000
|
[
{
"input": "6 3",
"output": "1 3"
},
{
"input": "10 1",
"output": "1 2"
},
{
"input": "10 9",
"output": "1 1"
},
{
"input": "8 0",
"output": "0 0"
},
{
"input": "8 8",
"output": "0 0"
},
{
"input": "966871928 890926970",
"output": "1 75944958"
},
{
"input": "20 2",
"output": "1 4"
},
{
"input": "1 0",
"output": "0 0"
},
{
"input": "1 1",
"output": "0 0"
},
{
"input": "2 0",
"output": "0 0"
},
{
"input": "2 1",
"output": "1 1"
},
{
"input": "2 2",
"output": "0 0"
},
{
"input": "7 2",
"output": "1 4"
},
{
"input": "8 3",
"output": "1 5"
},
{
"input": "9 4",
"output": "1 5"
},
{
"input": "10 3",
"output": "1 6"
},
{
"input": "10 4",
"output": "1 6"
},
{
"input": "10 5",
"output": "1 5"
},
{
"input": "1000 1000",
"output": "0 0"
},
{
"input": "1000 333",
"output": "1 666"
},
{
"input": "1000 334",
"output": "1 666"
},
{
"input": "999 333",
"output": "1 666"
},
{
"input": "999 334",
"output": "1 665"
},
{
"input": "998 332",
"output": "1 664"
},
{
"input": "998 333",
"output": "1 665"
},
{
"input": "89 4",
"output": "1 8"
},
{
"input": "66 50",
"output": "1 16"
},
{
"input": "88 15",
"output": "1 30"
},
{
"input": "95 43",
"output": "1 52"
},
{
"input": "900 344",
"output": "1 556"
},
{
"input": "777 113",
"output": "1 226"
},
{
"input": "964 42",
"output": "1 84"
},
{
"input": "982 867",
"output": "1 115"
},
{
"input": "1000000000 0",
"output": "0 0"
},
{
"input": "1000000000 1000000000",
"output": "0 0"
},
{
"input": "1000000000 333333333",
"output": "1 666666666"
},
{
"input": "1000000000 333333334",
"output": "1 666666666"
},
{
"input": "999999999 333333333",
"output": "1 666666666"
},
{
"input": "999999999 333333334",
"output": "1 666666665"
},
{
"input": "999999998 333333332",
"output": "1 666666664"
},
{
"input": "999999998 333333333",
"output": "1 666666665"
},
{
"input": "78602604 42160832",
"output": "1 36441772"
},
{
"input": "35679021 9137902",
"output": "1 18275804"
},
{
"input": "41949373 13173511",
"output": "1 26347022"
},
{
"input": "77855558 49163875",
"output": "1 28691683"
},
{
"input": "87187123 2851901",
"output": "1 5703802"
},
{
"input": "66849627 25004217",
"output": "1 41845410"
},
{
"input": "873046672 517064947",
"output": "1 355981725"
},
{
"input": "639857373 1393427",
"output": "1 2786854"
},
{
"input": "637563683 69636269",
"output": "1 139272538"
},
{
"input": "911669737 141068293",
"output": "1 282136586"
},
{
"input": "547575919 313272818",
"output": "1 234303101"
},
{
"input": "955020006 297895809",
"output": "1 595791618"
},
{
"input": "10 4",
"output": "1 6"
},
{
"input": "11 3",
"output": "1 6"
},
{
"input": "10 3",
"output": "1 6"
},
{
"input": "4 1",
"output": "1 2"
},
{
"input": "9 3",
"output": "1 6"
},
{
"input": "7 2",
"output": "1 4"
},
{
"input": "7 3",
"output": "1 4"
},
{
"input": "12 5",
"output": "1 7"
},
{
"input": "8 3",
"output": "1 5"
},
{
"input": "1000 8",
"output": "1 16"
}
] | 1,505,743,173
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 62
| 0
|
n, k = list(map(int, input().split()))
if k == 0:
print(0, 0)
elif n - k < k:
print(1, n - k)
else:
for i in range(k + 1):
if (k - i) * 2 + i <= n - k:
print(1, (k - i) * 2 + i)
break
|
Title: Maxim Buys an Apartment
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Maxim wants to buy an apartment in a new house at Line Avenue of Metropolis. The house has *n* apartments that are numbered from 1 to *n* and are arranged in a row. Two apartments are adjacent if their indices differ by 1. Some of the apartments can already be inhabited, others are available for sale.
Maxim often visits his neighbors, so apartment is good for him if it is available for sale and there is at least one already inhabited apartment adjacent to it. Maxim knows that there are exactly *k* already inhabited apartments, but he doesn't know their indices yet.
Find out what could be the minimum possible and the maximum possible number of apartments that are good for Maxim.
Input Specification:
The only line of the input contains two integers: *n* and *k* (1<=≤<=*n*<=≤<=109, 0<=≤<=*k*<=≤<=*n*).
Output Specification:
Print the minimum possible and the maximum possible number of apartments good for Maxim.
Demo Input:
['6 3\n']
Demo Output:
['1 3\n']
Note:
In the sample test, the number of good apartments could be minimum possible if, for example, apartments with indices 1, 2 and 3 were inhabited. In this case only apartment 4 is good. The maximum possible number could be, for example, if apartments with indices 1, 3 and 5 were inhabited. In this case all other apartments: 2, 4 and 6 are good.
|
```python
n, k = list(map(int, input().split()))
if k == 0:
print(0, 0)
elif n - k < k:
print(1, n - k)
else:
for i in range(k + 1):
if (k - i) * 2 + i <= n - k:
print(1, (k - i) * 2 + i)
break
```
| 0
|
|
50
|
A
|
Domino piling
|
PROGRAMMING
| 800
|
[
"greedy",
"math"
] |
A. Domino piling
|
2
|
256
|
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
|
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
|
Output one number — the maximal number of dominoes, which can be placed.
|
[
"2 4\n",
"3 3\n"
] |
[
"4\n",
"4\n"
] |
none
| 500
|
[
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "2 6",
"output": "6"
},
{
"input": "2 7",
"output": "7"
},
{
"input": "2 14",
"output": "14"
},
{
"input": "2 15",
"output": "15"
},
{
"input": "1 4",
"output": "2"
},
{
"input": "2 16",
"output": "16"
},
{
"input": "3 5",
"output": "7"
},
{
"input": "3 6",
"output": "9"
},
{
"input": "3 10",
"output": "15"
},
{
"input": "3 14",
"output": "21"
},
{
"input": "3 15",
"output": "22"
},
{
"input": "3 16",
"output": "24"
},
{
"input": "5 7",
"output": "17"
},
{
"input": "16 16",
"output": "128"
},
{
"input": "15 16",
"output": "120"
},
{
"input": "2 3",
"output": "3"
},
{
"input": "15 15",
"output": "112"
},
{
"input": "14 16",
"output": "112"
},
{
"input": "11 13",
"output": "71"
},
{
"input": "5 16",
"output": "40"
},
{
"input": "8 15",
"output": "60"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "3 4",
"output": "6"
},
{
"input": "4 4",
"output": "8"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "1 3",
"output": "1"
},
{
"input": "14 15",
"output": "105"
}
] | 1,625,023,441
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 35
| 154
| 0
|
import math as mt
[m, n] = input().split(" ")
area = int(m) * int(n)
print(mt.floor(area/2))
|
Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
Input Specification:
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output Specification:
Output one number — the maximal number of dominoes, which can be placed.
Demo Input:
['2 4\n', '3 3\n']
Demo Output:
['4\n', '4\n']
Note:
none
|
```python
import math as mt
[m, n] = input().split(" ")
area = int(m) * int(n)
print(mt.floor(area/2))
```
| 3.9615
|
886
|
B
|
Vlad and Cafes
|
PROGRAMMING
| 1,000
|
[] | 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<=≤<=*n*<=≤<=2·105) — number of cafes indices written by Vlad.
In second line, *n* numbers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=2·105) are written — 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 integer — 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.
| 1,000
|
[
{
"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"
}
] | 1,513,883,598
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 62
| 5,529,600
|
a=input().split()
b={}
for c in range(len(a)):
a[c]=int(a[c])
b[a[c]]=c
x=list(b.keys())
y=list((b.values()))
mi=len(y)
for p in range(len(y)):
if y[p]<=mi:
mi=y[p]
m=p
print(x[m])
|
Title: Vlad and Cafes
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<=≤<=*n*<=≤<=2·105) — number of cafes indices written by Vlad.
In second line, *n* numbers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=2·105) are written — 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 integer — 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
a=input().split()
b={}
for c in range(len(a)):
a[c]=int(a[c])
b[a[c]]=c
x=list(b.keys())
y=list((b.values()))
mi=len(y)
for p in range(len(y)):
if y[p]<=mi:
mi=y[p]
m=p
print(x[m])
```
| 0
|
|
34
|
A
|
Reconnaissance 2
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Reconnaissance 2
|
2
|
256
|
*n* soldiers stand in a circle. For each soldier his height *a**i* is known. A reconnaissance unit can be made of such two neighbouring soldiers, whose heights difference is minimal, i.e. |*a**i*<=-<=*a**j*| is minimal. So each of them will be less noticeable with the other. Output any pair of soldiers that can form a reconnaissance unit.
|
The first line contains integer *n* (2<=≤<=*n*<=≤<=100) — amount of soldiers. Then follow the heights of the soldiers in their order in the circle — *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000). The soldier heights are given in clockwise or counterclockwise direction.
|
Output two integers — indexes of neighbouring soldiers, who should form a reconnaissance unit. If there are many optimum solutions, output any of them. Remember, that the soldiers stand in a circle.
|
[
"5\n10 12 13 15 10\n",
"4\n10 20 30 40\n"
] |
[
"5 1\n",
"1 2\n"
] |
none
| 500
|
[
{
"input": "5\n10 12 13 15 10",
"output": "5 1"
},
{
"input": "4\n10 20 30 40",
"output": "1 2"
},
{
"input": "6\n744 359 230 586 944 442",
"output": "2 3"
},
{
"input": "5\n826 747 849 687 437",
"output": "1 2"
},
{
"input": "5\n999 999 993 969 999",
"output": "1 2"
},
{
"input": "5\n4 24 6 1 15",
"output": "3 4"
},
{
"input": "2\n511 32",
"output": "1 2"
},
{
"input": "3\n907 452 355",
"output": "2 3"
},
{
"input": "4\n303 872 764 401",
"output": "4 1"
},
{
"input": "10\n684 698 429 694 956 812 594 170 937 764",
"output": "1 2"
},
{
"input": "20\n646 840 437 946 640 564 936 917 487 752 844 734 468 969 674 646 728 642 514 695",
"output": "7 8"
},
{
"input": "30\n996 999 998 984 989 1000 996 993 1000 983 992 999 999 1000 979 992 987 1000 996 1000 1000 989 981 996 995 999 999 989 999 1000",
"output": "12 13"
},
{
"input": "50\n93 27 28 4 5 78 59 24 19 134 31 128 118 36 90 32 32 1 44 32 33 13 31 10 12 25 38 50 25 12 4 22 28 53 48 83 4 25 57 31 71 24 8 7 28 86 23 80 101 58",
"output": "16 17"
},
{
"input": "88\n1000 1000 1000 1000 1000 998 998 1000 1000 1000 1000 999 999 1000 1000 1000 999 1000 997 999 997 1000 999 998 1000 999 1000 1000 1000 999 1000 999 999 1000 1000 999 1000 999 1000 1000 998 1000 1000 1000 998 998 1000 1000 999 1000 1000 1000 1000 1000 1000 1000 998 1000 1000 1000 999 1000 1000 999 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 998 1000 1000 1000 998 1000 1000 998 1000 999 1000 1000 1000 1000",
"output": "1 2"
},
{
"input": "99\n4 4 21 6 5 3 13 2 6 1 3 4 1 3 1 9 11 1 6 17 4 5 20 4 1 9 5 11 3 4 14 1 3 3 1 4 3 5 27 1 1 2 10 7 11 4 19 7 11 6 11 13 3 1 10 7 2 1 16 1 9 4 29 13 2 12 14 2 21 1 9 8 26 12 12 5 2 14 7 8 8 8 9 4 12 2 6 6 7 16 8 14 2 10 20 15 3 7 4",
"output": "1 2"
},
{
"input": "100\n713 572 318 890 577 657 646 146 373 783 392 229 455 871 20 593 573 336 26 381 280 916 907 732 820 713 111 840 570 446 184 711 481 399 788 647 492 15 40 530 549 506 719 782 126 20 778 996 712 761 9 74 812 418 488 175 103 585 900 3 604 521 109 513 145 708 990 361 682 827 791 22 596 780 596 385 450 643 158 496 876 975 319 783 654 895 891 361 397 81 682 899 347 623 809 557 435 279 513 438",
"output": "86 87"
},
{
"input": "100\n31 75 86 68 111 27 22 22 26 30 54 163 107 75 160 122 14 23 17 26 27 20 43 58 59 71 21 148 9 32 43 91 133 286 132 70 90 156 84 14 77 93 23 18 13 72 18 131 33 28 72 175 30 86 249 20 14 208 28 57 63 199 6 10 24 30 62 267 43 479 60 28 138 1 45 3 19 47 7 166 116 117 50 140 28 14 95 85 93 43 61 15 2 70 10 51 7 95 9 25",
"output": "7 8"
},
{
"input": "100\n896 898 967 979 973 709 961 968 806 967 896 967 826 975 936 903 986 856 851 931 852 971 786 837 949 978 686 936 952 909 965 749 908 916 943 973 983 975 939 886 964 928 960 976 907 788 994 773 949 871 947 980 945 985 726 981 887 943 907 990 931 874 840 867 948 951 961 904 888 901 976 967 994 921 828 970 972 722 755 970 860 855 914 869 714 899 969 978 898 862 642 939 904 936 819 934 884 983 955 964",
"output": "1 2"
},
{
"input": "100\n994 927 872 970 815 986 952 996 965 1000 877 986 978 999 950 990 936 997 993 960 921 860 895 869 943 998 983 968 973 953 999 990 995 871 853 979 973 963 953 938 997 989 993 964 960 973 946 975 1000 962 920 746 989 957 904 965 920 979 966 961 1000 993 975 952 846 971 991 979 985 969 984 973 956 1000 952 778 983 974 956 927 995 997 980 997 1000 970 960 970 988 983 947 904 935 972 1000 863 992 996 932 967",
"output": "81 82"
},
{
"input": "100\n48 108 63 21 27 8 49 21 75 8 24 42 149 18 8 28 21 18 25 35 59 70 59 33 40 1 67 34 120 82 4 115 72 87 3 15 15 63 37 12 40 27 83 14 38 20 14 58 93 10 31 3 39 6 197 77 54 16 31 146 9 49 14 8 77 82 5 11 80 116 8 61 50 24 7 103 29 11 3 3 1 12 46 24 21 131 39 29 36 2 107 40 16 99 31 41 29 48 17 17",
"output": "36 37"
}
] | 1,579,794,650
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 20
| 280
| 0
|
n = int(input())
a = [int(x) for x in input().split()]
min = 99999
ind = -1
for i in range(0, len(a)):
if abs(a[(i + 1) % n] - a[i]) < min:
min = abs(a[(i + 1) % n] - a[i])
ind = i
print(ind + 1, end = " ")
print(((ind + 1) % n) + 1)
|
Title: Reconnaissance 2
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
*n* soldiers stand in a circle. For each soldier his height *a**i* is known. A reconnaissance unit can be made of such two neighbouring soldiers, whose heights difference is minimal, i.e. |*a**i*<=-<=*a**j*| is minimal. So each of them will be less noticeable with the other. Output any pair of soldiers that can form a reconnaissance unit.
Input Specification:
The first line contains integer *n* (2<=≤<=*n*<=≤<=100) — amount of soldiers. Then follow the heights of the soldiers in their order in the circle — *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000). The soldier heights are given in clockwise or counterclockwise direction.
Output Specification:
Output two integers — indexes of neighbouring soldiers, who should form a reconnaissance unit. If there are many optimum solutions, output any of them. Remember, that the soldiers stand in a circle.
Demo Input:
['5\n10 12 13 15 10\n', '4\n10 20 30 40\n']
Demo Output:
['5 1\n', '1 2\n']
Note:
none
|
```python
n = int(input())
a = [int(x) for x in input().split()]
min = 99999
ind = -1
for i in range(0, len(a)):
if abs(a[(i + 1) % n] - a[i]) < min:
min = abs(a[(i + 1) % n] - a[i])
ind = i
print(ind + 1, end = " ")
print(((ind + 1) % n) + 1)
```
| 3.93
|
551
|
A
|
GukiZ and Contest
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation",
"sortings"
] | null | null |
Professor GukiZ likes programming contests. He especially likes to rate his students on the contests he prepares. Now, he has decided to prepare a new contest.
In total, *n* students will attend, and before the start, every one of them has some positive integer rating. Students are indexed from 1 to *n*. Let's denote the rating of *i*-th student as *a**i*. After the contest ends, every student will end up with some positive integer position. GukiZ expects that his students will take places according to their ratings.
He thinks that each student will take place equal to . In particular, if student *A* has rating strictly lower then student *B*, *A* will get the strictly better position than *B*, and if two students have equal ratings, they will share the same position.
GukiZ would like you to reconstruct the results by following his expectations. Help him and determine the position after the end of the contest for each of his students if everything goes as expected.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=2000), number of GukiZ's students.
The second line contains *n* numbers *a*1,<=*a*2,<=... *a**n* (1<=≤<=*a**i*<=≤<=2000) where *a**i* is the rating of *i*-th student (1<=≤<=*i*<=≤<=*n*).
|
In a single line, print the position after the end of the contest for each of *n* students in the same order as they appear in the input.
|
[
"3\n1 3 3\n",
"1\n1\n",
"5\n3 5 3 4 5\n"
] |
[
"3 1 1\n",
"1\n",
"4 1 4 3 1\n"
] |
In the first sample, students 2 and 3 are positioned first (there is no other student with higher rating), and student 1 is positioned third since there are two students with higher rating.
In the second sample, first student is the only one on the contest.
In the third sample, students 2 and 5 share the first position with highest rating, student 4 is next with third position, and students 1 and 3 are the last sharing fourth position.
| 500
|
[
{
"input": "3\n1 3 3",
"output": "3 1 1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "5\n3 5 3 4 5",
"output": "4 1 4 3 1"
},
{
"input": "7\n1 3 5 4 2 2 1",
"output": "6 3 1 2 4 4 6"
},
{
"input": "11\n5 6 4 2 9 7 6 6 6 6 7",
"output": "9 4 10 11 1 2 4 4 4 4 2"
},
{
"input": "1\n2000",
"output": "1"
},
{
"input": "2\n2000 2000",
"output": "1 1"
},
{
"input": "3\n500 501 502",
"output": "3 2 1"
},
{
"input": "10\n105 106 1 1 1 11 1000 999 1000 999",
"output": "6 5 8 8 8 7 1 3 1 3"
},
{
"input": "6\n1 2 3 4 5 6",
"output": "6 5 4 3 2 1"
},
{
"input": "7\n6 5 4 3 2 1 1",
"output": "1 2 3 4 5 6 6"
},
{
"input": "8\n153 100 87 14 10 8 6 5",
"output": "1 2 3 4 5 6 7 8"
},
{
"input": "70\n11 54 37 62 1 46 13 17 38 47 28 15 63 5 61 34 49 66 32 59 3 41 58 28 23 62 41 64 20 5 14 41 10 37 51 32 65 46 61 8 15 19 16 44 31 42 19 46 66 25 26 58 60 5 19 18 69 53 20 40 45 27 24 41 32 23 57 56 62 10",
"output": "62 18 35 7 70 23 61 56 34 22 42 58 6 66 10 37 21 2 38 13 69 29 14 42 48 7 29 5 50 66 60 29 63 35 20 38 4 23 10 65 58 52 57 27 41 28 52 23 2 46 45 14 12 66 52 55 1 19 50 33 26 44 47 29 38 48 16 17 7 63"
},
{
"input": "5\n1 2000 1 1 2000",
"output": "3 1 3 3 1"
}
] | 1,568,534,612
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 36
| 140
| 307,200
|
n = int(input())
scores = [score for score in map(int, input().split())]
sorted_scores = sorted(scores, reverse=True)
ranks = []
for score in scores:
ranks.append(str(sorted_scores.index(score) + 1))
print(' '.join(ranks))
|
Title: GukiZ and Contest
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Professor GukiZ likes programming contests. He especially likes to rate his students on the contests he prepares. Now, he has decided to prepare a new contest.
In total, *n* students will attend, and before the start, every one of them has some positive integer rating. Students are indexed from 1 to *n*. Let's denote the rating of *i*-th student as *a**i*. After the contest ends, every student will end up with some positive integer position. GukiZ expects that his students will take places according to their ratings.
He thinks that each student will take place equal to . In particular, if student *A* has rating strictly lower then student *B*, *A* will get the strictly better position than *B*, and if two students have equal ratings, they will share the same position.
GukiZ would like you to reconstruct the results by following his expectations. Help him and determine the position after the end of the contest for each of his students if everything goes as expected.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=2000), number of GukiZ's students.
The second line contains *n* numbers *a*1,<=*a*2,<=... *a**n* (1<=≤<=*a**i*<=≤<=2000) where *a**i* is the rating of *i*-th student (1<=≤<=*i*<=≤<=*n*).
Output Specification:
In a single line, print the position after the end of the contest for each of *n* students in the same order as they appear in the input.
Demo Input:
['3\n1 3 3\n', '1\n1\n', '5\n3 5 3 4 5\n']
Demo Output:
['3 1 1\n', '1\n', '4 1 4 3 1\n']
Note:
In the first sample, students 2 and 3 are positioned first (there is no other student with higher rating), and student 1 is positioned third since there are two students with higher rating.
In the second sample, first student is the only one on the contest.
In the third sample, students 2 and 5 share the first position with highest rating, student 4 is next with third position, and students 1 and 3 are the last sharing fourth position.
|
```python
n = int(input())
scores = [score for score in map(int, input().split())]
sorted_scores = sorted(scores, reverse=True)
ranks = []
for score in scores:
ranks.append(str(sorted_scores.index(score) + 1))
print(' '.join(ranks))
```
| 3
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
Kevin has just recevied his disappointing results on the USA Identification of Cows Olympiad (USAICO) in the form of a binary string of length *n*. Each character of Kevin's string represents Kevin's score on one of the *n* questions of the olympiad—'1' for a correctly identified cow and '0' otherwise.
However, all is not lost. Kevin is a big proponent of alternative thinking and believes that his score, instead of being the sum of his points, should be the length of the longest alternating subsequence of his string. Here, we define an alternating subsequence of a string as a not-necessarily contiguous subsequence where no two consecutive elements are equal. For example, {0,<=1,<=0,<=1}, {1,<=0,<=1}, and {1,<=0,<=1,<=0} are alternating sequences, while {1,<=0,<=0} and {0,<=1,<=0,<=1,<=1} are not.
Kevin, being the sneaky little puffball that he is, is willing to hack into the USAICO databases to improve his score. In order to be subtle, he decides that he will flip exactly one substring—that is, take a contiguous non-empty substring of his score and change all '0's in that substring to '1's and vice versa. After such an operation, Kevin wants to know the length of the longest possible alternating subsequence that his string could have.
|
The first line contains the number of questions on the olympiad *n* (1<=≤<=*n*<=≤<=100<=000).
The following line contains a binary string of length *n* representing Kevin's results on the USAICO.
|
Output a single integer, the length of the longest possible alternating subsequence that Kevin can create in his string after flipping a single substring.
|
[
"8\n10000011\n",
"2\n01\n"
] |
[
"5\n",
"2\n"
] |
In the first sample, Kevin can flip the bolded substring '10000011' and turn his string into '10011011', which has an alternating subsequence of length 5: '10011011'.
In the second sample, Kevin can flip the entire string and still have the same score.
| 0
|
[
{
"input": "8\n10000011",
"output": "5"
},
{
"input": "2\n01",
"output": "2"
},
{
"input": "5\n10101",
"output": "5"
},
{
"input": "75\n010101010101010101010101010101010101010101010101010101010101010101010101010",
"output": "75"
},
{
"input": "11\n00000000000",
"output": "3"
},
{
"input": "56\n10101011010101010101010101010101010101011010101010101010",
"output": "56"
},
{
"input": "50\n01011010110101010101010101010101010101010101010100",
"output": "49"
},
{
"input": "7\n0110100",
"output": "7"
},
{
"input": "8\n11011111",
"output": "5"
},
{
"input": "6\n000000",
"output": "3"
},
{
"input": "5\n01000",
"output": "5"
},
{
"input": "59\n10101010101010101010101010101010101010101010101010101010101",
"output": "59"
},
{
"input": "88\n1010101010101010101010101010101010101010101010101010101010101010101010101010101010101010",
"output": "88"
},
{
"input": "93\n010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010",
"output": "93"
},
{
"input": "70\n0101010101010101010101010101010101010101010101010101010101010101010101",
"output": "70"
},
{
"input": "78\n010101010101010101010101010101101010101010101010101010101010101010101010101010",
"output": "78"
},
{
"input": "83\n10101010101010101010101010101010101010101010101010110101010101010101010101010101010",
"output": "83"
},
{
"input": "87\n101010101010101010101010101010101010101010101010101010101010101010101010101010010101010",
"output": "87"
},
{
"input": "65\n01010101101010101010101010101010101010101010101010101010101010101",
"output": "65"
},
{
"input": "69\n010101010101010101101010101010101010101010101010101010101010101010101",
"output": "69"
},
{
"input": "74\n01010101010101010101010101010101010101010101010101010101010101000101010101",
"output": "74"
},
{
"input": "77\n01010101010101001010101010101010100101010101010101010101010101010101010101010",
"output": "77"
},
{
"input": "60\n101010110101010101010101010110101010101010101010101010101010",
"output": "60"
},
{
"input": "89\n01010101010101010101010101010101010101010101010101010101101010101010101010100101010101010",
"output": "89"
},
{
"input": "68\n01010101010101010101010101010101010100101010100101010101010100101010",
"output": "67"
},
{
"input": "73\n0101010101010101010101010101010101010101010111011010101010101010101010101",
"output": "72"
},
{
"input": "55\n1010101010101010010101010101101010101010101010100101010",
"output": "54"
},
{
"input": "85\n1010101010101010101010101010010101010101010101101010101010101010101011010101010101010",
"output": "84"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "10\n1111111111",
"output": "3"
},
{
"input": "2\n10",
"output": "2"
},
{
"input": "2\n11",
"output": "2"
},
{
"input": "2\n00",
"output": "2"
},
{
"input": "3\n000",
"output": "3"
},
{
"input": "3\n001",
"output": "3"
},
{
"input": "3\n010",
"output": "3"
},
{
"input": "3\n011",
"output": "3"
},
{
"input": "3\n100",
"output": "3"
},
{
"input": "3\n101",
"output": "3"
},
{
"input": "3\n110",
"output": "3"
},
{
"input": "3\n111",
"output": "3"
},
{
"input": "4\n0000",
"output": "3"
},
{
"input": "4\n0001",
"output": "4"
},
{
"input": "4\n0010",
"output": "4"
},
{
"input": "4\n0011",
"output": "4"
},
{
"input": "4\n0100",
"output": "4"
},
{
"input": "4\n0101",
"output": "4"
},
{
"input": "4\n0110",
"output": "4"
},
{
"input": "4\n0111",
"output": "4"
},
{
"input": "4\n1000",
"output": "4"
},
{
"input": "4\n1001",
"output": "4"
},
{
"input": "4\n1010",
"output": "4"
},
{
"input": "4\n1011",
"output": "4"
},
{
"input": "4\n1100",
"output": "4"
},
{
"input": "4\n1101",
"output": "4"
},
{
"input": "4\n1110",
"output": "4"
},
{
"input": "4\n1111",
"output": "3"
},
{
"input": "5\n00000",
"output": "3"
},
{
"input": "5\n00001",
"output": "4"
},
{
"input": "5\n00010",
"output": "5"
},
{
"input": "5\n00011",
"output": "4"
},
{
"input": "5\n00100",
"output": "5"
},
{
"input": "5\n00101",
"output": "5"
},
{
"input": "5\n00110",
"output": "5"
},
{
"input": "5\n00111",
"output": "4"
},
{
"input": "5\n01000",
"output": "5"
},
{
"input": "5\n01001",
"output": "5"
},
{
"input": "5\n01010",
"output": "5"
},
{
"input": "5\n01011",
"output": "5"
},
{
"input": "5\n01100",
"output": "5"
},
{
"input": "5\n01101",
"output": "5"
},
{
"input": "5\n01110",
"output": "5"
},
{
"input": "5\n01111",
"output": "4"
},
{
"input": "5\n10000",
"output": "4"
},
{
"input": "5\n10001",
"output": "5"
},
{
"input": "5\n10010",
"output": "5"
},
{
"input": "5\n10100",
"output": "5"
},
{
"input": "5\n10101",
"output": "5"
},
{
"input": "5\n10110",
"output": "5"
},
{
"input": "5\n10111",
"output": "5"
},
{
"input": "5\n11000",
"output": "4"
},
{
"input": "5\n11001",
"output": "5"
},
{
"input": "5\n11010",
"output": "5"
},
{
"input": "5\n11011",
"output": "5"
},
{
"input": "5\n11100",
"output": "4"
},
{
"input": "5\n11101",
"output": "5"
},
{
"input": "5\n11110",
"output": "4"
},
{
"input": "5\n11111",
"output": "3"
}
] | 1,537,384,355
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 116
| 155
| 921,600
|
n=int(input())
s=input().strip('\n')
cur=s[0]
l=1
for i in range(1,n):
if s[i]!=cur:
l+=1
cur=s[i]
print(min(l+2,n))
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kevin has just recevied his disappointing results on the USA Identification of Cows Olympiad (USAICO) in the form of a binary string of length *n*. Each character of Kevin's string represents Kevin's score on one of the *n* questions of the olympiad—'1' for a correctly identified cow and '0' otherwise.
However, all is not lost. Kevin is a big proponent of alternative thinking and believes that his score, instead of being the sum of his points, should be the length of the longest alternating subsequence of his string. Here, we define an alternating subsequence of a string as a not-necessarily contiguous subsequence where no two consecutive elements are equal. For example, {0,<=1,<=0,<=1}, {1,<=0,<=1}, and {1,<=0,<=1,<=0} are alternating sequences, while {1,<=0,<=0} and {0,<=1,<=0,<=1,<=1} are not.
Kevin, being the sneaky little puffball that he is, is willing to hack into the USAICO databases to improve his score. In order to be subtle, he decides that he will flip exactly one substring—that is, take a contiguous non-empty substring of his score and change all '0's in that substring to '1's and vice versa. After such an operation, Kevin wants to know the length of the longest possible alternating subsequence that his string could have.
Input Specification:
The first line contains the number of questions on the olympiad *n* (1<=≤<=*n*<=≤<=100<=000).
The following line contains a binary string of length *n* representing Kevin's results on the USAICO.
Output Specification:
Output a single integer, the length of the longest possible alternating subsequence that Kevin can create in his string after flipping a single substring.
Demo Input:
['8\n10000011\n', '2\n01\n']
Demo Output:
['5\n', '2\n']
Note:
In the first sample, Kevin can flip the bolded substring '10000011' and turn his string into '10011011', which has an alternating subsequence of length 5: '10011011'.
In the second sample, Kevin can flip the entire string and still have the same score.
|
```python
n=int(input())
s=input().strip('\n')
cur=s[0]
l=1
for i in range(1,n):
if s[i]!=cur:
l+=1
cur=s[i]
print(min(l+2,n))
```
| 3
|
|
71
|
A
|
Way Too Long Words
|
PROGRAMMING
| 800
|
[
"strings"
] |
A. Way Too Long Words
|
1
|
256
|
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
|
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
|
[
"4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n"
] |
[
"word\nl10n\ni18n\np43s\n"
] |
none
| 500
|
[
{
"input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis",
"output": "word\nl10n\ni18n\np43s"
},
{
"input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm",
"output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m"
},
{
"input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd",
"output": "n20n\nj27b\nk15d"
},
{
"input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt",
"output": "t41t"
},
{
"input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings",
"output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz",
"output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz"
},
{
"input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij",
"output": "a98j"
},
{
"input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco",
"output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco"
},
{
"input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt",
"output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t"
},
{
"input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr",
"output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr"
}
] | 1,688,516,735
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 20
| 46
| 0
|
for i in range(int(input())):
string = input()
print(f"{string[0]}{len(string)-2}{string[len(string)-1]}") if len(string) >10 else print(string)
|
Title: Way Too Long Words
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output Specification:
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
Demo Input:
['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n']
Demo Output:
['word\nl10n\ni18n\np43s\n']
Note:
none
|
```python
for i in range(int(input())):
string = input()
print(f"{string[0]}{len(string)-2}{string[len(string)-1]}") if len(string) >10 else print(string)
```
| 3.977
|
220
|
C
|
Little Elephant and Shifts
|
PROGRAMMING
| 2,100
|
[
"data structures"
] | null | null |
The Little Elephant has two permutations *a* and *b* of length *n*, consisting of numbers from 1 to *n*, inclusive. Let's denote the *i*-th (1<=≤<=*i*<=≤<=*n*) element of the permutation *a* as *a**i*, the *j*-th (1<=≤<=*j*<=≤<=*n*) element of the permutation *b* — as *b**j*.
