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3.99
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
958
|
B1
|
Maximum Control (easy)
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
The Resistance is trying to take control over all planets in a particular solar system. This solar system is shaped like a tree. More precisely, some planets are connected by bidirectional hyperspace tunnels in such a way that there is a path between every pair of the planets, but removing any tunnel would disconnect some of them.
The Resistance already has measures in place that will, when the time is right, enable them to control every planet that is not remote. A planet is considered to be remote if it is connected to the rest of the planets only via a single hyperspace tunnel.
How much work is there left to be done: that is, how many remote planets are there?
|
The first line of the input contains an integer *N* (2<=≤<=*N*<=≤<=1000) – the number of planets in the galaxy.
The next *N*<=-<=1 lines describe the hyperspace tunnels between the planets. Each of the *N*<=-<=1 lines contains two space-separated integers *u* and *v* (1<=≤<=*u*,<=*v*<=≤<=*N*) indicating that there is a bidirectional hyperspace tunnel between the planets *u* and *v*. It is guaranteed that every two planets are connected by a path of tunnels, and that each tunnel connects a different pair of planets.
|
A single integer denoting the number of remote planets.
|
[
"5\n4 1\n4 2\n1 3\n1 5\n",
"4\n1 2\n4 3\n1 4\n"
] |
[
"3\n",
"2\n"
] |
In the first example, only planets 2, 3 and 5 are connected by a single tunnel.
In the second example, the remote planets are 2 and 3.
Note that this problem has only two versions – easy and medium.
| 0
|
[
{
"input": "5\n4 1\n4 2\n1 3\n1 5",
"output": "3"
},
{
"input": "4\n1 2\n4 3\n1 4",
"output": "2"
},
{
"input": "10\n4 3\n2 6\n10 1\n5 7\n5 8\n10 6\n5 9\n9 3\n2 9",
"output": "4"
}
] | 1,611,501,024
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 9
| 77
| 307,200
|
resole = []
result = 0
for i in range(int(input()) - 1):
resole.extend(list(map(lambda x: int(x), input().split(' '))))
resole.sort()
for i in range(1, len(resole) - 1):
if not resole[i + 1] == resole[i] and not resole[i - 1] == resole[i]:
result += 1
if not resole[len(resole) - 1] == resole[len(resole) - 2]:
result += 1
if not resole[0] == resole[1]:
result += 1
print(result)
|
Title: Maximum Control (easy)
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Resistance is trying to take control over all planets in a particular solar system. This solar system is shaped like a tree. More precisely, some planets are connected by bidirectional hyperspace tunnels in such a way that there is a path between every pair of the planets, but removing any tunnel would disconnect some of them.
The Resistance already has measures in place that will, when the time is right, enable them to control every planet that is not remote. A planet is considered to be remote if it is connected to the rest of the planets only via a single hyperspace tunnel.
How much work is there left to be done: that is, how many remote planets are there?
Input Specification:
The first line of the input contains an integer *N* (2<=≤<=*N*<=≤<=1000) – the number of planets in the galaxy.
The next *N*<=-<=1 lines describe the hyperspace tunnels between the planets. Each of the *N*<=-<=1 lines contains two space-separated integers *u* and *v* (1<=≤<=*u*,<=*v*<=≤<=*N*) indicating that there is a bidirectional hyperspace tunnel between the planets *u* and *v*. It is guaranteed that every two planets are connected by a path of tunnels, and that each tunnel connects a different pair of planets.
Output Specification:
A single integer denoting the number of remote planets.
Demo Input:
['5\n4 1\n4 2\n1 3\n1 5\n', '4\n1 2\n4 3\n1 4\n']
Demo Output:
['3\n', '2\n']
Note:
In the first example, only planets 2, 3 and 5 are connected by a single tunnel.
In the second example, the remote planets are 2 and 3.
Note that this problem has only two versions – easy and medium.
|
```python
resole = []
result = 0
for i in range(int(input()) - 1):
resole.extend(list(map(lambda x: int(x), input().split(' '))))
resole.sort()
for i in range(1, len(resole) - 1):
if not resole[i + 1] == resole[i] and not resole[i - 1] == resole[i]:
result += 1
if not resole[len(resole) - 1] == resole[len(resole) - 2]:
result += 1
if not resole[0] == resole[1]:
result += 1
print(result)
```
| 3
|
|
884
|
B
|
Japanese Crosswords Strike Back
|
PROGRAMMING
| 1,100
|
[
"implementation"
] | null | null |
A one-dimensional Japanese crossword can be represented as a binary string of length *x*. An encoding of this crossword is an array *a* of size *n*, where *n* is the number of segments formed completely of 1's, and *a**i* is the length of *i*-th segment. No two segments touch or intersect.
For example:
- If *x*<==<=6 and the crossword is 111011, then its encoding is an array {3,<=2}; - If *x*<==<=8 and the crossword is 01101010, then its encoding is an array {2,<=1,<=1}; - If *x*<==<=5 and the crossword is 11111, then its encoding is an array {5}; - If *x*<==<=5 and the crossword is 00000, then its encoding is an empty array.
Mishka wants to create a new one-dimensional Japanese crossword. He has already picked the length and the encoding for this crossword. And now he needs to check if there is exactly one crossword such that its length and encoding are equal to the length and encoding he picked. Help him to check it!
|
The first line contains two integer numbers *n* and *x* (1<=≤<=*n*<=≤<=100000, 1<=≤<=*x*<=≤<=109) — the number of elements in the encoding and the length of the crossword Mishka picked.
The second line contains *n* integer numbers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=10000) — the encoding.
|
Print YES if there exists exaclty one crossword with chosen length and encoding. Otherwise, print NO.
|
[
"2 4\n1 3\n",
"3 10\n3 3 2\n",
"2 10\n1 3\n"
] |
[
"NO\n",
"YES\n",
"NO\n"
] |
none
| 0
|
[
{
"input": "2 4\n1 3",
"output": "NO"
},
{
"input": "3 10\n3 3 2",
"output": "YES"
},
{
"input": "2 10\n1 3",
"output": "NO"
},
{
"input": "1 1\n1",
"output": "YES"
},
{
"input": "1 10\n10",
"output": "YES"
},
{
"input": "1 10000\n10000",
"output": "YES"
},
{
"input": "10 1\n5 78 3 87 4 9 5 8 9 1235",
"output": "NO"
},
{
"input": "3 12\n3 3 3",
"output": "NO"
},
{
"input": "3 9\n2 2 2",
"output": "NO"
},
{
"input": "2 5\n1 1",
"output": "NO"
},
{
"input": "1 2\n1",
"output": "NO"
},
{
"input": "3 13\n3 3 3",
"output": "NO"
},
{
"input": "3 6\n1 1 1",
"output": "NO"
},
{
"input": "1 6\n5",
"output": "NO"
},
{
"input": "3 11\n3 3 2",
"output": "NO"
},
{
"input": "2 6\n1 3",
"output": "NO"
},
{
"input": "3 10\n2 2 2",
"output": "NO"
},
{
"input": "3 8\n2 1 1",
"output": "NO"
},
{
"input": "1 5\n2",
"output": "NO"
},
{
"input": "1 3\n1",
"output": "NO"
},
{
"input": "5 5\n1 1 1 1 1",
"output": "NO"
},
{
"input": "2 10\n4 4",
"output": "NO"
},
{
"input": "2 8\n2 3",
"output": "NO"
},
{
"input": "2 4\n1 1",
"output": "NO"
},
{
"input": "3 10\n1 2 4",
"output": "NO"
},
{
"input": "3 10\n2 1 3",
"output": "NO"
},
{
"input": "2 6\n1 2",
"output": "NO"
},
{
"input": "3 4\n1 1 1",
"output": "NO"
},
{
"input": "3 11\n1 2 4",
"output": "NO"
},
{
"input": "3 12\n3 3 2",
"output": "NO"
},
{
"input": "4 9\n1 1 1 1",
"output": "NO"
},
{
"input": "1 10\n9",
"output": "NO"
},
{
"input": "1 7\n5",
"output": "NO"
},
{
"input": "2 5\n1 2",
"output": "NO"
},
{
"input": "3 8\n1 1 2",
"output": "NO"
},
{
"input": "3 7\n1 1 1",
"output": "NO"
},
{
"input": "1 10\n1",
"output": "NO"
},
{
"input": "3 10\n2 3 4",
"output": "NO"
},
{
"input": "3 9\n1 2 3",
"output": "NO"
},
{
"input": "3 9\n3 3 2",
"output": "NO"
},
{
"input": "3 6\n3 3 2",
"output": "NO"
},
{
"input": "1 1\n3",
"output": "NO"
},
{
"input": "1 3\n2",
"output": "NO"
},
{
"input": "3 10\n3 3 3",
"output": "NO"
},
{
"input": "3 5\n1 1 1",
"output": "YES"
},
{
"input": "2 1\n100 100",
"output": "NO"
},
{
"input": "3 3\n3 3 3",
"output": "NO"
},
{
"input": "4 17\n3 3 9 1",
"output": "NO"
},
{
"input": "4 1660\n1505 13 37 100",
"output": "NO"
},
{
"input": "3 5\n3 3 2",
"output": "NO"
},
{
"input": "4 10\n3 3 2 5",
"output": "NO"
},
{
"input": "5 5\n5 5 5 5 5",
"output": "NO"
},
{
"input": "1 1\n2",
"output": "NO"
},
{
"input": "5 10\n1 2 2 4 5",
"output": "NO"
},
{
"input": "2 1\n1 1",
"output": "NO"
}
] | 1,509,115,350
| 2,250
|
Python 3
|
OK
|
TESTS
| 66
| 109
| 7,372,800
|
(n, length) = map(int, input().split())
lst = []
for x in input().split():
lst.append(int(x))
k = sum(lst) + len(lst) - 1
if k == length:
print("YES")
else:
print("NO")
|
Title: Japanese Crosswords Strike Back
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A one-dimensional Japanese crossword can be represented as a binary string of length *x*. An encoding of this crossword is an array *a* of size *n*, where *n* is the number of segments formed completely of 1's, and *a**i* is the length of *i*-th segment. No two segments touch or intersect.
For example:
- If *x*<==<=6 and the crossword is 111011, then its encoding is an array {3,<=2}; - If *x*<==<=8 and the crossword is 01101010, then its encoding is an array {2,<=1,<=1}; - If *x*<==<=5 and the crossword is 11111, then its encoding is an array {5}; - If *x*<==<=5 and the crossword is 00000, then its encoding is an empty array.
Mishka wants to create a new one-dimensional Japanese crossword. He has already picked the length and the encoding for this crossword. And now he needs to check if there is exactly one crossword such that its length and encoding are equal to the length and encoding he picked. Help him to check it!
Input Specification:
The first line contains two integer numbers *n* and *x* (1<=≤<=*n*<=≤<=100000, 1<=≤<=*x*<=≤<=109) — the number of elements in the encoding and the length of the crossword Mishka picked.
The second line contains *n* integer numbers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=10000) — the encoding.
Output Specification:
Print YES if there exists exaclty one crossword with chosen length and encoding. Otherwise, print NO.
Demo Input:
['2 4\n1 3\n', '3 10\n3 3 2\n', '2 10\n1 3\n']
Demo Output:
['NO\n', 'YES\n', 'NO\n']
Note:
none
|
```python
(n, length) = map(int, input().split())
lst = []
for x in input().split():
lst.append(int(x))
k = sum(lst) + len(lst) - 1
if k == length:
print("YES")
else:
print("NO")
```
| 3
|
|
25
|
A
|
IQ test
|
PROGRAMMING
| 1,300
|
[
"brute force"
] |
A. IQ test
|
2
|
256
|
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
|
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
|
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
|
[
"5\n2 4 7 8 10\n",
"4\n1 2 1 1\n"
] |
[
"3\n",
"2\n"
] |
none
| 0
|
[
{
"input": "5\n2 4 7 8 10",
"output": "3"
},
{
"input": "4\n1 2 1 1",
"output": "2"
},
{
"input": "3\n1 2 2",
"output": "1"
},
{
"input": "3\n100 99 100",
"output": "2"
},
{
"input": "3\n5 3 2",
"output": "3"
},
{
"input": "4\n43 28 1 91",
"output": "2"
},
{
"input": "4\n75 13 94 77",
"output": "3"
},
{
"input": "4\n97 8 27 3",
"output": "2"
},
{
"input": "10\n95 51 12 91 85 3 1 31 25 7",
"output": "3"
},
{
"input": "20\n88 96 66 51 14 88 2 92 18 72 18 88 20 30 4 82 90 100 24 46",
"output": "4"
},
{
"input": "30\n20 94 56 50 10 98 52 32 14 22 24 60 4 8 98 46 34 68 82 82 98 90 50 20 78 49 52 94 64 36",
"output": "26"
},
{
"input": "50\n79 27 77 57 37 45 27 49 65 33 57 21 71 19 75 85 65 61 23 97 85 9 23 1 9 3 99 77 77 21 79 69 15 37 15 7 93 81 13 89 91 31 45 93 15 97 55 80 85 83",
"output": "48"
},
{
"input": "60\n46 11 73 65 3 69 3 53 43 53 97 47 55 93 31 75 35 3 9 73 23 31 3 81 91 79 61 21 15 11 11 11 81 7 83 75 39 87 83 59 89 55 93 27 49 67 67 29 1 93 11 17 9 19 35 21 63 31 31 25",
"output": "1"
},
{
"input": "70\n28 42 42 92 64 54 22 38 38 78 62 38 4 38 14 66 4 92 66 58 94 26 4 44 41 88 48 82 44 26 74 44 48 4 16 92 34 38 26 64 94 4 30 78 50 54 12 90 8 16 80 98 28 100 74 50 36 42 92 18 76 98 8 22 2 50 58 50 64 46",
"output": "25"
},
{
"input": "100\n43 35 79 53 13 91 91 45 65 83 57 9 42 39 85 45 71 51 61 59 31 13 63 39 25 21 79 39 91 67 21 61 97 75 93 83 29 79 59 97 11 37 63 51 39 55 91 23 21 17 47 23 35 75 49 5 69 99 5 7 41 17 25 89 15 79 21 63 53 81 43 91 59 91 69 99 85 15 91 51 49 37 65 7 89 81 21 93 61 63 97 93 45 17 13 69 57 25 75 73",
"output": "13"
},
{
"input": "100\n50 24 68 60 70 30 52 22 18 74 68 98 20 82 4 46 26 68 100 78 84 58 74 98 38 88 68 86 64 80 82 100 20 22 98 98 52 6 94 10 48 68 2 18 38 22 22 82 44 20 66 72 36 58 64 6 36 60 4 96 76 64 12 90 10 58 64 60 74 28 90 26 24 60 40 58 2 16 76 48 58 36 82 60 24 44 4 78 28 38 8 12 40 16 38 6 66 24 31 76",
"output": "99"
},
{
"input": "100\n47 48 94 48 14 18 94 36 96 22 12 30 94 20 48 98 40 58 2 94 8 36 98 18 98 68 2 60 76 38 18 100 8 72 100 68 2 86 92 72 58 16 48 14 6 58 72 76 6 88 80 66 20 28 74 62 86 68 90 86 2 56 34 38 56 90 4 8 76 44 32 86 12 98 38 34 54 92 70 94 10 24 82 66 90 58 62 2 32 58 100 22 58 72 2 22 68 72 42 14",
"output": "1"
},
{
"input": "99\n38 20 68 60 84 16 28 88 60 48 80 28 4 92 70 60 46 46 20 34 12 100 76 2 40 10 8 86 6 80 50 66 12 34 14 28 26 70 46 64 34 96 10 90 98 96 56 88 50 74 70 94 2 94 24 66 68 46 22 30 6 10 64 32 88 14 98 100 64 58 50 18 50 50 8 38 8 16 54 2 60 54 62 84 92 98 4 72 66 26 14 88 99 16 10 6 88 56 22",
"output": "93"
},
{
"input": "99\n50 83 43 89 53 47 69 1 5 37 63 87 95 15 55 95 75 89 33 53 89 75 93 75 11 85 49 29 11 97 49 67 87 11 25 37 97 73 67 49 87 43 53 97 43 29 53 33 45 91 37 73 39 49 59 5 21 43 87 35 5 63 89 57 63 47 29 99 19 85 13 13 3 13 43 19 5 9 61 51 51 57 15 89 13 97 41 13 99 79 13 27 97 95 73 33 99 27 23",
"output": "1"
},
{
"input": "98\n61 56 44 30 58 14 20 24 88 28 46 56 96 52 58 42 94 50 46 30 46 80 72 88 68 16 6 60 26 90 10 98 76 20 56 40 30 16 96 20 88 32 62 30 74 58 36 76 60 4 24 36 42 54 24 92 28 14 2 74 86 90 14 52 34 82 40 76 8 64 2 56 10 8 78 16 70 86 70 42 70 74 22 18 76 98 88 28 62 70 36 72 20 68 34 48 80 98",
"output": "1"
},
{
"input": "98\n66 26 46 42 78 32 76 42 26 82 8 12 4 10 24 26 64 44 100 46 94 64 30 18 88 28 8 66 30 82 82 28 74 52 62 80 80 60 94 86 64 32 44 88 92 20 12 74 94 28 34 58 4 22 16 10 94 76 82 58 40 66 22 6 30 32 92 54 16 76 74 98 18 48 48 30 92 2 16 42 84 74 30 60 64 52 50 26 16 86 58 96 79 60 20 62 82 94",
"output": "93"
},
{
"input": "95\n9 31 27 93 17 77 75 9 9 53 89 39 51 99 5 1 11 39 27 49 91 17 27 79 81 71 37 75 35 13 93 4 99 55 85 11 23 57 5 43 5 61 15 35 23 91 3 81 99 85 43 37 39 27 5 67 7 33 75 59 13 71 51 27 15 93 51 63 91 53 43 99 25 47 17 71 81 15 53 31 59 83 41 23 73 25 91 91 13 17 25 13 55 57 29",
"output": "32"
},
{
"input": "100\n91 89 81 45 53 1 41 3 77 93 55 97 55 97 87 27 69 95 73 41 93 21 75 35 53 56 5 51 87 59 91 67 33 3 99 45 83 17 97 47 75 97 7 89 17 99 23 23 81 25 55 97 27 35 69 5 77 35 93 19 55 59 37 21 31 37 49 41 91 53 73 69 7 37 37 39 17 71 7 97 55 17 47 23 15 73 31 39 57 37 9 5 61 41 65 57 77 79 35 47",
"output": "26"
},
{
"input": "99\n38 56 58 98 80 54 26 90 14 16 78 92 52 74 40 30 84 14 44 80 16 90 98 68 26 24 78 72 42 16 84 40 14 44 2 52 50 2 12 96 58 66 8 80 44 52 34 34 72 98 74 4 66 74 56 21 8 38 76 40 10 22 48 32 98 34 12 62 80 68 64 82 22 78 58 74 20 22 48 56 12 38 32 72 6 16 74 24 94 84 26 38 18 24 76 78 98 94 72",
"output": "56"
},
{
"input": "100\n44 40 6 40 56 90 98 8 36 64 76 86 98 76 36 92 6 30 98 70 24 98 96 60 24 82 88 68 86 96 34 42 58 10 40 26 56 10 88 58 70 32 24 28 14 82 52 12 62 36 70 60 52 34 74 30 78 76 10 16 42 94 66 90 70 38 52 12 58 22 98 96 14 68 24 70 4 30 84 98 8 50 14 52 66 34 100 10 28 100 56 48 38 12 38 14 91 80 70 86",
"output": "97"
},
{
"input": "100\n96 62 64 20 90 46 56 90 68 36 30 56 70 28 16 64 94 34 6 32 34 50 94 22 90 32 40 2 72 10 88 38 28 92 20 26 56 80 4 100 100 90 16 74 74 84 8 2 30 20 80 32 16 46 92 56 42 12 96 64 64 42 64 58 50 42 74 28 2 4 36 32 70 50 54 92 70 16 45 76 28 16 18 50 48 2 62 94 4 12 52 52 4 100 70 60 82 62 98 42",
"output": "79"
},
{
"input": "99\n14 26 34 68 90 58 50 36 8 16 18 6 2 74 54 20 36 84 32 50 52 2 26 24 3 64 20 10 54 26 66 44 28 72 4 96 78 90 96 86 68 28 94 4 12 46 100 32 22 36 84 32 44 94 76 94 4 52 12 30 74 4 34 64 58 72 44 16 70 56 54 8 14 74 8 6 58 62 98 54 14 40 80 20 36 72 28 98 20 58 40 52 90 64 22 48 54 70 52",
"output": "25"
},
{
"input": "95\n82 86 30 78 6 46 80 66 74 72 16 24 18 52 52 38 60 36 86 26 62 28 22 46 96 26 94 84 20 46 66 88 76 32 12 86 74 18 34 88 4 48 94 6 58 6 100 82 4 24 88 32 54 98 34 48 6 76 42 88 42 28 100 4 22 2 10 66 82 54 98 20 60 66 38 98 32 47 86 58 6 100 12 46 2 42 8 84 78 28 24 70 34 28 86",
"output": "78"
},
{
"input": "90\n40 50 8 42 76 24 58 42 26 68 20 48 54 12 34 84 14 36 32 88 6 50 96 56 20 92 48 16 40 34 96 46 20 84 30 50 20 98 8 44 96 42 8 76 70 38 84 30 40 88 84 72 2 22 52 58 16 62 100 66 80 40 50 32 14 62 88 72 22 99 76 50 84 82 8 82 98 46 26 40 2 98 18 78 30 72 70 18 34 68",
"output": "70"
},
{
"input": "80\n81 43 87 1 55 43 53 61 27 19 43 13 89 9 33 83 75 55 97 71 91 37 95 5 21 69 81 93 95 69 31 83 55 7 97 7 79 57 8 61 27 85 49 1 15 97 63 79 29 73 41 85 5 41 31 93 67 11 63 59 15 99 91 77 43 69 23 23 81 73 19 1 67 51 1 75 99 67 3 81",
"output": "39"
},
{
"input": "98\n13 83 61 27 35 1 85 95 97 73 95 65 73 45 5 43 27 83 91 19 11 3 85 59 9 39 69 23 45 7 51 85 5 71 5 95 1 51 75 3 43 57 3 11 33 71 21 99 47 41 87 39 71 87 31 85 91 49 83 5 49 85 47 91 55 99 33 23 31 23 23 73 29 77 55 31 25 5 81 49 91 15 15 39 87 5 9 40 69 47 29 33 11 21 49 79 51 83",
"output": "88"
},
{
"input": "3\n100 100 1",
"output": "3"
}
] | 1,617,869,299
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 32
| 154
| 0
|
evens=[]
odds=[]
n=int(input())
li=list(map(int,input().split(" ")))
for x in li:
if x%2==0:
evens.append(x)
else:
odds.append(x)
if len(evens)==1:
print(li.index(evens[0])+1)
else:
print(li.index(odds[0])+1)
|
Title: IQ test
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
Output Specification:
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
Demo Input:
['5\n2 4 7 8 10\n', '4\n1 2 1 1\n']
Demo Output:
['3\n', '2\n']
Note:
none
|
```python
evens=[]
odds=[]
n=int(input())
li=list(map(int,input().split(" ")))
for x in li:
if x%2==0:
evens.append(x)
else:
odds.append(x)
if len(evens)==1:
print(li.index(evens[0])+1)
else:
print(li.index(odds[0])+1)
```
| 3.9615
|
122
|
A
|
Lucky Division
|
PROGRAMMING
| 1,000
|
[
"brute force",
"number theory"
] | null | null |
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya calls a number almost lucky if it could be evenly divided by some lucky number. Help him find out if the given number *n* is almost lucky.
|
The single line contains an integer *n* (1<=≤<=*n*<=≤<=1000) — the number that needs to be checked.
|
In the only line print "YES" (without the quotes), if number *n* is almost lucky. Otherwise, print "NO" (without the quotes).
|
[
"47\n",
"16\n",
"78\n"
] |
[
"YES\n",
"YES\n",
"NO\n"
] |
Note that all lucky numbers are almost lucky as any number is evenly divisible by itself.
In the first sample 47 is a lucky number. In the second sample 16 is divisible by 4.
| 500
|
[
{
"input": "47",
"output": "YES"
},
{
"input": "16",
"output": "YES"
},
{
"input": "78",
"output": "NO"
},
{
"input": "48",
"output": "YES"
},
{
"input": "100",
"output": "YES"
},
{
"input": "107",
"output": "NO"
},
{
"input": "77",
"output": "YES"
},
{
"input": "477",
"output": "YES"
},
{
"input": "480",
"output": "YES"
},
{
"input": "1",
"output": "NO"
},
{
"input": "3",
"output": "NO"
},
{
"input": "4",
"output": "YES"
},
{
"input": "49",
"output": "YES"
},
{
"input": "56",
"output": "YES"
},
{
"input": "124",
"output": "YES"
},
{
"input": "1000",
"output": "YES"
},
{
"input": "999",
"output": "NO"
},
{
"input": "298",
"output": "NO"
},
{
"input": "274",
"output": "NO"
},
{
"input": "998",
"output": "NO"
},
{
"input": "42",
"output": "YES"
},
{
"input": "788",
"output": "YES"
},
{
"input": "70",
"output": "YES"
},
{
"input": "444",
"output": "YES"
},
{
"input": "777",
"output": "YES"
},
{
"input": "799",
"output": "YES"
},
{
"input": "25",
"output": "NO"
},
{
"input": "882",
"output": "YES"
},
{
"input": "88",
"output": "YES"
},
{
"input": "11",
"output": "NO"
},
{
"input": "2",
"output": "NO"
},
{
"input": "7",
"output": "YES"
},
{
"input": "8",
"output": "YES"
},
{
"input": "94",
"output": "YES"
},
{
"input": "477",
"output": "YES"
},
{
"input": "141",
"output": "YES"
}
] | 1,695,417,016
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 36
| 186
| 2,867,200
|
import sys
import math
import bisect
import re
def input(): return sys.stdin.readline().rstrip("\r\n")
def main():
#length = int(input())
n = int(input())
#string = input()
ans = 0
#numbers = [*map(int, input().split())]
lucky = [4,7,44,47,74,77,444,447,474,477,744,747,777]
print(["NO","YES"][any(n%i==0 for i in lucky)])
if __name__ == "__main__":
main()
|
Title: Lucky Division
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya calls a number almost lucky if it could be evenly divided by some lucky number. Help him find out if the given number *n* is almost lucky.
Input Specification:
The single line contains an integer *n* (1<=≤<=*n*<=≤<=1000) — the number that needs to be checked.
Output Specification:
In the only line print "YES" (without the quotes), if number *n* is almost lucky. Otherwise, print "NO" (without the quotes).
Demo Input:
['47\n', '16\n', '78\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
Note that all lucky numbers are almost lucky as any number is evenly divisible by itself.
In the first sample 47 is a lucky number. In the second sample 16 is divisible by 4.
|
```python
import sys
import math
import bisect
import re
def input(): return sys.stdin.readline().rstrip("\r\n")
def main():
#length = int(input())
n = int(input())
#string = input()
ans = 0
#numbers = [*map(int, input().split())]
lucky = [4,7,44,47,74,77,444,447,474,477,744,747,777]
print(["NO","YES"][any(n%i==0 for i in lucky)])
if __name__ == "__main__":
main()
```
| 3
|
|
822
|
A
|
I'm bored with life
|
PROGRAMMING
| 800
|
[
"implementation",
"math",
"number theory"
] | null | null |
Holidays have finished. Thanks to the help of the hacker Leha, Noora managed to enter the university of her dreams which is located in a town Pavlopolis. It's well known that universities provide students with dormitory for the period of university studies. Consequently Noora had to leave Vičkopolis and move to Pavlopolis. Thus Leha was left completely alone in a quiet town Vičkopolis. He almost even fell into a depression from boredom!
Leha came up with a task for himself to relax a little. He chooses two integers *A* and *B* and then calculates the greatest common divisor of integers "*A* factorial" and "*B* factorial". Formally the hacker wants to find out GCD(*A*!,<=*B*!). It's well known that the factorial of an integer *x* is a product of all positive integers less than or equal to *x*. Thus *x*!<==<=1·2·3·...·(*x*<=-<=1)·*x*. For example 4!<==<=1·2·3·4<==<=24. Recall that GCD(*x*,<=*y*) is the largest positive integer *q* that divides (without a remainder) both *x* and *y*.
Leha has learned how to solve this task very effective. You are able to cope with it not worse, aren't you?
|
The first and single line contains two integers *A* and *B* (1<=≤<=*A*,<=*B*<=≤<=109,<=*min*(*A*,<=*B*)<=≤<=12).
|
Print a single integer denoting the greatest common divisor of integers *A*! and *B*!.
|
[
"4 3\n"
] |
[
"6\n"
] |
Consider the sample.
4! = 1·2·3·4 = 24. 3! = 1·2·3 = 6. The greatest common divisor of integers 24 and 6 is exactly 6.
| 500
|
[
{
"input": "4 3",
"output": "6"
},
{
"input": "10 399603090",
"output": "3628800"
},
{
"input": "6 973151934",
"output": "720"
},
{
"input": "2 841668075",
"output": "2"
},
{
"input": "7 415216919",
"output": "5040"
},
{
"input": "3 283733059",
"output": "6"
},
{
"input": "11 562314608",
"output": "39916800"
},
{
"input": "3 990639260",
"output": "6"
},
{
"input": "11 859155400",
"output": "39916800"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "5 3",
"output": "6"
},
{
"input": "1 4",
"output": "1"
},
{
"input": "5 4",
"output": "24"
},
{
"input": "1 12",
"output": "1"
},
{
"input": "9 7",
"output": "5040"
},
{
"input": "2 3",
"output": "2"
},
{
"input": "6 11",
"output": "720"
},
{
"input": "6 7",
"output": "720"
},
{
"input": "11 11",
"output": "39916800"
},
{
"input": "4 999832660",
"output": "24"
},
{
"input": "7 999228288",
"output": "5040"
},
{
"input": "11 999257105",
"output": "39916800"
},
{
"input": "11 999286606",
"output": "39916800"
},
{
"input": "3 999279109",
"output": "6"
},
{
"input": "999632727 11",
"output": "39916800"
},
{
"input": "999625230 7",
"output": "5040"
},
{
"input": "999617047 3",
"output": "6"
},
{
"input": "999646548 7",
"output": "5040"
},
{
"input": "999639051 3",
"output": "6"
},
{
"input": "12 12",
"output": "479001600"
},
{
"input": "12 1",
"output": "1"
},
{
"input": "1213 5",
"output": "120"
},
{
"input": "8 9",
"output": "40320"
},
{
"input": "12 9",
"output": "362880"
},
{
"input": "12 1000000000",
"output": "479001600"
},
{
"input": "1000000000 1",
"output": "1"
},
{
"input": "12 13",
"output": "479001600"
},
{
"input": "2 29845",
"output": "2"
},
{
"input": "10 21",
"output": "3628800"
},
{
"input": "12 20",
"output": "479001600"
},
{
"input": "15 12",
"output": "479001600"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1000000000 12",
"output": "479001600"
},
{
"input": "11 30",
"output": "39916800"
},
{
"input": "17 12",
"output": "479001600"
},
{
"input": "4 19",
"output": "24"
},
{
"input": "12 15",
"output": "479001600"
},
{
"input": "20 6",
"output": "720"
},
{
"input": "10 20",
"output": "3628800"
},
{
"input": "10 10",
"output": "3628800"
},
{
"input": "22 12",
"output": "479001600"
},
{
"input": "20 12",
"output": "479001600"
},
{
"input": "12 23",
"output": "479001600"
},
{
"input": "12 22",
"output": "479001600"
},
{
"input": "18 3",
"output": "6"
},
{
"input": "14 10",
"output": "3628800"
},
{
"input": "14 12",
"output": "479001600"
},
{
"input": "8 3",
"output": "6"
},
{
"input": "5 5",
"output": "120"
}
] | 1,668,577,187
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 59
| 46
| 0
|
import math
a,b=map(int,input().split())
x=min(a,b)
print(math.factorial(x))
|
Title: I'm bored with life
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Holidays have finished. Thanks to the help of the hacker Leha, Noora managed to enter the university of her dreams which is located in a town Pavlopolis. It's well known that universities provide students with dormitory for the period of university studies. Consequently Noora had to leave Vičkopolis and move to Pavlopolis. Thus Leha was left completely alone in a quiet town Vičkopolis. He almost even fell into a depression from boredom!
Leha came up with a task for himself to relax a little. He chooses two integers *A* and *B* and then calculates the greatest common divisor of integers "*A* factorial" and "*B* factorial". Formally the hacker wants to find out GCD(*A*!,<=*B*!). It's well known that the factorial of an integer *x* is a product of all positive integers less than or equal to *x*. Thus *x*!<==<=1·2·3·...·(*x*<=-<=1)·*x*. For example 4!<==<=1·2·3·4<==<=24. Recall that GCD(*x*,<=*y*) is the largest positive integer *q* that divides (without a remainder) both *x* and *y*.
Leha has learned how to solve this task very effective. You are able to cope with it not worse, aren't you?
Input Specification:
The first and single line contains two integers *A* and *B* (1<=≤<=*A*,<=*B*<=≤<=109,<=*min*(*A*,<=*B*)<=≤<=12).
Output Specification:
Print a single integer denoting the greatest common divisor of integers *A*! and *B*!.
Demo Input:
['4 3\n']
Demo Output:
['6\n']
Note:
Consider the sample.
4! = 1·2·3·4 = 24. 3! = 1·2·3 = 6. The greatest common divisor of integers 24 and 6 is exactly 6.
|
```python
import math
a,b=map(int,input().split())
x=min(a,b)
print(math.factorial(x))
```
| 3
|
|
676
|
C
|
Vasya and String
|
PROGRAMMING
| 1,500
|
[
"binary search",
"dp",
"strings",
"two pointers"
] | null | null |
High school student Vasya got a string of length *n* as a birthday present. This string consists of letters 'a' and 'b' only. Vasya denotes beauty of the string as the maximum length of a substring (consecutive subsequence) consisting of equal letters.
Vasya can change no more than *k* characters of the original string. What is the maximum beauty of the string he can achieve?
|
The first line of the input contains two integers *n* and *k* (1<=≤<=*n*<=≤<=100<=000,<=0<=≤<=*k*<=≤<=*n*) — the length of the string and the maximum number of characters to change.
The second line contains the string, consisting of letters 'a' and 'b' only.
|
Print the only integer — the maximum beauty of the string Vasya can achieve by changing no more than *k* characters.
|
[
"4 2\nabba\n",
"8 1\naabaabaa\n"
] |
[
"4\n",
"5\n"
] |
In the first sample, Vasya can obtain both strings "aaaa" and "bbbb".
In the second sample, the optimal answer is obtained with the string "aaaaabaa" or with the string "aabaaaaa".
| 1,500
|
[
{
"input": "4 2\nabba",
"output": "4"
},
{
"input": "8 1\naabaabaa",
"output": "5"
},
{
"input": "1 0\na",
"output": "1"
},
{
"input": "1 1\nb",
"output": "1"
},
{
"input": "1 0\nb",
"output": "1"
},
{
"input": "1 1\na",
"output": "1"
},
{
"input": "10 10\nbbbbbbbbbb",
"output": "10"
},
{
"input": "10 2\nbbbbbbbbbb",
"output": "10"
},
{
"input": "10 1\nbbabbabbba",
"output": "6"
},
{
"input": "10 10\nbbabbbaabb",
"output": "10"
},
{
"input": "10 9\nbabababbba",
"output": "10"
},
{
"input": "10 4\nbababbaaab",
"output": "9"
},
{
"input": "10 10\naabaaabaaa",
"output": "10"
},
{
"input": "10 10\naaaabbbaaa",
"output": "10"
},
{
"input": "10 1\nbaaaaaaaab",
"output": "9"
},
{
"input": "10 5\naaaaabaaaa",
"output": "10"
},
{
"input": "10 4\naaaaaaaaaa",
"output": "10"
},
{
"input": "100 10\nbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb",
"output": "100"
},
{
"input": "100 7\nbbbbabbbbbaabbbabbbbbbbbbbbabbbbbbbbbbbbbbbbbbbbbbbbbabbbbbbbbbbbabbabbbbbbbbbbbbbbbbbbbbbbbbbbbbbab",
"output": "93"
},
{
"input": "100 30\nbbaabaaabbbbbbbbbbaababababbbbbbaabaabbbbbbbbabbbbbabbbbabbbbbbbbaabbbbbbbbbabbbbbabbbbbbbbbaaaaabba",
"output": "100"
},
{
"input": "100 6\nbaababbbaabbabbaaabbabbaabbbbbbbbaabbbabbbbaabbabbbbbabababbbbabbbbbbabbbbbbbbbaaaabbabbbbaabbabaabb",
"output": "34"
},
{
"input": "100 45\naabababbabbbaaabbbbbbaabbbabbaabbbbbabbbbbbbbabbbbbbabbaababbaabbababbbbbbababbbbbaabbbbbbbaaaababab",
"output": "100"
},
{
"input": "100 2\nababaabababaaababbaaaabbaabbbababbbaaabbbbabababbbabababaababaaabaabbbbaaabbbabbbbbabbbbbbbaabbabbba",
"output": "17"
},
{
"input": "100 25\nbabbbaaababaaabbbaabaabaabbbabbabbbbaaaaaaabaaabaaaaaaaaaabaaaabaaabbbaaabaaababaaabaabbbbaaaaaaaaaa",
"output": "80"
},
{
"input": "100 14\naabaaaaabababbabbabaaaabbaaaabaaabbbaaabaaaaaaaabaaaaabbaaaaaaaaabaaaaaaabbaababaaaababbbbbabaaaabaa",
"output": "61"
},
{
"input": "100 8\naaaaabaaaabaabaaaaaaaabaaaabaaaaaaaaaaaaaabaaaaabaaaaaaaaaaaaaaaaabaaaababaabaaaaaaaaaaaaabbabaaaaaa",
"output": "76"
},
{
"input": "100 12\naaaaaaaaaaaaaaaabaaabaaaaaaaaaabbaaaabbabaaaaaaaaaaaaaaaaaaaaabbaaabaaaaaaaaaaaabaaaaaaaabaaaaaaaaaa",
"output": "100"
},
{
"input": "100 65\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "100"
},
{
"input": "10 0\nbbbbbbbbbb",
"output": "10"
},
{
"input": "10 0\nbbbbabbbbb",
"output": "5"
},
{
"input": "10 0\nbbabbbabba",
"output": "3"
},
{
"input": "10 0\nbaabbbbaba",
"output": "4"
},
{
"input": "10 0\naababbbbaa",
"output": "4"
},
{
"input": "10 2\nabbbbbaaba",
"output": "8"
},
{
"input": "10 0\nabbaaabaaa",
"output": "3"
},
{
"input": "10 0\naabbaaabaa",
"output": "3"
},
{
"input": "10 1\naaaaaababa",
"output": "8"
},
{
"input": "10 0\nbaaaaaaaaa",
"output": "9"
},
{
"input": "10 0\naaaaaaaaaa",
"output": "10"
},
{
"input": "100 0\nbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb",
"output": "100"
},
{
"input": "100 0\nbbbbbbbbbbabbbbaaabbbbbbbbbbbabbbabbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbabbbbbbbbbabbbbbbbbbbbbbab",
"output": "40"
},
{
"input": "100 11\nbaabbbbbababbbbabbbbbbbabbbbbbbbbbbbbbabbbbbbababbbbababbbbaaabbbbabbbbbabbbbbbbbabababbbabbbbbbbabb",
"output": "65"
},
{
"input": "100 8\nbbababbbbbaabbbaaababbbbababababbbbababbabbbabbbbbaabbbabbbababbabbbbabbbabbbbaabbbbabbbaabbbbaaaabb",
"output": "33"
},
{
"input": "100 21\nabbaaaabbbababaabbbababbbbbbbbabbaababababbbabbbaaabbaaabbbbabbabbbabbbabaababbbabbbbbabbbbbbabbbbab",
"output": "65"
},
{
"input": "100 9\nabbbaabaabaaaaaaabbabbbababbaaabbbaaabbaabaaaaabbbbbabbaabaabbbbbaaaaababbaaabbabaabaaabababbaababbb",
"output": "26"
},
{
"input": "100 5\naababababbaaaaaaaabbbabaaaabbabaaaabbaabaaaaabababbabaabaaabaaaaaaaabaababbabbaaabaabbabbaaaaabbabba",
"output": "22"
},
{
"input": "100 9\naababaabaaaaaaaaabbbaabaaaaaaabaaaaaaaaaaaaabaaabaabaabbbbabbaababbabbaaaabbababaabaababaabaaaaaaaaa",
"output": "49"
},
{
"input": "100 6\naaaaabbaaaaaaaaaaabaaaabaaaaaaaaabaaabaaaaaabaaaaaaaaaaabaabaaaabaaaaaaaaaaaaaaabaabbaaaaaaaaaaaaaaa",
"output": "56"
},
{
"input": "100 7\nabaaabaabaabaaaaaabaaaaaaaaaaaaaaaaaaaaaaaaaaaaabaaabaaaaaaabbabaaaaaaaaaaaaaaaaabaaaaaaaaaaaaaaaaba",
"output": "86"
},
{
"input": "100 0\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "100"
}
] | 1,677,237,495
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 117
| 140
| 204,800
|
n, k = map(int, input().split())
s = input()
dp = [0, 0]
ans = 0
for i in range(n):
dp[ord(s[i]) - ord('a')] += 1
if min(dp) > k:
dp[ord(s[i - ans]) - ord('a')] -= 1
else:
ans += 1
print(ans)
|
Title: Vasya and String
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
High school student Vasya got a string of length *n* as a birthday present. This string consists of letters 'a' and 'b' only. Vasya denotes beauty of the string as the maximum length of a substring (consecutive subsequence) consisting of equal letters.
Vasya can change no more than *k* characters of the original string. What is the maximum beauty of the string he can achieve?
Input Specification:
The first line of the input contains two integers *n* and *k* (1<=≤<=*n*<=≤<=100<=000,<=0<=≤<=*k*<=≤<=*n*) — the length of the string and the maximum number of characters to change.
The second line contains the string, consisting of letters 'a' and 'b' only.
Output Specification:
Print the only integer — the maximum beauty of the string Vasya can achieve by changing no more than *k* characters.
Demo Input:
['4 2\nabba\n', '8 1\naabaabaa\n']
Demo Output:
['4\n', '5\n']
Note:
In the first sample, Vasya can obtain both strings "aaaa" and "bbbb".
In the second sample, the optimal answer is obtained with the string "aaaaabaa" or with the string "aabaaaaa".
|
```python
n, k = map(int, input().split())
s = input()
dp = [0, 0]
ans = 0
for i in range(n):
dp[ord(s[i]) - ord('a')] += 1
if min(dp) > k:
dp[ord(s[i - ans]) - ord('a')] -= 1
else:
ans += 1
print(ans)
```
| 3
|
|
268
|
A
|
Games
|
PROGRAMMING
| 800
|
[
"brute force"
] | null | null |
Manao works on a sports TV. He's spent much time watching the football games of some country. After a while he began to notice different patterns. For example, each team has two sets of uniforms: home uniform and guest uniform. When a team plays a game at home, the players put on the home uniform. When a team plays as a guest on somebody else's stadium, the players put on the guest uniform. The only exception to that rule is: when the home uniform color of the host team matches the guests' uniform, the host team puts on its guest uniform as well. For each team the color of the home and guest uniform is different.
There are *n* teams taking part in the national championship. The championship consists of *n*·(*n*<=-<=1) games: each team invites each other team to its stadium. At this point Manao wondered: how many times during the championship is a host team going to put on the guest uniform? Note that the order of the games does not affect this number.
You know the colors of the home and guest uniform for each team. For simplicity, the colors are numbered by integers in such a way that no two distinct colors have the same number. Help Manao find the answer to his question.
|
The first line contains an integer *n* (2<=≤<=*n*<=≤<=30). Each of the following *n* lines contains a pair of distinct space-separated integers *h**i*, *a**i* (1<=≤<=*h**i*,<=*a**i*<=≤<=100) — the colors of the *i*-th team's home and guest uniforms, respectively.
|
In a single line print the number of games where the host team is going to play in the guest uniform.
|
[
"3\n1 2\n2 4\n3 4\n",
"4\n100 42\n42 100\n5 42\n100 5\n",
"2\n1 2\n1 2\n"
] |
[
"1\n",
"5\n",
"0\n"
] |
In the first test case the championship consists of 6 games. The only game with the event in question is the game between teams 2 and 1 on the stadium of team 2.
In the second test sample the host team will have to wear guest uniform in the games between teams: 1 and 2, 2 and 1, 2 and 3, 3 and 4, 4 and 2 (the host team is written first).
| 500
|
[
{
"input": "3\n1 2\n2 4\n3 4",
"output": "1"
},
{
"input": "4\n100 42\n42 100\n5 42\n100 5",
"output": "5"
},
{
"input": "2\n1 2\n1 2",
"output": "0"
},
{
"input": "7\n4 7\n52 55\n16 4\n55 4\n20 99\n3 4\n7 52",
"output": "6"
},
{
"input": "10\n68 42\n1 35\n25 70\n59 79\n65 63\n46 6\n28 82\n92 62\n43 96\n37 28",
"output": "1"
},
{
"input": "30\n10 39\n89 1\n78 58\n75 99\n36 13\n77 50\n6 97\n79 28\n27 52\n56 5\n93 96\n40 21\n33 74\n26 37\n53 59\n98 56\n61 65\n42 57\n9 7\n25 63\n74 34\n96 84\n95 47\n12 23\n34 21\n71 6\n27 13\n15 47\n64 14\n12 77",
"output": "6"
},
{
"input": "30\n46 100\n87 53\n34 84\n44 66\n23 20\n50 34\n90 66\n17 39\n13 22\n94 33\n92 46\n63 78\n26 48\n44 61\n3 19\n41 84\n62 31\n65 89\n23 28\n58 57\n19 85\n26 60\n75 66\n69 67\n76 15\n64 15\n36 72\n90 89\n42 69\n45 35",
"output": "4"
},
{
"input": "2\n46 6\n6 46",
"output": "2"
},
{
"input": "29\n8 18\n33 75\n69 22\n97 95\n1 97\n78 10\n88 18\n13 3\n19 64\n98 12\n79 92\n41 72\n69 15\n98 31\n57 74\n15 56\n36 37\n15 66\n63 100\n16 42\n47 56\n6 4\n73 15\n30 24\n27 71\n12 19\n88 69\n85 6\n50 11",
"output": "10"
},
{
"input": "23\n43 78\n31 28\n58 80\n66 63\n20 4\n51 95\n40 20\n50 14\n5 34\n36 39\n77 42\n64 97\n62 89\n16 56\n8 34\n58 16\n37 35\n37 66\n8 54\n50 36\n24 8\n68 48\n85 33",
"output": "6"
},
{
"input": "13\n76 58\n32 85\n99 79\n23 58\n96 59\n72 35\n53 43\n96 55\n41 78\n75 10\n28 11\n72 7\n52 73",
"output": "0"
},
{
"input": "18\n6 90\n70 79\n26 52\n67 81\n29 95\n41 32\n94 88\n18 58\n59 65\n51 56\n64 68\n34 2\n6 98\n95 82\n34 2\n40 98\n83 78\n29 2",
"output": "1"
},
{
"input": "18\n6 90\n100 79\n26 100\n67 100\n29 100\n100 32\n94 88\n18 58\n59 65\n51 56\n64 68\n34 2\n6 98\n95 82\n34 2\n40 98\n83 78\n29 100",
"output": "8"
},
{
"input": "30\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1",
"output": "450"
},
{
"input": "30\n100 99\n58 59\n56 57\n54 55\n52 53\n50 51\n48 49\n46 47\n44 45\n42 43\n40 41\n38 39\n36 37\n34 35\n32 33\n30 31\n28 29\n26 27\n24 25\n22 23\n20 21\n18 19\n16 17\n14 15\n12 13\n10 11\n8 9\n6 7\n4 5\n2 3",
"output": "0"
},
{
"input": "15\n9 3\n2 6\n7 6\n5 10\n9 5\n8 1\n10 5\n2 8\n4 5\n9 8\n5 3\n3 8\n9 8\n4 10\n8 5",
"output": "20"
},
{
"input": "15\n2 1\n1 2\n1 2\n1 2\n2 1\n2 1\n2 1\n1 2\n2 1\n2 1\n2 1\n1 2\n2 1\n2 1\n1 2",
"output": "108"
},
{
"input": "25\n2 1\n1 2\n1 2\n1 2\n2 1\n1 2\n1 2\n1 2\n2 1\n2 1\n2 1\n1 2\n1 2\n1 2\n2 1\n2 1\n2 1\n1 2\n2 1\n1 2\n2 1\n2 1\n2 1\n2 1\n1 2",
"output": "312"
},
{
"input": "25\n91 57\n2 73\n54 57\n2 57\n23 57\n2 6\n57 54\n57 23\n91 54\n91 23\n57 23\n91 57\n54 2\n6 91\n57 54\n2 57\n57 91\n73 91\n57 23\n91 57\n2 73\n91 2\n23 6\n2 73\n23 6",
"output": "96"
},
{
"input": "28\n31 66\n31 91\n91 31\n97 66\n31 66\n31 66\n66 91\n91 31\n97 31\n91 97\n97 31\n66 31\n66 97\n91 31\n31 66\n31 66\n66 31\n31 97\n66 97\n97 31\n31 91\n66 91\n91 66\n31 66\n91 66\n66 31\n66 31\n91 97",
"output": "210"
},
{
"input": "29\n78 27\n50 68\n24 26\n68 43\n38 78\n26 38\n78 28\n28 26\n27 24\n23 38\n24 26\n24 43\n61 50\n38 78\n27 23\n61 26\n27 28\n43 23\n28 78\n43 27\n43 78\n27 61\n28 38\n61 78\n50 26\n43 27\n26 78\n28 50\n43 78",
"output": "73"
},
{
"input": "29\n80 27\n69 80\n27 80\n69 80\n80 27\n80 27\n80 27\n80 69\n27 69\n80 69\n80 27\n27 69\n69 27\n80 69\n27 69\n69 80\n27 69\n80 69\n80 27\n69 27\n27 69\n27 80\n80 27\n69 80\n27 69\n80 69\n69 80\n69 80\n27 80",
"output": "277"
},
{
"input": "30\n19 71\n7 89\n89 71\n21 7\n19 21\n7 89\n19 71\n89 8\n89 21\n19 8\n21 7\n8 89\n19 89\n7 21\n19 8\n19 7\n7 19\n8 21\n71 21\n71 89\n7 19\n7 19\n21 7\n21 19\n21 19\n71 8\n21 8\n71 19\n19 71\n8 21",
"output": "154"
},
{
"input": "30\n44 17\n44 17\n44 17\n17 44\n44 17\n44 17\n17 44\n17 44\n17 44\n44 17\n44 17\n44 17\n44 17\n44 17\n17 44\n17 44\n17 44\n44 17\n44 17\n17 44\n44 17\n44 17\n44 17\n17 44\n17 44\n44 17\n17 44\n44 17\n44 17\n44 17",
"output": "418"
},
{
"input": "22\n78 92\n15 92\n92 78\n78 80\n92 16\n24 80\n92 16\n16 92\n78 16\n24 78\n80 78\n92 80\n16 80\n80 78\n15 78\n92 16\n24 15\n24 80\n80 16\n16 80\n92 80\n24 80",
"output": "74"
},
{
"input": "24\n9 83\n90 31\n83 3\n83 3\n21 31\n83 3\n32 31\n12 21\n31 21\n90 32\n32 21\n12 9\n12 31\n9 83\n83 12\n32 3\n32 83\n90 31\n9 32\n31 21\n83 90\n32 21\n21 3\n32 9",
"output": "59"
},
{
"input": "30\n67 21\n85 39\n85 87\n21 39\n66 85\n10 95\n10 21\n87 85\n82 21\n67 21\n95 10\n21 39\n82 21\n21 66\n66 39\n95 30\n67 85\n66 82\n85 82\n21 66\n10 39\n67 10\n21 85\n10 82\n85 95\n10 85\n21 39\n85 39\n39 10\n95 67",
"output": "100"
},
{
"input": "4\n8 7\n8 7\n7 8\n7 8",
"output": "8"
},
{
"input": "6\n1 2\n1 2\n1 2\n1 2\n1 2\n2 1",
"output": "10"
},
{
"input": "12\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1",
"output": "72"
},
{
"input": "4\n1 2\n1 2\n2 1\n2 1",
"output": "8"
}
] | 1,681,126,713
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 31
| 62
| 0
|
n = int(input())
team = []
res = 0
for _ in range(n):
team.append(input().split())
for i, array in enumerate(team):
color = array[0]
for j, elem in enumerate(team):
if i != j and elem[1] == color:
res += 1
print(res)
|
Title: Games
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Manao works on a sports TV. He's spent much time watching the football games of some country. After a while he began to notice different patterns. For example, each team has two sets of uniforms: home uniform and guest uniform. When a team plays a game at home, the players put on the home uniform. When a team plays as a guest on somebody else's stadium, the players put on the guest uniform. The only exception to that rule is: when the home uniform color of the host team matches the guests' uniform, the host team puts on its guest uniform as well. For each team the color of the home and guest uniform is different.
There are *n* teams taking part in the national championship. The championship consists of *n*·(*n*<=-<=1) games: each team invites each other team to its stadium. At this point Manao wondered: how many times during the championship is a host team going to put on the guest uniform? Note that the order of the games does not affect this number.
You know the colors of the home and guest uniform for each team. For simplicity, the colors are numbered by integers in such a way that no two distinct colors have the same number. Help Manao find the answer to his question.
Input Specification:
The first line contains an integer *n* (2<=≤<=*n*<=≤<=30). Each of the following *n* lines contains a pair of distinct space-separated integers *h**i*, *a**i* (1<=≤<=*h**i*,<=*a**i*<=≤<=100) — the colors of the *i*-th team's home and guest uniforms, respectively.
Output Specification:
In a single line print the number of games where the host team is going to play in the guest uniform.
Demo Input:
['3\n1 2\n2 4\n3 4\n', '4\n100 42\n42 100\n5 42\n100 5\n', '2\n1 2\n1 2\n']
Demo Output:
['1\n', '5\n', '0\n']
Note:
In the first test case the championship consists of 6 games. The only game with the event in question is the game between teams 2 and 1 on the stadium of team 2.
In the second test sample the host team will have to wear guest uniform in the games between teams: 1 and 2, 2 and 1, 2 and 3, 3 and 4, 4 and 2 (the host team is written first).
|
```python
n = int(input())
team = []
res = 0
for _ in range(n):
team.append(input().split())
for i, array in enumerate(team):
color = array[0]
for j, elem in enumerate(team):
if i != j and elem[1] == color:
res += 1
print(res)
```
| 3
|
|
224
|
A
|
Parallelepiped
|
PROGRAMMING
| 1,100
|
[
"brute force",
"geometry",
"math"
] | null | null |
You've got a rectangular parallelepiped with integer edge lengths. You know the areas of its three faces that have a common vertex. Your task is to find the sum of lengths of all 12 edges of this parallelepiped.
|
The first and the single line contains three space-separated integers — 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"
}
] | 1,648,293,353
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 27
| 62
| 0
|
import math
data = input().split()
A1, A2, A3 = int(data[0]), int(data[1]), int(data[2])
H = int(math.sqrt((A2 * A3) // A1))
W = A3 // H
L = A1 // W
print(4*L + 4*W + 4*H)
|
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
import math
data = input().split()
A1, A2, A3 = int(data[0]), int(data[1]), int(data[2])
H = int(math.sqrt((A2 * A3) // A1))
W = A3 // H
L = A1 // W
print(4*L + 4*W + 4*H)
```
| 3
|
|
804
|
B
|
Minimum number of steps
|
PROGRAMMING
| 1,400
|
[
"combinatorics",
"greedy",
"implementation",
"math"
] | null | null |
We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of substrings "ab" in the string and replace it with the string "bba". If we have no "ab" as a substring, our job is done. Print the minimum number of steps we should perform to make our job done modulo 109<=+<=7.
The string "ab" appears as a substring if there is a letter 'b' right after the letter 'a' somewhere in the string.
|
The first line contains the initial string consisting of letters 'a' and 'b' only with length from 1 to 106.
|
Print the minimum number of steps modulo 109<=+<=7.
|
[
"ab\n",
"aab\n"
] |
[
"1\n",
"3\n"
] |
The first example: "ab" → "bba".
The second example: "aab" → "abba" → "bbaba" → "bbbbaa".
| 1,000
|
[
{
"input": "ab",
"output": "1"
},
{
"input": "aab",
"output": "3"
},
{
"input": "aaaaabaabababaaaaaba",
"output": "17307"
},
{
"input": "abaabaaabbabaabab",
"output": "1795"
},
{
"input": "abbaa",
"output": "2"
},
{
"input": "abbaaabaabaaaaabbbbaababaaaaabaabbaaaaabbaabbaaaabbbabbbabb",
"output": "690283580"
},
{
"input": "aababbaaaabbaabbbbbbbbabbababbbaaabbaaabbabbba",
"output": "2183418"
},
{
"input": "aabbaababbabbbaabbaababaaaabbaaaabaaaaaababbaaaabaababbabbbb",
"output": "436420225"
},
{
"input": "aaabaaaabbababbaabbababbbbaaaaaaabbabbba",
"output": "8431094"
},
{
"input": "abbbbababbabbbbbabaabbbaabbbbbbbaaab",
"output": "8180"
},
{
"input": "bbababbbaabaaaaaaaabbabbbb",
"output": "40979"
},
{
"input": "abbbaaabbbbbabaabbaaabbbababbbaabaabababababa",
"output": "2065758"
},
{
"input": "abaaaaaabaaaabbabbaaabbbbabababaaaaabbaabbaaaaabbbaababaaaaaaabbbbbaaaaabaababbabababbabbbbaabbaabbabbbabaabbaabbaaaaaab",
"output": "235606597"
},
{
"input": "abbbbbbbbbbbbbbbbbbbbbbbbbbaababaaaaaaabaabaaababaabaababaaabababaababab",
"output": "7"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbaaaaaaaaabaabaaababaabaababaaabababaabbbbbbb",
"output": "557763786"
}
] | 1,668,698,134
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 32
| 327
| 2,867,200
|
ab = input()
value = 1
count = 0
mod = (10 ** 9) + 7
for e in ab:
if e == 'b':
count = (value + count - 1) % mod
else:
value = (value * 2) % mod
print(count)
|
Title: Minimum number of steps
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of substrings "ab" in the string and replace it with the string "bba". If we have no "ab" as a substring, our job is done. Print the minimum number of steps we should perform to make our job done modulo 109<=+<=7.
The string "ab" appears as a substring if there is a letter 'b' right after the letter 'a' somewhere in the string.
Input Specification:
The first line contains the initial string consisting of letters 'a' and 'b' only with length from 1 to 106.
Output Specification:
Print the minimum number of steps modulo 109<=+<=7.
Demo Input:
['ab\n', 'aab\n']
Demo Output:
['1\n', '3\n']
Note:
The first example: "ab" → "bba".
The second example: "aab" → "abba" → "bbaba" → "bbbbaa".
|
```python
ab = input()
value = 1
count = 0
mod = (10 ** 9) + 7
for e in ab:
if e == 'b':
count = (value + count - 1) % mod
else:
value = (value * 2) % mod
print(count)
```
| 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"
}
] | 1,621,378,004
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 58
| 77
| 0
|
cadena = input()
cadArr = [char for char in cadena]
search = 'heidi'
found = False
for i in cadArr:
if len(search)==0:
yes = print('YES')
found = True
break
cAct = search[0]
if i == cAct:
s1 = search[1:]
search = s1
if len(search) != 0 and found==False:
print('NO')
elif found == False and len(search) == 0:
print('YES')
|
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
cadena = input()
cadArr = [char for char in cadena]
search = 'heidi'
found = False
for i in cadArr:
if len(search)==0:
yes = print('YES')
found = True
break
cAct = search[0]
if i == cAct:
s1 = search[1:]
search = s1
if len(search) != 0 and found==False:
print('NO')
elif found == False and len(search) == 0:
print('YES')
```
| 3
|
|
554
|
A
|
Kyoya and Photobooks
|
PROGRAMMING
| 900
|
[
"brute force",
"math",
"strings"
] | null | null |
Kyoya Ootori is selling photobooks of the Ouran High School Host Club. He has 26 photos, labeled "a" to "z", and he has compiled them into a photo booklet with some photos in some order (possibly with some photos being duplicated). A photo booklet can be described as a string of lowercase letters, consisting of the photos in the booklet in order. He now wants to sell some "special edition" photobooks, each with one extra photo inserted anywhere in the book. He wants to make as many distinct photobooks as possible, so he can make more money. He asks Haruhi, how many distinct photobooks can he make by inserting one extra photo into the photobook he already has?
Please help Haruhi solve this problem.
|
The first line of input will be a single string *s* (1<=≤<=|*s*|<=≤<=20). String *s* consists only of lowercase English letters.
|
Output a single integer equal to the number of distinct photobooks Kyoya Ootori can make.
|
[
"a\n",
"hi\n"
] |
[
"51\n",
"76\n"
] |
In the first case, we can make 'ab','ac',...,'az','ba','ca',...,'za', and 'aa', producing a total of 51 distinct photo booklets.
| 250
|
[
{
"input": "a",
"output": "51"
},
{
"input": "hi",
"output": "76"
},
{
"input": "y",
"output": "51"
},
{
"input": "kgan",
"output": "126"
},
{
"input": "zoabkyuvus",
"output": "276"
},
{
"input": "spyemhyznjieyhhbk",
"output": "451"
},
{
"input": "xulsyfkuizjauadjjopu",
"output": "526"
},
{
"input": "e",
"output": "51"
},
{
"input": "zv",
"output": "76"
},
{
"input": "jgv",
"output": "101"
},
{
"input": "zsfo",
"output": "126"
},
{
"input": "jselr",
"output": "151"
},
{
"input": "dwemig",
"output": "176"
},
{
"input": "mddoxsf",
"output": "201"
},
{
"input": "jgirkrmi",
"output": "226"
},
{
"input": "spkxurcum",
"output": "251"
},
{
"input": "fykkiubdkt",
"output": "276"
},
{
"input": "fznbcxsxygs",
"output": "301"
},
{
"input": "qcrvrdqcbtou",
"output": "326"
},
{
"input": "qktrbjzrqgmlr",
"output": "351"
},
{
"input": "foamodbvptlxxg",
"output": "376"
},
{
"input": "ydzpjhsidipricw",
"output": "401"
},
{
"input": "lpfpndmjfvqejdgf",
"output": "426"
},
{
"input": "ofkvparuvjtggnmab",
"output": "451"
},
{
"input": "xxncfutrtxcwdzwbgs",
"output": "476"
},
{
"input": "zovhffccflkgqncsdte",
"output": "501"
},
{
"input": "cskgsxywlvfeicoueglr",
"output": "526"
},
{
"input": "gggggggggggggggggggg",
"output": "526"
},
{
"input": "qdqdddqddqqddqddqdqd",
"output": "526"
},
{
"input": "takttttaakaaktakttkt",
"output": "526"
},
{
"input": "coccoooogogcgocccmcg",
"output": "526"
},
{
"input": "kskkskkkssksssk",
"output": "401"
},
{
"input": "lllllllllllllll",
"output": "401"
}
] | 1,435,165,445
| 2,045
|
Python 3
|
OK
|
TESTS
| 33
| 62
| 0
|
assd=input()
print(((len(assd)+1)*26-len(assd)))
|
Title: Kyoya and Photobooks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kyoya Ootori is selling photobooks of the Ouran High School Host Club. He has 26 photos, labeled "a" to "z", and he has compiled them into a photo booklet with some photos in some order (possibly with some photos being duplicated). A photo booklet can be described as a string of lowercase letters, consisting of the photos in the booklet in order. He now wants to sell some "special edition" photobooks, each with one extra photo inserted anywhere in the book. He wants to make as many distinct photobooks as possible, so he can make more money. He asks Haruhi, how many distinct photobooks can he make by inserting one extra photo into the photobook he already has?
Please help Haruhi solve this problem.
Input Specification:
The first line of input will be a single string *s* (1<=≤<=|*s*|<=≤<=20). String *s* consists only of lowercase English letters.
Output Specification:
Output a single integer equal to the number of distinct photobooks Kyoya Ootori can make.
Demo Input:
['a\n', 'hi\n']
Demo Output:
['51\n', '76\n']
Note:
In the first case, we can make 'ab','ac',...,'az','ba','ca',...,'za', and 'aa', producing a total of 51 distinct photo booklets.
|
```python
assd=input()
print(((len(assd)+1)*26-len(assd)))
```
| 3
|
|
580
|
E
|
Kefa and Watch
|
PROGRAMMING
| 2,500
|
[
"data structures",
"hashing",
"strings"
] | null | null |
One day Kefa the parrot was walking down the street as he was on the way home from the restaurant when he saw something glittering by the road. As he came nearer he understood that it was a watch. He decided to take it to the pawnbroker to earn some money.
The pawnbroker said that each watch contains a serial number represented by a string of digits from 0 to 9, and the more quality checks this number passes, the higher is the value of the watch. The check is defined by three positive integers *l*, *r* and *d*. The watches pass a check if a substring of the serial number from *l* to *r* has period *d*. Sometimes the pawnbroker gets distracted and Kefa changes in some substring of the serial number all digits to *c* in order to increase profit from the watch.
The seller has a lot of things to do to begin with and with Kefa messing about, he gave you a task: to write a program that determines the value of the watch.
Let us remind you that number *x* is called a period of string *s* (1<=≤<=*x*<=≤<=|*s*|), if *s**i*<=<==<=<=*s**i*<=+<=*x* for all *i* from 1 to |*s*|<=<=-<=<=*x*.
|
The first line of the input contains three positive integers *n*, *m* and *k* (1<=≤<=*n*<=≤<=105, 1<=≤<=*m*<=+<=*k*<=≤<=105) — the length of the serial number, the number of change made by Kefa and the number of quality checks.
The second line contains a serial number consisting of *n* digits.
Then *m*<=+<=*k* lines follow, containing either checks or changes.
The changes are given as 1 *l* *r* *c* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*, 0<=≤<=*c*<=≤<=9). That means that Kefa changed all the digits from the *l*-th to the *r*-th to be *c*.
The checks are given as 2 *l* *r* *d* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*, 1<=≤<=*d*<=≤<=*r*<=-<=*l*<=+<=1).
|
For each check on a single line print "YES" if the watch passed it, otherwise print "NO".
|
[
"3 1 2\n112\n2 2 3 1\n1 1 3 8\n2 1 2 1\n",
"6 2 3\n334934\n2 2 5 2\n1 4 4 3\n2 1 6 3\n1 2 3 8\n2 3 6 1\n"
] |
[
"NO\nYES\n",
"NO\nYES\nNO\n"
] |
In the first sample test two checks will be made. In the first one substring "12" is checked on whether or not it has period 1, so the answer is "NO". In the second one substring "88", is checked on whether or not it has period 1, and it has this period, so the answer is "YES".
In the second statement test three checks will be made. The first check processes substring "3493", which doesn't have period 2. Before the second check the string looks as "334334", so the answer to it is "YES". And finally, the third check processes substring "8334", which does not have period 1.
| 2,500
|
[
{
"input": "3 1 2\n112\n2 2 3 1\n1 1 3 8\n2 1 2 1",
"output": "NO\nYES"
},
{
"input": "6 2 3\n334934\n2 2 5 2\n1 4 4 3\n2 1 6 3\n1 2 3 8\n2 3 6 1",
"output": "NO\nYES\nNO"
},
{
"input": "1 0 1\n5\n2 1 1 1",
"output": "YES"
},
{
"input": "20 1 2\n34075930750342906718\n2 1 20 20\n1 1 20 6\n2 1 20 1",
"output": "YES\nYES"
},
{
"input": "10 1 4\n4545454545\n2 1 10 2\n2 2 4 2\n2 2 9 4\n1 2 9 6\n2 3 8 3",
"output": "YES\nYES\nYES\nYES"
},
{
"input": "15 1 5\n234072305423089\n2 1 15 1\n2 5 6 1\n2 8 11 2\n2 2 13 6\n1 5 12 4\n2 5 13 3",
"output": "NO\nNO\nNO\nNO\nNO"
},
{
"input": "9 7 5\n622851212\n2 1 9 3\n1 1 4 2\n1 6 9 7\n2 2 8 1\n1 2 3 9\n1 7 8 5\n2 1 9 9\n1 2 3 7\n1 7 7 2\n2 4 9 3\n1 2 2 5\n2 1 9 3",
"output": "NO\nNO\nYES\nYES\nYES"
},
{
"input": "18 0 6\n000000000000000000\n2 1 18 1\n2 1 18 18\n2 1 18 6\n2 1 18 3\n2 1 18 9\n2 1 18 2",
"output": "YES\nYES\nYES\nYES\nYES\nYES"
},
{
"input": "8 3 4\n90925761\n2 5 8 2\n1 2 4 5\n2 2 5 2\n1 6 7 5\n2 2 7 3\n1 3 4 9\n2 1 4 2",
"output": "NO\nYES\nYES\nNO"
},
{
"input": "10 10 7\n8888888888\n1 1 1 4\n1 2 2 5\n1 3 3 7\n1 4 4 7\n1 5 5 7\n1 6 6 7\n1 7 7 5\n1 8 8 6\n1 9 9 3\n1 10 10 7\n2 5 6 1\n2 8 8 1\n2 5 6 1\n2 7 9 3\n2 5 6 1\n2 4 4 1\n2 9 10 1",
"output": "YES\nYES\nYES\nYES\nYES\nYES\nNO"
},
{
"input": "20 5 5\n23655146364900318111\n1 5 19 9\n2 1 3 3\n2 4 5 1\n1 2 17 9\n2 4 5 1\n1 8 9 0\n2 4 5 1\n1 4 15 2\n2 1 3 3\n1 20 20 6",
"output": "YES\nNO\nYES\nYES\nYES"
},
{
"input": "20 10 15\n00137794455431057085\n2 1 20 1\n2 8 10 3\n2 1 20 1\n1 2 2 6\n1 11 13 0\n2 1 20 1\n2 1 2 1\n1 14 16 0\n1 5 9 0\n1 5 8 3\n1 10 11 7\n1 17 19 5\n2 1 20 1\n2 8 10 3\n2 1 20 1\n1 17 20 0\n1 7 10 7\n1 7 12 7\n2 1 20 1\n2 1 2 1\n2 8 10 3\n2 1 2 1\n2 8 10 3\n2 8 10 3\n2 8 10 3",
"output": "NO\nYES\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nNO\nYES\nNO\nYES\nYES\nYES"
},
{
"input": "50 10 9\n78117811831783178317831700000000000000000000117773\n1 5 22 4\n1 11 24 0\n1 35 37 5\n2 45 46 1\n2 45 46 1\n1 41 41 3\n1 24 27 1\n2 9 24 4\n1 2 21 5\n2 45 46 1\n1 3 9 1\n1 11 23 5\n1 25 32 1\n2 47 49 1\n2 9 24 4\n1 34 45 0\n2 9 24 4\n2 1 8 4\n2 9 24 4",
"output": "YES\nYES\nNO\nYES\nYES\nNO\nNO\nNO\nNO"
},
{
"input": "52 5 30\n0073971598462524060181848948785829847120611111998011\n2 43 46 1\n1 25 28 2\n1 1 30 4\n2 1 52 1\n2 1 3 3\n2 1 3 3\n1 11 15 2\n2 1 52 1\n2 43 46 1\n1 3 7 9\n2 1 3 3\n1 26 49 3\n2 1 3 3\n2 1 52 1\n2 43 46 1\n2 1 3 3\n2 1 52 1\n2 1 52 1\n2 1 3 3\n2 1 52 1\n2 1 3 3\n2 1 3 3\n2 1 52 1\n2 1 3 3\n2 1 52 1\n2 43 46 1\n2 1 52 1\n2 43 46 1\n2 1 3 3\n2 43 46 1\n2 43 46 1\n2 43 46 1\n2 1 3 3\n2 1 52 1\n2 43 46 1",
"output": "YES\nNO\nYES\nYES\nNO\nYES\nYES\nYES\nNO\nYES\nYES\nNO\nNO\nYES\nNO\nYES\nYES\nNO\nYES\nNO\nYES\nNO\nYES\nYES\nYES\nYES\nYES\nYES\nNO\nYES"
},
{
"input": "314 0 1\n12121112111122221121111111212111122212111111112211111111111211121121212112222211222222112222121112121112211211111211111221211112111122212121112221111112111111121122122111111211121112111111121112121222222111211212221212111221112121111112111111112111121121121222112211212212121111112112122111112121212111222221111111\n2 1 314 157",
"output": "NO"
},
{
"input": "153 0 16\n000000000961748941961748947961748951961748969961748987961748993961749023961749037961749043961749067961749079961749091961749097961749101961749121961749157\n2 1 18 9\n2 1 27 18\n2 1 36 27\n2 1 45 36\n2 1 54 45\n2 1 63 54\n2 1 72 63\n2 1 81 72\n2 1 90 81\n2 1 99 90\n2 1 108 99\n2 1 117 108\n2 1 126 117\n2 1 135 126\n2 1 144 135\n2 1 153 144",
"output": "NO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO"
},
{
"input": "20 1 1\n52018731676138902386\n2 1 20 10\n1 1 20 8",
"output": "NO"
}
] | 1,697,628,262
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 80
| 763
| 23,449,600
|
if True:
from io import BytesIO, IOBase
import math
import random
import sys
import os
import bisect
import typing
from collections import Counter, defaultdict, deque
from copy import deepcopy
from functools import cmp_to_key, lru_cache, reduce
from heapq import heapify, heappop, heappush, heappushpop, nlargest, nsmallest
from itertools import accumulate, combinations, permutations, count
from operator import add, iand, ior, itemgetter, mul, xor
from string import ascii_lowercase, ascii_uppercase, ascii_letters
from typing import *
BUFSIZE = 4096
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin = IOWrapper(sys.stdin)
sys.stdout = IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def I():
return input()
def II():
return int(input())
def MII():
return map(int, input().split())
def LI():
return list(input().split())
def LII():
return list(map(int, input().split()))
def LFI():
return list(map(float, input().split()))
def GMI():
return map(lambda x: int(x) - 1, input().split())
def LGMI():
return list(map(lambda x: int(x) - 1, input().split()))
inf = float('inf')
dfs, hashing = True, False
if dfs:
from types import GeneratorType
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
if hashing:
RANDOM = random.getrandbits(20)
class Wrapper(int):
def __init__(self, x):
int.__init__(x)
def __hash__(self):
return super(Wrapper, self).__hash__() ^ RANDOM
class LazySegTree:
def __init__(
self,
op: typing.Callable[[typing.Any, typing.Any], typing.Any],
e: typing.Any,
mapping: typing.Callable[[typing.Any, typing.Any], typing.Any],
composition: typing.Callable[[typing.Any, typing.Any], typing.Any],
id_: typing.Any,
v: typing.Union[int, typing.List[typing.Any]]) -> None:
self._op = op
self._e = e
self._mapping = mapping
self._composition = composition
self._id = id_
if isinstance(v, int):
v = [e] * v
self._n = len(v)
self._log = (self._n - 1).bit_length()
self._size = 1 << self._log
self._d = [e] * (2 * self._size)
self._lz = [self._id] * self._size
for i in range(self._n):
self._d[self._size + i] = v[i]
for i in range(self._size - 1, 0, -1):
self._update(i)
def set(self, p: int, x: typing.Any) -> None:
assert 0 <= p < self._n
p += self._size
for i in range(self._log, 0, -1):
self._push(p >> i)
self._d[p] = x
for i in range(1, self._log + 1):
self._update(p >> i)
def get(self, p: int) -> typing.Any:
assert 0 <= p < self._n
p += self._size
for i in range(self._log, 0, -1):
self._push(p >> i)
return self._d[p]
def prod(self, left: int, right: int) -> typing.Any:
assert 0 <= left <= right <= self._n
if left == right:
return self._e
left += self._size
right += self._size
for i in range(self._log, 0, -1):
if ((left >> i) << i) != left:
self._push(left >> i)
if ((right >> i) << i) != right:
self._push(right >> i)
sml = self._e
smr = self._e
while left < right:
if left & 1:
sml = self._op(sml, self._d[left])
left += 1
if right & 1:
right -= 1
smr = self._op(self._d[right], smr)
left >>= 1
right >>= 1
return self._op(sml, smr)
def all_prod(self) -> typing.Any:
return self._d[1]
def apply(self, left: int, right: typing.Optional[int] = None,
f: typing.Optional[typing.Any] = None) -> None:
assert f is not None
if right is None:
p = left
assert 0 <= left < self._n
p += self._size
for i in range(self._log, 0, -1):
self._push(p >> i)
self._d[p] = self._mapping(f, self._d[p])
for i in range(1, self._log + 1):
self._update(p >> i)
else:
assert 0 <= left <= right <= self._n
if left == right:
return
left += self._size
right += self._size
for i in range(self._log, 0, -1):
if ((left >> i) << i) != left:
self._push(left >> i)
if ((right >> i) << i) != right:
self._push((right - 1) >> i)
l2 = left
r2 = right
while left < right:
if left & 1:
self._all_apply(left, f)
left += 1
if right & 1:
right -= 1
self._all_apply(right, f)
left >>= 1
right >>= 1
left = l2
right = r2
for i in range(1, self._log + 1):
if ((left >> i) << i) != left:
self._update(left >> i)
if ((right >> i) << i) != right:
self._update((right - 1) >> i)
def max_right(
self, left: int, g: typing.Callable[[typing.Any], bool]) -> int:
assert 0 <= left <= self._n
assert g(self._e)
if left == self._n:
return self._n
left += self._size
for i in range(self._log, 0, -1):
self._push(left >> i)
sm = self._e
first = True
while first or (left & -left) != left:
first = False
while left % 2 == 0:
left >>= 1
if not g(self._op(sm, self._d[left])):
while left < self._size:
self._push(left)
left *= 2
if g(self._op(sm, self._d[left])):
sm = self._op(sm, self._d[left])
left += 1
return left - self._size
sm = self._op(sm, self._d[left])
left += 1
return self._n
def min_left(self, right: int, g: typing.Any) -> int:
assert 0 <= right <= self._n
assert g(self._e)
if right == 0:
return 0
right += self._size
for i in range(self._log, 0, -1):
self._push((right - 1) >> i)
sm = self._e
first = True
while first or (right & -right) != right:
first = False
right -= 1
while right > 1 and right % 2:
right >>= 1
if not g(self._op(self._d[right], sm)):
while right < self._size:
self._push(right)
right = 2 * right + 1
if g(self._op(self._d[right], sm)):
sm = self._op(self._d[right], sm)
right -= 1
return right + 1 - self._size
sm = self._op(self._d[right], sm)
return 0
def _update(self, k: int) -> None:
self._d[k] = self._op(self._d[2 * k], self._d[2 * k + 1])
def _all_apply(self, k: int, f: typing.Any) -> None:
self._d[k] = self._mapping(f, self._d[k])
if k < self._size:
self._lz[k] = self._composition(f, self._lz[k])
def _push(self, k: int) -> None:
self._all_apply(2 * k, self._lz[k])
self._all_apply(2 * k + 1, self._lz[k])
self._lz[k] = self._id
n, m, k = MII()
q = m + k
mod = random.getrandbits(32)
pow10 = [1] * (n + 1)
for i in range(1, n + 1):
pow10[i] = pow10[i-1] * 10 % mod
pow10_acc = list(accumulate(pow10, initial=0))
for i in range(n + 1):
pow10_acc[i] %= mod
mask = (1 << 32) - 1
def op(x, y):
l1, v1 = x >> 32, x & mask
l2, v2 = y >> 32, y & mask
l = l1 + l2
v = (v1 * pow10[l2] + v2) % mod
return l << 32 | v
def mapping(f, x):
if f == -1: return x
l = x >> 32
return l << 32 | (pow10_acc[l] * f % mod)
def composition(f1, f2):
return f2 if f1 == -1 else f1
seg = LazySegTree(op, 0, mapping, composition, -1, [int(x) + mask + 1 for x in I()])
for _ in range(q):
t, l, r, x = GMI()
x += 1
if t: print('YES' if seg.prod(l, r-x+1) == seg.prod(l+x, r+1) else 'NO')
else: seg.apply(l, r + 1, x)
|
Title: Kefa and Watch
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Kefa the parrot was walking down the street as he was on the way home from the restaurant when he saw something glittering by the road. As he came nearer he understood that it was a watch. He decided to take it to the pawnbroker to earn some money.
The pawnbroker said that each watch contains a serial number represented by a string of digits from 0 to 9, and the more quality checks this number passes, the higher is the value of the watch. The check is defined by three positive integers *l*, *r* and *d*. The watches pass a check if a substring of the serial number from *l* to *r* has period *d*. Sometimes the pawnbroker gets distracted and Kefa changes in some substring of the serial number all digits to *c* in order to increase profit from the watch.
The seller has a lot of things to do to begin with and with Kefa messing about, he gave you a task: to write a program that determines the value of the watch.
Let us remind you that number *x* is called a period of string *s* (1<=≤<=*x*<=≤<=|*s*|), if *s**i*<=<==<=<=*s**i*<=+<=*x* for all *i* from 1 to |*s*|<=<=-<=<=*x*.
Input Specification:
The first line of the input contains three positive integers *n*, *m* and *k* (1<=≤<=*n*<=≤<=105, 1<=≤<=*m*<=+<=*k*<=≤<=105) — the length of the serial number, the number of change made by Kefa and the number of quality checks.
The second line contains a serial number consisting of *n* digits.
Then *m*<=+<=*k* lines follow, containing either checks or changes.
The changes are given as 1 *l* *r* *c* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*, 0<=≤<=*c*<=≤<=9). That means that Kefa changed all the digits from the *l*-th to the *r*-th to be *c*.
The checks are given as 2 *l* *r* *d* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*, 1<=≤<=*d*<=≤<=*r*<=-<=*l*<=+<=1).
Output Specification:
For each check on a single line print "YES" if the watch passed it, otherwise print "NO".
Demo Input:
['3 1 2\n112\n2 2 3 1\n1 1 3 8\n2 1 2 1\n', '6 2 3\n334934\n2 2 5 2\n1 4 4 3\n2 1 6 3\n1 2 3 8\n2 3 6 1\n']
Demo Output:
['NO\nYES\n', 'NO\nYES\nNO\n']
Note:
In the first sample test two checks will be made. In the first one substring "12" is checked on whether or not it has period 1, so the answer is "NO". In the second one substring "88", is checked on whether or not it has period 1, and it has this period, so the answer is "YES".
In the second statement test three checks will be made. The first check processes substring "3493", which doesn't have period 2. Before the second check the string looks as "334334", so the answer to it is "YES". And finally, the third check processes substring "8334", which does not have period 1.
|
```python
if True:
from io import BytesIO, IOBase
import math
import random
import sys
import os
import bisect
import typing
from collections import Counter, defaultdict, deque
from copy import deepcopy
from functools import cmp_to_key, lru_cache, reduce
from heapq import heapify, heappop, heappush, heappushpop, nlargest, nsmallest
from itertools import accumulate, combinations, permutations, count
from operator import add, iand, ior, itemgetter, mul, xor
from string import ascii_lowercase, ascii_uppercase, ascii_letters
from typing import *
BUFSIZE = 4096
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin = IOWrapper(sys.stdin)
sys.stdout = IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def I():
return input()
def II():
return int(input())
def MII():
return map(int, input().split())
def LI():
return list(input().split())
def LII():
return list(map(int, input().split()))
def LFI():
return list(map(float, input().split()))
def GMI():
return map(lambda x: int(x) - 1, input().split())
def LGMI():
return list(map(lambda x: int(x) - 1, input().split()))
inf = float('inf')
dfs, hashing = True, False
if dfs:
from types import GeneratorType
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
if hashing:
RANDOM = random.getrandbits(20)
class Wrapper(int):
def __init__(self, x):
int.__init__(x)
def __hash__(self):
return super(Wrapper, self).__hash__() ^ RANDOM
class LazySegTree:
def __init__(
self,
op: typing.Callable[[typing.Any, typing.Any], typing.Any],
e: typing.Any,
mapping: typing.Callable[[typing.Any, typing.Any], typing.Any],
composition: typing.Callable[[typing.Any, typing.Any], typing.Any],
id_: typing.Any,
v: typing.Union[int, typing.List[typing.Any]]) -> None:
self._op = op
self._e = e
self._mapping = mapping
self._composition = composition
self._id = id_
if isinstance(v, int):
v = [e] * v
self._n = len(v)
self._log = (self._n - 1).bit_length()
self._size = 1 << self._log
self._d = [e] * (2 * self._size)
self._lz = [self._id] * self._size
for i in range(self._n):
self._d[self._size + i] = v[i]
for i in range(self._size - 1, 0, -1):
self._update(i)
def set(self, p: int, x: typing.Any) -> None:
assert 0 <= p < self._n
p += self._size
for i in range(self._log, 0, -1):
self._push(p >> i)
self._d[p] = x
for i in range(1, self._log + 1):
self._update(p >> i)
def get(self, p: int) -> typing.Any:
assert 0 <= p < self._n
p += self._size
for i in range(self._log, 0, -1):
self._push(p >> i)
return self._d[p]
def prod(self, left: int, right: int) -> typing.Any:
assert 0 <= left <= right <= self._n
if left == right:
return self._e
left += self._size
right += self._size
for i in range(self._log, 0, -1):
if ((left >> i) << i) != left:
self._push(left >> i)
if ((right >> i) << i) != right:
self._push(right >> i)
sml = self._e
smr = self._e
while left < right:
if left & 1:
sml = self._op(sml, self._d[left])
left += 1
if right & 1:
right -= 1
smr = self._op(self._d[right], smr)
left >>= 1
right >>= 1
return self._op(sml, smr)
def all_prod(self) -> typing.Any:
return self._d[1]
def apply(self, left: int, right: typing.Optional[int] = None,
f: typing.Optional[typing.Any] = None) -> None:
assert f is not None
if right is None:
p = left
assert 0 <= left < self._n
p += self._size
for i in range(self._log, 0, -1):
self._push(p >> i)
self._d[p] = self._mapping(f, self._d[p])
for i in range(1, self._log + 1):
self._update(p >> i)
else:
assert 0 <= left <= right <= self._n
if left == right:
return
left += self._size
right += self._size
for i in range(self._log, 0, -1):
if ((left >> i) << i) != left:
self._push(left >> i)
if ((right >> i) << i) != right:
self._push((right - 1) >> i)
l2 = left
r2 = right
while left < right:
if left & 1:
self._all_apply(left, f)
left += 1
if right & 1:
right -= 1
self._all_apply(right, f)
left >>= 1
right >>= 1
left = l2
right = r2
for i in range(1, self._log + 1):
if ((left >> i) << i) != left:
self._update(left >> i)
if ((right >> i) << i) != right:
self._update((right - 1) >> i)
def max_right(
self, left: int, g: typing.Callable[[typing.Any], bool]) -> int:
assert 0 <= left <= self._n
assert g(self._e)
if left == self._n:
return self._n
left += self._size
for i in range(self._log, 0, -1):
self._push(left >> i)
sm = self._e
first = True
while first or (left & -left) != left:
first = False
while left % 2 == 0:
left >>= 1
if not g(self._op(sm, self._d[left])):
while left < self._size:
self._push(left)
left *= 2
if g(self._op(sm, self._d[left])):
sm = self._op(sm, self._d[left])
left += 1
return left - self._size
sm = self._op(sm, self._d[left])
left += 1
return self._n
def min_left(self, right: int, g: typing.Any) -> int:
assert 0 <= right <= self._n
assert g(self._e)
if right == 0:
return 0
right += self._size
for i in range(self._log, 0, -1):
self._push((right - 1) >> i)
sm = self._e
first = True
while first or (right & -right) != right:
first = False
right -= 1
while right > 1 and right % 2:
right >>= 1
if not g(self._op(self._d[right], sm)):
while right < self._size:
self._push(right)
right = 2 * right + 1
if g(self._op(self._d[right], sm)):
sm = self._op(self._d[right], sm)
right -= 1
return right + 1 - self._size
sm = self._op(self._d[right], sm)
return 0
def _update(self, k: int) -> None:
self._d[k] = self._op(self._d[2 * k], self._d[2 * k + 1])
def _all_apply(self, k: int, f: typing.Any) -> None:
self._d[k] = self._mapping(f, self._d[k])
if k < self._size:
self._lz[k] = self._composition(f, self._lz[k])
def _push(self, k: int) -> None:
self._all_apply(2 * k, self._lz[k])
self._all_apply(2 * k + 1, self._lz[k])
self._lz[k] = self._id
n, m, k = MII()
q = m + k
mod = random.getrandbits(32)
pow10 = [1] * (n + 1)
for i in range(1, n + 1):
pow10[i] = pow10[i-1] * 10 % mod
pow10_acc = list(accumulate(pow10, initial=0))
for i in range(n + 1):
pow10_acc[i] %= mod
mask = (1 << 32) - 1
def op(x, y):
l1, v1 = x >> 32, x & mask
l2, v2 = y >> 32, y & mask
l = l1 + l2
v = (v1 * pow10[l2] + v2) % mod
return l << 32 | v
def mapping(f, x):
if f == -1: return x
l = x >> 32
return l << 32 | (pow10_acc[l] * f % mod)
def composition(f1, f2):
return f2 if f1 == -1 else f1
seg = LazySegTree(op, 0, mapping, composition, -1, [int(x) + mask + 1 for x in I()])
for _ in range(q):
t, l, r, x = GMI()
x += 1
if t: print('YES' if seg.prod(l, r-x+1) == seg.prod(l+x, r+1) else 'NO')
else: seg.apply(l, r + 1, x)
```
| 3
|
|
151
|
A
|
Soft Drinking
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
This winter is so cold in Nvodsk! A group of *n* friends decided to buy *k* bottles of a soft drink called "Take-It-Light" to warm up a bit. Each bottle has *l* milliliters of the drink. Also they bought *c* limes and cut each of them into *d* slices. After that they found *p* grams of salt.
To make a toast, each friend needs *nl* milliliters of the drink, a slice of lime and *np* grams of salt. The friends want to make as many toasts as they can, provided they all drink the same amount. How many toasts can each friend make?
|
The first and only line contains positive integers *n*, *k*, *l*, *c*, *d*, *p*, *nl*, *np*, not exceeding 1000 and no less than 1. The numbers are separated by exactly one space.
|
Print a single integer — the number of toasts each friend can make.
|
[
"3 4 5 10 8 100 3 1\n",
"5 100 10 1 19 90 4 3\n",
"10 1000 1000 25 23 1 50 1\n"
] |
[
"2\n",
"3\n",
"0\n"
] |
A comment to the first sample:
Overall the friends have 4 * 5 = 20 milliliters of the drink, it is enough to make 20 / 3 = 6 toasts. The limes are enough for 10 * 8 = 80 toasts and the salt is enough for 100 / 1 = 100 toasts. However, there are 3 friends in the group, so the answer is *min*(6, 80, 100) / 3 = 2.
| 500
|
[
{
"input": "3 4 5 10 8 100 3 1",
"output": "2"
},
{
"input": "5 100 10 1 19 90 4 3",
"output": "3"
},
{
"input": "10 1000 1000 25 23 1 50 1",
"output": "0"
},
{
"input": "1 7 4 5 5 8 3 2",
"output": "4"
},
{
"input": "2 3 3 5 5 10 1 3",
"output": "1"
},
{
"input": "2 6 4 5 6 5 1 3",
"output": "0"
},
{
"input": "1 7 3 5 3 6 2 1",
"output": "6"
},
{
"input": "2 4 5 4 5 7 3 2",
"output": "1"
},
{
"input": "2 3 6 5 7 8 2 1",
"output": "4"
},
{
"input": "1 4 5 5 3 10 3 1",
"output": "6"
},
{
"input": "1 4 6 7 3 5 1 3",
"output": "1"
},
{
"input": "1 6 5 5 5 8 3 1",
"output": "8"
},
{
"input": "1 7 5 3 3 9 2 1",
"output": "9"
},
{
"input": "3 5 3 7 6 10 3 1",
"output": "1"
},
{
"input": "3 6 3 5 3 6 3 1",
"output": "2"
},
{
"input": "1 7 5 5 5 5 2 2",
"output": "2"
},
{
"input": "2 5 3 5 6 9 2 1",
"output": "3"
},
{
"input": "3 4 3 5 3 6 2 1",
"output": "2"
},
{
"input": "1 5 5 4 7 6 3 1",
"output": "6"
},
{
"input": "2 3 7 6 5 9 3 1",
"output": "3"
},
{
"input": "2 6 5 3 3 8 1 1",
"output": "4"
},
{
"input": "2 4 7 3 4 10 2 1",
"output": "5"
},
{
"input": "1 1000 1000 1000 1000 1000 1 1",
"output": "1000"
},
{
"input": "17 1000 1000 1000 1000 1000 3 7",
"output": "8"
},
{
"input": "115 1000 1000 1000 1000 1000 17 15",
"output": "0"
},
{
"input": "1 587 981 1 2 1 1 1",
"output": "1"
},
{
"input": "1 1 2 1 2 2 1 1",
"output": "2"
}
] | 1,672,746,866
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 27
| 92
| 0
|
n, k, l, c, d, p, nl, np = [int(i) for i in input().split(" ")]
print(min(k*l//nl, c*d, p//np)//n)
|
Title: Soft Drinking
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
This winter is so cold in Nvodsk! A group of *n* friends decided to buy *k* bottles of a soft drink called "Take-It-Light" to warm up a bit. Each bottle has *l* milliliters of the drink. Also they bought *c* limes and cut each of them into *d* slices. After that they found *p* grams of salt.
To make a toast, each friend needs *nl* milliliters of the drink, a slice of lime and *np* grams of salt. The friends want to make as many toasts as they can, provided they all drink the same amount. How many toasts can each friend make?
Input Specification:
The first and only line contains positive integers *n*, *k*, *l*, *c*, *d*, *p*, *nl*, *np*, not exceeding 1000 and no less than 1. The numbers are separated by exactly one space.
Output Specification:
Print a single integer — the number of toasts each friend can make.
Demo Input:
['3 4 5 10 8 100 3 1\n', '5 100 10 1 19 90 4 3\n', '10 1000 1000 25 23 1 50 1\n']
Demo Output:
['2\n', '3\n', '0\n']
Note:
A comment to the first sample:
Overall the friends have 4 * 5 = 20 milliliters of the drink, it is enough to make 20 / 3 = 6 toasts. The limes are enough for 10 * 8 = 80 toasts and the salt is enough for 100 / 1 = 100 toasts. However, there are 3 friends in the group, so the answer is *min*(6, 80, 100) / 3 = 2.
|
```python
n, k, l, c, d, p, nl, np = [int(i) for i in input().split(" ")]
print(min(k*l//nl, c*d, p//np)//n)
```
| 3
|
|
252
|
A
|
Little Xor
|
PROGRAMMING
| 1,100
|
[
"brute force",
"implementation"
] | null | null |
Little Petya likes arrays that consist of non-negative integers a lot. Recently his mom has presented him one such array consisting of *n* elements. Petya immediately decided to find there a segment of consecutive elements, such that the *xor* of all numbers from this segment was maximal possible. Help him with that.
The *xor* operation is the bitwise exclusive "OR", that is denoted as "xor" in Pascal and "^" in C/C++/Java.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of elements in the array. The second line contains the space-separated integers from the array. All numbers are non-negative integers strictly less than 230.
|
Print a single integer — the required maximal *xor* of a segment of consecutive elements.
|
[
"5\n1 2 1 1 2\n",
"3\n1 2 7\n",
"4\n4 2 4 8\n"
] |
[
"3\n",
"7\n",
"14\n"
] |
In the first sample one of the optimal segments is the segment that consists of the first and the second array elements, if we consider the array elements indexed starting from one.
The second sample contains only one optimal segment, which contains exactly one array element (element with index three).
| 500
|
[
{
"input": "5\n1 2 1 1 2",
"output": "3"
},
{
"input": "3\n1 2 7",
"output": "7"
},
{
"input": "4\n4 2 4 8",
"output": "14"
},
{
"input": "5\n1 1 1 1 1",
"output": "1"
},
{
"input": "16\n0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15",
"output": "15"
},
{
"input": "20\n1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10",
"output": "15"
},
{
"input": "100\n28 20 67 103 72 81 82 83 7 109 122 30 50 118 83 89 108 82 92 17 97 3 62 12 9 100 14 11 99 106 10 8 60 101 88 119 104 62 76 6 5 57 32 94 60 50 58 97 1 97 107 108 80 24 45 20 112 1 98 106 49 98 25 57 47 90 74 68 14 35 22 10 61 80 10 4 53 13 90 99 57 100 40 84 22 116 60 61 98 57 74 127 61 73 49 51 20 19 56 111",
"output": "127"
},
{
"input": "99\n87 67 4 84 13 20 35 7 11 86 25 1 58 1 74 64 74 86 98 74 72 46 63 78 84 13 60 38 30 45 45 60 9 44 36 70 33 22 82 15 71 7 43 47 23 2 20 49 42 43 54 27 51 51 53 23 27 37 17 66 90 89 61 0 18 20 49 30 84 20 13 32 64 69 56 68 59 10 91 96 43 64 19 10 2 57 62 23 100 39 32 19 95 55 77 19 24 4 77",
"output": "127"
},
{
"input": "1\n100",
"output": "100"
},
{
"input": "2\n1 1",
"output": "1"
},
{
"input": "2\n4 10",
"output": "14"
},
{
"input": "99\n3511 2076 9314 3598 7737 271 4110 4454 5830 8661 8584 8227 2236 2626 1025 3806 3162 2837 6071 9627 1836 7729 6629 4054 9377 3261 1325 8859 3610 7616 5052 9736 1249 6224 7031 6074 4196 8075 4005 5419 5664 8214 4391 731 8714 8622 6122 6976 4057 1770 8025 3778 2792 994 5204 1826 6083 5438 8409 8029 1962 3133 9836 8884 1234 1776 824 1630 6523 2869 9950 8609 5397 1472 7120 5012 6490 5958 8030 7741 5077 2771 7925 9784 9456 8596 6288 8939 4779 4415 3743 5188 5574 9686 1483 4245 4769 4917 1887",
"output": "16383"
},
{
"input": "27\n78 918 443 3900 591 12 4 10 1 24 70 88 429 2 3257 65 275 2 258 62 587 625 25 26 853 728 765",
"output": "4027"
},
{
"input": "2\n3 1",
"output": "3"
},
{
"input": "2\n3 2",
"output": "3"
},
{
"input": "3\n7 3 1",
"output": "7"
},
{
"input": "3\n1 7 2",
"output": "7"
},
{
"input": "2\n7 1",
"output": "7"
},
{
"input": "4\n1 2 7 1",
"output": "7"
},
{
"input": "3\n10 5 1",
"output": "15"
}
] | 1,564,900,313
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 56
| 248
| 0
|
n=int(input())
a=list(map(int,input().split()))
m=0
for i in range(0,n-1):
x=a[i]
m=max(m,x)
for j in range(i+1,n):
x=x^a[j]
m=max(m,x)
print(max(m,a[n-1]))
|
Title: Little Xor
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Petya likes arrays that consist of non-negative integers a lot. Recently his mom has presented him one such array consisting of *n* elements. Petya immediately decided to find there a segment of consecutive elements, such that the *xor* of all numbers from this segment was maximal possible. Help him with that.
The *xor* operation is the bitwise exclusive "OR", that is denoted as "xor" in Pascal and "^" in C/C++/Java.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of elements in the array. The second line contains the space-separated integers from the array. All numbers are non-negative integers strictly less than 230.
Output Specification:
Print a single integer — the required maximal *xor* of a segment of consecutive elements.
Demo Input:
['5\n1 2 1 1 2\n', '3\n1 2 7\n', '4\n4 2 4 8\n']
Demo Output:
['3\n', '7\n', '14\n']
Note:
In the first sample one of the optimal segments is the segment that consists of the first and the second array elements, if we consider the array elements indexed starting from one.
The second sample contains only one optimal segment, which contains exactly one array element (element with index three).
|
```python
n=int(input())
a=list(map(int,input().split()))
m=0
for i in range(0,n-1):
x=a[i]
m=max(m,x)
for j in range(i+1,n):
x=x^a[j]
m=max(m,x)
print(max(m,a[n-1]))
```
| 3
|
|
218
|
B
|
Airport
|
PROGRAMMING
| 1,100
|
[
"implementation"
] | null | null |
Lolek and Bolek are about to travel abroad by plane. The local airport has a special "Choose Your Plane" offer. The offer's conditions are as follows:
- it is up to a passenger to choose a plane to fly on; - if the chosen plane has *x* (*x*<=><=0) empty seats at the given moment, then the ticket for such a plane costs *x* zlotys (units of Polish currency).
The only ticket office of the airport already has a queue of *n* passengers in front of it. Lolek and Bolek have not stood in the queue yet, but they are already wondering what is the maximum and the minimum number of zlotys the airport administration can earn if all *n* passengers buy tickets according to the conditions of this offer?
The passengers buy tickets in turn, the first person in the queue goes first, then goes the second one, and so on up to *n*-th person.
|
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the number of passengers in the queue and the number of planes in the airport, correspondingly. The next line contains *m* integers *a*1,<=*a*2,<=...,<=*a**m* (1<=≤<=*a**i*<=≤<=1000) — *a**i* stands for the number of empty seats in the *i*-th plane before the ticket office starts selling tickets.
The numbers in the lines are separated by a space. It is guaranteed that there are at least *n* empty seats in total.
|
Print two integers — the maximum and the minimum number of zlotys that the airport administration can earn, correspondingly.
|
[
"4 3\n2 1 1\n",
"4 3\n2 2 2\n"
] |
[
"5 5\n",
"7 6\n"
] |
In the first test sample the number of passengers is equal to the number of empty seats, so regardless of the way the planes are chosen, the administration will earn the same sum.
In the second sample the sum is maximized if the 1-st person in the queue buys a ticket to the 1-st plane, the 2-nd person — to the 2-nd plane, the 3-rd person — to the 3-rd plane, the 4-th person — to the 1-st plane. The sum is minimized if the 1-st person in the queue buys a ticket to the 1-st plane, the 2-nd person — to the 1-st plane, the 3-rd person — to the 2-nd plane, the 4-th person — to the 2-nd plane.
| 500
|
[
{
"input": "4 3\n2 1 1",
"output": "5 5"
},
{
"input": "4 3\n2 2 2",
"output": "7 6"
},
{
"input": "10 5\n10 3 3 1 2",
"output": "58 26"
},
{
"input": "10 1\n10",
"output": "55 55"
},
{
"input": "10 1\n100",
"output": "955 955"
},
{
"input": "10 2\n4 7",
"output": "37 37"
},
{
"input": "40 10\n1 2 3 4 5 6 7 10 10 10",
"output": "223 158"
},
{
"input": "1 1\n6",
"output": "6 6"
},
{
"input": "1 2\n10 9",
"output": "10 9"
},
{
"input": "2 1\n7",
"output": "13 13"
},
{
"input": "2 2\n7 2",
"output": "13 3"
},
{
"input": "3 2\n4 7",
"output": "18 9"
},
{
"input": "3 3\n2 1 1",
"output": "4 4"
},
{
"input": "3 3\n2 1 1",
"output": "4 4"
},
{
"input": "10 10\n3 1 2 2 1 1 2 1 2 3",
"output": "20 13"
},
{
"input": "10 2\n7 3",
"output": "34 34"
},
{
"input": "10 1\n19",
"output": "145 145"
},
{
"input": "100 3\n29 36 35",
"output": "1731 1731"
},
{
"input": "100 5\n3 38 36 35 2",
"output": "2019 1941"
},
{
"input": "510 132\n50 76 77 69 94 30 47 65 14 62 18 121 26 35 49 17 105 93 47 16 78 3 7 74 7 37 30 36 30 83 71 113 7 58 86 10 65 57 34 102 55 44 43 47 106 44 115 75 109 70 47 45 16 57 62 55 20 88 74 40 45 84 41 1 9 53 65 25 67 31 115 2 63 51 123 70 65 65 18 14 75 14 103 26 117 105 36 104 81 37 35 61 44 90 71 70 88 89 26 21 64 77 89 16 87 99 13 79 27 3 46 120 116 11 14 17 32 70 113 94 108 57 29 100 53 48 44 29 70 30 32 62",
"output": "50279 5479"
},
{
"input": "510 123\n5 2 3 2 5 7 2 3 1 3 6 6 3 1 5 3 5 6 2 2 1 5 5 5 2 2 3 1 6 3 5 8 4 6 1 5 4 5 1 6 5 5 3 6 4 1 6 1 3 5 2 7 5 2 4 4 5 6 5 5 4 3 4 6 5 4 4 3 5 8 5 5 6 3 1 7 4 4 3 3 5 3 6 3 3 6 2 5 3 2 4 5 4 5 2 2 4 4 4 7 3 4 6 5 3 6 4 7 1 6 5 7 6 5 7 3 7 4 4 1 6 6 4",
"output": "1501 1501"
},
{
"input": "610 33\n15 44 8 8 17 11 39 39 38 25 17 36 17 25 21 37 10 11 34 30 29 50 29 50 4 20 32 13 41 14 2 11 2",
"output": "12204 8871"
}
] | 1,622,116,067
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 33
| 186
| 0
|
def maximum(lst,n):
amount = 0
for _ in range(n):
maxempty = max(lst)
amount += maxempty
lst[lst.index(maxempty)] -= 1
return amount
def minimum(anotherlst,n):
amount = 0
count = n
while count:
minempty = min(anotherlst)
if minempty<=0:
anotherlst.remove(minempty)
continue
amount += minempty
anotherlst[anotherlst.index(minempty)] -= 1
count -= 1
return amount
n,m = tuple([int(item) for item in input().split()])
lst = [int(item) for item in input().split()]
anotherlst = lst.copy()
print(maximum(lst,n),end=' ')
print(minimum(anotherlst,n))
|
Title: Airport
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Lolek and Bolek are about to travel abroad by plane. The local airport has a special "Choose Your Plane" offer. The offer's conditions are as follows:
- it is up to a passenger to choose a plane to fly on; - if the chosen plane has *x* (*x*<=><=0) empty seats at the given moment, then the ticket for such a plane costs *x* zlotys (units of Polish currency).
The only ticket office of the airport already has a queue of *n* passengers in front of it. Lolek and Bolek have not stood in the queue yet, but they are already wondering what is the maximum and the minimum number of zlotys the airport administration can earn if all *n* passengers buy tickets according to the conditions of this offer?
The passengers buy tickets in turn, the first person in the queue goes first, then goes the second one, and so on up to *n*-th person.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the number of passengers in the queue and the number of planes in the airport, correspondingly. The next line contains *m* integers *a*1,<=*a*2,<=...,<=*a**m* (1<=≤<=*a**i*<=≤<=1000) — *a**i* stands for the number of empty seats in the *i*-th plane before the ticket office starts selling tickets.
The numbers in the lines are separated by a space. It is guaranteed that there are at least *n* empty seats in total.
Output Specification:
Print two integers — the maximum and the minimum number of zlotys that the airport administration can earn, correspondingly.
Demo Input:
['4 3\n2 1 1\n', '4 3\n2 2 2\n']
Demo Output:
['5 5\n', '7 6\n']
Note:
In the first test sample the number of passengers is equal to the number of empty seats, so regardless of the way the planes are chosen, the administration will earn the same sum.
In the second sample the sum is maximized if the 1-st person in the queue buys a ticket to the 1-st plane, the 2-nd person — to the 2-nd plane, the 3-rd person — to the 3-rd plane, the 4-th person — to the 1-st plane. The sum is minimized if the 1-st person in the queue buys a ticket to the 1-st plane, the 2-nd person — to the 1-st plane, the 3-rd person — to the 2-nd plane, the 4-th person — to the 2-nd plane.
|
```python
def maximum(lst,n):
amount = 0
for _ in range(n):
maxempty = max(lst)
amount += maxempty
lst[lst.index(maxempty)] -= 1
return amount
def minimum(anotherlst,n):
amount = 0
count = n
while count:
minempty = min(anotherlst)
if minempty<=0:
anotherlst.remove(minempty)
continue
amount += minempty
anotherlst[anotherlst.index(minempty)] -= 1
count -= 1
return amount
n,m = tuple([int(item) for item in input().split()])
lst = [int(item) for item in input().split()]
anotherlst = lst.copy()
print(maximum(lst,n),end=' ')
print(minimum(anotherlst,n))
```
| 3
|
|
835
|
A
|
Key races
|
PROGRAMMING
| 800
|
[
"math"
] | null | null |
Two boys decided to compete in text typing on the site "Key races". During the competition, they have to type a text consisting of *s* characters. The first participant types one character in *v*1 milliseconds and has ping *t*1 milliseconds. The second participant types one character in *v*2 milliseconds and has ping *t*2 milliseconds.
If connection ping (delay) is *t* milliseconds, the competition passes for a participant as follows:
1. Exactly after *t* milliseconds after the start of the competition the participant receives the text to be entered. 1. Right after that he starts to type it. 1. Exactly *t* milliseconds after he ends typing all the text, the site receives information about it.
The winner is the participant whose information on the success comes earlier. If the information comes from both participants at the same time, it is considered that there is a draw.
Given the length of the text and the information about participants, determine the result of the game.
|
The first line contains five integers *s*, *v*1, *v*2, *t*1, *t*2 (1<=≤<=*s*,<=*v*1,<=*v*2,<=*t*1,<=*t*2<=≤<=1000) — the number of characters in the text, the time of typing one character for the first participant, the time of typing one character for the the second participant, the ping of the first participant and the ping of the second participant.
|
If the first participant wins, print "First". If the second participant wins, print "Second". In case of a draw print "Friendship".
|
[
"5 1 2 1 2\n",
"3 3 1 1 1\n",
"4 5 3 1 5\n"
] |
[
"First\n",
"Second\n",
"Friendship\n"
] |
In the first example, information on the success of the first participant comes in 7 milliseconds, of the second participant — in 14 milliseconds. So, the first wins.
In the second example, information on the success of the first participant comes in 11 milliseconds, of the second participant — in 5 milliseconds. So, the second wins.