The distance between permutations *a* and *b* is the minimum absolute value of the difference between the positions of the occurrences of some number in *a* and in *b*. More formally, it's such minimum |*i*<=-<=*j*|, that *a**i*<==<=*b**j*.
A cyclic shift number *i* (1<=≤<=*i*<=≤<=*n*) of permutation *b* consisting from *n* elements is a permutation *b**i**b**i*<=+<=1... *b**n**b*1*b*2... *b**i*<=-<=1. Overall a permutation has *n* cyclic shifts.
The Little Elephant wonders, for all cyclic shifts of permutation *b*, what is the distance between the cyclic shift and permutation *a*?
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the size of the permutations. The second line contains permutation *a* as *n* distinct numbers from 1 to *n*, inclusive. The numbers are separated with single spaces. The third line contains permutation *b* in the same format.
|
In *n* lines print *n* integers — the answers for cyclic shifts. Print the answers to the shifts in the order of the shifts' numeration in permutation *b*, that is, first for the 1-st cyclic shift, then for the 2-nd, and so on.
|
[
"2\n1 2\n2 1\n",
"4\n2 1 3 4\n3 4 2 1\n"
] |
[
"1\n0\n",
"2\n1\n0\n1\n"
] |
none
| 1,500
|
[] | 1,689,591,144
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 92
| 0
|
print("_RANDOM_GUESS_1689591144.008161")# 1689591144.0081792
|
Title: Little Elephant and Shifts
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Little Elephant has two permutations *a* and *b* of length *n*, consisting of numbers from 1 to *n*, inclusive. Let's denote the *i*-th (1<=≤<=*i*<=≤<=*n*) element of the permutation *a* as *a**i*, the *j*-th (1<=≤<=*j*<=≤<=*n*) element of the permutation *b* — as *b**j*.
The distance between permutations *a* and *b* is the minimum absolute value of the difference between the positions of the occurrences of some number in *a* and in *b*. More formally, it's such minimum |*i*<=-<=*j*|, that *a**i*<==<=*b**j*.
A cyclic shift number *i* (1<=≤<=*i*<=≤<=*n*) of permutation *b* consisting from *n* elements is a permutation *b**i**b**i*<=+<=1... *b**n**b*1*b*2... *b**i*<=-<=1. Overall a permutation has *n* cyclic shifts.
The Little Elephant wonders, for all cyclic shifts of permutation *b*, what is the distance between the cyclic shift and permutation *a*?
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the size of the permutations. The second line contains permutation *a* as *n* distinct numbers from 1 to *n*, inclusive. The numbers are separated with single spaces. The third line contains permutation *b* in the same format.
Output Specification:
In *n* lines print *n* integers — the answers for cyclic shifts. Print the answers to the shifts in the order of the shifts' numeration in permutation *b*, that is, first for the 1-st cyclic shift, then for the 2-nd, and so on.
Demo Input:
['2\n1 2\n2 1\n', '4\n2 1 3 4\n3 4 2 1\n']
Demo Output:
['1\n0\n', '2\n1\n0\n1\n']
Note:
none
|
```python
print("_RANDOM_GUESS_1689591144.008161")# 1689591144.0081792
```
| 0
|
|
38
|
A
|
Army
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Army
|
2
|
256
|
The Berland Armed Forces System consists of *n* ranks that are numbered using natural numbers from 1 to *n*, where 1 is the lowest rank and *n* is the highest rank.
One needs exactly *d**i* years to rise from rank *i* to rank *i*<=+<=1. Reaching a certain rank *i* having not reached all the previous *i*<=-<=1 ranks is impossible.
Vasya has just reached a new rank of *a*, but he dreams of holding the rank of *b*. Find for how many more years Vasya should serve in the army until he can finally realize his dream.
|
The first input line contains an integer *n* (2<=≤<=*n*<=≤<=100). The second line contains *n*<=-<=1 integers *d**i* (1<=≤<=*d**i*<=≤<=100). The third input line contains two integers *a* and *b* (1<=≤<=*a*<=<<=*b*<=≤<=*n*). The numbers on the lines are space-separated.
|
Print the single number which is the number of years that Vasya needs to rise from rank *a* to rank *b*.
|
[
"3\n5 6\n1 2\n",
"3\n5 6\n1 3\n"
] |
[
"5\n",
"11\n"
] |
none
| 0
|
[
{
"input": "3\n5 6\n1 2",
"output": "5"
},
{
"input": "3\n5 6\n1 3",
"output": "11"
},
{
"input": "2\n55\n1 2",
"output": "55"
},
{
"input": "3\n85 78\n1 3",
"output": "163"
},
{
"input": "4\n63 4 49\n2 3",
"output": "4"
},
{
"input": "5\n93 83 42 56\n2 5",
"output": "181"
},
{
"input": "6\n22 9 87 89 57\n1 6",
"output": "264"
},
{
"input": "7\n52 36 31 23 74 78\n2 7",
"output": "242"
},
{
"input": "8\n82 14 24 5 91 49 94\n3 8",
"output": "263"
},
{
"input": "9\n12 40 69 39 59 21 59 5\n4 6",
"output": "98"
},
{
"input": "10\n95 81 32 59 71 30 50 61 100\n1 6",
"output": "338"
},
{
"input": "15\n89 55 94 4 15 69 19 60 91 77 3 94 91 62\n3 14",
"output": "617"
},
{
"input": "20\n91 1 41 51 95 67 92 35 23 70 44 91 57 50 21 8 9 71 40\n8 17",
"output": "399"
},
{
"input": "25\n70 95 21 84 97 39 12 98 53 24 78 29 84 65 70 22 100 17 69 27 62 48 35 80\n8 23",
"output": "846"
},
{
"input": "30\n35 69 50 44 19 56 86 56 98 24 21 2 61 24 85 30 2 22 57 35 59 84 12 77 92 53 50 92 9\n1 16",
"output": "730"
},
{
"input": "35\n2 34 47 15 27 61 6 88 67 20 53 65 29 68 77 5 78 86 44 98 32 81 91 79 54 84 95 23 65 97 22 33 42 87\n8 35",
"output": "1663"
},
{
"input": "40\n32 88 59 36 95 45 28 78 73 30 97 13 13 47 48 100 43 21 22 45 88 25 15 13 63 25 72 92 29 5 25 11 50 5 54 51 48 84 23\n7 26",
"output": "862"
},
{
"input": "45\n83 74 73 95 10 31 100 26 29 15 80 100 22 70 31 88 9 56 19 70 2 62 48 30 27 47 52 50 94 44 21 94 23 85 15 3 95 72 43 62 94 89 68 88\n17 40",
"output": "1061"
},
{
"input": "50\n28 8 16 29 19 82 70 51 96 84 74 72 17 69 12 21 37 21 39 3 18 66 19 49 86 96 94 93 2 90 96 84 59 88 58 15 61 33 55 22 35 54 51 29 64 68 29 38 40\n23 28",
"output": "344"
},
{
"input": "60\n24 28 25 21 43 71 64 73 71 90 51 83 69 43 75 43 78 72 56 61 99 7 23 86 9 16 16 94 23 74 18 56 20 72 13 31 75 34 35 86 61 49 4 72 84 7 65 70 66 52 21 38 6 43 69 40 73 46 5\n28 60",
"output": "1502"
},
{
"input": "70\n69 95 34 14 67 61 6 95 94 44 28 94 73 66 39 13 19 71 73 71 28 48 26 22 32 88 38 95 43 59 88 77 80 55 17 95 40 83 67 1 38 95 58 63 56 98 49 2 41 4 73 8 78 41 64 71 60 71 41 61 67 4 4 19 97 14 39 20 27\n9 41",
"output": "1767"
},
{
"input": "80\n65 15 43 6 43 98 100 16 69 98 4 54 25 40 2 35 12 23 38 29 10 89 30 6 4 8 7 96 64 43 11 49 89 38 20 59 54 85 46 16 16 89 60 54 28 37 32 34 67 9 78 30 50 87 58 53 99 48 77 3 5 6 19 99 16 20 31 10 80 76 82 56 56 83 72 81 84 60 28\n18 24",
"output": "219"
},
{
"input": "90\n61 35 100 99 67 87 42 90 44 4 81 65 29 63 66 56 53 22 55 87 39 30 34 42 27 80 29 97 85 28 81 22 50 22 24 75 67 86 78 79 94 35 13 97 48 76 68 66 94 13 82 1 22 85 5 36 86 73 65 97 43 56 35 26 87 25 74 47 81 67 73 75 99 75 53 38 70 21 66 78 38 17 57 40 93 57 68 55 1\n12 44",
"output": "1713"
},
{
"input": "95\n37 74 53 96 65 84 65 72 95 45 6 77 91 35 58 50 51 51 97 30 51 20 79 81 92 10 89 34 40 76 71 54 26 34 73 72 72 28 53 19 95 64 97 10 44 15 12 38 5 63 96 95 86 8 36 96 45 53 81 5 18 18 47 97 65 9 33 53 41 86 37 53 5 40 15 76 83 45 33 18 26 5 19 90 46 40 100 42 10 90 13 81 40 53\n6 15",
"output": "570"
},
{
"input": "96\n51 32 95 75 23 54 70 89 67 3 1 51 4 100 97 30 9 35 56 38 54 77 56 98 43 17 60 43 72 46 87 61 100 65 81 22 74 38 16 96 5 10 54 22 23 22 10 91 9 54 49 82 29 73 33 98 75 8 4 26 24 90 71 42 90 24 94 74 94 10 41 98 56 63 18 43 56 21 26 64 74 33 22 38 67 66 38 60 64 76 53 10 4 65 76\n21 26",
"output": "328"
},
{
"input": "97\n18 90 84 7 33 24 75 55 86 10 96 72 16 64 37 9 19 71 62 97 5 34 85 15 46 72 82 51 52 16 55 68 27 97 42 72 76 97 32 73 14 56 11 86 2 81 59 95 60 93 1 22 71 37 77 100 6 16 78 47 78 62 94 86 16 91 56 46 47 35 93 44 7 86 70 10 29 45 67 62 71 61 74 39 36 92 24 26 65 14 93 92 15 28 79 59\n6 68",
"output": "3385"
},
{
"input": "98\n32 47 26 86 43 42 79 72 6 68 40 46 29 80 24 89 29 7 21 56 8 92 13 33 50 79 5 7 84 85 24 23 1 80 51 21 26 55 96 51 24 2 68 98 81 88 57 100 64 84 54 10 14 2 74 1 89 71 1 20 84 85 17 31 42 58 69 67 48 60 97 90 58 10 21 29 2 21 60 61 68 89 77 39 57 18 61 44 67 100 33 74 27 40 83 29 6\n8 77",
"output": "3319"
},
{
"input": "99\n46 5 16 66 53 12 84 89 26 27 35 68 41 44 63 17 88 43 80 15 59 1 42 50 53 34 75 16 16 55 92 30 28 11 12 71 27 65 11 28 86 47 24 10 60 47 7 53 16 75 6 49 56 66 70 3 20 78 75 41 38 57 89 23 16 74 30 39 1 32 49 84 9 33 25 95 75 45 54 59 17 17 29 40 79 96 47 11 69 86 73 56 91 4 87 47 31 24\n23 36",
"output": "514"
},
{
"input": "100\n63 65 21 41 95 23 3 4 12 23 95 50 75 63 58 34 71 27 75 31 23 94 96 74 69 34 43 25 25 55 44 19 43 86 68 17 52 65 36 29 72 96 84 25 84 23 71 54 6 7 71 7 21 100 99 58 93 35 62 47 36 70 68 9 75 13 35 70 76 36 62 22 52 51 2 87 66 41 54 35 78 62 30 35 65 44 74 93 78 37 96 70 26 32 71 27 85 85 63\n43 92",
"output": "2599"
},
{
"input": "51\n85 38 22 38 42 36 55 24 36 80 49 15 66 91 88 61 46 82 1 61 89 92 6 56 28 8 46 80 56 90 91 38 38 17 69 64 57 68 13 44 45 38 8 72 61 39 87 2 73 88\n15 27",
"output": "618"
},
{
"input": "2\n3\n1 2",
"output": "3"
},
{
"input": "5\n6 8 22 22\n2 3",
"output": "8"
},
{
"input": "6\n3 12 27 28 28\n3 4",
"output": "27"
},
{
"input": "9\n1 2 2 2 2 3 3 5\n3 7",
"output": "9"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1\n6 8",
"output": "2"
},
{
"input": "20\n1 1 1 1 1 1 1 1 2 2 2 2 2 3 3 3 3 3 3\n5 17",
"output": "23"
},
{
"input": "25\n1 1 1 4 5 6 8 11 11 11 11 12 13 14 14 14 15 16 16 17 17 17 19 19\n4 8",
"output": "23"
},
{
"input": "35\n1 1 1 1 1 1 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\n30 31",
"output": "2"
},
{
"input": "45\n1 1 1 1 2 2 2 2 2 2 2 3 3 3 3 3 3 4 5 5 5 5 6 6 6 6 6 6 6 7 7 7 7 8 8 8 9 9 9 9 9 10 10 10\n42 45",
"output": "30"
},
{
"input": "50\n1 8 8 13 14 15 15 16 19 21 22 24 26 31 32 37 45 47 47 47 50 50 51 54 55 56 58 61 61 61 63 63 64 66 66 67 67 70 71 80 83 84 85 92 92 94 95 95 100\n4 17",
"output": "285"
},
{
"input": "60\n1 2 4 4 4 6 6 8 9 10 10 13 14 18 20 20 21 22 23 23 26 29 30 32 33 34 35 38 40 42 44 44 46 48 52 54 56 56 60 60 66 67 68 68 69 73 73 74 80 80 81 81 82 84 86 86 87 89 89\n56 58",
"output": "173"
},
{
"input": "70\n1 2 3 3 4 5 5 7 7 7 8 8 8 8 9 9 10 12 12 12 12 13 16 16 16 16 16 16 17 17 18 18 20 20 21 23 24 25 25 26 29 29 29 29 31 32 32 34 35 36 36 37 37 38 39 39 40 40 40 40 41 41 42 43 44 44 44 45 45\n62 65",
"output": "126"
},
{
"input": "80\n1 1 1 1 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 4 4 4 4 5 5 5 5 5 5 5 6 7 7 7 7 7 7 8 8 8 8 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 11 11 11 11 11 11 11 12 12 12 12 12 12 12 12\n17 65",
"output": "326"
},
{
"input": "90\n1 1 3 5 8 9 10 11 11 11 11 12 13 14 15 15 15 16 16 19 19 20 22 23 24 25 25 28 29 29 30 31 33 34 35 37 37 38 41 43 43 44 45 47 51 54 55 56 58 58 59 59 60 62 66 67 67 67 68 68 69 70 71 72 73 73 76 77 77 78 78 78 79 79 79 82 83 84 85 85 87 87 89 93 93 93 95 99 99\n28 48",
"output": "784"
},
{
"input": "95\n2 2 3 3 4 6 6 7 7 7 9 10 12 12 12 12 13 14 15 16 17 18 20 20 20 20 21 21 21 21 22 22 22 22 22 23 23 23 25 26 26 27 27 27 28 29 29 30 30 31 32 33 34 36 37 37 38 39 39 39 42 43 43 43 45 47 48 50 50 51 52 53 54 54 54 55 55 55 58 59 60 61 61 61 61 62 62 63 64 65 66 67 67 67\n64 93",
"output": "1636"
},
{
"input": "96\n1 1 2 3 3 5 8 9 9 10 10 10 11 11 11 11 11 12 13 13 13 14 15 15 16 16 17 17 17 17 18 18 20 20 20 21 21 21 23 24 24 25 25 26 27 27 27 27 29 29 29 30 30 30 32 32 32 32 32 32 33 33 34 34 34 35 35 35 36 36 37 37 37 38 39 40 41 41 41 41 42 42 43 43 45 45 45 46 46 47 47 49 50 52 52\n76 96",
"output": "898"
},
{
"input": "98\n2 3 4 4 5 7 8 10 10 10 11 11 12 12 12 12 13 14 15 15 16 16 18 19 19 20 21 21 21 21 22 23 24 25 26 26 27 27 27 27 29 29 30 30 31 31 37 40 40 40 41 41 41 42 43 44 44 44 46 46 47 49 49 50 50 50 51 53 55 55 56 56 56 56 56 57 57 58 59 60 60 60 62 62 63 64 64 64 65 66 66 67 68 70 70 71 71\n8 90",
"output": "3016"
},
{
"input": "99\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\n66 95",
"output": "29"
},
{
"input": "100\n1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 3 3 3 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 8 8 8 8 9 9 9 9 10 10 10 10 11 11 11 11 12 12 12 13 13 13 13 13 13 13 13 13 13 14 14 14 14 14 14 15 15 15 15 15 15 16 16 16 17 17\n39 52",
"output": "82"
},
{
"input": "51\n5 7 9 16 19 25 26 29 29 30 31 32 32 41 43 44 47 49 50 50 51 52 54 55 56 63 64 66 67 70 74 74 77 78 79 80 80 85 86 87 89 89 90 92 93 94 94 95 95 97\n3 44",
"output": "2268"
}
] | 1,607,618,229
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 0
| 216
| 0
|
n=int(input())
d=list(map(int,input().split()))
a,b=map(int,input().split())
print(sum(d[a-1:b]))
|
Title: Army
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The Berland Armed Forces System consists of *n* ranks that are numbered using natural numbers from 1 to *n*, where 1 is the lowest rank and *n* is the highest rank.
One needs exactly *d**i* years to rise from rank *i* to rank *i*<=+<=1. Reaching a certain rank *i* having not reached all the previous *i*<=-<=1 ranks is impossible.
Vasya has just reached a new rank of *a*, but he dreams of holding the rank of *b*. Find for how many more years Vasya should serve in the army until he can finally realize his dream.
Input Specification:
The first input line contains an integer *n* (2<=≤<=*n*<=≤<=100). The second line contains *n*<=-<=1 integers *d**i* (1<=≤<=*d**i*<=≤<=100). The third input line contains two integers *a* and *b* (1<=≤<=*a*<=<<=*b*<=≤<=*n*). The numbers on the lines are space-separated.
Output Specification:
Print the single number which is the number of years that Vasya needs to rise from rank *a* to rank *b*.
Demo Input:
['3\n5 6\n1 2\n', '3\n5 6\n1 3\n']
Demo Output:
['5\n', '11\n']
Note:
none
|
```python
n=int(input())
d=list(map(int,input().split()))
a,b=map(int,input().split())
print(sum(d[a-1:b]))
```
| 0
|
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 — 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<=≤<=*n*<=≤<=1000) — the number of contests where the coder participated.
The next line contains *n* space-separated non-negative integer numbers — 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 — 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"
}
] | 1,684,645,981
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 37
| 92
| 0
|
n = int(input())
points = list(map(int, input().split()))
best_performance = points[0]
worst_performance = points[0]
amazing_performances = 0
for i in range(1, n):
if points[i] > best_performance:
best_performance = points[i]
amazing_performances += 1
elif points[i] < worst_performance:
worst_performance = points[i]
amazing_performances += 1
print(amazing_performances)
|
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 — 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<=≤<=*n*<=≤<=1000) — the number of contests where the coder participated.
The next line contains *n* space-separated non-negative integer numbers — 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 — 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())
points = list(map(int, input().split()))
best_performance = points[0]
worst_performance = points[0]
amazing_performances = 0
for i in range(1, n):
if points[i] > best_performance:
best_performance = points[i]
amazing_performances += 1
elif points[i] < worst_performance:
worst_performance = points[i]
amazing_performances += 1
print(amazing_performances)
```
| 3
|
|
63
|
B
|
Settlers' Training
|
PROGRAMMING
| 1,200
|
[
"implementation"
] |
B. Settlers' Training
|
2
|
256
|
In a strategic computer game "Settlers II" one has to build defense structures to expand and protect the territory. Let's take one of these buildings. At the moment the defense structure accommodates exactly *n* soldiers. Within this task we can assume that the number of soldiers in the defense structure won't either increase or decrease.
Every soldier has a rank — some natural number from 1 to *k*. 1 stands for a private and *k* stands for a general. The higher the rank of the soldier is, the better he fights. Therefore, the player profits from having the soldiers of the highest possible rank.
To increase the ranks of soldiers they need to train. But the soldiers won't train for free, and each training session requires one golden coin. On each training session all the *n* soldiers are present.
At the end of each training session the soldiers' ranks increase as follows. First all the soldiers are divided into groups with the same rank, so that the least possible number of groups is formed. Then, within each of the groups where the soldiers below the rank *k* are present, exactly one soldier increases his rank by one.
You know the ranks of all *n* soldiers at the moment. Determine the number of golden coins that are needed to increase the ranks of all the soldiers to the rank *k*.
|
The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100). They represent the number of soldiers and the number of different ranks correspondingly. The second line contains *n* numbers in the non-decreasing order. The *i*-th of them, *a**i*, represents the rank of the *i*-th soldier in the defense building (1<=≤<=*i*<=≤<=*n*, 1<=≤<=*a**i*<=≤<=*k*).
|
Print a single integer — the number of golden coins needed to raise all the soldiers to the maximal rank.
|
[
"4 4\n1 2 2 3\n",
"4 3\n1 1 1 1\n"
] |
[
"4",
"5"
] |
In the first example the ranks will be raised in the following manner:
1 2 2 3 → 2 2 3 4 → 2 3 4 4 → 3 4 4 4 → 4 4 4 4
Thus totals to 4 training sessions that require 4 golden coins.
| 1,000
|
[
{
"input": "4 4\n1 2 2 3",
"output": "4"
},
{
"input": "4 3\n1 1 1 1",
"output": "5"
},
{
"input": "3 3\n1 2 3",
"output": "2"
},
{
"input": "1 1\n1",
"output": "0"
},
{
"input": "1 5\n1",
"output": "4"
},
{
"input": "1 5\n4",
"output": "1"
},
{
"input": "2 6\n2 5",
"output": "4"
},
{
"input": "6 10\n1 1 3 4 9 9",
"output": "10"
},
{
"input": "7 7\n1 1 1 1 1 1 7",
"output": "11"
},
{
"input": "10 10\n1 1 1 3 3 4 7 8 8 8",
"output": "11"
},
{
"input": "10 13\n1 1 1 1 1 1 1 1 1 1",
"output": "21"
},
{
"input": "10 13\n2 6 6 7 9 9 9 10 12 12",
"output": "11"
},
{
"input": "17 9\n2 3 4 5 5 5 5 5 6 6 7 7 8 8 8 8 8",
"output": "17"
},
{
"input": "18 24\n3 3 3 4 5 7 8 8 9 9 9 9 10 10 11 11 11 11",
"output": "30"
},
{
"input": "23 2\n1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2",
"output": "12"
},
{
"input": "37 42\n1 1 1 1 1 2 2 2 2 2 3 4 4 4 4 5 5 5 5 6 6 6 6 6 6 6 6 7 7 7 7 7 8 8 8 8 8",
"output": "70"
},
{
"input": "44 50\n38 38 38 38 38 38 38 39 39 39 39 39 39 39 40 40 40 40 40 41 41 41 41 41 41 41 42 42 42 43 43 43 44 44 44 44 45 45 45 46 46 46 46 46",
"output": "47"
},
{
"input": "57 100\n2 2 4 7 8 10 12 12 14 15 16 18 19 21 21 22 25 26 26 33 38 40 44 44 44 45 47 47 50 51 51 54 54 54 54 55 56 58 61 65 67 68 68 70 74 75 78 79 83 86 89 90 92 95 96 96 97",
"output": "99"
},
{
"input": "78 10\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9",
"output": "78"
},
{
"input": "96 78\n20 20 20 20 20 21 21 21 22 23 23 24 24 25 25 27 28 29 30 30 30 32 32 32 33 33 33 33 34 34 35 36 37 37 39 39 41 41 41 41 42 42 43 43 43 44 44 45 46 46 48 48 49 50 51 51 51 52 53 55 55 56 56 56 56 57 58 59 60 61 61 61 62 62 62 63 63 64 64 64 65 65 65 66 66 67 68 69 71 72 72 73 73 75 75 75",
"output": "98"
},
{
"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": "0"
},
{
"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": "198"
},
{
"input": "100 100\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": "0"
},
{
"input": "100 100\n1 1 4 4 5 5 7 9 10 10 11 11 12 12 12 13 14 15 16 16 16 17 18 18 19 20 22 25 26 27 29 32 33 34 34 35 35 35 36 36 37 37 38 39 39 40 41 42 44 44 46 47 47 47 47 50 53 53 53 55 56 56 57 57 58 58 59 59 62 64 64 64 64 68 68 68 69 70 70 71 74 77 77 77 79 80 80 81 84 86 88 88 91 93 94 96 96 99 99 99",
"output": "108"
},
{
"input": "100 100\n1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 3 3 3 3 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6 7 7 7 7 8 8 8 8 8 9 9 9 9 9 9 9 10 10 10 10 10 11 11 11 11 11 12 12 12 12 12 12 13 13 13 13 13 13 13 14 14 14 14 14 14 14 14 14 14 14 14 14 15 15 15 15 15",
"output": "184"
},
{
"input": "100 100\n20 20 20 21 21 21 21 21 22 23 23 23 23 23 23 24 24 25 25 26 26 26 26 26 27 27 27 27 28 28 28 28 29 29 29 29 29 30 30 30 30 31 32 32 34 34 34 34 34 34 34 34 35 35 35 36 36 37 37 37 37 37 37 38 38 38 39 40 41 41 42 42 42 42 42 43 43 43 44 44 44 44 44 45 45 45 45 45 46 46 46 46 46 47 47 47 48 48 48 50",
"output": "150"
},
{
"input": "100 2\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 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",
"output": "59"
},
{
"input": "30 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 49",
"output": "77"
},
{
"input": "40 20\n5 5 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 20 20 20 20 20 20 20 20 20 20",
"output": "31"
},
{
"input": "81 90\n1 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": "89"
},
{
"input": "100 20\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 13 13 13 13 13 13 13 13 13",
"output": "106"
},
{
"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 1 100",
"output": "197"
},
{
"input": "100 100\n49 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 51",
"output": "148"
},
{
"input": "1 100\n1",
"output": "99"
},
{
"input": "4 3\n1 1 2 2",
"output": "4"
},
{
"input": "10 100\n98 99 99 99 99 99 99 100 100 100",
"output": "7"
},
{
"input": "5 100\n1 2 2 100 100",
"output": "100"
}
] | 1,601,735,186
| 1,886
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 154
| 307,200
|
n, k = [int(x) for x in input().split(' ')]
a = [int(x) for x in input().split(' ')]
a_dict = {i: 0 for i in range(n + 1)}
for x in a:
if x in a_dict.keys():
a_dict[x] += 1
else:
a_dict[x] = 1
ops = 0
while n not in a_dict.keys() or a_dict[n] < n:
a_dict = {i: int(a_dict[max(i - 1, 0)] > 0) + max(0, a_dict[i] - 1) * int(i != n) + a_dict[i] * int(i == n) for i in range(n + 1)}
ops += 1
print(ops)
|
Title: Settlers' Training
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
In a strategic computer game "Settlers II" one has to build defense structures to expand and protect the territory. Let's take one of these buildings. At the moment the defense structure accommodates exactly *n* soldiers. Within this task we can assume that the number of soldiers in the defense structure won't either increase or decrease.
Every soldier has a rank — some natural number from 1 to *k*. 1 stands for a private and *k* stands for a general. The higher the rank of the soldier is, the better he fights. Therefore, the player profits from having the soldiers of the highest possible rank.
To increase the ranks of soldiers they need to train. But the soldiers won't train for free, and each training session requires one golden coin. On each training session all the *n* soldiers are present.
At the end of each training session the soldiers' ranks increase as follows. First all the soldiers are divided into groups with the same rank, so that the least possible number of groups is formed. Then, within each of the groups where the soldiers below the rank *k* are present, exactly one soldier increases his rank by one.
You know the ranks of all *n* soldiers at the moment. Determine the number of golden coins that are needed to increase the ranks of all the soldiers to the rank *k*.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100). They represent the number of soldiers and the number of different ranks correspondingly. The second line contains *n* numbers in the non-decreasing order. The *i*-th of them, *a**i*, represents the rank of the *i*-th soldier in the defense building (1<=≤<=*i*<=≤<=*n*, 1<=≤<=*a**i*<=≤<=*k*).
Output Specification:
Print a single integer — the number of golden coins needed to raise all the soldiers to the maximal rank.
Demo Input:
['4 4\n1 2 2 3\n', '4 3\n1 1 1 1\n']
Demo Output:
['4', '5']
Note:
In the first example the ranks will be raised in the following manner:
1 2 2 3 → 2 2 3 4 → 2 3 4 4 → 3 4 4 4 → 4 4 4 4
Thus totals to 4 training sessions that require 4 golden coins.
|
```python
n, k = [int(x) for x in input().split(' ')]
a = [int(x) for x in input().split(' ')]
a_dict = {i: 0 for i in range(n + 1)}
for x in a:
if x in a_dict.keys():
a_dict[x] += 1
else:
a_dict[x] = 1
ops = 0
while n not in a_dict.keys() or a_dict[n] < n:
a_dict = {i: int(a_dict[max(i - 1, 0)] > 0) + max(0, a_dict[i] - 1) * int(i != n) + a_dict[i] * int(i == n) for i in range(n + 1)}
ops += 1
print(ops)
```
| 0
|
120
|
A
|
Elevator
|
PROGRAMMING
| 1,000
|
[
"brute force",
"implementation",
"math"
] | null | null |
A sky scraper with 1000 floors has been built in the city of N. It has modern superfast elevators to help to travel from one floor to another. Each elevator has two doors, the front one and the back one. If one goes in through the front door, he goes out through the back one and vice versa. The elevator has two rails numbered with numbers 1 and 2. Rail 1 is located to the left of the entrance to the front door (or correspondingly, to the right of the entrance to the back door). Rail 2 is located opposite it, to the right of the entrance to the front door and to the left of the entrance to the back door. We know that each person in the city of N holds at a rail with the strongest hand.
One day a VIP person visited the city and of course, he took a look at the skyscraper and took a ride in the elevator. We know the door through which he entered and the rail he was holding at. Now we need to determine as soon as possible whether he is left-handed or right-handed.
|
The first line indicates the door through which the very important person entered the elevator. It contains "front" if the person enters the elevator through the front door and "back" if he entered the elevator through the back door. The second line contains integer *a* (1<=≤<=*a*<=≤<=2) which denotes the number of the rail at which the person was holding.
|
Print character "R" if the VIP is right-handed or "L" if he is left-handed.
|
[
"front\n1\n"
] |
[
"L\n"
] |
none
| 0
|
[
{
"input": "front\n1",
"output": "L"
},
{
"input": "back\n1",
"output": "R"
},
{
"input": "front\n2",
"output": "R"
},
{
"input": "back\n2",
"output": "L"
}
] | 1,576,920,241
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 186
| 0
|
with open('input.txt', 'r') as r, open('output.txt', 'w') as w:
a = readln(r);
b = int(readln(r))
if (a == "front" and b == 2) or (a == 'back' and b == 1):
print("R", w)
else:
print("L", w)
# Sat Dec 21 2019 12:24:01 GMT+0300 (MSK)
|
Title: Elevator
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A sky scraper with 1000 floors has been built in the city of N. It has modern superfast elevators to help to travel from one floor to another. Each elevator has two doors, the front one and the back one. If one goes in through the front door, he goes out through the back one and vice versa. The elevator has two rails numbered with numbers 1 and 2. Rail 1 is located to the left of the entrance to the front door (or correspondingly, to the right of the entrance to the back door). Rail 2 is located opposite it, to the right of the entrance to the front door and to the left of the entrance to the back door. We know that each person in the city of N holds at a rail with the strongest hand.
One day a VIP person visited the city and of course, he took a look at the skyscraper and took a ride in the elevator. We know the door through which he entered and the rail he was holding at. Now we need to determine as soon as possible whether he is left-handed or right-handed.
Input Specification:
The first line indicates the door through which the very important person entered the elevator. It contains "front" if the person enters the elevator through the front door and "back" if he entered the elevator through the back door. The second line contains integer *a* (1<=≤<=*a*<=≤<=2) which denotes the number of the rail at which the person was holding.
Output Specification:
Print character "R" if the VIP is right-handed or "L" if he is left-handed.