In the third example, information on the success of the first participant comes in 22 milliseconds, of the second participant — in 22 milliseconds. So, it is be a draw.
| 500
|
[
{
"input": "5 1 2 1 2",
"output": "First"
},
{
"input": "3 3 1 1 1",
"output": "Second"
},
{
"input": "4 5 3 1 5",
"output": "Friendship"
},
{
"input": "1000 1000 1000 1000 1000",
"output": "Friendship"
},
{
"input": "1 1 1 1 1",
"output": "Friendship"
},
{
"input": "8 8 1 1 1",
"output": "Second"
},
{
"input": "15 14 32 65 28",
"output": "First"
},
{
"input": "894 197 325 232 902",
"output": "First"
},
{
"input": "1 2 8 8 5",
"output": "Friendship"
},
{
"input": "37 261 207 1 1000",
"output": "Friendship"
},
{
"input": "29 344 406 900 1",
"output": "Friendship"
},
{
"input": "1 2 8 9 8",
"output": "First"
},
{
"input": "2 9 8 8 9",
"output": "Friendship"
},
{
"input": "213 480 811 134 745",
"output": "First"
},
{
"input": "2 313 856 964 421",
"output": "Friendship"
},
{
"input": "1 10 2 6 10",
"output": "Friendship"
},
{
"input": "2 7 6 2 3",
"output": "Friendship"
},
{
"input": "637 324 69 612 998",
"output": "Second"
},
{
"input": "13 849 819 723 918",
"output": "Friendship"
},
{
"input": "9 5 7 8 7",
"output": "First"
},
{
"input": "6 5 7 10 4",
"output": "Friendship"
},
{
"input": "61 464 623 89 548",
"output": "First"
},
{
"input": "641 31 29 161 802",
"output": "Friendship"
},
{
"input": "3 3 1 6 9",
"output": "Friendship"
},
{
"input": "2 3 9 8 2",
"output": "Friendship"
},
{
"input": "485 117 368 567 609",
"output": "First"
},
{
"input": "4 202 512 995 375",
"output": "Friendship"
},
{
"input": "424 41 41 909 909",
"output": "Friendship"
},
{
"input": "884 913 263 641 265",
"output": "Second"
},
{
"input": "12 462 8 311 327",
"output": "Second"
},
{
"input": "436 306 266 493 580",
"output": "Second"
},
{
"input": "69 1 2 1 2",
"output": "First"
}
] | 1,592,837,884
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 32
| 109
| 0
|
s,v1,v2,t1,t2 = list(map(int,input().split()))
a = 2*t1+s*v1
b = 2*t2+s*v2
if a < b:
print("First")
elif a > b:
print("Second")
elif a == b:
print("Friendship")
|
Title: Key races
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Two boys decided to compete in text typing on the site "Key races". During the competition, they have to type a text consisting of *s* characters. The first participant types one character in *v*1 milliseconds and has ping *t*1 milliseconds. The second participant types one character in *v*2 milliseconds and has ping *t*2 milliseconds.
If connection ping (delay) is *t* milliseconds, the competition passes for a participant as follows:
1. Exactly after *t* milliseconds after the start of the competition the participant receives the text to be entered. 1. Right after that he starts to type it. 1. Exactly *t* milliseconds after he ends typing all the text, the site receives information about it.
The winner is the participant whose information on the success comes earlier. If the information comes from both participants at the same time, it is considered that there is a draw.
Given the length of the text and the information about participants, determine the result of the game.
Input Specification:
The first line contains five integers *s*, *v*1, *v*2, *t*1, *t*2 (1<=≤<=*s*,<=*v*1,<=*v*2,<=*t*1,<=*t*2<=≤<=1000) — the number of characters in the text, the time of typing one character for the first participant, the time of typing one character for the the second participant, the ping of the first participant and the ping of the second participant.
Output Specification:
If the first participant wins, print "First". If the second participant wins, print "Second". In case of a draw print "Friendship".
Demo Input:
['5 1 2 1 2\n', '3 3 1 1 1\n', '4 5 3 1 5\n']
Demo Output:
['First\n', 'Second\n', 'Friendship\n']
Note:
In the first example, information on the success of the first participant comes in 7 milliseconds, of the second participant — in 14 milliseconds. So, the first wins.
In the second example, information on the success of the first participant comes in 11 milliseconds, of the second participant — in 5 milliseconds. So, the second wins.
In the third example, information on the success of the first participant comes in 22 milliseconds, of the second participant — in 22 milliseconds. So, it is be a draw.
|
```python
s,v1,v2,t1,t2 = list(map(int,input().split()))
a = 2*t1+s*v1
b = 2*t2+s*v2
if a < b:
print("First")
elif a > b:
print("Second")
elif a == b:
print("Friendship")
```
| 3
|
|
680
|
B
|
Bear and Finding Criminals
|
PROGRAMMING
| 1,000
|
[
"constructive algorithms",
"implementation"
] | null | null |
There are *n* cities in Bearland, numbered 1 through *n*. Cities are arranged in one long row. The distance between cities *i* and *j* is equal to |*i*<=-<=*j*|.
Limak is a police officer. He lives in a city *a*. His job is to catch criminals. It's hard because he doesn't know in which cities criminals are. Though, he knows that there is at most one criminal in each city.
Limak is going to use a BCD (Bear Criminal Detector). The BCD will tell Limak how many criminals there are for every distance from a city *a*. After that, Limak can catch a criminal in each city for which he is sure that there must be a criminal.
You know in which cities criminals are. Count the number of criminals Limak will catch, after he uses the BCD.
|
The first line of the input contains two integers *n* and *a* (1<=≤<=*a*<=≤<=*n*<=≤<=100) — the number of cities and the index of city where Limak lives.
The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (0<=≤<=*t**i*<=≤<=1). There are *t**i* criminals in the *i*-th city.
|
Print the number of criminals Limak will catch.
|
[
"6 3\n1 1 1 0 1 0\n",
"5 2\n0 0 0 1 0\n"
] |
[
"3\n",
"1\n"
] |
In the first sample, there are six cities and Limak lives in the third one (blue arrow below). Criminals are in cities marked red.
Using the BCD gives Limak the following information:
- There is one criminal at distance 0 from the third city — Limak is sure that this criminal is exactly in the third city. - There is one criminal at distance 1 from the third city — Limak doesn't know if a criminal is in the second or fourth city. - There are two criminals at distance 2 from the third city — Limak is sure that there is one criminal in the first city and one in the fifth city. - There are zero criminals for every greater distance.
So, Limak will catch criminals in cities 1, 3 and 5, that is 3 criminals in total.
In the second sample (drawing below), the BCD gives Limak the information that there is one criminal at distance 2 from Limak's city. There is only one city at distance 2 so Limak is sure where a criminal is.
| 1,000
|
[
{
"input": "6 3\n1 1 1 0 1 0",
"output": "3"
},
{
"input": "5 2\n0 0 0 1 0",
"output": "1"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "1 1\n0",
"output": "0"
},
{
"input": "9 3\n1 1 1 1 1 1 1 1 0",
"output": "8"
},
{
"input": "9 5\n1 0 1 0 1 0 1 0 1",
"output": "5"
},
{
"input": "20 17\n1 1 0 1 1 1 1 0 1 0 1 1 1 0 1 1 0 0 0 0",
"output": "10"
},
{
"input": "100 60\n1 1 1 1 1 1 0 1 0 0 1 1 0 1 1 1 1 1 0 0 1 1 0 0 0 0 0 1 0 1 1 0 1 0 1 0 1 0 1 1 0 0 0 0 0 1 1 1 0 1 1 0 0 0 1 0 0 0 1 1 1 0 1 0 0 1 1 1 0 1 1 1 0 0 1 1 0 1 0 0 0 1 0 0 0 0 0 0 1 1 1 0 0 1 1 1 0 1 0 0",
"output": "27"
},
{
"input": "8 1\n1 0 1 1 0 0 1 0",
"output": "4"
},
{
"input": "11 11\n0 1 0 0 1 1 1 0 0 0 0",
"output": "4"
},
{
"input": "19 10\n0 1 1 0 1 0 0 1 1 0 0 1 0 1 0 0 1 0 1",
"output": "4"
},
{
"input": "100 38\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 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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 38\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 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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": "1"
},
{
"input": "100 38\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 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "3"
},
{
"input": "99 38\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 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "25"
},
{
"input": "99 38\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 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "24"
},
{
"input": "99 38\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 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "24"
},
{
"input": "98 70\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 0 0 0 0 0 0 0 0 0 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": "41"
},
{
"input": "99 70\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 1 1 1 1 1 1 1 1 1 1 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": "9"
},
{
"input": "99 60\n0 0 0 1 1 0 1 1 0 0 1 0 0 0 0 0 1 0 1 0 0 0 1 0 1 1 1 0 1 1 1 1 0 0 1 1 1 0 1 1 1 1 1 0 1 1 0 0 0 0 0 1 0 0 1 0 1 1 1 1 1 0 1 0 1 1 0 0 1 0 1 0 0 1 0 0 1 1 1 0 0 0 0 1 1 0 1 1 1 1 0 0 0 0 0 0 1 1 1",
"output": "34"
},
{
"input": "98 24\n0 0 0 1 1 0 1 1 0 0 0 0 1 0 0 1 0 0 1 1 1 0 0 1 0 0 1 0 1 0 0 1 1 0 1 1 1 0 1 0 0 1 0 0 0 1 1 1 1 0 1 1 1 0 1 1 1 1 0 0 1 0 0 0 1 0 1 1 0 1 0 1 1 1 0 0 1 0 0 0 1 1 0 0 1 1 1 1 1 1 0 1 0 1 0 0 1 1",
"output": "39"
},
{
"input": "100 100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "100"
},
{
"input": "100 1\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": "2 1\n0 1",
"output": "1"
}
] | 1,619,101,316
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 24
| 62
| 204,800
|
n,k=map(int,input().split())
k-=1
x=list(map(int,input().split()))
y=[0]*n
z=[0]*n
for i in range(n):
if x[i]==1:
y[abs(i-k)]+=1
z[abs(i-k)]+=1
c=0
for i in range(n):
if z[i]==y[i]:
c+=z[i]
print(c)
|
Title: Bear and Finding Criminals
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* cities in Bearland, numbered 1 through *n*. Cities are arranged in one long row. The distance between cities *i* and *j* is equal to |*i*<=-<=*j*|.
Limak is a police officer. He lives in a city *a*. His job is to catch criminals. It's hard because he doesn't know in which cities criminals are. Though, he knows that there is at most one criminal in each city.
Limak is going to use a BCD (Bear Criminal Detector). The BCD will tell Limak how many criminals there are for every distance from a city *a*. After that, Limak can catch a criminal in each city for which he is sure that there must be a criminal.
You know in which cities criminals are. Count the number of criminals Limak will catch, after he uses the BCD.
Input Specification:
The first line of the input contains two integers *n* and *a* (1<=≤<=*a*<=≤<=*n*<=≤<=100) — the number of cities and the index of city where Limak lives.
The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (0<=≤<=*t**i*<=≤<=1). There are *t**i* criminals in the *i*-th city.
Output Specification:
Print the number of criminals Limak will catch.
Demo Input:
['6 3\n1 1 1 0 1 0\n', '5 2\n0 0 0 1 0\n']
Demo Output:
['3\n', '1\n']
Note:
In the first sample, there are six cities and Limak lives in the third one (blue arrow below). Criminals are in cities marked red.
Using the BCD gives Limak the following information:
- There is one criminal at distance 0 from the third city — Limak is sure that this criminal is exactly in the third city. - There is one criminal at distance 1 from the third city — Limak doesn't know if a criminal is in the second or fourth city. - There are two criminals at distance 2 from the third city — Limak is sure that there is one criminal in the first city and one in the fifth city. - There are zero criminals for every greater distance.
So, Limak will catch criminals in cities 1, 3 and 5, that is 3 criminals in total.
In the second sample (drawing below), the BCD gives Limak the information that there is one criminal at distance 2 from Limak's city. There is only one city at distance 2 so Limak is sure where a criminal is.
|
```python
n,k=map(int,input().split())
k-=1
x=list(map(int,input().split()))
y=[0]*n
z=[0]*n
for i in range(n):
if x[i]==1:
y[abs(i-k)]+=1
z[abs(i-k)]+=1
c=0
for i in range(n):
if z[i]==y[i]:
c+=z[i]
print(c)
```
| 3
|
|
2
|
A
|
Winner
|
PROGRAMMING
| 1,500
|
[
"hashing",
"implementation"
] |
A. Winner
|
1
|
64
|
The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number of such players is more than one. During each round a player gains or loses a particular number of points. In the course of the game the number of points is registered in the line "name score", where name is a player's name, and score is the number of points gained in this round, which is an integer number. If score is negative, this means that the player has lost in the round. So, if two or more players have the maximum number of points (say, it equals to *m*) at the end of the game, than wins the one of them who scored at least *m* points first. Initially each player has 0 points. It's guaranteed that at the end of the game at least one player has a positive number of points.
|
The first line contains an integer number *n* (1<=<=≤<=<=*n*<=<=≤<=<=1000), *n* is the number of rounds played. Then follow *n* lines, containing the information about the rounds in "name score" format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is an integer number between -1000 and 1000, inclusive.
|
Print the name of the winner.
|
[
"3\nmike 3\nandrew 5\nmike 2\n",
"3\nandrew 3\nandrew 2\nmike 5\n"
] |
[
"andrew\n",
"andrew\n"
] |
none
| 0
|
[
{
"input": "3\nmike 3\nandrew 5\nmike 2",
"output": "andrew"
},
{
"input": "3\nandrew 3\nandrew 2\nmike 5",
"output": "andrew"
},
{
"input": "5\nkaxqybeultn -352\nmgochgrmeyieyskhuourfg -910\nkaxqybeultn 691\nmgochgrmeyieyskhuourfg -76\nkaxqybeultn -303",
"output": "kaxqybeultn"
},
{
"input": "7\nksjuuerbnlklcfdjeyq 312\ndthjlkrvvbyahttifpdewvyslsh -983\nksjuuerbnlklcfdjeyq 268\ndthjlkrvvbyahttifpdewvyslsh 788\nksjuuerbnlklcfdjeyq -79\nksjuuerbnlklcfdjeyq -593\nksjuuerbnlklcfdjeyq 734",
"output": "ksjuuerbnlklcfdjeyq"
},
{
"input": "12\natrtthfpcvishmqbakprquvnejr 185\natrtthfpcvishmqbakprquvnejr -699\natrtthfpcvishmqbakprquvnejr -911\natrtthfpcvishmqbakprquvnejr -220\nfcgslzkicjrpbqaifgweyzreajjfdo 132\nfcgslzkicjrpbqaifgweyzreajjfdo -242\nm 177\nm -549\natrtthfpcvishmqbakprquvnejr -242\nm 38\natrtthfpcvishmqbakprquvnejr -761\nfcgslzkicjrpbqaifgweyzreajjfdo 879",
"output": "fcgslzkicjrpbqaifgweyzreajjfdo"
},
{
"input": "15\naawtvezfntstrcpgbzjbf 681\nzhahpvqiptvksnbjkdvmknb -74\naawtvezfntstrcpgbzjbf 661\njpdwmyke 474\naawtvezfntstrcpgbzjbf -547\naawtvezfntstrcpgbzjbf 600\nzhahpvqiptvksnbjkdvmknb -11\njpdwmyke 711\nbjmj 652\naawtvezfntstrcpgbzjbf -1000\naawtvezfntstrcpgbzjbf -171\nbjmj -302\naawtvezfntstrcpgbzjbf 961\nzhahpvqiptvksnbjkdvmknb 848\nbjmj -735",
"output": "aawtvezfntstrcpgbzjbf"
},
{
"input": "17\nqdplghhx -649\nivhgbxiv 424\nivhgbxiv -72\nivhgbxiv 479\nugowfouebkelargxylsxn 83\nugowfouebkelargxylsxn -840\nivhgbxiv 382\nqdplghhx -904\nivhgbxiv -710\nugowfouebkelargxylsxn -838\nqdplghhx -974\nqdplghhx 571\nivhgbxiv -145\nugowfouebkelargxylsxn -788\nivhgbxiv 209\nivhgbxiv 420\nqdplghhx -158",
"output": "ivhgbxiv"
}
] | 1,689,949,113
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 20
| 124
| 0
|
from sys import stdin
input = lambda: stdin.readline().rstrip('\r\n')
n=int(input())
p,q=[],{}
for _ in range(n):
a,b = input().split()
q[a] = q.get(a,0)+int(b)
p.append([a,q[a]])
maxi = max(q.values())
for i,j in p:
if q[i]==maxi and j>=maxi:
print(i)
break
|
Title: Winner
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number of such players is more than one. During each round a player gains or loses a particular number of points. In the course of the game the number of points is registered in the line "name score", where name is a player's name, and score is the number of points gained in this round, which is an integer number. If score is negative, this means that the player has lost in the round. So, if two or more players have the maximum number of points (say, it equals to *m*) at the end of the game, than wins the one of them who scored at least *m* points first. Initially each player has 0 points. It's guaranteed that at the end of the game at least one player has a positive number of points.
Input Specification:
The first line contains an integer number *n* (1<=<=≤<=<=*n*<=<=≤<=<=1000), *n* is the number of rounds played. Then follow *n* lines, containing the information about the rounds in "name score" format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is an integer number between -1000 and 1000, inclusive.
Output Specification:
Print the name of the winner.
Demo Input:
['3\nmike 3\nandrew 5\nmike 2\n', '3\nandrew 3\nandrew 2\nmike 5\n']
Demo Output:
['andrew\n', 'andrew\n']
Note:
none
|
```python
from sys import stdin
input = lambda: stdin.readline().rstrip('\r\n')
n=int(input())
p,q=[],{}
for _ in range(n):
a,b = input().split()
q[a] = q.get(a,0)+int(b)
p.append([a,q[a]])
maxi = max(q.values())
for i,j in p:
if q[i]==maxi and j>=maxi:
print(i)
break
```
| 3.938
|
266
|
A
|
Stones on the Table
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
There are *n* stones on the table in a row, each of them can be red, green or blue. Count the minimum number of stones to take from the table so that any two neighboring stones had different colors. Stones in a row are considered neighboring if there are no other stones between them.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of stones on the table.
The next line contains string *s*, which represents the colors of the stones. We'll consider the stones in the row numbered from 1 to *n* from left to right. Then the *i*-th character *s* equals "R", if the *i*-th stone is red, "G", if it's green and "B", if it's blue.
|
Print a single integer — the answer to the problem.
|
[
"3\nRRG\n",
"5\nRRRRR\n",
"4\nBRBG\n"
] |
[
"1\n",
"4\n",
"0\n"
] |
none
| 500
|
[
{
"input": "3\nRRG",
"output": "1"
},
{
"input": "5\nRRRRR",
"output": "4"
},
{
"input": "4\nBRBG",
"output": "0"
},
{
"input": "1\nB",
"output": "0"
},
{
"input": "2\nBG",
"output": "0"
},
{
"input": "3\nBGB",
"output": "0"
},
{
"input": "4\nRBBR",
"output": "1"
},
{
"input": "5\nRGGBG",
"output": "1"
},
{
"input": "10\nGGBRBRGGRB",
"output": "2"
},
{
"input": "50\nGRBGGRBRGRBGGBBBBBGGGBBBBRBRGBRRBRGBBBRBBRRGBGGGRB",
"output": "18"
},
{
"input": "15\nBRRBRGGBBRRRRGR",
"output": "6"
},
{
"input": "20\nRRGBBRBRGRGBBGGRGRRR",
"output": "6"
},
{
"input": "25\nBBGBGRBGGBRRBGRRBGGBBRBRB",
"output": "6"
},
{
"input": "30\nGRGGGBGGRGBGGRGRBGBGBRRRRRRGRB",
"output": "9"
},
{
"input": "35\nGBBGBRGBBGGRBBGBRRGGRRRRRRRBRBBRRGB",
"output": "14"
},
{
"input": "40\nGBBRRGBGGGRGGGRRRRBRBGGBBGGGBGBBBBBRGGGG",
"output": "20"
},
{
"input": "45\nGGGBBRBBRRGRBBGGBGRBRGGBRBRGBRRGBGRRBGRGRBRRG",
"output": "11"
},
{
"input": "50\nRBGGBGGRBGRBBBGBBGRBBBGGGRBBBGBBBGRGGBGGBRBGBGRRGG",
"output": "17"
},
{
"input": "50\nGGGBBRGGGGGRRGGRBGGRGBBRBRRBGRGBBBGBRBGRGBBGRGGBRB",
"output": "16"
},
{
"input": "50\nGBGRGRRBRRRRRGGBBGBRRRBBBRBBBRRGRBBRGBRBGGRGRBBGGG",
"output": "19"
},
{
"input": "10\nGRRBRBRBGR",
"output": "1"
},
{
"input": "10\nBRBGBGRRBR",
"output": "1"
},
{
"input": "20\nGBGBGGRRRRGRBBGRGRGR",
"output": "5"
},
{
"input": "20\nRRGGRBBGBBRBGRRBRRBG",
"output": "6"
},
{
"input": "30\nBGBRGBBBGRGBBRGBGRBBBRGGRRGRRB",
"output": "8"
},
{
"input": "30\nBBBBGGBRBGBBGBGBGBGGGRGRRGGBBB",
"output": "11"
},
{
"input": "40\nGBRRGRBGBRRGBRGGGBRGBGBRGBBRRGRGGBBGBGBB",
"output": "9"
},
{
"input": "40\nBRGRGGRGGRBBRRRBRBBGGGRRGBGBBGRBBRGBRRGG",
"output": "13"
},
{
"input": "50\nRBGBGGRRGGRGGBGBGRRBGGBGBRRBBGBBGBBBGBBRBBRBRBRGRG",
"output": "13"
},
{
"input": "50\nRBRRGBGRRRBGRRBGRRGRBBRBBRRBRGGBRBRRBGGRBGGBRBRGRB",
"output": "12"
},
{
"input": "2\nBB",
"output": "1"
},
{
"input": "50\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "49"
},
{
"input": "50\nRRRRRRRRGRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "47"
},
{
"input": "50\nRRRRRRRRRRRRGGRRRRRRRRRBRRRRRRRRRRRRRRBBRRRRRRRRRR",
"output": "43"
}
] | 1,694,529,094
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 34
| 92
| 0
|
a=b=0
input()
for x in input():
a+=b==x
b=x
print(a)
|
Title: Stones on the Table
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* stones on the table in a row, each of them can be red, green or blue. Count the minimum number of stones to take from the table so that any two neighboring stones had different colors. Stones in a row are considered neighboring if there are no other stones between them.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of stones on the table.
The next line contains string *s*, which represents the colors of the stones. We'll consider the stones in the row numbered from 1 to *n* from left to right. Then the *i*-th character *s* equals "R", if the *i*-th stone is red, "G", if it's green and "B", if it's blue.
Output Specification:
Print a single integer — the answer to the problem.
Demo Input:
['3\nRRG\n', '5\nRRRRR\n', '4\nBRBG\n']
Demo Output:
['1\n', '4\n', '0\n']
Note:
none
|
```python
a=b=0
input()
for x in input():
a+=b==x
b=x
print(a)
```
| 3
|
|
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"
}
] | 1,580,143,500
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 45
| 218
| 307,200
|
a,b=map(int,input().split());c=[];t=2
for i in list(range(b,a,-1)):
t=2
while((t**2)<=i):
if i%t==0:c.append(i);break
else:t+=1
print("NO" if b-a-len(c)>1 or b in c else "YES")
|
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
a,b=map(int,input().split());c=[];t=2
for i in list(range(b,a,-1)):
t=2
while((t**2)<=i):
if i%t==0:c.append(i);break
else:t+=1
print("NO" if b-a-len(c)>1 or b in c else "YES")
```
| 3.944928
|
158
|
A
|
Next Round
|
PROGRAMMING
| 800
|
[
"*special",
"implementation"
] | null | null |
"Contestant who earns a score equal to or greater than the *k*-th place finisher's score will advance to the next round, as long as the contestant earns a positive score..." — an excerpt from contest rules.
A total of *n* participants took part in the contest (*n*<=≥<=*k*), and you already know their scores. Calculate how many participants will advance to the next round.
|
The first line of the input contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=50) separated by a single space.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100), where *a**i* is the score earned by the participant who got the *i*-th place. The given sequence is non-increasing (that is, for all *i* from 1 to *n*<=-<=1 the following condition is fulfilled: *a**i*<=≥<=*a**i*<=+<=1).
|
Output the number of participants who advance to the next round.
|
[
"8 5\n10 9 8 7 7 7 5 5\n",
"4 2\n0 0 0 0\n"
] |
[
"6\n",
"0\n"
] |
In the first example the participant on the 5th place earned 7 points. As the participant on the 6th place also earned 7 points, there are 6 advancers.
In the second example nobody got a positive score.
| 500
|
[
{
"input": "8 5\n10 9 8 7 7 7 5 5",
"output": "6"
},
{
"input": "4 2\n0 0 0 0",
"output": "0"
},
{
"input": "5 1\n1 1 1 1 1",
"output": "5"
},
{
"input": "5 5\n1 1 1 1 1",
"output": "5"
},
{
"input": "1 1\n10",
"output": "1"
},
{
"input": "17 14\n16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0",
"output": "14"
},
{
"input": "5 5\n3 2 1 0 0",
"output": "3"
},
{
"input": "8 6\n10 9 8 7 7 7 5 5",
"output": "6"
},
{
"input": "8 7\n10 9 8 7 7 7 5 5",
"output": "8"
},
{
"input": "8 4\n10 9 8 7 7 7 5 5",
"output": "6"
},
{
"input": "8 3\n10 9 8 7 7 7 5 5",
"output": "3"
},
{
"input": "8 1\n10 9 8 7 7 7 5 5",
"output": "1"
},
{
"input": "8 2\n10 9 8 7 7 7 5 5",
"output": "2"
},
{
"input": "1 1\n100",
"output": "1"
},
{
"input": "1 1\n0",
"output": "0"
},
{
"input": "50 25\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": "50"
},
{
"input": "50 25\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 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": "25"
},
{
"input": "50 25\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 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": "26"
},
{
"input": "50 25\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 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": "50"
},
{
"input": "11 5\n100 99 98 97 96 95 94 93 92 91 90",
"output": "5"
},
{
"input": "10 4\n100 81 70 69 64 43 34 29 15 3",
"output": "4"
},
{
"input": "11 6\n87 71 62 52 46 46 43 35 32 25 12",
"output": "6"
},
{
"input": "17 12\n99 88 86 82 75 75 74 65 58 52 45 30 21 16 7 2 2",
"output": "12"
},
{
"input": "20 3\n98 98 96 89 87 82 82 80 76 74 74 68 61 60 43 32 30 22 4 2",
"output": "3"
},
{
"input": "36 12\n90 87 86 85 83 80 79 78 76 70 69 69 61 61 59 58 56 48 45 44 42 41 33 31 27 25 23 21 20 19 15 14 12 7 5 5",
"output": "12"
},
{
"input": "49 8\n99 98 98 96 92 92 90 89 89 86 86 85 83 80 79 76 74 69 67 67 58 56 55 51 49 47 47 46 45 41 41 40 39 34 34 33 25 23 18 15 13 13 11 9 5 4 3 3 1",
"output": "9"
},
{
"input": "49 29\n100 98 98 96 96 96 95 87 85 84 81 76 74 70 63 63 63 62 57 57 56 54 53 52 50 47 45 41 41 39 38 31 30 28 27 26 23 22 20 15 15 11 7 6 6 4 2 1 0",
"output": "29"
},
{
"input": "49 34\n99 98 96 96 93 92 90 89 88 86 85 85 82 76 73 69 66 64 63 63 60 59 57 57 56 55 54 54 51 48 47 44 42 42 40 39 38 36 33 26 24 23 19 17 17 14 12 7 4",
"output": "34"
},
{
"input": "50 44\n100 100 99 97 95 91 91 84 83 83 79 71 70 69 69 62 61 60 59 59 58 58 58 55 55 54 52 48 47 45 44 44 38 36 32 31 28 28 25 25 24 24 24 22 17 15 14 13 12 4",
"output": "44"
},
{
"input": "50 13\n99 95 94 94 88 87 81 79 78 76 74 72 72 69 68 67 67 67 66 63 62 61 58 57 55 55 54 51 50 50 48 48 42 41 38 35 34 32 31 30 26 24 13 13 12 6 5 4 3 3",
"output": "13"
},
{
"input": "50 30\n100 98 96 94 91 89 88 81 81 81 81 81 76 73 72 71 70 69 66 64 61 59 59 56 52 50 49 48 43 39 36 35 34 34 31 29 27 26 24 22 16 16 15 14 14 14 9 7 4 3",
"output": "30"
},
{
"input": "2 1\n10 10",
"output": "2"
},
{
"input": "2 2\n10 10",
"output": "2"
},
{
"input": "2 2\n10 0",
"output": "1"
},
{
"input": "2 2\n10 1",
"output": "2"
},
{
"input": "2 1\n10 0",
"output": "1"
},
{
"input": "2 1\n10 2",
"output": "1"
},
{
"input": "50 13\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": "0"
},
{
"input": "50 1\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": "0"
},
{
"input": "50 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": "0"
},
{
"input": "10 1\n5 5 5 3 3 3 0 0 0 0",
"output": "3"
},
{
"input": "10 2\n5 5 5 3 3 3 0 0 0 0",
"output": "3"
},
{
"input": "10 3\n5 5 5 3 3 3 0 0 0 0",
"output": "3"
},
{
"input": "10 4\n5 5 5 3 3 3 0 0 0 0",
"output": "6"
},
{
"input": "10 5\n5 5 5 3 3 3 0 0 0 0",
"output": "6"
},
{
"input": "10 6\n5 5 5 3 3 3 0 0 0 0",
"output": "6"
},
{
"input": "10 7\n5 5 5 3 3 3 0 0 0 0",
"output": "6"
},
{
"input": "10 8\n5 5 5 3 3 3 0 0 0 0",
"output": "6"
},
{
"input": "10 9\n5 5 5 3 3 3 0 0 0 0",
"output": "6"
},
{
"input": "10 10\n5 5 5 3 3 3 0 0 0 0",
"output": "6"
}
] | 1,699,542,908
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 50
| 92
| 0
|
n, k = map(int, input().split())
scores = list(map(int, input().split()))
mark = scores[k - 1]
count = sum(1 for m in scores if m >= mark and m > 0)
print(count)
|
Title: Next Round
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
"Contestant who earns a score equal to or greater than the *k*-th place finisher's score will advance to the next round, as long as the contestant earns a positive score..." — an excerpt from contest rules.
A total of *n* participants took part in the contest (*n*<=≥<=*k*), and you already know their scores. Calculate how many participants will advance to the next round.
Input Specification:
The first line of the input contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=50) separated by a single space.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100), where *a**i* is the score earned by the participant who got the *i*-th place. The given sequence is non-increasing (that is, for all *i* from 1 to *n*<=-<=1 the following condition is fulfilled: *a**i*<=≥<=*a**i*<=+<=1).
Output Specification:
Output the number of participants who advance to the next round.
Demo Input:
['8 5\n10 9 8 7 7 7 5 5\n', '4 2\n0 0 0 0\n']
Demo Output:
['6\n', '0\n']
Note:
In the first example the participant on the 5th place earned 7 points. As the participant on the 6th place also earned 7 points, there are 6 advancers.
In the second example nobody got a positive score.
|
```python
n, k = map(int, input().split())
scores = list(map(int, input().split()))
mark = scores[k - 1]
count = sum(1 for m in scores if m >= mark and m > 0)
print(count)
```
| 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"
}
] | 1,693,225,128
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 35
| 92
| 0
|
def max_dominoes(M, N):
if M % 2 == 1 or N % 2 == 1:
return (M * N) // 2
else:
return (M * N) // 2
# Read input
M, N = map(int, input().split())
# Calculate and print the maximum number of dominoes
print(max_dominoes(M, N))
|
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
def max_dominoes(M, N):
if M % 2 == 1 or N % 2 == 1:
return (M * N) // 2
else:
return (M * N) // 2
# Read input
M, N = map(int, input().split())
# Calculate and print the maximum number of dominoes
print(max_dominoes(M, N))
```
| 3.977
|
935
|
A
|
Fafa and his Company
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation"
] | null | null |
Fafa owns a company that works on huge projects. There are *n* employees in Fafa's company. Whenever the company has a new project to start working on, Fafa has to divide the tasks of this project among all the employees.
Fafa finds doing this every time is very tiring for him. So, he decided to choose the best *l* employees in his company as team leaders. Whenever there is a new project, Fafa will divide the tasks among only the team leaders and each team leader will be responsible of some positive number of employees to give them the tasks. To make this process fair for the team leaders, each one of them should be responsible for the same number of employees. Moreover, every employee, who is not a team leader, has to be under the responsibility of exactly one team leader, and no team leader is responsible for another team leader.
Given the number of employees *n*, find in how many ways Fafa could choose the number of team leaders *l* in such a way that it is possible to divide employees between them evenly.
|
The input consists of a single line containing a positive integer *n* (2<=≤<=*n*<=≤<=105) — the number of employees in Fafa's company.
|
Print a single integer representing the answer to the problem.
|
[
"2\n",
"10\n"
] |
[
"1\n",
"3\n"
] |
In the second sample Fafa has 3 ways:
- choose only 1 employee as a team leader with 9 employees under his responsibility. - choose 2 employees as team leaders with 4 employees under the responsibility of each of them. - choose 5 employees as team leaders with 1 employee under the responsibility of each of them.
| 500
|
[
{
"input": "2",
"output": "1"
},
{
"input": "10",
"output": "3"
},
{
"input": "3",
"output": "1"
},
{
"input": "4",
"output": "2"
},
{
"input": "6",
"output": "3"
},
{
"input": "13",
"output": "1"
},
{
"input": "100000",
"output": "35"
},
{
"input": "1024",
"output": "10"
},
{
"input": "99999",
"output": "11"
},
{
"input": "10007",
"output": "1"
},
{
"input": "4096",
"output": "12"
},
{
"input": "65536",
"output": "16"
},
{
"input": "40320",
"output": "95"
},
{
"input": "30030",
"output": "63"
},
{
"input": "161",
"output": "3"
},
{
"input": "1000",
"output": "15"
},
{
"input": "10000",
"output": "24"
},
{
"input": "777",
"output": "7"
},
{
"input": "121",
"output": "2"
},
{
"input": "25",
"output": "2"
},
{
"input": "40000",
"output": "34"
},
{
"input": "99990",
"output": "47"
},
{
"input": "98765",
"output": "3"
},
{
"input": "56789",
"output": "3"
},
{
"input": "13579",
"output": "3"
},
{
"input": "97531",
"output": "3"
},
{
"input": "12345",
"output": "7"
},
{
"input": "54321",
"output": "7"
},
{
"input": "83160",
"output": "127"
},
{
"input": "9",
"output": "2"
},
{
"input": "21",
"output": "3"
},
{
"input": "11",
"output": "1"
},
{
"input": "15",
"output": "3"
}
] | 1,637,335,792
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 33
| 61
| 0
|
n=int(input())
c=0
if n<=2:
print(1)
else:
for i in range(1,(n//2)+1):
if n%i==0:
c+=1
print(c)
|
Title: Fafa and his Company
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Fafa owns a company that works on huge projects. There are *n* employees in Fafa's company. Whenever the company has a new project to start working on, Fafa has to divide the tasks of this project among all the employees.
Fafa finds doing this every time is very tiring for him. So, he decided to choose the best *l* employees in his company as team leaders. Whenever there is a new project, Fafa will divide the tasks among only the team leaders and each team leader will be responsible of some positive number of employees to give them the tasks. To make this process fair for the team leaders, each one of them should be responsible for the same number of employees. Moreover, every employee, who is not a team leader, has to be under the responsibility of exactly one team leader, and no team leader is responsible for another team leader.
Given the number of employees *n*, find in how many ways Fafa could choose the number of team leaders *l* in such a way that it is possible to divide employees between them evenly.
Input Specification:
The input consists of a single line containing a positive integer *n* (2<=≤<=*n*<=≤<=105) — the number of employees in Fafa's company.
Output Specification:
Print a single integer representing the answer to the problem.
Demo Input:
['2\n', '10\n']
Demo Output:
['1\n', '3\n']
Note:
In the second sample Fafa has 3 ways:
- choose only 1 employee as a team leader with 9 employees under his responsibility. - choose 2 employees as team leaders with 4 employees under the responsibility of each of them. - choose 5 employees as team leaders with 1 employee under the responsibility of each of them.
|
```python
n=int(input())
c=0
if n<=2:
print(1)
else:
for i in range(1,(n//2)+1):
if n%i==0:
c+=1
print(c)
```
| 3
|
|
989
|
A
|
A Blend of Springtime
|
PROGRAMMING
| 900
|
[
"implementation",
"strings"
] | null | null |
"What a pity it's already late spring," sighs Mino with regret, "one more drizzling night and they'd be gone."
"But these blends are at their best, aren't they?" Absorbed in the landscape, Kanno remains optimistic.
The landscape can be expressed as a row of consecutive cells, each of which either contains a flower of colour amber or buff or canary yellow, or is empty.
When a flower withers, it disappears from the cell that it originally belonged to, and it spreads petals of its colour in its two neighbouring cells (or outside the field if the cell is on the side of the landscape). In case petals fall outside the given cells, they simply become invisible.
You are to help Kanno determine whether it's possible that after some (possibly none or all) flowers shed their petals, at least one of the cells contains all three colours, considering both petals and flowers. Note that flowers can wither in arbitrary order.
|
The first and only line of input contains a non-empty string $s$ consisting of uppercase English letters 'A', 'B', 'C' and characters '.' (dots) only ($\lvert s \rvert \leq 100$) — denoting cells containing an amber flower, a buff one, a canary yellow one, and no flowers, respectively.
|
Output "Yes" if it's possible that all three colours appear in some cell, and "No" otherwise.
You can print each letter in any case (upper or lower).
|
[
".BAC.\n",
"AA..CB\n"
] |
[
"Yes\n",
"No\n"
] |
In the first example, the buff and canary yellow flowers can leave their petals in the central cell, blending all three colours in it.
In the second example, it's impossible to satisfy the requirement because there is no way that amber and buff meet in any cell.
| 500
|
[
{
"input": ".BAC.",
"output": "Yes"
},
{
"input": "AA..CB",
"output": "No"
},
{
"input": ".",
"output": "No"
},
{
"input": "ACB.AAAAAA",
"output": "Yes"
},
{
"input": "B.BC.BBBCA",
"output": "Yes"
},
{
"input": "BA..CAB..B",
"output": "Yes"
},
{
"input": "CACCBAA.BC",
"output": "Yes"
},
{
"input": ".CAACCBBA.CBB.AC..BABCCBCCB..B.BC..CBC.CA.CC.C.CC.B.A.CC.BBCCBB..ACAACAC.CBCCB.AABAAC.CBCC.BA..CCBC.",
"output": "Yes"
},
{
"input": "A",
"output": "No"
},
{
"input": "..",
"output": "No"
},
{
"input": "BC",
"output": "No"
},
{
"input": "CAB",
"output": "Yes"
},
{
"input": "A.CB",
"output": "No"
},
{
"input": "B.ACAA.CA..CBCBBAA.B.CCBCB.CAC.ABC...BC.BCCC.BC.CB",
"output": "Yes"
},
{
"input": "B.B...CC.B..CCCB.CB..CBCB..CBCC.CCBC.B.CB..CA.C.C.",
"output": "No"
},
{
"input": "AA.CBAABABCCC..B..B.ABBABAB.B.B.CCA..CB.B...A..CBC",
"output": "Yes"
},
{
"input": "CA.ABB.CC.B.C.BBBABAAB.BBBAACACAAA.C.AACA.AAC.C.BCCB.CCBC.C..CCACA.CBCCB.CCAABAAB.AACAA..A.AAA.",
"output": "No"
},
{
"input": "CBC...AC.BBBB.BBABABA.CAAACC.AAABB..A.BA..BC.CBBBC.BBBBCCCAA.ACCBB.AB.C.BA..CC..AAAC...AB.A.AAABBA.A",
"output": "No"
},
{
"input": "CC.AAAC.BA.BBB.AABABBCCAA.A.CBCCB.B.BC.ABCBCBBAA.CACA.CCCA.CB.CCB.A.BCCCB...C.A.BCCBC..B.ABABB.C.BCB",
"output": "Yes"
},
{
"input": "CCC..A..CACACCA.CA.ABAAB.BBA..C.AAA...ACB.ACA.CA.B.AB.A..C.BC.BC.A.C....ABBCCACCCBCC.BBBAA.ACCACB.BB",
"output": "Yes"
},
{
"input": "BC.ABACAACC..AC.A..CCCAABBCCACAC.AA.CC.BAABABABBCBB.BA..C.C.C.A.BBA.C..BC.ACACCC.AAAACCCCC.AAC.AC.AB",
"output": "Yes"
},
{
"input": "ACAC.BAA.C..CAAC..ABBAACC..BAA...CC...ACCBBCA.BAABABAACCAC.A.BBCACCC..BCB.BABAAAACCBCB.BCAABBC.C.BBB",
"output": "Yes"
},
{
"input": "CCAC.BCBC.A.ABBAB.C.C.BC.CCABBCBCCBC..B.AA.C.BC...B..BAA.ACCCCBBB.AAAACA.CAACCB.CCB.CC.BCCAB.BBBBABB",
"output": "Yes"
},
{
"input": ".AACAA.AAAAC.BBBB.BC...CCACCACAAA.A..CCA..BCC.AB.ABAAB..AABA...B.C.CBAB.BAAB.A.C.AAC.BBBA.ACAAA.BB.C",
"output": "Yes"
},
{
"input": "CC.ACCC.BCCCCAA.BBAACB.ABABAAAA.A.CBAB.CBACBBC..C.CA.AAA..AA..ABBB.A.C..CBBCAAACC.B..CC.AC..CAABACB.",
"output": "Yes"
},
{
"input": ".BAB.",
"output": "No"
},
{
"input": "BBBBBBBBB",
"output": "No"
},
{
"input": "..AAC..",
"output": "No"
},
{
"input": ".AAABBBCCC.",
"output": "No"
},
{
"input": "AAABC",
"output": "Yes"
},
{
"input": "BBB",
"output": "No"
},
{
"input": "AAAAABABAAAAA",
"output": "No"
},
{
"input": "AABBCC",
"output": "No"
},
{
"input": ".BA",
"output": "No"
},
{
"input": "CAAAAB",
"output": "No"
}
] | 1,623,921,789
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 37
| 62
| 0
|
def f(l,s):
for i in l:
if i in s:
return "Yes"
return "No"
s=input()
l=["ABC","BAC","BCA","ACB","CBA","CAB"]
print(f(l,s))
|
Title: A Blend of Springtime
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
"What a pity it's already late spring," sighs Mino with regret, "one more drizzling night and they'd be gone."
"But these blends are at their best, aren't they?" Absorbed in the landscape, Kanno remains optimistic.
The landscape can be expressed as a row of consecutive cells, each of which either contains a flower of colour amber or buff or canary yellow, or is empty.
When a flower withers, it disappears from the cell that it originally belonged to, and it spreads petals of its colour in its two neighbouring cells (or outside the field if the cell is on the side of the landscape). In case petals fall outside the given cells, they simply become invisible.
You are to help Kanno determine whether it's possible that after some (possibly none or all) flowers shed their petals, at least one of the cells contains all three colours, considering both petals and flowers. Note that flowers can wither in arbitrary order.
Input Specification:
The first and only line of input contains a non-empty string $s$ consisting of uppercase English letters 'A', 'B', 'C' and characters '.' (dots) only ($\lvert s \rvert \leq 100$) — denoting cells containing an amber flower, a buff one, a canary yellow one, and no flowers, respectively.
Output Specification:
Output "Yes" if it's possible that all three colours appear in some cell, and "No" otherwise.
You can print each letter in any case (upper or lower).
Demo Input:
['.BAC.\n', 'AA..CB\n']
Demo Output:
['Yes\n', 'No\n']
Note:
In the first example, the buff and canary yellow flowers can leave their petals in the central cell, blending all three colours in it.
In the second example, it's impossible to satisfy the requirement because there is no way that amber and buff meet in any cell.
|
```python
def f(l,s):
for i in l:
if i in s:
return "Yes"
return "No"
s=input()
l=["ABC","BAC","BCA","ACB","CBA","CAB"]
print(f(l,s))
```
| 3
|
|
32
|
B
|
Borze
|
PROGRAMMING
| 800
|
[
"expression parsing",
"implementation"
] |
B. Borze
|
2
|
256
|
Ternary numeric notation is quite popular in Berland. To telegraph the ternary number the Borze alphabet is used. Digit 0 is transmitted as «.», 1 as «-.» and 2 as «--». You are to decode the Borze code, i.e. to find out the ternary number given its representation in Borze alphabet.
|
The first line contains a number in Borze code. The length of the string is between 1 and 200 characters. It's guaranteed that the given string is a valid Borze code of some ternary number (this number can have leading zeroes).
|
Output the decoded ternary number. It can have leading zeroes.
|
[
".-.--\n",
"--.\n",
"-..-.--\n"
] |
[
"012",
"20",
"1012"
] |
none
| 1,000
|
[
{
"input": ".-.--",
"output": "012"
},
{
"input": "--.",
"output": "20"
},
{
"input": "-..-.--",
"output": "1012"
},
{
"input": "---..",
"output": "210"
},
{
"input": "..--.---..",
"output": "0020210"
},
{
"input": "-.....----.",
"output": "10000220"
},
{
"input": ".",
"output": "0"
},
{
"input": "-.",
"output": "1"
},
{
"input": "--",
"output": "2"
},
{
"input": "..",
"output": "00"
},
{
"input": "--.",
"output": "20"
},
{
"input": ".--.",
"output": "020"
},
{
"input": ".-.-..",
"output": "0110"
},
{
"input": "----.-.",
"output": "2201"
},
{
"input": "-..--.-.",
"output": "10201"
},
{
"input": "..--..--.",
"output": "0020020"
},
{
"input": "-.-.---.--..-..-.-.-..-..-.--.",
"output": "112120010111010120"
},
{
"input": "---.-.-.------..-..-..-..-.-..-.--.-.-..-.-.-----..-.-.",
"output": "21112220010101011012011011221011"
},
{
"input": "-.-..--.-.-.-.-.-..-.-.-.---------.--.---..--...--.-----.-.-.-...--.-.-.---.------.--..-.--.-----.-...-..------",
"output": "11020111110111222212021020002022111100201121222020012022110010222"
},
{
"input": "-.-..-.--.---..---.-..---.-...-.-.----..-.---.-.---..-.--.---.-.-------.---.--....----.-.---.---.---.----.-----..---.-.-.-.-----.--.-------.-..",
"output": "110120210211021100112200121121012021122212120000220121212122022102111122120222110"
},
{
"input": ".-..-.-.---.-----.--.---...-.--.-.-....-..",
"output": "01011212212021001201100010"
},
{
"input": ".------.-.---..--...-..-..-.-.-.--.--.-..-.--...-.-.---.-.-.------..--..-.---..----.-..-.--.---.-.----.-.---...-.-.-.-----.-.-.---.---.-.....-.-...-----.-...-.---.-..-.-----.--...---.-.-..-.--.-.---..",
"output": "022201210200010101112020101200011211122200200121022010120211220121001112211121211000011002211001211012212000211101201210"
},
{
"input": ".-.--.---.-----.-.-----.-.-..-----..-..----..--.-.--.----..---.---..-.-.-----..-------.----..----.-..---...-----..-..-----...-..-.-.-----....---..---..-.-----...-.--...--.-.---.-.-.-.-.-...---..----.",
"output": "01202122112211102210102200201202200212101122102221220022010210022101022100101122100021021012210012000201211111100210220"
},
{
"input": "..-.-.-.---.-.-.-..-.-..-.-.---.-------.---..-----.---....-.---.--.--.-.---.---------.-..---.-.-.--..---.---.-.---.-.-..-.-..-.-.-.----.--.-....--------.-.---..----.------.-.-.--.--.-----.-----.----",
"output": "0011121111011011212221210221210001212020121222211021112002121121110110111220201000222201210220222011202022122122"
},
{
"input": "-..-------.------.-..--.-.-..--.-.-..-----..-.-.-..-..-..--.---..-----..---..-..--.-..-.-.---...-.....-------.---.-----.-...-.-...-.-.---.---.-----.--.--...-.--..-.-..-...-.-.-.-.---..---.-..-.-.-.-..",
"output": "102221222010201102011022101110101020210221021010201011210010000222121221100110011212122120200012001101001111210211011110"
},
{
"input": ".-.----.-.--..-.-.-.-..----..-.-...--.-.---.---.-------..-.--..-......--.------.--.----.--...-.--.--..-----..-.....--.--.-.-.------..--------.----------..-.---.----.---.-..--..-.....-..------.--.",
"output": "012201200111102200110020121212221012001000002022202022020001202002210100002020112220022220222220012122021102001000010222020"
},
{
"input": "------.-----.-....--.-.----.-.---.-.-..---.-.---.-----..-...-.-.---..-.-.-..-.-.-...-.-.-.----..--.------.----.-..-.--...-.-------...-.-..-.-.--.--.---..--..--------.--.-.-.---.-.-.-...----.--..-.--..",
"output": "222022110002012201211102112122101001121011101110011122002022202201012000122210011011202021020022220201121111002202001200"
},
{
"input": "-.---...----...--.--...-.--.----",
"output": "121002200020200012022"
},
{
"input": "--.--.--.---.--.-.---.-.-..-..--.-..---.-.....-..---.-----.--...-.-.-------.-.--.-.----.-..-.------.",
"output": "202020212012111010201021100001021221200011222112012201012220"
}
] | 1,683,557,556
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 92
| 0
|
s = input()
sol = ""
i = 0
while i < len(s):
if s[i] == ".":
sol += '0'
else:
i += 1
if s[i] == ".":
sol += '1'
else:
sol += '2'
i += 1
print(sol)
|
Title: Borze
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Ternary numeric notation is quite popular in Berland. To telegraph the ternary number the Borze alphabet is used. Digit 0 is transmitted as «.», 1 as «-.» and 2 as «--». You are to decode the Borze code, i.e. to find out the ternary number given its representation in Borze alphabet.
Input Specification:
The first line contains a number in Borze code. The length of the string is between 1 and 200 characters. It's guaranteed that the given string is a valid Borze code of some ternary number (this number can have leading zeroes).
Output Specification:
Output the decoded ternary number. It can have leading zeroes.
Demo Input:
['.-.--\n', '--.\n', '-..-.--\n']
Demo Output:
['012', '20', '1012']
Note:
none
|
```python
s = input()
sol = ""
i = 0
while i < len(s):
if s[i] == ".":
sol += '0'
else:
i += 1
if s[i] == ".":
sol += '1'
else:
sol += '2'
i += 1
print(sol)
```
| 3.977
|
270
|
A
|
Fancy Fence
|
PROGRAMMING
| 1,100
|
[
"geometry",
"implementation",
"math"
] | null | null |
Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot.
He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle *a*.
Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to *a*?
|
The first line of input contains an integer *t* (0<=<<=*t*<=<<=180) — the number of tests. Each of the following *t* lines contains a single integer *a* (0<=<<=*a*<=<<=180) — the angle the robot can make corners at measured in degrees.
|
For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible.
|
[
"3\n30\n60\n90\n"
] |
[
"NO\nYES\nYES\n"
] |
In the first test case, it is impossible to build the fence, since there is no regular polygon with angle <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/df5f4b07dd5316fde165b43657b2696e2919e791.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second test case, the fence is a regular triangle, and in the last test case — a square.
| 500
|
[
{
"input": "3\n30\n60\n90",
"output": "NO\nYES\nYES"
},
{
"input": "6\n1\n2\n3\n170\n179\n25",
"output": "NO\nNO\nNO\nYES\nYES\nNO"
}
] | 1,662,972,934
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 3
| 62
| 0
|
n = int(input())
ans = n * [0]
for i in range(n):
m = int(input())
if (360 % (180 - m) == 0):
ans[i] = 1
else:
continue
for i in range(n):
if ans[i] == 1:
print('YES')
else:
print('NO')
|
Title: Fancy Fence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot.
He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle *a*.
Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to *a*?
Input Specification:
The first line of input contains an integer *t* (0<=<<=*t*<=<<=180) — the number of tests. Each of the following *t* lines contains a single integer *a* (0<=<<=*a*<=<<=180) — the angle the robot can make corners at measured in degrees.
Output Specification:
For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible.
Demo Input:
['3\n30\n60\n90\n']
Demo Output:
['NO\nYES\nYES\n']
Note:
In the first test case, it is impossible to build the fence, since there is no regular polygon with angle <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/df5f4b07dd5316fde165b43657b2696e2919e791.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second test case, the fence is a regular triangle, and in the last test case — a square.
|
```python
n = int(input())
ans = n * [0]
for i in range(n):
m = int(input())
if (360 % (180 - m) == 0):
ans[i] = 1
else:
continue
for i in range(n):
if ans[i] == 1:
print('YES')
else:
print('NO')
```
| 3
|
|
61
|
A
|
Ultra-Fast Mathematician
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Ultra-Fast Mathematician
|
2
|
256
|
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
|
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
|
Write one line — the corresponding answer. Do not omit the leading 0s.
|
[
"1010100\n0100101\n",
"000\n111\n",
"1110\n1010\n",
"01110\n01100\n"
] |
[
"1110001\n",
"111\n",
"0100\n",
"00010\n"
] |
none
| 500
|
[
{
"input": "1010100\n0100101",
"output": "1110001"
},
{
"input": "000\n111",
"output": "111"
},
{
"input": "1110\n1010",
"output": "0100"
},
{
"input": "01110\n01100",
"output": "00010"
},
{
"input": "011101\n000001",
"output": "011100"
},
{
"input": "10\n01",
"output": "11"
},
{
"input": "00111111\n11011101",
"output": "11100010"
},
{
"input": "011001100\n101001010",
"output": "110000110"
},
{
"input": "1100100001\n0110101100",
"output": "1010001101"
},
{
"input": "00011101010\n10010100101",
"output": "10001001111"
},
{
"input": "100000101101\n111010100011",
"output": "011010001110"
},
{
"input": "1000001111010\n1101100110001",
"output": "0101101001011"
},
{
"input": "01011111010111\n10001110111010",
"output": "11010001101101"
},
{
"input": "110010000111100\n001100101011010",
"output": "111110101100110"
},
{
"input": "0010010111110000\n0000000011010110",
"output": "0010010100100110"
},
{
"input": "00111110111110000\n01111100001100000",
"output": "01000010110010000"
},
{
"input": "101010101111010001\n001001111101111101",
"output": "100011010010101100"
},
{
"input": "0110010101111100000\n0011000101000000110",
"output": "0101010000111100110"
},
{
"input": "11110100011101010111\n00001000011011000000",
"output": "11111100000110010111"
},
{
"input": "101010101111101101001\n111010010010000011111",
"output": "010000111101101110110"
},
{
"input": "0000111111100011000010\n1110110110110000001010",
"output": "1110001001010011001000"
},
{
"input": "10010010101000110111000\n00101110100110111000111",
"output": "10111100001110001111111"
},
{
"input": "010010010010111100000111\n100100111111100011001110",
"output": "110110101101011111001001"
},
{
"input": "0101110100100111011010010\n0101100011010111001010001",
"output": "0000010111110000010000011"
},
{
"input": "10010010100011110111111011\n10000110101100000001000100",
"output": "00010100001111110110111111"
},
{
"input": "000001111000000100001000000\n011100111101111001110110001",
"output": "011101000101111101111110001"
},
{
"input": "0011110010001001011001011100\n0000101101000011101011001010",
"output": "0011011111001010110010010110"
},
{
"input": "11111000000000010011001101111\n11101110011001010100010000000",
"output": "00010110011001000111011101111"
},
{
"input": "011001110000110100001100101100\n001010000011110000001000101001",
"output": "010011110011000100000100000101"
},
{
"input": "1011111010001100011010110101111\n1011001110010000000101100010101",
"output": "0000110100011100011111010111010"
},
{
"input": "10111000100001000001010110000001\n10111000001100101011011001011000",
"output": "00000000101101101010001111011001"
},
{
"input": "000001010000100001000000011011100\n111111111001010100100001100000111",
"output": "111110101001110101100001111011011"
},
{
"input": "1101000000000010011011101100000110\n1110000001100010011010000011011110",
"output": "0011000001100000000001101111011000"
},
{
"input": "01011011000010100001100100011110001\n01011010111000001010010100001110000",
"output": "00000001111010101011110000010000001"
},
{
"input": "000011111000011001000110111100000100\n011011000110000111101011100111000111",
"output": "011000111110011110101101011011000011"
},
{
"input": "1001000010101110001000000011111110010\n0010001011010111000011101001010110000",
"output": "1011001001111001001011101010101000010"
},
{
"input": "00011101011001100101111111000000010101\n10010011011011001011111000000011101011",
"output": "10001110000010101110000111000011111110"
},
{
"input": "111011100110001001101111110010111001010\n111111101101111001110010000101101000100",
"output": "000100001011110000011101110111010001110"
},
{
"input": "1111001001101000001000000010010101001010\n0010111100111110001011000010111110111001",
"output": "1101110101010110000011000000101011110011"
},
{
"input": "00100101111000000101011111110010100011010\n11101110001010010101001000111110101010100",
"output": "11001011110010010000010111001100001001110"
},
{
"input": "101011001110110100101001000111010101101111\n100111100110101011010100111100111111010110",
"output": "001100101000011111111101111011101010111001"
},
{
"input": "1111100001100101000111101001001010011100001\n1000110011000011110010001011001110001000001",
"output": "0111010010100110110101100010000100010100000"
},
{
"input": "01100111011111010101000001101110000001110101\n10011001011111110000000101011001001101101100",
"output": "11111110000000100101000100110111001100011001"
},
{
"input": "110010100111000100100101100000011100000011001\n011001111011100110000110111001110110100111011",
"output": "101011011100100010100011011001101010100100010"
},
{
"input": "0001100111111011010110100100111000000111000110\n1100101011000000000001010010010111001100110001",
"output": "1101001100111011010111110110101111001011110111"
},
{
"input": "00000101110110110001110010100001110100000100000\n10010000110011110001101000111111101010011010001",
"output": "10010101000101000000011010011110011110011110001"
},
{
"input": "110000100101011100100011001111110011111110010001\n101011111001011100110110111101110011010110101100",
"output": "011011011100000000010101110010000000101000111101"
},
{
"input": "0101111101011111010101011101000011101100000000111\n0000101010110110001110101011011110111001010100100",
"output": "0101010111101001011011110110011101010101010100011"
},
{
"input": "11000100010101110011101000011111001010110111111100\n00001111000111001011111110000010101110111001000011",
"output": "11001011010010111000010110011101100100001110111111"
},
{
"input": "101000001101111101101111111000001110110010101101010\n010011100111100001100000010001100101000000111011011",
"output": "111011101010011100001111101001101011110010010110001"
},
{
"input": "0011111110010001010100010110111000110011001101010100\n0111000000100010101010000100101000000100101000111001",
"output": "0100111110110011111110010010010000110111100101101101"
},
{
"input": "11101010000110000011011010000001111101000111011111100\n10110011110001010100010110010010101001010111100100100",
"output": "01011001110111010111001100010011010100010000111011000"
},
{
"input": "011000100001000001101000010110100110011110100111111011\n111011001000001001110011001111011110111110110011011111",
"output": "100011101001001000011011011001111000100000010100100100"
},
{
"input": "0111010110010100000110111011010110100000000111110110000\n1011100100010001101100000100111111101001110010000100110",
"output": "1100110010000101101010111111101001001001110101110010110"
},
{
"input": "10101000100111000111010001011011011011110100110101100011\n11101111000000001100100011111000100100000110011001101110",
"output": "01000111100111001011110010100011111111110010101100001101"
},
{
"input": "000000111001010001000000110001001011100010011101010011011\n110001101000010010000101000100001111101001100100001010010",
"output": "110001010001000011000101110101000100001011111001011001001"
},
{
"input": "0101011100111010000111110010101101111111000000111100011100\n1011111110000010101110111001000011100000100111111111000111",
"output": "1110100010111000101001001011101110011111100111000011011011"
},
{
"input": "11001000001100100111100111100100101011000101001111001001101\n10111110100010000011010100110100100011101001100000001110110",
"output": "01110110101110100100110011010000001000101100101111000111011"
},
{
"input": "010111011011101000000110000110100110001110100001110110111011\n101011110011101011101101011111010100100001100111100100111011",
"output": "111100101000000011101011011001110010101111000110010010000000"
},
{
"input": "1001011110110110000100011001010110000100011010010111010101110\n1101111100001000010111110011010101111010010100000001000010111",
"output": "0100100010111110010011101010000011111110001110010110010111001"
},
{
"input": "10000010101111100111110101111000010100110111101101111111111010\n10110110101100101010011001011010100110111011101100011001100111",
"output": "00110100000011001101101100100010110010001100000001100110011101"
},
{
"input": "011111010011111000001010101001101001000010100010111110010100001\n011111001011000011111001000001111001010110001010111101000010011",
"output": "000000011000111011110011101000010000010100101000000011010110010"
},
{
"input": "1111000000110001011101000100100100001111011100001111001100011111\n1101100110000101100001100000001001011011111011010101000101001010",
"output": "0010100110110100111100100100101101010100100111011010001001010101"
},
{
"input": "01100000101010010011001110100110110010000110010011011001100100011\n10110110010110111100100111000111000110010000000101101110000010111",
"output": "11010110111100101111101001100001110100010110010110110111100110100"
},
{
"input": "001111111010000100001100001010011001111110011110010111110001100111\n110000101001011000100010101100100110000111100000001101001110010111",
"output": "111111010011011100101110100110111111111001111110011010111111110000"
},
{
"input": "1011101011101101011110101101011101011000010011100101010101000100110\n0001000001001111010111100100111101100000000001110001000110000000110",
"output": "1010101010100010001001001001100000111000010010010100010011000100000"
},
{
"input": "01000001011001010011011100010000100100110101111011011011110000001110\n01011110000110011011000000000011000111100001010000000011111001110000",
"output": "00011111011111001000011100010011100011010100101011011000001001111110"
},
{
"input": "110101010100110101000001111110110100010010000100111110010100110011100\n111010010111111011100110101011001011001110110111110100000110110100111",
"output": "001111000011001110100111010101111111011100110011001010010010000111011"
},
{
"input": "1001101011000001011111100110010010000011010001001111011100010100110001\n1111100111110101001111010001010000011001001001010110001111000000100101",
"output": "0110001100110100010000110111000010011010011000011001010011010100010100"
},
{
"input": "00000111110010110001110110001010010101000111011001111111100110011110010\n00010111110100000100110101000010010001100001100011100000001100010100010",
"output": "00010000000110110101000011001000000100100110111010011111101010001010000"
},
{
"input": "100101011100101101000011010001011001101110101110001100010001010111001110\n100001111100101011011111110000001111000111001011111110000010101110111001",
"output": "000100100000000110011100100001010110101001100101110010010011111001110111"
},
{
"input": "1101100001000111001101001011101000111000011110000001001101101001111011010\n0101011101010100011011010110101000010010110010011110101100000110110001000",
"output": "1000111100010011010110011101000000101010101100011111100001101111001010010"
},
{
"input": "01101101010011110101100001110101111011100010000010001101111000011110111111\n00101111001101001100111010000101110000100101101111100111101110010100011011",
"output": "01000010011110111001011011110000001011000111101101101010010110001010100100"
},
{
"input": "101100101100011001101111110110110010100110110010100001110010110011001101011\n000001011010101011110011111101001110000111000010001101000010010000010001101",
"output": "101101110110110010011100001011111100100001110000101100110000100011011100110"
},
{
"input": "0010001011001010001100000010010011110110011000100000000100110000101111001110\n1100110100111000110100001110111001011101001100001010100001010011100110110001",
"output": "1110111111110010111000001100101010101011010100101010100101100011001001111111"
},
{
"input": "00101101010000000101011001101011001100010001100000101011101110000001111001000\n10010110010111000000101101000011101011001010000011011101101011010000000011111",
"output": "10111011000111000101110100101000100111011011100011110110000101010001111010111"
},
{
"input": "111100000100100000101001100001001111001010001000001000000111010000010101101011\n001000100010100101111011111011010110101100001111011000010011011011100010010110",
"output": "110100100110000101010010011010011001100110000111010000010100001011110111111101"
},
{
"input": "0110001101100100001111110101101000100101010010101010011001101001001101110000000\n0111011000000010010111011110010000000001000110001000011001101000000001110100111",
"output": "0001010101100110011000101011111000100100010100100010000000000001001100000100111"
},
{
"input": "10001111111001000101001011110101111010100001011010101100111001010001010010001000\n10000111010010011110111000111010101100000011110001101111001000111010100000000001",
"output": "00001000101011011011110011001111010110100010101011000011110001101011110010001001"
},
{
"input": "100110001110110000100101001110000011110110000110000000100011110100110110011001101\n110001110101110000000100101001101011111100100100001001000110000001111100011110110",
"output": "010111111011000000100001100111101000001010100010001001100101110101001010000111011"
},
{
"input": "0000010100100000010110111100011111111010011101000000100000011001001101101100111010\n0100111110011101010110101011110110010111001111000110101100101110111100101000111111",
"output": "0100101010111101000000010111101001101101010010000110001100110111110001000100000101"
},
{
"input": "11000111001010100001110000001001011010010010110000001110100101000001010101100110111\n11001100100100100001101010110100000111100011101110011010110100001001000011011011010",
"output": "00001011101110000000011010111101011101110001011110010100010001001000010110111101101"
},
{
"input": "010110100010001000100010101001101010011010111110100001000100101000111011100010100001\n110000011111101101010011111000101010111010100001001100001001100101000000111000000000",
"output": "100110111101100101110001010001000000100000011111101101001101001101111011011010100001"
},
{
"input": "0000011110101110010101110110110101100001011001101010101001000010000010000000101001101\n1100111111011100000110000111101110011111100111110001011001000010011111100001001100011",
"output": "1100100001110010010011110001011011111110111110011011110000000000011101100001100101110"
},
{
"input": "10100000101101110001100010010010100101100011010010101000110011100000101010110010000000\n10001110011011010010111011011101101111000111110000111000011010010101001100000001010011",
"output": "00101110110110100011011001001111001010100100100010010000101001110101100110110011010011"
},
{
"input": "001110000011111101101010011111000101010111010100001001100001001100101000000111000000000\n111010000000000000101001110011001000111011001100101010011001000011101001001011110000011",
"output": "110100000011111101000011101100001101101100011000100011111000001111000001001100110000011"
},
{
"input": "1110111100111011010101011011001110001010010010110011110010011111000010011111010101100001\n1001010101011001001010100010101100000110111101011000100010101111111010111100001110010010",
"output": "0111101001100010011111111001100010001100101111101011010000110000111000100011011011110011"
},
{
"input": "11100010001100010011001100001100010011010001101110011110100101110010101101011101000111111\n01110000000110111010110100001010000101011110100101010011000110101110101101110111011110001",
"output": "10010010001010101001111000000110010110001111001011001101100011011100000000101010011001110"
},
{
"input": "001101011001100101101100110000111000101011001001100100000100101000100000110100010111111101\n101001111110000010111101111110001001111001111101111010000110111000100100110010010001011111",
"output": "100100100111100111010001001110110001010010110100011110000010010000000100000110000110100010"
},
{
"input": "1010110110010101000110010010110101011101010100011001101011000110000000100011100100011000000\n0011011111100010001111101101000111001011101110100000110111100100101111010110101111011100011",
"output": "1001101001110111001001111111110010010110111010111001011100100010101111110101001011000100011"
},
{
"input": "10010010000111010111011111110010100101100000001100011100111011100010000010010001011100001100\n00111010100010110010000100010111010001111110100100100011101000101111111111001101101100100100",
"output": "10101000100101100101011011100101110100011110101000111111010011001101111101011100110000101000"
},
{
"input": "010101110001010101100000010111010000000111110011001101100011001000000011001111110000000010100\n010010111011100101010101111110110000000111000100001101101001001000001100101110001010000100001",
"output": "000111001010110000110101101001100000000000110111000000001010000000001111100001111010000110101"
},
{
"input": "1100111110011001000111101001001011000110011010111111100010111111001100111111011101100111101011\n1100000011001000110100110111000001011001010111101000010010100011000001100100111101101000010110",
"output": "0000111101010001110011011110001010011111001101010111110000011100001101011011100000001111111101"
},
{
"input": "00011000100100110111100101100100000000010011110111110010101110110011100001010111010011110100101\n00011011111011111011100101100111100101001110010111000010000111000100100100000001110101111011011",
"output": "00000011011111001100000000000011100101011101100000110000101001110111000101010110100110001111110"
},
{
"input": "000101011001001100000111100010110101111011110101111101000110001101011010111110110011100100000001\n011000101010011111011000111000100000000011011000000001111110001000001111101010110000011100001111",
"output": "011101110011010011011111011010010101111000101101111100111000000101010101010100000011111000001110"
},
{
"input": "1000101001011010000100100100010010011101011001110101111011101111111110010101001101010001010101001\n0110110010011100011111011111110111000000010001110100001010111110101011010011111011111110000110000",
"output": "1110011011000110011011111011100101011101001000000001110001010001010101000110110110101111010011001"
},
{
"input": "01111010010101100111110111111011011010100001011101010000111100101101101110111011001100101011100111\n00001100110110100001111011000010001001001100000010111101000001111011100000010111010010000011000010",
"output": "01110110100011000110001100111001010011101101011111101101111101010110001110101100011110101000100101"
},
{
"input": "111101011101111000001011001101111010110111001101110100100011111011011101001101010101011010111000110\n101111100101101101001100110011000001111010011101110111110110101110011011110011111100001001110101101",
"output": "010010111000010101000111111110111011001101010000000011010101010101000110111110101001010011001101011"
},
{
"input": "1010000011010110011101001101110001110010000111011101110010110110111111001001110100101100010101010001\n0111101000111100101100000101111010100100001000011101010100110011100011010011010101000100101011100011",
"output": "1101101011101010110001001000001011010110001111000000100110000101011100011010100001101000111110110010"
},
{
"input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001\n1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111010",
"output": "1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111011"
},
{
"input": "0\n0",
"output": "0"
},
{
"input": "0\n1",
"output": "1"
}
] | 1,655,463,776
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 102
| 62
| 0
|
from io import BytesIO
from os import fstat, read
from sys import stdout
def fast_input(file_no = 0):
byte_stream = BytesIO(read(file_no, fstat(file_no).st_size))
return byte_stream
#fi = open(PATH_INPUT, "r")
#io_byte_input = fast_input(fi.fileno())
io_byte_input = fast_input()
#fi.close()
f_input = lambda: io_byte_input.readline().decode().strip()
def f_print(*output, sep = "\n"):
for i in output:
stdout.write(str(i) + sep)
#if sep != "\n":
# stdout.write("\n")
s1 = f_input()
s2 = f_input()
ans = ""
for i in range(len(s1)):
if s1[i] != s2[i]:
ans += '1'
else:
ans += '0'
f_print(ans)
|
Title: Ultra-Fast Mathematician
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
Input Specification:
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
Output Specification:
Write one line — the corresponding answer. Do not omit the leading 0s.
Demo Input:
['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n']
Demo Output:
['1110001\n', '111\n', '0100\n', '00010\n']
Note:
none
|
```python
from io import BytesIO
from os import fstat, read
from sys import stdout
def fast_input(file_no = 0):
byte_stream = BytesIO(read(file_no, fstat(file_no).st_size))
return byte_stream
#fi = open(PATH_INPUT, "r")
#io_byte_input = fast_input(fi.fileno())
io_byte_input = fast_input()
#fi.close()
f_input = lambda: io_byte_input.readline().decode().strip()
def f_print(*output, sep = "\n"):
for i in output:
stdout.write(str(i) + sep)
#if sep != "\n":
# stdout.write("\n")
s1 = f_input()
s2 = f_input()
ans = ""
for i in range(len(s1)):
if s1[i] != s2[i]:
ans += '1'
else:
ans += '0'
f_print(ans)
```
| 3.9845
|
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"
}
] | 1,692,419,153
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 27
| 154
| 0
|
input_year = int( input() )
year = input_year + 1
searching = True
while searching:
searching = False
str_year = str(year)
for digit in str_year:
if str_year.count(digit) > 1:
searching = True
break
if not searching:
break
year += 1
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
input_year = int( input() )
year = input_year + 1
searching = True
while searching:
searching = False
str_year = str(year)
for digit in str_year:
if str_year.count(digit) > 1:
searching = True
break
if not searching:
break
year += 1
print(year)
```
| 3
|
|
653
|
A
|
Bear and Three Balls
|
PROGRAMMING
| 900
|
[
"brute force",
"implementation",
"sortings"
] | null | null |
Limak is a little polar bear. He has *n* balls, the *i*-th ball has size *t**i*.
Limak wants to give one ball to each of his three friends. Giving gifts isn't easy — there are two rules Limak must obey to make friends happy:
- No two friends can get balls of the same size. - No two friends can get balls of sizes that differ by more than 2.
For example, Limak can choose balls with sizes 4, 5 and 3, or balls with sizes 90, 91 and 92. But he can't choose balls with sizes 5, 5 and 6 (two friends would get balls of the same size), and he can't choose balls with sizes 30, 31 and 33 (because sizes 30 and 33 differ by more than 2).
Your task is to check whether Limak can choose three balls that satisfy conditions above.
|
The first line of the input contains one integer *n* (3<=≤<=*n*<=≤<=50) — the number of balls Limak has.
The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=1000) where *t**i* denotes the size of the *i*-th ball.
|
Print "YES" (without quotes) if Limak can choose three balls of distinct sizes, such that any two of them differ by no more than 2. Otherwise, print "NO" (without quotes).
|
[
"4\n18 55 16 17\n",
"6\n40 41 43 44 44 44\n",
"8\n5 972 3 4 1 4 970 971\n"
] |
[
"YES\n",
"NO\n",
"YES\n"
] |
In the first sample, there are 4 balls and Limak is able to choose three of them to satisfy the rules. He must must choose balls with sizes 18, 16 and 17.
In the second sample, there is no way to give gifts to three friends without breaking the rules.
In the third sample, there is even more than one way to choose balls:
1. Choose balls with sizes 3, 4 and 5. 1. Choose balls with sizes 972, 970, 971.
| 500
|
[
{
"input": "4\n18 55 16 17",
"output": "YES"
},
{
"input": "6\n40 41 43 44 44 44",
"output": "NO"
},
{
"input": "8\n5 972 3 4 1 4 970 971",
"output": "YES"
},
{
"input": "3\n959 747 656",
"output": "NO"
},
{
"input": "4\n1 2 2 3",
"output": "YES"
},
{
"input": "50\n998 30 384 289 505 340 872 223 663 31 929 625 864 699 735 589 676 399 745 635 963 381 75 97 324 612 597 797 103 382 25 894 219 458 337 572 201 355 294 275 278 311 586 573 965 704 936 237 715 543",
"output": "NO"
},
{
"input": "50\n941 877 987 982 966 979 984 810 811 909 872 980 957 897 845 995 924 905 984 914 824 840 868 910 815 808 872 858 883 952 823 835 860 874 959 972 931 867 866 987 982 837 800 921 887 910 982 980 828 869",
"output": "YES"
},
{
"input": "3\n408 410 409",
"output": "YES"
},
{
"input": "3\n903 902 904",
"output": "YES"
},
{
"input": "3\n399 400 398",
"output": "YES"
},
{
"input": "3\n450 448 449",
"output": "YES"
},
{
"input": "3\n390 389 388",
"output": "YES"
},
{
"input": "3\n438 439 440",
"output": "YES"
},
{
"input": "11\n488 688 490 94 564 615 641 170 489 517 669",
"output": "YES"
},
{
"input": "24\n102 672 983 82 720 501 81 721 982 312 207 897 159 964 611 956 118 984 37 271 596 403 772 954",
"output": "YES"
},
{
"input": "36\n175 551 70 479 875 480 979 32 465 402 640 116 76 687 874 678 359 785 753 401 978 629 162 963 886 641 39 845 132 930 2 372 478 947 407 318",
"output": "YES"
},
{
"input": "6\n10 79 306 334 304 305",
"output": "YES"
},
{
"input": "34\n787 62 26 683 486 364 684 891 846 801 969 837 359 800 836 359 471 637 732 91 841 836 7 799 959 405 416 841 737 803 615 483 323 365",
"output": "YES"
},
{
"input": "30\n860 238 14 543 669 100 428 789 576 484 754 274 849 850 586 377 711 386 510 408 520 693 23 477 266 851 728 711 964 73",
"output": "YES"
},
{
"input": "11\n325 325 324 324 324 325 325 324 324 324 324",
"output": "NO"
},
{
"input": "7\n517 517 518 517 518 518 518",
"output": "NO"
},
{
"input": "20\n710 710 711 711 711 711 710 710 710 710 711 710 710 710 710 710 710 711 711 710",
"output": "NO"
},
{
"input": "48\n29 30 29 29 29 30 29 30 30 30 30 29 30 30 30 29 29 30 30 29 30 29 29 30 29 30 29 30 30 29 30 29 29 30 30 29 29 30 30 29 29 30 30 30 29 29 30 29",
"output": "NO"
},
{
"input": "7\n880 880 514 536 881 881 879",
"output": "YES"
},
{
"input": "15\n377 432 262 376 261 375 377 262 263 263 261 376 262 262 375",
"output": "YES"
},
{
"input": "32\n305 426 404 961 426 425 614 304 404 425 615 403 303 304 615 303 305 405 427 614 403 303 425 615 404 304 427 403 206 616 405 404",
"output": "YES"
},
{
"input": "41\n115 686 988 744 762 519 745 519 518 83 85 115 520 44 687 686 685 596 988 687 989 988 114 745 84 519 519 746 988 84 745 744 115 114 85 115 520 746 745 116 987",
"output": "YES"
},
{
"input": "47\n1 2 483 28 7 109 270 651 464 162 353 521 224 989 721 499 56 69 197 716 313 446 580 645 828 197 100 138 789 499 147 677 384 711 783 937 300 543 540 93 669 604 739 122 632 822 116",
"output": "NO"
},
{
"input": "31\n1 2 1 373 355 692 750 920 578 666 615 232 141 129 663 929 414 704 422 559 568 731 354 811 532 618 39 879 292 602 995",
"output": "NO"
},
{
"input": "50\n5 38 41 4 15 40 27 39 20 3 44 47 30 6 36 29 35 12 19 26 10 2 21 50 11 46 48 49 17 16 33 13 32 28 31 18 23 34 7 14 24 45 9 37 1 8 42 25 43 22",
"output": "YES"
},
{
"input": "50\n967 999 972 990 969 978 963 987 954 955 973 970 959 981 995 983 986 994 979 957 965 982 992 977 953 975 956 961 993 997 998 958 980 962 960 951 996 991 1000 966 971 988 976 968 989 984 974 964 985 952",
"output": "YES"
},
{
"input": "50\n850 536 761 506 842 898 857 723 583 637 536 943 895 929 890 612 832 633 696 731 553 880 710 812 665 877 915 636 711 540 748 600 554 521 813 796 568 513 543 809 798 820 928 504 999 646 907 639 550 911",
"output": "NO"
},
{
"input": "3\n3 1 2",
"output": "YES"
},
{
"input": "3\n500 999 1000",
"output": "NO"
},
{
"input": "10\n101 102 104 105 107 109 110 112 113 115",
"output": "NO"
},
{
"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": "NO"
},
{
"input": "50\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",
"output": "NO"
},
{
"input": "3\n1000 999 998",
"output": "YES"
},
{
"input": "49\n343 322 248 477 53 156 245 493 209 141 370 66 229 184 434 137 276 472 216 456 147 180 140 114 493 323 393 262 380 314 222 124 98 441 129 346 48 401 347 460 122 125 114 106 189 260 374 165 456",
"output": "NO"
},
{
"input": "20\n1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 3 3 3 3 3",
"output": "YES"
},
{
"input": "3\n999 999 1000",
"output": "NO"
},
{
"input": "9\n2 4 5 13 25 100 200 300 400",
"output": "NO"
},
{
"input": "9\n1 1 1 2 2 2 3 3 3",
"output": "YES"
},
{
"input": "3\n1 1 2",
"output": "NO"
},
{
"input": "3\n998 999 1000",
"output": "YES"
},
{
"input": "12\n1 1 1 1 1 1 1 1 1 2 2 4",
"output": "NO"
},
{
"input": "4\n4 3 4 5",
"output": "YES"
},
{
"input": "6\n1 1 1 2 2 2",
"output": "NO"
},
{
"input": "3\n2 3 2",
"output": "NO"
},
{
"input": "5\n10 5 6 3 2",
"output": "NO"
},
{
"input": "3\n1 2 1",
"output": "NO"
},
{
"input": "3\n1 2 3",
"output": "YES"
},
{
"input": "4\n998 999 1000 1000",
"output": "YES"
},
{
"input": "5\n2 3 9 9 4",
"output": "YES"
},
{
"input": "4\n1 2 4 4",
"output": "NO"
},
{
"input": "3\n1 1 1",
"output": "NO"
},
{
"input": "3\n2 2 3",
"output": "NO"
},
{
"input": "7\n1 2 2 2 4 5 6",
"output": "YES"
},
{
"input": "5\n1 3 10 3 10",
"output": "NO"
},
{
"input": "3\n1 2 2",
"output": "NO"
},
{
"input": "4\n1000 1000 999 998",
"output": "YES"
},
{
"input": "3\n5 3 7",
"output": "NO"
},
{
"input": "6\n1 1 2 2 3 3",
"output": "YES"
},
{
"input": "9\n6 6 6 5 5 5 4 4 4",
"output": "YES"
},
{
"input": "7\n5 6 6 6 7 7 7",
"output": "YES"
},
{
"input": "5\n2 3 3 3 4",
"output": "YES"
},
{
"input": "5\n2 1 2 1 3",
"output": "YES"
},
{
"input": "3\n1 2 7",
"output": "NO"
},
{
"input": "3\n1000 1000 1000",
"output": "NO"
},
{
"input": "5\n1 100 2 100 3",
"output": "YES"
},
{
"input": "5\n5 4 6 5 5",
"output": "YES"
},
{
"input": "12\n1 1 1 1 2 2 2 2 3 3 3 3",
"output": "YES"
},
{
"input": "5\n9 9 1 2 3",
"output": "YES"
},
{
"input": "6\n1 2 3 1 2 3",
"output": "YES"
},
{
"input": "7\n1 1 1 1 2 3 3",
"output": "YES"
},
{
"input": "3\n13 13 13",
"output": "NO"
},
{
"input": "3\n42 42 42",
"output": "NO"
},
{
"input": "8\n1 1 1 1 2 2 2 2",
"output": "NO"
},
{
"input": "6\n1 1 1 1 2 3",
"output": "YES"
},
{
"input": "6\n1 1 2 2 6 6",
"output": "NO"
},
{
"input": "6\n1 2 5 5 5 5",
"output": "NO"
},
{
"input": "9\n1 2 3 1 2 3 1 2 3",
"output": "YES"
},
{
"input": "4\n1 2 1 100",
"output": "NO"
},
{
"input": "5\n1 1 2 2 3",
"output": "YES"
}
] | 1,661,101,088
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 84
| 46
| 0
|
input()
T = set(map(int, input().split()))
print('YES' if any(t-1 in T and t+1 in T for t in T) else 'NO')
|
Title: Bear and Three Balls
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Limak is a little polar bear. He has *n* balls, the *i*-th ball has size *t**i*.
Limak wants to give one ball to each of his three friends. Giving gifts isn't easy — there are two rules Limak must obey to make friends happy:
- No two friends can get balls of the same size. - No two friends can get balls of sizes that differ by more than 2.
For example, Limak can choose balls with sizes 4, 5 and 3, or balls with sizes 90, 91 and 92. But he can't choose balls with sizes 5, 5 and 6 (two friends would get balls of the same size), and he can't choose balls with sizes 30, 31 and 33 (because sizes 30 and 33 differ by more than 2).
Your task is to check whether Limak can choose three balls that satisfy conditions above.
Input Specification:
The first line of the input contains one integer *n* (3<=≤<=*n*<=≤<=50) — the number of balls Limak has.
The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=1000) where *t**i* denotes the size of the *i*-th ball.
Output Specification:
Print "YES" (without quotes) if Limak can choose three balls of distinct sizes, such that any two of them differ by no more than 2. Otherwise, print "NO" (without quotes).
Demo Input:
['4\n18 55 16 17\n', '6\n40 41 43 44 44 44\n', '8\n5 972 3 4 1 4 970 971\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
In the first sample, there are 4 balls and Limak is able to choose three of them to satisfy the rules. He must must choose balls with sizes 18, 16 and 17.
In the second sample, there is no way to give gifts to three friends without breaking the rules.
In the third sample, there is even more than one way to choose balls:
1. Choose balls with sizes 3, 4 and 5. 1. Choose balls with sizes 972, 970, 971.
|
```python
input()
T = set(map(int, input().split()))
print('YES' if any(t-1 in T and t+1 in T for t in T) else 'NO')
```
| 3
|
|
351
|
B
|
Jeff and Furik
|
PROGRAMMING
| 1,900
|
[
"combinatorics",
"dp",
"probabilities"
] | null | null |
Jeff has become friends with Furik. Now these two are going to play one quite amusing game.
At the beginning of the game Jeff takes a piece of paper and writes down a permutation consisting of *n* numbers: *p*1, *p*2, ..., *p**n*. Then the guys take turns to make moves, Jeff moves first. During his move, Jeff chooses two adjacent permutation elements and then the boy swaps them. During his move, Furic tosses a coin and if the coin shows "heads" he chooses a random pair of adjacent elements with indexes *i* and *i*<=+<=1, for which an inequality *p**i*<=><=*p**i*<=+<=1 holds, and swaps them. But if the coin shows "tails", Furik chooses a random pair of adjacent elements with indexes *i* and *i*<=+<=1, for which the inequality *p**i*<=<<=*p**i*<=+<=1 holds, and swaps them. If the coin shows "heads" or "tails" and Furik has multiple ways of adjacent pairs to take, then he uniformly takes one of the pairs. If Furik doesn't have any pair to take, he tosses a coin one more time. The game ends when the permutation is sorted in the increasing order.
Jeff wants the game to finish as quickly as possible (that is, he wants both players to make as few moves as possible). Help Jeff find the minimum mathematical expectation of the number of moves in the game if he moves optimally well.
You can consider that the coin shows the heads (or tails) with the probability of 50 percent.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=3000). The next line contains *n* distinct integers *p*1, *p*2, ..., *p**n* (1<=≤<=*p**i*<=≤<=*n*) — the permutation *p*. The numbers are separated by spaces.
|
In a single line print a single real value — the answer to the problem. The answer will be considered correct if the absolute or relative error doesn't exceed 10<=-<=6.
|
[
"2\n1 2\n",
"5\n3 5 2 4 1\n"
] |
[
"0.000000\n",
"13.000000\n"
] |
In the first test the sequence is already sorted, so the answer is 0.
| 1,000
|
[
{
"input": "2\n1 2",
"output": "0.000000"
},
{
"input": "5\n3 5 2 4 1",
"output": "13.000000"
},
{
"input": "16\n6 15 3 8 7 11 9 10 2 13 4 14 1 16 5 12",
"output": "108.000000"
},
{
"input": "9\n1 7 8 5 3 4 6 9 2",
"output": "33.000000"
},
{
"input": "5\n2 3 4 5 1",
"output": "8.000000"
},
{
"input": "9\n4 1 8 6 7 5 2 9 3",
"output": "33.000000"
},
{
"input": "10\n3 4 1 5 7 9 8 10 6 2",
"output": "29.000000"
},
{
"input": "13\n3 1 11 12 4 5 8 10 13 7 9 2 6",
"output": "69.000000"
},
{
"input": "10\n8 4 1 7 6 10 9 5 3 2",
"output": "53.000000"
},
{
"input": "2\n2 1",
"output": "1.000000"
},
{
"input": "95\n68 56 24 89 79 20 74 69 49 59 85 67 95 66 15 34 2 13 92 25 84 77 70 71 17 93 62 81 1 87 76 38 75 31 63 51 35 33 37 11 36 52 23 10 27 90 12 6 45 32 86 26 60 47 91 65 58 80 78 88 50 9 44 4 28 29 22 8 48 7 19 57 14 54 55 83 5 30 72 18 82 94 43 46 41 3 61 53 73 39 40 16 64 42 21",
"output": "5076.000000"
}
] | 1,692,331,690
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 25
| 156
| 2,048,000
|
import sys
input = sys.stdin.readline
MOD=10**9+7
if __name__ == '__main__':
n=int(input())
arr=list(map(int, input().split()))
cnt=0
for i in range(n-1):
for j in range(i+1,n):
if arr[j]<arr[i]: cnt +=1
print((cnt*2-(cnt%2))*1.0)
|
Title: Jeff and Furik
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Jeff has become friends with Furik. Now these two are going to play one quite amusing game.
At the beginning of the game Jeff takes a piece of paper and writes down a permutation consisting of *n* numbers: *p*1, *p*2, ..., *p**n*. Then the guys take turns to make moves, Jeff moves first. During his move, Jeff chooses two adjacent permutation elements and then the boy swaps them. During his move, Furic tosses a coin and if the coin shows "heads" he chooses a random pair of adjacent elements with indexes *i* and *i*<=+<=1, for which an inequality *p**i*<=><=*p**i*<=+<=1 holds, and swaps them. But if the coin shows "tails", Furik chooses a random pair of adjacent elements with indexes *i* and *i*<=+<=1, for which the inequality *p**i*<=<<=*p**i*<=+<=1 holds, and swaps them. If the coin shows "heads" or "tails" and Furik has multiple ways of adjacent pairs to take, then he uniformly takes one of the pairs. If Furik doesn't have any pair to take, he tosses a coin one more time. The game ends when the permutation is sorted in the increasing order.
Jeff wants the game to finish as quickly as possible (that is, he wants both players to make as few moves as possible). Help Jeff find the minimum mathematical expectation of the number of moves in the game if he moves optimally well.
You can consider that the coin shows the heads (or tails) with the probability of 50 percent.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=3000). The next line contains *n* distinct integers *p*1, *p*2, ..., *p**n* (1<=≤<=*p**i*<=≤<=*n*) — the permutation *p*. The numbers are separated by spaces.
Output Specification:
In a single line print a single real value — the answer to the problem. The answer will be considered correct if the absolute or relative error doesn't exceed 10<=-<=6.
Demo Input:
['2\n1 2\n', '5\n3 5 2 4 1\n']
Demo Output:
['0.000000\n', '13.000000\n']
Note:
In the first test the sequence is already sorted, so the answer is 0.
|
```python
import sys
input = sys.stdin.readline
MOD=10**9+7
if __name__ == '__main__':
n=int(input())
arr=list(map(int, input().split()))
cnt=0
for i in range(n-1):
for j in range(i+1,n):
if arr[j]<arr[i]: cnt +=1
print((cnt*2-(cnt%2))*1.0)
```
| 3
|
|
776
|
A
|
A Serial Killer
|
PROGRAMMING
| 900
|
[
"brute force",
"implementation",
"strings"
] | null | null |
Our beloved detective, Sherlock is currently trying to catch a serial killer who kills a person each day. Using his powers of deduction, he came to know that the killer has a strategy for selecting his next victim.
The killer starts with two potential victims on his first day, selects one of these two, kills selected victim and replaces him with a new person. He repeats this procedure each day. This way, each day he has two potential victims to choose from. Sherlock knows the initial two potential victims. Also, he knows the murder that happened on a particular day and the new person who replaced this victim.
You need to help him get all the pairs of potential victims at each day so that Sherlock can observe some pattern.
|
First line of input contains two names (length of each of them doesn't exceed 10), the two initials potential victims. Next line contains integer *n* (1<=≤<=*n*<=≤<=1000), the number of days.
Next *n* lines contains two names (length of each of them doesn't exceed 10), first being the person murdered on this day and the second being the one who replaced that person.
The input format is consistent, that is, a person murdered is guaranteed to be from the two potential victims at that time. Also, all the names are guaranteed to be distinct and consists of lowercase English letters.
|
Output *n*<=+<=1 lines, the *i*-th line should contain the two persons from which the killer selects for the *i*-th murder. The (*n*<=+<=1)-th line should contain the two persons from which the next victim is selected. In each line, the two names can be printed in any order.
|
[
"ross rachel\n4\nross joey\nrachel phoebe\nphoebe monica\nmonica chandler\n",
"icm codeforces\n1\ncodeforces technex\n"
] |
[
"ross rachel\njoey rachel\njoey phoebe\njoey monica\njoey chandler\n",
"icm codeforces\nicm technex\n"
] |
In first example, the killer starts with ross and rachel.
- After day 1, ross is killed and joey appears. - After day 2, rachel is killed and phoebe appears. - After day 3, phoebe is killed and monica appears. - After day 4, monica is killed and chandler appears.
| 500
|
[
{
"input": "ross rachel\n4\nross joey\nrachel phoebe\nphoebe monica\nmonica chandler",
"output": "ross rachel\njoey rachel\njoey phoebe\njoey monica\njoey chandler"
},
{
"input": "icm codeforces\n1\ncodeforces technex",
"output": "icm codeforces\nicm technex"
},
{
"input": "a b\n3\na c\nb d\nd e",
"output": "a b\nc b\nc d\nc e"
},
{
"input": "ze udggmyop\n4\nze szhrbmft\nudggmyop mjorab\nszhrbmft ojdtfnzxj\nojdtfnzxj yjlkg",
"output": "ze udggmyop\nszhrbmft udggmyop\nszhrbmft mjorab\nojdtfnzxj mjorab\nyjlkg mjorab"
},
{
"input": "q s\n10\nq b\nb j\ns g\nj f\nf m\ng c\nc a\nm d\nd z\nz o",
"output": "q s\nb s\nj s\nj g\nf g\nm g\nm c\nm a\nd a\nz a\no a"
},
{
"input": "iii iiiiii\n7\niii iiiiiiiiii\niiiiiiiiii iiii\niiii i\niiiiii iiiiiiii\niiiiiiii iiiiiiiii\ni iiiii\niiiii ii",
"output": "iii iiiiii\niiiiiiiiii iiiiii\niiii iiiiii\ni iiiiii\ni iiiiiiii\ni iiiiiiiii\niiiii iiiiiiiii\nii iiiiiiiii"
},
{
"input": "bwyplnjn zkms\n26\nzkms nzmcsytxh\nnzmcsytxh yujsb\nbwyplnjn gtbzhudpb\ngtbzhudpb hpk\nyujsb xvy\nhpk wrwnfokml\nwrwnfokml ndouuikw\nndouuikw ucgrja\nucgrja tgfmpldz\nxvy nycrfphn\nnycrfphn quvs\nquvs htdy\nhtdy k\ntgfmpldz xtdpkxm\nxtdpkxm suwqxs\nk fv\nsuwqxs qckllwy\nqckllwy diun\nfv lefa\nlefa gdoqjysx\ndiun dhpz\ngdoqjysx bdmqdyt\ndhpz dgz\ndgz v\nbdmqdyt aswy\naswy ydkayhlrnm",
"output": "bwyplnjn zkms\nbwyplnjn nzmcsytxh\nbwyplnjn yujsb\ngtbzhudpb yujsb\nhpk yujsb\nhpk xvy\nwrwnfokml xvy\nndouuikw xvy\nucgrja xvy\ntgfmpldz xvy\ntgfmpldz nycrfphn\ntgfmpldz quvs\ntgfmpldz htdy\ntgfmpldz k\nxtdpkxm k\nsuwqxs k\nsuwqxs fv\nqckllwy fv\ndiun fv\ndiun lefa\ndiun gdoqjysx\ndhpz gdoqjysx\ndhpz bdmqdyt\ndgz bdmqdyt\nv bdmqdyt\nv aswy\nv ydkayhlrnm"
},
{
"input": "wxz hbeqwqp\n7\nhbeqwqp cpieghnszh\ncpieghnszh tlqrpd\ntlqrpd ttwrtio\nttwrtio xapvds\nxapvds zk\nwxz yryk\nzk b",
"output": "wxz hbeqwqp\nwxz cpieghnszh\nwxz tlqrpd\nwxz ttwrtio\nwxz xapvds\nwxz zk\nyryk zk\nyryk b"
},
{
"input": "wced gnsgv\n23\ngnsgv japawpaf\njapawpaf nnvpeu\nnnvpeu a\na ddupputljq\nddupputljq qyhnvbh\nqyhnvbh pqwijl\nwced khuvs\nkhuvs bjkh\npqwijl ysacmboc\nbjkh srf\nsrf jknoz\njknoz hodf\nysacmboc xqtkoyh\nhodf rfp\nxqtkoyh bivgnwqvoe\nbivgnwqvoe nknf\nnknf wuig\nrfp e\ne bqqknq\nwuig sznhhhu\nbqqknq dhrtdld\ndhrtdld n\nsznhhhu bguylf",
"output": "wced gnsgv\nwced japawpaf\nwced nnvpeu\nwced a\nwced ddupputljq\nwced qyhnvbh\nwced pqwijl\nkhuvs pqwijl\nbjkh pqwijl\nbjkh ysacmboc\nsrf ysacmboc\njknoz ysacmboc\nhodf ysacmboc\nhodf xqtkoyh\nrfp xqtkoyh\nrfp bivgnwqvoe\nrfp nknf\nrfp wuig\ne wuig\nbqqknq wuig\nbqqknq sznhhhu\ndhrtdld sznhhhu\nn sznhhhu\nn bguylf"
},
{
"input": "qqqqqqqqqq qqqqqqqq\n3\nqqqqqqqq qqqqqqqqq\nqqqqqqqqq qqqqq\nqqqqq q",
"output": "qqqqqqqqqq qqqqqqqq\nqqqqqqqqqq qqqqqqqqq\nqqqqqqqqqq qqqqq\nqqqqqqqqqq q"
},
{
"input": "wwwww w\n8\nwwwww wwwwwwww\nwwwwwwww wwwwwwwww\nwwwwwwwww wwwwwwwwww\nw www\nwwwwwwwwww wwww\nwwww ww\nwww wwwwww\nwwwwww wwwwwww",
"output": "wwwww w\nwwwwwwww w\nwwwwwwwww w\nwwwwwwwwww w\nwwwwwwwwww www\nwwww www\nww www\nww wwwwww\nww wwwwwww"
},
{
"input": "k d\n17\nk l\nd v\nv z\nl r\nz i\nr s\ns p\np w\nw j\nj h\ni c\nh m\nm q\nc o\no g\nq x\nx n",
"output": "k d\nl d\nl v\nl z\nr z\nr i\ns i\np i\nw i\nj i\nh i\nh c\nm c\nq c\nq o\nq g\nx g\nn g"
}
] | 1,617,532,497
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 57
| 155
| 3,276,800
|
str1,str2 = map(str,input().split())
n = int(input())
print(str1,str2)
for each in range(n):
murdered,replace = map(str,input().split())
if(str1==murdered):
str1 = replace
print(str1,str2)
else:
str2 = replace
print(str1,str2)
|
Title: A Serial Killer
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Our beloved detective, Sherlock is currently trying to catch a serial killer who kills a person each day. Using his powers of deduction, he came to know that the killer has a strategy for selecting his next victim.
The killer starts with two potential victims on his first day, selects one of these two, kills selected victim and replaces him with a new person. He repeats this procedure each day. This way, each day he has two potential victims to choose from. Sherlock knows the initial two potential victims. Also, he knows the murder that happened on a particular day and the new person who replaced this victim.
You need to help him get all the pairs of potential victims at each day so that Sherlock can observe some pattern.
Input Specification:
First line of input contains two names (length of each of them doesn't exceed 10), the two initials potential victims. Next line contains integer *n* (1<=≤<=*n*<=≤<=1000), the number of days.
Next *n* lines contains two names (length of each of them doesn't exceed 10), first being the person murdered on this day and the second being the one who replaced that person.
The input format is consistent, that is, a person murdered is guaranteed to be from the two potential victims at that time. Also, all the names are guaranteed to be distinct and consists of lowercase English letters.
Output Specification:
Output *n*<=+<=1 lines, the *i*-th line should contain the two persons from which the killer selects for the *i*-th murder. The (*n*<=+<=1)-th line should contain the two persons from which the next victim is selected. In each line, the two names can be printed in any order.
Demo Input:
['ross rachel\n4\nross joey\nrachel phoebe\nphoebe monica\nmonica chandler\n', 'icm codeforces\n1\ncodeforces technex\n']
Demo Output:
['ross rachel\njoey rachel\njoey phoebe\njoey monica\njoey chandler\n', 'icm codeforces\nicm technex\n']
Note:
In first example, the killer starts with ross and rachel.
- After day 1, ross is killed and joey appears. - After day 2, rachel is killed and phoebe appears. - After day 3, phoebe is killed and monica appears. - After day 4, monica is killed and chandler appears.
|
```python
str1,str2 = map(str,input().split())
n = int(input())
print(str1,str2)
for each in range(n):
murdered,replace = map(str,input().split())
if(str1==murdered):
str1 = replace
print(str1,str2)
else:
str2 = replace
print(str1,str2)
```
| 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"
}
] | 1,662,430,670
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 35
| 92
| 0
|
k,n=map(int,input().split())
f=k*n
print(f//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
k,n=map(int,input().split())
f=k*n
print(f//2)
```
| 3.977
|
298
|
B
|
Sail
|
PROGRAMMING
| 1,200
|
[
"brute force",
"greedy",
"implementation"
] | null | null |
The polar bears are going fishing. They plan to sail from (*s**x*,<=*s**y*) to (*e**x*,<=*e**y*). However, the boat can only sail by wind. At each second, the wind blows in one of these directions: east, south, west or north. Assume the boat is currently at (*x*,<=*y*).
- If the wind blows to the east, the boat will move to (*x*<=+<=1,<=*y*). - If the wind blows to the south, the boat will move to (*x*,<=*y*<=-<=1). - If the wind blows to the west, the boat will move to (*x*<=-<=1,<=*y*). - If the wind blows to the north, the boat will move to (*x*,<=*y*<=+<=1).
Alternatively, they can hold the boat by the anchor. In this case, the boat stays at (*x*,<=*y*). Given the wind direction for *t* seconds, what is the earliest time they sail to (*e**x*,<=*e**y*)?
|
The first line contains five integers *t*,<=*s**x*,<=*s**y*,<=*e**x*,<=*e**y* (1<=≤<=*t*<=≤<=105,<=<=-<=109<=≤<=*s**x*,<=*s**y*,<=*e**x*,<=*e**y*<=≤<=109). The starting location and the ending location will be different.
The second line contains *t* characters, the *i*-th character is the wind blowing direction at the *i*-th second. It will be one of the four possibilities: "E" (east), "S" (south), "W" (west) and "N" (north).
|
If they can reach (*e**x*,<=*e**y*) within *t* seconds, print the earliest time they can achieve it. Otherwise, print "-1" (without quotes).
|
[
"5 0 0 1 1\nSESNW\n",
"10 5 3 3 6\nNENSWESNEE\n"
] |
[
"4\n",
"-1\n"
] |
In the first sample, they can stay at seconds 1, 3, and move at seconds 2, 4.
In the second sample, they cannot sail to the destination.
| 500
|
[
{
"input": "5 0 0 1 1\nSESNW",
"output": "4"
},
{
"input": "10 5 3 3 6\nNENSWESNEE",
"output": "-1"
},
{
"input": "19 -172106364 -468680119 -172106365 -468680119\nSSEEESSSESESWSEESSS",
"output": "13"
},
{
"input": "39 -1000000000 -1000000000 -999999997 -1000000000\nENEENWSWSSWESNSSEESNSESWSWNSWESNENWNWEE",
"output": "4"
},
{
"input": "41 -264908123 -86993764 -264908123 -86993723\nNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNN",
"output": "41"
},
{
"input": "34 -1000000000 -1000000000 -999999983 -1000000000\nEEEEESSEWNSSSESWEWSWESEWSEESNEWEEE",
"output": "-1"
},
{
"input": "1 0 0 0 -1\nS",
"output": "1"
},
{
"input": "1 5 5 5 6\nE",
"output": "-1"
},
{
"input": "15 1 1 1 2\nNNNNNNNNNNNNNNN",
"output": "1"
},
{
"input": "11 1 1 1 2\nNNNNNNNNNNN",
"output": "1"
}
] | 1,561,819,951
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 43
| 312
| 2,764,800
|
t, sx, sy, fx, fy = map(int, input().split())
m = input()
ans = 0
for i in m:
if sx < fx and i == 'E':
sx += 1
if sx > fx and i == 'W':
sx -= 1
if sy < fy and i == 'N':
sy += 1
if sy > fy and i == 'S':
sy -= 1
ans += 1
if sx == fx and sy == fy:
exit(print(ans))
print(-1)
|
Title: Sail
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The polar bears are going fishing. They plan to sail from (*s**x*,<=*s**y*) to (*e**x*,<=*e**y*). However, the boat can only sail by wind. At each second, the wind blows in one of these directions: east, south, west or north. Assume the boat is currently at (*x*,<=*y*).
- If the wind blows to the east, the boat will move to (*x*<=+<=1,<=*y*). - If the wind blows to the south, the boat will move to (*x*,<=*y*<=-<=1). - If the wind blows to the west, the boat will move to (*x*<=-<=1,<=*y*). - If the wind blows to the north, the boat will move to (*x*,<=*y*<=+<=1).
Alternatively, they can hold the boat by the anchor. In this case, the boat stays at (*x*,<=*y*). Given the wind direction for *t* seconds, what is the earliest time they sail to (*e**x*,<=*e**y*)?
Input Specification:
The first line contains five integers *t*,<=*s**x*,<=*s**y*,<=*e**x*,<=*e**y* (1<=≤<=*t*<=≤<=105,<=<=-<=109<=≤<=*s**x*,<=*s**y*,<=*e**x*,<=*e**y*<=≤<=109). The starting location and the ending location will be different.
The second line contains *t* characters, the *i*-th character is the wind blowing direction at the *i*-th second. It will be one of the four possibilities: "E" (east), "S" (south), "W" (west) and "N" (north).
Output Specification:
If they can reach (*e**x*,<=*e**y*) within *t* seconds, print the earliest time they can achieve it. Otherwise, print "-1" (without quotes).
Demo Input:
['5 0 0 1 1\nSESNW\n', '10 5 3 3 6\nNENSWESNEE\n']
Demo Output:
['4\n', '-1\n']
Note:
In the first sample, they can stay at seconds 1, 3, and move at seconds 2, 4.
In the second sample, they cannot sail to the destination.
|
```python
t, sx, sy, fx, fy = map(int, input().split())
m = input()
ans = 0
for i in m:
if sx < fx and i == 'E':
sx += 1
if sx > fx and i == 'W':
sx -= 1
if sy < fy and i == 'N':
sy += 1
if sy > fy and i == 'S':
sy -= 1
ans += 1
if sx == fx and sy == fy:
exit(print(ans))
print(-1)
```
| 3
|
|
820
|
A
|
Mister B and Book Reading
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Mister B once received a gift: it was a book about aliens, which he started read immediately. This book had *c* pages.
At first day Mister B read *v*0 pages, but after that he started to speed up. Every day, starting from the second, he read *a* pages more than on the previous day (at first day he read *v*0 pages, at second — *v*0<=+<=*a* pages, at third — *v*0<=+<=2*a* pages, and so on). But Mister B is just a human, so he physically wasn't able to read more than *v*1 pages per day.
Also, to refresh his memory, every day, starting from the second, Mister B had to reread last *l* pages he read on the previous day. Mister B finished the book when he read the last page for the first time.
Help Mister B to calculate how many days he needed to finish the book.
|
First and only line contains five space-separated integers: *c*, *v*0, *v*1, *a* and *l* (1<=≤<=*c*<=≤<=1000, 0<=≤<=*l*<=<<=*v*0<=≤<=*v*1<=≤<=1000, 0<=≤<=*a*<=≤<=1000) — the length of the book in pages, the initial reading speed, the maximum reading speed, the acceleration in reading speed and the number of pages for rereading.
|
Print one integer — the number of days Mister B needed to finish the book.
|
[
"5 5 10 5 4\n",
"12 4 12 4 1\n",
"15 1 100 0 0\n"
] |
[
"1\n",
"3\n",
"15\n"
] |
In the first sample test the book contains 5 pages, so Mister B read it right at the first day.
In the second sample test at first day Mister B read pages number 1 - 4, at second day — 4 - 11, at third day — 11 - 12 and finished the book.