Demo Input:
['front\n1\n']
Demo Output:
['L\n']
Note:
none
|
```python
with open('input.txt', 'r') as r, open('output.txt', 'w') as w:
a = readln(r);
b = int(readln(r))
if (a == "front" and b == 2) or (a == 'back' and b == 1):
print("R", w)
else:
print("L", w)
# Sat Dec 21 2019 12:24:01 GMT+0300 (MSK)
```
| -1
|
|
80
|
A
|
Panoramix's Prediction
|
PROGRAMMING
| 800
|
[
"brute force"
] |
A. Panoramix's Prediction
|
2
|
256
|
A prime number is a number which has exactly two distinct divisors: one and itself. For example, numbers 2, 7, 3 are prime, and 1, 6, 4 are not.
The next prime number after *x* is the smallest prime number greater than *x*. For example, the next prime number after 2 is 3, and the next prime number after 3 is 5. Note that there is exactly one next prime number after each number. So 5 is not the next prime number for 2.
One cold April morning Panoramix predicted that soon Kakofonix will break free from his straitjacket, and this will be a black day for the residents of the Gallic countryside.
Panoramix's prophecy tells that if some day Asterix and Obelix beat exactly *x* Roman soldiers, where *x* is a prime number, and next day they beat exactly *y* Roman soldiers, where *y* is the next prime number after *x*, then it's time to wait for Armageddon, for nothing can shut Kakofonix up while he sings his infernal song.
Yesterday the Gauls beat *n* Roman soldiers and it turned out that the number *n* was prime! Today their victims were a troop of *m* Romans (*m*<=><=*n*). Determine whether the Gauls should wait for the black day after today's victory of Asterix and Obelix?
|
The first and only input line contains two positive integers — *n* and *m* (2<=≤<=*n*<=<<=*m*<=≤<=50). It is guaranteed that *n* is prime.
Pretests contain all the cases with restrictions 2<=≤<=*n*<=<<=*m*<=≤<=4.
|
Print YES, if *m* is the next prime number after *n*, or NO otherwise.
|
[
"3 5\n",
"7 11\n",
"7 9\n"
] |
[
"YES",
"YES",
"NO"
] |
none
| 500
|
[
{
"input": "3 5",
"output": "YES"
},
{
"input": "7 11",
"output": "YES"
},
{
"input": "7 9",
"output": "NO"
},
{
"input": "2 3",
"output": "YES"
},
{
"input": "2 4",
"output": "NO"
},
{
"input": "3 4",
"output": "NO"
},
{
"input": "3 5",
"output": "YES"
},
{
"input": "5 7",
"output": "YES"
},
{
"input": "7 11",
"output": "YES"
},
{
"input": "11 13",
"output": "YES"
},
{
"input": "13 17",
"output": "YES"
},
{
"input": "17 19",
"output": "YES"
},
{
"input": "19 23",
"output": "YES"
},
{
"input": "23 29",
"output": "YES"
},
{
"input": "29 31",
"output": "YES"
},
{
"input": "31 37",
"output": "YES"
},
{
"input": "37 41",
"output": "YES"
},
{
"input": "41 43",
"output": "YES"
},
{
"input": "43 47",
"output": "YES"
},
{
"input": "2 5",
"output": "NO"
},
{
"input": "2 7",
"output": "NO"
},
{
"input": "2 6",
"output": "NO"
},
{
"input": "2 11",
"output": "NO"
},
{
"input": "3 6",
"output": "NO"
},
{
"input": "3 7",
"output": "NO"
},
{
"input": "3 9",
"output": "NO"
},
{
"input": "5 6",
"output": "NO"
},
{
"input": "5 9",
"output": "NO"
},
{
"input": "5 11",
"output": "NO"
},
{
"input": "5 13",
"output": "NO"
},
{
"input": "5 15",
"output": "NO"
},
{
"input": "7 8",
"output": "NO"
},
{
"input": "7 13",
"output": "NO"
},
{
"input": "13 15",
"output": "NO"
},
{
"input": "13 17",
"output": "YES"
},
{
"input": "19 21",
"output": "NO"
},
{
"input": "13 20",
"output": "NO"
},
{
"input": "41 49",
"output": "NO"
},
{
"input": "43 49",
"output": "NO"
},
{
"input": "47 50",
"output": "NO"
},
{
"input": "47 49",
"output": "NO"
},
{
"input": "47 48",
"output": "NO"
},
{
"input": "23 25",
"output": "NO"
},
{
"input": "2 50",
"output": "NO"
},
{
"input": "31 33",
"output": "NO"
}
] | 1,642,447,642
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 45
| 92
| 0
|
n, m = map(int, input().split())
prime = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]
for i in range(0, len(prime)):
if n==prime[i]:
if m==prime[i+1]:
print("YES")
else:
print("NO")
|
Title: Panoramix's Prediction
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A prime number is a number which has exactly two distinct divisors: one and itself. For example, numbers 2, 7, 3 are prime, and 1, 6, 4 are not.
The next prime number after *x* is the smallest prime number greater than *x*. For example, the next prime number after 2 is 3, and the next prime number after 3 is 5. Note that there is exactly one next prime number after each number. So 5 is not the next prime number for 2.
One cold April morning Panoramix predicted that soon Kakofonix will break free from his straitjacket, and this will be a black day for the residents of the Gallic countryside.
Panoramix's prophecy tells that if some day Asterix and Obelix beat exactly *x* Roman soldiers, where *x* is a prime number, and next day they beat exactly *y* Roman soldiers, where *y* is the next prime number after *x*, then it's time to wait for Armageddon, for nothing can shut Kakofonix up while he sings his infernal song.
Yesterday the Gauls beat *n* Roman soldiers and it turned out that the number *n* was prime! Today their victims were a troop of *m* Romans (*m*<=><=*n*). Determine whether the Gauls should wait for the black day after today's victory of Asterix and Obelix?
Input Specification:
The first and only input line contains two positive integers — *n* and *m* (2<=≤<=*n*<=<<=*m*<=≤<=50). It is guaranteed that *n* is prime.
Pretests contain all the cases with restrictions 2<=≤<=*n*<=<<=*m*<=≤<=4.
Output Specification:
Print YES, if *m* is the next prime number after *n*, or NO otherwise.
Demo Input:
['3 5\n', '7 11\n', '7 9\n']
Demo Output:
['YES', 'YES', 'NO']
Note:
none
|
```python
n, m = map(int, input().split())
prime = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]
for i in range(0, len(prime)):
if n==prime[i]:
if m==prime[i+1]:
print("YES")
else:
print("NO")
```
| 3.977
|
71
|
A
|
Way Too Long Words
|
PROGRAMMING
| 800
|
[
"strings"
] |
A. Way Too Long Words
|
1
|
256
|
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
|
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
|
[
"4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n"
] |
[
"word\nl10n\ni18n\np43s\n"
] |
none
| 500
|
[
{
"input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis",
"output": "word\nl10n\ni18n\np43s"
},
{
"input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm",
"output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m"
},
{
"input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd",
"output": "n20n\nj27b\nk15d"
},
{
"input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt",
"output": "t41t"
},
{
"input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings",
"output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz",
"output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz"
},
{
"input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij",
"output": "a98j"
},
{
"input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco",
"output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco"
},
{
"input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt",
"output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t"
},
{
"input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr",
"output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr"
}
] | 1,672,297,087
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 30
| 0
|
n = int(input())
for n in range(0,n):
a = input()
if len(a)>10:
s = len(a)-2
x = str(s)
print(a[0]+ x + [a-1])
"""
for j in[0]*int(input()):
a=input()
s=len(a)-2
print([a,a[0]+str(s)+a[-1]][s>8])
#include<stdio.h>
#include<ctype.h>
int main(){
int n,i,m;
char x[6000];
scanf("%d",&n);
for(i=1;i<=n;i++){
scanf("%s",&x);
int l=strlen(x);
if(l>10){
printf("%c",x[0]);
printf("%d",l-2);
printf("%c\n",x[l-1]);
}
else printf("%s\n",x);}
return 0;
}
for j in[0]*int(input()):
a=input()
s=len(a)-2
print([a,a[0]+str(s)+a[-1]][s>8])
"""
|
Title: Way Too Long Words
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output Specification:
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
Demo Input:
['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n']
Demo Output:
['word\nl10n\ni18n\np43s\n']
Note:
none
|
```python
n = int(input())
for n in range(0,n):
a = input()
if len(a)>10:
s = len(a)-2
x = str(s)
print(a[0]+ x + [a-1])
"""
for j in[0]*int(input()):
a=input()
s=len(a)-2
print([a,a[0]+str(s)+a[-1]][s>8])
#include<stdio.h>
#include<ctype.h>
int main(){
int n,i,m;
char x[6000];
scanf("%d",&n);
for(i=1;i<=n;i++){
scanf("%s",&x);
int l=strlen(x);
if(l>10){
printf("%c",x[0]);
printf("%d",l-2);
printf("%c\n",x[l-1]);
}
else printf("%s\n",x);}
return 0;
}
for j in[0]*int(input()):
a=input()
s=len(a)-2
print([a,a[0]+str(s)+a[-1]][s>8])
"""
```
| -1
|
979
|
A
|
Pizza, Pizza, Pizza!!!
|
PROGRAMMING
| 1,000
|
[
"math"
] | null | null |
Katie, Kuro and Shiro are best friends. They have known each other since kindergarten. That's why they often share everything with each other and work together on some very hard problems.
Today is Shiro's birthday. She really loves pizza so she wants to invite her friends to the pizza restaurant near her house to celebrate her birthday, including her best friends Katie and Kuro.
She has ordered a very big round pizza, in order to serve her many friends. Exactly $n$ of Shiro's friends are here. That's why she has to divide the pizza into $n + 1$ slices (Shiro also needs to eat). She wants the slices to be exactly the same size and shape. If not, some of her friends will get mad and go home early, and the party will be over.
Shiro is now hungry. She wants to cut the pizza with minimum of straight cuts. A cut is a straight segment, it might have ends inside or outside the pizza. But she is too lazy to pick up the calculator.
As usual, she will ask Katie and Kuro for help. But they haven't come yet. Could you help Shiro with this problem?
|
A single line contains one non-negative integer $n$ ($0 \le n \leq 10^{18}$) — the number of Shiro's friends. The circular pizza has to be sliced into $n + 1$ pieces.
|
A single integer — the number of straight cuts Shiro needs.
|
[
"3\n",
"4\n"
] |
[
"2",
"5"
] |
To cut the round pizza into quarters one has to make two cuts through the center with angle $90^{\circ}$ between them.
To cut the round pizza into five equal parts one has to make five cuts.
| 500
|
[
{
"input": "3",
"output": "2"
},
{
"input": "4",
"output": "5"
},
{
"input": "10",
"output": "11"
},
{
"input": "10000000000",
"output": "10000000001"
},
{
"input": "1234567891",
"output": "617283946"
},
{
"input": "7509213957",
"output": "3754606979"
},
{
"input": "99999999999999999",
"output": "50000000000000000"
},
{
"input": "21",
"output": "11"
},
{
"input": "712394453192",
"output": "712394453193"
},
{
"input": "172212168",
"output": "172212169"
},
{
"input": "822981260158260519",
"output": "411490630079130260"
},
{
"input": "28316250877914571",
"output": "14158125438957286"
},
{
"input": "779547116602436424",
"output": "779547116602436425"
},
{
"input": "578223540024979436",
"output": "578223540024979437"
},
{
"input": "335408917861648766",
"output": "335408917861648767"
},
{
"input": "74859962623690078",
"output": "74859962623690079"
},
{
"input": "252509054433933439",
"output": "126254527216966720"
},
{
"input": "760713016476190622",
"output": "760713016476190623"
},
{
"input": "919845426262703496",
"output": "919845426262703497"
},
{
"input": "585335723211047194",
"output": "585335723211047195"
},
{
"input": "522842184971407769",
"output": "261421092485703885"
},
{
"input": "148049062628894320",
"output": "148049062628894321"
},
{
"input": "84324828731963974",
"output": "84324828731963975"
},
{
"input": "354979173822804781",
"output": "177489586911402391"
},
{
"input": "1312150450968413",
"output": "656075225484207"
},
{
"input": "269587449430302150",
"output": "269587449430302151"
},
{
"input": "645762258982631926",
"output": "645762258982631927"
},
{
"input": "615812229161735895",
"output": "307906114580867948"
},
{
"input": "0",
"output": "0"
},
{
"input": "349993004923078531",
"output": "174996502461539266"
},
{
"input": "891351282707723851",
"output": "445675641353861926"
},
{
"input": "563324731189330734",
"output": "563324731189330735"
},
{
"input": "520974001910286909",
"output": "260487000955143455"
},
{
"input": "666729339802329204",
"output": "666729339802329205"
},
{
"input": "856674611404539671",
"output": "428337305702269836"
},
{
"input": "791809296303238499",
"output": "395904648151619250"
},
{
"input": "711066337317063338",
"output": "711066337317063339"
},
{
"input": "931356503492686566",
"output": "931356503492686567"
},
{
"input": "234122432773361866",
"output": "234122432773361867"
},
{
"input": "1000000000000000000",
"output": "1000000000000000001"
},
{
"input": "1",
"output": "1"
},
{
"input": "2",
"output": "3"
},
{
"input": "7",
"output": "4"
},
{
"input": "63",
"output": "32"
},
{
"input": "24",
"output": "25"
},
{
"input": "8",
"output": "9"
},
{
"input": "15",
"output": "8"
}
] | 1,526,566,936
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 10
| 78
| 0
|
n = int(input())
attendant = n+1
if attendant%2:
print (attendant)
else:
print (int(attendant/2))
|
Title: Pizza, Pizza, Pizza!!!
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Katie, Kuro and Shiro are best friends. They have known each other since kindergarten. That's why they often share everything with each other and work together on some very hard problems.
Today is Shiro's birthday. She really loves pizza so she wants to invite her friends to the pizza restaurant near her house to celebrate her birthday, including her best friends Katie and Kuro.
She has ordered a very big round pizza, in order to serve her many friends. Exactly $n$ of Shiro's friends are here. That's why she has to divide the pizza into $n + 1$ slices (Shiro also needs to eat). She wants the slices to be exactly the same size and shape. If not, some of her friends will get mad and go home early, and the party will be over.
Shiro is now hungry. She wants to cut the pizza with minimum of straight cuts. A cut is a straight segment, it might have ends inside or outside the pizza. But she is too lazy to pick up the calculator.
As usual, she will ask Katie and Kuro for help. But they haven't come yet. Could you help Shiro with this problem?
Input Specification:
A single line contains one non-negative integer $n$ ($0 \le n \leq 10^{18}$) — the number of Shiro's friends. The circular pizza has to be sliced into $n + 1$ pieces.
Output Specification:
A single integer — the number of straight cuts Shiro needs.
Demo Input:
['3\n', '4\n']
Demo Output:
['2', '5']
Note:
To cut the round pizza into quarters one has to make two cuts through the center with angle $90^{\circ}$ between them.
To cut the round pizza into five equal parts one has to make five cuts.
|
```python
n = int(input())
attendant = n+1
if attendant%2:
print (attendant)
else:
print (int(attendant/2))
```
| 0
|
|
347
|
A
|
Difference Row
|
PROGRAMMING
| 1,300
|
[
"constructive algorithms",
"implementation",
"sortings"
] | null | null |
You want to arrange *n* integers *a*1,<=*a*2,<=...,<=*a**n* in some order in a row. Let's define the value of an arrangement as the sum of differences between all pairs of adjacent integers.
More formally, let's denote some arrangement as a sequence of integers *x*1,<=*x*2,<=...,<=*x**n*, where sequence *x* is a permutation of sequence *a*. The value of such an arrangement is (*x*1<=-<=*x*2)<=+<=(*x*2<=-<=*x*3)<=+<=...<=+<=(*x**n*<=-<=1<=-<=*x**n*).
Find the largest possible value of an arrangement. Then, output the lexicographically smallest sequence *x* that corresponds to an arrangement of the largest possible value.
|
The first line of the input contains integer *n* (2<=≤<=*n*<=≤<=100). The second line contains *n* space-separated integers *a*1, *a*2, ..., *a**n* (|*a**i*|<=≤<=1000).
|
Print the required sequence *x*1,<=*x*2,<=...,<=*x**n*. Sequence *x* should be the lexicographically smallest permutation of *a* that corresponds to an arrangement of the largest possible value.
|
[
"5\n100 -100 50 0 -50\n"
] |
[
"100 -50 0 50 -100 \n"
] |
In the sample test case, the value of the output arrangement is (100 - ( - 50)) + (( - 50) - 0) + (0 - 50) + (50 - ( - 100)) = 200. No other arrangement has a larger value, and among all arrangements with the value of 200, the output arrangement is the lexicographically smallest one.
Sequence *x*<sub class="lower-index">1</sub>, *x*<sub class="lower-index">2</sub>, ... , *x*<sub class="lower-index">*p*</sub> is lexicographically smaller than sequence *y*<sub class="lower-index">1</sub>, *y*<sub class="lower-index">2</sub>, ... , *y*<sub class="lower-index">*p*</sub> if there exists an integer *r* (0 ≤ *r* < *p*) such that *x*<sub class="lower-index">1</sub> = *y*<sub class="lower-index">1</sub>, *x*<sub class="lower-index">2</sub> = *y*<sub class="lower-index">2</sub>, ... , *x*<sub class="lower-index">*r*</sub> = *y*<sub class="lower-index">*r*</sub> and *x*<sub class="lower-index">*r* + 1</sub> < *y*<sub class="lower-index">*r* + 1</sub>.
| 500
|
[
{
"input": "5\n100 -100 50 0 -50",
"output": "100 -50 0 50 -100 "
},
{
"input": "10\n764 -367 0 963 -939 -795 -26 -49 948 -282",
"output": "963 -795 -367 -282 -49 -26 0 764 948 -939 "
},
{
"input": "20\n262 -689 -593 161 -678 -555 -633 -697 369 258 673 50 833 737 -650 198 -651 -621 -396 939",
"output": "939 -689 -678 -651 -650 -633 -621 -593 -555 -396 50 161 198 258 262 369 673 737 833 -697 "
},
{
"input": "50\n-262 -377 -261 903 547 759 -800 -53 670 92 758 109 547 877 152 -901 -318 -527 -388 24 139 -227 413 -135 811 -886 -22 -526 -643 -431 284 609 -745 -62 323 -441 743 -800 86 862 587 -513 -468 -651 -760 197 141 -414 -909 438",
"output": "903 -901 -886 -800 -800 -760 -745 -651 -643 -527 -526 -513 -468 -441 -431 -414 -388 -377 -318 -262 -261 -227 -135 -62 -53 -22 24 86 92 109 139 141 152 197 284 323 413 438 547 547 587 609 670 743 758 759 811 862 877 -909 "
},
{
"input": "100\n144 -534 -780 -1 -259 -945 -992 -967 -679 -239 -22 387 130 -908 140 -270 16 646 398 599 -631 -231 687 -505 89 77 584 162 124 132 33 271 212 734 350 -678 969 43 487 -689 -432 -225 -603 801 -828 -684 349 318 109 723 33 -247 719 368 -286 217 260 77 -618 955 408 994 -313 -341 578 609 60 900 222 -779 -507 464 -147 -789 -477 -235 -407 -432 35 300 -53 -896 -476 927 -293 -869 -852 -566 -759 95 506 -914 -405 -621 319 -622 -49 -334 328 -104",
"output": "994 -967 -945 -914 -908 -896 -869 -852 -828 -789 -780 -779 -759 -689 -684 -679 -678 -631 -622 -621 -618 -603 -566 -534 -507 -505 -477 -476 -432 -432 -407 -405 -341 -334 -313 -293 -286 -270 -259 -247 -239 -235 -231 -225 -147 -104 -53 -49 -22 -1 16 33 33 35 43 60 77 77 89 95 109 124 130 132 140 144 162 212 217 222 260 271 300 318 319 328 349 350 368 387 398 408 464 487 506 578 584 599 609 646 687 719 723 734 801 900 927 955 969 -992 "
},
{
"input": "100\n-790 341 910 905 -779 279 696 -375 525 -21 -2 751 -887 764 520 -844 850 -537 -882 -183 139 -397 561 -420 -991 691 587 -93 -701 -957 -89 227 233 545 934 309 -26 454 -336 -994 -135 -840 -320 -387 -943 650 628 -583 701 -708 -881 287 -932 -265 -312 -757 695 985 -165 -329 -4 -462 -627 798 -124 -539 843 -492 -967 -782 879 -184 -351 -385 -713 699 -477 828 219 961 -170 -542 877 -718 417 152 -905 181 301 920 685 -502 518 -115 257 998 -112 -234 -223 -396",
"output": "998 -991 -967 -957 -943 -932 -905 -887 -882 -881 -844 -840 -790 -782 -779 -757 -718 -713 -708 -701 -627 -583 -542 -539 -537 -502 -492 -477 -462 -420 -397 -396 -387 -385 -375 -351 -336 -329 -320 -312 -265 -234 -223 -184 -183 -170 -165 -135 -124 -115 -112 -93 -89 -26 -21 -4 -2 139 152 181 219 227 233 257 279 287 301 309 341 417 454 518 520 525 545 561 587 628 650 685 691 695 696 699 701 751 764 798 828 843 850 877 879 905 910 920 934 961 985 -994 "
},
{
"input": "100\n720 331 -146 -935 399 248 525 -669 614 -245 320 229 842 -894 -73 584 -458 -975 -604 -78 607 -120 -377 409 -743 862 -969 980 105 841 -795 996 696 -759 -482 624 -578 421 -717 -553 -652 -268 405 426 642 870 -650 -812 178 -882 -237 -737 -724 358 407 714 759 779 -899 -726 398 -663 -56 -736 -825 313 -746 117 -457 330 -925 497 332 -794 -506 -811 -990 -799 -343 -380 598 926 671 967 -573 -687 741 484 -641 -698 -251 -391 23 692 337 -639 126 8 -915 -386",
"output": "996 -975 -969 -935 -925 -915 -899 -894 -882 -825 -812 -811 -799 -795 -794 -759 -746 -743 -737 -736 -726 -724 -717 -698 -687 -669 -663 -652 -650 -641 -639 -604 -578 -573 -553 -506 -482 -458 -457 -391 -386 -380 -377 -343 -268 -251 -245 -237 -146 -120 -78 -73 -56 8 23 105 117 126 178 229 248 313 320 330 331 332 337 358 398 399 405 407 409 421 426 484 497 525 584 598 607 614 624 642 671 692 696 714 720 741 759 779 841 842 862 870 926 967 980 -990 "
},
{
"input": "100\n-657 320 -457 -472 -423 -227 -902 -520 702 -27 -103 149 268 -922 307 -292 377 730 117 1000 935 459 -502 796 -494 892 -523 866 166 -248 57 -606 -96 -948 988 194 -687 832 -425 28 -356 -884 688 353 225 204 -68 960 -929 -312 -479 381 512 -274 -505 -260 -506 572 226 -822 -13 325 -370 403 -714 494 339 283 356 327 159 -151 -13 -760 -159 -991 498 19 -159 583 178 -50 -421 -679 -978 334 688 -99 117 -988 371 693 946 -58 -699 -133 62 693 535 -375",
"output": "1000 -988 -978 -948 -929 -922 -902 -884 -822 -760 -714 -699 -687 -679 -657 -606 -523 -520 -506 -505 -502 -494 -479 -472 -457 -425 -423 -421 -375 -370 -356 -312 -292 -274 -260 -248 -227 -159 -159 -151 -133 -103 -99 -96 -68 -58 -50 -27 -13 -13 19 28 57 62 117 117 149 159 166 178 194 204 225 226 268 283 307 320 325 327 334 339 353 356 371 377 381 403 459 494 498 512 535 572 583 688 688 693 693 702 730 796 832 866 892 935 946 960 988 -991 "
},
{
"input": "100\n853 752 931 -453 -943 -118 -772 -814 791 191 -83 -373 -748 -136 -286 250 627 292 -48 -896 -296 736 -628 -376 -246 -495 366 610 228 664 -951 -952 811 192 -730 -377 319 799 753 166 827 501 157 -834 -776 424 655 -827 549 -487 608 -643 419 349 -88 95 231 -520 -508 -105 -727 568 -241 286 586 -956 -880 892 866 22 658 832 -216 -54 491 -500 -687 393 24 129 946 303 931 563 -269 -203 -251 647 -824 -163 248 -896 -133 749 -619 -212 -2 491 287 219",
"output": "946 -952 -951 -943 -896 -896 -880 -834 -827 -824 -814 -776 -772 -748 -730 -727 -687 -643 -628 -619 -520 -508 -500 -495 -487 -453 -377 -376 -373 -296 -286 -269 -251 -246 -241 -216 -212 -203 -163 -136 -133 -118 -105 -88 -83 -54 -48 -2 22 24 95 129 157 166 191 192 219 228 231 248 250 286 287 292 303 319 349 366 393 419 424 491 491 501 549 563 568 586 608 610 627 647 655 658 664 736 749 752 753 791 799 811 827 832 853 866 892 931 931 -956 "
},
{
"input": "100\n9 857 227 -593 -983 -439 17 -523 -354 -189 780 -267 771 -981 943 620 -832 79 761 -943 218 -966 75 131 -596 534 51 796 -612 -381 -690 -353 -170 648 804 -256 257 -16 964 -728 310 50 453 737 -228 -625 618 841 -102 974 -850 -641 -788 231 -982 -84 -917 942 -913 -768 -83 298 388 447 -490 271 -949 976 -820 -876 -822 -188 -306 877 219 854 561 -307 -920 916 -925 -591 -149 -166 -572 860 -217 -831 -552 822 355 -150 203 -710 530 910 889 964 -125 -597",
"output": "976 -982 -981 -966 -949 -943 -925 -920 -917 -913 -876 -850 -832 -831 -822 -820 -788 -768 -728 -710 -690 -641 -625 -612 -597 -596 -593 -591 -572 -552 -523 -490 -439 -381 -354 -353 -307 -306 -267 -256 -228 -217 -189 -188 -170 -166 -150 -149 -125 -102 -84 -83 -16 9 17 50 51 75 79 131 203 218 219 227 231 257 271 298 310 355 388 447 453 530 534 561 618 620 648 737 761 771 780 796 804 822 841 854 857 860 877 889 910 916 942 943 964 964 974 -983 "
},
{
"input": "2\n-1000 1000",
"output": "1000 -1000 "
},
{
"input": "2\n1000 -1000",
"output": "1000 -1000 "
},
{
"input": "2\n0 0",
"output": "0 0 "
},
{
"input": "5\n1 2 3 4 5",
"output": "5 2 3 4 1 "
},
{
"input": "6\n1 1 1 2 2 2",
"output": "2 1 1 2 2 1 "
},
{
"input": "3\n-1 -1 -1",
"output": "-1 -1 -1 "
}
] | 1,629,135,813
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 16
| 124
| 6,963,200
|
input()
a=list(map(int,input().split()))
a=sorted(a)
b=a[1:len(a)-1]
b=sorted(b)
#print(b,a)
ans=[a[-1]]+b+[a[0]]
answer=""
for i in ans:
answer=answer+str(i)+" "
print(answer)
|
Title: Difference Row
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You want to arrange *n* integers *a*1,<=*a*2,<=...,<=*a**n* in some order in a row. Let's define the value of an arrangement as the sum of differences between all pairs of adjacent integers.
More formally, let's denote some arrangement as a sequence of integers *x*1,<=*x*2,<=...,<=*x**n*, where sequence *x* is a permutation of sequence *a*. The value of such an arrangement is (*x*1<=-<=*x*2)<=+<=(*x*2<=-<=*x*3)<=+<=...<=+<=(*x**n*<=-<=1<=-<=*x**n*).
Find the largest possible value of an arrangement. Then, output the lexicographically smallest sequence *x* that corresponds to an arrangement of the largest possible value.
Input Specification:
The first line of the input contains integer *n* (2<=≤<=*n*<=≤<=100). The second line contains *n* space-separated integers *a*1, *a*2, ..., *a**n* (|*a**i*|<=≤<=1000).
Output Specification:
Print the required sequence *x*1,<=*x*2,<=...,<=*x**n*. Sequence *x* should be the lexicographically smallest permutation of *a* that corresponds to an arrangement of the largest possible value.
Demo Input:
['5\n100 -100 50 0 -50\n']
Demo Output:
['100 -50 0 50 -100 \n']
Note:
In the sample test case, the value of the output arrangement is (100 - ( - 50)) + (( - 50) - 0) + (0 - 50) + (50 - ( - 100)) = 200. No other arrangement has a larger value, and among all arrangements with the value of 200, the output arrangement is the lexicographically smallest one.
Sequence *x*<sub class="lower-index">1</sub>, *x*<sub class="lower-index">2</sub>, ... , *x*<sub class="lower-index">*p*</sub> is lexicographically smaller than sequence *y*<sub class="lower-index">1</sub>, *y*<sub class="lower-index">2</sub>, ... , *y*<sub class="lower-index">*p*</sub> if there exists an integer *r* (0 ≤ *r* < *p*) such that *x*<sub class="lower-index">1</sub> = *y*<sub class="lower-index">1</sub>, *x*<sub class="lower-index">2</sub> = *y*<sub class="lower-index">2</sub>, ... , *x*<sub class="lower-index">*r*</sub> = *y*<sub class="lower-index">*r*</sub> and *x*<sub class="lower-index">*r* + 1</sub> < *y*<sub class="lower-index">*r* + 1</sub>.
|
```python
input()
a=list(map(int,input().split()))
a=sorted(a)
b=a[1:len(a)-1]
b=sorted(b)
#print(b,a)
ans=[a[-1]]+b+[a[0]]
answer=""
for i in ans:
answer=answer+str(i)+" "
print(answer)
```
| 3
|
|
931
|
A
|
Friends Meeting
|
PROGRAMMING
| 800
|
[
"brute force",
"greedy",
"implementation",
"math"
] | null | null |
Two friends are on the coordinate axis *Ox* in points with integer coordinates. One of them is in the point *x*1<==<=*a*, another one is in the point *x*2<==<=*b*.
Each of the friends can move by one along the line in any direction unlimited number of times. When a friend moves, the tiredness of a friend changes according to the following rules: the first move increases the tiredness by 1, the second move increases the tiredness by 2, the third — by 3 and so on. For example, if a friend moves first to the left, then to the right (returning to the same point), and then again to the left his tiredness becomes equal to 1<=+<=2<=+<=3<==<=6.
The friends want to meet in a integer point. Determine the minimum total tiredness they should gain, if they meet in the same point.
|
The first line contains a single integer *a* (1<=≤<=*a*<=≤<=1000) — the initial position of the first friend.
The second line contains a single integer *b* (1<=≤<=*b*<=≤<=1000) — the initial position of the second friend.
It is guaranteed that *a*<=≠<=*b*.
|
Print the minimum possible total tiredness if the friends meet in the same point.
|
[
"3\n4\n",
"101\n99\n",
"5\n10\n"
] |
[
"1\n",
"2\n",
"9\n"
] |
In the first example the first friend should move by one to the right (then the meeting happens at point 4), or the second friend should move by one to the left (then the meeting happens at point 3). In both cases, the total tiredness becomes 1.
In the second example the first friend should move by one to the left, and the second friend should move by one to the right. Then they meet in the point 100, and the total tiredness becomes 1 + 1 = 2.
In the third example one of the optimal ways is the following. The first friend should move three times to the right, and the second friend — two times to the left. Thus the friends meet in the point 8, and the total tiredness becomes 1 + 2 + 3 + 1 + 2 = 9.
| 500
|
[
{
"input": "3\n4",
"output": "1"
},
{
"input": "101\n99",
"output": "2"
},
{
"input": "5\n10",
"output": "9"
},
{
"input": "1\n2",
"output": "1"
},
{
"input": "1\n1000",
"output": "250000"
},
{
"input": "999\n1000",
"output": "1"
},
{
"input": "1000\n999",
"output": "1"
},
{
"input": "1000\n1",
"output": "250000"
},
{
"input": "2\n1",
"output": "1"
},
{
"input": "2\n999",
"output": "249001"
},
{
"input": "2\n998",
"output": "248502"
},
{
"input": "999\n2",
"output": "249001"
},
{
"input": "998\n2",
"output": "248502"
},
{
"input": "2\n1000",
"output": "249500"
},
{
"input": "1000\n2",
"output": "249500"
},
{
"input": "1\n999",
"output": "249500"
},
{
"input": "999\n1",
"output": "249500"
},
{
"input": "188\n762",
"output": "82656"
},
{
"input": "596\n777",
"output": "8281"
},
{
"input": "773\n70",
"output": "123904"
},
{
"input": "825\n729",
"output": "2352"
},
{
"input": "944\n348",
"output": "89102"
},
{
"input": "352\n445",
"output": "2209"
},
{
"input": "529\n656",
"output": "4096"
},
{
"input": "19\n315",
"output": "22052"
},
{
"input": "138\n370",
"output": "13572"
},
{
"input": "546\n593",
"output": "576"
},
{
"input": "285\n242",
"output": "484"
},
{
"input": "773\n901",
"output": "4160"
},
{
"input": "892\n520",
"output": "34782"
},
{
"input": "864\n179",
"output": "117649"
},
{
"input": "479\n470",
"output": "25"
},
{
"input": "967\n487",
"output": "57840"
},
{
"input": "648\n106",
"output": "73712"
},
{
"input": "58\n765",
"output": "125316"
},
{
"input": "235\n56",
"output": "8100"
},
{
"input": "285\n153",
"output": "4422"
},
{
"input": "943\n13",
"output": "216690"
},
{
"input": "675\n541",
"output": "4556"
},
{
"input": "4\n912",
"output": "206570"
}
] | 1,579,874,860
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 40
| 109
| 307,200
|
import sys
import math
#to read string
get_string = lambda: sys.stdin.readline().strip()
#to read list of integers
get_list = lambda: list( map(int,sys.stdin.readline().strip().split()) )
#to read non spaced string and elements are integers to list of int
get_intList_from_str = lambda: list(map(int,list(sys.stdin.readline().strip())))
#to read non spaced string and elements are character to list of character
get_strList_from_str = lambda: list(sys.stdin.readline().strip())
#to read integers
get_int = lambda: int(sys.stdin.readline().strip())
#to print faster
pt = lambda x: sys.stdout.write(str(x))
#--------------------------------WhiteHat010--------------------------------------#
a = get_int()
b = get_int()
d = (a+b)//2
first = abs(d-a)
second = abs(b-d)
print( first*(first+1)//2 + second*(second+1)//2)
|
Title: Friends Meeting
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Two friends are on the coordinate axis *Ox* in points with integer coordinates. One of them is in the point *x*1<==<=*a*, another one is in the point *x*2<==<=*b*.
Each of the friends can move by one along the line in any direction unlimited number of times. When a friend moves, the tiredness of a friend changes according to the following rules: the first move increases the tiredness by 1, the second move increases the tiredness by 2, the third — by 3 and so on. For example, if a friend moves first to the left, then to the right (returning to the same point), and then again to the left his tiredness becomes equal to 1<=+<=2<=+<=3<==<=6.
The friends want to meet in a integer point. Determine the minimum total tiredness they should gain, if they meet in the same point.
Input Specification:
The first line contains a single integer *a* (1<=≤<=*a*<=≤<=1000) — the initial position of the first friend.
The second line contains a single integer *b* (1<=≤<=*b*<=≤<=1000) — the initial position of the second friend.
It is guaranteed that *a*<=≠<=*b*.
Output Specification:
Print the minimum possible total tiredness if the friends meet in the same point.
Demo Input:
['3\n4\n', '101\n99\n', '5\n10\n']
Demo Output:
['1\n', '2\n', '9\n']
Note:
In the first example the first friend should move by one to the right (then the meeting happens at point 4), or the second friend should move by one to the left (then the meeting happens at point 3). In both cases, the total tiredness becomes 1.
In the second example the first friend should move by one to the left, and the second friend should move by one to the right. Then they meet in the point 100, and the total tiredness becomes 1 + 1 = 2.
In the third example one of the optimal ways is the following. The first friend should move three times to the right, and the second friend — two times to the left. Thus the friends meet in the point 8, and the total tiredness becomes 1 + 2 + 3 + 1 + 2 = 9.
|
```python
import sys
import math
#to read string
get_string = lambda: sys.stdin.readline().strip()
#to read list of integers
get_list = lambda: list( map(int,sys.stdin.readline().strip().split()) )
#to read non spaced string and elements are integers to list of int
get_intList_from_str = lambda: list(map(int,list(sys.stdin.readline().strip())))
#to read non spaced string and elements are character to list of character
get_strList_from_str = lambda: list(sys.stdin.readline().strip())
#to read integers
get_int = lambda: int(sys.stdin.readline().strip())
#to print faster
pt = lambda x: sys.stdout.write(str(x))
#--------------------------------WhiteHat010--------------------------------------#
a = get_int()
b = get_int()
d = (a+b)//2
first = abs(d-a)
second = abs(b-d)
print( first*(first+1)//2 + second*(second+1)//2)
```
| 3
|
|
630
|
A
|
Again Twenty Five!
|
PROGRAMMING
| 800
|
[
"number theory"
] | null | null |
The HR manager was disappointed again. The last applicant failed the interview the same way as 24 previous ones. "Do I give such a hard task?" — the HR manager thought. "Just raise number 5 to the power of *n* and get last two digits of the number. Yes, of course, *n* can be rather big, and one cannot find the power using a calculator, but we need people who are able to think, not just follow the instructions."
Could you pass the interview in the machine vision company in IT City?
|
The only line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=2·1018) — the power in which you need to raise number 5.
|
Output the last two digits of 5*n* without spaces between them.
|
[
"2\n"
] |
[
"25"
] |
none
| 0
|
[
{
"input": "2",
"output": "25"
},
{
"input": "7",
"output": "25"
},
{
"input": "1000000000000000000",
"output": "25"
},
{
"input": "2000000000000000000",
"output": "25"
},
{
"input": "987654321012345678",
"output": "25"
}
] | 1,699,322,471
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 5
| 46
| 0
|
if int(input()) == 1:
print(5)
else:
print(25)
|
Title: Again Twenty Five!
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The HR manager was disappointed again. The last applicant failed the interview the same way as 24 previous ones. "Do I give such a hard task?" — the HR manager thought. "Just raise number 5 to the power of *n* and get last two digits of the number. Yes, of course, *n* can be rather big, and one cannot find the power using a calculator, but we need people who are able to think, not just follow the instructions."
Could you pass the interview in the machine vision company in IT City?
Input Specification:
The only line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=2·1018) — the power in which you need to raise number 5.
Output Specification:
Output the last two digits of 5*n* without spaces between them.
Demo Input:
['2\n']
Demo Output:
['25']
Note:
none
|
```python
if int(input()) == 1:
print(5)
else:
print(25)
```
| 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<=≤<=*n*<=≤<=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 — 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: 991 → 19 → 10 → 1. 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"
}
] | 1,676,466,285
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 51
| 530
| 0
|
# LUOGU_RID: 102214607
def f(x):
t=0
for i in x:t+=int(i)
return t
while True:
try:
s=input().strip()
n=0
while int(s)>=10:
s=f(str(s))
n+=1
print(n)
except:
break
|
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<=≤<=*n*<=≤<=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 — 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: 991 → 19 → 10 → 1. After three transformations the number becomes one-digit.
|
```python
# LUOGU_RID: 102214607
def f(x):
t=0
for i in x:t+=int(i)
return t
while True:
try:
s=input().strip()
n=0
while int(s)>=10:
s=f(str(s))
n+=1
print(n)
except:
break
```
| 3.8675
|
69
|
A
|
Young Physicist
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] |
A. Young Physicist
|
2
|
256
|
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
|
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
|
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
|
[
"3\n4 1 7\n-2 4 -1\n1 -5 -3\n",
"3\n3 -1 7\n-5 2 -4\n2 -1 -3\n"
] |
[
"NO",
"YES"
] |
none
| 500
|
[
{
"input": "3\n4 1 7\n-2 4 -1\n1 -5 -3",
"output": "NO"
},
{
"input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3",
"output": "YES"
},
{
"input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41",
"output": "NO"
},
{
"input": "10\n25 -33 43\n-27 -42 28\n-35 -20 19\n41 -42 -1\n49 -39 -4\n-49 -22 7\n-19 29 41\n8 -27 -43\n8 34 9\n-11 -3 33",
"output": "NO"
},
{
"input": "10\n-6 21 18\n20 -11 -8\n37 -11 41\n-5 8 33\n29 23 32\n30 -33 -11\n39 -49 -36\n28 34 -49\n22 29 -34\n-18 -6 7",
"output": "NO"
},
{
"input": "10\n47 -2 -27\n0 26 -14\n5 -12 33\n2 18 3\n45 -30 -49\n4 -18 8\n-46 -44 -41\n-22 -10 -40\n-35 -21 26\n33 20 38",
"output": "NO"
},
{
"input": "13\n-3 -36 -46\n-11 -50 37\n42 -11 -15\n9 42 44\n-29 -12 24\n3 9 -40\n-35 13 50\n14 43 18\n-13 8 24\n-48 -15 10\n50 9 -50\n21 0 -50\n0 0 -6",
"output": "YES"
},
{
"input": "14\n43 23 17\n4 17 44\n5 -5 -16\n-43 -7 -6\n47 -48 12\n50 47 -45\n2 14 43\n37 -30 15\n4 -17 -11\n17 9 -45\n-50 -3 -8\n-50 0 0\n-50 0 0\n-16 0 0",
"output": "YES"
},
{
"input": "13\n29 49 -11\n38 -11 -20\n25 1 -40\n-11 28 11\n23 -19 1\n45 -41 -17\n-3 0 -19\n-13 -33 49\n-30 0 28\n34 17 45\n-50 9 -27\n-50 0 0\n-37 0 0",
"output": "YES"
},
{
"input": "12\n3 28 -35\n-32 -44 -17\n9 -25 -6\n-42 -22 20\n-19 15 38\n-21 38 48\n-1 -37 -28\n-10 -13 -50\n-5 21 29\n34 28 50\n50 11 -49\n34 0 0",
"output": "YES"
},
{
"input": "37\n-64 -79 26\n-22 59 93\n-5 39 -12\n77 -9 76\n55 -86 57\n83 100 -97\n-70 94 84\n-14 46 -94\n26 72 35\n14 78 -62\n17 82 92\n-57 11 91\n23 15 92\n-80 -1 1\n12 39 18\n-23 -99 -75\n-34 50 19\n-39 84 -7\n45 -30 -39\n-60 49 37\n45 -16 -72\n33 -51 -56\n-48 28 5\n97 91 88\n45 -82 -11\n-21 -15 -90\n-53 73 -26\n-74 85 -90\n-40 23 38\n100 -13 49\n32 -100 -100\n0 -100 -70\n0 -100 0\n0 -100 0\n0 -100 0\n0 -100 0\n0 -37 0",
"output": "YES"
},
{
"input": "4\n68 3 100\n68 21 -100\n-100 -24 0\n-36 0 0",
"output": "YES"
},
{
"input": "33\n-1 -46 -12\n45 -16 -21\n-11 45 -21\n-60 -42 -93\n-22 -45 93\n37 96 85\n-76 26 83\n-4 9 55\n7 -52 -9\n66 8 -85\n-100 -54 11\n-29 59 74\n-24 12 2\n-56 81 85\n-92 69 -52\n-26 -97 91\n54 59 -51\n58 21 -57\n7 68 56\n-47 -20 -51\n-59 77 -13\n-85 27 91\n79 60 -56\n66 -80 5\n21 -99 42\n-31 -29 98\n66 93 76\n-49 45 61\n100 -100 -100\n100 -100 -100\n66 -75 -100\n0 0 -100\n0 0 -87",
"output": "YES"
},
{
"input": "3\n1 2 3\n3 2 1\n0 0 0",
"output": "NO"
},
{
"input": "2\n5 -23 12\n0 0 0",
"output": "NO"
},
{
"input": "1\n0 0 0",
"output": "YES"
},
{
"input": "1\n1 -2 0",
"output": "NO"
},
{
"input": "2\n-23 77 -86\n23 -77 86",
"output": "YES"
},
{
"input": "26\n86 7 20\n-57 -64 39\n-45 6 -93\n-44 -21 100\n-11 -49 21\n73 -71 -80\n-2 -89 56\n-65 -2 7\n5 14 84\n57 41 13\n-12 69 54\n40 -25 27\n-17 -59 0\n64 -91 -30\n-53 9 42\n-54 -8 14\n-35 82 27\n-48 -59 -80\n88 70 79\n94 57 97\n44 63 25\n84 -90 -40\n-100 100 -100\n-92 100 -100\n0 10 -100\n0 0 -82",
"output": "YES"
},
{
"input": "42\n11 27 92\n-18 -56 -57\n1 71 81\n33 -92 30\n82 83 49\n-87 -61 -1\n-49 45 49\n73 26 15\n-22 22 -77\n29 -93 87\n-68 44 -90\n-4 -84 20\n85 67 -6\n-39 26 77\n-28 -64 20\n65 -97 24\n-72 -39 51\n35 -75 -91\n39 -44 -8\n-25 -27 -57\n91 8 -46\n-98 -94 56\n94 -60 59\n-9 -95 18\n-53 -37 98\n-8 -94 -84\n-52 55 60\n15 -14 37\n65 -43 -25\n94 12 66\n-8 -19 -83\n29 81 -78\n-58 57 33\n24 86 -84\n-53 32 -88\n-14 7 3\n89 97 -53\n-5 -28 -91\n-100 100 -6\n-84 100 0\n0 100 0\n0 70 0",
"output": "YES"
},
{
"input": "3\n96 49 -12\n2 -66 28\n-98 17 -16",
"output": "YES"
},
{
"input": "5\n70 -46 86\n-100 94 24\n-27 63 -63\n57 -100 -47\n0 -11 0",
"output": "YES"
},
{
"input": "18\n-86 -28 70\n-31 -89 42\n31 -48 -55\n95 -17 -43\n24 -95 -85\n-21 -14 31\n68 -18 81\n13 31 60\n-15 28 99\n-42 15 9\n28 -61 -62\n-16 71 29\n-28 75 -48\n-77 -67 36\n-100 83 89\n100 100 -100\n57 34 -100\n0 0 -53",
"output": "YES"
},
{
"input": "44\n52 -54 -29\n-82 -5 -94\n-54 43 43\n91 16 71\n7 80 -91\n3 15 29\n-99 -6 -77\n-3 -77 -64\n73 67 34\n25 -10 -18\n-29 91 63\n-72 86 -16\n-68 85 -81\n-3 36 44\n-74 -14 -80\n34 -96 -97\n-76 -78 -33\n-24 44 -58\n98 12 77\n95 -63 -6\n-51 3 -90\n-92 -10 72\n7 3 -68\n57 -53 71\n29 57 -48\n35 -60 10\n79 -70 -61\n-20 77 55\n-86 -15 -35\n84 -88 -18\n100 -42 77\n-20 46 8\n-41 -43 -65\n38 -98 -23\n-100 65 45\n-7 -91 -63\n46 88 -85\n48 59 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 1",
"output": "YES"
},
{
"input": "18\n-14 -64 -91\n-8 -66 -86\n-23 92 -40\n6 -3 -53\n57 41 78\n-79 42 -22\n-88 -17 45\n4 -45 44\n83 -18 -25\n34 86 -92\n75 -30 12\n44 99 11\n-67 -13 72\n22 83 -56\n-37 71 72\n-9 -100 100\n0 -100 31\n0 -58 0",
"output": "YES"
},
{
"input": "23\n-70 37 78\n42 84 6\n28 -94 -24\n-49 76 95\n-67 18 84\n-53 78 -5\n65 -63 -64\n-66 -64 -69\n81 -93 95\n10 52 -79\n-89 -61 -64\n-64 -47 43\n-81 -35 55\n80 82 73\n-60 -81 -18\n49 -10 -19\n-58 70 12\n-24 -15 -93\n98 -93 -54\n-28 -75 11\n100 100 -63\n100 100 0\n56 34 0",
"output": "YES"
},
{
"input": "3\n0 2 -2\n1 -1 3\n-3 0 0",
"output": "NO"
}
] | 1,573,391,362
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 122
| 0
|
n = int(input())
x = list(map(int,input().split()))[:n]
y = list(map(int,input().split()))[:n]
z = list(map(int,input().split()))[:n]
for i in range (len(x)):
count = x[i]
for j in range (len(y)):
zeb = y[j]
for k in range (len(z)):
bez = z[k]
el = count+zeb+bez
if el == 0 :
print("yes")
else:
print("no")
|
Title: Young Physicist
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
Output Specification:
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Demo Input:
['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n']
Demo Output:
['NO', 'YES']
Note:
none
|
```python
n = int(input())
x = list(map(int,input().split()))[:n]
y = list(map(int,input().split()))[:n]
z = list(map(int,input().split()))[:n]
for i in range (len(x)):
count = x[i]
for j in range (len(y)):
zeb = y[j]
for k in range (len(z)):
bez = z[k]
el = count+zeb+bez
if el == 0 :
print("yes")
else:
print("no")
```
| 0
|
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<=≤<=*n*<=≤<=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 999 999 999.
| 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"
}
] | 1,677,250,129
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 87
| 93
| 17,920,000
|
Z = int(input())
A = list(map(int, input().split()))
s = sum(A)
if s % 2 == 0:
print(s)
else:
print(s - min([x for x in A if x % 2 == 1]))
|
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<=≤<=*n*<=≤<=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 999 999 999.
|
```python
Z = int(input())
A = list(map(int, input().split()))
s = sum(A)
if s % 2 == 0:
print(s)
else:
print(s - min([x for x in A if x % 2 == 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*<=×<=*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*<=×<=*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<=≤<=<=*n*,<=*m*,<=*a*<=≤<=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"
}
] | 1,693,754,928
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 6
| 46
| 0
|
n ,m ,a= map(int,input().split())
c=(n*m) / (a*a)
if int(c)== c:
print(int(c))
elif int(c)+0.5 <= c:
print(int(c)+1)
else:
print(int(c)+2)
|
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*<=×<=*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*<=×<=*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<=≤<=<=*n*,<=*m*,<=*a*<=≤<=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())
c=(n*m) / (a*a)
if int(c)== c:
print(int(c))
elif int(c)+0.5 <= c:
print(int(c)+1)
else:
print(int(c)+2)
```
| 0
|
604
|
B
|
More Cowbell
|
PROGRAMMING
| 1,400
|
[
"binary search",
"greedy"
] | null | null |
Kevin Sun wants to move his precious collection of *n* cowbells from Naperthrill to Exeter, where there is actually grass instead of corn. Before moving, he must pack his cowbells into *k* boxes of a fixed size. In order to keep his collection safe during transportation, he won't place more than two cowbells into a single box. Since Kevin wishes to minimize expenses, he is curious about the smallest size box he can use to pack his entire collection.
Kevin is a meticulous cowbell collector and knows that the size of his *i*-th (1<=≤<=*i*<=≤<=*n*) cowbell is an integer *s**i*. In fact, he keeps his cowbells sorted by size, so *s**i*<=-<=1<=≤<=*s**i* for any *i*<=><=1. Also an expert packer, Kevin can fit one or two cowbells into a box of size *s* if and only if the sum of their sizes does not exceed *s*. Given this information, help Kevin determine the smallest *s* for which it is possible to put all of his cowbells into *k* boxes of size *s*.
|
The first line of the input contains two space-separated integers *n* and *k* (1<=≤<=*n*<=≤<=2·*k*<=≤<=100<=000), denoting the number of cowbells and the number of boxes, respectively.
The next line contains *n* space-separated integers *s*1,<=*s*2,<=...,<=*s**n* (1<=≤<=*s*1<=≤<=*s*2<=≤<=...<=≤<=*s**n*<=≤<=1<=000<=000), the sizes of Kevin's cowbells. It is guaranteed that the sizes *s**i* are given in non-decreasing order.
|
Print a single integer, the smallest *s* for which it is possible for Kevin to put all of his cowbells into *k* boxes of size *s*.
|
[
"2 1\n2 5\n",
"4 3\n2 3 5 9\n",
"3 2\n3 5 7\n"
] |
[
"7\n",
"9\n",
"8\n"
] |
In the first sample, Kevin must pack his two cowbells into the same box.
In the second sample, Kevin can pack together the following sets of cowbells: {2, 3}, {5} and {9}.
In the third sample, the optimal solution is {3, 5} and {7}.
| 1,000
|
[
{
"input": "2 1\n2 5",
"output": "7"
},
{
"input": "4 3\n2 3 5 9",
"output": "9"
},
{
"input": "3 2\n3 5 7",
"output": "8"
},
{
"input": "20 11\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "2"
},
{
"input": "10 10\n3 15 31 61 63 63 68 94 98 100",
"output": "100"
},
{
"input": "100 97\n340 402 415 466 559 565 649 689 727 771 774 776 789 795 973 1088 1212 1293 1429 1514 1587 1599 1929 1997 2278 2529 2656 2677 2839 2894 2951 3079 3237 3250 3556 3568 3569 3578 3615 3641 3673 3892 4142 4418 4515 4766 4846 4916 5225 5269 5352 5460 5472 5635 5732 5886 5941 5976 5984 6104 6113 6402 6409 6460 6550 6563 6925 7006 7289 7401 7441 7451 7709 7731 7742 7750 7752 7827 8101 8154 8376 8379 8432 8534 8578 8630 8706 8814 8882 8972 9041 9053 9109 9173 9473 9524 9547 9775 9791 9983",
"output": "9983"
},
{
"input": "10 9\n7 29 35 38 41 47 54 56 73 74",
"output": "74"
},
{
"input": "1 2342\n12345",
"output": "12345"
},
{
"input": "10 5\n15 15 20 28 38 44 46 52 69 94",
"output": "109"
},
{
"input": "10 9\n6 10 10 32 36 38 69 80 82 93",
"output": "93"
},
{
"input": "10 10\n4 19 22 24 25 43 49 56 78 88",
"output": "88"
},
{
"input": "100 89\n474 532 759 772 803 965 1043 1325 1342 1401 1411 1452 1531 1707 1906 1928 2034 2222 2335 2606 2757 2968 2978 3211 3513 3734 3772 3778 3842 3948 3976 4038 4055 4113 4182 4267 4390 4408 4478 4595 4668 4792 4919 5133 5184 5255 5312 5341 5476 5628 5683 5738 5767 5806 5973 6051 6134 6254 6266 6279 6314 6342 6599 6676 6747 6777 6827 6842 7057 7097 7259 7340 7378 7405 7510 7520 7698 7796 8148 8351 8507 8601 8805 8814 8826 8978 9116 9140 9174 9338 9394 9403 9407 9423 9429 9519 9764 9784 9838 9946",
"output": "9946"
},
{
"input": "100 74\n10 211 323 458 490 592 979 981 1143 1376 1443 1499 1539 1612 1657 1874 2001 2064 2123 2274 2346 2471 2522 2589 2879 2918 2933 2952 3160 3164 3167 3270 3382 3404 3501 3522 3616 3802 3868 3985 4007 4036 4101 4580 4687 4713 4714 4817 4955 5257 5280 5343 5428 5461 5566 5633 5727 5874 5925 6233 6309 6389 6500 6701 6731 6847 6916 7088 7088 7278 7296 7328 7564 7611 7646 7887 7887 8065 8075 8160 8300 8304 8316 8355 8404 8587 8758 8794 8890 9038 9163 9235 9243 9339 9410 9587 9868 9916 9923 9986",
"output": "9986"
},
{
"input": "100 61\n82 167 233 425 432 456 494 507 562 681 683 921 1218 1323 1395 1531 1586 1591 1675 1766 1802 1842 2116 2625 2697 2735 2739 3337 3349 3395 3406 3596 3610 3721 4059 4078 4305 4330 4357 4379 4558 4648 4651 4784 4819 4920 5049 5312 5361 5418 5440 5463 5547 5594 5821 5951 5972 6141 6193 6230 6797 6842 6853 6854 7017 7026 7145 7322 7391 7460 7599 7697 7756 7768 7872 7889 8094 8215 8408 8440 8462 8714 8756 8760 8881 9063 9111 9184 9281 9373 9406 9417 9430 9511 9563 9634 9660 9788 9883 9927",
"output": "9927"
},
{
"input": "100 84\n53 139 150 233 423 570 786 861 995 1017 1072 1196 1276 1331 1680 1692 1739 1748 1826 2067 2280 2324 2368 2389 2607 2633 2760 2782 2855 2996 3030 3093 3513 3536 3557 3594 3692 3707 3823 3832 4009 4047 4088 4095 4408 4537 4565 4601 4784 4878 4935 5029 5252 5322 5389 5407 5511 5567 5857 6182 6186 6198 6280 6290 6353 6454 6458 6567 6843 7166 7216 7257 7261 7375 7378 7539 7542 7762 7771 7797 7980 8363 8606 8612 8663 8801 8808 8823 8918 8975 8997 9240 9245 9259 9356 9755 9759 9760 9927 9970",
"output": "9970"
},
{
"input": "100 50\n130 248 312 312 334 589 702 916 921 1034 1047 1346 1445 1500 1585 1744 1951 2123 2273 2362 2400 2455 2496 2530 2532 2944 3074 3093 3094 3134 3698 3967 4047 4102 4109 4260 4355 4466 4617 4701 4852 4892 4915 4917 4936 4981 4999 5106 5152 5203 5214 5282 5412 5486 5525 5648 5897 5933 5969 6251 6400 6421 6422 6558 6805 6832 6908 6924 6943 6980 7092 7206 7374 7417 7479 7546 7672 7756 7973 8020 8028 8079 8084 8085 8137 8153 8178 8239 8639 8667 8829 9263 9333 9370 9420 9579 9723 9784 9841 9993",
"output": "11103"
},
{
"input": "100 50\n156 182 208 409 496 515 659 761 772 794 827 912 1003 1236 1305 1388 1412 1422 1428 1465 1613 2160 2411 2440 2495 2684 2724 2925 3033 3035 3155 3260 3378 3442 3483 3921 4031 4037 4091 4113 4119 4254 4257 4442 4559 4614 4687 4839 4896 5054 5246 5316 5346 5859 5928 5981 6148 6250 6422 6433 6448 6471 6473 6485 6503 6779 6812 7050 7064 7074 7141 7378 7424 7511 7574 7651 7808 7858 8286 8291 8446 8536 8599 8628 8636 8768 8900 8981 9042 9055 9114 9146 9186 9411 9480 9590 9681 9749 9757 9983",
"output": "10676"
},
{
"input": "100 50\n145 195 228 411 577 606 629 775 1040 1040 1058 1187 1307 1514 1784 1867 1891 2042 2042 2236 2549 2555 2560 2617 2766 2807 2829 2917 3070 3072 3078 3095 3138 3147 3149 3196 3285 3287 3309 3435 3531 3560 3563 3769 3830 3967 4081 4158 4315 4387 4590 4632 4897 4914 5128 5190 5224 5302 5402 5416 5420 5467 5517 5653 5820 5862 5941 6053 6082 6275 6292 6316 6490 6530 6619 6632 6895 7071 7234 7323 7334 7412 7626 7743 8098 8098 8136 8158 8264 8616 8701 8718 8770 8803 8809 8983 9422 9530 9811 9866",
"output": "10011"
},
{
"input": "100 50\n56 298 387 456 518 532 589 792 870 1041 1055 1122 1141 1166 1310 1329 1523 1548 1626 1730 1780 1833 1850 1911 2006 2157 2303 2377 2403 2442 2450 2522 2573 2822 2994 3200 3238 3252 3280 3311 3345 3422 3429 3506 3526 3617 3686 3791 4134 4467 4525 4614 4633 4792 5017 5220 5243 5338 5445 5536 5639 5675 5763 5875 6129 6220 6228 6287 6385 6616 6789 6822 6940 6959 6985 7297 7304 7391 7443 7580 7824 7884 7981 8055 8063 8223 8280 8322 8346 8473 8688 8986 9087 9323 9477 9489 9518 9621 9699 9804",
"output": "10009"
},
{
"input": "100 49990\n221 257 543 613 818 855 884 1195 1233 1293 1327 1474 1567 1568 1728 1789 1809 1957 1994 2198 2217 2289 2298 2360 2441 2454 2517 2519 2597 2652 2909 2958 2970 3271 3310 3409 3450 3646 3950 3965 4016 4116 4134 4159 4462 4624 4631 4783 4804 5031 5276 5538 5729 5845 5880 5894 5959 5961 6028 6105 6184 6335 6370 6523 6532 6677 6762 6868 6998 7066 7245 7360 7490 7492 7645 7677 7786 7804 8038 8131 8207 8400 8507 8556 8575 8599 8673 8676 8712 8816 9123 9158 9216 9369 9628 9707 9768 9800 9876 9934",
"output": "9934"
},
{
"input": "1 50000\n2",
"output": "2"
},
{
"input": "1 40000\n23456",
"output": "23456"
},
{
"input": "1 2\n10",
"output": "10"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "2 2\n2 3",
"output": "3"
},
{
"input": "4 2\n1 2 8 9",
"output": "10"
}
] | 1,490,131,524
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 10
| 77
| 12,288,000
|
n,k = map(int, input().split())
a = list(map(int, input().split()))
print(a[-1] if k >= n else max(a[-1], a[n - 1 - k] + a[n-k]))
|
Title: More Cowbell
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kevin Sun wants to move his precious collection of *n* cowbells from Naperthrill to Exeter, where there is actually grass instead of corn. Before moving, he must pack his cowbells into *k* boxes of a fixed size. In order to keep his collection safe during transportation, he won't place more than two cowbells into a single box. Since Kevin wishes to minimize expenses, he is curious about the smallest size box he can use to pack his entire collection.
Kevin is a meticulous cowbell collector and knows that the size of his *i*-th (1<=≤<=*i*<=≤<=*n*) cowbell is an integer *s**i*. In fact, he keeps his cowbells sorted by size, so *s**i*<=-<=1<=≤<=*s**i* for any *i*<=><=1. Also an expert packer, Kevin can fit one or two cowbells into a box of size *s* if and only if the sum of their sizes does not exceed *s*. Given this information, help Kevin determine the smallest *s* for which it is possible to put all of his cowbells into *k* boxes of size *s*.
Input Specification:
The first line of the input contains two space-separated integers *n* and *k* (1<=≤<=*n*<=≤<=2·*k*<=≤<=100<=000), denoting the number of cowbells and the number of boxes, respectively.
The next line contains *n* space-separated integers *s*1,<=*s*2,<=...,<=*s**n* (1<=≤<=*s*1<=≤<=*s*2<=≤<=...<=≤<=*s**n*<=≤<=1<=000<=000), the sizes of Kevin's cowbells. It is guaranteed that the sizes *s**i* are given in non-decreasing order.
Output Specification:
Print a single integer, the smallest *s* for which it is possible for Kevin to put all of his cowbells into *k* boxes of size *s*.
Demo Input:
['2 1\n2 5\n', '4 3\n2 3 5 9\n', '3 2\n3 5 7\n']
Demo Output:
['7\n', '9\n', '8\n']
Note:
In the first sample, Kevin must pack his two cowbells into the same box.
In the second sample, Kevin can pack together the following sets of cowbells: {2, 3}, {5} and {9}.
In the third sample, the optimal solution is {3, 5} and {7}.
|
```python
n,k = map(int, input().split())
a = list(map(int, input().split()))
print(a[-1] if k >= n else max(a[-1], a[n - 1 - k] + a[n-k]))
```
| 0
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
Gerald has been selling state secrets at leisure. All the secrets cost the same: *n* marks. The state which secrets Gerald is selling, has no paper money, only coins. But there are coins of all positive integer denominations that are powers of three: 1 mark, 3 marks, 9 marks, 27 marks and so on. There are no coins of other denominations. Of course, Gerald likes it when he gets money without the change. And all buyers respect him and try to give the desired sum without change, if possible. But this does not always happen.
One day an unlucky buyer came. He did not have the desired sum without change. Then he took out all his coins and tried to give Gerald a larger than necessary sum with as few coins as possible. What is the maximum number of coins he could get?
The formal explanation of the previous paragraph: we consider all the possible combinations of coins for which the buyer can not give Gerald the sum of *n* marks without change. For each such combination calculate the minimum number of coins that can bring the buyer at least *n* marks. Among all combinations choose the maximum of the minimum number of coins. This is the number we want.
|
The single line contains a single integer *n* (1<=≤<=*n*<=≤<=1017).
Please, do not use the %lld specifier to read or write 64 bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
|
In a single line print an integer: the maximum number of coins the unlucky buyer could have paid with.
|
[
"1\n",
"4\n"
] |
[
"1\n",
"2\n"
] |
In the first test case, if a buyer has exactly one coin of at least 3 marks, then, to give Gerald one mark, he will have to give this coin. In this sample, the customer can not have a coin of one mark, as in this case, he will be able to give the money to Gerald without any change.
In the second test case, if the buyer had exactly three coins of 3 marks, then, to give Gerald 4 marks, he will have to give two of these coins. The buyer cannot give three coins as he wants to minimize the number of coins that he gives.
| 0
|
[
{
"input": "1",
"output": "1"
},
{
"input": "4",
"output": "2"
},
{
"input": "3",
"output": "1"
},
{
"input": "8",
"output": "3"
},
{
"input": "10",
"output": "4"
},
{
"input": "100000000000000000",
"output": "33333333333333334"
},
{
"input": "99999999999999999",
"output": "3703703703703704"
},
{
"input": "50031545098999707",
"output": "1"
},
{
"input": "16677181699666569",
"output": "1"
},
{
"input": "72900000000000",
"output": "33333333334"
},
{
"input": "99999999999999997",
"output": "33333333333333333"
},
{
"input": "58061299250691018",
"output": "32"
},
{
"input": "49664023559436051",
"output": "128191526"
},
{
"input": "66708726798666276",
"output": "2"
},
{
"input": "29442431889534807",
"output": "48"
},
{
"input": "70414767176369958",
"output": "13"
},
{
"input": "93886356235159944",
"output": "51"
},
{
"input": "97626528902553453",
"output": "551104613133"
},
{
"input": "52013157885656046",
"output": "880847395988"
},
{
"input": "37586570003500923",
"output": "548"
},
{
"input": "34391854792828422",
"output": "582429080812"
},
{
"input": "205891132094649",
"output": "1"
},
{
"input": "243",
"output": "1"
},
{
"input": "5559060566555523",
"output": "1"
},
{
"input": "81",
"output": "1"
},
{
"input": "108",
"output": "2"
},
{
"input": "2",
"output": "1"
},
{
"input": "1129718145924",
"output": "2"
}
] | 1,379,491,327
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 92
| 0
|
import math
n = int(math.ceil(math.log(int(input()))/math.log(3)))
if(n == 0) : n = n+1
print(n)
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Gerald has been selling state secrets at leisure. All the secrets cost the same: *n* marks. The state which secrets Gerald is selling, has no paper money, only coins. But there are coins of all positive integer denominations that are powers of three: 1 mark, 3 marks, 9 marks, 27 marks and so on. There are no coins of other denominations. Of course, Gerald likes it when he gets money without the change. And all buyers respect him and try to give the desired sum without change, if possible. But this does not always happen.