In third sample test every day Mister B read 1 page of the book, so he finished in 15 days.
| 500
|
[
{
"input": "5 5 10 5 4",
"output": "1"
},
{
"input": "12 4 12 4 1",
"output": "3"
},
{
"input": "15 1 100 0 0",
"output": "15"
},
{
"input": "1 1 1 0 0",
"output": "1"
},
{
"input": "1000 999 1000 1000 998",
"output": "2"
},
{
"input": "1000 2 2 5 1",
"output": "999"
},
{
"input": "1000 1 1 1000 0",
"output": "1000"
},
{
"input": "737 41 74 12 11",
"output": "13"
},
{
"input": "1000 1000 1000 0 999",
"output": "1"
},
{
"input": "765 12 105 5 7",
"output": "17"
},
{
"input": "15 2 2 1000 0",
"output": "8"
},
{
"input": "1000 1 1000 1000 0",
"output": "2"
},
{
"input": "20 3 7 1 2",
"output": "6"
},
{
"input": "1000 500 500 1000 499",
"output": "501"
},
{
"input": "1 1000 1000 1000 0",
"output": "1"
},
{
"input": "1000 2 1000 56 0",
"output": "7"
},
{
"input": "1000 2 1000 802 0",
"output": "3"
},
{
"input": "16 1 8 2 0",
"output": "4"
},
{
"input": "20 6 10 2 2",
"output": "3"
},
{
"input": "8 2 12 4 1",
"output": "3"
},
{
"input": "8 6 13 2 5",
"output": "2"
},
{
"input": "70 4 20 87 0",
"output": "5"
},
{
"input": "97 8 13 234 5",
"output": "13"
},
{
"input": "16 4 23 8 3",
"output": "3"
},
{
"input": "65 7 22 7 4",
"output": "5"
},
{
"input": "93 10 18 11 7",
"output": "9"
},
{
"input": "86 13 19 15 9",
"output": "9"
},
{
"input": "333 17 50 10 16",
"output": "12"
},
{
"input": "881 16 55 10 12",
"output": "23"
},
{
"input": "528 11 84 3 9",
"output": "19"
},
{
"input": "896 2 184 8 1",
"output": "16"
},
{
"input": "236 10 930 9 8",
"output": "8"
},
{
"input": "784 1 550 14 0",
"output": "12"
},
{
"input": "506 1 10 4 0",
"output": "53"
},
{
"input": "460 1 3 2 0",
"output": "154"
},
{
"input": "701 1 3 1 0",
"output": "235"
},
{
"input": "100 49 50 1000 2",
"output": "3"
},
{
"input": "100 1 100 100 0",
"output": "2"
},
{
"input": "12 1 4 2 0",
"output": "4"
},
{
"input": "22 10 12 0 0",
"output": "3"
},
{
"input": "20 10 15 1 4",
"output": "3"
},
{
"input": "1000 5 10 1 4",
"output": "169"
},
{
"input": "1000 1 1000 1 0",
"output": "45"
},
{
"input": "4 1 2 2 0",
"output": "3"
},
{
"input": "1 5 5 1 1",
"output": "1"
},
{
"input": "19 10 11 0 2",
"output": "3"
},
{
"input": "1 2 3 0 0",
"output": "1"
},
{
"input": "10 1 4 10 0",
"output": "4"
},
{
"input": "20 3 100 1 1",
"output": "5"
},
{
"input": "1000 5 9 5 0",
"output": "112"
},
{
"input": "1 11 12 0 10",
"output": "1"
},
{
"input": "1 1 1 1 0",
"output": "1"
},
{
"input": "1000 1 20 1 0",
"output": "60"
},
{
"input": "9 1 4 2 0",
"output": "4"
},
{
"input": "129 2 3 4 0",
"output": "44"
},
{
"input": "4 2 2 0 1",
"output": "3"
},
{
"input": "1000 1 10 100 0",
"output": "101"
},
{
"input": "100 1 100 1 0",
"output": "14"
},
{
"input": "8 3 4 2 0",
"output": "3"
},
{
"input": "20 1 6 4 0",
"output": "5"
},
{
"input": "8 2 4 2 0",
"output": "3"
},
{
"input": "11 5 6 7 2",
"output": "3"
},
{
"input": "100 120 130 120 0",
"output": "1"
},
{
"input": "7 1 4 1 0",
"output": "4"
},
{
"input": "5 3 10 0 2",
"output": "3"
},
{
"input": "5 2 2 0 0",
"output": "3"
},
{
"input": "1000 10 1000 10 0",
"output": "14"
},
{
"input": "25 3 50 4 2",
"output": "4"
},
{
"input": "9 10 10 10 9",
"output": "1"
},
{
"input": "17 10 12 6 5",
"output": "2"
},
{
"input": "15 5 10 3 0",
"output": "3"
},
{
"input": "8 3 5 1 0",
"output": "3"
},
{
"input": "19 1 12 5 0",
"output": "4"
},
{
"input": "1000 10 1000 1 0",
"output": "37"
},
{
"input": "100 1 2 1000 0",
"output": "51"
},
{
"input": "20 10 11 1000 9",
"output": "6"
},
{
"input": "16 2 100 1 1",
"output": "5"
},
{
"input": "18 10 13 2 5",
"output": "3"
},
{
"input": "12 3 5 3 1",
"output": "4"
},
{
"input": "17 3 11 2 0",
"output": "4"
},
{
"input": "4 2 100 1 1",
"output": "2"
},
{
"input": "7 4 5 2 3",
"output": "3"
},
{
"input": "100 1 2 2 0",
"output": "51"
},
{
"input": "50 4 5 5 0",
"output": "11"
},
{
"input": "1 2 2 0 1",
"output": "1"
},
{
"input": "1000 2 3 10 1",
"output": "500"
},
{
"input": "500 10 500 1000 0",
"output": "2"
},
{
"input": "1000 4 12 1 0",
"output": "87"
},
{
"input": "18 10 13 1 5",
"output": "3"
},
{
"input": "7 3 6 2 2",
"output": "3"
},
{
"input": "15 5 100 1 2",
"output": "4"
},
{
"input": "100 1 10 1 0",
"output": "15"
},
{
"input": "8 2 7 5 1",
"output": "2"
},
{
"input": "11 2 4 1 1",
"output": "5"
},
{
"input": "1000 500 900 100 300",
"output": "3"
},
{
"input": "7 1 2 5 0",
"output": "4"
},
{
"input": "7 3 5 3 2",
"output": "3"
},
{
"input": "7 3 10 2 1",
"output": "2"
},
{
"input": "1000 501 510 1 499",
"output": "50"
},
{
"input": "1000 1 1000 2 0",
"output": "32"
},
{
"input": "1 5 5 0 0",
"output": "1"
},
{
"input": "18 10 15 1 5",
"output": "3"
},
{
"input": "100 4 1000 1 2",
"output": "13"
},
{
"input": "20 2 40 1 1",
"output": "6"
},
{
"input": "1 11 1000 100 1",
"output": "1"
},
{
"input": "6 4 4 1 2",
"output": "2"
},
{
"input": "8 3 5 3 1",
"output": "3"
},
{
"input": "10 5 7 1 2",
"output": "3"
},
{
"input": "400 100 198 1 99",
"output": "25"
},
{
"input": "3 1 2 5 0",
"output": "2"
}
] | 1,498,574,536
| 436
|
Python 3
|
OK
|
TESTS
| 110
| 77
| 5,529,600
|
c, v, v1, a, l = map(int, input().split())
for i in range(1, 1000000):
c -= v
if i > 1:
c += l
if c <= 0:
print(i)
break
v += a
v = min(v, v1)
|
Title: Mister B and Book Reading
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mister B once received a gift: it was a book about aliens, which he started read immediately. This book had *c* pages.
At first day Mister B read *v*0 pages, but after that he started to speed up. Every day, starting from the second, he read *a* pages more than on the previous day (at first day he read *v*0 pages, at second — *v*0<=+<=*a* pages, at third — *v*0<=+<=2*a* pages, and so on). But Mister B is just a human, so he physically wasn't able to read more than *v*1 pages per day.
Also, to refresh his memory, every day, starting from the second, Mister B had to reread last *l* pages he read on the previous day. Mister B finished the book when he read the last page for the first time.
Help Mister B to calculate how many days he needed to finish the book.
Input Specification:
First and only line contains five space-separated integers: *c*, *v*0, *v*1, *a* and *l* (1<=≤<=*c*<=≤<=1000, 0<=≤<=*l*<=<<=*v*0<=≤<=*v*1<=≤<=1000, 0<=≤<=*a*<=≤<=1000) — the length of the book in pages, the initial reading speed, the maximum reading speed, the acceleration in reading speed and the number of pages for rereading.
Output Specification:
Print one integer — the number of days Mister B needed to finish the book.
Demo Input:
['5 5 10 5 4\n', '12 4 12 4 1\n', '15 1 100 0 0\n']
Demo Output:
['1\n', '3\n', '15\n']
Note:
In the first sample test the book contains 5 pages, so Mister B read it right at the first day.
In the second sample test at first day Mister B read pages number 1 - 4, at second day — 4 - 11, at third day — 11 - 12 and finished the book.
In third sample test every day Mister B read 1 page of the book, so he finished in 15 days.
|
```python
c, v, v1, a, l = map(int, input().split())
for i in range(1, 1000000):
c -= v
if i > 1:
c += l
if c <= 0:
print(i)
break
v += a
v = min(v, v1)
```
| 3
|
|
540
|
A
|
Combination Lock
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Scrooge McDuck keeps his most treasured savings in a home safe with a combination lock. Each time he wants to put there the treasures that he's earned fair and square, he has to open the lock.
The combination lock is represented by *n* rotating disks with digits from 0 to 9 written on them. Scrooge McDuck has to turn some disks so that the combination of digits on the disks forms a secret combination. In one move, he can rotate one disk one digit forwards or backwards. In particular, in one move he can go from digit 0 to digit 9 and vice versa. What minimum number of actions does he need for that?
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of disks on the combination lock.
The second line contains a string of *n* digits — the original state of the disks.
The third line contains a string of *n* digits — Scrooge McDuck's combination that opens the lock.
|
Print a single integer — the minimum number of moves Scrooge McDuck needs to open the lock.
|
[
"5\n82195\n64723\n"
] |
[
"13\n"
] |
In the sample he needs 13 moves:
- 1 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b8967f65a723782358b93eff9ce69f336817cf70.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 2 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/07fa58573ece0d32c4d555e498d2b24d2f70f36a.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 3 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/cc2275d9252aae35a6867c6a5b4ba7596e9a7626.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 4 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b100aea470fcaaab4e9529b234ba0d7875943c10.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 5 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/eb2cbe4324cebca65b85816262a85e473cd65967.png" style="max-width: 100.0%;max-height: 100.0%;"/>
| 500
|
[
{
"input": "5\n82195\n64723",
"output": "13"
},
{
"input": "12\n102021090898\n010212908089",
"output": "16"
},
{
"input": "1\n8\n1",
"output": "3"
},
{
"input": "2\n83\n57",
"output": "7"
},
{
"input": "10\n0728592530\n1362615763",
"output": "27"
},
{
"input": "100\n4176196363694273682807653052945037727131821799902563705176501742060696655282954944720643131654235909\n3459912084922154505910287499879975659298239371519889866585472674423008837878123067103005344986554746",
"output": "245"
},
{
"input": "1\n8\n1",
"output": "3"
},
{
"input": "2\n83\n57",
"output": "7"
},
{
"input": "3\n607\n684",
"output": "5"
},
{
"input": "4\n0809\n0636",
"output": "8"
},
{
"input": "5\n84284\n08941",
"output": "16"
},
{
"input": "25\n8037856825987124762280548\n9519431339078678836940020",
"output": "72"
},
{
"input": "125\n23269567683904664184142384849516523616863461607751021071772615078579713054027902974007001544768640273491193035874486891541257\n47635110303703399505805044019026243695451609639556649012447370081552870340011971572363458960190590266459684717415349529509024",
"output": "305"
},
{
"input": "5\n84284\n08941",
"output": "16"
},
{
"input": "25\n8037856825987124762285484\n9519431339078678836940202",
"output": "74"
},
{
"input": "125\n23269567689466418414238845152168634610771021717726157879713054270294007001544768647391193035874486891412573389247025830678706\n47635110307339950580504010224954516093956649124473708152870340117152363458960190596659684717415349529090241694059599629136831",
"output": "357"
},
{
"input": "5\n84284\n08941",
"output": "16"
},
{
"input": "25\n8378525987476228048406972\n9194339078883694020217816",
"output": "55"
},
{
"input": "125\n23269576839046618414238484916523616863461607750210717761078579713054027902974007015447686027349193035874486891541257338624472\n47635103037033950580504401926243695451609639556490124437081552870340011971572363489601905026645984717415349529509024169604599",
"output": "274"
},
{
"input": "1\n0\n0",
"output": "0"
},
{
"input": "1\n7\n7",
"output": "0"
},
{
"input": "1\n0\n5",
"output": "5"
},
{
"input": "1\n2\n7",
"output": "5"
},
{
"input": "1\n7\n9",
"output": "2"
},
{
"input": "1\n9\n7",
"output": "2"
},
{
"input": "1\n2\n9",
"output": "3"
},
{
"input": "1\n9\n2",
"output": "3"
},
{
"input": "25\n3164978461316464614169874\n9413979197249127496597357",
"output": "66"
},
{
"input": "4\n9999\n9999",
"output": "0"
},
{
"input": "2\n11\n11",
"output": "0"
}
] | 1,596,387,325
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 43
| 109
| 6,656,000
|
n=int(input())
s=input()
o=input()
f="01234567890123456789"
ans=0
for i in range(n):
si=int(s[i])
oi=int(o[i])
ans=ans+min(abs(si-oi),abs(si-(10+oi)),abs(si+10-oi))
print(ans)
|
Title: Combination Lock
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Scrooge McDuck keeps his most treasured savings in a home safe with a combination lock. Each time he wants to put there the treasures that he's earned fair and square, he has to open the lock.
The combination lock is represented by *n* rotating disks with digits from 0 to 9 written on them. Scrooge McDuck has to turn some disks so that the combination of digits on the disks forms a secret combination. In one move, he can rotate one disk one digit forwards or backwards. In particular, in one move he can go from digit 0 to digit 9 and vice versa. What minimum number of actions does he need for that?
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of disks on the combination lock.
The second line contains a string of *n* digits — the original state of the disks.
The third line contains a string of *n* digits — Scrooge McDuck's combination that opens the lock.
Output Specification:
Print a single integer — the minimum number of moves Scrooge McDuck needs to open the lock.
Demo Input:
['5\n82195\n64723\n']
Demo Output:
['13\n']
Note:
In the sample he needs 13 moves:
- 1 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b8967f65a723782358b93eff9ce69f336817cf70.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 2 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/07fa58573ece0d32c4d555e498d2b24d2f70f36a.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 3 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/cc2275d9252aae35a6867c6a5b4ba7596e9a7626.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 4 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b100aea470fcaaab4e9529b234ba0d7875943c10.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 5 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/eb2cbe4324cebca65b85816262a85e473cd65967.png" style="max-width: 100.0%;max-height: 100.0%;"/>
|
```python
n=int(input())
s=input()
o=input()
f="01234567890123456789"
ans=0
for i in range(n):
si=int(s[i])
oi=int(o[i])
ans=ans+min(abs(si-oi),abs(si-(10+oi)),abs(si+10-oi))
print(ans)
```
| 3
|
|
29
|
C
|
Mail Stamps
|
PROGRAMMING
| 1,700
|
[
"data structures",
"dfs and similar",
"graphs",
"implementation"
] |
C. Mail Stamps
|
2
|
256
|
One day Bob got a letter in an envelope. Bob knows that when Berland's post officers send a letter directly from city «A» to city «B», they stamp it with «A B», or «B A». Unfortunately, often it is impossible to send a letter directly from the city of the sender to the city of the receiver, that's why the letter is sent via some intermediate cities. Post officers never send a letter in such a way that the route of this letter contains some city more than once. Bob is sure that the post officers stamp the letters accurately.
There are *n* stamps on the envelope of Bob's letter. He understands that the possible routes of this letter are only two. But the stamps are numerous, and Bob can't determine himself none of these routes. That's why he asks you to help him. Find one of the possible routes of the letter.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — amount of mail stamps on the envelope. Then there follow *n* lines with two integers each — description of the stamps. Each stamp is described with indexes of the cities between which a letter is sent. The indexes of cities are integers from 1 to 109. Indexes of all the cities are different. Every time the letter is sent from one city to another, exactly one stamp is put on the envelope. It is guaranteed that the given stamps correspond to some valid route from some city to some other city.
|
Output *n*<=+<=1 numbers — indexes of cities in one of the two possible routes of the letter.
|
[
"2\n1 100\n100 2\n",
"3\n3 1\n100 2\n3 2\n"
] |
[
"2 100 1 ",
"100 2 3 1 "
] |
none
| 1,500
|
[
{
"input": "2\n1 100\n100 2",
"output": "2 100 1 "
},
{
"input": "3\n3 1\n100 2\n3 2",
"output": "100 2 3 1 "
},
{
"input": "3\n458744979 589655889\n248228386 824699605\n458744979 824699605",
"output": "589655889 458744979 824699605 248228386 "
},
{
"input": "4\n90104473 221011623\n18773664 221011623\n90104473 74427905\n74427905 186329050",
"output": "186329050 74427905 90104473 221011623 18773664 "
},
{
"input": "5\n695442143 421284135\n641835294 542627184\n852367357 120042890\n641835294 852367357\n542627184 421284135",
"output": "695442143 421284135 542627184 641835294 852367357 120042890 "
},
{
"input": "6\n264896923 2497658\n57071588 447086061\n2497658 483723090\n57071588 264896923\n158310110 483723090\n158310110 72866107",
"output": "447086061 57071588 264896923 2497658 483723090 158310110 72866107 "
},
{
"input": "1\n1 1000000000",
"output": "1000000000 1 "
},
{
"input": "1\n1000000000 999999999",
"output": "1000000000 999999999 "
},
{
"input": "10\n661239801 721746596\n225324231 116454751\n687002568 865423160\n799202882 865423160\n661239801 116454751\n387882517 687002568\n748798833 721746596\n179630172 225324231\n945958362 387882517\n179630172 945958362",
"output": "799202882 865423160 687002568 387882517 945958362 179630172 225324231 116454751 661239801 721746596 748798833 "
},
{
"input": "21\n280810160 291988863\n760364563 140163983\n16417017 364832782\n400253359 677358550\n597688496 794948223\n400253359 603304541\n589408417 603304541\n385039298 307729574\n293170375 805849550\n140163983 219301181\n732214548 760364563\n307729574 280810160\n131915938 219301181\n4615652 347722938\n396478457 805849550\n16417017 732214548\n4615652 677358550\n131915938 589408417\n291988863 364832782\n396478457 794948223\n385039298 597688496",
"output": "347722938 4615652 677358550 400253359 603304541 589408417 131915938 219301181 140163983 760364563 732214548 16417017 364832782 291988863 280810160 307729574 385039298 597688496 794948223 396478457 805849550 293170375 "
},
{
"input": "1\n2105127 227379126",
"output": "227379126 2105127 "
}
] | 1,630,674,913
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 20
| 1,310
| 28,672,000
|
from _collections import deque
n=int(input())-1
x,y=map(int,input().split())
q=deque([x,y])
w={x:{y},y:{x}}
l={x,y}
for i in range(n):
x,y=map(int,input().split())
if x in w:w[x].add(y)
else:w[x]={y}
if y in w:w[y].add(x)
else:w[y]={x}
while len(w[q[-1]])>1:
for i in w[q[-1]]:
if i not in l:
q.append(i)
l.add(i)
while len(w[q[0]])>1:
for i in w[q[0]]:
if i not in l:
q.appendleft(i)
l.add(i)
print(*q)
|
Title: Mail Stamps
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
One day Bob got a letter in an envelope. Bob knows that when Berland's post officers send a letter directly from city «A» to city «B», they stamp it with «A B», or «B A». Unfortunately, often it is impossible to send a letter directly from the city of the sender to the city of the receiver, that's why the letter is sent via some intermediate cities. Post officers never send a letter in such a way that the route of this letter contains some city more than once. Bob is sure that the post officers stamp the letters accurately.
There are *n* stamps on the envelope of Bob's letter. He understands that the possible routes of this letter are only two. But the stamps are numerous, and Bob can't determine himself none of these routes. That's why he asks you to help him. Find one of the possible routes of the letter.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — amount of mail stamps on the envelope. Then there follow *n* lines with two integers each — description of the stamps. Each stamp is described with indexes of the cities between which a letter is sent. The indexes of cities are integers from 1 to 109. Indexes of all the cities are different. Every time the letter is sent from one city to another, exactly one stamp is put on the envelope. It is guaranteed that the given stamps correspond to some valid route from some city to some other city.
Output Specification:
Output *n*<=+<=1 numbers — indexes of cities in one of the two possible routes of the letter.
Demo Input:
['2\n1 100\n100 2\n', '3\n3 1\n100 2\n3 2\n']
Demo Output:
['2 100 1 ', '100 2 3 1 ']
Note:
none
|
```python
from _collections import deque
n=int(input())-1
x,y=map(int,input().split())
q=deque([x,y])
w={x:{y},y:{x}}
l={x,y}
for i in range(n):
x,y=map(int,input().split())
if x in w:w[x].add(y)
else:w[x]={y}
if y in w:w[y].add(x)
else:w[y]={x}
while len(w[q[-1]])>1:
for i in w[q[-1]]:
if i not in l:
q.append(i)
l.add(i)
while len(w[q[0]])>1:
for i in w[q[0]]:
if i not in l:
q.appendleft(i)
l.add(i)
print(*q)
```
| 3.619094
|
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"
}
] | 1,584,694,031
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 45
| 310
| 0
|
import sys
import math
#to read string
get_string = lambda: sys.stdin.readline().strip()
#to read list of integers
get_int_list = lambda: list( map(int,sys.stdin.readline().strip().split()) )
#to read integers
get_int = lambda: int(sys.stdin.readline())
#to print fast
pt = lambda x: sys.stdout.write(str(x)+'\n')
#--------------------------------WhiteHat010--------------------------------------#
a,b,n = get_int_list()
turn = 0
while True:
if turn == 0:
m = math.gcd(n,a)
if n >= m:
n -= m
turn = 1
else:
print(1)
break
else:
m = math.gcd(n,b)
if n >= m:
n -= m
turn = 0
else:
print(0)
break
|
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
import math
#to read string
get_string = lambda: sys.stdin.readline().strip()
#to read list of integers
get_int_list = lambda: list( map(int,sys.stdin.readline().strip().split()) )
#to read integers
get_int = lambda: int(sys.stdin.readline())
#to print fast
pt = lambda x: sys.stdout.write(str(x)+'\n')
#--------------------------------WhiteHat010--------------------------------------#
a,b,n = get_int_list()
turn = 0
while True:
if turn == 0:
m = math.gcd(n,a)
if n >= m:
n -= m
turn = 1
else:
print(1)
break
else:
m = math.gcd(n,b)
if n >= m:
n -= m
turn = 0
else:
print(0)
break
```
| 3
|
|
928
|
B
|
Chat
|
PROGRAMMING
| 1,400
|
[
"*special",
"dp"
] | null | null |
There are times you recall a good old friend and everything you've come through together. Luckily there are social networks — they store all your message history making it easy to know what you argued over 10 years ago.
More formal, your message history is a sequence of messages ordered by time sent numbered from 1 to *n* where *n* is the total number of messages in the chat.
Each message might contain a link to an earlier message which it is a reply to. When opening a message *x* or getting a link to it, the dialogue is shown in such a way that *k* previous messages, message *x* and *k* next messages are visible (with respect to message *x*). In case there are less than *k* messages somewhere, they are yet all shown.
Digging deep into your message history, you always read all visible messages and then go by the link in the current message *x* (if there is one) and continue reading in the same manner.
Determine the number of messages you'll read if your start from message number *t* for all *t* from 1 to *n*. Calculate these numbers independently. If you start with message *x*, the initial configuration is *x* itself, *k* previous and *k* next messages. Messages read multiple times are considered as one.
|
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=105, 0<=≤<=*k*<=≤<=*n*) — the total amount of messages and the number of previous and next messages visible.
The second line features a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=<<=*i*), where *a**i* denotes the *i*-th message link destination or zero, if there's no link from *i*. All messages are listed in chronological order. It's guaranteed that the link from message *x* goes to message with number strictly less than *x*.
|
Print *n* integers with *i*-th denoting the number of distinct messages you can read starting from message *i* and traversing the links while possible.
|
[
"6 0\n0 1 1 2 3 2\n",
"10 1\n0 1 0 3 4 5 2 3 7 0\n",
"2 2\n0 1\n"
] |
[
"1 2 2 3 3 3 \n",
"2 3 3 4 5 6 6 6 8 2 \n",
"2 2 \n"
] |
Consider *i* = 6 in sample case one. You will read message 6, then 2, then 1 and then there will be no link to go.
In the second sample case *i* = 6 gives you messages 5, 6, 7 since *k* = 1, then 4, 5, 6, then 2, 3, 4 and then the link sequence breaks. The number of distinct messages here is equal to 6.
| 1,250
|
[
{
"input": "6 0\n0 1 1 2 3 2",
"output": "1 2 2 3 3 3 "
},
{
"input": "10 1\n0 1 0 3 4 5 2 3 7 0",
"output": "2 3 3 4 5 6 6 6 8 2 "
},
{
"input": "2 2\n0 1",
"output": "2 2 "
},
{
"input": "1 1\n0",
"output": "1 "
},
{
"input": "5 2\n0 1 2 3 1",
"output": "3 4 5 5 5 "
},
{
"input": "30 1\n0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 2 0 0 0 0 0 2 1 0",
"output": "2 3 3 3 3 3 3 3 3 3 3 3 3 5 5 5 3 3 3 3 3 6 3 3 3 3 3 6 5 2 "
},
{
"input": "100 5\n0 1 1 1 0 5 6 6 8 8 9 11 12 11 8 0 0 14 6 16 7 21 15 23 15 24 0 0 0 28 0 29 26 27 19 0 0 21 37 32 40 30 37 34 39 38 34 38 0 0 41 24 45 47 0 33 46 26 31 0 21 57 57 31 63 63 25 59 65 56 68 0 30 55 55 0 70 43 59 49 59 79 66 74 0 11 65 0 80 63 0 84 73 49 73 81 0 86 76 98",
"output": "6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 11 11 23 22 15 23 24 28 29 30 31 11 11 11 13 11 14 38 18 33 11 11 34 13 22 23 24 17 28 19 42 29 44 11 11 33 40 27 36 11 49 53 42 22 11 34 58 59 22 61 62 41 31 65 60 34 11 24 22 22 11 67 28 33 22 33 36 73 32 11 27 72 11 31 70 11 40 35 22 35 43 9 35 18 35 "
},
{
"input": "2 2\n0 0",
"output": "2 2 "
},
{
"input": "2 1\n0 0",
"output": "2 2 "
},
{
"input": "2 1\n0 1",
"output": "2 2 "
},
{
"input": "2 0\n0 0",
"output": "1 1 "
},
{
"input": "2 0\n0 1",
"output": "1 2 "
},
{
"input": "3 0\n0 0 0",
"output": "1 1 1 "
},
{
"input": "3 0\n0 0 1",
"output": "1 1 2 "
},
{
"input": "3 0\n0 0 2",
"output": "1 1 2 "
},
{
"input": "3 0\n0 1 0",
"output": "1 2 1 "
},
{
"input": "3 0\n0 1 1",
"output": "1 2 2 "
},
{
"input": "3 0\n0 1 2",
"output": "1 2 3 "
},
{
"input": "3 1\n0 0 0",
"output": "2 3 2 "
},
{
"input": "3 1\n0 0 1",
"output": "2 3 3 "
},
{
"input": "3 1\n0 0 2",
"output": "2 3 3 "
},
{
"input": "3 1\n0 1 0",
"output": "2 3 2 "
},
{
"input": "3 1\n0 1 1",
"output": "2 3 3 "
},
{
"input": "3 1\n0 1 2",
"output": "2 3 3 "
},
{
"input": "3 2\n0 0 0",
"output": "3 3 3 "
},
{
"input": "3 2\n0 0 1",
"output": "3 3 3 "
},
{
"input": "3 2\n0 0 2",
"output": "3 3 3 "
},
{
"input": "3 2\n0 1 0",
"output": "3 3 3 "
},
{
"input": "3 2\n0 1 1",
"output": "3 3 3 "
},
{
"input": "3 2\n0 1 2",
"output": "3 3 3 "
},
{
"input": "3 3\n0 0 0",
"output": "3 3 3 "
},
{
"input": "3 3\n0 0 1",
"output": "3 3 3 "
},
{
"input": "3 3\n0 0 2",
"output": "3 3 3 "
},
{
"input": "3 3\n0 1 0",
"output": "3 3 3 "
},
{
"input": "3 3\n0 1 1",
"output": "3 3 3 "
},
{
"input": "3 3\n0 1 2",
"output": "3 3 3 "
},
{
"input": "10 3\n0 0 0 0 0 0 0 4 0 4",
"output": "4 5 6 7 7 7 7 10 5 10 "
},
{
"input": "20 2\n0 0 0 0 2 1 0 3 0 1 1 11 0 10 0 0 9 17 9 0",
"output": "3 4 5 5 7 8 5 10 5 8 8 9 5 12 5 5 10 11 9 3 "
},
{
"input": "40 0\n0 1 2 3 4 5 0 0 0 0 0 11 12 0 14 10 0 16 15 0 19 21 22 0 23 25 25 24 24 29 29 0 0 31 0 35 31 36 34 29",
"output": "1 2 3 4 5 6 1 1 1 1 1 2 3 1 2 2 1 3 3 1 4 5 6 1 7 8 8 2 2 3 3 1 1 4 1 2 4 3 5 3 "
}
] | 1,673,329,625
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 118
| 109
| 18,636,800
|
n, k = map(int, input().split())
a = list(map(int, input().split()))
dp = [0] * n
for i in range(n):
if a[i] == 0:
s = min(k, i) + 1
else:
s = dp[a[i] - 1] + min(2 * k, i - a[i]) + 1
dp[i] = s
an = []
for i in range(n):
an.append(str(dp[i] + min(k, n - i - 1)))
print(" ".join(an))
|
Title: Chat
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are times you recall a good old friend and everything you've come through together. Luckily there are social networks — they store all your message history making it easy to know what you argued over 10 years ago.
More formal, your message history is a sequence of messages ordered by time sent numbered from 1 to *n* where *n* is the total number of messages in the chat.
Each message might contain a link to an earlier message which it is a reply to. When opening a message *x* or getting a link to it, the dialogue is shown in such a way that *k* previous messages, message *x* and *k* next messages are visible (with respect to message *x*). In case there are less than *k* messages somewhere, they are yet all shown.
Digging deep into your message history, you always read all visible messages and then go by the link in the current message *x* (if there is one) and continue reading in the same manner.
Determine the number of messages you'll read if your start from message number *t* for all *t* from 1 to *n*. Calculate these numbers independently. If you start with message *x*, the initial configuration is *x* itself, *k* previous and *k* next messages. Messages read multiple times are considered as one.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=105, 0<=≤<=*k*<=≤<=*n*) — the total amount of messages and the number of previous and next messages visible.
The second line features a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=<<=*i*), where *a**i* denotes the *i*-th message link destination or zero, if there's no link from *i*. All messages are listed in chronological order. It's guaranteed that the link from message *x* goes to message with number strictly less than *x*.
Output Specification:
Print *n* integers with *i*-th denoting the number of distinct messages you can read starting from message *i* and traversing the links while possible.
Demo Input:
['6 0\n0 1 1 2 3 2\n', '10 1\n0 1 0 3 4 5 2 3 7 0\n', '2 2\n0 1\n']
Demo Output:
['1 2 2 3 3 3 \n', '2 3 3 4 5 6 6 6 8 2 \n', '2 2 \n']
Note:
Consider *i* = 6 in sample case one. You will read message 6, then 2, then 1 and then there will be no link to go.
In the second sample case *i* = 6 gives you messages 5, 6, 7 since *k* = 1, then 4, 5, 6, then 2, 3, 4 and then the link sequence breaks. The number of distinct messages here is equal to 6.
|
```python
n, k = map(int, input().split())
a = list(map(int, input().split()))
dp = [0] * n
for i in range(n):
if a[i] == 0:
s = min(k, i) + 1
else:
s = dp[a[i] - 1] + min(2 * k, i - a[i]) + 1
dp[i] = s
an = []
for i in range(n):
an.append(str(dp[i] + min(k, n - i - 1)))
print(" ".join(an))
```
| 3
|
|
554
|
A
|
Kyoya and Photobooks
|
PROGRAMMING
| 900
|
[
"brute force",
"math",
"strings"
] | null | null |
Kyoya Ootori is selling photobooks of the Ouran High School Host Club. He has 26 photos, labeled "a" to "z", and he has compiled them into a photo booklet with some photos in some order (possibly with some photos being duplicated). A photo booklet can be described as a string of lowercase letters, consisting of the photos in the booklet in order. He now wants to sell some "special edition" photobooks, each with one extra photo inserted anywhere in the book. He wants to make as many distinct photobooks as possible, so he can make more money. He asks Haruhi, how many distinct photobooks can he make by inserting one extra photo into the photobook he already has?
Please help Haruhi solve this problem.
|
The first line of input will be a single string *s* (1<=≤<=|*s*|<=≤<=20). String *s* consists only of lowercase English letters.
|
Output a single integer equal to the number of distinct photobooks Kyoya Ootori can make.
|
[
"a\n",
"hi\n"
] |
[
"51\n",
"76\n"
] |
In the first case, we can make 'ab','ac',...,'az','ba','ca',...,'za', and 'aa', producing a total of 51 distinct photo booklets.
| 250
|
[
{
"input": "a",
"output": "51"
},
{
"input": "hi",
"output": "76"
},
{
"input": "y",
"output": "51"
},
{
"input": "kgan",
"output": "126"
},
{
"input": "zoabkyuvus",
"output": "276"
},
{
"input": "spyemhyznjieyhhbk",
"output": "451"
},
{
"input": "xulsyfkuizjauadjjopu",
"output": "526"
},
{
"input": "e",
"output": "51"
},
{
"input": "zv",
"output": "76"
},
{
"input": "jgv",
"output": "101"
},
{
"input": "zsfo",
"output": "126"
},
{
"input": "jselr",
"output": "151"
},
{
"input": "dwemig",
"output": "176"
},
{
"input": "mddoxsf",
"output": "201"
},
{
"input": "jgirkrmi",
"output": "226"
},
{
"input": "spkxurcum",
"output": "251"
},
{
"input": "fykkiubdkt",
"output": "276"
},
{
"input": "fznbcxsxygs",
"output": "301"
},
{
"input": "qcrvrdqcbtou",
"output": "326"
},
{
"input": "qktrbjzrqgmlr",
"output": "351"
},
{
"input": "foamodbvptlxxg",
"output": "376"
},
{
"input": "ydzpjhsidipricw",
"output": "401"
},
{
"input": "lpfpndmjfvqejdgf",
"output": "426"
},
{
"input": "ofkvparuvjtggnmab",
"output": "451"
},
{
"input": "xxncfutrtxcwdzwbgs",
"output": "476"
},
{
"input": "zovhffccflkgqncsdte",
"output": "501"
},
{
"input": "cskgsxywlvfeicoueglr",
"output": "526"
},
{
"input": "gggggggggggggggggggg",
"output": "526"
},
{
"input": "qdqdddqddqqddqddqdqd",
"output": "526"
},
{
"input": "takttttaakaaktakttkt",
"output": "526"
},
{
"input": "coccoooogogcgocccmcg",
"output": "526"
},
{
"input": "kskkskkkssksssk",
"output": "401"
},
{
"input": "lllllllllllllll",
"output": "401"
}
] | 1,548,389,828
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 33
| 109
| 0
|
book=input()
print((len(book)*26)+(26-len(book)))
|
Title: Kyoya and Photobooks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kyoya Ootori is selling photobooks of the Ouran High School Host Club. He has 26 photos, labeled "a" to "z", and he has compiled them into a photo booklet with some photos in some order (possibly with some photos being duplicated). A photo booklet can be described as a string of lowercase letters, consisting of the photos in the booklet in order. He now wants to sell some "special edition" photobooks, each with one extra photo inserted anywhere in the book. He wants to make as many distinct photobooks as possible, so he can make more money. He asks Haruhi, how many distinct photobooks can he make by inserting one extra photo into the photobook he already has?
Please help Haruhi solve this problem.
Input Specification:
The first line of input will be a single string *s* (1<=≤<=|*s*|<=≤<=20). String *s* consists only of lowercase English letters.
Output Specification:
Output a single integer equal to the number of distinct photobooks Kyoya Ootori can make.
Demo Input:
['a\n', 'hi\n']
Demo Output:
['51\n', '76\n']
Note:
In the first case, we can make 'ab','ac',...,'az','ba','ca',...,'za', and 'aa', producing a total of 51 distinct photo booklets.
|
```python
book=input()
print((len(book)*26)+(26-len(book)))
```
| 3
|
|
555
|
A
|
Case of Matryoshkas
|
PROGRAMMING
| 1,500
|
[
"implementation"
] | null | null |
Andrewid the Android is a galaxy-famous detective. He is now investigating the case of vandalism at the exhibition of contemporary art.
The main exhibit is a construction of *n* matryoshka dolls that can be nested one into another. The matryoshka dolls are numbered from 1 to *n*. A matryoshka with a smaller number can be nested in a matryoshka with a higher number, two matryoshkas can not be directly nested in the same doll, but there may be chain nestings, for example, 1<=→<=2<=→<=4<=→<=5.
In one second, you can perform one of the two following operations:
- Having a matryoshka *a* that isn't nested in any other matryoshka and a matryoshka *b*, such that *b* doesn't contain any other matryoshka and is not nested in any other matryoshka, you may put *a* in *b*; - Having a matryoshka *a* directly contained in matryoshka *b*, such that *b* is not nested in any other matryoshka, you may get *a* out of *b*.
According to the modern aesthetic norms the matryoshka dolls on display were assembled in a specific configuration, i.e. as several separate chains of nested matryoshkas, but the criminal, following the mysterious plan, took out all the dolls and assembled them into a single large chain (1<=→<=2<=→<=...<=→<=*n*). In order to continue the investigation Andrewid needs to know in what minimum time it is possible to perform this action.
|
The first line contains integers *n* (1<=≤<=*n*<=≤<=105) and *k* (1<=≤<=*k*<=≤<=105) — the number of matryoshkas and matryoshka chains in the initial configuration.
The next *k* lines contain the descriptions of the chains: the *i*-th line first contains number *m**i* (1<=≤<=*m**i*<=≤<=*n*), and then *m**i* numbers *a**i*1,<=*a**i*2,<=...,<=*a**im**i* — the numbers of matryoshkas in the chain (matryoshka *a**i*1 is nested into matryoshka *a**i*2, that is nested into matryoshka *a**i*3, and so on till the matryoshka *a**im**i* that isn't nested into any other matryoshka).
It is guaranteed that *m*1<=+<=*m*2<=+<=...<=+<=*m**k*<==<=*n*, the numbers of matryoshkas in all the chains are distinct, in each chain the numbers of matryoshkas follow in the ascending order.
|
In the single line print the minimum number of seconds needed to assemble one large chain from the initial configuration.
|
[
"3 2\n2 1 2\n1 3\n",
"7 3\n3 1 3 7\n2 2 5\n2 4 6\n"
] |
[
"1\n",
"10\n"
] |
In the first sample test there are two chains: 1 → 2 and 3. In one second you can nest the first chain into the second one and get 1 → 2 → 3.
In the second sample test you need to disassemble all the three chains into individual matryoshkas in 2 + 1 + 1 = 4 seconds and then assemble one big chain in 6 seconds.
| 250
|
[
{
"input": "3 2\n2 1 2\n1 3",
"output": "1"
},
{
"input": "7 3\n3 1 3 7\n2 2 5\n2 4 6",
"output": "10"
},
{
"input": "1 1\n1 1",
"output": "0"
},
{
"input": "3 2\n1 2\n2 1 3",
"output": "3"
},
{
"input": "5 3\n1 4\n3 1 2 3\n1 5",
"output": "2"
},
{
"input": "8 5\n2 1 2\n2 3 4\n1 5\n2 6 7\n1 8",
"output": "8"
},
{
"input": "10 10\n1 5\n1 4\n1 10\n1 3\n1 7\n1 1\n1 8\n1 6\n1 9\n1 2",
"output": "9"
},
{
"input": "20 6\n3 8 9 13\n3 4 14 20\n2 15 17\n3 2 5 11\n5 7 10 12 18 19\n4 1 3 6 16",
"output": "33"
},
{
"input": "50 10\n6 17 21 31 42 45 49\n6 11 12 15 22 26 38\n3 9 29 36\n3 10 23 43\n5 14 19 28 46 48\n2 30 39\n6 13 20 24 33 37 47\n8 1 2 3 4 5 6 7 8\n7 16 18 25 27 34 40 44\n4 32 35 41 50",
"output": "75"
},
{
"input": "13 8\n1 5\n2 8 10\n1 13\n4 1 2 3 11\n1 7\n2 6 12\n1 4\n1 9",
"output": "13"
},
{
"input": "21 13\n1 18\n2 8 13\n1 21\n1 17\n2 7 9\n1 20\n1 19\n1 4\n1 16\n2 5 6\n3 12 14 15\n3 1 2 3\n2 10 11",
"output": "24"
},
{
"input": "50 50\n1 2\n1 5\n1 28\n1 46\n1 42\n1 24\n1 3\n1 37\n1 33\n1 50\n1 23\n1 40\n1 43\n1 26\n1 49\n1 34\n1 8\n1 45\n1 15\n1 1\n1 22\n1 18\n1 27\n1 25\n1 13\n1 39\n1 38\n1 10\n1 44\n1 6\n1 17\n1 47\n1 7\n1 35\n1 20\n1 36\n1 31\n1 21\n1 32\n1 29\n1 4\n1 12\n1 19\n1 16\n1 11\n1 41\n1 9\n1 14\n1 30\n1 48",
"output": "49"
},
{
"input": "100 3\n45 1 2 3 4 5 6 7 8 9 19 21 24 27 28 30 34 35 37 39 40 41 42 43 46 47 48 51 52 55 58 59 61 63 64 66 69 71 76 80 85 86 88 89 94 99\n26 10 11 15 18 23 29 31 33 36 38 44 49 54 56 60 62 65 75 78 82 83 84 95 96 97 98\n29 12 13 14 16 17 20 22 25 26 32 45 50 53 57 67 68 70 72 73 74 77 79 81 87 90 91 92 93 100",
"output": "180"
},
{
"input": "100 19\n6 62 72 83 91 94 97\n3 61 84 99\n1 63\n5 46 53 56 69 78\n5 41 43 49 74 89\n5 55 57 79 85 87\n3 47 59 98\n3 64 76 82\n3 48 66 75\n2 60 88\n2 67 77\n4 40 51 73 95\n41 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 44 71 81\n4 58 65 90 93\n1 100\n5 39 45 52 80 86\n2 50 68\n1 92\n4 42 54 70 96",
"output": "106"
}
] | 1,539,739,640
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 47
| 374
| 7,270,400
|
n, k = list(map(int,input().split(" ")))
cost = 0
for i in range(k):
inp = list(map(int,input().split(" ")))
m = inp[0]
a = inp[1:]
if a[0] != 1:
cost += m * 2 - 1
else:
for j in range(m):
if a[j] == j+1:
continue
j -= 1
break
cost += (m - j) * 2 - 1
print(cost - 1)
|
Title: Case of Matryoshkas
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Andrewid the Android is a galaxy-famous detective. He is now investigating the case of vandalism at the exhibition of contemporary art.
The main exhibit is a construction of *n* matryoshka dolls that can be nested one into another. The matryoshka dolls are numbered from 1 to *n*. A matryoshka with a smaller number can be nested in a matryoshka with a higher number, two matryoshkas can not be directly nested in the same doll, but there may be chain nestings, for example, 1<=→<=2<=→<=4<=→<=5.
In one second, you can perform one of the two following operations:
- Having a matryoshka *a* that isn't nested in any other matryoshka and a matryoshka *b*, such that *b* doesn't contain any other matryoshka and is not nested in any other matryoshka, you may put *a* in *b*; - Having a matryoshka *a* directly contained in matryoshka *b*, such that *b* is not nested in any other matryoshka, you may get *a* out of *b*.
According to the modern aesthetic norms the matryoshka dolls on display were assembled in a specific configuration, i.e. as several separate chains of nested matryoshkas, but the criminal, following the mysterious plan, took out all the dolls and assembled them into a single large chain (1<=→<=2<=→<=...<=→<=*n*). In order to continue the investigation Andrewid needs to know in what minimum time it is possible to perform this action.
Input Specification:
The first line contains integers *n* (1<=≤<=*n*<=≤<=105) and *k* (1<=≤<=*k*<=≤<=105) — the number of matryoshkas and matryoshka chains in the initial configuration.
The next *k* lines contain the descriptions of the chains: the *i*-th line first contains number *m**i* (1<=≤<=*m**i*<=≤<=*n*), and then *m**i* numbers *a**i*1,<=*a**i*2,<=...,<=*a**im**i* — the numbers of matryoshkas in the chain (matryoshka *a**i*1 is nested into matryoshka *a**i*2, that is nested into matryoshka *a**i*3, and so on till the matryoshka *a**im**i* that isn't nested into any other matryoshka).
It is guaranteed that *m*1<=+<=*m*2<=+<=...<=+<=*m**k*<==<=*n*, the numbers of matryoshkas in all the chains are distinct, in each chain the numbers of matryoshkas follow in the ascending order.
Output Specification:
In the single line print the minimum number of seconds needed to assemble one large chain from the initial configuration.
Demo Input:
['3 2\n2 1 2\n1 3\n', '7 3\n3 1 3 7\n2 2 5\n2 4 6\n']
Demo Output:
['1\n', '10\n']
Note:
In the first sample test there are two chains: 1 → 2 and 3. In one second you can nest the first chain into the second one and get 1 → 2 → 3.
In the second sample test you need to disassemble all the three chains into individual matryoshkas in 2 + 1 + 1 = 4 seconds and then assemble one big chain in 6 seconds.
|
```python
n, k = list(map(int,input().split(" ")))
cost = 0
for i in range(k):
inp = list(map(int,input().split(" ")))
m = inp[0]
a = inp[1:]
if a[0] != 1:
cost += m * 2 - 1
else:
for j in range(m):
if a[j] == j+1:
continue
j -= 1
break
cost += (m - j) * 2 - 1
print(cost - 1)
```
| 3
|
|
629
|
A
|
Far Relative’s Birthday Cake
|
PROGRAMMING
| 800
|
[
"brute force",
"combinatorics",
"constructive algorithms",
"implementation"
] | null | null |
Door's family is going celebrate Famil Doors's birthday party. They love Famil Door so they are planning to make his birthday cake weird!
The cake is a *n*<=×<=*n* square consisting of equal squares with side length 1. Each square is either empty or consists of a single chocolate. They bought the cake and randomly started to put the chocolates on the cake. The value of Famil Door's happiness will be equal to the number of pairs of cells with chocolates that are in the same row or in the same column of the cake. Famil Doors's family is wondering what is the amount of happiness of Famil going to be?
Please, note that any pair can be counted no more than once, as two different cells can't share both the same row and the same column.
|
In the first line of the input, you are given a single integer *n* (1<=≤<=*n*<=≤<=100) — the length of the side of the cake.
Then follow *n* lines, each containing *n* characters. Empty cells are denoted with '.', while cells that contain chocolates are denoted by 'C'.
|
Print the value of Famil Door's happiness, i.e. the number of pairs of chocolate pieces that share the same row or the same column.
|
[
"3\n.CC\nC..\nC.C\n",
"4\nCC..\nC..C\n.CC.\n.CC.\n"
] |
[
"4\n",
"9\n"
] |
If we number rows from top to bottom and columns from left to right, then, pieces that share the same row in the first sample are:
1. (1, 2) and (1, 3) 1. (3, 1) and (3, 3) 1. (2, 1) and (3, 1) 1. (1, 3) and (3, 3)
| 500
|
[
{
"input": "3\n.CC\nC..\nC.C",
"output": "4"
},
{
"input": "4\nCC..\nC..C\n.CC.\n.CC.",
"output": "9"
},
{
"input": "5\n.CCCC\nCCCCC\n.CCC.\nCC...\n.CC.C",
"output": "46"
},
{
"input": "7\n.CC..CC\nCC.C..C\nC.C..C.\nC...C.C\nCCC.CCC\n.CC...C\n.C.CCC.",
"output": "84"
},
{
"input": "8\n..C....C\nC.CCC.CC\n.C..C.CC\nCC......\nC..C..CC\nC.C...C.\nC.C..C..\nC...C.C.",
"output": "80"
},
{
"input": "9\n.C...CCCC\nC.CCCC...\n....C..CC\n.CC.CCC..\n.C.C..CC.\nC...C.CCC\nCCC.C...C\nCCCC....C\n..C..C..C",
"output": "144"
},
{
"input": "10\n..C..C.C..\n..CC..C.CC\n.C.C...C.C\n..C.CC..CC\n....C..C.C\n...C..C..C\nCC.CC....C\n..CCCC.C.C\n..CC.CCC..\nCCCC..C.CC",
"output": "190"
},
{
"input": "11\nC.CC...C.CC\nCC.C....C.C\n.....C..CCC\n....C.CC.CC\nC..C..CC...\nC...C...C..\nCC..CCC.C.C\n..C.CC.C..C\nC...C.C..CC\n.C.C..CC..C\n.C.C.CC.C..",
"output": "228"
},
{
"input": "21\n...CCC.....CC..C..C.C\n..CCC...CC...CC.CCC.C\n....C.C.C..CCC..C.C.C\n....CCC..C..C.CC.CCC.\n...CCC.C..C.C.....CCC\n.CCC.....CCC..C...C.C\nCCCC.C...CCC.C...C.CC\nC..C...C.CCC..CC..C..\nC...CC..C.C.CC..C.CC.\nCC..CCCCCCCCC..C....C\n.C..CCCC.CCCC.CCC...C\nCCC...CCC...CCC.C..C.\n.CCCCCCCC.CCCC.CC.C..\n.C.C..C....C.CCCCCC.C\n...C...C.CCC.C.CC..C.\nCCC...CC..CC...C..C.C\n.CCCCC...C.C..C.CC.C.\n..CCC.C.C..CCC.CCC...\n..C..C.C.C.....CC.C..\n.CC.C...C.CCC.C....CC\n...C..CCCC.CCC....C..",
"output": "2103"
},
{
"input": "20\nC.C.CCC.C....C.CCCCC\nC.CC.C..CCC....CCCC.\n.CCC.CC...CC.CCCCCC.\n.C...CCCC..C....CCC.\n.C..CCCCCCC.C.C.....\nC....C.C..CCC.C..CCC\n...C.C.CC..CC..CC...\nC...CC.C.CCCCC....CC\n.CC.C.CCC....C.CCC.C\nCC...CC...CC..CC...C\nC.C..CC.C.CCCC.C.CC.\n..CCCCC.C.CCC..CCCC.\n....C..C..C.CC...C.C\nC..CCC..CC..C.CC..CC\n...CC......C.C..C.C.\nCC.CCCCC.CC.CC...C.C\n.C.CC..CC..CCC.C.CCC\nC..C.CC....C....C...\n..CCC..CCC...CC..C.C\n.C.CCC.CCCCCCCCC..CC",
"output": "2071"
},
{
"input": "17\nCCC..C.C....C.C.C\n.C.CC.CC...CC..C.\n.CCCC.CC.C..CCC.C\n...CCC.CC.CCC.C.C\nCCCCCCCC..C.CC.CC\n...C..C....C.CC.C\nCC....CCC...C.CC.\n.CC.C.CC..C......\n.CCCCC.C.CC.CCCCC\n..CCCC...C..CC..C\nC.CC.C.CC..C.C.C.\nC..C..C..CCC.C...\n.C..CCCC..C......\n.CC.C...C..CC.CC.\nC..C....CC...CC..\nC.CC.CC..C.C..C..\nCCCC...C.C..CCCC.",
"output": "1160"
},
{
"input": "15\nCCCC.C..CCC....\nCCCCCC.CC.....C\n...C.CC.C.C.CC.\nCCCCCCC..C..C..\nC..CCC..C.CCCC.\n.CC..C.C.C.CC.C\n.C.C..C..C.C..C\n...C...C..CCCC.\n.....C.C..CC...\nCC.C.C..CC.C..C\n..CCCCC..CCC...\nCC.CC.C..CC.CCC\n..CCC...CC.C..C\nCC..C.C..CCC..C\n.C.C....CCC...C",
"output": "789"
},
{
"input": "1\n.",
"output": "0"
},
{
"input": "3\n.CC\nC..\nC.C",
"output": "4"
},
{
"input": "13\nC.C...C.C.C..\nCC.CCCC.CC..C\n.C.CCCCC.CC..\nCCCC..C...C..\n...CC.C.C...C\n.CC.CCC...CC.\nCC.CCCCCC....\n.C...C..CC..C\nCCCC.CC...C..\n.C.CCC..C.CC.\n..C...CC..C.C\n..C.CCC..CC.C\n.C...CCC.CC.C",
"output": "529"
},
{
"input": "16\n.C.C.C.C.C...C.C\n..C..C.CCCCCC...\n..C.C.C.C..C..C.\n.CC....C.CCC..C.\n.C.CCC..C....CCC\nCC..C.CC..C.C.CC\n...C..C..CC..CC.\n.CCC..C.CC.C.C..\n.CC.C..........C\nC...C....CC..C..\nC.CCC.C..C..C...\n.CCCCCCCCCCCC..C\n..C.C.CC.CC.CCC.\nCC..C.C....C..CC\nC.CCC..C..C.C.CC\n.C.CCC.CC..CCC.C",
"output": "874"
},
{
"input": "2\nCC\nCC",
"output": "4"
},
{
"input": "3\nC..\nC..\nC..",
"output": "3"
}
] | 1,597,667,661
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 48
| 109
| 307,200
|
fact = [1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800, 39916800, 479001600, 6227020800, 87178291200, 1307674368000, 20922789888000, 355687428096000, 6402373705728000, 121645100408832000, 2432902008176640000, 51090942171709440000, 1124000727777607680000, 25852016738884976640000, 620448401733239439360000, 15511210043330985984000000, 403291461126605635584000000, 10888869450418352160768000000, 304888344611713860501504000000, 8841761993739701954543616000000, 265252859812191058636308480000000, 8222838654177922817725562880000000, 263130836933693530167218012160000000, 8683317618811886495518194401280000000, 295232799039604140847618609643520000000, 10333147966386144929666651337523200000000, 371993326789901217467999448150835200000000, 13763753091226345046315979581580902400000000, 523022617466601111760007224100074291200000000, 20397882081197443358640281739902897356800000000, 815915283247897734345611269596115894272000000000, 33452526613163807108170062053440751665152000000000, 1405006117752879898543142606244511569936384000000000, 60415263063373835637355132068513997507264512000000000, 2658271574788448768043625811014615890319638528000000000, 119622220865480194561963161495657715064383733760000000000, 5502622159812088949850305428800254892961651752960000000000, 258623241511168180642964355153611979969197632389120000000000, 12413915592536072670862289047373375038521486354677760000000000, 608281864034267560872252163321295376887552831379210240000000000, 30414093201713378043612608166064768844377641568960512000000000000, 1551118753287382280224243016469303211063259720016986112000000000000, 80658175170943878571660636856403766975289505440883277824000000000000, 4274883284060025564298013753389399649690343788366813724672000000000000, 230843697339241380472092742683027581083278564571807941132288000000000000, 12696403353658275925965100847566516959580321051449436762275840000000000000, 710998587804863451854045647463724949736497978881168458687447040000000000000, 40526919504877216755680601905432322134980384796226602145184481280000000000000, 2350561331282878571829474910515074683828862318181142924420699914240000000000000, 138683118545689835737939019720389406345902876772687432540821294940160000000000000, 8320987112741390144276341183223364380754172606361245952449277696409600000000000000, 507580213877224798800856812176625227226004528988036003099405939480985600000000000000, 31469973260387937525653122354950764088012280797258232192163168247821107200000000000000, 1982608315404440064116146708361898137544773690227268628106279599612729753600000000000000, 126886932185884164103433389335161480802865516174545192198801894375214704230400000000000000, 8247650592082470666723170306785496252186258551345437492922123134388955774976000000000000000, 544344939077443064003729240247842752644293064388798874532860126869671081148416000000000000000, 36471110918188685288249859096605464427167635314049524593701628500267962436943872000000000000000, 2480035542436830599600990418569171581047399201355367672371710738018221445712183296000000000000000, 171122452428141311372468338881272839092270544893520369393648040923257279754140647424000000000000000, 11978571669969891796072783721689098736458938142546425857555362864628009582789845319680000000000000000, 850478588567862317521167644239926010288584608120796235886430763388588680378079017697280000000000000000, 61234458376886086861524070385274672740778091784697328983823014963978384987221689274204160000000000000000, 4470115461512684340891257138125051110076800700282905015819080092370422104067183317016903680000000000000000, 330788544151938641225953028221253782145683251820934971170611926835411235700971565459250872320000000000000000, 24809140811395398091946477116594033660926243886570122837795894512655842677572867409443815424000000000000000000, 1885494701666050254987932260861146558230394535379329335672487982961844043495537923117729972224000000000000000000, 145183092028285869634070784086308284983740379224208358846781574688061991349156420080065207861248000000000000000000, 11324281178206297831457521158732046228731749579488251990048962825668835325234200766245086213177344000000000000000000, 894618213078297528685144171539831652069808216779571907213868063227837990693501860533361810841010176000000000000000000, 71569457046263802294811533723186532165584657342365752577109445058227039255480148842668944867280814080000000000000000000, 5797126020747367985879734231578109105412357244731625958745865049716390179693892056256184534249745940480000000000000000000, 475364333701284174842138206989404946643813294067993328617160934076743994734899148613007131808479167119360000000000000000000, 39455239697206586511897471180120610571436503407643446275224357528369751562996629334879591940103770870906880000000000000000000, 3314240134565353266999387579130131288000666286242049487118846032383059131291716864129885722968716753156177920000000000000000000, 281710411438055027694947944226061159480056634330574206405101912752560026159795933451040286452340924018275123200000000000000000000, 24227095383672732381765523203441259715284870552429381750838764496720162249742450276789464634901319465571660595200000000000000000000, 2107757298379527717213600518699389595229783738061356212322972511214654115727593174080683423236414793504734471782400000000000000000000, 185482642257398439114796845645546284380220968949399346684421580986889562184028199319100141244804501828416633516851200000000000000000000, 16507955160908461081216919262453619309839666236496541854913520707833171034378509739399912570787600662729080382999756800000000000000000000, 1485715964481761497309522733620825737885569961284688766942216863704985393094065876545992131370884059645617234469978112000000000000000000000, 135200152767840296255166568759495142147586866476906677791741734597153670771559994765685283954750449427751168336768008192000000000000000000000, 12438414054641307255475324325873553077577991715875414356840239582938137710983519518443046123837041347353107486982656753664000000000000000000000, 1156772507081641574759205162306240436214753229576413535186142281213246807121467315215203289516844845303838996289387078090752000000000000000000000, 108736615665674308027365285256786601004186803580182872307497374434045199869417927630229109214583415458560865651202385340530688000000000000000000000, 10329978488239059262599702099394727095397746340117372869212250571234293987594703124871765375385424468563282236864226607350415360000000000000000000000, 991677934870949689209571401541893801158183648651267795444376054838492222809091499987689476037000748982075094738965754305639874560000000000000000000000, 96192759682482119853328425949563698712343813919172976158104477319333745612481875498805879175589072651261284189679678167647067832320000000000000000000000, 9426890448883247745626185743057242473809693764078951663494238777294707070023223798882976159207729119823605850588608460429412647567360000000000000000000000, 933262154439441526816992388562667004907159682643816214685929638952175999932299156089414639761565182862536979208272237582511852109168640000000000000000000000, 93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000]
def nCr(n, r):
return fact[n]//(fact[r]*fact[n-r])
n = int(input().strip())
arr = []
for _ in range(n):
arr += [list(input().strip())]
happiness = 0
for row in arr:
happiness+=nCr(row.count("C"),2)
for j in range(n):
col = []
for i in range(n):
col += [arr[i][j]]
happiness+=nCr(col.count("C"),2)
print(happiness)
|
Title: Far Relative’s Birthday Cake
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Door's family is going celebrate Famil Doors's birthday party. They love Famil Door so they are planning to make his birthday cake weird!