One day an unlucky buyer came. He did not have the desired sum without change. Then he took out all his coins and tried to give Gerald a larger than necessary sum with as few coins as possible. What is the maximum number of coins he could get?
The formal explanation of the previous paragraph: we consider all the possible combinations of coins for which the buyer can not give Gerald the sum of *n* marks without change. For each such combination calculate the minimum number of coins that can bring the buyer at least *n* marks. Among all combinations choose the maximum of the minimum number of coins. This is the number we want.
Input Specification:
The single line contains a single integer *n* (1<=≤<=*n*<=≤<=1017).
Please, do not use the %lld specifier to read or write 64 bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output Specification:
In a single line print an integer: the maximum number of coins the unlucky buyer could have paid with.
Demo Input:
['1\n', '4\n']
Demo Output:
['1\n', '2\n']
Note:
In the first test case, if a buyer has exactly one coin of at least 3 marks, then, to give Gerald one mark, he will have to give this coin. In this sample, the customer can not have a coin of one mark, as in this case, he will be able to give the money to Gerald without any change.
In the second test case, if the buyer had exactly three coins of 3 marks, then, to give Gerald 4 marks, he will have to give two of these coins. The buyer cannot give three coins as he wants to minimize the number of coins that he gives.
|
```python
import math
n = int(math.ceil(math.log(int(input()))/math.log(3)))
if(n == 0) : n = n+1
print(n)
```
| 0
|
|
432
|
A
|
Choosing Teams
|
PROGRAMMING
| 800
|
[
"greedy",
"implementation",
"sortings"
] | null | null |
The Saratov State University Olympiad Programmers Training Center (SSU OPTC) has *n* students. For each student you know the number of times he/she has participated in the ACM ICPC world programming championship. According to the ACM ICPC rules, each person can participate in the world championship at most 5 times.
The head of the SSU OPTC is recently gathering teams to participate in the world championship. Each team must consist of exactly three people, at that, any person cannot be a member of two or more teams. What maximum number of teams can the head make if he wants each team to participate in the world championship with the same members at least *k* times?
|
The first line contains two integers, *n* and *k* (1<=≤<=*n*<=≤<=2000; 1<=≤<=*k*<=≤<=5). The next line contains *n* integers: *y*1,<=*y*2,<=...,<=*y**n* (0<=≤<=*y**i*<=≤<=5), where *y**i* shows the number of times the *i*-th person participated in the ACM ICPC world championship.
|
Print a single number — the answer to the problem.
|
[
"5 2\n0 4 5 1 0\n",
"6 4\n0 1 2 3 4 5\n",
"6 5\n0 0 0 0 0 0\n"
] |
[
"1\n",
"0\n",
"2\n"
] |
In the first sample only one team could be made: the first, the fourth and the fifth participants.
In the second sample no teams could be created.
In the third sample two teams could be created. Any partition into two teams fits.
| 500
|
[
{
"input": "5 2\n0 4 5 1 0",
"output": "1"
},
{
"input": "6 4\n0 1 2 3 4 5",
"output": "0"
},
{
"input": "6 5\n0 0 0 0 0 0",
"output": "2"
},
{
"input": "3 4\n0 1 0",
"output": "1"
},
{
"input": "3 4\n0 2 0",
"output": "0"
},
{
"input": "6 5\n0 0 0 0 0 0",
"output": "2"
},
{
"input": "12 2\n0 1 2 3 4 0 1 2 3 4 0 1",
"output": "3"
},
{
"input": "15 2\n0 1 2 3 4 0 1 2 3 4 0 1 2 3 4",
"output": "4"
},
{
"input": "13 1\n5 0 5 0 1 5 0 4 1 1 1 4 1",
"output": "3"
},
{
"input": "20 1\n5 0 4 2 2 3 2 1 2 4 1 3 5 5 5 4 4 1 3 0",
"output": "5"
},
{
"input": "20 2\n1 4 5 1 3 4 0 5 0 3 3 0 4 0 0 1 4 2 0 4",
"output": "4"
},
{
"input": "20 2\n1 4 5 1 3 4 0 5 0 3 3 0 4 0 0 1 4 2 0 4",
"output": "4"
},
{
"input": "20 2\n1 4 5 1 3 4 0 5 0 3 3 0 4 0 0 1 4 2 0 4",
"output": "4"
},
{
"input": "20 2\n1 4 5 1 3 4 0 5 0 3 3 0 4 0 0 1 4 2 0 4",
"output": "4"
},
{
"input": "6 3\n4 4 4 4 4 4",
"output": "0"
},
{
"input": "1 4\n1",
"output": "0"
},
{
"input": "1 1\n0",
"output": "0"
},
{
"input": "2 3\n0 0",
"output": "0"
}
] | 1,647,845,844
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 35
| 46
| 0
|
n,m=map(int,input().split())
count=0
l=list(map(int,input().split()))
for i in range(n):
l[i]=l[i]+m
for i in range(n):
if l[i]<=5:
count+=1
s=count//3
print(s)
|
Title: Choosing Teams
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Saratov State University Olympiad Programmers Training Center (SSU OPTC) has *n* students. For each student you know the number of times he/she has participated in the ACM ICPC world programming championship. According to the ACM ICPC rules, each person can participate in the world championship at most 5 times.
The head of the SSU OPTC is recently gathering teams to participate in the world championship. Each team must consist of exactly three people, at that, any person cannot be a member of two or more teams. What maximum number of teams can the head make if he wants each team to participate in the world championship with the same members at least *k* times?
Input Specification:
The first line contains two integers, *n* and *k* (1<=≤<=*n*<=≤<=2000; 1<=≤<=*k*<=≤<=5). The next line contains *n* integers: *y*1,<=*y*2,<=...,<=*y**n* (0<=≤<=*y**i*<=≤<=5), where *y**i* shows the number of times the *i*-th person participated in the ACM ICPC world championship.
Output Specification:
Print a single number — the answer to the problem.
Demo Input:
['5 2\n0 4 5 1 0\n', '6 4\n0 1 2 3 4 5\n', '6 5\n0 0 0 0 0 0\n']
Demo Output:
['1\n', '0\n', '2\n']
Note:
In the first sample only one team could be made: the first, the fourth and the fifth participants.
In the second sample no teams could be created.
In the third sample two teams could be created. Any partition into two teams fits.
|
```python
n,m=map(int,input().split())
count=0
l=list(map(int,input().split()))
for i in range(n):
l[i]=l[i]+m
for i in range(n):
if l[i]<=5:
count+=1
s=count//3
print(s)
```
| 3
|
|
678
|
A
|
Johny Likes Numbers
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
Johny likes numbers *n* and *k* very much. Now Johny wants to find the smallest integer *x* greater than *n*, so it is divisible by the number *k*.
|
The only line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=109).
|
Print the smallest integer *x*<=><=*n*, so it is divisible by the number *k*.
|
[
"5 3\n",
"25 13\n",
"26 13\n"
] |
[
"6\n",
"26\n",
"39\n"
] |
none
| 0
|
[
{
"input": "5 3",
"output": "6"
},
{
"input": "25 13",
"output": "26"
},
{
"input": "26 13",
"output": "39"
},
{
"input": "1 1",
"output": "2"
},
{
"input": "8 8",
"output": "16"
},
{
"input": "14 15",
"output": "15"
},
{
"input": "197 894",
"output": "894"
},
{
"input": "6058 8581",
"output": "8581"
},
{
"input": "97259 41764",
"output": "125292"
},
{
"input": "453145 333625",
"output": "667250"
},
{
"input": "2233224 4394826",
"output": "4394826"
},
{
"input": "76770926 13350712",
"output": "80104272"
},
{
"input": "687355301 142098087",
"output": "710490435"
},
{
"input": "1000000000 999999999",
"output": "1999999998"
},
{
"input": "1000000000 1000000000",
"output": "2000000000"
},
{
"input": "999999999 1000000000",
"output": "1000000000"
},
{
"input": "1000000000 1",
"output": "1000000001"
},
{
"input": "1000000000 2",
"output": "1000000002"
},
{
"input": "999999999 1",
"output": "1000000000"
},
{
"input": "100000000 1",
"output": "100000001"
},
{
"input": "999999999 500000000",
"output": "1000000000"
},
{
"input": "999999990 10",
"output": "1000000000"
},
{
"input": "1000000000 999999997",
"output": "1999999994"
},
{
"input": "999999999 2",
"output": "1000000000"
},
{
"input": "999999984 1",
"output": "999999985"
},
{
"input": "999999983 1",
"output": "999999984"
},
{
"input": "666666666 1",
"output": "666666667"
},
{
"input": "1000000000 990000000",
"output": "1980000000"
},
{
"input": "41 48",
"output": "48"
},
{
"input": "123456 2",
"output": "123458"
},
{
"input": "111 111",
"output": "222"
},
{
"input": "878787 1",
"output": "878788"
},
{
"input": "121 1",
"output": "122"
},
{
"input": "114514 114514",
"output": "229028"
},
{
"input": "500000001 1000000000",
"output": "1000000000"
},
{
"input": "999999997 1",
"output": "999999998"
},
{
"input": "100000000 10",
"output": "100000010"
}
] | 1,676,580,403
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 11
| 500
| 0
|
#task8
n,k=map(int,input().split())
b=n+1
while True:
if b%k==0:
print(b)
break
else:
b+=1
|
Title: Johny Likes Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Johny likes numbers *n* and *k* very much. Now Johny wants to find the smallest integer *x* greater than *n*, so it is divisible by the number *k*.
Input Specification:
The only line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=109).
Output Specification:
Print the smallest integer *x*<=><=*n*, so it is divisible by the number *k*.
Demo Input:
['5 3\n', '25 13\n', '26 13\n']
Demo Output:
['6\n', '26\n', '39\n']
Note:
none
|
```python
#task8
n,k=map(int,input().split())
b=n+1
while True:
if b%k==0:
print(b)
break
else:
b+=1
```
| 0
|
|
547
|
B
|
Mike and Feet
|
PROGRAMMING
| 1,900
|
[
"binary search",
"data structures",
"dp",
"dsu"
] | null | null |
Mike is the president of country What-The-Fatherland. There are *n* bears living in this country besides Mike. All of them are standing in a line and they are numbered from 1 to *n* from left to right. *i*-th bear is exactly *a**i* feet high.
A group of bears is a non-empty contiguous segment of the line. The size of a group is the number of bears in that group. The strength of a group is the minimum height of the bear in that group.
Mike is a curious to know for each *x* such that 1<=≤<=*x*<=≤<=*n* the maximum strength among all groups of size *x*.
|
The first line of input contains integer *n* (1<=≤<=*n*<=≤<=2<=×<=105), the number of bears.
The second line contains *n* integers separated by space, *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109), heights of bears.
|
Print *n* integers in one line. For each *x* from 1 to *n*, print the maximum strength among all groups of size *x*.
|
[
"10\n1 2 3 4 5 4 3 2 1 6\n"
] |
[
"6 4 4 3 3 2 2 1 1 1 \n"
] |
none
| 1,000
|
[
{
"input": "10\n1 2 3 4 5 4 3 2 1 6",
"output": "6 4 4 3 3 2 2 1 1 1 "
},
{
"input": "3\n524125987 923264237 374288891",
"output": "923264237 524125987 374288891 "
},
{
"input": "5\n585325539 365329221 412106895 291882089 564718673",
"output": "585325539 365329221 365329221 291882089 291882089 "
},
{
"input": "20\n452405440 586588704 509061481 552472140 16115810 148658854 66743034 628305150 677780684 519361360 208050516 401554301 954478790 346543678 387546138 832279893 641889899 80960260 717802881 588066499",
"output": "954478790 641889899 519361360 452405440 346543678 346543678 208050516 208050516 208050516 208050516 80960260 80960260 80960260 66743034 66743034 16115810 16115810 16115810 16115810 16115810 "
},
{
"input": "1\n1376",
"output": "1376 "
},
{
"input": "2\n10 10",
"output": "10 10 "
},
{
"input": "2\n10 9",
"output": "10 9 "
},
{
"input": "3\n1 2 3",
"output": "3 2 1 "
},
{
"input": "3\n1 3 2",
"output": "3 2 1 "
},
{
"input": "10\n802030518 598196518 640274071 983359971 71550121 96204862 799843967 446173607 796619138 402690754",
"output": "983359971 640274071 598196518 598196518 96204862 71550121 71550121 71550121 71550121 71550121 "
},
{
"input": "19\n519879446 764655030 680293934 914539062 744988123 317088317 653721289 239862203 605157354 943428394 261437390 821695238 312192823 432992892 547139308 408916833 829654733 223751525 672158759",
"output": "943428394 744988123 680293934 680293934 519879446 317088317 317088317 261437390 261437390 239862203 239862203 239862203 239862203 239862203 239862203 239862203 239862203 223751525 223751525 "
}
] | 1,695,756,343
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
n = int(input())
queue = list(map(int,input().split()))
answer = 0
aux = float('inf')
for i in range(n-1,-1,-1):
aux = min(aux - 1, queue[i])
aux = max(0, aux)
ans += aux
print(answer)
|
Title: Mike and Feet
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mike is the president of country What-The-Fatherland. There are *n* bears living in this country besides Mike. All of them are standing in a line and they are numbered from 1 to *n* from left to right. *i*-th bear is exactly *a**i* feet high.
A group of bears is a non-empty contiguous segment of the line. The size of a group is the number of bears in that group. The strength of a group is the minimum height of the bear in that group.
Mike is a curious to know for each *x* such that 1<=≤<=*x*<=≤<=*n* the maximum strength among all groups of size *x*.
Input Specification:
The first line of input contains integer *n* (1<=≤<=*n*<=≤<=2<=×<=105), the number of bears.
The second line contains *n* integers separated by space, *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109), heights of bears.
Output Specification:
Print *n* integers in one line. For each *x* from 1 to *n*, print the maximum strength among all groups of size *x*.
Demo Input:
['10\n1 2 3 4 5 4 3 2 1 6\n']
Demo Output:
['6 4 4 3 3 2 2 1 1 1 \n']
Note:
none
|
```python
n = int(input())
queue = list(map(int,input().split()))
answer = 0
aux = float('inf')
for i in range(n-1,-1,-1):
aux = min(aux - 1, queue[i])
aux = max(0, aux)
ans += aux
print(answer)
```
| -1
|
|
320
|
A
|
Magic Numbers
|
PROGRAMMING
| 900
|
[
"brute force",
"greedy"
] | null | null |
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if it is a magic number or not.
|
The first line of input contains an integer *n*, (1<=≤<=*n*<=≤<=109). This number doesn't contain leading zeros.
|
Print "YES" if *n* is a magic number or print "NO" if it's not.
|
[
"114114\n",
"1111\n",
"441231\n"
] |
[
"YES\n",
"YES\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "114114",
"output": "YES"
},
{
"input": "1111",
"output": "YES"
},
{
"input": "441231",
"output": "NO"
},
{
"input": "1",
"output": "YES"
},
{
"input": "14",
"output": "YES"
},
{
"input": "114",
"output": "YES"
},
{
"input": "9",
"output": "NO"
},
{
"input": "414",
"output": "NO"
},
{
"input": "1000000000",
"output": "NO"
},
{
"input": "144144144",
"output": "YES"
},
{
"input": "1444",
"output": "NO"
},
{
"input": "11",
"output": "YES"
},
{
"input": "141414141",
"output": "YES"
},
{
"input": "11110111",
"output": "NO"
},
{
"input": "114114144",
"output": "YES"
},
{
"input": "444",
"output": "NO"
},
{
"input": "9999",
"output": "NO"
},
{
"input": "111444",
"output": "NO"
},
{
"input": "11114",
"output": "YES"
},
{
"input": "41111",
"output": "NO"
},
{
"input": "114414441",
"output": "NO"
},
{
"input": "144414441",
"output": "NO"
},
{
"input": "144244144",
"output": "NO"
},
{
"input": "111111111",
"output": "YES"
},
{
"input": "144444444",
"output": "NO"
},
{
"input": "444444444",
"output": "NO"
},
{
"input": "141441441",
"output": "YES"
},
{
"input": "441",
"output": "NO"
},
{
"input": "15",
"output": "NO"
},
{
"input": "14444",
"output": "NO"
},
{
"input": "11444",
"output": "NO"
},
{
"input": "144",
"output": "YES"
},
{
"input": "1414414",
"output": "YES"
},
{
"input": "141444",
"output": "NO"
},
{
"input": "14144",
"output": "YES"
},
{
"input": "4",
"output": "NO"
},
{
"input": "1144",
"output": "YES"
},
{
"input": "141111444",
"output": "NO"
},
{
"input": "14414414",
"output": "YES"
},
{
"input": "141414144",
"output": "YES"
},
{
"input": "1414",
"output": "YES"
},
{
"input": "1441",
"output": "YES"
},
{
"input": "12",
"output": "NO"
},
{
"input": "144144141",
"output": "YES"
},
{
"input": "144144",
"output": "YES"
},
{
"input": "14414411",
"output": "YES"
},
{
"input": "14414",
"output": "YES"
}
] | 1,683,187,590
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 92
| 0
|
n = int(input(""))
y = str(n)
if "1" in y and "4" in y :
if "2" in y or "3" in y or "5" in y or "6" in y or "7" in y or "8" in y or "9" in y or "0" in y :
print("NO")
elif "1" in y :
print("YES")
elif "14" in y :
print("YES")
elif "144" in y :
print("YES")
elif "14" in y and "1" in y :
print("YES")
elif "144" in y and "1" in y :
print("YES")
elif "14" in y and "144" in y :
print("YES")
elif "2" in n:
print("NO")
|
Title: Magic Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if it is a magic number or not.
Input Specification:
The first line of input contains an integer *n*, (1<=≤<=*n*<=≤<=109). This number doesn't contain leading zeros.
Output Specification:
Print "YES" if *n* is a magic number or print "NO" if it's not.
Demo Input:
['114114\n', '1111\n', '441231\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
none
|
```python
n = int(input(""))
y = str(n)
if "1" in y and "4" in y :
if "2" in y or "3" in y or "5" in y or "6" in y or "7" in y or "8" in y or "9" in y or "0" in y :
print("NO")
elif "1" in y :
print("YES")
elif "14" in y :
print("YES")
elif "144" in y :
print("YES")
elif "14" in y and "1" in y :
print("YES")
elif "144" in y and "1" in y :
print("YES")
elif "14" in y and "144" in y :
print("YES")
elif "2" in n:
print("NO")
```
| 0
|
|
981
|
A
|
Antipalindrome
|
PROGRAMMING
| 900
|
[
"brute force",
"implementation",
"strings"
] | null | null |
A string is a palindrome if it reads the same from the left to the right and from the right to the left. For example, the strings "kek", "abacaba", "r" and "papicipap" are palindromes, while the strings "abb" and "iq" are not.
A substring $s[l \ldots r]$ ($1<=\leq<=l<=\leq<=r<=\leq<=|s|$) of a string $s<==<=s_{1}s_{2} \ldots s_{|s|}$ is the string $s_{l}s_{l<=+<=1} \ldots s_{r}$.
Anna does not like palindromes, so she makes her friends call her Ann. She also changes all the words she reads in a similar way. Namely, each word $s$ is changed into its longest substring that is not a palindrome. If all the substrings of $s$ are palindromes, she skips the word at all.
Some time ago Ann read the word $s$. What is the word she changed it into?
|
The first line contains a non-empty string $s$ with length at most $50$ characters, containing lowercase English letters only.
|
If there is such a substring in $s$ that is not a palindrome, print the maximum length of such a substring. Otherwise print $0$.
Note that there can be multiple longest substrings that are not palindromes, but their length is unique.
|
[
"mew\n",
"wuffuw\n",
"qqqqqqqq\n"
] |
[
"3\n",
"5\n",
"0\n"
] |
"mew" is not a palindrome, so the longest substring of it that is not a palindrome, is the string "mew" itself. Thus, the answer for the first example is $3$.
The string "uffuw" is one of the longest non-palindrome substrings (of length $5$) of the string "wuffuw", so the answer for the second example is $5$.
All substrings of the string "qqqqqqqq" consist of equal characters so they are palindromes. This way, there are no non-palindrome substrings. Thus, the answer for the third example is $0$.
| 500
|
[
{
"input": "mew",
"output": "3"
},
{
"input": "wuffuw",
"output": "5"
},
{
"input": "qqqqqqqq",
"output": "0"
},
{
"input": "ijvji",
"output": "4"
},
{
"input": "iiiiiii",
"output": "0"
},
{
"input": "wobervhvvkihcuyjtmqhaaigvvgiaahqmtjyuchikvvhvrebow",
"output": "49"
},
{
"input": "wwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww",
"output": "0"
},
{
"input": "wobervhvvkihcuyjtmqhaaigvahheoqleromusrartldojsjvy",
"output": "50"
},
{
"input": "ijvxljt",
"output": "7"
},
{
"input": "fyhcncnchyf",
"output": "10"
},
{
"input": "ffffffffffff",
"output": "0"
},
{
"input": "fyhcncfsepqj",
"output": "12"
},
{
"input": "ybejrrlbcinttnicblrrjeby",
"output": "23"
},
{
"input": "yyyyyyyyyyyyyyyyyyyyyyyyy",
"output": "0"
},
{
"input": "ybejrrlbcintahovgjddrqatv",
"output": "25"
},
{
"input": "oftmhcmclgyqaojljoaqyglcmchmtfo",
"output": "30"
},
{
"input": "oooooooooooooooooooooooooooooooo",
"output": "0"
},
{
"input": "oftmhcmclgyqaojllbotztajglsmcilv",
"output": "32"
},
{
"input": "gxandbtgpbknxvnkjaajknvxnkbpgtbdnaxg",
"output": "35"
},
{
"input": "gggggggggggggggggggggggggggggggggggg",
"output": "0"
},
{
"input": "gxandbtgpbknxvnkjaygommzqitqzjfalfkk",
"output": "36"
},
{
"input": "fcliblymyqckxvieotjooojtoeivxkcqymylbilcf",
"output": "40"
},
{
"input": "fffffffffffffffffffffffffffffffffffffffffff",
"output": "0"
},
{
"input": "fcliblymyqckxvieotjootiqwtyznhhvuhbaixwqnsy",
"output": "43"
},
{
"input": "rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr",
"output": "0"
},
{
"input": "rajccqwqnqmshmerpvjyfepxwpxyldzpzhctqjnstxyfmlhiy",
"output": "49"
},
{
"input": "a",
"output": "0"
},
{
"input": "abca",
"output": "4"
},
{
"input": "aaaaabaaaaa",
"output": "10"
},
{
"input": "aba",
"output": "2"
},
{
"input": "asaa",
"output": "4"
},
{
"input": "aabaa",
"output": "4"
},
{
"input": "aabbaa",
"output": "5"
},
{
"input": "abcdaaa",
"output": "7"
},
{
"input": "aaholaa",
"output": "7"
},
{
"input": "abcdefghijka",
"output": "12"
},
{
"input": "aaadcba",
"output": "7"
},
{
"input": "aaaabaaaa",
"output": "8"
},
{
"input": "abaa",
"output": "4"
},
{
"input": "abcbaa",
"output": "6"
},
{
"input": "ab",
"output": "2"
},
{
"input": "l",
"output": "0"
},
{
"input": "aaaabcaaaa",
"output": "10"
},
{
"input": "abbaaaaaabba",
"output": "11"
},
{
"input": "abaaa",
"output": "5"
},
{
"input": "baa",
"output": "3"
},
{
"input": "aaaaaaabbba",
"output": "11"
},
{
"input": "ccbcc",
"output": "4"
},
{
"input": "bbbaaab",
"output": "7"
},
{
"input": "abaaaaaaaa",
"output": "10"
},
{
"input": "abaaba",
"output": "5"
},
{
"input": "aabsdfaaaa",
"output": "10"
},
{
"input": "aaaba",
"output": "5"
},
{
"input": "aaabaaa",
"output": "6"
},
{
"input": "baaabbb",
"output": "7"
},
{
"input": "ccbbabbcc",
"output": "8"
},
{
"input": "cabc",
"output": "4"
},
{
"input": "aabcd",
"output": "5"
},
{
"input": "abcdea",
"output": "6"
},
{
"input": "bbabb",
"output": "4"
},
{
"input": "aaaaabababaaaaa",
"output": "14"
},
{
"input": "bbabbb",
"output": "6"
},
{
"input": "aababd",
"output": "6"
},
{
"input": "abaaaa",
"output": "6"
},
{
"input": "aaaaaaaabbba",
"output": "12"
},
{
"input": "aabca",
"output": "5"
},
{
"input": "aaabccbaaa",
"output": "9"
},
{
"input": "aaaaaaaaaaaaaaaaaaaab",
"output": "21"
},
{
"input": "babb",
"output": "4"
},
{
"input": "abcaa",
"output": "5"
},
{
"input": "qwqq",
"output": "4"
},
{
"input": "aaaaaaaaaaabbbbbbbbbbbbbbbaaaaaaaaaaaaaaaaaaaaaa",
"output": "48"
},
{
"input": "aaab",
"output": "4"
},
{
"input": "aaaaaabaaaaa",
"output": "12"
},
{
"input": "wwuww",
"output": "4"
},
{
"input": "aaaaabcbaaaaa",
"output": "12"
},
{
"input": "aaabbbaaa",
"output": "8"
},
{
"input": "aabcbaa",
"output": "6"
},
{
"input": "abccdefccba",
"output": "11"
},
{
"input": "aabbcbbaa",
"output": "8"
},
{
"input": "aaaabbaaaa",
"output": "9"
},
{
"input": "aabcda",
"output": "6"
},
{
"input": "abbca",
"output": "5"
},
{
"input": "aaaaaabbaaa",
"output": "11"
},
{
"input": "sssssspssssss",
"output": "12"
},
{
"input": "sdnmsdcs",
"output": "8"
},
{
"input": "aaabbbccbbbaaa",
"output": "13"
},
{
"input": "cbdbdc",
"output": "6"
},
{
"input": "abb",
"output": "3"
},
{
"input": "abcdefaaaa",
"output": "10"
},
{
"input": "abbbaaa",
"output": "7"
},
{
"input": "v",
"output": "0"
},
{
"input": "abccbba",
"output": "7"
},
{
"input": "axyza",
"output": "5"
},
{
"input": "abcdefgaaaa",
"output": "11"
},
{
"input": "aaabcdaaa",
"output": "9"
},
{
"input": "aaaacaaaa",
"output": "8"
},
{
"input": "aaaaaaaaaaaaaaaaaaaabaaaaaaaaaaaaaaaaaaaaa",
"output": "42"
},
{
"input": "abbbaa",
"output": "6"
},
{
"input": "abcdee",
"output": "6"
},
{
"input": "oom",
"output": "3"
},
{
"input": "aabcaa",
"output": "6"
},
{
"input": "abba",
"output": "3"
},
{
"input": "aaca",
"output": "4"
},
{
"input": "aacbca",
"output": "6"
},
{
"input": "ababa",
"output": "4"
},
{
"input": "abcda",
"output": "5"
},
{
"input": "cccaaccc",
"output": "7"
},
{
"input": "aaabcda",
"output": "7"
},
{
"input": "aa",
"output": "0"
},
{
"input": "aabaaaa",
"output": "7"
},
{
"input": "abbaaaa",
"output": "7"
},
{
"input": "aaabcbaaa",
"output": "8"
},
{
"input": "aabba",
"output": "5"
},
{
"input": "xyxx",
"output": "4"
},
{
"input": "aaaaaaaaaaaabc",
"output": "14"
},
{
"input": "bbaaaabb",
"output": "7"
},
{
"input": "aaabaa",
"output": "6"
},
{
"input": "sssssabsssss",
"output": "12"
},
{
"input": "bbbaaaabbb",
"output": "9"
},
{
"input": "abbbbaaaa",
"output": "9"
},
{
"input": "wwufuww",
"output": "6"
},
{
"input": "oowoo",
"output": "4"
},
{
"input": "cccaccc",
"output": "6"
},
{
"input": "aaa",
"output": "0"
},
{
"input": "bbbcc",
"output": "5"
},
{
"input": "abcdef",
"output": "6"
},
{
"input": "abbba",
"output": "4"
},
{
"input": "aab",
"output": "3"
},
{
"input": "aaba",
"output": "4"
},
{
"input": "azbyaaa",
"output": "7"
},
{
"input": "oooooiooooo",
"output": "10"
},
{
"input": "aabbbbbaaaaaa",
"output": "13"
}
] | 1,694,206,701
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 133
| 46
| 0
|
s = input()
l = 0
r = len(s)-1
flag = False
while l < r:
if(s[l] == s[r]):
l+=1
r-=1
continue
flag = True
break
if(flag):
print(len(s))
else:
dif = False
for i in range(len(s)-1):
if(s[i] != s[i+1]):
dif = True
if(dif):
print(len(s)-1)
else:
print(0)
|
Title: Antipalindrome
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A string is a palindrome if it reads the same from the left to the right and from the right to the left. For example, the strings "kek", "abacaba", "r" and "papicipap" are palindromes, while the strings "abb" and "iq" are not.
A substring $s[l \ldots r]$ ($1<=\leq<=l<=\leq<=r<=\leq<=|s|$) of a string $s<==<=s_{1}s_{2} \ldots s_{|s|}$ is the string $s_{l}s_{l<=+<=1} \ldots s_{r}$.
Anna does not like palindromes, so she makes her friends call her Ann. She also changes all the words she reads in a similar way. Namely, each word $s$ is changed into its longest substring that is not a palindrome. If all the substrings of $s$ are palindromes, she skips the word at all.
Some time ago Ann read the word $s$. What is the word she changed it into?
Input Specification:
The first line contains a non-empty string $s$ with length at most $50$ characters, containing lowercase English letters only.
Output Specification:
If there is such a substring in $s$ that is not a palindrome, print the maximum length of such a substring. Otherwise print $0$.
Note that there can be multiple longest substrings that are not palindromes, but their length is unique.
Demo Input:
['mew\n', 'wuffuw\n', 'qqqqqqqq\n']
Demo Output:
['3\n', '5\n', '0\n']
Note:
"mew" is not a palindrome, so the longest substring of it that is not a palindrome, is the string "mew" itself. Thus, the answer for the first example is $3$.
The string "uffuw" is one of the longest non-palindrome substrings (of length $5$) of the string "wuffuw", so the answer for the second example is $5$.
All substrings of the string "qqqqqqqq" consist of equal characters so they are palindromes. This way, there are no non-palindrome substrings. Thus, the answer for the third example is $0$.
|
```python
s = input()
l = 0
r = len(s)-1
flag = False
while l < r:
if(s[l] == s[r]):
l+=1
r-=1
continue
flag = True
break
if(flag):
print(len(s))
else:
dif = False
for i in range(len(s)-1):
if(s[i] != s[i+1]):
dif = True
if(dif):
print(len(s)-1)
else:
print(0)
```
| 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 — 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* — 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"
}
] | 1,649,156,233
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 60
| 0
|
word = str(input())
i = 0
num_low = 0
num_up = 0
for i in word:
if (i >= 'a' and i <= 'z'):
num_low += 1
elif (i>= 'A' and i <= 'Z'):
num_up += 1
if num_low >= num_up:
print(num_low)
print(num_up)
word = word.lower()
print(word)
else:
word = word.upper()
print(word)
|
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 — 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* — 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
word = str(input())
i = 0
num_low = 0
num_up = 0
for i in word:
if (i >= 'a' and i <= 'z'):
num_low += 1
elif (i>= 'A' and i <= 'Z'):
num_up += 1
if num_low >= num_up:
print(num_low)
print(num_up)
word = word.lower()
print(word)
else:
word = word.upper()
print(word)
```
| 0
|
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. 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 $n$ ($1 \le n \le 1000$) — 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"
}
] | 1,593,169,619
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 35
| 155
| 21,708,800
|
n=int(input())
li=list(map(int,input().split()))
i=1
while i<n:
k=max(li)
li.remove(k)
i+=1
if i<n:
k1=min(li)
li.remove(k1)
i+=1
print(*li)
|
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. 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 $n$ ($1 \le n \le 1000$) — 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
n=int(input())
li=list(map(int,input().split()))
i=1
while i<n:
k=max(li)
li.remove(k)
i+=1
if i<n:
k1=min(li)
li.remove(k1)
i+=1
print(*li)
```
| 3
|
|
600
|
B
|
Queries about less or equal elements
|
PROGRAMMING
| 1,300
|
[
"binary search",
"data structures",
"sortings",
"two pointers"
] | null | null |
You are given two arrays of integers *a* and *b*. For each element of the second array *b**j* you should find the number of elements in array *a* that are less than or equal to the value *b**j*.
|
The first line contains two integers *n*,<=*m* (1<=≤<=*n*,<=*m*<=≤<=2·105) — the sizes of arrays *a* and *b*.
The second line contains *n* integers — the elements of array *a* (<=-<=109<=≤<=*a**i*<=≤<=109).
The third line contains *m* integers — the elements of array *b* (<=-<=109<=≤<=*b**j*<=≤<=109).
|
Print *m* integers, separated by spaces: the *j*-th of which is equal to the number of such elements in array *a* that are less than or equal to the value *b**j*.
|
[
"5 4\n1 3 5 7 9\n6 4 2 8\n",
"5 5\n1 2 1 2 5\n3 1 4 1 5\n"
] |
[
"3 2 1 4\n",
"4 2 4 2 5\n"
] |
none
| 0
|
[
{
"input": "5 4\n1 3 5 7 9\n6 4 2 8",
"output": "3 2 1 4"
},
{
"input": "5 5\n1 2 1 2 5\n3 1 4 1 5",
"output": "4 2 4 2 5"
},
{
"input": "1 1\n-1\n-2",
"output": "0"
},
{
"input": "1 1\n-80890826\n686519510",
"output": "1"
},
{
"input": "11 11\n237468511 -779187544 -174606592 193890085 404563196 -71722998 -617934776 170102710 -442808289 109833389 953091341\n994454001 322957429 216874735 -606986750 -455806318 -663190696 3793295 41395397 -929612742 -787653860 -684738874",
"output": "11 9 8 2 2 1 5 5 0 0 1"
},
{
"input": "20 22\n858276994 -568758442 -918490847 -983345984 -172435358 389604931 200224783 486556113 413281867 -258259500 -627945379 -584563643 444685477 -602481243 -370745158 965672503 630955806 -626138773 -997221880 633102929\n-61330638 -977252080 -212144219 385501731 669589742 954357160 563935906 584468977 -895883477 405774444 853372186 186056475 -964575261 -952431965 632332084 -388829939 -23011650 310957048 -770695392 977376693 321435214 199223897",
"output": "11 2 10 12 18 19 16 16 3 13 18 11 2 2 17 8 11 12 3 20 12 11"
},
{
"input": "5 9\n1 3 5 7 9\n1 2 3 4 5 6 7 8 9",
"output": "1 1 2 2 3 3 4 4 5"
},
{
"input": "22 1\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22\n1",
"output": "1"
},
{
"input": "5 1\n1 3 3 3 5\n3",
"output": "4"
},
{
"input": "4 5\n1 1 1 4\n1 5 5 4 3",
"output": "3 4 4 4 3"
},
{
"input": "5 4\n0 5 5 5 6\n5 1 6 3",
"output": "4 1 5 1"
},
{
"input": "1 3\n0\n-1 0 1",
"output": "0 1 1"
},
{
"input": "96 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\n1",
"output": "96"
},
{
"input": "7 1\n1 2 3 4 5 6 7\n1",
"output": "1"
},
{
"input": "13 13\n-1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000\n-1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000",
"output": "7 13 7 13 7 13 7 13 7 13 7 13 7"
},
{
"input": "9 5\n1 2 3 4 5 6 7 8 9\n1 2 3 4 5",
"output": "1 2 3 4 5"
},
{
"input": "3 8\n1 1 1\n1 1 1 1 1 1 1 1",
"output": "3 3 3 3 3 3 3 3"
},
{
"input": "1 1\n-11111\n-5938",
"output": "1"
},
{
"input": "1 1\n1\n400000009",
"output": "1"
},
{
"input": "1 1\n1\n300000009",
"output": "1"
},
{
"input": "1 1\n1\n200000009",
"output": "1"
},
{
"input": "1 1\n1\n200000003",
"output": "1"
}
] | 1,639,602,754
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 124
| 0
|
n, m = [int(num) for num in input().split()]
a = sorted([int(elem) for elem in input().split()])
b = [int(num) for num in input().split()]
resp = ''
i = 0
j = n-1
for elem in b:
if a[i] <= elem:
resp += str(i+1)
if a[j] <= elem:
resp += str(j+1)
if elem > a[i]:
i += 1
if elem < a[j]:
j -= 1
print(resp)
|
Title: Queries about less or equal elements
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given two arrays of integers *a* and *b*. For each element of the second array *b**j* you should find the number of elements in array *a* that are less than or equal to the value *b**j*.
Input Specification:
The first line contains two integers *n*,<=*m* (1<=≤<=*n*,<=*m*<=≤<=2·105) — the sizes of arrays *a* and *b*.
The second line contains *n* integers — the elements of array *a* (<=-<=109<=≤<=*a**i*<=≤<=109).
The third line contains *m* integers — the elements of array *b* (<=-<=109<=≤<=*b**j*<=≤<=109).
Output Specification:
Print *m* integers, separated by spaces: the *j*-th of which is equal to the number of such elements in array *a* that are less than or equal to the value *b**j*.
Demo Input:
['5 4\n1 3 5 7 9\n6 4 2 8\n', '5 5\n1 2 1 2 5\n3 1 4 1 5\n']
Demo Output:
['3 2 1 4\n', '4 2 4 2 5\n']
Note:
none
|
```python
n, m = [int(num) for num in input().split()]
a = sorted([int(elem) for elem in input().split()])
b = [int(num) for num in input().split()]
resp = ''
i = 0
j = n-1
for elem in b:
if a[i] <= elem:
resp += str(i+1)
if a[j] <= elem:
resp += str(j+1)
if elem > a[i]:
i += 1
if elem < a[j]:
j -= 1
print(resp)
```
| 0
|
|
721
|
B
|
Passwords
|
PROGRAMMING
| 1,100
|
[
"implementation",
"math",
"sortings",
"strings"
] | null | null |
Vanya is managed to enter his favourite site Codehorses. Vanya uses *n* distinct passwords for sites at all, however he can't remember which one exactly he specified during Codehorses registration.
Vanya will enter passwords in order of non-decreasing their lengths, and he will enter passwords of same length in arbitrary order. Just when Vanya will have entered the correct password, he is immediately authorized on the site. Vanya will not enter any password twice.
Entering any passwords takes one second for Vanya. But if Vanya will enter wrong password *k* times, then he is able to make the next try only 5 seconds after that. Vanya makes each try immediately, that is, at each moment when Vanya is able to enter password, he is doing that.
Determine how many seconds will Vanya need to enter Codehorses in the best case for him (if he spends minimum possible number of second) and in the worst case (if he spends maximum possible amount of seconds).
|
The first line of the input contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100) — the number of Vanya's passwords and the number of failed tries, after which the access to the site is blocked for 5 seconds.
The next *n* lines contains passwords, one per line — pairwise distinct non-empty strings consisting of latin letters and digits. Each password length does not exceed 100 characters.
The last line of the input contains the Vanya's Codehorses password. It is guaranteed that the Vanya's Codehorses password is equal to some of his *n* passwords.
|
Print two integers — time (in seconds), Vanya needs to be authorized to Codehorses in the best case for him and in the worst case respectively.
|
[
"5 2\ncba\nabc\nbb1\nabC\nABC\nabc\n",
"4 100\n11\n22\n1\n2\n22\n"
] |
[
"1 15\n",
"3 4\n"
] |
Consider the first sample case. As soon as all passwords have the same length, Vanya can enter the right password at the first try as well as at the last try. If he enters it at the first try, he spends exactly 1 second. Thus in the best case the answer is 1. If, at the other hand, he enters it at the last try, he enters another 4 passwords before. He spends 2 seconds to enter first 2 passwords, then he waits 5 seconds as soon as he made 2 wrong tries. Then he spends 2 more seconds to enter 2 wrong passwords, again waits 5 seconds and, finally, enters the correct password spending 1 more second. In summary in the worst case he is able to be authorized in 15 seconds.
Consider the second sample case. There is no way of entering passwords and get the access to the site blocked. As soon as the required password has length of 2, Vanya enters all passwords of length 1 anyway, spending 2 seconds for that. Then, in the best case, he immediately enters the correct password and the answer for the best case is 3, but in the worst case he enters wrong password of length 2 and only then the right one, spending 4 seconds at all.
| 1,000
|
[
{
"input": "5 2\ncba\nabc\nbb1\nabC\nABC\nabc",
"output": "1 15"
},
{
"input": "4 100\n11\n22\n1\n2\n22",
"output": "3 4"
},
{
"input": "1 1\na1\na1",
"output": "1 1"
},
{
"input": "1 100\na1\na1",
"output": "1 1"
},
{
"input": "2 1\nabc\nAbc\nAbc",
"output": "1 7"
},
{
"input": "2 2\nabc\nAbc\nabc",
"output": "1 2"
},
{
"input": "2 1\nab\nabc\nab",
"output": "1 1"
},
{
"input": "2 2\nab\nabc\nab",
"output": "1 1"
},
{
"input": "2 1\nab\nabc\nabc",
"output": "7 7"
},
{
"input": "2 2\nab\nabc\nabc",
"output": "2 2"
},
{
"input": "10 3\nOIbV1igi\no\nZS\nQM\n9woLzI\nWreboD\nQ7yl\nA5Rb\nS9Lno72TkP\nfT97o\no",
"output": "1 1"
},
{
"input": "10 3\nHJZNMsT\nLaPcH2C\nlrhqIO\n9cxw\noTC1XwjW\nGHL9Ul6\nUyIs\nPuzwgR4ZKa\nyIByoKR5\nd3QA\nPuzwgR4ZKa",
"output": "25 25"
},
{
"input": "20 5\nvSyC787KlIL8kZ2Uv5sw\nWKWOP\n7i8J3E8EByIq\nNW2VyGweL\nmyR2sRNu\nmXusPP0\nf4jgGxra\n4wHRzRhOCpEt\npPz9kybGb\nOtSpePCRoG5nkjZ2VxRy\nwHYsSttWbJkg\nKBOP9\nQfiOiFyHPPsw3GHo8J8\nxB8\nqCpehZEeEhdq\niOLjICK6\nQ91\nHmCsfMGTFKoFFnv238c\nJKjhg\ngkEUh\nKBOP9",
"output": "3 11"
},
{
"input": "15 2\nw6S9WyU\nMVh\nkgUhQHW\nhGQNOF\nUuym\n7rGQA\nBM8vLPRB\n9E\nDs32U\no\nz1aV2C5T\n8\nzSXjrqQ\n1FO\n3kIt\nBM8vLPRB",
"output": "44 50"
},
{
"input": "20 2\ni\n5Rp6\nE4vsr\nSY\nORXx\nh13C\nk6tzC\ne\nN\nKQf4C\nWZcdL\ndiA3v\n0InQT\nuJkAr\nGCamp\nBuIRd\nY\nM\nxZYx7\n0a5A\nWZcdL",
"output": "36 65"
},
{
"input": "20 2\naWLQ6\nSgQ9r\nHcPdj\n2BNaO\n3TjNb\nnvwFM\nqsKt7\nFnb6N\nLoc0p\njxuLq\nBKAjf\nEKgZB\nBfOSa\nsMIvr\nuIWcR\nIura3\nLAqSf\ntXq3G\n8rQ8I\n8otAO\nsMIvr",
"output": "1 65"
},
{
"input": "20 15\n0ZpQugVlN7\nm0SlKGnohN\nRFXTqhNGcn\n1qm2ZbB\nQXtJWdf78P\nbc2vH\nP21dty2Z1P\nm2c71LFhCk\n23EuP1Dvh3\nanwri5RhQN\n55v6HYv288\n1u5uKOjM5r\n6vg0GC1\nDAPYiA3ns1\nUZaaJ3Gmnk\nwB44x7V4Zi\n4hgB2oyU8P\npYFQpy8gGK\ndbz\nBv\n55v6HYv288",
"output": "6 25"
},
{
"input": "3 1\na\nb\naa\naa",
"output": "13 13"
},
{
"input": "6 3\nab\nac\nad\nabc\nabd\nabe\nabc",
"output": "9 11"
},
{
"input": "4 2\n1\n2\n11\n22\n22",
"output": "8 9"
},
{
"input": "2 1\n1\n12\n12",
"output": "7 7"
},
{
"input": "3 1\nab\nabc\nabd\nabc",
"output": "7 13"
},
{
"input": "2 1\na\nab\nab",
"output": "7 7"
},
{
"input": "5 2\na\nb\nc\nab\naa\naa",
"output": "9 15"
},
{
"input": "6 1\n1\n2\n11\n22\n111\n2222\n22",
"output": "13 19"
},
{
"input": "3 1\n1\n2\n11\n11",
"output": "13 13"
},
{
"input": "10 4\na\nb\nc\nd\ne\nf\nab\ncd\nac\nad\nac",
"output": "12 20"
},
{
"input": "4 2\na\nb\nc\nd\na",
"output": "1 9"
},
{
"input": "4 1\n1\n2\n3\n4\n4",
"output": "1 19"
},
{
"input": "5 1\na\nb\nc\nd\nef\nef",
"output": "25 25"
},
{
"input": "6 4\n1\n2\n22\n33\n44\n555\n555",
"output": "11 11"
},
{
"input": "5 2\na\nb\nc\nd\nab\nab",
"output": "15 15"
},
{
"input": "6 2\n1\n2\n3\n4\n5\n23\n23",
"output": "16 16"
},
{
"input": "4 2\na\nb\naa\nbb\naa",
"output": "8 9"
},
{
"input": "5 4\na\nbb\ncc\ndd\nee\nbb",
"output": "2 10"
},
{
"input": "4 1\na\nb\nc\nab\nab",
"output": "19 19"
},
{
"input": "7 100\na\nb\nc\nd\ne\ng\nab\nab",
"output": "7 7"
},
{
"input": "6 1\na\nb\nc\nd\ne\naa\naa",
"output": "31 31"
},
{
"input": "4 1\na\nas\nasd\nasde\nasde",
"output": "19 19"
},
{
"input": "5 2\n1\n2\n3\n11\n22\n22",
"output": "9 15"
},
{
"input": "10 2\na\nb\nc\nd\nee\nff\ngg\nhh\nii\njj\nii",
"output": "15 30"
},
{
"input": "3 1\na\nab\nbc\nab",
"output": "7 13"
},
{
"input": "6 4\na\nb\nc\nbb\nbc\ncc\ncc",
"output": "4 11"
}
] | 1,566,986,682
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 66
| 124
| 0
|
import math
n,k= map(int,input().split())
l=[]
for _ in range(n):
w= input()
l.append(w)
r= input()
l.sort(key=len)
le= len(r)
m=[x for x in l if len(x)<le]
print((len(m)+1)+(len(m)//k)*5,end=" ") #best
o= [x for x in l if len(x)<=le]
print(((len(o)-1)//k)*5+ len(o)) #worst
|
Title: Passwords
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vanya is managed to enter his favourite site Codehorses. Vanya uses *n* distinct passwords for sites at all, however he can't remember which one exactly he specified during Codehorses registration.
Vanya will enter passwords in order of non-decreasing their lengths, and he will enter passwords of same length in arbitrary order. Just when Vanya will have entered the correct password, he is immediately authorized on the site. Vanya will not enter any password twice.
Entering any passwords takes one second for Vanya. But if Vanya will enter wrong password *k* times, then he is able to make the next try only 5 seconds after that. Vanya makes each try immediately, that is, at each moment when Vanya is able to enter password, he is doing that.
Determine how many seconds will Vanya need to enter Codehorses in the best case for him (if he spends minimum possible number of second) and in the worst case (if he spends maximum possible amount of seconds).
Input Specification:
The first line of the input contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100) — the number of Vanya's passwords and the number of failed tries, after which the access to the site is blocked for 5 seconds.
The next *n* lines contains passwords, one per line — pairwise distinct non-empty strings consisting of latin letters and digits. Each password length does not exceed 100 characters.
The last line of the input contains the Vanya's Codehorses password. It is guaranteed that the Vanya's Codehorses password is equal to some of his *n* passwords.
Output Specification:
Print two integers — time (in seconds), Vanya needs to be authorized to Codehorses in the best case for him and in the worst case respectively.
Demo Input:
['5 2\ncba\nabc\nbb1\nabC\nABC\nabc\n', '4 100\n11\n22\n1\n2\n22\n']
Demo Output:
['1 15\n', '3 4\n']
Note:
Consider the first sample case. As soon as all passwords have the same length, Vanya can enter the right password at the first try as well as at the last try. If he enters it at the first try, he spends exactly 1 second. Thus in the best case the answer is 1. If, at the other hand, he enters it at the last try, he enters another 4 passwords before. He spends 2 seconds to enter first 2 passwords, then he waits 5 seconds as soon as he made 2 wrong tries. Then he spends 2 more seconds to enter 2 wrong passwords, again waits 5 seconds and, finally, enters the correct password spending 1 more second. In summary in the worst case he is able to be authorized in 15 seconds.
Consider the second sample case. There is no way of entering passwords and get the access to the site blocked. As soon as the required password has length of 2, Vanya enters all passwords of length 1 anyway, spending 2 seconds for that. Then, in the best case, he immediately enters the correct password and the answer for the best case is 3, but in the worst case he enters wrong password of length 2 and only then the right one, spending 4 seconds at all.
|
```python
import math
n,k= map(int,input().split())
l=[]
for _ in range(n):
w= input()
l.append(w)
r= input()
l.sort(key=len)
le= len(r)
m=[x for x in l if len(x)<le]
print((len(m)+1)+(len(m)//k)*5,end=" ") #best
o= [x for x in l if len(x)<=le]
print(((len(o)-1)//k)*5+ len(o)) #worst
```
| 3
|
|
711
|
A
|
Bus to Udayland
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation"
] | null | null |
ZS the Coder and Chris the Baboon are travelling to Udayland! To get there, they have to get on the special IOI bus. The IOI bus has *n* rows of seats. There are 4 seats in each row, and the seats are separated into pairs by a walkway. When ZS and Chris came, some places in the bus was already occupied.
ZS and Chris are good friends. They insist to get a pair of neighbouring empty seats. Two seats are considered neighbouring if they are in the same row and in the same pair. Given the configuration of the bus, can you help ZS and Chris determine where they should sit?
|
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of rows of seats in the bus.
Then, *n* lines follow. Each line contains exactly 5 characters, the first two of them denote the first pair of seats in the row, the third character denotes the walkway (it always equals '|') and the last two of them denote the second pair of seats in the row.
Each character, except the walkway, equals to 'O' or to 'X'. 'O' denotes an empty seat, 'X' denotes an occupied seat. See the sample cases for more details.
|
If it is possible for Chris and ZS to sit at neighbouring empty seats, print "YES" (without quotes) in the first line. In the next *n* lines print the bus configuration, where the characters in the pair of seats for Chris and ZS is changed with characters '+'. Thus the configuration should differ from the input one by exactly two charaters (they should be equal to 'O' in the input and to '+' in the output).
If there is no pair of seats for Chris and ZS, print "NO" (without quotes) in a single line.
If there are multiple solutions, you may print any of them.
|
[
"6\nOO|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX\n",
"4\nXO|OX\nXO|XX\nOX|OX\nXX|OX\n",
"5\nXX|XX\nXX|XX\nXO|OX\nXO|OO\nOX|XO\n"
] |
[
"YES\n++|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX\n",
"NO\n",
"YES\nXX|XX\nXX|XX\nXO|OX\nXO|++\nOX|XO\n"
] |
Note that the following is an incorrect configuration for the first sample case because the seats must be in the same pair.
O+|+X
XO|XX
OX|OO
XX|OX
OO|OO
OO|XX
| 500
|
[
{
"input": "6\nOO|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX",
"output": "YES\n++|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX"
},
{
"input": "4\nXO|OX\nXO|XX\nOX|OX\nXX|OX",
"output": "NO"
},
{
"input": "5\nXX|XX\nXX|XX\nXO|OX\nXO|OO\nOX|XO",
"output": "YES\nXX|XX\nXX|XX\nXO|OX\nXO|++\nOX|XO"
},
{
"input": "1\nXO|OX",
"output": "NO"
},
{
"input": "1\nOO|OO",
"output": "YES\n++|OO"
},
{
"input": "4\nXO|XX\nXX|XO\nOX|XX\nXO|XO",
"output": "NO"
},
{
"input": "9\nOX|XO\nOX|XO\nXO|OX\nOX|OX\nXO|OX\nXX|OO\nOX|OX\nOX|XO\nOX|OX",
"output": "YES\nOX|XO\nOX|XO\nXO|OX\nOX|OX\nXO|OX\nXX|++\nOX|OX\nOX|XO\nOX|OX"
},
{
"input": "61\nOX|XX\nOX|XX\nOX|XX\nXO|XO\nXX|XO\nXX|XX\nXX|XX\nOX|XX\nXO|XO\nOX|XO\nXO|OX\nXX|XX\nXX|XX\nOX|OX\nXX|OX\nOX|XO\nOX|XO\nXO|OX\nXO|XX\nOX|XX\nOX|XX\nXO|OX\nXO|XX\nXO|XX\nOX|XX\nXX|XX\nXX|XO\nXO|XX\nXX|XX\nXO|OX\nXX|XO\nXO|XX\nXO|XO\nXO|OX\nXX|OX\nXO|OX\nOX|XX\nXX|OX\nXX|XX\nOX|XO\nOX|XX\nXO|OX\nOX|XX\nOX|XX\nXO|XO\nXO|XX\nOX|XX\nXO|XO\nOX|XX\nXX|XX\nOX|XO\nXO|XO\nXO|XO\nOX|OX\nXX|OX\nXX|OX\nOX|XO\nOX|XX\nOX|OX\nXO|XX\nOX|XX",
"output": "NO"
},
{
"input": "1\nOO|XX",
"output": "YES\n++|XX"
},
{
"input": "10\nOO|XX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXX|XX",
"output": "YES\n++|XX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXX|XX"
},
{
"input": "10\nXX|XX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXX|OO",
"output": "YES\nXX|XX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXX|++"
},
{
"input": "5\nXX|XX\nXX|XX\nXO|OX\nOO|OX\nOX|XO",
"output": "YES\nXX|XX\nXX|XX\nXO|OX\n++|OX\nOX|XO"
},
{
"input": "6\nOO|XX\nXO|XX\nOX|OO\nXX|OX\nOO|XX\nOX|XX",
"output": "YES\n++|XX\nXO|XX\nOX|OO\nXX|OX\nOO|XX\nOX|XX"
}
] | 1,617,021,442
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 71
| 77
| 307,200
|
n=int(input())
s=""
c=0
for i in range(n):
x,y=map(str,input().split("|"))
if x=="OO" and c==0:
s+="++|"+y+"\n"
c+=1
elif y=="OO" and c==0:
s+=x+"|++"+"\n"
c+=1
else:
s+=x+"|"+y+"\n"
if c>=1:
print("YES")
print(s)
else:
print("No")
|
Title: Bus to Udayland
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
ZS the Coder and Chris the Baboon are travelling to Udayland! To get there, they have to get on the special IOI bus. The IOI bus has *n* rows of seats. There are 4 seats in each row, and the seats are separated into pairs by a walkway. When ZS and Chris came, some places in the bus was already occupied.
ZS and Chris are good friends. They insist to get a pair of neighbouring empty seats. Two seats are considered neighbouring if they are in the same row and in the same pair. Given the configuration of the bus, can you help ZS and Chris determine where they should sit?
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of rows of seats in the bus.
Then, *n* lines follow. Each line contains exactly 5 characters, the first two of them denote the first pair of seats in the row, the third character denotes the walkway (it always equals '|') and the last two of them denote the second pair of seats in the row.
Each character, except the walkway, equals to 'O' or to 'X'. 'O' denotes an empty seat, 'X' denotes an occupied seat. See the sample cases for more details.
Output Specification:
If it is possible for Chris and ZS to sit at neighbouring empty seats, print "YES" (without quotes) in the first line. In the next *n* lines print the bus configuration, where the characters in the pair of seats for Chris and ZS is changed with characters '+'. Thus the configuration should differ from the input one by exactly two charaters (they should be equal to 'O' in the input and to '+' in the output).
If there is no pair of seats for Chris and ZS, print "NO" (without quotes) in a single line.
If there are multiple solutions, you may print any of them.
Demo Input:
['6\nOO|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX\n', '4\nXO|OX\nXO|XX\nOX|OX\nXX|OX\n', '5\nXX|XX\nXX|XX\nXO|OX\nXO|OO\nOX|XO\n']
Demo Output:
['YES\n++|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX\n', 'NO\n', 'YES\nXX|XX\nXX|XX\nXO|OX\nXO|++\nOX|XO\n']
Note:
Note that the following is an incorrect configuration for the first sample case because the seats must be in the same pair.
O+|+X
XO|XX
OX|OO
XX|OX
OO|OO
OO|XX
|
```python
n=int(input())
s=""
c=0
for i in range(n):
x,y=map(str,input().split("|"))
if x=="OO" and c==0:
s+="++|"+y+"\n"
c+=1
elif y=="OO" and c==0:
s+=x+"|++"+"\n"
c+=1
else:
s+=x+"|"+y+"\n"
if c>=1:
print("YES")
print(s)
else:
print("No")
```
| 3
|
|
620
|
B
|
Grandfather Dovlet’s calculator
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
Once Max found an electronic calculator from his grandfather Dovlet's chest. He noticed that the numbers were written with seven-segment indicators ([https://en.wikipedia.org/wiki/Seven-segment_display](https://en.wikipedia.org/wiki/Seven-segment_display)).
Max starts to type all the values from *a* to *b*. After typing each number Max resets the calculator. Find the total number of segments printed on the calculator.
For example if *a*<==<=1 and *b*<==<=3 then at first the calculator will print 2 segments, then — 5 segments and at last it will print 5 segments. So the total number of printed segments is 12.
|
The only line contains two integers *a*,<=*b* (1<=≤<=*a*<=≤<=*b*<=≤<=106) — the first and the last number typed by Max.
|
Print the only integer *a* — the total number of printed segments.
|
[
"1 3\n",
"10 15\n"
] |
[
"12\n",
"39\n"
] |
none
| 0
|
[
{
"input": "1 3",
"output": "12"
},
{
"input": "10 15",
"output": "39"
},
{
"input": "1 100",
"output": "928"
},
{
"input": "100 10000",
"output": "188446"
},
{
"input": "213 221442",
"output": "5645356"
},
{
"input": "1 1000000",
"output": "28733372"
},
{
"input": "1000000 1000000",
"output": "38"
},
{
"input": "222145 353252",
"output": "3860750"
},
{
"input": "2 1000000",
"output": "28733370"
},
{
"input": "1 999999",
"output": "28733334"
},
{
"input": "192 200",
"output": "122"
}
] | 1,453,389,161
| 761
|
Python 3
|
OK
|
TESTS
| 11
| 966
| 0
|
def main():
v = {
"1" : 2,
"2" : 5,
"3" : 5,
"4" : 4,
"5" : 5,
"6" : 6,
"7" : 3,
"8" : 7,
"9" : 6,
"0" : 6
}
a, b = map(int, input().split())
answer = 0
for i in range(a, b + 1):
s = str(i)
for e in s:
answer += v[e]
print(answer)
if __name__ == "__main__":
main()
|
Title: Grandfather Dovlet’s calculator
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Once Max found an electronic calculator from his grandfather Dovlet's chest. He noticed that the numbers were written with seven-segment indicators ([https://en.wikipedia.org/wiki/Seven-segment_display](https://en.wikipedia.org/wiki/Seven-segment_display)).
Max starts to type all the values from *a* to *b*. After typing each number Max resets the calculator. Find the total number of segments printed on the calculator.
For example if *a*<==<=1 and *b*<==<=3 then at first the calculator will print 2 segments, then — 5 segments and at last it will print 5 segments. So the total number of printed segments is 12.
Input Specification:
The only line contains two integers *a*,<=*b* (1<=≤<=*a*<=≤<=*b*<=≤<=106) — the first and the last number typed by Max.
Output Specification:
Print the only integer *a* — the total number of printed segments.
Demo Input:
['1 3\n', '10 15\n']
Demo Output:
['12\n', '39\n']
Note:
none
|
```python
def main():
v = {
"1" : 2,
"2" : 5,
"3" : 5,
"4" : 4,
"5" : 5,
"6" : 6,
"7" : 3,
"8" : 7,
"9" : 6,
"0" : 6
}
a, b = map(int, input().split())
answer = 0
for i in range(a, b + 1):
s = str(i)
for e in s:
answer += v[e]
print(answer)
if __name__ == "__main__":
main()
```
| 3
|
|
758
|
A
|
Holiday Of Equality
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
In Berland it is the holiday of equality. In honor of the holiday the king decided to equalize the welfare of all citizens in Berland by the expense of the state treasury.
Totally in Berland there are *n* citizens, the welfare of each of them is estimated as the integer in *a**i* burles (burle is the currency in Berland).
You are the royal treasurer, which needs to count the minimum charges of the kingdom on the king's present. The king can only give money, he hasn't a power to take away them.
|
The first line contains the integer *n* (1<=≤<=*n*<=≤<=100) — the number of citizens in the kingdom.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n*, where *a**i* (0<=≤<=*a**i*<=≤<=106) — the welfare of the *i*-th citizen.
|
In the only line print the integer *S* — the minimum number of burles which are had to spend.
|
[
"5\n0 1 2 3 4\n",
"5\n1 1 0 1 1\n",
"3\n1 3 1\n",
"1\n12\n"
] |
[
"10",
"1",
"4",
"0"
] |
In the first example if we add to the first citizen 4 burles, to the second 3, to the third 2 and to the fourth 1, then the welfare of all citizens will equal 4.
In the second example it is enough to give one burle to the third citizen.
In the third example it is necessary to give two burles to the first and the third citizens to make the welfare of citizens equal 3.
In the fourth example it is possible to give nothing to everyone because all citizens have 12 burles.
| 500
|
[
{
"input": "5\n0 1 2 3 4",
"output": "10"
},
{
"input": "5\n1 1 0 1 1",
"output": "1"
},
{
"input": "3\n1 3 1",
"output": "4"
},
{
"input": "1\n12",
"output": "0"
},
{
"input": "3\n1 2 3",
"output": "3"
},
{
"input": "14\n52518 718438 358883 462189 853171 592966 225788 46977 814826 295697 676256 561479 56545 764281",
"output": "5464380"
},
{
"input": "21\n842556 216391 427181 626688 775504 168309 851038 448402 880826 73697 593338 519033 135115 20128 424606 939484 846242 756907 377058 241543 29353",
"output": "9535765"
},
{
"input": "3\n1 3 2",
"output": "3"
},
{
"input": "3\n2 1 3",
"output": "3"
},
{
"input": "3\n2 3 1",
"output": "3"
},
{
"input": "3\n3 1 2",
"output": "3"
},
{
"input": "3\n3 2 1",
"output": "3"
},
{
"input": "1\n228503",
"output": "0"
},
{
"input": "2\n32576 550340",
"output": "517764"
},
{
"input": "3\n910648 542843 537125",
"output": "741328"
},
{
"input": "4\n751720 572344 569387 893618",
"output": "787403"
},
{
"input": "6\n433864 631347 597596 794426 713555 231193",
"output": "1364575"
},
{
"input": "9\n31078 645168 695751 126111 375934 150495 838412 434477 993107",
"output": "4647430"
},
{
"input": "30\n315421 772664 560686 654312 151528 356749 351486 707462 820089 226682 546700 136028 824236 842130 578079 337807 665903 764100 617900 822937 992759 591749 651310 742085 767695 695442 17967 515106 81059 186025",
"output": "13488674"
},
{
"input": "45\n908719 394261 815134 419990 926993 383792 772842 277695 527137 655356 684956 695716 273062 550324 106247 399133 442382 33076 462920 294674 846052 817752 421365 474141 290471 358990 109812 74492 543281 169434 919692 786809 24028 197184 310029 801476 699355 429672 51343 374128 776726 850380 293868 981569 550763",
"output": "21993384"
},
{
"input": "56\n100728 972537 13846 385421 756708 184642 259487 319707 376662 221694 675284 972837 499419 13846 38267 289898 901299 831197 954715 197515 514102 910423 127555 883934 362472 870788 538802 741008 973434 448124 391526 363321 947321 544618 68006 782313 955075 741981 815027 723297 585059 718114 700739 413489 454091 736144 308999 98065 3716 347323 9635 289003 986510 607065 60236 273351",
"output": "26984185"
},
{
"input": "70\n644488 5444 150441 714420 602059 335330 510670 196555 546346 740011 509449 850947 692874 524857 750434 952985 223744 374727 896124 753037 367352 679050 560202 172728 569291 778616 332116 286927 843598 372698 244906 498046 900681 709791 420904 724593 864493 813094 791377 39998 296710 625656 403891 579231 706693 984045 16901 574259 562265 761104 930361 256045 124461 538980 573508 372148 988722 108592 784354 55302 232524 277205 782251 299943 436488 743389 324618 742543 266915 99642",
"output": "32816391"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "1\n1000000",
"output": "0"
},
{
"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": "0"
},
{
"input": "100\n1000000 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": "99000000"
},
{
"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 1000000",
"output": "99000000"
},
{
"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 1000000 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": "99000000"
},
{
"input": "3\n0 0 0",
"output": "0"
},
{
"input": "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",
"output": "0"
},
{
"input": "3\n5 0 0",
"output": "10"
},
{
"input": "5\n2 10 0 0 0",
"output": "38"
}
] | 1,662,619,532
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 41
| 46
| 0
|
x=int(input());l=list(map(int,input().split()));m=max(l);c=0
for i in l: c+=m-i
print(c)
|
Title: Holiday Of Equality
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In Berland it is the holiday of equality. In honor of the holiday the king decided to equalize the welfare of all citizens in Berland by the expense of the state treasury.