The cake is a *n*<=×<=*n* square consisting of equal squares with side length 1. Each square is either empty or consists of a single chocolate. They bought the cake and randomly started to put the chocolates on the cake. The value of Famil Door's happiness will be equal to the number of pairs of cells with chocolates that are in the same row or in the same column of the cake. Famil Doors's family is wondering what is the amount of happiness of Famil going to be?
Please, note that any pair can be counted no more than once, as two different cells can't share both the same row and the same column.
Input Specification:
In the first line of the input, you are given a single integer *n* (1<=≤<=*n*<=≤<=100) — the length of the side of the cake.
Then follow *n* lines, each containing *n* characters. Empty cells are denoted with '.', while cells that contain chocolates are denoted by 'C'.
Output Specification:
Print the value of Famil Door's happiness, i.e. the number of pairs of chocolate pieces that share the same row or the same column.
Demo Input:
['3\n.CC\nC..\nC.C\n', '4\nCC..\nC..C\n.CC.\n.CC.\n']
Demo Output:
['4\n', '9\n']
Note:
If we number rows from top to bottom and columns from left to right, then, pieces that share the same row in the first sample are:
1. (1, 2) and (1, 3) 1. (3, 1) and (3, 3) 1. (2, 1) and (3, 1) 1. (1, 3) and (3, 3)
|
```python
fact = [1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800, 39916800, 479001600, 6227020800, 87178291200, 1307674368000, 20922789888000, 355687428096000, 6402373705728000, 121645100408832000, 2432902008176640000, 51090942171709440000, 1124000727777607680000, 25852016738884976640000, 620448401733239439360000, 15511210043330985984000000, 403291461126605635584000000, 10888869450418352160768000000, 304888344611713860501504000000, 8841761993739701954543616000000, 265252859812191058636308480000000, 8222838654177922817725562880000000, 263130836933693530167218012160000000, 8683317618811886495518194401280000000, 295232799039604140847618609643520000000, 10333147966386144929666651337523200000000, 371993326789901217467999448150835200000000, 13763753091226345046315979581580902400000000, 523022617466601111760007224100074291200000000, 20397882081197443358640281739902897356800000000, 815915283247897734345611269596115894272000000000, 33452526613163807108170062053440751665152000000000, 1405006117752879898543142606244511569936384000000000, 60415263063373835637355132068513997507264512000000000, 2658271574788448768043625811014615890319638528000000000, 119622220865480194561963161495657715064383733760000000000, 5502622159812088949850305428800254892961651752960000000000, 258623241511168180642964355153611979969197632389120000000000, 12413915592536072670862289047373375038521486354677760000000000, 608281864034267560872252163321295376887552831379210240000000000, 30414093201713378043612608166064768844377641568960512000000000000, 1551118753287382280224243016469303211063259720016986112000000000000, 80658175170943878571660636856403766975289505440883277824000000000000, 4274883284060025564298013753389399649690343788366813724672000000000000, 230843697339241380472092742683027581083278564571807941132288000000000000, 12696403353658275925965100847566516959580321051449436762275840000000000000, 710998587804863451854045647463724949736497978881168458687447040000000000000, 40526919504877216755680601905432322134980384796226602145184481280000000000000, 2350561331282878571829474910515074683828862318181142924420699914240000000000000, 138683118545689835737939019720389406345902876772687432540821294940160000000000000, 8320987112741390144276341183223364380754172606361245952449277696409600000000000000, 507580213877224798800856812176625227226004528988036003099405939480985600000000000000, 31469973260387937525653122354950764088012280797258232192163168247821107200000000000000, 1982608315404440064116146708361898137544773690227268628106279599612729753600000000000000, 126886932185884164103433389335161480802865516174545192198801894375214704230400000000000000, 8247650592082470666723170306785496252186258551345437492922123134388955774976000000000000000, 544344939077443064003729240247842752644293064388798874532860126869671081148416000000000000000, 36471110918188685288249859096605464427167635314049524593701628500267962436943872000000000000000, 2480035542436830599600990418569171581047399201355367672371710738018221445712183296000000000000000, 171122452428141311372468338881272839092270544893520369393648040923257279754140647424000000000000000, 11978571669969891796072783721689098736458938142546425857555362864628009582789845319680000000000000000, 850478588567862317521167644239926010288584608120796235886430763388588680378079017697280000000000000000, 61234458376886086861524070385274672740778091784697328983823014963978384987221689274204160000000000000000, 4470115461512684340891257138125051110076800700282905015819080092370422104067183317016903680000000000000000, 330788544151938641225953028221253782145683251820934971170611926835411235700971565459250872320000000000000000, 24809140811395398091946477116594033660926243886570122837795894512655842677572867409443815424000000000000000000, 1885494701666050254987932260861146558230394535379329335672487982961844043495537923117729972224000000000000000000, 145183092028285869634070784086308284983740379224208358846781574688061991349156420080065207861248000000000000000000, 11324281178206297831457521158732046228731749579488251990048962825668835325234200766245086213177344000000000000000000, 894618213078297528685144171539831652069808216779571907213868063227837990693501860533361810841010176000000000000000000, 71569457046263802294811533723186532165584657342365752577109445058227039255480148842668944867280814080000000000000000000, 5797126020747367985879734231578109105412357244731625958745865049716390179693892056256184534249745940480000000000000000000, 475364333701284174842138206989404946643813294067993328617160934076743994734899148613007131808479167119360000000000000000000, 39455239697206586511897471180120610571436503407643446275224357528369751562996629334879591940103770870906880000000000000000000, 3314240134565353266999387579130131288000666286242049487118846032383059131291716864129885722968716753156177920000000000000000000, 281710411438055027694947944226061159480056634330574206405101912752560026159795933451040286452340924018275123200000000000000000000, 24227095383672732381765523203441259715284870552429381750838764496720162249742450276789464634901319465571660595200000000000000000000, 2107757298379527717213600518699389595229783738061356212322972511214654115727593174080683423236414793504734471782400000000000000000000, 185482642257398439114796845645546284380220968949399346684421580986889562184028199319100141244804501828416633516851200000000000000000000, 16507955160908461081216919262453619309839666236496541854913520707833171034378509739399912570787600662729080382999756800000000000000000000, 1485715964481761497309522733620825737885569961284688766942216863704985393094065876545992131370884059645617234469978112000000000000000000000, 135200152767840296255166568759495142147586866476906677791741734597153670771559994765685283954750449427751168336768008192000000000000000000000, 12438414054641307255475324325873553077577991715875414356840239582938137710983519518443046123837041347353107486982656753664000000000000000000000, 1156772507081641574759205162306240436214753229576413535186142281213246807121467315215203289516844845303838996289387078090752000000000000000000000, 108736615665674308027365285256786601004186803580182872307497374434045199869417927630229109214583415458560865651202385340530688000000000000000000000, 10329978488239059262599702099394727095397746340117372869212250571234293987594703124871765375385424468563282236864226607350415360000000000000000000000, 991677934870949689209571401541893801158183648651267795444376054838492222809091499987689476037000748982075094738965754305639874560000000000000000000000, 96192759682482119853328425949563698712343813919172976158104477319333745612481875498805879175589072651261284189679678167647067832320000000000000000000000, 9426890448883247745626185743057242473809693764078951663494238777294707070023223798882976159207729119823605850588608460429412647567360000000000000000000000, 933262154439441526816992388562667004907159682643816214685929638952175999932299156089414639761565182862536979208272237582511852109168640000000000000000000000, 93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000]
def nCr(n, r):
return fact[n]//(fact[r]*fact[n-r])
n = int(input().strip())
arr = []
for _ in range(n):
arr += [list(input().strip())]
happiness = 0
for row in arr:
happiness+=nCr(row.count("C"),2)
for j in range(n):
col = []
for i in range(n):
col += [arr[i][j]]
happiness+=nCr(col.count("C"),2)
print(happiness)
```
| 3
|
|
651
|
A
|
Joysticks
|
PROGRAMMING
| 1,100
|
[
"dp",
"greedy",
"implementation",
"math"
] | null | null |
Friends are going to play console. They have two joysticks and only one charger for them. Initially first joystick is charged at *a*1 percent and second one is charged at *a*2 percent. You can connect charger to a joystick only at the beginning of each minute. In one minute joystick either discharges by 2 percent (if not connected to a charger) or charges by 1 percent (if connected to a charger).
Game continues while both joysticks have a positive charge. Hence, if at the beginning of minute some joystick is charged by 1 percent, it has to be connected to a charger, otherwise the game stops. If some joystick completely discharges (its charge turns to 0), the game also stops.
Determine the maximum number of minutes that game can last. It is prohibited to pause the game, i. e. at each moment both joysticks should be enabled. It is allowed for joystick to be charged by more than 100 percent.
|
The first line of the input contains two positive integers *a*1 and *a*2 (1<=≤<=*a*1,<=*a*2<=≤<=100), the initial charge level of first and second joystick respectively.
|
Output the only integer, the maximum number of minutes that the game can last. Game continues until some joystick is discharged.
|
[
"3 5\n",
"4 4\n"
] |
[
"6\n",
"5\n"
] |
In the first sample game lasts for 6 minute by using the following algorithm:
- at the beginning of the first minute connect first joystick to the charger, by the end of this minute first joystick is at 4%, second is at 3%; - continue the game without changing charger, by the end of the second minute the first joystick is at 5%, second is at 1%; - at the beginning of the third minute connect second joystick to the charger, after this minute the first joystick is at 3%, the second one is at 2%; - continue the game without changing charger, by the end of the fourth minute first joystick is at 1%, second one is at 3%; - at the beginning of the fifth minute connect first joystick to the charger, after this minute the first joystick is at 2%, the second one is at 1%; - at the beginning of the sixth minute connect second joystick to the charger, after this minute the first joystick is at 0%, the second one is at 2%.
After that the first joystick is completely discharged and the game is stopped.
| 500
|
[
{
"input": "3 5",
"output": "6"
},
{
"input": "4 4",
"output": "5"
},
{
"input": "100 100",
"output": "197"
},
{
"input": "1 100",
"output": "98"
},
{
"input": "100 1",
"output": "98"
},
{
"input": "1 4",
"output": "2"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "8 8",
"output": "13"
},
{
"input": "7 2",
"output": "7"
},
{
"input": "24 15",
"output": "36"
},
{
"input": "19 30",
"output": "47"
},
{
"input": "15 31",
"output": "44"
},
{
"input": "14 15",
"output": "27"
},
{
"input": "58 33",
"output": "89"
},
{
"input": "15 25",
"output": "38"
},
{
"input": "59 45",
"output": "102"
},
{
"input": "3 73",
"output": "74"
},
{
"input": "48 1",
"output": "47"
},
{
"input": "100 25",
"output": "122"
},
{
"input": "40 49",
"output": "86"
},
{
"input": "85 73",
"output": "155"
},
{
"input": "29 1",
"output": "28"
},
{
"input": "74 25",
"output": "97"
},
{
"input": "24 57",
"output": "78"
},
{
"input": "23 12",
"output": "33"
},
{
"input": "2 99",
"output": "99"
},
{
"input": "98 2",
"output": "97"
},
{
"input": "2 97",
"output": "97"
},
{
"input": "30 54",
"output": "81"
},
{
"input": "32 53",
"output": "82"
},
{
"input": "32 54",
"output": "84"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "2 1",
"output": "1"
},
{
"input": "2 2",
"output": "1"
},
{
"input": "1 3",
"output": "2"
},
{
"input": "3 1",
"output": "2"
},
{
"input": "1 4",
"output": "2"
},
{
"input": "2 3",
"output": "3"
},
{
"input": "3 2",
"output": "3"
}
] | 1,592,325,730
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 39
| 109
| 0
|
a1,a2=[int(x) for x in input().split()]
k=0
flag=0
if a1==a2==1:
flag=1
else:
while a1>0 and a2>0:
if a1>=a2:
a2+=1
a1-=2
else:
a2-=2
a1+=1
k+=1
if flag==1:
print(0)
else:
print(k)
|
Title: Joysticks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Friends are going to play console. They have two joysticks and only one charger for them. Initially first joystick is charged at *a*1 percent and second one is charged at *a*2 percent. You can connect charger to a joystick only at the beginning of each minute. In one minute joystick either discharges by 2 percent (if not connected to a charger) or charges by 1 percent (if connected to a charger).
Game continues while both joysticks have a positive charge. Hence, if at the beginning of minute some joystick is charged by 1 percent, it has to be connected to a charger, otherwise the game stops. If some joystick completely discharges (its charge turns to 0), the game also stops.
Determine the maximum number of minutes that game can last. It is prohibited to pause the game, i. e. at each moment both joysticks should be enabled. It is allowed for joystick to be charged by more than 100 percent.
Input Specification:
The first line of the input contains two positive integers *a*1 and *a*2 (1<=≤<=*a*1,<=*a*2<=≤<=100), the initial charge level of first and second joystick respectively.
Output Specification:
Output the only integer, the maximum number of minutes that the game can last. Game continues until some joystick is discharged.
Demo Input:
['3 5\n', '4 4\n']
Demo Output:
['6\n', '5\n']
Note:
In the first sample game lasts for 6 minute by using the following algorithm:
- at the beginning of the first minute connect first joystick to the charger, by the end of this minute first joystick is at 4%, second is at 3%; - continue the game without changing charger, by the end of the second minute the first joystick is at 5%, second is at 1%; - at the beginning of the third minute connect second joystick to the charger, after this minute the first joystick is at 3%, the second one is at 2%; - continue the game without changing charger, by the end of the fourth minute first joystick is at 1%, second one is at 3%; - at the beginning of the fifth minute connect first joystick to the charger, after this minute the first joystick is at 2%, the second one is at 1%; - at the beginning of the sixth minute connect second joystick to the charger, after this minute the first joystick is at 0%, the second one is at 2%.
After that the first joystick is completely discharged and the game is stopped.
|
```python
a1,a2=[int(x) for x in input().split()]
k=0
flag=0
if a1==a2==1:
flag=1
else:
while a1>0 and a2>0:
if a1>=a2:
a2+=1
a1-=2
else:
a2-=2
a1+=1
k+=1
if flag==1:
print(0)
else:
print(k)
```
| 3
|
|
186
|
A
|
Comparing Strings
|
PROGRAMMING
| 1,100
|
[
"implementation",
"strings"
] | null | null |
Some dwarves that are finishing the StUDY (State University for Dwarven Youngsters) Bachelor courses, have been told "no genome, no degree". That means that all dwarves should write a thesis on genome. Dwarven genome is far from simple. It is represented by a string that consists of lowercase Latin letters.
Dwarf Misha has already chosen the subject for his thesis: determining by two dwarven genomes, whether they belong to the same race. Two dwarves belong to the same race if we can swap two characters in the first dwarf's genome and get the second dwarf's genome as a result. Help Dwarf Misha and find out whether two gnomes belong to the same race or not.
|
The first line contains the first dwarf's genome: a non-empty string, consisting of lowercase Latin letters.
The second line contains the second dwarf's genome: a non-empty string, consisting of lowercase Latin letters.
The number of letters in each genome doesn't exceed 105. It is guaranteed that the strings that correspond to the genomes are different. The given genomes may have different length.
|
Print "YES", if the dwarves belong to the same race. Otherwise, print "NO".
|
[
"ab\nba\n",
"aa\nab\n"
] |
[
"YES\n",
"NO\n"
] |
- First example: you can simply swap two letters in string "ab". So we get "ba". - Second example: we can't change string "aa" into string "ab", because "aa" does not contain letter "b".
| 500
|
[
{
"input": "ab\nba",
"output": "YES"
},
{
"input": "aa\nab",
"output": "NO"
},
{
"input": "a\nza",
"output": "NO"
},
{
"input": "vvea\nvvae",
"output": "YES"
},
{
"input": "rtfabanpc\natfabrnpc",
"output": "YES"
},
{
"input": "mt\ntm",
"output": "YES"
},
{
"input": "qxolmbkkt\naovlajmlf",
"output": "NO"
},
{
"input": "b\ng",
"output": "NO"
},
{
"input": "ab\naba",
"output": "NO"
},
{
"input": "ba\na",
"output": "NO"
},
{
"input": "a\nab",
"output": "NO"
},
{
"input": "a\naa",
"output": "NO"
},
{
"input": "a\nz",
"output": "NO"
},
{
"input": "aabb\nbbaa",
"output": "NO"
},
{
"input": "ab\nbd",
"output": "NO"
},
{
"input": "bac\ndae",
"output": "NO"
},
{
"input": "abc\nakl",
"output": "NO"
},
{
"input": "cb\naa",
"output": "NO"
},
{
"input": "abaab\naabba",
"output": "NO"
},
{
"input": "aab\naaa",
"output": "NO"
},
{
"input": "abcde\nedcba",
"output": "NO"
},
{
"input": "abab\nbaba",
"output": "NO"
},
{
"input": "ab\nbac",
"output": "NO"
},
{
"input": "abcd\naqcb",
"output": "NO"
},
{
"input": "abc\nbad",
"output": "NO"
},
{
"input": "ab\nca",
"output": "NO"
},
{
"input": "abc\nab",
"output": "NO"
},
{
"input": "ab\nbae",
"output": "NO"
},
{
"input": "aaaabcccca\naaaadccccb",
"output": "NO"
},
{
"input": "acaa\nabca",
"output": "NO"
},
{
"input": "aba\naab",
"output": "YES"
},
{
"input": "ah\nla",
"output": "NO"
},
{
"input": "aab\naa",
"output": "NO"
},
{
"input": "abab\nabcd",
"output": "NO"
},
{
"input": "ed\nab",
"output": "NO"
}
] | 1,592,500,483
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 53
| 248
| 716,800
|
g1=input()
g2=input()
if len(g2)!=len(g1): print("NO"); exit()
dif=0; j=k=-1
for i in range(len(g1)):
if g1[i]!=g2[i]:
dif+=1
if j>-1: k=i
else: j=i
if dif>2: break
if dif!=2: print("NO")
elif g1[k]==g2[j] and g1[j]==g2[k]: print("YES")
else: print("NO")
|
Title: Comparing Strings
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Some dwarves that are finishing the StUDY (State University for Dwarven Youngsters) Bachelor courses, have been told "no genome, no degree". That means that all dwarves should write a thesis on genome. Dwarven genome is far from simple. It is represented by a string that consists of lowercase Latin letters.
Dwarf Misha has already chosen the subject for his thesis: determining by two dwarven genomes, whether they belong to the same race. Two dwarves belong to the same race if we can swap two characters in the first dwarf's genome and get the second dwarf's genome as a result. Help Dwarf Misha and find out whether two gnomes belong to the same race or not.
Input Specification:
The first line contains the first dwarf's genome: a non-empty string, consisting of lowercase Latin letters.
The second line contains the second dwarf's genome: a non-empty string, consisting of lowercase Latin letters.
The number of letters in each genome doesn't exceed 105. It is guaranteed that the strings that correspond to the genomes are different. The given genomes may have different length.
Output Specification:
Print "YES", if the dwarves belong to the same race. Otherwise, print "NO".
Demo Input:
['ab\nba\n', 'aa\nab\n']
Demo Output:
['YES\n', 'NO\n']
Note:
- First example: you can simply swap two letters in string "ab". So we get "ba". - Second example: we can't change string "aa" into string "ab", because "aa" does not contain letter "b".
|
```python
g1=input()
g2=input()
if len(g2)!=len(g1): print("NO"); exit()
dif=0; j=k=-1
for i in range(len(g1)):
if g1[i]!=g2[i]:
dif+=1
if j>-1: k=i
else: j=i
if dif>2: break
if dif!=2: print("NO")
elif g1[k]==g2[j] and g1[j]==g2[k]: print("YES")
else: print("NO")
```
| 3
|
|
92
|
A
|
Chips
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] |
A. Chips
|
2
|
256
|
There are *n* walruses sitting in a circle. All of them are numbered in the clockwise order: the walrus number 2 sits to the left of the walrus number 1, the walrus number 3 sits to the left of the walrus number 2, ..., the walrus number 1 sits to the left of the walrus number *n*.
The presenter has *m* chips. The presenter stands in the middle of the circle and starts giving the chips to the walruses starting from walrus number 1 and moving clockwise. The walrus number *i* gets *i* chips. If the presenter can't give the current walrus the required number of chips, then the presenter takes the remaining chips and the process ends. Determine by the given *n* and *m* how many chips the presenter will get in the end.
|
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=50, 1<=≤<=*m*<=≤<=104) — the number of walruses and the number of chips correspondingly.
|
Print the number of chips the presenter ended up with.
|
[
"4 11\n",
"17 107\n",
"3 8\n"
] |
[
"0\n",
"2\n",
"1\n"
] |
In the first sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, four chips to the walrus number 4, then again one chip to the walrus number 1. After that the presenter runs out of chips. He can't give anything to the walrus number 2 and the process finishes.
In the third sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, then again one chip to the walrus number 1. The presenter has one chip left and he can't give two chips to the walrus number 2, that's why the presenter takes the last chip.
| 500
|
[
{
"input": "4 11",
"output": "0"
},
{
"input": "17 107",
"output": "2"
},
{
"input": "3 8",
"output": "1"
},
{
"input": "46 7262",
"output": "35"
},
{
"input": "32 6864",
"output": "0"
},
{
"input": "36 6218",
"output": "14"
},
{
"input": "25 9712",
"output": "11"
},
{
"input": "9 7601",
"output": "5"
},
{
"input": "1 9058",
"output": "0"
},
{
"input": "29 7772",
"output": "26"
},
{
"input": "45 9465",
"output": "14"
},
{
"input": "46 866",
"output": "5"
},
{
"input": "29 1241",
"output": "20"
},
{
"input": "17 4248",
"output": "12"
},
{
"input": "20 8082",
"output": "11"
},
{
"input": "50 9555",
"output": "0"
},
{
"input": "4 7455",
"output": "2"
},
{
"input": "36 880",
"output": "4"
},
{
"input": "24 7440",
"output": "9"
},
{
"input": "44 7888",
"output": "12"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "50 10000",
"output": "40"
},
{
"input": "1 10000",
"output": "0"
},
{
"input": "50 1",
"output": "0"
},
{
"input": "50 50",
"output": "5"
}
] | 1,634,217,984
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 25
| 154
| 0
|
n,m=map(int,input().split())
kk=int((n+1)*n/2)
if(m>=kk):
a=m%kk
for i in range(n):
if(a-i<0):
break
a-=i
print(a)
else:
for i in range(n):
if(m-i<0):
break
m-=i
print(m)
|
Title: Chips
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
There are *n* walruses sitting in a circle. All of them are numbered in the clockwise order: the walrus number 2 sits to the left of the walrus number 1, the walrus number 3 sits to the left of the walrus number 2, ..., the walrus number 1 sits to the left of the walrus number *n*.
The presenter has *m* chips. The presenter stands in the middle of the circle and starts giving the chips to the walruses starting from walrus number 1 and moving clockwise. The walrus number *i* gets *i* chips. If the presenter can't give the current walrus the required number of chips, then the presenter takes the remaining chips and the process ends. Determine by the given *n* and *m* how many chips the presenter will get in the end.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=50, 1<=≤<=*m*<=≤<=104) — the number of walruses and the number of chips correspondingly.
Output Specification:
Print the number of chips the presenter ended up with.
Demo Input:
['4 11\n', '17 107\n', '3 8\n']
Demo Output:
['0\n', '2\n', '1\n']
Note:
In the first sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, four chips to the walrus number 4, then again one chip to the walrus number 1. After that the presenter runs out of chips. He can't give anything to the walrus number 2 and the process finishes.
In the third sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, then again one chip to the walrus number 1. The presenter has one chip left and he can't give two chips to the walrus number 2, that's why the presenter takes the last chip.
|
```python
n,m=map(int,input().split())
kk=int((n+1)*n/2)
if(m>=kk):
a=m%kk
for i in range(n):
if(a-i<0):
break
a-=i
print(a)
else:
for i in range(n):
if(m-i<0):
break
m-=i
print(m)
```
| 3.9615
|
227
|
B
|
Effective Approach
|
PROGRAMMING
| 1,100
|
[
"implementation"
] | null | null |
Once at a team training Vasya, Petya and Sasha got a problem on implementing linear search in an array.
According to the boys, linear search works as follows. The array elements in a pre-selected order are in turn compared with the number that you need to find. Once you find the array element that is equal to the required one, the search ends. The efficiency of the algorithm is the number of performed comparisons. The fewer comparisons the linear search has made, the more effective it is.
Vasya believes that a linear search would work better if it sequentially iterates through the elements, starting with the 1-st one (in this problem we consider the elements of the array indexed from 1 to *n*) and ending with the *n*-th one. And Petya says that Vasya is wrong: the search will need less comparisons if it sequentially iterates the elements starting from the *n*-th and ending with the 1-st one. Sasha argues that the two approaches are equivalent.
To finally begin the task, the teammates decided to settle the debate and compare the two approaches on an example. For this, they took an array that is a permutation of integers from 1 to *n*, and generated *m* queries of the form: find element with value *b**i* in the array. They want to calculate for both approaches how many comparisons in total the linear search will need to respond to all queries. If the first search needs fewer comparisons, then the winner of the dispute is Vasya. If the second one does, then the winner is Petya. If both approaches make the same number of comparisons, then Sasha's got the upper hand.
But the problem is, linear search is too slow. That's why the boys aren't going to find out who is right before the end of the training, unless you come in here. Help them to determine who will win the dispute.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of elements in the array. The second line contains *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the elements of array.
The third line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. The last line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*) — the search queries. Note that the queries can repeat.
|
Print two integers, showing how many comparisons Vasya's approach needs and how many comparisons Petya's approach needs. Separate the numbers by spaces.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.
|
[
"2\n1 2\n1\n1\n",
"2\n2 1\n1\n1\n",
"3\n3 1 2\n3\n1 2 3\n"
] |
[
"1 2\n",
"2 1\n",
"6 6\n"
] |
In the first sample Vasya's approach will make one comparison (it starts with the 1-st element and immediately finds the required number), and Petya's approach makes two comparisons (first he compares with the 2-nd array element, doesn't find the search item and compares with the 1-st element).
In the second sample, on the contrary, Vasya's approach will need two comparisons (first with 1-st element, and then with the 2-nd), and Petya's approach will find the required value in one comparison (the first comparison with the 2-nd element).
| 1,000
|
[
{
"input": "2\n1 2\n1\n1",
"output": "1 2"
},
{
"input": "2\n2 1\n1\n1",
"output": "2 1"
},
{
"input": "3\n3 1 2\n3\n1 2 3",
"output": "6 6"
},
{
"input": "9\n2 9 3 1 6 4 7 8 5\n9\n5 1 5 2 8 4 4 4 5",
"output": "58 32"
},
{
"input": "10\n3 10 9 2 7 6 5 8 4 1\n1\n4",
"output": "9 2"
},
{
"input": "10\n5 2 10 8 3 1 9 7 6 4\n9\n2 5 9 2 3 2 5 9 8",
"output": "31 68"
},
{
"input": "9\n3 8 4 7 1 2 5 6 9\n3\n2 7 1",
"output": "15 15"
},
{
"input": "9\n5 3 8 4 2 6 1 7 9\n4\n6 1 9 2",
"output": "27 13"
},
{
"input": "4\n1 3 2 4\n4\n3 1 2 3",
"output": "8 12"
},
{
"input": "3\n1 2 3\n8\n3 2 1 1 2 3 1 2",
"output": "15 17"
}
] | 1,679,272,107
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 34
| 342
| 14,848,000
|
n = int(input())
dic = {}
l1 = list(map(int,input().split()))
for i in range(n):
dic[l1[i]] = i+1
n2 = int(input())
l2 = list(map(int,input().split()))
s1 = 0
s2 = 0
for i in l2 :
s1+=dic[i]
for i in l2 :
s2+= (n-dic[i]+1)
print(f"{s1} {s2}")
|
Title: Effective Approach
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Once at a team training Vasya, Petya and Sasha got a problem on implementing linear search in an array.
According to the boys, linear search works as follows. The array elements in a pre-selected order are in turn compared with the number that you need to find. Once you find the array element that is equal to the required one, the search ends. The efficiency of the algorithm is the number of performed comparisons. The fewer comparisons the linear search has made, the more effective it is.
Vasya believes that a linear search would work better if it sequentially iterates through the elements, starting with the 1-st one (in this problem we consider the elements of the array indexed from 1 to *n*) and ending with the *n*-th one. And Petya says that Vasya is wrong: the search will need less comparisons if it sequentially iterates the elements starting from the *n*-th and ending with the 1-st one. Sasha argues that the two approaches are equivalent.
To finally begin the task, the teammates decided to settle the debate and compare the two approaches on an example. For this, they took an array that is a permutation of integers from 1 to *n*, and generated *m* queries of the form: find element with value *b**i* in the array. They want to calculate for both approaches how many comparisons in total the linear search will need to respond to all queries. If the first search needs fewer comparisons, then the winner of the dispute is Vasya. If the second one does, then the winner is Petya. If both approaches make the same number of comparisons, then Sasha's got the upper hand.
But the problem is, linear search is too slow. That's why the boys aren't going to find out who is right before the end of the training, unless you come in here. Help them to determine who will win the dispute.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of elements in the array. The second line contains *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the elements of array.
The third line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. The last line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*) — the search queries. Note that the queries can repeat.
Output Specification:
Print two integers, showing how many comparisons Vasya's approach needs and how many comparisons Petya's approach needs. Separate the numbers by spaces.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.
Demo Input:
['2\n1 2\n1\n1\n', '2\n2 1\n1\n1\n', '3\n3 1 2\n3\n1 2 3\n']
Demo Output:
['1 2\n', '2 1\n', '6 6\n']
Note:
In the first sample Vasya's approach will make one comparison (it starts with the 1-st element and immediately finds the required number), and Petya's approach makes two comparisons (first he compares with the 2-nd array element, doesn't find the search item and compares with the 1-st element).
In the second sample, on the contrary, Vasya's approach will need two comparisons (first with 1-st element, and then with the 2-nd), and Petya's approach will find the required value in one comparison (the first comparison with the 2-nd element).
|
```python
n = int(input())
dic = {}
l1 = list(map(int,input().split()))
for i in range(n):
dic[l1[i]] = i+1
n2 = int(input())
l2 = list(map(int,input().split()))
s1 = 0
s2 = 0
for i in l2 :
s1+=dic[i]
for i in l2 :
s2+= (n-dic[i]+1)
print(f"{s1} {s2}")
```
| 3
|
|
20
|
A
|
BerOS file system
|
PROGRAMMING
| 1,700
|
[
"implementation"
] |
A. BerOS file system
|
2
|
64
|
The new operating system BerOS has a nice feature. It is possible to use any number of characters '/' as a delimiter in path instead of one traditional '/'. For example, strings //usr///local//nginx/sbin// and /usr/local/nginx///sbin are equivalent. The character '/' (or some sequence of such characters) at the end of the path is required only in case of the path to the root directory, which can be represented as single character '/'.
A path called normalized if it contains the smallest possible number of characters '/'.
Your task is to transform a given path to the normalized form.
|
The first line of the input contains only lowercase Latin letters and character '/' — the path to some directory. All paths start with at least one character '/'. The length of the given line is no more than 100 characters, it is not empty.
|
The path in normalized form.
|
[
"//usr///local//nginx/sbin\n"
] |
[
"/usr/local/nginx/sbin\n"
] |
none
| 500
|
[
{
"input": "//usr///local//nginx/sbin",
"output": "/usr/local/nginx/sbin"
},
{
"input": "////a//b/////g",
"output": "/a/b/g"
},
{
"input": "/a/b/c",
"output": "/a/b/c"
},
{
"input": "/",
"output": "/"
},
{
"input": "////",
"output": "/"
},
{
"input": "/a//aa/a//",
"output": "/a/aa/a"
},
{
"input": "/aa//b/aa",
"output": "/aa/b/aa"
},
{
"input": "////////////////////////////////////////////////////////////////////////////////////////////////////",
"output": "/"
},
{
"input": "/opt///pokerband///srvb/opt///pokerband///srvb////pokerband///srvb/",
"output": "/opt/pokerband/srvb/opt/pokerband/srvb/pokerband/srvb"
},
{
"input": "/root/pokerband/trunk/server/game/code/src/main/java/com/uosipa/pokerband/gameserver/game/",
"output": "/root/pokerband/trunk/server/game/code/src/main/java/com/uosipa/pokerband/gameserver/game"
},
{
"input": "/root//pokerband/trunk/server/game/code/src///main/java/com/uosipa/pokerband//gameserver/game//",
"output": "/root/pokerband/trunk/server/game/code/src/main/java/com/uosipa/pokerband/gameserver/game"
},
{
"input": "/root//pokerband/trunk/server/game/code/src///main/java/com/uosipa/pokerband//gameserver/game//test/",
"output": "/root/pokerband/trunk/server/game/code/src/main/java/com/uosipa/pokerband/gameserver/game/test"
},
{
"input": "/root//pokerband/trunk/server/game/code/src///main/java/com/uosipa/pokerband//gameserver/game//testt",
"output": "/root/pokerband/trunk/server/game/code/src/main/java/com/uosipa/pokerband/gameserver/game/testt"
},
{
"input": "///a//a////a/a//a//a//a/////",
"output": "/a/a/a/a/a/a/a"
},
{
"input": "/zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz",
"output": "/zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz"
},
{
"input": "/zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz/",
"output": "/zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz"
},
{
"input": "/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a",
"output": "/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a"
},
{
"input": "///////////////////////////////////////////////////////////////////////////////////////////////////z",
"output": "/z"
},
{
"input": "/z//////////////////////////////////////////////////////////////////////////////////////////////////",
"output": "/z"
}
] | 1,647,599,649
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 19
| 92
| 0
|
x=input()
str1=''
if x.count('/')==len(x):
print('/')
else:
for i in range(len(x)):
if i==0:
str1+=x[i]
elif x[i]=='/' and str1[-1]=='/':
continue
else:
str1+=x[i]
for i in range(100000000000000000):
if str1[-1]=='/':
str1=str1[:-1]
else:
break
print(str1)
|
Title: BerOS file system
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
The new operating system BerOS has a nice feature. It is possible to use any number of characters '/' as a delimiter in path instead of one traditional '/'. For example, strings //usr///local//nginx/sbin// and /usr/local/nginx///sbin are equivalent. The character '/' (or some sequence of such characters) at the end of the path is required only in case of the path to the root directory, which can be represented as single character '/'.
A path called normalized if it contains the smallest possible number of characters '/'.
Your task is to transform a given path to the normalized form.
Input Specification:
The first line of the input contains only lowercase Latin letters and character '/' — the path to some directory. All paths start with at least one character '/'. The length of the given line is no more than 100 characters, it is not empty.
Output Specification:
The path in normalized form.
Demo Input:
['//usr///local//nginx/sbin\n']
Demo Output:
['/usr/local/nginx/sbin\n']
Note:
none
|
```python
x=input()
str1=''
if x.count('/')==len(x):
print('/')
else:
for i in range(len(x)):
if i==0:
str1+=x[i]
elif x[i]=='/' and str1[-1]=='/':
continue
else:
str1+=x[i]
for i in range(100000000000000000):
if str1[-1]=='/':
str1=str1[:-1]
else:
break
print(str1)
```
| 3.977
|
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"
}
] | 1,669,095,048
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 18
| 92
| 0
|
inp1 = input("")
inp1_lst = [int(i) for i in inp1.split()]
n = inp1_lst[0]
m = inp1_lst[1]
mth = input("")
m_lst = [int(i) for i in mth.split()]
m_lst.sort()
lowest = None
j = n-1
for i in range(len(m_lst)-1):
if j >= len(m_lst):
break
dif = m_lst[j] - m_lst[i]
j += 1
if i == 0:
lowest = dif
elif dif < lowest:
lowest = dif
print(lowest)
|
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
inp1 = input("")
inp1_lst = [int(i) for i in inp1.split()]
n = inp1_lst[0]
m = inp1_lst[1]
mth = input("")
m_lst = [int(i) for i in mth.split()]
m_lst.sort()
lowest = None
j = n-1
for i in range(len(m_lst)-1):
if j >= len(m_lst):
break
dif = m_lst[j] - m_lst[i]
j += 1
if i == 0:
lowest = dif
elif dif < lowest:
lowest = dif
print(lowest)
```
| 3
|
|
431
|
A
|
Black Square
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Quite recently, a very smart student named Jury decided that lectures are boring, so he downloaded a game called "Black Square" on his super cool touchscreen phone.
In this game, the phone's screen is divided into four vertical strips. Each second, a black square appears on some of the strips. According to the rules of the game, Jury must use this second to touch the corresponding strip to make the square go away. As Jury is both smart and lazy, he counted that he wastes exactly *a**i* calories on touching the *i*-th strip.
You've got a string *s*, describing the process of the game and numbers *a*1,<=*a*2,<=*a*3,<=*a*4. Calculate how many calories Jury needs to destroy all the squares?
|
The first line contains four space-separated integers *a*1, *a*2, *a*3, *a*4 (0<=≤<=*a*1,<=*a*2,<=*a*3,<=*a*4<=≤<=104).
The second line contains string *s* (1<=≤<=|*s*|<=≤<=105), where the *і*-th character of the string equals "1", if on the *i*-th second of the game the square appears on the first strip, "2", if it appears on the second strip, "3", if it appears on the third strip, "4", if it appears on the fourth strip.
|
Print a single integer — the total number of calories that Jury wastes.
|
[
"1 2 3 4\n123214\n",
"1 5 3 2\n11221\n"
] |
[
"13\n",
"13\n"
] |
none
| 500
|
[
{
"input": "1 2 3 4\n123214",
"output": "13"
},
{
"input": "1 5 3 2\n11221",
"output": "13"
},
{
"input": "5 5 5 1\n3422",
"output": "16"
},
{
"input": "4 3 2 1\n2",
"output": "3"
},
{
"input": "5651 6882 6954 4733\n2442313421",
"output": "60055"
},
{
"input": "0 0 0 0\n4132",
"output": "0"
},
{
"input": "3163 5778 83 7640\n11141442444",
"output": "64270"
},
{
"input": "1809 1302 7164 6122\n3144121413113111223311232232114144321414421243443243422322144324121433444342231344234443332241322442",
"output": "420780"
},
{
"input": "0 0 0 0\n1",
"output": "0"
},
{
"input": "1 2 3 4\n4",
"output": "4"
},
{
"input": "2343 7653 1242 5432\n1",
"output": "2343"
},
{
"input": "2343 7653 1242 5432\n2",
"output": "7653"
},
{
"input": "2343 7653 1242 5432\n3",
"output": "1242"
},
{
"input": "2343 7653 1242 5432\n4",
"output": "5432"
},
{
"input": "1 2 3 4\n123412",
"output": "13"
},
{
"input": "50 50 50 50\n11111111111111111111111111111111111111111111111111111",
"output": "2650"
},
{
"input": "1 2 3 4\n11111111111111111111111111111111111111111111111111",
"output": "50"
},
{
"input": "1 2 3 4\n23123231321231231231231231221232123121312321",
"output": "87"
},
{
"input": "1 2 3 4\n1111111111111222222222233333333333444444444444444",
"output": "126"
},
{
"input": "2 3 1 4\n121321232412342112312313213123123412131231231232",
"output": "105"
}
] | 1,667,286,795
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 49
| 62
| 204,800
|
a1, a2, a3, a4 = map(int, input().split())
s = str(input())
kal = 0
for i in range(len(s)):
if s[i] == '1':
kal += a1
elif s[i] == '2':
kal += a2
elif s[i] == '3':
kal += a3
elif s[i] == '4':
kal += a4
print(kal)
|
Title: Black Square
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Quite recently, a very smart student named Jury decided that lectures are boring, so he downloaded a game called "Black Square" on his super cool touchscreen phone.
In this game, the phone's screen is divided into four vertical strips. Each second, a black square appears on some of the strips. According to the rules of the game, Jury must use this second to touch the corresponding strip to make the square go away. As Jury is both smart and lazy, he counted that he wastes exactly *a**i* calories on touching the *i*-th strip.
You've got a string *s*, describing the process of the game and numbers *a*1,<=*a*2,<=*a*3,<=*a*4. Calculate how many calories Jury needs to destroy all the squares?
Input Specification:
The first line contains four space-separated integers *a*1, *a*2, *a*3, *a*4 (0<=≤<=*a*1,<=*a*2,<=*a*3,<=*a*4<=≤<=104).
The second line contains string *s* (1<=≤<=|*s*|<=≤<=105), where the *і*-th character of the string equals "1", if on the *i*-th second of the game the square appears on the first strip, "2", if it appears on the second strip, "3", if it appears on the third strip, "4", if it appears on the fourth strip.
Output Specification:
Print a single integer — the total number of calories that Jury wastes.
Demo Input:
['1 2 3 4\n123214\n', '1 5 3 2\n11221\n']
Demo Output:
['13\n', '13\n']
Note:
none
|
```python
a1, a2, a3, a4 = map(int, input().split())
s = str(input())
kal = 0
for i in range(len(s)):
if s[i] == '1':
kal += a1
elif s[i] == '2':
kal += a2
elif s[i] == '3':
kal += a3
elif s[i] == '4':
kal += a4
print(kal)
```
| 3
|
|
886
|
A
|
ACM ICPC
|
PROGRAMMING
| 1,000
|
[
"brute force"
] | null | null |
In a small but very proud high school it was decided to win ACM ICPC. This goal requires to compose as many teams of three as possible, but since there were only 6 students who wished to participate, the decision was to build exactly two teams.
After practice competition, participant number *i* got a score of *a**i*. Team score is defined as sum of scores of its participants. High school management is interested if it's possible to build two teams with equal scores. Your task is to answer that question.
|
The single line contains six integers *a*1,<=...,<=*a*6 (0<=≤<=*a**i*<=≤<=1000) — scores of the participants
|
Print "YES" (quotes for clarity), if it is possible to build teams with equal score, and "NO" otherwise.
You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES").
|
[
"1 3 2 1 2 1\n",
"1 1 1 1 1 99\n"
] |
[
"YES\n",
"NO\n"
] |
In the first sample, first team can be composed of 1st, 2nd and 6th participant, second — of 3rd, 4th and 5th: team scores are 1 + 3 + 1 = 2 + 1 + 2 = 5.
In the second sample, score of participant number 6 is too high: his team score will be definitely greater.
| 500
|
[
{
"input": "1 3 2 1 2 1",
"output": "YES"
},
{
"input": "1 1 1 1 1 99",
"output": "NO"
},
{
"input": "1000 1000 1000 1000 1000 1000",
"output": "YES"
},
{
"input": "0 0 0 0 0 0",
"output": "YES"
},
{
"input": "633 609 369 704 573 416",
"output": "NO"
},
{
"input": "353 313 327 470 597 31",
"output": "NO"
},
{
"input": "835 638 673 624 232 266",
"output": "NO"
},
{
"input": "936 342 19 398 247 874",
"output": "NO"
},
{
"input": "417 666 978 553 271 488",
"output": "NO"
},
{
"input": "71 66 124 199 67 147",
"output": "YES"
},
{
"input": "54 26 0 171 239 12",
"output": "YES"
},
{
"input": "72 8 186 92 267 69",
"output": "YES"
},
{
"input": "180 179 188 50 75 214",
"output": "YES"
},
{
"input": "16 169 110 136 404 277",
"output": "YES"
},
{
"input": "101 400 9 200 300 10",
"output": "YES"
},
{
"input": "101 400 200 9 300 10",
"output": "YES"
},
{
"input": "101 200 400 9 300 10",
"output": "YES"
},
{
"input": "101 400 200 300 9 10",
"output": "YES"
},
{
"input": "101 200 400 300 9 10",
"output": "YES"
},
{
"input": "4 4 4 4 5 4",
"output": "NO"
},
{
"input": "2 2 2 2 2 1",
"output": "NO"
},
{
"input": "1000 1000 999 1000 1000 1000",
"output": "NO"
},
{
"input": "129 1 10 29 8 111",
"output": "NO"
},
{
"input": "1000 1000 1000 999 999 1000",
"output": "YES"
},
{
"input": "101 200 300 400 9 10",
"output": "YES"
},
{
"input": "101 400 200 300 10 9",
"output": "YES"
},
{
"input": "101 200 400 300 10 9",
"output": "YES"
},
{
"input": "101 200 300 400 10 9",
"output": "YES"
},
{
"input": "101 200 300 10 400 9",
"output": "YES"
},
{
"input": "1 1 1 1 1 5",
"output": "NO"
},
{
"input": "8 1 1 3 3 0",
"output": "NO"
},
{
"input": "1 1 2 2 3 3",
"output": "YES"
},
{
"input": "1 2 2 5 2 5",
"output": "NO"
},
{
"input": "1 2 3 6 6 6",
"output": "NO"
},
{
"input": "36 91 7 86 51 89",
"output": "NO"
},
{
"input": "10 1 1 1 23 24",
"output": "NO"
},
{
"input": "1 1 1 10 23 24",
"output": "NO"
},
{
"input": "20 10 1 2 3 44",
"output": "NO"
},
{
"input": "7 0 14 11 8 6",
"output": "NO"
},
{
"input": "100 496 1 1 1 1",
"output": "NO"
},
{
"input": "5 4 2 5 11 3",
"output": "NO"
},
{
"input": "1 3 7 8 8 9",
"output": "YES"
},
{
"input": "1 3 4 5 18 19",
"output": "YES"
},
{
"input": "5 5 1 2 2 15",
"output": "NO"
},
{
"input": "2 1 0 0 0 5",
"output": "NO"
},
{
"input": "1 6 6 1 20 2",
"output": "NO"
},
{
"input": "2 10 0 0 0 0",
"output": "NO"
},
{
"input": "1 1 3 1 1 11",
"output": "NO"
},
{
"input": "10 10 1 1 1 37",
"output": "NO"
},
{
"input": "1 1 0 0 0 4",
"output": "NO"
},
{
"input": "1 1 10 1 1 28",
"output": "NO"
},
{
"input": "1 5 5 5 6 8",
"output": "YES"
},
{
"input": "0 2 3 4 4 5",
"output": "YES"
}
] | 1,606,984,105
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 53
| 109
| 0
|
num=list(map(int,input().split()))
total=0
total=sum(num)
#print(total)
#if total%2!=0:
# print("NO")
flag=0
for i in range(0,4):
for j in range(i+1,5):
for k in range(j+1,6):
if(num[i]+num[j]+num[k]==total/2):
flag=1
break
if(flag):
print("YES")
else:
print("NO")
|
Title: ACM ICPC
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In a small but very proud high school it was decided to win ACM ICPC. This goal requires to compose as many teams of three as possible, but since there were only 6 students who wished to participate, the decision was to build exactly two teams.
After practice competition, participant number *i* got a score of *a**i*. Team score is defined as sum of scores of its participants. High school management is interested if it's possible to build two teams with equal scores. Your task is to answer that question.
Input Specification:
The single line contains six integers *a*1,<=...,<=*a*6 (0<=≤<=*a**i*<=≤<=1000) — scores of the participants
Output Specification:
Print "YES" (quotes for clarity), if it is possible to build teams with equal score, and "NO" otherwise.
You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES").
Demo Input:
['1 3 2 1 2 1\n', '1 1 1 1 1 99\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample, first team can be composed of 1st, 2nd and 6th participant, second — of 3rd, 4th and 5th: team scores are 1 + 3 + 1 = 2 + 1 + 2 = 5.
In the second sample, score of participant number 6 is too high: his team score will be definitely greater.
|
```python
num=list(map(int,input().split()))
total=0
total=sum(num)
#print(total)
#if total%2!=0:
# print("NO")
flag=0
for i in range(0,4):
for j in range(i+1,5):
for k in range(j+1,6):
if(num[i]+num[j]+num[k]==total/2):
flag=1
break
if(flag):
print("YES")
else:
print("NO")
```
| 3
|
|
499
|
A
|
Watching a movie
|
PROGRAMMING
| 1,000
|
[
"greedy",
"implementation"
] | null | null |
You have decided to watch the best moments of some movie. There are two buttons on your player:
1. Watch the current minute of the movie. By pressing this button, you watch the current minute of the movie and the player automatically proceeds to the next minute of the movie. 1. Skip exactly *x* minutes of the movie (*x* is some fixed positive integer). If the player is now at the *t*-th minute of the movie, then as a result of pressing this button, it proceeds to the minute (*t*<=+<=*x*).
Initially the movie is turned on in the player on the first minute, and you want to watch exactly *n* best moments of the movie, the *i*-th best moment starts at the *l**i*-th minute and ends at the *r**i*-th minute (more formally, the *i*-th best moment consists of minutes: *l**i*,<=*l**i*<=+<=1,<=...,<=*r**i*).
Determine, what is the minimum number of minutes of the movie you have to watch if you want to watch all the best moments?
|
The first line contains two space-separated integers *n*, *x* (1<=≤<=*n*<=≤<=50, 1<=≤<=*x*<=≤<=105) — the number of the best moments of the movie and the value of *x* for the second button.
The following *n* lines contain the descriptions of the best moments of the movie, the *i*-th line of the description contains two integers separated by a space *l**i*, *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=105).
It is guaranteed that for all integers *i* from 2 to *n* the following condition holds: *r**i*<=-<=1<=<<=*l**i*.
|
Output a single number — the answer to the problem.
|
[
"2 3\n5 6\n10 12\n",
"1 1\n1 100000\n"
] |
[
"6\n",
"100000\n"
] |
In the first sample, the player was initially standing on the first minute. As the minutes from the 1-st to the 4-th one don't contain interesting moments, we press the second button. Now we can not press the second button and skip 3 more minutes, because some of them contain interesting moments. Therefore, we watch the movie from the 4-th to the 6-th minute, after that the current time is 7. Similarly, we again skip 3 minutes and then watch from the 10-th to the 12-th minute of the movie. In total, we watch 6 minutes of the movie.
In the second sample, the movie is very interesting, so you'll have to watch all 100000 minutes of the movie.
| 500
|
[
{
"input": "2 3\n5 6\n10 12",
"output": "6"
},
{
"input": "1 1\n1 100000",
"output": "100000"
},
{
"input": "10 1\n2156 3497\n4784 7775\n14575 31932\n33447 35902\n36426 47202\n48772 60522\n63982 68417\n78537 79445\n90081 90629\n94325 95728",
"output": "53974"
},
{
"input": "10 3\n2156 3497\n4784 7775\n14575 31932\n33447 35902\n36426 47202\n48772 60522\n63982 68417\n78537 79445\n90081 90629\n94325 95728",
"output": "53983"
},
{
"input": "10 10\n2156 3497\n4784 7775\n14575 31932\n33447 35902\n36426 47202\n48772 60522\n63982 68417\n78537 79445\n90081 90629\n94325 95728",
"output": "54038"
},
{
"input": "10 1000\n2156 3497\n4784 7775\n14575 31932\n33447 35902\n36426 47202\n48772 60522\n63982 68417\n78537 79445\n90081 90629\n94325 95728",
"output": "58728"
},
{
"input": "12 14\n2156 3497\n4784 7775\n14575 23857\n29211 30739\n31932 33447\n35902 36426\n47202 48772\n60522 63982\n68417 78537\n79445 86918\n90081 90629\n94325 95728",
"output": "41870"
},
{
"input": "12 17\n2156 3497\n4784 7775\n14575 23857\n29211 30739\n31932 33447\n35902 36426\n47202 48772\n60522 63982\n68417 78537\n79445 86918\n90081 90629\n94325 95728",
"output": "41872"
},
{
"input": "18 111\n1449 2156\n3497 4784\n7775 14575\n23857 24593\n29211 30739\n31932 33447\n35902 36426\n36991 38506\n39679 47202\n48772 60016\n60522 61868\n63340 63982\n68417 78537\n79445 86568\n86918 87940\n90081 90310\n90629 94325\n95728 96263",
"output": "58967"
},
{
"input": "1 1\n14575 60522",
"output": "45948"
},
{
"input": "1 10\n14575 60522",
"output": "45952"
},
{
"input": "2 10\n14575 36426\n60522 79445",
"output": "40785"
},
{
"input": "3 1\n14575 36426\n48772 60522\n79445 90081",
"output": "44240"
},
{
"input": "1 100000\n1 100000",
"output": "100000"
},
{
"input": "1 2\n2 100000",
"output": "100000"
},
{
"input": "1 3\n4 100000",
"output": "99997"
},
{
"input": "2 100\n5 6\n7 7",
"output": "7"
},
{
"input": "1 10000\n500 600",
"output": "600"
},
{
"input": "2 3\n1 3\n4 8",
"output": "8"
},
{
"input": "3 1\n1 2\n3 4\n6 7",
"output": "6"
},
{
"input": "1 1\n1 1",
"output": "1"
},
{
"input": "1 3\n5 6",
"output": "3"
},
{
"input": "1 5\n7 10",
"output": "5"
},
{
"input": "1 100\n5 6",
"output": "6"
},
{
"input": "1 2\n1 1",
"output": "1"
},
{
"input": "1 100000\n100000 100000",
"output": "100000"
},
{
"input": "2 3\n1 1\n5 6",
"output": "3"
}
] | 1,649,047,196
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 41
| 46
| 0
|
n, x = input().split()
n = int(n)
x = int(x)
sm = 0
j = 1
for i in range(n):
l, r = input().split()
l = int(l)
r = int(r)
while(1):
if j+x-1 >= l or j+x-1 >= r:
sm = sm+r-j+1
j = r+1
break
else:
j = j+x
print(sm)
|
Title: Watching a movie
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have decided to watch the best moments of some movie. There are two buttons on your player:
1. Watch the current minute of the movie. By pressing this button, you watch the current minute of the movie and the player automatically proceeds to the next minute of the movie. 1. Skip exactly *x* minutes of the movie (*x* is some fixed positive integer). If the player is now at the *t*-th minute of the movie, then as a result of pressing this button, it proceeds to the minute (*t*<=+<=*x*).