Totally in Berland there are *n* citizens, the welfare of each of them is estimated as the integer in *a**i* burles (burle is the currency in Berland).
You are the royal treasurer, which needs to count the minimum charges of the kingdom on the king's present. The king can only give money, he hasn't a power to take away them.
Input Specification:
The first line contains the integer *n* (1<=≤<=*n*<=≤<=100) — the number of citizens in the kingdom.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n*, where *a**i* (0<=≤<=*a**i*<=≤<=106) — the welfare of the *i*-th citizen.
Output Specification:
In the only line print the integer *S* — the minimum number of burles which are had to spend.
Demo Input:
['5\n0 1 2 3 4\n', '5\n1 1 0 1 1\n', '3\n1 3 1\n', '1\n12\n']
Demo Output:
['10', '1', '4', '0']
Note:
In the first example if we add to the first citizen 4 burles, to the second 3, to the third 2 and to the fourth 1, then the welfare of all citizens will equal 4.
In the second example it is enough to give one burle to the third citizen.
In the third example it is necessary to give two burles to the first and the third citizens to make the welfare of citizens equal 3.
In the fourth example it is possible to give nothing to everyone because all citizens have 12 burles.
|
```python
x=int(input());l=list(map(int,input().split()));m=max(l);c=0
for i in l: c+=m-i
print(c)
```
| 3
|
|
867
|
A
|
Between the Offices
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
As you may know, MemSQL has American offices in both San Francisco and Seattle. Being a manager in the company, you travel a lot between the two cities, always by plane.
You prefer flying from Seattle to San Francisco than in the other direction, because it's warmer in San Francisco. You are so busy that you don't remember the number of flights you have made in either direction. However, for each of the last *n* days you know whether you were in San Francisco office or in Seattle office. You always fly at nights, so you never were at both offices on the same day. Given this information, determine if you flew more times from Seattle to San Francisco during the last *n* days, or not.
|
The first line of input contains single integer *n* (2<=≤<=*n*<=≤<=100) — the number of days.
The second line contains a string of length *n* consisting of only capital 'S' and 'F' letters. If the *i*-th letter is 'S', then you were in Seattle office on that day. Otherwise you were in San Francisco. The days are given in chronological order, i.e. today is the last day in this sequence.
|
Print "YES" if you flew more times from Seattle to San Francisco, and "NO" otherwise.
You can print each letter in any case (upper or lower).
|
[
"4\nFSSF\n",
"2\nSF\n",
"10\nFFFFFFFFFF\n",
"10\nSSFFSFFSFF\n"
] |
[
"NO\n",
"YES\n",
"NO\n",
"YES\n"
] |
In the first example you were initially at San Francisco, then flew to Seattle, were there for two days and returned to San Francisco. You made one flight in each direction, so the answer is "NO".
In the second example you just flew from Seattle to San Francisco, so the answer is "YES".
In the third example you stayed the whole period in San Francisco, so the answer is "NO".
In the fourth example if you replace 'S' with ones, and 'F' with zeros, you'll get the first few digits of π in binary representation. Not very useful information though.
| 500
|
[
{
"input": "4\nFSSF",
"output": "NO"
},
{
"input": "2\nSF",
"output": "YES"
},
{
"input": "10\nFFFFFFFFFF",
"output": "NO"
},
{
"input": "10\nSSFFSFFSFF",
"output": "YES"
},
{
"input": "20\nSFSFFFFSSFFFFSSSSFSS",
"output": "NO"
},
{
"input": "20\nSSFFFFFSFFFFFFFFFFFF",
"output": "YES"
},
{
"input": "20\nSSFSFSFSFSFSFSFSSFSF",
"output": "YES"
},
{
"input": "20\nSSSSFSFSSFSFSSSSSSFS",
"output": "NO"
},
{
"input": "100\nFFFSFSFSFSSFSFFSSFFFFFSSSSFSSFFFFSFFFFFSFFFSSFSSSFFFFSSFFSSFSFFSSFSSSFSFFSFSFFSFSFFSSFFSFSSSSFSFSFSS",
"output": "NO"
},
{
"input": "100\nFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF",
"output": "NO"
},
{
"input": "100\nFFFFFFFFFFFFFFFFFFFFFFFFFFSFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFSFFFFFFFFFFFFFFFFFSS",
"output": "NO"
},
{
"input": "100\nFFFFFFFFFFFFFSFFFFFFFFFSFSSFFFFFFFFFFFFFFFFFFFFFFSFFSFFFFFSFFFFFFFFSFFFFFFFFFFFFFSFFFFFFFFSFFFFFFFSF",
"output": "NO"
},
{
"input": "100\nSFFSSFFFFFFSSFFFSSFSFFFFFSSFFFSFFFFFFSFSSSFSFSFFFFSFSSFFFFFFFFSFFFFFSFFFFFSSFFFSFFSFSFFFFSFFSFFFFFFF",
"output": "YES"
},
{
"input": "100\nFFFFSSSSSFFSSSFFFSFFFFFSFSSFSFFSFFSSFFSSFSFFFFFSFSFSFSFFFFFFFFFSFSFFSFFFFSFSFFFFFFFFFFFFSFSSFFSSSSFF",
"output": "NO"
},
{
"input": "100\nFFFFFFFFFFFFSSFFFFSFSFFFSFSSSFSSSSSFSSSSFFSSFFFSFSFSSFFFSSSFFSFSFSSFSFSSFSFFFSFFFFFSSFSFFFSSSFSSSFFS",
"output": "NO"
},
{
"input": "100\nFFFSSSFSFSSSSFSSFSFFSSSFFSSFSSFFSSFFSFSSSSFFFSFFFSFSFSSSFSSFSFSFSFFSSSSSFSSSFSFSFFSSFSFSSFFSSFSFFSFS",
"output": "NO"
},
{
"input": "100\nFFSSSSFSSSFSSSSFSSSFFSFSSFFSSFSSSFSSSFFSFFSSSSSSSSSSSSFSSFSSSSFSFFFSSFFFFFFSFSFSSSSSSFSSSFSFSSFSSFSS",
"output": "NO"
},
{
"input": "100\nSSSFFFSSSSFFSSSSSFSSSSFSSSFSSSSSFSSSSSSSSFSFFSSSFFSSFSSSSFFSSSSSSFFSSSSFSSSSSSFSSSFSSSSSSSFSSSSFSSSS",
"output": "NO"
},
{
"input": "100\nFSSSSSSSSSSSFSSSSSSSSSSSSSSSSFSSSSSSFSSSSSSSSSSSSSFSSFSSSSSFSSFSSSSSSSSSFFSSSSSFSFSSSFFSSSSSSSSSSSSS",
"output": "NO"
},
{
"input": "100\nSSSSSSSSSSSSSFSSSSSSSSSSSSFSSSFSSSSSSSSSSSSSSSSSSSSSSSSSSSSSFSSSSSSSSSSSSSSSSFSFSSSSSSSSSSSSSSSSSSFS",
"output": "NO"
},
{
"input": "100\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS",
"output": "NO"
},
{
"input": "100\nSFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF",
"output": "YES"
},
{
"input": "100\nSFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFSFSFFFFFFFFFFFSFSFFFFFFFFFFFFFSFFFFFFFFFFFFFFFFFFFFFFFFF",
"output": "YES"
},
{
"input": "100\nSFFFFFFFFFFFFSSFFFFSFFFFFFFFFFFFFFFFFFFSFFFSSFFFFSFSFFFSFFFFFFFFFFFFFFFSSFFFFFFFFSSFFFFFFFFFFFFFFSFF",
"output": "YES"
},
{
"input": "100\nSFFSSSFFSFSFSFFFFSSFFFFSFFFFFFFFSFSFFFSFFFSFFFSFFFFSFSFFFFFFFSFFFFFFFFFFSFFSSSFFSSFFFFSFFFFSFFFFSFFF",
"output": "YES"
},
{
"input": "100\nSFFFSFFFFSFFFSSFFFSFSFFFSFFFSSFSFFFFFSFFFFFFFFSFSFSFFSFFFSFSSFSFFFSFSFFSSFSFSSSFFFFFFSSFSFFSFFFFFFFF",
"output": "YES"
},
{
"input": "100\nSSSSFFFFSFFFFFFFSFFFFSFSFFFFSSFFFFFFFFFSFFSSFFFFFFSFSFSSFSSSFFFFFFFSFSFFFSSSFFFFFFFSFFFSSFFFFSSFFFSF",
"output": "YES"
},
{
"input": "100\nSSSFSSFFFSFSSSSFSSFSSSSFSSFFFFFSFFSSSSFFSSSFSSSFSSSSFSSSSFSSSSSSSFSFSSFFFSSFFSFFSSSSFSSFFSFSSFSFFFSF",
"output": "YES"
},
{
"input": "100\nSFFSFSSSSSSSFFSSSFSSSSFSFSSFFFSSSSSSFSSSSFSSFSSSFSSSSSSSFSSFSFFFSSFSSFSFSFSSSSSSSSSSSSFFFFSSSSSFSFFF",
"output": "YES"
},
{
"input": "100\nSSSFSFFSFSFFSSSSSFSSSFSSSFFFSSSSSSSSSFSFSFSSSSFSFSSFFFFFSSSSSSSSSSSSSSSSSSSFFSSSSSFSFSSSSFFSSSSFSSSF",
"output": "YES"
},
{
"input": "100\nSSSFSSSSSSSSSSFSSSSFSSSSSSFSSSSSSFSSSSSSSSSSSSSSFSSSFSSSFSSSSSSSSSSSFSSSSSSFSFSSSSFSSSSSSFSSSSSSSSFF",
"output": "YES"
},
{
"input": "100\nSSSSSSSSSSSSSSSFSFSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSFFSSSSSSSSSFSSSSSSSSSSSSSSSSSF",
"output": "YES"
},
{
"input": "100\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSF",
"output": "YES"
},
{
"input": "2\nSS",
"output": "NO"
}
] | 1,591,036,104
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
#include <iostream>
#include <string>
#include <math.h>
#include <vector>
#include <iomanip>
#include <algorithm>
using namespace std;
int main() {
int n, stof = 0, ftos = 0;
string s;
cin >> n >> s;
for(int i=0; i<n-1; i++) {
if(s.substr(i,2) == "SF") { stof++; }
else if(s.substr(i,2) == "FS") { ftos++; }
}
if(stof > ftos) { cout << "YES" << endl; }
else { cout << " NO" << endl; }
return 0;
}
|
Title: Between the Offices
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As you may know, MemSQL has American offices in both San Francisco and Seattle. Being a manager in the company, you travel a lot between the two cities, always by plane.
You prefer flying from Seattle to San Francisco than in the other direction, because it's warmer in San Francisco. You are so busy that you don't remember the number of flights you have made in either direction. However, for each of the last *n* days you know whether you were in San Francisco office or in Seattle office. You always fly at nights, so you never were at both offices on the same day. Given this information, determine if you flew more times from Seattle to San Francisco during the last *n* days, or not.
Input Specification:
The first line of input contains single integer *n* (2<=≤<=*n*<=≤<=100) — the number of days.
The second line contains a string of length *n* consisting of only capital 'S' and 'F' letters. If the *i*-th letter is 'S', then you were in Seattle office on that day. Otherwise you were in San Francisco. The days are given in chronological order, i.e. today is the last day in this sequence.
Output Specification:
Print "YES" if you flew more times from Seattle to San Francisco, and "NO" otherwise.
You can print each letter in any case (upper or lower).
Demo Input:
['4\nFSSF\n', '2\nSF\n', '10\nFFFFFFFFFF\n', '10\nSSFFSFFSFF\n']
Demo Output:
['NO\n', 'YES\n', 'NO\n', 'YES\n']
Note:
In the first example you were initially at San Francisco, then flew to Seattle, were there for two days and returned to San Francisco. You made one flight in each direction, so the answer is "NO".
In the second example you just flew from Seattle to San Francisco, so the answer is "YES".
In the third example you stayed the whole period in San Francisco, so the answer is "NO".
In the fourth example if you replace 'S' with ones, and 'F' with zeros, you'll get the first few digits of π in binary representation. Not very useful information though.
|
```python
#include <iostream>
#include <string>
#include <math.h>
#include <vector>
#include <iomanip>
#include <algorithm>
using namespace std;
int main() {
int n, stof = 0, ftos = 0;
string s;
cin >> n >> s;
for(int i=0; i<n-1; i++) {
if(s.substr(i,2) == "SF") { stof++; }
else if(s.substr(i,2) == "FS") { ftos++; }
}
if(stof > ftos) { cout << "YES" << endl; }
else { cout << " NO" << endl; }
return 0;
}
```
| -1
|
|
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<=≤<=*i*<=<<=*j*<=≤<=*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<=≤<=*n*<=≤<=200<=000) — the number of watchmen.
Each of the following *n* lines contains two integers *x**i* and *y**i* (|*x**i*|,<=|*y**i*|<=≤<=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 |1 - 7| + |1 - 5| = 10 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, 1), (1, 5) and (7, 5), (1, 5) 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"
}
] | 1,568,928,031
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 71
| 1,091
| 28,774,400
|
import collections
xx = collections.Counter()
yy = collections.Counter()
zz = collections.Counter()
ans = 0
for _ in range(int(input())):
x, y = map(int, input().split())
ans += xx[x] + yy[y] - zz[(x, y)]
xx[x] += 1
yy[y] += 1
zz[(x, y)] += 1
print(ans)
|
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<=≤<=*i*<=<<=*j*<=≤<=*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<=≤<=*n*<=≤<=200<=000) — the number of watchmen.
Each of the following *n* lines contains two integers *x**i* and *y**i* (|*x**i*|,<=|*y**i*|<=≤<=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 |1 - 7| + |1 - 5| = 10 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, 1), (1, 5) and (7, 5), (1, 5) Doctor Manhattan and Daniel will calculate the same distances.
|
```python
import collections
xx = collections.Counter()
yy = collections.Counter()
zz = collections.Counter()
ans = 0
for _ in range(int(input())):
x, y = map(int, input().split())
ans += xx[x] + yy[y] - zz[(x, y)]
xx[x] += 1
yy[y] += 1
zz[(x, y)] += 1
print(ans)
```
| 3
|
|
231
|
A
|
Team
|
PROGRAMMING
| 800
|
[
"brute force",
"greedy"
] | null | null |
One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.
This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution.
|
The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces.
|
Print a single integer — the number of problems the friends will implement on the contest.
|
[
"3\n1 1 0\n1 1 1\n1 0 0\n",
"2\n1 0 0\n0 1 1\n"
] |
[
"2\n",
"1\n"
] |
In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.
In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution.
| 500
|
[
{
"input": "3\n1 1 0\n1 1 1\n1 0 0",
"output": "2"
},
{
"input": "2\n1 0 0\n0 1 1",
"output": "1"
},
{
"input": "1\n1 0 0",
"output": "0"
},
{
"input": "2\n1 0 0\n1 1 1",
"output": "1"
},
{
"input": "5\n1 0 0\n0 1 0\n1 1 1\n0 0 1\n0 0 0",
"output": "1"
},
{
"input": "10\n0 1 0\n0 1 0\n1 1 0\n1 0 0\n0 0 1\n0 1 1\n1 1 1\n1 1 0\n0 0 0\n0 0 0",
"output": "4"
},
{
"input": "15\n0 1 0\n1 0 0\n1 1 0\n1 1 1\n0 1 0\n0 0 1\n1 0 1\n1 0 1\n1 0 1\n0 0 0\n1 1 1\n1 1 0\n0 1 1\n1 1 0\n1 1 1",
"output": "10"
},
{
"input": "50\n0 0 0\n0 1 1\n1 1 1\n0 1 0\n1 0 1\n1 1 1\n0 0 1\n1 0 0\n1 1 0\n1 0 1\n0 1 0\n0 0 1\n1 1 0\n0 1 0\n1 1 0\n0 0 0\n1 1 1\n1 0 1\n0 0 1\n1 1 0\n1 1 1\n0 1 1\n1 1 0\n0 0 0\n0 0 0\n1 1 1\n0 0 0\n1 1 1\n0 1 1\n0 0 1\n0 0 0\n0 0 0\n1 1 0\n1 1 0\n1 0 1\n1 0 0\n1 0 1\n1 0 1\n0 1 1\n1 1 0\n1 1 0\n0 1 0\n1 0 1\n0 0 0\n0 0 0\n0 0 0\n0 0 1\n1 1 1\n0 1 1\n1 0 1",
"output": "29"
},
{
"input": "1\n1 1 1",
"output": "1"
},
{
"input": "8\n0 0 0\n0 0 1\n0 0 0\n0 1 1\n1 0 0\n1 0 1\n1 1 0\n1 1 1",
"output": "4"
},
{
"input": "16\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1",
"output": "16"
}
] | 1,699,470,360
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 21
| 62
| 0
|
n = int(input())
a = 0
for i in range(n):
j,f,k = map(int, input().split())
if j+f+k>=2:
a+=1
print(a)
|
Title: Team
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.
This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution.
Input Specification:
The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces.
Output Specification:
Print a single integer — the number of problems the friends will implement on the contest.
Demo Input:
['3\n1 1 0\n1 1 1\n1 0 0\n', '2\n1 0 0\n0 1 1\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.
In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution.
|
```python
n = int(input())
a = 0
for i in range(n):
j,f,k = map(int, input().split())
if j+f+k>=2:
a+=1
print(a)
```
| 3
|
|
950
|
A
|
Left-handers, Right-handers and Ambidexters
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
You are at a water bowling training. There are *l* people who play with their left hand, *r* people, who play with their right hand, and *a* ambidexters, who can play with left or right hand.
The coach decided to form a team of even number of players, exactly half of the players should play with their right hand, and exactly half of the players should play with their left hand. One player should use only on of his hands.
Ambidexters play as well with their right hand as with their left hand. In the team, an ambidexter can play with their left hand, or with their right hand.
Please find the maximum possible size of the team, where equal number of players use their left and right hands, respectively.
|
The only line contains three integers *l*, *r* and *a* (0<=≤<=*l*,<=*r*,<=*a*<=≤<=100) — the number of left-handers, the number of right-handers and the number of ambidexters at the training.
|
Print a single even integer — the maximum number of players in the team. It is possible that the team can only have zero number of players.
|
[
"1 4 2\n",
"5 5 5\n",
"0 2 0\n"
] |
[
"6\n",
"14\n",
"0\n"
] |
In the first example you can form a team of 6 players. You should take the only left-hander and two ambidexters to play with left hand, and three right-handers to play with right hand. The only person left can't be taken into the team.
In the second example you can form a team of 14 people. You have to take all five left-handers, all five right-handers, two ambidexters to play with left hand and two ambidexters to play with right hand.
| 500
|
[
{
"input": "1 4 2",
"output": "6"
},
{
"input": "5 5 5",
"output": "14"
},
{
"input": "0 2 0",
"output": "0"
},
{
"input": "30 70 34",
"output": "128"
},
{
"input": "89 32 24",
"output": "112"
},
{
"input": "89 44 77",
"output": "210"
},
{
"input": "0 0 0",
"output": "0"
},
{
"input": "100 100 100",
"output": "300"
},
{
"input": "1 1 1",
"output": "2"
},
{
"input": "30 70 35",
"output": "130"
},
{
"input": "89 44 76",
"output": "208"
},
{
"input": "0 100 100",
"output": "200"
},
{
"input": "100 0 100",
"output": "200"
},
{
"input": "100 1 100",
"output": "200"
},
{
"input": "1 100 100",
"output": "200"
},
{
"input": "100 100 0",
"output": "200"
},
{
"input": "100 100 1",
"output": "200"
},
{
"input": "1 2 1",
"output": "4"
},
{
"input": "0 0 100",
"output": "100"
},
{
"input": "0 100 0",
"output": "0"
},
{
"input": "100 0 0",
"output": "0"
},
{
"input": "10 8 7",
"output": "24"
},
{
"input": "45 47 16",
"output": "108"
},
{
"input": "59 43 100",
"output": "202"
},
{
"input": "34 1 30",
"output": "62"
},
{
"input": "14 81 1",
"output": "30"
},
{
"input": "53 96 94",
"output": "242"
},
{
"input": "62 81 75",
"output": "218"
},
{
"input": "21 71 97",
"output": "188"
},
{
"input": "49 82 73",
"output": "204"
},
{
"input": "88 19 29",
"output": "96"
},
{
"input": "89 4 62",
"output": "132"
},
{
"input": "58 3 65",
"output": "126"
},
{
"input": "27 86 11",
"output": "76"
},
{
"input": "35 19 80",
"output": "134"
},
{
"input": "4 86 74",
"output": "156"
},
{
"input": "32 61 89",
"output": "182"
},
{
"input": "68 60 98",
"output": "226"
},
{
"input": "37 89 34",
"output": "142"
},
{
"input": "92 9 28",
"output": "74"
},
{
"input": "79 58 98",
"output": "234"
},
{
"input": "35 44 88",
"output": "166"
},
{
"input": "16 24 19",
"output": "58"
},
{
"input": "74 71 75",
"output": "220"
},
{
"input": "83 86 99",
"output": "268"
},
{
"input": "97 73 15",
"output": "176"
},
{
"input": "77 76 73",
"output": "226"
},
{
"input": "48 85 55",
"output": "188"
},
{
"input": "1 2 2",
"output": "4"
},
{
"input": "2 2 2",
"output": "6"
},
{
"input": "2 1 2",
"output": "4"
},
{
"input": "2 2 1",
"output": "4"
},
{
"input": "3 2 1",
"output": "6"
},
{
"input": "1 2 3",
"output": "6"
},
{
"input": "1 3 2",
"output": "6"
},
{
"input": "2 1 3",
"output": "6"
},
{
"input": "2 3 1",
"output": "6"
},
{
"input": "3 1 2",
"output": "6"
},
{
"input": "99 99 99",
"output": "296"
},
{
"input": "99 99 100",
"output": "298"
},
{
"input": "99 100 99",
"output": "298"
},
{
"input": "99 100 100",
"output": "298"
},
{
"input": "100 99 99",
"output": "298"
},
{
"input": "100 99 100",
"output": "298"
},
{
"input": "100 100 99",
"output": "298"
},
{
"input": "89 32 23",
"output": "110"
},
{
"input": "4 5 0",
"output": "8"
},
{
"input": "3 0 3",
"output": "6"
},
{
"input": "0 0 2",
"output": "2"
},
{
"input": "97 97 0",
"output": "194"
},
{
"input": "1 4 0",
"output": "2"
},
{
"input": "5 2 0",
"output": "4"
},
{
"input": "0 5 10",
"output": "14"
},
{
"input": "0 1 2",
"output": "2"
},
{
"input": "5 2 3",
"output": "10"
},
{
"input": "5 5 0",
"output": "10"
},
{
"input": "0 0 10",
"output": "10"
},
{
"input": "0 1 1",
"output": "2"
},
{
"input": "0 0 1",
"output": "0"
}
] | 1,571,407,723
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
l,r,a=input().split()
l=int(l)
r=int(r)
a=int(a)
if r>l and l!=0:
l=l+a
if r!=l:
b=r-l
r=r-b
print("l+r")
if l>r and r!=0:
r=r+a
if l!=r:
p=l-r
l=l-p
print("l+r")
if l=rand :
l=l+a//2
r=r+a//2
print("l+r")
|
Title: Left-handers, Right-handers and Ambidexters
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are at a water bowling training. There are *l* people who play with their left hand, *r* people, who play with their right hand, and *a* ambidexters, who can play with left or right hand.
The coach decided to form a team of even number of players, exactly half of the players should play with their right hand, and exactly half of the players should play with their left hand. One player should use only on of his hands.
Ambidexters play as well with their right hand as with their left hand. In the team, an ambidexter can play with their left hand, or with their right hand.
Please find the maximum possible size of the team, where equal number of players use their left and right hands, respectively.
Input Specification:
The only line contains three integers *l*, *r* and *a* (0<=≤<=*l*,<=*r*,<=*a*<=≤<=100) — the number of left-handers, the number of right-handers and the number of ambidexters at the training.
Output Specification:
Print a single even integer — the maximum number of players in the team. It is possible that the team can only have zero number of players.
Demo Input:
['1 4 2\n', '5 5 5\n', '0 2 0\n']
Demo Output:
['6\n', '14\n', '0\n']
Note:
In the first example you can form a team of 6 players. You should take the only left-hander and two ambidexters to play with left hand, and three right-handers to play with right hand. The only person left can't be taken into the team.
In the second example you can form a team of 14 people. You have to take all five left-handers, all five right-handers, two ambidexters to play with left hand and two ambidexters to play with right hand.
|
```python
l,r,a=input().split()
l=int(l)
r=int(r)
a=int(a)
if r>l and l!=0:
l=l+a
if r!=l:
b=r-l
r=r-b
print("l+r")
if l>r and r!=0:
r=r+a
if l!=r:
p=l-r
l=l-p
print("l+r")
if l=rand :
l=l+a//2
r=r+a//2
print("l+r")
```
| -1
|
|
660
|
B
|
Seating On Bus
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
Consider 2*n* rows of the seats in a bus. *n* rows of the seats on the left and *n* rows of the seats on the right. Each row can be filled by two people. So the total capacity of the bus is 4*n*.
Consider that *m* (*m*<=≤<=4*n*) people occupy the seats in the bus. The passengers entering the bus are numbered from 1 to *m* (in the order of their entering the bus). The pattern of the seat occupation is as below:
1-st row left window seat, 1-st row right window seat, 2-nd row left window seat, 2-nd row right window seat, ... , *n*-th row left window seat, *n*-th row right window seat.
After occupying all the window seats (for *m*<=><=2*n*) the non-window seats are occupied:
1-st row left non-window seat, 1-st row right non-window seat, ... , *n*-th row left non-window seat, *n*-th row right non-window seat.
All the passengers go to a single final destination. In the final destination, the passengers get off in the given order.
1-st row left non-window seat, 1-st row left window seat, 1-st row right non-window seat, 1-st row right window seat, ... , *n*-th row left non-window seat, *n*-th row left window seat, *n*-th row right non-window seat, *n*-th row right window seat.
You are given the values *n* and *m*. Output *m* numbers from 1 to *m*, the order in which the passengers will get off the bus.
|
The only line contains two integers, *n* and *m* (1<=≤<=*n*<=≤<=100,<=1<=≤<=*m*<=≤<=4*n*) — the number of pairs of rows and the number of passengers.
|
Print *m* distinct integers from 1 to *m* — the order in which the passengers will get off the bus.
|
[
"2 7\n",
"9 36\n"
] |
[
"5 1 6 2 7 3 4\n",
"19 1 20 2 21 3 22 4 23 5 24 6 25 7 26 8 27 9 28 10 29 11 30 12 31 13 32 14 33 15 34 16 35 17 36 18\n"
] |
none
| 0
|
[
{
"input": "2 7",
"output": "5 1 6 2 7 3 4"
},
{
"input": "9 36",
"output": "19 1 20 2 21 3 22 4 23 5 24 6 25 7 26 8 27 9 28 10 29 11 30 12 31 13 32 14 33 15 34 16 35 17 36 18"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 4",
"output": "3 1 4 2"
},
{
"input": "10 1",
"output": "1"
},
{
"input": "10 10",
"output": "1 2 3 4 5 6 7 8 9 10"
},
{
"input": "10 40",
"output": "21 1 22 2 23 3 24 4 25 5 26 6 27 7 28 8 29 9 30 10 31 11 32 12 33 13 34 14 35 15 36 16 37 17 38 18 39 19 40 20"
},
{
"input": "10 39",
"output": "21 1 22 2 23 3 24 4 25 5 26 6 27 7 28 8 29 9 30 10 31 11 32 12 33 13 34 14 35 15 36 16 37 17 38 18 39 19 20"
},
{
"input": "77 1",
"output": "1"
},
{
"input": "77 13",
"output": "1 2 3 4 5 6 7 8 9 10 11 12 13"
},
{
"input": "77 53",
"output": "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 51 52 53"
},
{
"input": "77 280",
"output": "155 1 156 2 157 3 158 4 159 5 160 6 161 7 162 8 163 9 164 10 165 11 166 12 167 13 168 14 169 15 170 16 171 17 172 18 173 19 174 20 175 21 176 22 177 23 178 24 179 25 180 26 181 27 182 28 183 29 184 30 185 31 186 32 187 33 188 34 189 35 190 36 191 37 192 38 193 39 194 40 195 41 196 42 197 43 198 44 199 45 200 46 201 47 202 48 203 49 204 50 205 51 206 52 207 53 208 54 209 55 210 56 211 57 212 58 213 59 214 60 215 61 216 62 217 63 218 64 219 65 220 66 221 67 222 68 223 69 224 70 225 71 226 72 227 73 228 74 22..."
},
{
"input": "100 1",
"output": "1"
},
{
"input": "100 13",
"output": "1 2 3 4 5 6 7 8 9 10 11 12 13"
},
{
"input": "100 77",
"output": "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 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"
},
{
"input": "100 103",
"output": "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 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103"
},
{
"input": "100 200",
"output": "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 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..."
},
{
"input": "100 199",
"output": "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 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..."
},
{
"input": "100 201",
"output": "201 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 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..."
},
{
"input": "100 300",
"output": "201 1 202 2 203 3 204 4 205 5 206 6 207 7 208 8 209 9 210 10 211 11 212 12 213 13 214 14 215 15 216 16 217 17 218 18 219 19 220 20 221 21 222 22 223 23 224 24 225 25 226 26 227 27 228 28 229 29 230 30 231 31 232 32 233 33 234 34 235 35 236 36 237 37 238 38 239 39 240 40 241 41 242 42 243 43 244 44 245 45 246 46 247 47 248 48 249 49 250 50 251 51 252 52 253 53 254 54 255 55 256 56 257 57 258 58 259 59 260 60 261 61 262 62 263 63 264 64 265 65 266 66 267 67 268 68 269 69 270 70 271 71 272 72 273 73 274 74 27..."
},
{
"input": "100 399",
"output": "201 1 202 2 203 3 204 4 205 5 206 6 207 7 208 8 209 9 210 10 211 11 212 12 213 13 214 14 215 15 216 16 217 17 218 18 219 19 220 20 221 21 222 22 223 23 224 24 225 25 226 26 227 27 228 28 229 29 230 30 231 31 232 32 233 33 234 34 235 35 236 36 237 37 238 38 239 39 240 40 241 41 242 42 243 43 244 44 245 45 246 46 247 47 248 48 249 49 250 50 251 51 252 52 253 53 254 54 255 55 256 56 257 57 258 58 259 59 260 60 261 61 262 62 263 63 264 64 265 65 266 66 267 67 268 68 269 69 270 70 271 71 272 72 273 73 274 74 27..."
},
{
"input": "100 400",
"output": "201 1 202 2 203 3 204 4 205 5 206 6 207 7 208 8 209 9 210 10 211 11 212 12 213 13 214 14 215 15 216 16 217 17 218 18 219 19 220 20 221 21 222 22 223 23 224 24 225 25 226 26 227 27 228 28 229 29 230 30 231 31 232 32 233 33 234 34 235 35 236 36 237 37 238 38 239 39 240 40 241 41 242 42 243 43 244 44 245 45 246 46 247 47 248 48 249 49 250 50 251 51 252 52 253 53 254 54 255 55 256 56 257 57 258 58 259 59 260 60 261 61 262 62 263 63 264 64 265 65 266 66 267 67 268 68 269 69 270 70 271 71 272 72 273 73 274 74 27..."
},
{
"input": "3 9",
"output": "7 1 8 2 9 3 4 5 6"
}
] | 1,510,075,081
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 23
| 62
| 0
|
def main():
n, m = map(int, input().split())
bus = [[None for _ in range(4)] for _ in range(n)]
half = 2 * n
for i in range(0, 2 * n, 2):
row = i // 2
bus[row][0] = i + 1
bus[row][3] = i + 2
bus[row][1] = half + i + 1
bus[row][2] = half + i + 2
out = []
for row in bus:
if row[1] <= m:
out.append(row[1])
if row[0] <= m:
out.append(row[0])
if row[2] <= m:
out.append(row[2])
if row[3] <= m:
out.append(row[3])
print(' '.join(str(x) for x in out))
main()
|
Title: Seating On Bus
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Consider 2*n* rows of the seats in a bus. *n* rows of the seats on the left and *n* rows of the seats on the right. Each row can be filled by two people. So the total capacity of the bus is 4*n*.
Consider that *m* (*m*<=≤<=4*n*) people occupy the seats in the bus. The passengers entering the bus are numbered from 1 to *m* (in the order of their entering the bus). The pattern of the seat occupation is as below:
1-st row left window seat, 1-st row right window seat, 2-nd row left window seat, 2-nd row right window seat, ... , *n*-th row left window seat, *n*-th row right window seat.
After occupying all the window seats (for *m*<=><=2*n*) the non-window seats are occupied:
1-st row left non-window seat, 1-st row right non-window seat, ... , *n*-th row left non-window seat, *n*-th row right non-window seat.
All the passengers go to a single final destination. In the final destination, the passengers get off in the given order.
1-st row left non-window seat, 1-st row left window seat, 1-st row right non-window seat, 1-st row right window seat, ... , *n*-th row left non-window seat, *n*-th row left window seat, *n*-th row right non-window seat, *n*-th row right window seat.
You are given the values *n* and *m*. Output *m* numbers from 1 to *m*, the order in which the passengers will get off the bus.
Input Specification:
The only line contains two integers, *n* and *m* (1<=≤<=*n*<=≤<=100,<=1<=≤<=*m*<=≤<=4*n*) — the number of pairs of rows and the number of passengers.
Output Specification:
Print *m* distinct integers from 1 to *m* — the order in which the passengers will get off the bus.
Demo Input:
['2 7\n', '9 36\n']
Demo Output:
['5 1 6 2 7 3 4\n', '19 1 20 2 21 3 22 4 23 5 24 6 25 7 26 8 27 9 28 10 29 11 30 12 31 13 32 14 33 15 34 16 35 17 36 18\n']
Note:
none
|
```python
def main():
n, m = map(int, input().split())
bus = [[None for _ in range(4)] for _ in range(n)]
half = 2 * n
for i in range(0, 2 * n, 2):
row = i // 2
bus[row][0] = i + 1
bus[row][3] = i + 2
bus[row][1] = half + i + 1
bus[row][2] = half + i + 2
out = []
for row in bus:
if row[1] <= m:
out.append(row[1])
if row[0] <= m:
out.append(row[0])
if row[2] <= m:
out.append(row[2])
if row[3] <= m:
out.append(row[3])
print(' '.join(str(x) for x in out))
main()
```
| 3
|
|
337
|
A
|
Puzzles
|
PROGRAMMING
| 900
|
[
"greedy"
] | null | null |
The end of the school year is near and Ms. Manana, the teacher, will soon have to say goodbye to a yet another class. She decided to prepare a goodbye present for her *n* students and give each of them a jigsaw puzzle (which, as wikipedia states, is a tiling puzzle that requires the assembly of numerous small, often oddly shaped, interlocking and tessellating pieces).
The shop assistant told the teacher that there are *m* puzzles in the shop, but they might differ in difficulty and size. Specifically, the first jigsaw puzzle consists of *f*1 pieces, the second one consists of *f*2 pieces and so on.
Ms. Manana doesn't want to upset the children, so she decided that the difference between the numbers of pieces in her presents must be as small as possible. Let *A* be the number of pieces in the largest puzzle that the teacher buys and *B* be the number of pieces in the smallest such puzzle. She wants to choose such *n* puzzles that *A*<=-<=*B* is minimum possible. Help the teacher and find the least possible value of *A*<=-<=*B*.
|
The first line contains space-separated integers *n* and *m* (2<=≤<=*n*<=≤<=*m*<=≤<=50). The second line contains *m* space-separated integers *f*1,<=*f*2,<=...,<=*f**m* (4<=≤<=*f**i*<=≤<=1000) — the quantities of pieces in the puzzles sold in the shop.
|
Print a single integer — the least possible difference the teacher can obtain.
|
[
"4 6\n10 12 10 7 5 22\n"
] |
[
"5\n"
] |
Sample 1. The class has 4 students. The shop sells 6 puzzles. If Ms. Manana buys the first four puzzles consisting of 10, 12, 10 and 7 pieces correspondingly, then the difference between the sizes of the largest and the smallest puzzle will be equal to 5. It is impossible to obtain a smaller difference. Note that the teacher can also buy puzzles 1, 3, 4 and 5 to obtain the difference 5.
| 500
|
[
{
"input": "4 6\n10 12 10 7 5 22",
"output": "5"
},
{
"input": "2 2\n4 4",
"output": "0"
},
{
"input": "2 10\n4 5 6 7 8 9 10 11 12 12",
"output": "0"
},
{
"input": "4 5\n818 136 713 59 946",
"output": "759"
},
{
"input": "3 20\n446 852 783 313 549 965 40 88 86 617 479 118 768 34 47 826 366 957 463 903",
"output": "13"
},
{
"input": "2 25\n782 633 152 416 432 825 115 97 386 357 836 310 530 413 354 373 847 882 913 682 729 582 671 674 94",
"output": "3"
},
{
"input": "4 25\n226 790 628 528 114 64 239 279 619 39 894 763 763 847 525 93 882 697 999 643 650 244 159 884 190",
"output": "31"
},
{
"input": "2 50\n971 889 628 39 253 157 925 694 129 516 660 272 738 319 611 816 142 717 514 392 41 105 132 676 958 118 306 768 600 685 103 857 704 346 857 309 23 718 618 161 176 379 846 834 640 468 952 878 164 997",
"output": "0"
},
{
"input": "25 50\n582 146 750 905 313 509 402 21 488 512 32 898 282 64 579 869 37 996 377 929 975 697 666 837 311 205 116 992 533 298 648 268 54 479 792 595 152 69 267 417 184 433 894 603 988 712 24 414 301 176",
"output": "412"
},
{
"input": "49 50\n58 820 826 960 271 294 473 102 925 318 729 672 244 914 796 646 868 6 893 882 726 203 528 498 271 195 355 459 721 680 547 147 631 116 169 804 145 996 133 559 110 257 771 476 576 251 607 314 427 886",
"output": "938"
},
{
"input": "50 50\n374 573 323 744 190 806 485 247 628 336 491 606 702 321 991 678 337 579 86 240 993 208 668 686 855 205 363 177 719 249 896 919 782 434 59 647 787 996 286 216 636 212 546 903 958 559 544 126 608 993",
"output": "937"
},
{
"input": "6 50\n6 8 7 8 5 4 4 5 7 8 6 5 7 4 7 7 7 8 6 4 6 6 8 8 7 7 8 7 5 8 5 4 4 7 8 4 4 6 6 6 8 7 4 7 6 6 5 8 4 7",
"output": "0"
},
{
"input": "37 50\n14 5 11 17 8 20 19 16 20 11 17 20 16 9 14 14 13 18 11 20 8 8 8 5 19 17 6 18 10 20 9 7 12 6 14 17 4 4 10 13 7 4 11 6 20 19 12 12 15 19",
"output": "12"
},
{
"input": "40 50\n4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4",
"output": "0"
},
{
"input": "40 50\n17 20 43 26 41 37 14 8 30 35 30 24 43 8 42 9 41 50 41 35 27 32 35 43 28 36 31 16 5 7 23 16 14 29 8 39 12 16 36 18 49 39 33 37 38 6 6 27 23 17",
"output": "31"
},
{
"input": "2 2\n1000 4",
"output": "996"
},
{
"input": "2 3\n4 502 1000",
"output": "498"
},
{
"input": "3 3\n4 1000 4",
"output": "996"
}
] | 1,698,050,343
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 62
| 0
|
sandar = list(map(int, input().split()))
sandar2 = list(map(int, input().split()))
print(min(sandar2))
|
Title: Puzzles
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The end of the school year is near and Ms. Manana, the teacher, will soon have to say goodbye to a yet another class. She decided to prepare a goodbye present for her *n* students and give each of them a jigsaw puzzle (which, as wikipedia states, is a tiling puzzle that requires the assembly of numerous small, often oddly shaped, interlocking and tessellating pieces).
The shop assistant told the teacher that there are *m* puzzles in the shop, but they might differ in difficulty and size. Specifically, the first jigsaw puzzle consists of *f*1 pieces, the second one consists of *f*2 pieces and so on.
Ms. Manana doesn't want to upset the children, so she decided that the difference between the numbers of pieces in her presents must be as small as possible. Let *A* be the number of pieces in the largest puzzle that the teacher buys and *B* be the number of pieces in the smallest such puzzle. She wants to choose such *n* puzzles that *A*<=-<=*B* is minimum possible. Help the teacher and find the least possible value of *A*<=-<=*B*.
Input Specification:
The first line contains space-separated integers *n* and *m* (2<=≤<=*n*<=≤<=*m*<=≤<=50). The second line contains *m* space-separated integers *f*1,<=*f*2,<=...,<=*f**m* (4<=≤<=*f**i*<=≤<=1000) — the quantities of pieces in the puzzles sold in the shop.
Output Specification:
Print a single integer — the least possible difference the teacher can obtain.
Demo Input:
['4 6\n10 12 10 7 5 22\n']
Demo Output:
['5\n']
Note:
Sample 1. The class has 4 students. The shop sells 6 puzzles. If Ms. Manana buys the first four puzzles consisting of 10, 12, 10 and 7 pieces correspondingly, then the difference between the sizes of the largest and the smallest puzzle will be equal to 5. It is impossible to obtain a smaller difference. Note that the teacher can also buy puzzles 1, 3, 4 and 5 to obtain the difference 5.
|
```python
sandar = list(map(int, input().split()))
sandar2 = list(map(int, input().split()))
print(min(sandar2))
```
| 0
|
|
463
|
B
|
Caisa and Pylons
|
PROGRAMMING
| 1,100
|
[
"brute force",
"implementation",
"math"
] | null | null |
Caisa solved the problem with the sugar and now he is on the way back to home.
Caisa is playing a mobile game during his path. There are (*n*<=+<=1) pylons numbered from 0 to *n* in this game. The pylon with number 0 has zero height, the pylon with number *i* (*i*<=><=0) has height *h**i*. The goal of the game is to reach *n*-th pylon, and the only move the player can do is to jump from the current pylon (let's denote its number as *k*) to the next one (its number will be *k*<=+<=1). When the player have made such a move, its energy increases by *h**k*<=-<=*h**k*<=+<=1 (if this value is negative the player loses energy). The player must have non-negative amount of energy at any moment of the time.
Initially Caisa stand at 0 pylon and has 0 energy. The game provides a special opportunity: one can pay a single dollar and increase the height of anyone pylon by one. Caisa may use that opportunity several times, but he doesn't want to spend too much money. What is the minimal amount of money he must paid to reach the goal of the game?
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=105). The next line contains *n* integers *h*1, *h*2,<=..., *h**n* (1<=<=≤<=<=*h**i*<=<=≤<=<=105) representing the heights of the pylons.
|
Print a single number representing the minimum number of dollars paid by Caisa.
|
[
"5\n3 4 3 2 4\n",
"3\n4 4 4\n"
] |
[
"4\n",
"4\n"
] |
In the first sample he can pay 4 dollars and increase the height of pylon with number 0 by 4 units. Then he can safely pass to the last pylon.
| 1,000
|
[
{
"input": "5\n3 4 3 2 4",
"output": "4"
},
{
"input": "3\n4 4 4",
"output": "4"
},
{
"input": "99\n1401 2019 1748 3785 3236 3177 3443 3772 2138 1049 353 908 310 2388 1322 88 2160 2783 435 2248 1471 706 2468 2319 3156 3506 2794 1999 1983 2519 2597 3735 537 344 3519 3772 3872 2961 3895 2010 10 247 3269 671 2986 942 758 1146 77 1545 3745 1547 2250 2565 217 1406 2070 3010 3404 404 1528 2352 138 2065 3047 3656 2188 2919 2616 2083 1280 2977 2681 548 4000 1667 1489 1109 3164 1565 2653 3260 3463 903 1824 3679 2308 245 2689 2063 648 568 766 785 2984 3812 440 1172 2730",
"output": "4000"
},
{
"input": "68\n477 1931 3738 3921 2306 1823 3328 2057 661 3993 2967 3520 171 1739 1525 1817 209 3475 1902 2666 518 3283 3412 3040 3383 2331 1147 1460 1452 1800 1327 2280 82 1416 2200 2388 3238 1879 796 250 1872 114 121 2042 1853 1645 211 2061 1472 2464 726 1989 1746 489 1380 1128 2819 2527 2939 622 678 265 2902 1111 2032 1453 3850 1621",
"output": "3993"
},
{
"input": "30\n30 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": "30"
},
{
"input": "3\n3 2 1",
"output": "3"
},
{
"input": "1\n69",
"output": "69"
}
] | 1,568,876,929
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 124
| 0
|
# Caisa and Pylons
def calculate(length, numbers):
cost = 0
current_power = numbers[0]
for num in numbers[1:]:
if current_power - num < 0:
diff = num - current_power
cost += diff
current_power += diff
else:
current_power -= num
return cost
if __name__ == "__main__":
num = int(input())
numbers = list(map(lambda x: int(x), input().split()))
result = calculate(num, numbers)
print(result)
|
Title: Caisa and Pylons
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Caisa solved the problem with the sugar and now he is on the way back to home.
Caisa is playing a mobile game during his path. There are (*n*<=+<=1) pylons numbered from 0 to *n* in this game. The pylon with number 0 has zero height, the pylon with number *i* (*i*<=><=0) has height *h**i*. The goal of the game is to reach *n*-th pylon, and the only move the player can do is to jump from the current pylon (let's denote its number as *k*) to the next one (its number will be *k*<=+<=1). When the player have made such a move, its energy increases by *h**k*<=-<=*h**k*<=+<=1 (if this value is negative the player loses energy). The player must have non-negative amount of energy at any moment of the time.
Initially Caisa stand at 0 pylon and has 0 energy. The game provides a special opportunity: one can pay a single dollar and increase the height of anyone pylon by one. Caisa may use that opportunity several times, but he doesn't want to spend too much money. What is the minimal amount of money he must paid to reach the goal of the game?
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105). The next line contains *n* integers *h*1, *h*2,<=..., *h**n* (1<=<=≤<=<=*h**i*<=<=≤<=<=105) representing the heights of the pylons.
Output Specification:
Print a single number representing the minimum number of dollars paid by Caisa.
Demo Input:
['5\n3 4 3 2 4\n', '3\n4 4 4\n']
Demo Output:
['4\n', '4\n']
Note:
In the first sample he can pay 4 dollars and increase the height of pylon with number 0 by 4 units. Then he can safely pass to the last pylon.
|
```python
# Caisa and Pylons
def calculate(length, numbers):
cost = 0
current_power = numbers[0]
for num in numbers[1:]:
if current_power - num < 0:
diff = num - current_power
cost += diff
current_power += diff
else:
current_power -= num
return cost
if __name__ == "__main__":
num = int(input())
numbers = list(map(lambda x: int(x), input().split()))
result = calculate(num, numbers)
print(result)
```
| 0
|
|
200
|
B
|
Drinks
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
Little Vasya loves orange juice very much. That's why any food and drink in his kitchen necessarily contains orange juice. There are *n* drinks in his fridge, the volume fraction of orange juice in the *i*-th drink equals *p**i* percent.
One day Vasya decided to make himself an orange cocktail. He took equal proportions of each of the *n* drinks and mixed them. Then he wondered, how much orange juice the cocktail has.
Find the volume fraction of orange juice in the final drink.
|
The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of orange-containing drinks in Vasya's fridge. The second line contains *n* integers *p**i* (0<=≤<=*p**i*<=≤<=100) — the volume fraction of orange juice in the *i*-th drink, in percent. The numbers are separated by a space.
|
Print the volume fraction in percent of orange juice in Vasya's cocktail. The answer will be considered correct if the absolute or relative error does not exceed 10<=<=-<=4.
|
[
"3\n50 50 100\n",
"4\n0 25 50 75\n"
] |
[
"66.666666666667\n",
"37.500000000000\n"
] |
Note to the first sample: let's assume that Vasya takes *x* milliliters of each drink from the fridge. Then the volume of pure juice in the cocktail will equal <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c1fac6e64d3a8ee6a5ac138cbe51e60039b22473.png" style="max-width: 100.0%;max-height: 100.0%;"/> milliliters. The total cocktail's volume equals 3·*x* milliliters, so the volume fraction of the juice in the cocktail equals <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ceb0664e55a1f9f5fa1243ec74680a4665a4d58d.png" style="max-width: 100.0%;max-height: 100.0%;"/>, that is, 66.(6) percent.
| 500
|
[
{
"input": "3\n50 50 100",
"output": "66.666666666667"
},
{
"input": "4\n0 25 50 75",
"output": "37.500000000000"
},
{
"input": "3\n0 1 8",
"output": "3.000000000000"
},
{
"input": "5\n96 89 93 95 70",
"output": "88.600000000000"
},
{
"input": "7\n62 41 78 4 38 39 75",
"output": "48.142857142857"
},
{
"input": "13\n2 22 7 0 1 17 3 17 11 2 21 26 22",
"output": "11.615384615385"
},
{
"input": "21\n5 4 11 7 0 5 45 21 0 14 51 6 0 16 10 19 8 9 7 12 18",
"output": "12.761904761905"
},
{
"input": "26\n95 70 93 74 94 70 91 70 39 79 80 57 87 75 37 93 48 67 51 90 85 26 23 64 66 84",
"output": "69.538461538462"
},
{
"input": "29\n84 99 72 96 83 92 95 98 97 93 76 84 99 93 81 76 93 99 99 100 95 100 96 95 97 100 71 98 94",
"output": "91.551724137931"
},
{
"input": "33\n100 99 100 100 99 99 99 100 100 100 99 99 99 100 100 100 100 99 100 99 100 100 97 100 100 100 100 100 100 100 98 98 100",
"output": "99.515151515152"
},
{
"input": "34\n14 9 10 5 4 26 18 23 0 1 0 20 18 15 2 2 3 5 14 1 9 4 2 15 7 1 7 19 10 0 0 11 0 2",
"output": "8.147058823529"
},
{
"input": "38\n99 98 100 100 99 92 99 99 98 84 88 94 86 99 93 100 98 99 65 98 85 84 64 97 96 89 79 96 91 84 99 93 72 96 94 97 96 93",
"output": "91.921052631579"
},
{
"input": "52\n100 94 99 98 99 99 99 95 97 97 98 100 100 98 97 100 98 90 100 99 97 94 90 98 100 100 90 99 100 95 98 95 94 85 97 94 96 94 99 99 99 98 100 100 94 99 99 100 98 87 100 100",
"output": "97.019230769231"
},
{
"input": "58\n10 70 12 89 1 82 100 53 40 100 21 69 92 91 67 66 99 77 25 48 8 63 93 39 46 79 82 14 44 42 1 79 0 69 56 73 67 17 59 4 65 80 20 60 77 52 3 61 16 76 33 18 46 100 28 59 9 6",
"output": "50.965517241379"
},
{
"input": "85\n7 8 1 16 0 15 1 7 0 11 15 6 2 12 2 8 9 8 2 0 3 7 15 7 1 8 5 7 2 26 0 3 11 1 8 10 31 0 7 6 1 8 1 0 9 14 4 8 7 16 9 1 0 16 10 9 6 1 1 4 2 7 4 5 4 1 20 6 16 16 1 1 10 17 8 12 14 19 3 8 1 7 10 23 10",
"output": "7.505882352941"
},
{
"input": "74\n5 3 0 7 13 10 12 10 18 5 0 18 2 13 7 17 2 7 5 2 40 19 0 2 2 3 0 45 4 20 0 4 2 8 1 19 3 9 17 1 15 0 16 1 9 4 0 9 32 2 6 18 11 18 1 15 16 12 7 19 5 3 9 28 26 8 3 10 33 29 4 13 28 6",
"output": "10.418918918919"
},
{
"input": "98\n42 9 21 11 9 11 22 12 52 20 10 6 56 9 26 27 1 29 29 14 38 17 41 21 7 45 15 5 29 4 51 20 6 8 34 17 13 53 30 45 0 10 16 41 4 5 6 4 14 2 31 6 0 11 13 3 3 43 13 36 51 0 7 16 28 23 8 36 30 22 8 54 21 45 39 4 50 15 1 30 17 8 18 10 2 20 16 50 6 68 15 6 38 7 28 8 29 41",
"output": "20.928571428571"
},
{
"input": "99\n60 65 40 63 57 44 30 84 3 10 39 53 40 45 72 20 76 11 61 32 4 26 97 55 14 57 86 96 34 69 52 22 26 79 31 4 21 35 82 47 81 28 72 70 93 84 40 4 69 39 83 58 30 7 32 73 74 12 92 23 61 88 9 58 70 32 75 40 63 71 46 55 39 36 14 97 32 16 95 41 28 20 85 40 5 50 50 50 75 6 10 64 38 19 77 91 50 72 96",
"output": "49.191919191919"
},
{
"input": "99\n100 88 40 30 81 80 91 98 69 73 88 96 79 58 14 100 87 84 52 91 83 88 72 83 99 35 54 80 46 79 52 72 85 32 99 39 79 79 45 83 88 50 75 75 50 59 65 75 97 63 92 58 89 46 93 80 89 33 69 86 99 99 66 85 72 74 79 98 85 95 46 63 77 97 49 81 89 39 70 76 68 91 90 56 31 93 51 87 73 95 74 69 87 95 57 68 49 95 92",
"output": "73.484848484848"
},
{
"input": "100\n18 15 17 0 3 3 0 4 1 8 2 22 7 21 5 0 0 8 3 16 1 0 2 9 9 3 10 8 17 20 5 4 8 12 2 3 1 1 3 2 23 0 1 0 5 7 4 0 1 3 3 4 25 2 2 14 8 4 9 3 0 11 0 3 12 3 14 16 7 7 14 1 17 9 0 35 42 12 3 1 25 9 3 8 5 3 2 8 22 14 11 6 3 9 6 8 7 7 4 6",
"output": "7.640000000000"
},
{
"input": "100\n88 77 65 87 100 63 91 96 92 89 77 95 76 80 84 83 100 71 85 98 26 54 74 78 69 59 96 86 88 91 95 26 52 88 64 70 84 81 76 84 94 82 100 66 97 98 43 94 59 94 100 80 98 73 69 83 94 70 74 79 91 31 62 88 69 55 62 97 40 64 62 83 87 85 50 90 69 72 67 49 100 51 69 96 81 90 83 91 86 34 79 69 100 66 97 98 47 97 74 100",
"output": "77.660000000000"
},
{
"input": "100\n91 92 90 91 98 84 85 96 83 98 99 87 94 70 87 75 86 90 89 88 82 83 91 94 88 86 90 99 100 98 97 75 95 99 95 100 91 92 76 93 95 97 88 93 95 81 96 89 88 100 98 87 90 96 100 99 58 90 96 77 92 82 100 100 93 93 98 99 79 88 97 95 98 66 96 83 96 100 99 92 98 98 92 93 100 97 98 100 98 97 100 100 94 90 99 100 98 79 80 81",
"output": "91.480000000000"
},
{
"input": "1\n0",
"output": "0.000000000000"
},
{
"input": "1\n100",
"output": "100.000000000000"
},
{
"input": "1\n78",
"output": "78.000000000000"
},
{
"input": "2\n0 100",
"output": "50.000000000000"
},
{
"input": "2\n100 100",
"output": "100.000000000000"
},
{
"input": "5\n0 0 0 0 1",
"output": "0.200000000000"
},
{
"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 99",
"output": "99.990000000000"
},
{
"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 1",
"output": "0.010000000000"
},
{
"input": "5\n100 100 100 100 100",
"output": "100.000000000000"
}
] | 1,689,867,590
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 31
| 92
| 0
|
n=int(input())
a=list(map(int,input().split()))
b = sum(a)/n
print('{:.12f}'.format(b))
|
Title: Drinks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Vasya loves orange juice very much. That's why any food and drink in his kitchen necessarily contains orange juice. There are *n* drinks in his fridge, the volume fraction of orange juice in the *i*-th drink equals *p**i* percent.
One day Vasya decided to make himself an orange cocktail. He took equal proportions of each of the *n* drinks and mixed them. Then he wondered, how much orange juice the cocktail has.
Find the volume fraction of orange juice in the final drink.
Input Specification:
The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of orange-containing drinks in Vasya's fridge. The second line contains *n* integers *p**i* (0<=≤<=*p**i*<=≤<=100) — the volume fraction of orange juice in the *i*-th drink, in percent. The numbers are separated by a space.
Output Specification:
Print the volume fraction in percent of orange juice in Vasya's cocktail. The answer will be considered correct if the absolute or relative error does not exceed 10<=<=-<=4.
Demo Input:
['3\n50 50 100\n', '4\n0 25 50 75\n']
Demo Output:
['66.666666666667\n', '37.500000000000\n']
Note:
Note to the first sample: let's assume that Vasya takes *x* milliliters of each drink from the fridge. Then the volume of pure juice in the cocktail will equal <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c1fac6e64d3a8ee6a5ac138cbe51e60039b22473.png" style="max-width: 100.0%;max-height: 100.0%;"/> milliliters. The total cocktail's volume equals 3·*x* milliliters, so the volume fraction of the juice in the cocktail equals <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ceb0664e55a1f9f5fa1243ec74680a4665a4d58d.png" style="max-width: 100.0%;max-height: 100.0%;"/>, that is, 66.(6) percent.
|
```python
n=int(input())
a=list(map(int,input().split()))
b = sum(a)/n
print('{:.12f}'.format(b))
```
| 3
|
|
1,003
|
B
|
Binary String Constructing
|
PROGRAMMING
| 1,300
|
[
"constructive algorithms"
] | null | null |
You are given three integers $a$, $b$ and $x$. Your task is to construct a binary string $s$ of length $n = a + b$ such that there are exactly $a$ zeroes, exactly $b$ ones and exactly $x$ indices $i$ (where $1 \le i < n$) such that $s_i \ne s_{i + 1}$. It is guaranteed that the answer always exists.
For example, for the string "01010" there are four indices $i$ such that $1 \le i < n$ and $s_i \ne s_{i + 1}$ ($i = 1, 2, 3, 4$). For the string "111001" there are two such indices $i$ ($i = 3, 5$).
Recall that binary string is a non-empty sequence of characters where each character is either 0 or 1.
|
The first line of the input contains three integers $a$, $b$ and $x$ ($1 \le a, b \le 100, 1 \le x < a + b)$.
|
Print only one string $s$, where $s$ is any binary string satisfying conditions described above. It is guaranteed that the answer always exists.
|
[
"2 2 1\n",
"3 3 3\n",
"5 3 6\n"
] |
[
"1100\n",
"101100\n",
"01010100\n"
] |
All possible answers for the first example:
- 1100; - 0011.
All possible answers for the second example:
- 110100; - 101100; - 110010; - 100110; - 011001; - 001101; - 010011; - 001011.
| 0
|
[
{
"input": "2 2 1",
"output": "1100"
},
{
"input": "3 3 3",
"output": "101100"
},
{
"input": "5 3 6",
"output": "01010100"
},
{
"input": "100 1 2",
"output": "01000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 1 1",
"output": "00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001"
},
{
"input": "1 100 1",
"output": "11111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111110"
},
{
"input": "1 100 2",
"output": "10111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111"
},
{
"input": "7 8 7",
"output": "101010111110000"
},
{
"input": "100 100 199",
"output": "10101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010"
},
{
"input": "50 47 18",
"output": "0101010101010101011111111111111111111111111111111111111100000000000000000000000000000000000000000"
},
{
"input": "2 3 3",
"output": "10110"
},
{
"input": "100 100 100",
"output": "10101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010100000000000000000000000000000000000000000000000000011111111111111111111111111111111111111111111111111"
},
{
"input": "2 2 2",
"output": "1001"
},
{
"input": "3 4 6",
"output": "1010101"
},
{
"input": "1 1 1",
"output": "10"
},
{
"input": "5 6 2",
"output": "10000011111"
},
{
"input": "5 4 2",
"output": "011110000"
},
{
"input": "2 3 4",
"output": "10101"
},
{
"input": "3 3 2",
"output": "100011"
},
{
"input": "100 99 100",
"output": "0101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101111111111111111111111111111111111111111111111111100000000000000000000000000000000000000000000000000"
},
{
"input": "3 2 1",
"output": "00011"
},
{
"input": "12 74 22",
"output": "10101010101010101010100111111111111111111111111111111111111111111111111111111111111111"
},
{
"input": "6 84 12",
"output": "101010101010111111111111111111111111111111111111111111111111111111111111111111111111111111"
},
{
"input": "3 2 4",
"output": "01010"
},
{
"input": "66 11 22",
"output": "01010101010101010101010000000000000000000000000000000000000000000000000000000"
},
{
"input": "83 83 83",
"output": "1010101010101010101010101010101010101010101010101010101010101010101010101010101010111111111111111111111111111111111111111111000000000000000000000000000000000000000000"
},
{
"input": "9 89 18",
"output": "10101010101010101011111111111111111111111111111111111111111111111111111111111111111111111111111111"
},
{
"input": "2 1 2",
"output": "010"
},
{
"input": "52 12 17",
"output": "0101010101010101000000000000000000000000000000000000000000001111"
},
{
"input": "55 56 110",
"output": "101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101"
},
{
"input": "67 81 40",
"output": "1010101010101010101010101010101010101010000000000000000000000000000000000000000000000001111111111111111111111111111111111111111111111111111111111111"
},
{
"input": "15 26 24",
"output": "10101010101010101010101000011111111111111"
},
{
"input": "7 99 14",
"output": "1010101010101011111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111"
},
{
"input": "99 41 17",
"output": "01010101010101010000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000111111111111111111111111111111111"
},
{
"input": "91 87 11",
"output": "0101010101000000000000000000000000000000000000000000000000000000000000000000000000000000000000001111111111111111111111111111111111111111111111111111111111111111111111111111111111"
},
{
"input": "73 61 122",
"output": "01010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101000000000000"
},
{
"input": "10 40 1",
"output": "11111111111111111111111111111111111111110000000000"
},
{
"input": "10 6 10",
"output": "0101010101100000"
},
{
"input": "78 67 117",
"output": "0101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010100000000000000000000111111111"
},
{
"input": "3 5 6",
"output": "10101011"
},
{
"input": "30 34 44",
"output": "1010101010101010101010101010101010101010101000000000111111111111"
},
{
"input": "2 15 4",
"output": "10101111111111111"
},
{
"input": "4 9 6",
"output": "1010100111111"
}
] | 1,594,104,149
| 689
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 6
| 124
| 20,172,800
|
a,b,x = map(int,input().split())
if(x%2==0):
s = "01"*(x//2)
a-=x//2;b-=x//2
print(s+"1"*b+"0"*a)
else:
s="01"*((x+1)//2)
a-=(x+1)//2;b-=(x+1)//2
print("0"*a+s+"1"*b)
|
Title: Binary String Constructing
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given three integers $a$, $b$ and $x$. Your task is to construct a binary string $s$ of length $n = a + b$ such that there are exactly $a$ zeroes, exactly $b$ ones and exactly $x$ indices $i$ (where $1 \le i < n$) such that $s_i \ne s_{i + 1}$. It is guaranteed that the answer always exists.
For example, for the string "01010" there are four indices $i$ such that $1 \le i < n$ and $s_i \ne s_{i + 1}$ ($i = 1, 2, 3, 4$). For the string "111001" there are two such indices $i$ ($i = 3, 5$).
Recall that binary string is a non-empty sequence of characters where each character is either 0 or 1.
Input Specification:
The first line of the input contains three integers $a$, $b$ and $x$ ($1 \le a, b \le 100, 1 \le x < a + b)$.
Output Specification:
Print only one string $s$, where $s$ is any binary string satisfying conditions described above. It is guaranteed that the answer always exists.
Demo Input:
['2 2 1\n', '3 3 3\n', '5 3 6\n']
Demo Output:
['1100\n', '101100\n', '01010100\n']
Note:
All possible answers for the first example:
- 1100; - 0011.
All possible answers for the second example:
- 110100; - 101100; - 110010; - 100110; - 011001; - 001101; - 010011; - 001011.
|
```python
a,b,x = map(int,input().split())
if(x%2==0):
s = "01"*(x//2)
a-=x//2;b-=x//2
print(s+"1"*b+"0"*a)
else:
s="01"*((x+1)//2)
a-=(x+1)//2;b-=(x+1)//2
print("0"*a+s+"1"*b)
```
| 0
|
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