Initially the movie is turned on in the player on the first minute, and you want to watch exactly *n* best moments of the movie, the *i*-th best moment starts at the *l**i*-th minute and ends at the *r**i*-th minute (more formally, the *i*-th best moment consists of minutes: *l**i*,<=*l**i*<=+<=1,<=...,<=*r**i*).
Determine, what is the minimum number of minutes of the movie you have to watch if you want to watch all the best moments?
Input Specification:
The first line contains two space-separated integers *n*, *x* (1<=≤<=*n*<=≤<=50, 1<=≤<=*x*<=≤<=105) — the number of the best moments of the movie and the value of *x* for the second button.
The following *n* lines contain the descriptions of the best moments of the movie, the *i*-th line of the description contains two integers separated by a space *l**i*, *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=105).
It is guaranteed that for all integers *i* from 2 to *n* the following condition holds: *r**i*<=-<=1<=<<=*l**i*.
Output Specification:
Output a single number — the answer to the problem.
Demo Input:
['2 3\n5 6\n10 12\n', '1 1\n1 100000\n']
Demo Output:
['6\n', '100000\n']
Note:
In the first sample, the player was initially standing on the first minute. As the minutes from the 1-st to the 4-th one don't contain interesting moments, we press the second button. Now we can not press the second button and skip 3 more minutes, because some of them contain interesting moments. Therefore, we watch the movie from the 4-th to the 6-th minute, after that the current time is 7. Similarly, we again skip 3 minutes and then watch from the 10-th to the 12-th minute of the movie. In total, we watch 6 minutes of the movie.
In the second sample, the movie is very interesting, so you'll have to watch all 100000 minutes of the movie.
|
```python
n, x = input().split()
n = int(n)
x = int(x)
sm = 0
j = 1
for i in range(n):
l, r = input().split()
l = int(l)
r = int(r)
while(1):
if j+x-1 >= l or j+x-1 >= r:
sm = sm+r-j+1
j = r+1
break
else:
j = j+x
print(sm)
```
| 3
|
|
493
|
C
|
Vasya and Basketball
|
PROGRAMMING
| 1,600
|
[
"binary search",
"brute force",
"data structures",
"implementation",
"sortings",
"two pointers"
] | null | null |
Vasya follows a basketball game and marks the distances from which each team makes a throw. He knows that each successful throw has value of either 2 or 3 points. A throw is worth 2 points if the distance it was made from doesn't exceed some value of *d* meters, and a throw is worth 3 points if the distance is larger than *d* meters, where *d* is some non-negative integer.
Vasya would like the advantage of the points scored by the first team (the points of the first team minus the points of the second team) to be maximum. For that he can mentally choose the value of *d*. Help him to do that.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=2·105) — the number of throws of the first team. Then follow *n* integer numbers — the distances of throws *a**i* (1<=≤<=*a**i*<=≤<=2·109).
Then follows number *m* (1<=≤<=*m*<=≤<=2·105) — the number of the throws of the second team. Then follow *m* integer numbers — the distances of throws of *b**i* (1<=≤<=*b**i*<=≤<=2·109).
|
Print two numbers in the format a:b — the score that is possible considering the problem conditions where the result of subtraction *a*<=-<=*b* is maximum. If there are several such scores, find the one in which number *a* is maximum.
|
[
"3\n1 2 3\n2\n5 6\n",
"5\n6 7 8 9 10\n5\n1 2 3 4 5\n"
] |
[
"9:6\n",
"15:10\n"
] |
none
| 2,000
|
[
{
"input": "3\n1 2 3\n2\n5 6",
"output": "9:6"
},
{
"input": "5\n6 7 8 9 10\n5\n1 2 3 4 5",
"output": "15:10"
},
{
"input": "5\n1 2 3 4 5\n5\n6 7 8 9 10",
"output": "15:15"
},
{
"input": "3\n1 2 3\n3\n6 4 5",
"output": "9:9"
},
{
"input": "10\n1 2 3 4 5 6 7 8 9 10\n1\n11",
"output": "30:3"
},
{
"input": "10\n1 2 3 4 5 6 7 8 9 11\n1\n10",
"output": "30:3"
},
{
"input": "3\n1 2 3\n3\n1 2 3",
"output": "9:9"
},
{
"input": "3\n1 2 3\n3\n3 4 5",
"output": "9:9"
},
{
"input": "4\n2 5 3 2\n4\n1 5 6 2",
"output": "12:11"
},
{
"input": "2\n3 3\n3\n1 3 3",
"output": "6:8"
},
{
"input": "3\n1 1 1\n4\n1 3 1 1",
"output": "6:8"
},
{
"input": "4\n4 2 1 1\n4\n3 2 2 2",
"output": "9:8"
},
{
"input": "3\n3 9 4\n2\n10 1",
"output": "9:5"
},
{
"input": "14\n4336 24047 24846 25681 28597 30057 32421 34446 48670 67750 68185 69661 85721 89013\n30\n8751 10576 14401 22336 22689 35505 38649 43073 43176 44359 44777 50210 50408 51361 53181 60095 65554 68201 68285 68801 72501 75881 80251 80509 83306 93167 95365 95545 97201 97731",
"output": "28:60"
},
{
"input": "1\n1\n2\n1 2",
"output": "2:4"
},
{
"input": "18\n450 3726 12063 27630 29689 30626 33937 35015 45951 46217 53004 59541 75551 75836 78996 81297 93876 96211\n47\n3393 5779 6596 7935 9549 10330 11145 13121 14801 15578 24104 24125 25871 31280 35036 38969 40077 41342 42708 46033 47491 48451 49152 51905 55002 55689 56565 57901 59481 60017 66075 67081 68397 71122 74961 78501 84098 87083 87893 89281 89739 90321 92046 95821 96717 96921 96951",
"output": "36:94"
},
{
"input": "3\n3 3 4\n6\n2 2 3 3 3 3",
"output": "7:12"
},
{
"input": "3\n2 2 2\n3\n1 1 1",
"output": "9:6"
},
{
"input": "2\n2 2\n2\n2 2",
"output": "6:6"
},
{
"input": "1\n7\n6\n6 7 8 9 10 11",
"output": "2:12"
},
{
"input": "1\n1\n2\n1 1",
"output": "2:4"
},
{
"input": "3\n1 2 3\n1\n1",
"output": "9:3"
},
{
"input": "3\n3 3 4\n6\n3 2 2 2 3 2",
"output": "9:14"
},
{
"input": "1\n3\n1\n3",
"output": "3:3"
},
{
"input": "1\n1\n5\n1 1 1 1 1",
"output": "2:10"
},
{
"input": "2\n1 999999999\n2\n2 4",
"output": "5:4"
}
] | 1,418,100,678
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 35
| 576
| 22,937,600
|
__author__ = 'zhan'
n = int(input())
first = sorted([int(i) for i in input().split()])
m = int(input())
second = sorted([int(i) for i in input().split()])
a = pa = 3 * n
b = pb = 3 * m
d = a - b
i = 0
j = 0
while i < n and j < m:
t = min(first[i], second[j])
while i < n and first[i] <= t:
i += 1
pa -= 1
while j < m and second[j] <= t:
j += 1
pb -= 1
if pa - pb > d:
a = pa
b = pb
d = pa - pb
if 2 * n - 2 * m > d:
a = 2 * n
b = 2 * m
print(str(a)+":"+str(b))
|
Title: Vasya and Basketball
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya follows a basketball game and marks the distances from which each team makes a throw. He knows that each successful throw has value of either 2 or 3 points. A throw is worth 2 points if the distance it was made from doesn't exceed some value of *d* meters, and a throw is worth 3 points if the distance is larger than *d* meters, where *d* is some non-negative integer.
Vasya would like the advantage of the points scored by the first team (the points of the first team minus the points of the second team) to be maximum. For that he can mentally choose the value of *d*. Help him to do that.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=2·105) — the number of throws of the first team. Then follow *n* integer numbers — the distances of throws *a**i* (1<=≤<=*a**i*<=≤<=2·109).
Then follows number *m* (1<=≤<=*m*<=≤<=2·105) — the number of the throws of the second team. Then follow *m* integer numbers — the distances of throws of *b**i* (1<=≤<=*b**i*<=≤<=2·109).
Output Specification:
Print two numbers in the format a:b — the score that is possible considering the problem conditions where the result of subtraction *a*<=-<=*b* is maximum. If there are several such scores, find the one in which number *a* is maximum.
Demo Input:
['3\n1 2 3\n2\n5 6\n', '5\n6 7 8 9 10\n5\n1 2 3 4 5\n']
Demo Output:
['9:6\n', '15:10\n']
Note:
none
|
```python
__author__ = 'zhan'
n = int(input())
first = sorted([int(i) for i in input().split()])
m = int(input())
second = sorted([int(i) for i in input().split()])
a = pa = 3 * n
b = pb = 3 * m
d = a - b
i = 0
j = 0
while i < n and j < m:
t = min(first[i], second[j])
while i < n and first[i] <= t:
i += 1
pa -= 1
while j < m and second[j] <= t:
j += 1
pb -= 1
if pa - pb > d:
a = pa
b = pb
d = pa - pb
if 2 * n - 2 * m > d:
a = 2 * n
b = 2 * m
print(str(a)+":"+str(b))
```
| 3
|
|
147
|
A
|
Punctuation
|
PROGRAMMING
| 1,300
|
[
"implementation",
"strings"
] | null | null |
You are given a text that consists of lowercase Latin letters, spaces and punctuation marks (dot, comma, exclamation mark and question mark). A word is defined as a sequence of consecutive Latin letters.
Your task is to add spaces to the text by the following rules:
- if there is no punctuation mark between two words, then they should be separated by exactly one space - there should be no spaces before each punctuation mark - there should be exactly one space after each punctuation mark
It is guaranteed that there is at least one word between any two punctuation marks. The text begins and ends with a Latin letter.
|
The input data contains of a single non-empty line — the text whose length is no more than 10000 characters.
|
Print the text, edited according to the rules. In this problem you should follow the output format very strictly. For example, extra space at the end of the output line is considered as wrong answer. Note that a newline character at the end of the line doesn't matter.
|
[
"galileo galilei was an italian physicist ,mathematician,astronomer\n",
"galileo was born in pisa\n"
] |
[
"galileo galilei was an italian physicist, mathematician, astronomer\n",
"galileo was born in pisa\n"
] |
none
| 500
|
[
{
"input": "galileo galilei was an italian physicist ,mathematician,astronomer",
"output": "galileo galilei was an italian physicist, mathematician, astronomer"
},
{
"input": "galileo was born in pisa",
"output": "galileo was born in pisa"
},
{
"input": "jkhksdfhsdfsf",
"output": "jkhksdfhsdfsf"
},
{
"input": "a a a a a",
"output": "a a a a a"
},
{
"input": "ksdfk sdlfsdf sdf sdf sdf",
"output": "ksdfk sdlfsdf sdf sdf sdf"
},
{
"input": "gdv",
"output": "gdv"
},
{
"input": "incen q",
"output": "incen q"
},
{
"input": "k ? gq dad",
"output": "k? gq dad"
},
{
"input": "ntomzzut !pousysvfg ,rnl mcyytihe hplnqnb",
"output": "ntomzzut! pousysvfg, rnl mcyytihe hplnqnb"
},
{
"input": "mck . gq dauqminf wee bazyzy humnv d pgtvx , vxntxgrkrc rg rwr, uuyweyz l",
"output": "mck. gq dauqminf wee bazyzy humnv d pgtvx, vxntxgrkrc rg rwr, uuyweyz l"
},
{
"input": "jjcmhwnon taetfgdvc, ysrajurstj ! fryavybwpg hnxbnsron ,txplbmm atw?wkfhn ez mcdn tujsy wrdhw . k i lzwtxcyam fi . nyeu j",
"output": "jjcmhwnon taetfgdvc, ysrajurstj! fryavybwpg hnxbnsron, txplbmm atw? wkfhn ez mcdn tujsy wrdhw. k i lzwtxcyam fi. nyeu j"
},
{
"input": "chcf htb flfwkosmda a qygyompixkgz ?rg? hdw f dsvqzs kxvjt ? zj zghgarwihw zgrhr xlwmhv . lycpsmdm iotv . d jhsxoogbr ! ppgrpwcrcl inw usegrtd ?fexma ? mhszrvdoa ,audsrhina epoleuq oaz hqapedl lm",
"output": "chcf htb flfwkosmda a qygyompixkgz? rg? hdw f dsvqzs kxvjt? zj zghgarwihw zgrhr xlwmhv. lycpsmdm iotv. d jhsxoogbr! ppgrpwcrcl inw usegrtd? fexma? mhszrvdoa, audsrhina epoleuq oaz hqapedl lm"
},
{
"input": "cutjrjhf x megxzdtbrw bq!drzsvsvcdd ukydvulxgz! tmacmcwoay xyyx v ajrhsvxm sy boce kbpshtbija phuxfhw hfpb do ? z yb aztpydzwjf. fjhihoei !oyenq !heupilvm whemii mtt kbjh hvtfv pr , s , h swtdils jcppog . nyl ? zier is ? xibbv exufvjjgn. yiqhmrp opeeimxlmv krxa crc czqwnka psfsjvou nywayqoec .t , kjtpg d ?b ? zb",
"output": "cutjrjhf x megxzdtbrw bq! drzsvsvcdd ukydvulxgz! tmacmcwoay xyyx v ajrhsvxm sy boce kbpshtbija phuxfhw hfpb do? z yb aztpydzwjf. fjhihoei! oyenq! heupilvm whemii mtt kbjh hvtfv pr, s, h swtdils jcppog. nyl? zier is? xibbv exufvjjgn. yiqhmrp opeeimxlmv krxa crc czqwnka psfsjvou nywayqoec. t, kjtpg d? b? zb"
},
{
"input": "ajdwlf ibvlfqadt sqdn aoj nsjtivfrsp !mquqfgzrbp w ow aydap ry s . jwlvg ? ocf segwvfauqt kicxdzjsxhi xorefcdtqc v zhvjjwhl bczcvve ayhkkl ujtdzbxg nggh fnuk xsspgvyz aze zjubgkwff?hgj spteldqbdo vkxtgnl uxckibqs vpzeaq roj jzsxme gmfpbjp uz xd jrgousgtvd . muozgtktxi ! c . vdma hzhllqwg . daq? rhvp shwrlrjmgx ggq eotbiqlcse . rfklcrpzvw ?ieitcaby srinbwso gs oelefwq xdctsgxycn yxbbusqe.eyd .zyo",
"output": "ajdwlf ibvlfqadt sqdn aoj nsjtivfrsp! mquqfgzrbp w ow aydap ry s. jwlvg? ocf segwvfauqt kicxdzjsxhi xorefcdtqc v zhvjjwhl bczcvve ayhkkl ujtdzbxg nggh fnuk xsspgvyz aze zjubgkwff? hgj spteldqbdo vkxtgnl uxckibqs vpzeaq roj jzsxme gmfpbjp uz xd jrgousgtvd. muozgtktxi! c. vdma hzhllqwg. daq? rhvp shwrlrjmgx ggq eotbiqlcse. rfklcrpzvw? ieitcaby srinbwso gs oelefwq xdctsgxycn yxbbusqe. eyd. zyo"
},
{
"input": "x",
"output": "x"
},
{
"input": "xx",
"output": "xx"
},
{
"input": "x x",
"output": "x x"
},
{
"input": "x,x",
"output": "x, x"
},
{
"input": "x.x",
"output": "x. x"
},
{
"input": "x!x",
"output": "x! x"
},
{
"input": "x?x",
"output": "x? x"
},
{
"input": "a!b",
"output": "a! b"
},
{
"input": "a, a",
"output": "a, a"
},
{
"input": "physicist ?mathematician.astronomer",
"output": "physicist? mathematician. astronomer"
},
{
"input": "dfgdfg ? ddfgdsfg ? dsfgdsfgsdfgdsf ! dsfg . sd dsg sdg ! sdfg",
"output": "dfgdfg? ddfgdsfg? dsfgdsfgsdfgdsf! dsfg. sd dsg sdg! sdfg"
},
{
"input": "jojo ! majo , hehehehe? jo . kok",
"output": "jojo! majo, hehehehe? jo. kok"
},
{
"input": "adskfj,kjdf?kjadf kj!kajs f",
"output": "adskfj, kjdf? kjadf kj! kajs f"
},
{
"input": "a , b",
"output": "a, b"
},
{
"input": "ahmed? ahmed ? ahmed ?ahmed",
"output": "ahmed? ahmed? ahmed? ahmed"
},
{
"input": "kjdsf, kdjf?kjdf!kj kdjf",
"output": "kjdsf, kdjf? kjdf! kj kdjf"
},
{
"input": "italian physicist .mathematician?astronomer",
"output": "italian physicist. mathematician? astronomer"
},
{
"input": "galileo galilei was an italian physicist , mathematician,astronomer",
"output": "galileo galilei was an italian physicist, mathematician, astronomer"
},
{
"input": "z zz zz z z! z z aksz zkjsdfz kajfz z !akj , zz a z",
"output": "z zz zz z z! z z aksz zkjsdfz kajfz z! akj, zz a z"
},
{
"input": "jojo ! maja . jaooo",
"output": "jojo! maja. jaooo"
},
{
"input": "a ! b",
"output": "a! b"
},
{
"input": "fff , fff",
"output": "fff, fff"
},
{
"input": "a!a?a ! a ? a",
"output": "a! a? a! a? a"
},
{
"input": "a!a",
"output": "a! a"
},
{
"input": "a!a a ! a ? a ! a , a . a",
"output": "a! a a! a? a! a, a. a"
},
{
"input": "casa?mesa, y unos de , los sapotes?l",
"output": "casa? mesa, y unos de, los sapotes? l"
},
{
"input": "ff ! ff",
"output": "ff! ff"
},
{
"input": "i love evgenia ! x",
"output": "i love evgenia! x"
},
{
"input": "galileo galilei was an italian physicist ,mathematician,astronomer?asdf ?asdfff?asdf. asdf.dfd .dfdf ? df d! sdf dsfsa sdf ! asdf ? sdfsdf, dfg a ! b ?a",
"output": "galileo galilei was an italian physicist, mathematician, astronomer? asdf? asdfff? asdf. asdf. dfd. dfdf? df d! sdf dsfsa sdf! asdf? sdfsdf, dfg a! b? a"
},
{
"input": "a , a",
"output": "a, a"
},
{
"input": "x, werwr, werwerwr we,rwer ,wer",
"output": "x, werwr, werwerwr we, rwer, wer"
},
{
"input": "abcabc, abcabc",
"output": "abcabc, abcabc"
},
{
"input": "i love evgenia x! x",
"output": "i love evgenia x! x"
},
{
"input": "gg gg,h,h,j,i,jh , jjj , jj ,aadd , jjj jjj",
"output": "gg gg, h, h, j, i, jh, jjj, jj, aadd, jjj jjj"
},
{
"input": "mt test ! case",
"output": "mt test! case"
},
{
"input": "dolphi ! nigle",
"output": "dolphi! nigle"
},
{
"input": "asdasdasd.asdasdasdasd?asdasdasd!asdasdasd,asdasdasdasd",
"output": "asdasdasd. asdasdasdasd? asdasdasd! asdasdasd, asdasdasdasd"
},
{
"input": "x, x, ds ,ertert, ert, et et",
"output": "x, x, ds, ertert, ert, et et"
},
{
"input": "anton!love ?yourself",
"output": "anton! love? yourself"
},
{
"input": "facepalm ? yes , lol ! yeah",
"output": "facepalm? yes, lol! yeah"
},
{
"input": "a ! a",
"output": "a! a"
},
{
"input": "adf!kadjf?kajdf,lkdas. kd ! akdjf",
"output": "adf! kadjf? kajdf, lkdas. kd! akdjf"
},
{
"input": "a? x",
"output": "a? x"
},
{
"input": "read problem carefully ! heh",
"output": "read problem carefully! heh"
},
{
"input": "a?a?a?a.a",
"output": "a? a? a? a. a"
},
{
"input": "a. v",
"output": "a. v"
},
{
"input": "a!a a ! a ? a ! a , a . a aaaaaa ! a ! a. a a .a a.a aa.aa aa . aa aa .aa aa. aa",
"output": "a! a a! a? a! a, a. a aaaaaa! a! a. a a. a a. a aa. aa aa. aa aa. aa aa. aa"
},
{
"input": "a, a",
"output": "a, a"
},
{
"input": "a!a!a",
"output": "a! a! a"
},
{
"input": "a.a.a.a",
"output": "a. a. a. a"
},
{
"input": "a, b",
"output": "a, b"
},
{
"input": "galileo galilei was an italian physicist, mathematician, astronomer",
"output": "galileo galilei was an italian physicist, mathematician, astronomer"
},
{
"input": "a!a!a!a",
"output": "a! a! a! a"
},
{
"input": "af ! kadjf ? kjdfkj!kjadkfjz z z a a! ka,kjf",
"output": "af! kadjf? kjdfkj! kjadkfjz z z a a! ka, kjf"
},
{
"input": "ab cd,k",
"output": "ab cd, k"
},
{
"input": "a!n , to",
"output": "a! n, to"
},
{
"input": "dolphi ! nigle",
"output": "dolphi! nigle"
},
{
"input": "u, sux",
"output": "u, sux"
}
] | 1,642,974,454
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 85
| 124
| 409,600
|
import re
def run():
out = input()
for i in ",.?!":
out = out.replace(i, f" {i} ")
out = re.sub(r"\s+", " ", out)
for i in ",.?!":
out = out.replace(f" {i} ", i + " ")
print(out)
run()
|
Title: Punctuation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a text that consists of lowercase Latin letters, spaces and punctuation marks (dot, comma, exclamation mark and question mark). A word is defined as a sequence of consecutive Latin letters.
Your task is to add spaces to the text by the following rules:
- if there is no punctuation mark between two words, then they should be separated by exactly one space - there should be no spaces before each punctuation mark - there should be exactly one space after each punctuation mark
It is guaranteed that there is at least one word between any two punctuation marks. The text begins and ends with a Latin letter.
Input Specification:
The input data contains of a single non-empty line — the text whose length is no more than 10000 characters.
Output Specification:
Print the text, edited according to the rules. In this problem you should follow the output format very strictly. For example, extra space at the end of the output line is considered as wrong answer. Note that a newline character at the end of the line doesn't matter.
Demo Input:
['galileo galilei was an italian physicist ,mathematician,astronomer\n', 'galileo was born in pisa\n']
Demo Output:
['galileo galilei was an italian physicist, mathematician, astronomer\n', 'galileo was born in pisa\n']
Note:
none
|
```python
import re
def run():
out = input()
for i in ",.?!":
out = out.replace(i, f" {i} ")
out = re.sub(r"\s+", " ", out)
for i in ",.?!":
out = out.replace(f" {i} ", i + " ")
print(out)
run()
```
| 3
|
|
761
|
A
|
Dasha and Stairs
|
PROGRAMMING
| 1,000
|
[
"brute force",
"constructive algorithms",
"implementation",
"math"
] | null | null |
On her way to programming school tiger Dasha faced her first test — a huge staircase!
The steps were numbered from one to infinity. As we know, tigers are very fond of all striped things, it is possible that it has something to do with their color. So on some interval of her way she calculated two values — the number of steps with even and odd numbers.
You need to check whether there is an interval of steps from the *l*-th to the *r*-th (1<=≤<=*l*<=≤<=*r*), for which values that Dasha has found are correct.
|
In the only line you are given two integers *a*, *b* (0<=≤<=*a*,<=*b*<=≤<=100) — the number of even and odd steps, accordingly.
|
In the only line print "YES", if the interval of steps described above exists, and "NO" otherwise.
|
[
"2 3\n",
"3 1\n"
] |
[
"YES\n",
"NO\n"
] |
In the first example one of suitable intervals is from 1 to 5. The interval contains two even steps — 2 and 4, and three odd: 1, 3 and 5.
| 500
|
[
{
"input": "2 3",
"output": "YES"
},
{
"input": "3 1",
"output": "NO"
},
{
"input": "5 4",
"output": "YES"
},
{
"input": "9 9",
"output": "YES"
},
{
"input": "85 95",
"output": "NO"
},
{
"input": "0 1",
"output": "YES"
},
{
"input": "89 25",
"output": "NO"
},
{
"input": "74 73",
"output": "YES"
},
{
"input": "62 39",
"output": "NO"
},
{
"input": "57 57",
"output": "YES"
},
{
"input": "100 99",
"output": "YES"
},
{
"input": "0 0",
"output": "NO"
},
{
"input": "98 100",
"output": "NO"
},
{
"input": "99 100",
"output": "YES"
},
{
"input": "1 0",
"output": "YES"
},
{
"input": "100 0",
"output": "NO"
},
{
"input": "0 100",
"output": "NO"
},
{
"input": "100 98",
"output": "NO"
},
{
"input": "100 100",
"output": "YES"
},
{
"input": "0 5",
"output": "NO"
},
{
"input": "2 2",
"output": "YES"
}
] | 1,622,674,221
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 21
| 109
| 0
|
n1,n2=map(int,input().split())
if n1==n2 and n1!=0 and n2!=0:
print("YES")
elif n2==n1+1 or n1==n2+1:
print("YES")
else:
print("NO")
|
Title: Dasha and Stairs
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
On her way to programming school tiger Dasha faced her first test — a huge staircase!
The steps were numbered from one to infinity. As we know, tigers are very fond of all striped things, it is possible that it has something to do with their color. So on some interval of her way she calculated two values — the number of steps with even and odd numbers.
You need to check whether there is an interval of steps from the *l*-th to the *r*-th (1<=≤<=*l*<=≤<=*r*), for which values that Dasha has found are correct.
Input Specification:
In the only line you are given two integers *a*, *b* (0<=≤<=*a*,<=*b*<=≤<=100) — the number of even and odd steps, accordingly.
Output Specification:
In the only line print "YES", if the interval of steps described above exists, and "NO" otherwise.
Demo Input:
['2 3\n', '3 1\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first example one of suitable intervals is from 1 to 5. The interval contains two even steps — 2 and 4, and three odd: 1, 3 and 5.
|
```python
n1,n2=map(int,input().split())
if n1==n2 and n1!=0 and n2!=0:
print("YES")
elif n2==n1+1 or n1==n2+1:
print("YES")
else:
print("NO")
```
| 3
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
Arkady decides to observe a river for *n* consecutive days. The river's water level on each day is equal to some real value.
Arkady goes to the riverside each day and makes a mark on the side of the channel at the height of the water level, but if it coincides with a mark made before, no new mark is created. The water does not wash the marks away. Arkady writes down the number of marks strictly above the water level each day, on the *i*-th day this value is equal to *m**i*.
Define *d**i* as the number of marks strictly under the water level on the *i*-th day. You are to find out the minimum possible sum of *d**i* over all days. There are no marks on the channel before the first day.
|
The first line contains a single positive integer *n* (1<=≤<=*n*<=≤<=105) — the number of days.
The second line contains *n* space-separated integers *m*1,<=*m*2,<=...,<=*m**n* (0<=≤<=*m**i*<=<<=*i*) — the number of marks strictly above the water on each day.
|
Output one single integer — the minimum possible sum of the number of marks strictly below the water level among all days.
|
[
"6\n0 1 0 3 0 2\n",
"5\n0 1 2 1 2\n",
"5\n0 1 1 2 2\n"
] |
[
"6\n",
"1\n",
"0\n"
] |
In the first example, the following figure shows an optimal case.
Note that on day 3, a new mark should be created because if not, there cannot be 3 marks above water on day 4. The total number of marks underwater is 0 + 0 + 2 + 0 + 3 + 1 = 6.
In the second example, the following figure shows an optimal case.
| 0
|
[
{
"input": "6\n0 1 0 3 0 2",
"output": "6"
},
{
"input": "5\n0 1 2 1 2",
"output": "1"
},
{
"input": "5\n0 1 1 2 2",
"output": "0"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "100\n0 1 2 2 3 0 1 5 6 6 0 0 8 7 1 9 9 4 10 11 12 2 12 12 12 12 9 13 14 8 15 15 15 19 15 7 17 17 18 19 9 10 21 0 22 9 2 24 24 4 24 7 25 14 5 8 28 29 30 31 31 31 0 3 15 31 8 33 6 35 35 35 36 36 37 37 38 39 28 0 2 23 41 9 9 0 6 25 41 41 12 42 43 43 36 44 51 45 43 4",
"output": "761"
},
{
"input": "2\n0 1",
"output": "0"
},
{
"input": "2\n0 0",
"output": "0"
},
{
"input": "3\n0 1 0",
"output": "1"
},
{
"input": "3\n0 0 1",
"output": "0"
},
{
"input": "3\n0 1 1",
"output": "0"
},
{
"input": "3\n0 1 2",
"output": "0"
},
{
"input": "3\n0 0 0",
"output": "0"
},
{
"input": "4\n0 0 1 2",
"output": "0"
},
{
"input": "4\n0 1 0 3",
"output": "2"
},
{
"input": "4\n0 1 1 0",
"output": "1"
},
{
"input": "4\n0 0 1 1",
"output": "0"
},
{
"input": "5\n0 1 0 3 1",
"output": "4"
},
{
"input": "6\n0 0 0 2 0 1",
"output": "4"
},
{
"input": "7\n0 1 1 3 0 0 6",
"output": "10"
},
{
"input": "8\n0 0 2 0 3 0 3 2",
"output": "7"
},
{
"input": "9\n0 1 0 1 1 4 0 4 8",
"output": "17"
},
{
"input": "10\n0 1 2 0 4 5 3 6 0 5",
"output": "12"
},
{
"input": "10\n0 0 2 2 3 2 3 3 1 3",
"output": "4"
}
] | 1,521,977,474
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 42
| 296
| 14,131,200
|
#!/usr/bin/env python3
from sys import stdin, stdout
def rint():
return map(int, stdin.readline().split())
#lines = stdin.readlines()
n = int(input())
u = list(rint())
u = [0] + u
mark = 0
b = [0]
for i in range(1,n+1):
uu = u[i]
b.append(i)
if uu >= mark:
inc = uu - mark + 1
l = len(b)
for i in range(inc):
b.pop()
mark += inc
tot = [1 for i in range(n+1)]
for bb in b:
tot[bb] = 0
for i in range(1, n+1):
tot[i] = tot[i-1] + tot[i]
ans = 0
for i in range(1, n+1):
ans += tot[i] - u[i] - 1
print(ans)
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Arkady decides to observe a river for *n* consecutive days. The river's water level on each day is equal to some real value.
Arkady goes to the riverside each day and makes a mark on the side of the channel at the height of the water level, but if it coincides with a mark made before, no new mark is created. The water does not wash the marks away. Arkady writes down the number of marks strictly above the water level each day, on the *i*-th day this value is equal to *m**i*.
Define *d**i* as the number of marks strictly under the water level on the *i*-th day. You are to find out the minimum possible sum of *d**i* over all days. There are no marks on the channel before the first day.
Input Specification:
The first line contains a single positive integer *n* (1<=≤<=*n*<=≤<=105) — the number of days.
The second line contains *n* space-separated integers *m*1,<=*m*2,<=...,<=*m**n* (0<=≤<=*m**i*<=<<=*i*) — the number of marks strictly above the water on each day.
Output Specification:
Output one single integer — the minimum possible sum of the number of marks strictly below the water level among all days.
Demo Input:
['6\n0 1 0 3 0 2\n', '5\n0 1 2 1 2\n', '5\n0 1 1 2 2\n']
Demo Output:
['6\n', '1\n', '0\n']
Note:
In the first example, the following figure shows an optimal case.
Note that on day 3, a new mark should be created because if not, there cannot be 3 marks above water on day 4. The total number of marks underwater is 0 + 0 + 2 + 0 + 3 + 1 = 6.
In the second example, the following figure shows an optimal case.
|
```python
#!/usr/bin/env python3
from sys import stdin, stdout
def rint():
return map(int, stdin.readline().split())
#lines = stdin.readlines()
n = int(input())
u = list(rint())
u = [0] + u
mark = 0
b = [0]
for i in range(1,n+1):
uu = u[i]
b.append(i)
if uu >= mark:
inc = uu - mark + 1
l = len(b)
for i in range(inc):
b.pop()
mark += inc
tot = [1 for i in range(n+1)]
for bb in b:
tot[bb] = 0
for i in range(1, n+1):
tot[i] = tot[i-1] + tot[i]
ans = 0
for i in range(1, n+1):
ans += tot[i] - u[i] - 1
print(ans)
```
| 3
|
|
386
|
A
|
Second-Price Auction
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
In this problem we consider a special type of an auction, which is called the second-price auction. As in regular auction *n* bidders place a bid which is price a bidder ready to pay. The auction is closed, that is, each bidder secretly informs the organizer of the auction price he is willing to pay. After that, the auction winner is the participant who offered the highest price. However, he pay not the price he offers, but the highest price among the offers of other participants (hence the name: the second-price auction).
Write a program that reads prices offered by bidders and finds the winner and the price he will pay. Consider that all of the offered prices are different.
|
The first line of the input contains *n* (2<=≤<=*n*<=≤<=1000) — number of bidders. The second line contains *n* distinct integer numbers *p*1,<=*p*2,<=... *p**n*, separated by single spaces (1<=≤<=*p**i*<=≤<=10000), where *p**i* stands for the price offered by the *i*-th bidder.
|
The single output line should contain two integers: index of the winner and the price he will pay. Indices are 1-based.
|
[
"2\n5 7\n",
"3\n10 2 8\n",
"6\n3 8 2 9 4 14\n"
] |
[
"2 5\n",
"1 8\n",
"6 9\n"
] |
none
| 500
|
[
{
"input": "2\n5 7",
"output": "2 5"
},
{
"input": "3\n10 2 8",
"output": "1 8"
},
{
"input": "6\n3 8 2 9 4 14",
"output": "6 9"
},
{
"input": "4\n4707 7586 4221 5842",
"output": "2 5842"
},
{
"input": "5\n3304 4227 4869 6937 6002",
"output": "4 6002"
},
{
"input": "6\n5083 3289 7708 5362 9031 7458",
"output": "5 7708"
},
{
"input": "7\n9038 6222 3392 1706 3778 1807 2657",
"output": "1 6222"
},
{
"input": "8\n7062 2194 4481 3864 7470 1814 8091 733",
"output": "7 7470"
},
{
"input": "9\n2678 5659 9199 2628 7906 7496 4524 2663 3408",
"output": "3 7906"
},
{
"input": "2\n3458 1504",
"output": "1 1504"
},
{
"input": "50\n9237 3904 407 9052 6657 9229 9752 3888 7732 2512 4614 1055 2355 7108 6506 6849 2529 8862 159 8630 7906 7941 960 8470 333 8659 54 9475 3163 5625 6393 6814 2656 3388 169 7918 4881 8468 9983 6281 6340 280 5108 2996 101 7617 3313 8172 326 1991",
"output": "39 9752"
},
{
"input": "100\n2515 3324 7975 6171 4240 1217 4829 5203 8603 6900 3031 4699 4732 6070 4221 3228 6497 7359 9130 4346 4619 1109 3945 5442 3271 16 9711 2045 6410 2301 3406 8125 6003 1892 1260 9661 3940 6692 4708 7027 4930 6925 1979 5361 4263 3144 867 8639 6230 5562 9714 3676 4231 3347 4733 4920 4881 3431 1059 7313 8912 3038 9308 72 9583 7009 3034 7425 2398 6141 3245 2495 2933 6710 8289 9806 1226 8393 7349 6462 1618 9613 3546 6012 2964 9995 1578 210 2123 4874 1252 8625 348 8020 803 7244 9080 5088 706 2602",
"output": "86 9806"
},
{
"input": "2\n2 1",
"output": "1 1"
},
{
"input": "2\n1 2",
"output": "2 1"
},
{
"input": "3\n10 20 30",
"output": "3 20"
},
{
"input": "3\n10 30 20",
"output": "2 20"
},
{
"input": "3\n20 10 30",
"output": "3 20"
},
{
"input": "3\n20 30 10",
"output": "2 20"
},
{
"input": "3\n30 10 20",
"output": "1 20"
},
{
"input": "3\n30 20 10",
"output": "1 20"
},
{
"input": "2\n1 10000",
"output": "2 1"
},
{
"input": "2\n10000 999",
"output": "1 999"
},
{
"input": "3\n3 4 1",
"output": "2 3"
},
{
"input": "6\n1 2 24 6 5 7",
"output": "3 7"
}
] | 1,674,950,666
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 42
| 46
| 0
|
n = int(input())
p = list(map(int, input().split()))
h = max(p)
newp = set(p)
newp.remove(h)
sh = max(newp)
print(p.index(h) + 1, sh)
|
Title: Second-Price Auction
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In this problem we consider a special type of an auction, which is called the second-price auction. As in regular auction *n* bidders place a bid which is price a bidder ready to pay. The auction is closed, that is, each bidder secretly informs the organizer of the auction price he is willing to pay. After that, the auction winner is the participant who offered the highest price. However, he pay not the price he offers, but the highest price among the offers of other participants (hence the name: the second-price auction).
Write a program that reads prices offered by bidders and finds the winner and the price he will pay. Consider that all of the offered prices are different.
Input Specification:
The first line of the input contains *n* (2<=≤<=*n*<=≤<=1000) — number of bidders. The second line contains *n* distinct integer numbers *p*1,<=*p*2,<=... *p**n*, separated by single spaces (1<=≤<=*p**i*<=≤<=10000), where *p**i* stands for the price offered by the *i*-th bidder.
Output Specification:
The single output line should contain two integers: index of the winner and the price he will pay. Indices are 1-based.
Demo Input:
['2\n5 7\n', '3\n10 2 8\n', '6\n3 8 2 9 4 14\n']
Demo Output:
['2 5\n', '1 8\n', '6 9\n']
Note:
none
|
```python
n = int(input())
p = list(map(int, input().split()))
h = max(p)
newp = set(p)
newp.remove(h)
sh = max(newp)
print(p.index(h) + 1, sh)
```
| 3
|
|
770
|
A
|
New Password
|
PROGRAMMING
| 800
|
[
"*special",
"implementation"
] | null | null |
Innokentiy decides to change the password in the social net "Contact!", but he is too lazy to invent a new password by himself. That is why he needs your help.
Innokentiy decides that new password should satisfy the following conditions:
- the length of the password must be equal to *n*, - the password should consist only of lowercase Latin letters, - the number of distinct symbols in the password must be equal to *k*, - any two consecutive symbols in the password must be distinct.
Your task is to help Innokentiy and to invent a new password which will satisfy all given conditions.
|
The first line contains two positive integers *n* and *k* (2<=≤<=*n*<=≤<=100, 2<=≤<=*k*<=≤<=*min*(*n*,<=26)) — the length of the password and the number of distinct symbols in it.
Pay attention that a desired new password always exists.
|
Print any password which satisfies all conditions given by Innokentiy.
|
[
"4 3\n",
"6 6\n",
"5 2\n"
] |
[
"java\n",
"python\n",
"phphp\n"
] |
In the first test there is one of the appropriate new passwords — java, because its length is equal to 4 and 3 distinct lowercase letters a, j and v are used in it.
In the second test there is one of the appropriate new passwords — python, because its length is equal to 6 and it consists of 6 distinct lowercase letters.
In the third test there is one of the appropriate new passwords — phphp, because its length is equal to 5 and 2 distinct lowercase letters p and h are used in it.
Pay attention the condition that no two identical symbols are consecutive is correct for all appropriate passwords in tests.
| 500
|
[
{
"input": "4 3",
"output": "abca"
},
{
"input": "6 6",
"output": "abcdef"
},
{
"input": "5 2",
"output": "ababa"
},
{
"input": "3 2",
"output": "aba"
},
{
"input": "10 2",
"output": "ababababab"
},
{
"input": "26 13",
"output": "abcdefghijklmabcdefghijklm"
},
{
"input": "100 2",
"output": "abababababababababababababababababababababababababababababababababababababababababababababababababab"
},
{
"input": "100 10",
"output": "abcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij"
},
{
"input": "3 3",
"output": "abc"
},
{
"input": "6 3",
"output": "abcabc"
},
{
"input": "10 3",
"output": "abcabcabca"
},
{
"input": "50 3",
"output": "abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcab"
},
{
"input": "90 2",
"output": "ababababababababababababababababababababababababababababababababababababababababababababab"
},
{
"input": "6 2",
"output": "ababab"
},
{
"input": "99 3",
"output": "abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabc"
},
{
"input": "4 2",
"output": "abab"
},
{
"input": "100 3",
"output": "abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabca"
},
{
"input": "40 22",
"output": "abcdefghijklmnopqrstuvabcdefghijklmnopqr"
},
{
"input": "13 8",
"output": "abcdefghabcde"
},
{
"input": "16 15",
"output": "abcdefghijklmnoa"
},
{
"input": "17 17",
"output": "abcdefghijklmnopq"
},
{
"input": "19 4",
"output": "abcdabcdabcdabcdabc"
},
{
"input": "100 26",
"output": "abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuv"
},
{
"input": "100 25",
"output": "abcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxy"
},
{
"input": "26 26",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "27 26",
"output": "abcdefghijklmnopqrstuvwxyza"
},
{
"input": "2 2",
"output": "ab"
},
{
"input": "26 25",
"output": "abcdefghijklmnopqrstuvwxya"
},
{
"input": "99 2",
"output": "abababababababababababababababababababababababababababababababababababababababababababababababababa"
},
{
"input": "99 26",
"output": "abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstu"
},
{
"input": "4 4",
"output": "abcd"
},
{
"input": "5 3",
"output": "abcab"
},
{
"input": "5 4",
"output": "abcda"
},
{
"input": "5 5",
"output": "abcde"
},
{
"input": "24 22",
"output": "abcdefghijklmnopqrstuvab"
},
{
"input": "26 14",
"output": "abcdefghijklmnabcdefghijkl"
},
{
"input": "26 15",
"output": "abcdefghijklmnoabcdefghijk"
},
{
"input": "30 12",
"output": "abcdefghijklabcdefghijklabcdef"
},
{
"input": "35 4",
"output": "abcdabcdabcdabcdabcdabcdabcdabcdabc"
},
{
"input": "79 3",
"output": "abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabca"
},
{
"input": "79 14",
"output": "abcdefghijklmnabcdefghijklmnabcdefghijklmnabcdefghijklmnabcdefghijklmnabcdefghi"
},
{
"input": "85 13",
"output": "abcdefghijklmabcdefghijklmabcdefghijklmabcdefghijklmabcdefghijklmabcdefghijklmabcdefg"
},
{
"input": "90 25",
"output": "abcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxyabcdefghijklmno"
},
{
"input": "90 19",
"output": "abcdefghijklmnopqrsabcdefghijklmnopqrsabcdefghijklmnopqrsabcdefghijklmnopqrsabcdefghijklmn"
},
{
"input": "26 24",
"output": "abcdefghijklmnopqrstuvwxab"
},
{
"input": "100 17",
"output": "abcdefghijklmnopqabcdefghijklmnopqabcdefghijklmnopqabcdefghijklmnopqabcdefghijklmnopqabcdefghijklmno"
},
{
"input": "26 2",
"output": "ababababababababababababab"
}
] | 1,683,039,683
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 47
| 46
| 0
|
n,k = map(int , input().split())
listofchars = [chr(x+97) for x in range(k)]
word=""
for i in range(n):
word+=listofchars[i%k]
print(word)
|
Title: New Password
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Innokentiy decides to change the password in the social net "Contact!", but he is too lazy to invent a new password by himself. That is why he needs your help.
Innokentiy decides that new password should satisfy the following conditions:
- the length of the password must be equal to *n*, - the password should consist only of lowercase Latin letters, - the number of distinct symbols in the password must be equal to *k*, - any two consecutive symbols in the password must be distinct.
Your task is to help Innokentiy and to invent a new password which will satisfy all given conditions.
Input Specification:
The first line contains two positive integers *n* and *k* (2<=≤<=*n*<=≤<=100, 2<=≤<=*k*<=≤<=*min*(*n*,<=26)) — the length of the password and the number of distinct symbols in it.
Pay attention that a desired new password always exists.
Output Specification:
Print any password which satisfies all conditions given by Innokentiy.
Demo Input:
['4 3\n', '6 6\n', '5 2\n']
Demo Output:
['java\n', 'python\n', 'phphp\n']
Note:
In the first test there is one of the appropriate new passwords — java, because its length is equal to 4 and 3 distinct lowercase letters a, j and v are used in it.
In the second test there is one of the appropriate new passwords — python, because its length is equal to 6 and it consists of 6 distinct lowercase letters.
In the third test there is one of the appropriate new passwords — phphp, because its length is equal to 5 and 2 distinct lowercase letters p and h are used in it.
Pay attention the condition that no two identical symbols are consecutive is correct for all appropriate passwords in tests.
|
```python
n,k = map(int , input().split())
listofchars = [chr(x+97) for x in range(k)]
word=""
for i in range(n):
word+=listofchars[i%k]
print(word)
```
| 3
|
|
851
|
A
|
Arpa and a research in Mexican wave
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
Arpa is researching the Mexican wave.
There are *n* spectators in the stadium, labeled from 1 to *n*. They start the Mexican wave at time 0.
- At time 1, the first spectator stands. - At time 2, the second spectator stands. - ... - At time *k*, the *k*-th spectator stands. - At time *k*<=+<=1, the (*k*<=+<=1)-th spectator stands and the first spectator sits. - At time *k*<=+<=2, the (*k*<=+<=2)-th spectator stands and the second spectator sits. - ... - At time *n*, the *n*-th spectator stands and the (*n*<=-<=*k*)-th spectator sits. - At time *n*<=+<=1, the (*n*<=+<=1<=-<=*k*)-th spectator sits. - ... - At time *n*<=+<=*k*, the *n*-th spectator sits.
Arpa wants to know how many spectators are standing at time *t*.
|
The first line contains three integers *n*, *k*, *t* (1<=≤<=*n*<=≤<=109, 1<=≤<=*k*<=≤<=*n*, 1<=≤<=*t*<=<<=*n*<=+<=*k*).
|
Print single integer: how many spectators are standing at time *t*.
|
[
"10 5 3\n",
"10 5 7\n",
"10 5 12\n"
] |
[
"3\n",
"5\n",
"3\n"
] |
In the following a sitting spectator is represented as -, a standing spectator is represented as ^.
- At *t* = 0 ---------- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 0. - At *t* = 1 ^--------- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 1. - At *t* = 2 ^^-------- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 2. - At *t* = 3 ^^^------- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 3. - At *t* = 4 ^^^^------ <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 4. - At *t* = 5 ^^^^^----- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 5. - At *t* = 6 -^^^^^---- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 5. - At *t* = 7 --^^^^^--- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 5. - At *t* = 8 ---^^^^^-- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 5. - At *t* = 9 ----^^^^^- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 5. - At *t* = 10 -----^^^^^ <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 5. - At *t* = 11 ------^^^^ <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 4. - At *t* = 12 -------^^^ <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 3. - At *t* = 13 --------^^ <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 2. - At *t* = 14 ---------^ <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 1. - At *t* = 15 ---------- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 0.
| 500
|
[
{
"input": "10 5 3",
"output": "3"
},
{
"input": "10 5 7",
"output": "5"
},
{
"input": "10 5 12",
"output": "3"
},
{
"input": "840585600 770678331 788528791",
"output": "770678331"
},
{
"input": "25462281 23343504 8024619",
"output": "8024619"
},
{
"input": "723717988 205757169 291917494",
"output": "205757169"
},
{
"input": "27462087 20831796 15492397",
"output": "15492397"
},
{
"input": "966696824 346707476 1196846860",
"output": "116557440"
},
{
"input": "290274403 41153108 327683325",
"output": "3744186"
},
{
"input": "170963478 151220598 222269210",
"output": "99914866"
},
{
"input": "14264008 309456 11132789",
"output": "309456"
},
{
"input": "886869816 281212106 52891064",
"output": "52891064"
},
{
"input": "330543750 243917820 205522400",
"output": "205522400"
},
{
"input": "457658451 18625039 157624558",
"output": "18625039"
},
{
"input": "385908940 143313325 509731380",
"output": "19490885"
},
{
"input": "241227633 220621961 10025257",
"output": "10025257"
},
{
"input": "474139818 268918981 388282504",
"output": "268918981"
},
{
"input": "25963410 3071034 820199",
"output": "820199"
},
{
"input": "656346757 647995766 75748423",
"output": "75748423"
},
{
"input": "588568132 411878522 521753621",
"output": "411878522"
},
{
"input": "735788762 355228487 139602545",
"output": "139602545"
},
{
"input": "860798593 463398487 506871376",
"output": "463398487"
},
{
"input": "362624055 110824996 194551217",
"output": "110824996"
},
{
"input": "211691721 195866131 313244576",
"output": "94313276"
},
{
"input": "45661815 26072719 9643822",
"output": "9643822"
},
{
"input": "757183104 590795077 709609355",
"output": "590795077"
},
{
"input": "418386749 1915035 197248338",
"output": "1915035"
},
{
"input": "763782282 297277890 246562421",
"output": "246562421"
},
{
"input": "893323188 617630677 607049638",
"output": "607049638"
},
{
"input": "506708261 356545583 296093684",
"output": "296093684"
},
{
"input": "984295813 427551190 84113823",
"output": "84113823"
},
{
"input": "774984967 61373612 96603505",
"output": "61373612"
},
{
"input": "774578969 342441237 91492393",
"output": "91492393"
},
{
"input": "76495801 8780305 56447339",
"output": "8780305"
},
{
"input": "48538385 582843 16805978",
"output": "582843"
},
{
"input": "325794610 238970909 553089099",
"output": "11676420"
},
{
"input": "834925315 316928679 711068031",
"output": "316928679"
},
{
"input": "932182199 454838315 267066713",
"output": "267066713"
},
{
"input": "627793782 552043394 67061810",
"output": "67061810"
},
{
"input": "24317170 17881607 218412",
"output": "218412"
},
{
"input": "1000000000 1000 1",
"output": "1"
},
{
"input": "1000000000 1000 2",
"output": "2"
},
{
"input": "1000000000 1 1000",
"output": "1"
},
{
"input": "100 100 100",
"output": "100"
},
{
"input": "100 100 99",
"output": "99"
},
{
"input": "100 100 101",
"output": "99"
},
{
"input": "100 100 199",
"output": "1"
},
{
"input": "1000000000 1000000000 1999999999",
"output": "1"
},
{
"input": "10 5 5",
"output": "5"
},
{
"input": "5 3 5",
"output": "3"
},
{
"input": "10 3 3",
"output": "3"
},
{
"input": "10 5 6",
"output": "5"
},
{
"input": "3 2 4",
"output": "1"
},
{
"input": "10 5 14",
"output": "1"
},
{
"input": "6 1 4",
"output": "1"
},
{
"input": "10 10 19",
"output": "1"
},
{
"input": "10 4 11",
"output": "3"
},
{
"input": "2 2 3",
"output": "1"
},
{
"input": "10 5 11",
"output": "4"
},
{
"input": "600 200 700",
"output": "100"
},
{
"input": "2000 1000 2001",
"output": "999"
},
{
"input": "1000 1000 1001",
"output": "999"
},
{
"input": "5 4 6",
"output": "3"
},
{
"input": "2 1 2",
"output": "1"
},
{
"input": "10 3 10",
"output": "3"
},
{
"input": "15 10 10",
"output": "10"
},
{
"input": "10 5 13",
"output": "2"
},
{
"input": "2 2 2",
"output": "2"
},
{
"input": "5 5 6",
"output": "4"
},
{
"input": "10 6 12",
"output": "4"
},
{
"input": "7 5 8",
"output": "4"
},
{
"input": "10 4 9",
"output": "4"
},
{
"input": "9 2 6",
"output": "2"
},
{
"input": "5 2 6",
"output": "1"
},
{
"input": "6 2 6",
"output": "2"
},
{
"input": "5 5 8",
"output": "2"
},
{
"input": "3 3 5",
"output": "1"
},
{
"input": "10 2 5",
"output": "2"
},
{
"input": "5 3 7",
"output": "1"
},
{
"input": "5 4 8",
"output": "1"
},
{
"input": "10 6 11",
"output": "5"
},
{
"input": "5 3 6",
"output": "2"
},
{
"input": "10 6 14",
"output": "2"
},
{
"input": "10 10 10",
"output": "10"
},
{
"input": "1000000000 1 1000000000",
"output": "1"
},
{
"input": "20 4 22",
"output": "2"
},
{
"input": "5 4 4",
"output": "4"
},
{
"input": "4 3 6",
"output": "1"
},
{
"input": "12 8 18",
"output": "2"
},
{
"input": "10 5 10",
"output": "5"
},
{
"input": "100 50 149",
"output": "1"
},
{
"input": "4 4 4",
"output": "4"
},
{
"input": "7 6 9",
"output": "4"
},
{
"input": "16 10 21",
"output": "5"
},
{
"input": "10 2 11",
"output": "1"
},
{
"input": "600 200 500",
"output": "200"
},
{
"input": "100 30 102",
"output": "28"
},
{
"input": "10 10 18",
"output": "2"
},
{
"input": "15 3 10",
"output": "3"
},
{
"input": "1000000000 1000000000 1000000000",
"output": "1000000000"
},
{
"input": "5 5 5",
"output": "5"
},
{
"input": "10 3 12",
"output": "1"
},
{
"input": "747 457 789",
"output": "415"
},
{
"input": "5 4 7",
"output": "2"
},
{
"input": "15 5 11",
"output": "5"
},
{
"input": "3 2 2",
"output": "2"
},
{
"input": "7 6 8",
"output": "5"
},
{
"input": "7 4 8",
"output": "3"
},
{
"input": "10 4 13",
"output": "1"
},
{
"input": "10 3 9",
"output": "3"
},
{
"input": "20 2 21",
"output": "1"
},
{
"input": "6 5 9",
"output": "2"
},
{
"input": "10 9 18",
"output": "1"
},
{
"input": "12 4 9",
"output": "4"
},
{
"input": "10 7 15",
"output": "2"
},
{
"input": "999999999 999999998 1500000000",
"output": "499999997"
},
{
"input": "20 5 20",
"output": "5"
},
{
"input": "4745 4574 4757",
"output": "4562"
},
{
"input": "10 7 12",
"output": "5"
},
{
"input": "17 15 18",
"output": "14"
},
{
"input": "3 1 3",
"output": "1"
},
{
"input": "100 3 7",
"output": "3"
},
{
"input": "6 2 7",
"output": "1"
},
{
"input": "8 5 10",
"output": "3"
},
{
"input": "3 3 3",
"output": "3"
},
{
"input": "9 5 10",
"output": "4"
},
{
"input": "10 6 13",
"output": "3"
},
{
"input": "13 10 14",
"output": "9"
},
{
"input": "13 12 15",
"output": "10"
},
{
"input": "10 4 12",
"output": "2"
},
{
"input": "41 3 3",
"output": "3"
},
{
"input": "1000000000 1000000000 1400000000",
"output": "600000000"
},
{
"input": "10 3 11",
"output": "2"
},
{
"input": "12 7 18",
"output": "1"
},
{
"input": "15 3 17",
"output": "1"
},
{
"input": "10 2 8",
"output": "2"
},
{
"input": "1000000000 1000 1000000999",
"output": "1"
},
{
"input": "5 5 9",
"output": "1"
},
{
"input": "100 3 6",
"output": "3"
},
{
"input": "100 5 50",
"output": "5"
},
{
"input": "10000 10 10000",
"output": "10"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "6 4 4",
"output": "4"
},
{
"input": "9979797 555554 10101010",
"output": "434341"
},
{
"input": "13 5 12",
"output": "5"
},
{
"input": "9 4 10",
"output": "3"
},
{
"input": "7 5 10",
"output": "2"
},
{
"input": "100000000 10000000 100005000",
"output": "9995000"
},
{
"input": "100000 50000 100001",
"output": "49999"
},
{
"input": "15 10 20",
"output": "5"
},
{
"input": "4 4 5",
"output": "3"
},
{
"input": "5 3 3",
"output": "3"
},
{
"input": "30 5 30",
"output": "5"
},
{
"input": "200000 10 200005",
"output": "5"
},
{
"input": "10 9 12",
"output": "7"
},
{
"input": "10 6 15",
"output": "1"
},
{
"input": "1000000000 10 1000000000",
"output": "10"
},
{
"input": "7 5 11",
"output": "1"
},
{
"input": "9 4 4",
"output": "4"
},
{
"input": "14 3 15",
"output": "2"
},
{
"input": "1000000000 100000000 1000000000",
"output": "100000000"
},
{
"input": "40 10 22",
"output": "10"
},
{
"input": "50 10 51",
"output": "9"
},
{
"input": "999999997 999999995 1999999991",
"output": "1"
},
{
"input": "92 79 144",
"output": "27"
},
{
"input": "8 4 4",
"output": "4"
}
] | 1,568,734,507
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 166
| 140
| 102,400
|
n,k,t= input().split()
n= int(n)
k= int(k)
t= int(t)
if t<=k:
print(t)
elif t>k and t<n:
print(k)
else:
print(k-t+n)
|
Title: Arpa and a research in Mexican wave
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Arpa is researching the Mexican wave.
There are *n* spectators in the stadium, labeled from 1 to *n*. They start the Mexican wave at time 0.
- At time 1, the first spectator stands. - At time 2, the second spectator stands. - ... - At time *k*, the *k*-th spectator stands. - At time *k*<=+<=1, the (*k*<=+<=1)-th spectator stands and the first spectator sits. - At time *k*<=+<=2, the (*k*<=+<=2)-th spectator stands and the second spectator sits. - ... - At time *n*, the *n*-th spectator stands and the (*n*<=-<=*k*)-th spectator sits. - At time *n*<=+<=1, the (*n*<=+<=1<=-<=*k*)-th spectator sits. - ... - At time *n*<=+<=*k*, the *n*-th spectator sits.
Arpa wants to know how many spectators are standing at time *t*.
Input Specification:
The first line contains three integers *n*, *k*, *t* (1<=≤<=*n*<=≤<=109, 1<=≤<=*k*<=≤<=*n*, 1<=≤<=*t*<=<<=*n*<=+<=*k*).
Output Specification:
Print single integer: how many spectators are standing at time *t*.
Demo Input:
['10 5 3\n', '10 5 7\n', '10 5 12\n']
Demo Output:
['3\n', '5\n', '3\n']
Note:
In the following a sitting spectator is represented as -, a standing spectator is represented as ^.
- At *t* = 0 ---------- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 0. - At *t* = 1 ^--------- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 1. - At *t* = 2 ^^-------- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 2. - At *t* = 3 ^^^------- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 3. - At *t* = 4 ^^^^------ <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 4. - At *t* = 5 ^^^^^----- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 5. - At *t* = 6 -^^^^^---- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 5. - At *t* = 7 --^^^^^--- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 5. - At *t* = 8 ---^^^^^-- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 5. - At *t* = 9 ----^^^^^- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 5. - At *t* = 10 -----^^^^^ <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 5. - At *t* = 11 ------^^^^ <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 4. - At *t* = 12 -------^^^ <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 3. - At *t* = 13 --------^^ <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 2. - At *t* = 14 ---------^ <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 1. - At *t* = 15 ---------- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 0.
|
```python
n,k,t= input().split()
n= int(n)
k= int(k)
t= int(t)
if t<=k:
print(t)
elif t>k and t<n:
print(k)
else:
print(k-t+n)
```
| 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"
}
] | 1,694,164,837
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 27
| 92
| 0
|
def is_distinct_year(year):
year_str = str(year)
return len(year_str) == len(set(year_str))
def find_next_distinct_year(y):
while True:
y += 1
if is_distinct_year(y):
return y
y = int(input())
next_distinct_year = find_next_distinct_year(y)
print(next_distinct_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
def is_distinct_year(year):
year_str = str(year)
return len(year_str) == len(set(year_str))
def find_next_distinct_year(y):
while True:
y += 1
if is_distinct_year(y):
return y
y = int(input())
next_distinct_year = find_next_distinct_year(y)
print(next_distinct_year)
```
| 3
|
|
780
|
A
|
Andryusha and Socks
|
PROGRAMMING
| 800
|
[
"implementation"
] | 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.
| 500
|
[
{
"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"
}
] | 1,563,430,089
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 56
| 218
| 13,619,200
|
n = int(input())
x = map(int, input().split())
Max = 0
tmp = set()
counter = 0
for i in x:
if i in tmp:
counter -= 1
else:
counter += 1
tmp.add(i)
if Max < counter:
Max = counter
print(Max)
# CodeForcesian
# ♥
# تا خدا بنده نواز است
|
Title: Andryusha and Socks
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
n = int(input())
x = map(int, input().split())
Max = 0
tmp = set()
counter = 0
for i in x:
if i in tmp:
counter -= 1
else:
counter += 1
tmp.add(i)
if Max < counter:
Max = counter
print(Max)
# CodeForcesian
# ♥
# تا خدا بنده نواز است
```
| 3
|
|
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": "//"
}
] | 1,676,361,369
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 31
| 31
| 0
|
str1 = input().strip()
str2 = input().strip()
s = "qwertyuiopasdfghjkl;zxcvbnm,./"
msg = ""
idx = 1 if str1 == 'L' else -1
for ch in str2:
current = s.index(ch)
new = s[idx + current]
msg += new
print(msg)
|
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
str1 = input().strip()
str2 = input().strip()
s = "qwertyuiopasdfghjkl;zxcvbnm,./"
msg = ""
idx = 1 if str1 == 'L' else -1
for ch in str2:
current = s.index(ch)
new = s[idx + current]
msg += new
print(msg)
```
| 3
|
|
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"
}
] | 1,663,946,507
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 45
| 92
| 0
|
x, y = map(int, input().split())
sieve = [True]*(y+1)
for i in range(2, y+1, 1):
if not sieve[i]: continue
t = 2*i
while(t < y+1):
sieve[t] = False
t += i
a = True
p = 0
for i in range(x+1, y+1, 1):
if sieve[i]:
p = i
break
if p == y:
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
x, y = map(int, input().split())
sieve = [True]*(y+1)
for i in range(2, y+1, 1):
if not sieve[i]: continue
t = 2*i
while(t < y+1):
sieve[t] = False
t += i
a = True
p = 0
for i in range(x+1, y+1, 1):
if sieve[i]:
p = i
break
if p == y:
print("YES")
else:
print("NO")
```
| 3.977
|
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"
}
] | 1,681,449,290
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 28
| 46
| 0
|
n, m = map(lambda x: int(x), input().split())
d = {}
while n != 0:
n -= 1
s = input().split()
d[s[1]] = s[0]
while m != 0:
m -= 1
l = input().split()
l.append('#' + d[l[1][:-1]])
print(' '.join(l))
|
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, m = map(lambda x: int(x), input().split())
d = {}
while n != 0:
n -= 1
s = input().split()
d[s[1]] = s[0]
while m != 0:
m -= 1
l = input().split()
l.append('#' + d[l[1][:-1]])
print(' '.join(l))
```
| 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"
}
] | 1,666,768,141
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 58
| 46
| 0
|
s = input()
heidi = 'heidi#'
cnt = 0
n = len(s)
for i in range(n):
if heidi[cnt] == s[i]:
cnt += 1
if cnt ==5:
print("YES")
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()
heidi = 'heidi#'
cnt = 0
n = len(s)
for i in range(n):
if heidi[cnt] == s[i]:
cnt += 1
if cnt ==5:
print("YES")
else:
print("NO")
```
| 3
|
|
586
|
A
|
Alena's Schedule
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Alena has successfully passed the entrance exams to the university and is now looking forward to start studying.
One two-hour lesson at the Russian university is traditionally called a pair, it lasts for two academic hours (an academic hour is equal to 45 minutes).
The University works in such a way that every day it holds exactly *n* lessons. Depending on the schedule of a particular group of students, on a given day, some pairs may actually contain classes, but some may be empty (such pairs are called breaks).
The official website of the university has already published the schedule for tomorrow for Alena's group. Thus, for each of the *n* pairs she knows if there will be a class at that time or not.
Alena's House is far from the university, so if there are breaks, she doesn't always go home. Alena has time to go home only if the break consists of at least two free pairs in a row, otherwise she waits for the next pair at the university.
Of course, Alena does not want to be sleepy during pairs, so she will sleep as long as possible, and will only come to the first pair that is presented in her schedule. Similarly, if there are no more pairs, then Alena immediately goes home.
Alena appreciates the time spent at home, so she always goes home when it is possible, and returns to the university only at the beginning of the next pair. Help Alena determine for how many pairs she will stay at the university. Note that during some pairs Alena may be at the university waiting for the upcoming pair.
|
The first line of the input contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of lessons at the university.
The second line contains *n* numbers *a**i* (0<=≤<=*a**i*<=≤<=1). Number *a**i* equals 0, if Alena doesn't have the *i*-th pairs, otherwise it is equal to 1. Numbers *a*1,<=*a*2,<=...,<=*a**n* are separated by spaces.
|
Print a single number — the number of pairs during which Alena stays at the university.
|
[
"5\n0 1 0 1 1\n",
"7\n1 0 1 0 0 1 0\n",
"1\n0\n"
] |
[
"4\n",
"4\n",
"0\n"
] |
In the first sample Alena stays at the university from the second to the fifth pair, inclusive, during the third pair she will be it the university waiting for the next pair.
In the last sample Alena doesn't have a single pair, so she spends all the time at home.
| 500
|
[
{
"input": "5\n0 1 0 1 1",
"output": "4"
},
{
"input": "7\n1 0 1 0 0 1 0",
"output": "4"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "2\n0 0",
"output": "0"
},
{
"input": "2\n0 1",
"output": "1"
},
{
"input": "2\n1 0",
"output": "1"
},
{
"input": "2\n1 1",
"output": "2"
},
{
"input": "10\n0 0 0 0 0 0 0 0 0 0",
"output": "0"
},
{
"input": "9\n1 1 1 1 1 1 1 1 1",
"output": "9"
},
{
"input": "11\n0 0 0 0 0 0 0 0 0 0 1",
"output": "1"
},
{
"input": "12\n1 0 0 0 0 0 0 0 0 0 0 0",
"output": "1"
},
{
"input": "20\n1 1 0 1 1 1 1 1 1 1 1 0 1 1 1 0 0 1 0 0",
"output": "16"
},
{
"input": "41\n1 1 0 1 0 1 0 0 1 0 1 1 1 0 0 0 1 1 1 0 1 0 1 1 0 1 0 1 0 0 0 0 0 0 1 0 0 1 0 1 1",
"output": "28"
},
{
"input": "63\n1 1 0 1 1 0 0 0 1 1 0 0 1 1 1 1 0 1 1 0 1 0 1 1 1 1 1 0 0 0 0 0 0 1 0 0 1 0 0 1 0 1 1 0 0 1 1 0 0 1 1 1 1 0 0 1 1 0 0 1 0 1 0",
"output": "39"
},
{
"input": "80\n0 1 1 1 0 1 1 1 1 1 0 0 1 0 1 1 0 1 1 1 0 1 1 1 1 0 1 0 1 0 0 0 1 1 0 1 1 0 0 0 0 1 1 1 0 0 0 1 0 0 1 1 1 0 0 0 0 0 0 1 0 1 0 0 1 0 1 1 1 1 1 0 0 0 1 1 0 0 1 1",
"output": "52"
},
{
"input": "99\n1 1 0 0 0 1 0 0 1 1 1 1 0 0 0 1 0 1 1 0 1 1 1 1 0 0 0 0 1 1 1 1 0 1 0 1 0 1 1 1 0 0 1 1 1 0 0 0 1 1 1 0 1 1 1 0 1 0 0 1 1 0 1 0 0 1 1 1 1 0 0 1 1 0 0 1 0 1 1 1 0 1 1 0 1 0 0 1 0 1 0 1 0 1 1 0 1 0 1",
"output": "72"
},
{
"input": "100\n0 1 1 0 1 1 0 0 1 1 0 1 1 1 1 1 0 0 1 1 1 0 0 0 0 1 1 0 0 1 0 0 1 0 0 0 0 1 1 1 1 1 1 0 0 1 1 0 0 0 0 1 0 1 1 1 0 1 1 0 1 0 0 0 0 0 1 0 1 1 0 0 1 1 0 1 1 0 0 1 1 1 0 1 1 1 1 1 1 0 0 1 1 1 1 0 1 1 1 0",
"output": "65"
},
{
"input": "11\n0 1 1 0 0 0 0 0 0 0 0",
"output": "2"
},
{
"input": "11\n0 1 0 1 0 0 1 1 0 1 1",
"output": "8"
},
{
"input": "11\n1 0 1 0 1 1 0 1 1 1 0",
"output": "10"
},
{
"input": "11\n1 0 0 0 0 0 1 0 1 1 1",
"output": "6"
},
{
"input": "22\n0 1 1 0 0 0 0 0 1 0 0 0 0 1 1 0 0 1 0 0 1 0",
"output": "7"
},
{
"input": "22\n0 1 0 1 0 1 1 1 1 0 0 1 1 1 0 1 1 1 0 0 0 1",
"output": "16"
},
{
"input": "22\n1 0 1 0 1 0 0 0 0 0 0 1 0 0 0 0 1 1 0 1 1 0",
"output": "11"
},
{
"input": "22\n1 0 1 0 0 0 1 0 0 1 1 0 1 0 1 1 0 0 0 1 0 1",
"output": "14"
},
{
"input": "33\n0 1 1 0 1 1 0 1 0 1 1 0 1 1 1 1 0 1 1 1 0 0 1 1 0 0 1 1 0 1 1 0 0",
"output": "26"
},
{
"input": "33\n0 1 0 1 0 1 1 0 0 0 1 1 1 0 1 0 1 1 0 1 0 1 0 0 1 1 1 0 1 1 1 0 1",
"output": "27"
},
{
"input": "33\n1 0 1 0 1 0 0 0 1 0 1 1 1 0 0 0 0 1 1 0 1 0 1 1 0 1 0 1 1 1 1 1 0",
"output": "25"
},
{
"input": "33\n1 0 1 0 1 1 1 1 1 0 1 0 1 1 0 0 1 0 1 0 0 0 1 0 1 0 1 0 0 0 0 1 1",
"output": "24"
},
{
"input": "44\n0 1 1 0 1 0 0 0 0 1 1 0 0 0 0 0 1 1 1 0 0 0 0 0 1 0 0 1 1 0 0 0 0 1 1 1 0 0 1 0 1 1 0 0",
"output": "19"
},
{
"input": "44\n0 1 1 1 1 0 1 0 0 1 0 1 0 0 1 1 0 1 1 0 0 1 0 1 0 1 1 0 1 0 1 0 1 0 1 0 0 0 0 0 1 0 1 1",
"output": "32"
},
{
"input": "44\n1 0 1 0 0 1 0 1 0 1 0 1 0 1 1 1 1 1 0 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 1 0 0 1 0 0 1 0",
"output": "23"
},
{
"input": "44\n1 0 1 0 1 1 1 0 0 0 0 0 0 1 0 0 0 1 1 1 0 0 0 1 0 1 0 1 1 0 1 1 0 1 0 1 1 0 1 0 1 1 1 1",
"output": "32"
},
{
"input": "55\n0 1 1 0 1 0 0 0 1 0 0 0 1 0 1 0 0 1 0 0 0 0 1 1 1 0 0 1 1 0 0 0 0 0 1 0 1 1 1 0 0 0 0 0 1 0 0 0 0 1 1 0 0 1 0",
"output": "23"
},
{
"input": "55\n0 1 1 0 1 0 1 1 1 1 0 1 1 0 0 1 1 1 0 0 0 1 1 0 0 1 0 1 0 1 0 0 1 1 1 1 1 0 0 0 1 1 1 1 1 1 1 1 0 1 1 0 0 0 1",
"output": "39"
},
{
"input": "55\n1 0 1 0 0 1 0 0 1 1 0 1 0 1 0 0 1 1 0 0 0 0 1 1 1 0 0 0 1 1 1 1 0 1 0 1 0 0 0 1 0 1 1 0 0 0 1 0 1 0 0 1 1 0 0",
"output": "32"
},
{
"input": "55\n1 0 1 0 1 0 1 0 1 1 0 0 1 1 1 1 0 1 0 0 0 1 1 0 0 1 0 1 0 1 1 1 0 0 0 0 0 0 1 0 0 0 1 1 1 0 0 0 1 0 1 0 1 1 1",
"output": "36"
},
{
"input": "66\n0 1 1 0 0 1 0 1 0 1 0 1 1 0 1 1 0 0 1 1 0 1 0 0 0 0 0 0 0 1 0 0 0 1 1 1 1 1 1 0 0 1 1 1 0 1 0 1 1 0 1 0 0 1 1 0 1 1 1 0 0 0 0 0 1 0",
"output": "41"
},
{
"input": "66\n0 1 1 0 1 1 1 0 0 0 1 1 0 1 1 0 0 1 1 1 1 1 0 1 1 1 0 1 1 1 0 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 0 0 1 0 0 0 0 0 1 0 1 0 0 1 0 0 1 1 0 1",
"output": "42"
},
{
"input": "66\n1 0 1 0 0 0 1 0 1 0 1 0 1 1 0 1 0 1 1 0 0 0 1 1 1 0 1 0 0 1 0 1 0 0 0 0 1 1 0 1 1 0 1 0 0 0 1 1 0 1 0 1 1 0 0 0 1 1 0 1 1 0 1 1 0 0",
"output": "46"
},
{
"input": "66\n1 0 1 0 0 0 1 1 1 1 1 0 1 0 0 0 1 1 1 0 1 1 0 1 0 1 0 0 1 0 0 1 0 1 0 1 0 0 1 0 0 1 0 1 1 1 1 1 0 1 1 1 1 1 1 0 0 0 1 0 1 1 0 0 0 1",
"output": "46"
},
{
"input": "77\n0 0 1 0 0 1 0 0 1 1 1 1 0 1 0 0 1 0 0 0 0 1 0 0 0 0 1 0 1 0 1 0 0 0 1 0 0 1 1 0 1 0 1 1 0 0 0 1 0 0 1 1 1 0 1 0 1 1 0 1 0 0 0 1 0 1 1 0 1 1 1 0 1 1 0 1 0",
"output": "47"
},
{
"input": "77\n0 0 1 0 0 0 1 0 1 1 1 1 0 1 1 1 0 1 1 0 1 1 1 0 1 1 0 1 0 0 0 0 1 1 0 0 0 1 1 0 0 1 1 0 1 0 0 1 0 0 0 1 0 0 1 0 0 0 1 0 0 0 1 0 0 1 0 1 1 0 1 0 0 0 0 1 1",
"output": "44"
},
{
"input": "77\n1 0 0 0 1 0 1 1 0 0 1 0 0 0 1 1 1 1 0 1 0 0 0 0 0 0 1 1 0 0 0 1 0 1 0 1 1 1 0 1 1 1 0 0 0 1 1 0 1 1 1 0 1 1 0 0 1 0 0 1 1 1 1 0 1 0 0 0 1 0 1 1 0 0 0 0 0",
"output": "45"
},
{
"input": "77\n1 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 0 1 1 1 0 0 1 1 1 0 1 1 0 1 0 0 0 0 1 1 1 0 1 0 0 1 1 0 1 0 1 1 1 1 1 1 1 0 0 1 1 0 0 1 0 1 1 1 1 1 1 1 1 0 0 1 0 1 1",
"output": "51"
},
{
"input": "88\n0 0 1 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 1 0 1 1 1 0 0 1 1 0 0 1 0 1 1 1 0 1 1 1 0 1 1 1 1 0 0 0 0 1 0 0 0 1 0 1 1 0 1 0 1 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 1 1 1 0",
"output": "44"
},
{
"input": "88\n0 0 1 0 0 0 1 1 0 1 0 1 1 1 0 1 1 1 0 1 1 1 1 1 0 1 1 0 1 1 1 0 1 0 0 1 0 1 1 0 0 0 0 0 1 1 0 0 1 0 1 1 1 0 1 1 0 1 1 0 0 0 1 1 1 1 1 1 1 0 0 1 0 1 0 0 0 1 0 1 0 0 0 0 0 0 0 1",
"output": "59"
},
{
"input": "88\n1 0 0 0 1 1 1 0 1 1 0 0 0 0 0 0 1 0 1 0 0 0 1 1 0 0 1 1 1 1 1 1 0 0 0 0 0 1 0 1 0 0 0 0 0 1 1 1 0 1 1 0 0 1 1 1 0 0 1 0 0 1 1 1 1 0 0 1 0 1 1 1 0 1 0 1 1 1 1 0 1 0 1 1 1 0 0 0",
"output": "53"
},
{
"input": "88\n1 1 1 0 0 1 1 0 1 0 0 0 1 0 1 1 1 1 1 0 1 1 1 1 1 1 1 0 0 1 0 1 1 1 0 0 0 1 1 0 1 1 0 1 0 0 1 0 0 1 0 0 1 0 1 1 0 1 0 1 0 1 0 0 1 1 1 0 0 0 1 0 0 1 0 0 1 1 0 1 1 1 1 0 1 1 0 1",
"output": "63"
},
{
"input": "99\n0 0 0 0 1 0 0 1 0 0 0 1 1 1 1 1 1 0 1 1 0 1 0 0 1 0 1 1 1 1 1 0 1 0 1 1 1 0 0 0 1 0 0 1 0 1 0 1 0 0 0 0 1 0 1 1 0 0 1 1 1 0 0 1 1 0 0 0 0 0 0 1 0 0 0 1 1 0 0 0 1 1 1 0 1 1 0 1 0 1 0 0 0 1 1 0 0 0 0",
"output": "56"
},
{
"input": "99\n0 0 1 0 0 1 1 0 0 0 1 1 0 0 1 0 0 0 1 1 1 1 0 0 0 1 1 0 0 0 1 0 1 1 0 0 1 1 1 0 1 1 0 0 0 0 0 1 0 0 1 0 1 1 0 1 0 1 0 0 1 0 1 1 1 1 1 1 0 1 0 0 1 1 0 0 1 0 1 0 1 0 1 1 1 0 0 1 1 1 0 0 0 0 0 0 1 1 1",
"output": "58"
},
{
"input": "99\n1 1 0 0 1 1 1 0 0 0 1 0 1 0 1 1 0 0 0 0 0 0 0 0 1 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 0 1 0 1 1 1 1 0 1 1 1 0 0 1 0 0 1 1 0 0 0 0 1 0 0 1 0 1 1 0 1 1 0 0 1 0 0 1 0 1 0 1 1 0 1 0 1 1 1 1 0 0 1 0",
"output": "65"
},
{
"input": "99\n1 1 1 0 1 0 1 1 0 1 1 0 0 1 0 0 1 1 1 0 1 1 0 0 0 1 1 1 1 0 1 1 1 0 1 1 0 1 1 0 1 0 1 0 0 1 1 1 1 1 0 1 1 0 1 1 0 0 0 1 0 1 0 1 0 1 0 0 0 1 1 1 1 0 0 1 1 0 1 0 0 0 1 0 1 1 1 0 0 1 1 1 1 1 0 1 1 1 1",
"output": "77"
},
{
"input": "90\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",
"output": "0"
},
{
"input": "90\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",
"output": "90"
},
{
"input": "95\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",
"output": "0"
},
{
"input": "95\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",
"output": "95"
},
{
"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\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"
}
] | 1,444,641,506
| 506
|
Python 3
|
OK
|
TESTS
| 60
| 62
| 0
|
input()
pprev, prev = None, None
ans = 0
for i in input().split():
if i == '1':
ans += 1
if prev == '0' and pprev == '1':
ans += 1
pprev, prev = prev, i
print(ans)
|
Title: Alena's Schedule
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Alena has successfully passed the entrance exams to the university and is now looking forward to start studying.
One two-hour lesson at the Russian university is traditionally called a pair, it lasts for two academic hours (an academic hour is equal to 45 minutes).
The University works in such a way that every day it holds exactly *n* lessons. Depending on the schedule of a particular group of students, on a given day, some pairs may actually contain classes, but some may be empty (such pairs are called breaks).
The official website of the university has already published the schedule for tomorrow for Alena's group. Thus, for each of the *n* pairs she knows if there will be a class at that time or not.
Alena's House is far from the university, so if there are breaks, she doesn't always go home. Alena has time to go home only if the break consists of at least two free pairs in a row, otherwise she waits for the next pair at the university.
Of course, Alena does not want to be sleepy during pairs, so she will sleep as long as possible, and will only come to the first pair that is presented in her schedule. Similarly, if there are no more pairs, then Alena immediately goes home.
Alena appreciates the time spent at home, so she always goes home when it is possible, and returns to the university only at the beginning of the next pair. Help Alena determine for how many pairs she will stay at the university. Note that during some pairs Alena may be at the university waiting for the upcoming pair.
Input Specification:
The first line of the input contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of lessons at the university.
The second line contains *n* numbers *a**i* (0<=≤<=*a**i*<=≤<=1). Number *a**i* equals 0, if Alena doesn't have the *i*-th pairs, otherwise it is equal to 1. Numbers *a*1,<=*a*2,<=...,<=*a**n* are separated by spaces.
Output Specification:
Print a single number — the number of pairs during which Alena stays at the university.
Demo Input:
['5\n0 1 0 1 1\n', '7\n1 0 1 0 0 1 0\n', '1\n0\n']
Demo Output:
['4\n', '4\n', '0\n']
Note:
In the first sample Alena stays at the university from the second to the fifth pair, inclusive, during the third pair she will be it the university waiting for the next pair.
In the last sample Alena doesn't have a single pair, so she spends all the time at home.
|
```python
input()
pprev, prev = None, None
ans = 0
for i in input().split():
if i == '1':
ans += 1
if prev == '0' and pprev == '1':
ans += 1
pprev, prev = prev, i
print(ans)
```
| 3
|
|
1,006
|
B
|
Polycarp's Practice
|
PROGRAMMING
| 1,200
|
[
"greedy",
"implementation",
"sortings"
] | null | null |
Polycarp is practicing his problem solving skill. He has a list of $n$ problems with difficulties $a_1, a_2, \dots, a_n$, respectively. His plan is to practice for exactly $k$ days. Each day he has to solve at least one problem from his list. Polycarp solves the problems in the order they are given in his list, he cannot skip any problem from his list. He has to solve all $n$ problems in exactly $k$ days.
Thus, each day Polycarp solves a contiguous sequence of (consecutive) problems from the start of the list. He can't skip problems or solve them multiple times. As a result, in $k$ days he will solve all the $n$ problems.
The profit of the $j$-th day of Polycarp's practice is the maximum among all the difficulties of problems Polycarp solves during the $j$-th day (i.e. if he solves problems with indices from $l$ to $r$ during a day, then the profit of the day is $\max\limits_{l \le i \le r}a_i$). The total profit of his practice is the sum of the profits over all $k$ days of his practice.
You want to help Polycarp to get the maximum possible total profit over all valid ways to solve problems. Your task is to distribute all $n$ problems between $k$ days satisfying the conditions above in such a way, that the total profit is maximum.
For example, if $n = 8, k = 3$ and $a = [5, 4, 2, 6, 5, 1, 9, 2]$, one of the possible distributions with maximum total profit is: $[5, 4, 2], [6, 5], [1, 9, 2]$. Here the total profit equals $5 + 6 + 9 = 20$.
|
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2000$) — the number of problems and the number of days, respectively.
The second line of the input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 2000$) — difficulties of problems in Polycarp's list, in the order they are placed in the list (i.e. in the order Polycarp will solve them).
|
In the first line of the output print the maximum possible total profit.
In the second line print exactly $k$ positive integers $t_1, t_2, \dots, t_k$ ($t_1 + t_2 + \dots + t_k$ must equal $n$), where $t_j$ means the number of problems Polycarp will solve during the $j$-th day in order to achieve the maximum possible total profit of his practice.
If there are many possible answers, you may print any of them.
|
[
"8 3\n5 4 2 6 5 1 9 2\n",
"5 1\n1 1 1 1 1\n",
"4 2\n1 2000 2000 2\n"
] |
[
"20\n3 2 3",
"1\n5\n",
"4000\n2 2\n"
] |
The first example is described in the problem statement.
In the second example there is only one possible distribution.
In the third example the best answer is to distribute problems in the following way: $[1, 2000], [2000, 2]$. The total profit of this distribution is $2000 + 2000 = 4000$.
| 0
|
[
{
"input": "8 3\n5 4 2 6 5 1 9 2",
"output": "20\n4 1 3"
},
{
"input": "5 1\n1 1 1 1 1",
"output": "1\n5"
},
{
"input": "4 2\n1 2000 2000 2",
"output": "4000\n2 2"
},
{
"input": "1 1\n2000",
"output": "2000\n1"
},
{
"input": "1 1\n1234",
"output": "1234\n1"
},
{
"input": "3 2\n1 1 1",
"output": "2\n2 1"
},
{
"input": "4 2\n3 5 1 1",
"output": "8\n1 3"
},
{
"input": "5 3\n5 5 6 7 1",
"output": "18\n2 1 2"
},
{
"input": "6 4\n1 1 1 1 2 2",
"output": "6\n3 1 1 1"
},
{
"input": "5 3\n5 5 6 6 4",
"output": "17\n2 1 2"
},
{
"input": "16 15\n14 4 9 12 17 1 1 8 12 13 6 9 17 2 18 12",
"output": "154\n1 1 1 1 1 2 1 1 1 1 1 1 1 1 1"
},
{
"input": "1 1\n1996",
"output": "1996\n1"
},
{
"input": "5 3\n5 5 5 9 10",
"output": "24\n3 1 1"
},
{
"input": "18 15\n18 2 13 1 18 3 2 18 18 20 9 2 20 20 4 20 9 12",
"output": "204\n1 2 2 1 2 1 1 1 1 1 1 1 1 1 1"
},
{
"input": "5 3\n1 20 20 50 50",
"output": "120\n3 1 1"
},
{
"input": "8 3\n15 14 11 19 17 14 14 8",
"output": "51\n1 3 4"
},
{
"input": "5 2\n15 20 6 19 6",
"output": "39\n2 3"
},
{
"input": "6 3\n5 5 5 5 6 9",
"output": "20\n4 1 1"
},
{
"input": "5 3\n2 2 2 3 3",
"output": "8\n3 1 1"
},
{
"input": "7 3\n2 2 2 2 2 3 3",
"output": "8\n5 1 1"
},
{
"input": "6 5\n1 1 6 6 6 6",
"output": "25\n2 1 1 1 1"
},
{
"input": "8 4\n1 2 2 2 2 3 4 5",
"output": "14\n5 1 1 1"
},
{
"input": "6 4\n1 1 1 5 5 5",
"output": "16\n3 1 1 1"
},
{
"input": "6 3\n1 2 2 2 4 5",
"output": "11\n4 1 1"
},
{
"input": "18 6\n17 17 19 14 10 20 18 16 6 7 2 15 14 16 13 6 12 11",
"output": "107\n1 1 1 3 1 11"
},
{
"input": "6 3\n1 1 2 2 3 4",
"output": "9\n4 1 1"
},
{
"input": "8 3\n5 4 2 5 6 1 9 2",
"output": "20\n4 1 3"
}
] | 1,666,270,204
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 37
| 93
| 4,505,600
|
n, k = map(int, input().split())
numbers = list(map(int, input().split()))
maximums = sorted(numbers, reverse=True)[:k]
print(sum(maximums))
left_point = -1
right_point = 0
while right_point < n:
if numbers[right_point] in maximums:
maximums.remove(numbers[right_point])
if len(maximums) == 0:
right_point = n-1
# print(left_point, right_point)
print(right_point - left_point, end = " ")
left_point = right_point
right_point += 1
|
Title: Polycarp's Practice
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp is practicing his problem solving skill. He has a list of $n$ problems with difficulties $a_1, a_2, \dots, a_n$, respectively. His plan is to practice for exactly $k$ days. Each day he has to solve at least one problem from his list. Polycarp solves the problems in the order they are given in his list, he cannot skip any problem from his list. He has to solve all $n$ problems in exactly $k$ days.
Thus, each day Polycarp solves a contiguous sequence of (consecutive) problems from the start of the list. He can't skip problems or solve them multiple times. As a result, in $k$ days he will solve all the $n$ problems.
The profit of the $j$-th day of Polycarp's practice is the maximum among all the difficulties of problems Polycarp solves during the $j$-th day (i.e. if he solves problems with indices from $l$ to $r$ during a day, then the profit of the day is $\max\limits_{l \le i \le r}a_i$). The total profit of his practice is the sum of the profits over all $k$ days of his practice.
You want to help Polycarp to get the maximum possible total profit over all valid ways to solve problems. Your task is to distribute all $n$ problems between $k$ days satisfying the conditions above in such a way, that the total profit is maximum.
For example, if $n = 8, k = 3$ and $a = [5, 4, 2, 6, 5, 1, 9, 2]$, one of the possible distributions with maximum total profit is: $[5, 4, 2], [6, 5], [1, 9, 2]$. Here the total profit equals $5 + 6 + 9 = 20$.
Input Specification:
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2000$) — the number of problems and the number of days, respectively.
The second line of the input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 2000$) — difficulties of problems in Polycarp's list, in the order they are placed in the list (i.e. in the order Polycarp will solve them).
Output Specification:
In the first line of the output print the maximum possible total profit.
In the second line print exactly $k$ positive integers $t_1, t_2, \dots, t_k$ ($t_1 + t_2 + \dots + t_k$ must equal $n$), where $t_j$ means the number of problems Polycarp will solve during the $j$-th day in order to achieve the maximum possible total profit of his practice.
If there are many possible answers, you may print any of them.
Demo Input:
['8 3\n5 4 2 6 5 1 9 2\n', '5 1\n1 1 1 1 1\n', '4 2\n1 2000 2000 2\n']
Demo Output:
['20\n3 2 3', '1\n5\n', '4000\n2 2\n']
Note:
The first example is described in the problem statement.
In the second example there is only one possible distribution.
In the third example the best answer is to distribute problems in the following way: $[1, 2000], [2000, 2]$. The total profit of this distribution is $2000 + 2000 = 4000$.
|
```python
n, k = map(int, input().split())
numbers = list(map(int, input().split()))
maximums = sorted(numbers, reverse=True)[:k]
print(sum(maximums))
left_point = -1
right_point = 0
while right_point < n:
if numbers[right_point] in maximums:
maximums.remove(numbers[right_point])
if len(maximums) == 0:
right_point = n-1
# print(left_point, right_point)
print(right_point - left_point, end = " ")
left_point = right_point
right_point += 1
```
| 3
|
|
344
|
A
|
Magnets
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.
Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own.
Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed.
|
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100000) — the number of magnets. Then *n* lines follow. The *i*-th line (1<=≤<=*i*<=≤<=*n*) contains either characters "01", if Mike put the *i*-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position.
|
On the single line of the output print the number of groups of magnets.
|
[
"6\n10\n10\n10\n01\n10\n10\n",
"4\n01\n01\n10\n10\n"
] |
[
"3\n",
"2\n"
] |
The first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.
The second testcase has two groups, each consisting of two magnets.
| 500
|
[
{
"input": "6\n10\n10\n10\n01\n10\n10",
"output": "3"
},
{
"input": "4\n01\n01\n10\n10",
"output": "2"
},
{
"input": "1\n10",
"output": "1"
},
{
"input": "2\n01\n10",
"output": "2"
},
{
"input": "2\n10\n10",
"output": "1"
},
{
"input": "3\n10\n01\n10",
"output": "3"
},
{
"input": "1\n01",
"output": "1"
},
{
"input": "2\n01\n01",
"output": "1"
},
{
"input": "2\n10\n01",
"output": "2"
},
{
"input": "3\n01\n01\n01",
"output": "1"
},
{
"input": "3\n10\n10\n01",
"output": "2"
},
{
"input": "3\n01\n10\n10",
"output": "2"
},
{
"input": "115\n10\n10\n10\n10\n01\n01\n10\n10\n10\n01\n01\n10\n01\n01\n10\n10\n10\n01\n10\n01\n10\n10\n01\n01\n10\n10\n10\n10\n01\n10\n01\n01\n10\n10\n10\n10\n01\n10\n10\n10\n01\n10\n01\n10\n10\n10\n10\n01\n01\n01\n10\n10\n01\n01\n01\n10\n10\n01\n10\n01\n01\n01\n01\n10\n10\n01\n10\n01\n01\n01\n01\n01\n10\n01\n10\n10\n01\n01\n01\n10\n01\n01\n10\n10\n01\n01\n01\n01\n01\n10\n01\n10\n01\n10\n01\n01\n01\n10\n01\n10\n10\n01\n10\n10\n01\n01\n01\n10\n10\n10\n10\n10\n10\n10\n10",
"output": "55"
}
] | 1,696,006,574
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 31
| 374
| 614,400
|
n = int(input())
c = 0
s = ''
for i in range(n):
a = input()
s += a
for j in range(1,len(s)):
if s[j] == s[j-1]:
c += 1
print(c+1)
|
Title: Magnets
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.
Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own.
Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed.
Input Specification:
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100000) — the number of magnets. Then *n* lines follow. The *i*-th line (1<=≤<=*i*<=≤<=*n*) contains either characters "01", if Mike put the *i*-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position.
Output Specification:
On the single line of the output print the number of groups of magnets.
Demo Input:
['6\n10\n10\n10\n01\n10\n10\n', '4\n01\n01\n10\n10\n']
Demo Output:
['3\n', '2\n']
Note:
The first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.
The second testcase has two groups, each consisting of two magnets.
|
```python
n = int(input())
c = 0
s = ''
for i in range(n):
a = input()
s += a
for j in range(1,len(s)):
if s[j] == s[j-1]:
c += 1
print(c+1)
```
| 3
|
|
557
|
A
|
Ilya and Diplomas
|
PROGRAMMING
| 1,100
|
[
"greedy",
"implementation",
"math"
] | null | null |
Soon a school Olympiad in Informatics will be held in Berland, *n* schoolchildren will participate there.
At a meeting of the jury of the Olympiad it was decided that each of the *n* participants, depending on the results, will get a diploma of the first, second or third degree. Thus, each student will receive exactly one diploma.
They also decided that there must be given at least *min*1 and at most *max*1 diplomas of the first degree, at least *min*2 and at most *max*2 diplomas of the second degree, and at least *min*3 and at most *max*3 diplomas of the third degree.
After some discussion it was decided to choose from all the options of distributing diplomas satisfying these limitations the one that maximizes the number of participants who receive diplomas of the first degree. Of all these options they select the one which maximizes the number of the participants who receive diplomas of the second degree. If there are multiple of these options, they select the option that maximizes the number of diplomas of the third degree.
Choosing the best option of distributing certificates was entrusted to Ilya, one of the best programmers of Berland. However, he found more important things to do, so it is your task now to choose the best option of distributing of diplomas, based on the described limitations.
It is guaranteed that the described limitations are such that there is a way to choose such an option of distributing diplomas that all *n* participants of the Olympiad will receive a diploma of some degree.
|
The first line of the input contains a single integer *n* (3<=≤<=*n*<=≤<=3·106) — the number of schoolchildren who will participate in the Olympiad.
The next line of the input contains two integers *min*1 and *max*1 (1<=≤<=*min*1<=≤<=*max*1<=≤<=106) — the minimum and maximum limits on the number of diplomas of the first degree that can be distributed.
The third line of the input contains two integers *min*2 and *max*2 (1<=≤<=*min*2<=≤<=*max*2<=≤<=106) — the minimum and maximum limits on the number of diplomas of the second degree that can be distributed.
The next line of the input contains two integers *min*3 and *max*3 (1<=≤<=*min*3<=≤<=*max*3<=≤<=106) — the minimum and maximum limits on the number of diplomas of the third degree that can be distributed.
It is guaranteed that *min*1<=+<=*min*2<=+<=*min*3<=≤<=*n*<=≤<=*max*1<=+<=*max*2<=+<=*max*3.
|
In the first line of the output print three numbers, showing how many diplomas of the first, second and third degree will be given to students in the optimal variant of distributing diplomas.
The optimal variant of distributing diplomas is the one that maximizes the number of students who receive diplomas of the first degree. Of all the suitable options, the best one is the one which maximizes the number of participants who receive diplomas of the second degree. If there are several of these options, the best one is the one that maximizes the number of diplomas of the third degree.
|
[
"6\n1 5\n2 6\n3 7\n",
"10\n1 2\n1 3\n1 5\n",
"6\n1 3\n2 2\n2 2\n"
] |
[
"1 2 3 \n",
"2 3 5 \n",
"2 2 2 \n"
] |
none
| 500
|
[
{
"input": "6\n1 5\n2 6\n3 7",
"output": "1 2 3 "
},
{
"input": "10\n1 2\n1 3\n1 5",
"output": "2 3 5 "
},
{
"input": "6\n1 3\n2 2\n2 2",
"output": "2 2 2 "
},
{
"input": "55\n1 1000000\n40 50\n10 200",
"output": "5 40 10 "
},
{
"input": "3\n1 1\n1 1\n1 1",
"output": "1 1 1 "
},
{
"input": "3\n1 1000000\n1 1000000\n1 1000000",
"output": "1 1 1 "
},
{
"input": "1000\n100 400\n300 500\n400 1200",
"output": "300 300 400 "
},
{
"input": "3000000\n1 1000000\n1 1000000\n1 1000000",
"output": "1000000 1000000 1000000 "
},
{
"input": "11\n3 5\n3 5\n3 5",
"output": "5 3 3 "
},
{
"input": "12\n3 5\n3 5\n3 5",
"output": "5 4 3 "
},
{
"input": "13\n3 5\n3 5\n3 5",
"output": "5 5 3 "
},
{
"input": "3000000\n1000000 1000000\n1000000 1000000\n1000000 1000000",
"output": "1000000 1000000 1000000 "
},
{
"input": "50\n1 100\n1 100\n1 100",
"output": "48 1 1 "
},
{
"input": "1279\n123 670\n237 614\n846 923",
"output": "196 237 846 "
},
{
"input": "1589\n213 861\n5 96\n506 634",
"output": "861 96 632 "
},
{
"input": "2115\n987 987\n112 483\n437 959",
"output": "987 483 645 "
},
{
"input": "641\n251 960\n34 370\n149 149",
"output": "458 34 149 "
},
{
"input": "1655\n539 539\n10 425\n605 895",
"output": "539 425 691 "
},
{
"input": "1477\n210 336\n410 837\n448 878",
"output": "336 693 448 "
},
{
"input": "1707\n149 914\n190 422\n898 899",
"output": "619 190 898 "
},
{
"input": "1529\n515 515\n563 869\n169 451",
"output": "515 845 169 "
},
{
"input": "1543\n361 994\n305 407\n102 197",
"output": "994 407 142 "
},
{
"input": "1107\n471 849\n360 741\n71 473",
"output": "676 360 71 "
},
{
"input": "1629279\n267360 999930\n183077 674527\n202618 786988",
"output": "999930 426731 202618 "
},
{
"input": "1233589\n2850 555444\n500608 921442\n208610 607343",
"output": "524371 500608 208610 "
},
{
"input": "679115\n112687 183628\n101770 982823\n81226 781340",
"output": "183628 414261 81226 "
},
{
"input": "1124641\n117999 854291\n770798 868290\n76651 831405",
"output": "277192 770798 76651 "
},
{
"input": "761655\n88152 620061\n60403 688549\n79370 125321",
"output": "620061 62224 79370 "
},
{
"input": "2174477\n276494 476134\n555283 954809\n319941 935631",
"output": "476134 954809 743534 "
},
{
"input": "1652707\n201202 990776\n34796 883866\n162979 983308",
"output": "990776 498952 162979 "
},
{
"input": "2065529\n43217 891429\n434379 952871\n650231 855105",
"output": "891429 523869 650231 "
},
{
"input": "1702543\n405042 832833\n50931 747750\n381818 796831",
"output": "832833 487892 381818 "
},
{
"input": "501107\n19061 859924\n126478 724552\n224611 489718",
"output": "150018 126478 224611 "
},
{
"input": "1629279\n850831 967352\n78593 463906\n452094 885430",
"output": "967352 209833 452094 "
},
{
"input": "1233589\n2850 157021\n535109 748096\n392212 475634",
"output": "157021 684356 392212 "
},
{
"input": "679115\n125987 786267\n70261 688983\n178133 976789",
"output": "430721 70261 178133 "
},
{
"input": "1124641\n119407 734250\n213706 860770\n102149 102149",
"output": "734250 288242 102149 "
},
{
"input": "761655\n325539 325539\n280794 792505\n18540 106895",
"output": "325539 417576 18540 "
},
{
"input": "2174477\n352351 791072\n365110 969163\n887448 955610",
"output": "791072 495957 887448 "
},
{
"input": "1652707\n266774 638522\n65688 235422\n924898 992826",
"output": "638522 89287 924898 "
},
{
"input": "2065529\n608515 608515\n751563 864337\n614898 705451",
"output": "608515 842116 614898 "
},
{
"input": "1702543\n5784 996578\n47395 300407\n151614 710197",
"output": "996578 300407 405558 "
},
{
"input": "501107\n8073 390048\n190494 647328\n274071 376923",
"output": "36542 190494 274071 "
},
{
"input": "200\n50 50\n100 100\n50 50",
"output": "50 100 50 "
},
{
"input": "14\n1 100\n1 100\n8 9",
"output": "5 1 8 "
},
{
"input": "300\n200 400\n50 100\n40 80",
"output": "210 50 40 "
},
{
"input": "10\n3 6\n3 6\n3 6",
"output": "4 3 3 "
},
{
"input": "14\n3 6\n3 6\n3 6",
"output": "6 5 3 "
},
{
"input": "17\n3 6\n3 6\n3 6",
"output": "6 6 5 "
},
{
"input": "1000000\n300000 600000\n300000 600000\n300000 600000",
"output": "400000 300000 300000 "
},
{
"input": "1400000\n300000 600000\n300000 600000\n300000 600000",
"output": "600000 500000 300000 "
},
{
"input": "1700000\n300000 600000\n300000 600000\n300000 600000",
"output": "600000 600000 500000 "
},
{
"input": "561\n400 400\n80 80\n81 81",
"output": "400 80 81 "
},
{
"input": "2000\n100 1000\n1 1\n1 2000",
"output": "1000 1 999 "
},
{
"input": "1000002\n1 1000000\n1 1000000\n999999 1000000",
"output": "2 1 999999 "
},
{
"input": "1000002\n1 1000000\n1 1000000\n1000000 1000000",
"output": "1 1 1000000 "
}
] | 1,568,190,156
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 56
| 171
| 1,843,200
|
import math
import itertools
import collections
def getdict(n):
d = {}
if type(n) is list:
for i in n:
if i in d:
d[i] += 1
else:
d[i] = 1
else:
for i in range(n):
t = ii()
if t in d:
d[t] += 1
else:
d[t] = 1
return d
def cdiv(n, k): return n // k + (n % k != 0)
def ii(): return int(input())
def mi(): return map(int, input().split())
def li(): return list(map(int, input().split()))
def lcm(a, b): return abs(a*b) // math.gcd(a, b)
def wr(arr): return ' '.join(map(str, arr))
def prime(n):
if n == 2:
return True
if n % 2 == 0 or n <= 1:
return False
sqr = int(math.sqrt(n)) + 1
for d in range(3, sqr, 2):
if n % d == 0:
return False
return True
def revn(n):
m = 0
while n > 0:
m = m * 10 + n % 10
n = n // 10
return m
n = ii()
mi1, ma1 = mi()
mi2, ma2 = mi()
mi3, ma3 = mi()
d = [mi1, mi2, mi3]
t = n - mi1 - mi2 - mi3
while t > 0 and d[0] < ma1:
d[0] += 1
t -= 1
while t > 0 and d[1] < ma2:
d[1] += 1
t -= 1
while t > 0 and d[2] < ma3:
d[2] += 1
t -= 1
print(wr(d))
|
Title: Ilya and Diplomas
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Soon a school Olympiad in Informatics will be held in Berland, *n* schoolchildren will participate there.
At a meeting of the jury of the Olympiad it was decided that each of the *n* participants, depending on the results, will get a diploma of the first, second or third degree. Thus, each student will receive exactly one diploma.
They also decided that there must be given at least *min*1 and at most *max*1 diplomas of the first degree, at least *min*2 and at most *max*2 diplomas of the second degree, and at least *min*3 and at most *max*3 diplomas of the third degree.
After some discussion it was decided to choose from all the options of distributing diplomas satisfying these limitations the one that maximizes the number of participants who receive diplomas of the first degree. Of all these options they select the one which maximizes the number of the participants who receive diplomas of the second degree. If there are multiple of these options, they select the option that maximizes the number of diplomas of the third degree.
Choosing the best option of distributing certificates was entrusted to Ilya, one of the best programmers of Berland. However, he found more important things to do, so it is your task now to choose the best option of distributing of diplomas, based on the described limitations.
It is guaranteed that the described limitations are such that there is a way to choose such an option of distributing diplomas that all *n* participants of the Olympiad will receive a diploma of some degree.
Input Specification:
The first line of the input contains a single integer *n* (3<=≤<=*n*<=≤<=3·106) — the number of schoolchildren who will participate in the Olympiad.
The next line of the input contains two integers *min*1 and *max*1 (1<=≤<=*min*1<=≤<=*max*1<=≤<=106) — the minimum and maximum limits on the number of diplomas of the first degree that can be distributed.
The third line of the input contains two integers *min*2 and *max*2 (1<=≤<=*min*2<=≤<=*max*2<=≤<=106) — the minimum and maximum limits on the number of diplomas of the second degree that can be distributed.
The next line of the input contains two integers *min*3 and *max*3 (1<=≤<=*min*3<=≤<=*max*3<=≤<=106) — the minimum and maximum limits on the number of diplomas of the third degree that can be distributed.
It is guaranteed that *min*1<=+<=*min*2<=+<=*min*3<=≤<=*n*<=≤<=*max*1<=+<=*max*2<=+<=*max*3.
Output Specification:
In the first line of the output print three numbers, showing how many diplomas of the first, second and third degree will be given to students in the optimal variant of distributing diplomas.
The optimal variant of distributing diplomas is the one that maximizes the number of students who receive diplomas of the first degree. Of all the suitable options, the best one is the one which maximizes the number of participants who receive diplomas of the second degree. If there are several of these options, the best one is the one that maximizes the number of diplomas of the third degree.
Demo Input:
['6\n1 5\n2 6\n3 7\n', '10\n1 2\n1 3\n1 5\n', '6\n1 3\n2 2\n2 2\n']
Demo Output:
['1 2 3 \n', '2 3 5 \n', '2 2 2 \n']
Note:
none
|
```python
import math
import itertools
import collections
def getdict(n):
d = {}
if type(n) is list:
for i in n:
if i in d:
d[i] += 1
else:
d[i] = 1
else:
for i in range(n):
t = ii()
if t in d:
d[t] += 1
else:
d[t] = 1
return d
def cdiv(n, k): return n // k + (n % k != 0)
def ii(): return int(input())
def mi(): return map(int, input().split())
def li(): return list(map(int, input().split()))
def lcm(a, b): return abs(a*b) // math.gcd(a, b)
def wr(arr): return ' '.join(map(str, arr))
def prime(n):
if n == 2:
return True
if n % 2 == 0 or n <= 1:
return False
sqr = int(math.sqrt(n)) + 1
for d in range(3, sqr, 2):
if n % d == 0:
return False
return True
def revn(n):
m = 0
while n > 0:
m = m * 10 + n % 10
n = n // 10
return m
n = ii()
mi1, ma1 = mi()
mi2, ma2 = mi()
mi3, ma3 = mi()
d = [mi1, mi2, mi3]
t = n - mi1 - mi2 - mi3
while t > 0 and d[0] < ma1:
d[0] += 1
t -= 1
while t > 0 and d[1] < ma2:
d[1] += 1
t -= 1
while t > 0 and d[2] < ma3:
d[2] += 1
t -= 1
print(wr(d))
```
| 3
|
|
11
|
B
|
Jumping Jack
|
PROGRAMMING
| 1,600
|
[
"math"
] |
B. Jumping Jack
|
1
|
64
|
Jack is working on his jumping skills recently. Currently he's located at point zero of the number line. He would like to get to the point *x*. In order to train, he has decided that he'll first jump by only one unit, and each subsequent jump will be exactly one longer than the previous one. He can go either left or right with each jump. He wonders how many jumps he needs to reach *x*.
|
The input data consists of only one integer *x* (<=-<=109<=≤<=*x*<=≤<=109).
|
Output the minimal number of jumps that Jack requires to reach *x*.
|
[
"2\n",
"6\n",
"0\n"
] |
[
"3\n",
"3\n",
"0\n"
] |
none
| 0
|
[
{
"input": "2",
"output": "3"
},
{
"input": "6",
"output": "3"
},
{
"input": "0",
"output": "0"
},
{
"input": "-1000000000",
"output": "44723"
},
{
"input": "999961560",
"output": "44720"
},
{
"input": "999961561",
"output": "44721"
},
{
"input": "999961559",
"output": "44721"
},
{
"input": "-488979819",
"output": "31273"
},
{
"input": "-865918189",
"output": "41617"
},
{
"input": "-278253406",
"output": "23591"
},
{
"input": "-87275307",
"output": "13213"
},
{
"input": "103702792",
"output": "14403"
},
{
"input": "294680891",
"output": "24277"
},
{
"input": "485658989",
"output": "31166"
},
{
"input": "676637088",
"output": "36787"
},
{
"input": "867615187",
"output": "41657"
},
{
"input": "-941406715",
"output": "43393"
},
{
"input": "-897912263",
"output": "42377"
},
{
"input": "-706934164",
"output": "37603"
},
{
"input": "-790244010",
"output": "39755"
},
{
"input": "16417015",
"output": "5730"
},
{
"input": "823078041",
"output": "40573"
},
{
"input": "-517744582",
"output": "32179"
},
{
"input": "288916443",
"output": "24038"
},
{
"input": "-904422532",
"output": "42531"
},
{
"input": "-245245154",
"output": "22147"
},
{
"input": "561415872",
"output": "33511"
},
{
"input": "-779406751",
"output": "39482"
},
{
"input": "27254274",
"output": "7383"
},
{
"input": "845249034",
"output": "41116"
}
] | 1,562,536,310
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 218
| 0
|
"""
Jack is working on his jumping skills recently. Currently he's located at point zero of the number line. He would like to get to the point x. In order to train, he has decided that he'll first jump by only one unit, and each subsequent jump will be exactly one longer than the previous one. He can go either left or right with each jump. He wonders how many jumps he needs to reach x.
Input
The input data consists of only one integer x ( - 109 ≤ x ≤ 109).
Output
Output the minimal number of jumps that Jack requires to reach x.
"""
from math import *
def is_even(a):
return False if a%2 else True
def calc_sum(n):
return n * (n + 1)/2
def check(a, x):
return (not is_even(a) and not is_even(x)) or (is_even(a) and is_even(x))
x = abs(int(input()))
if x == 0:
print(0)
else:
n = (-1+pow(1+4*2*x, 0.5))/2
ceil_n = ceil(n)
if check(calc_sum(ceil_n), x):
print(ceil_n)
else:
if check(calc_sum(ceil_n + 1), x):
print(ceil_n + 1)
else:
print(ceil_n + 2)
|
Title: Jumping Jack
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
Jack is working on his jumping skills recently. Currently he's located at point zero of the number line. He would like to get to the point *x*. In order to train, he has decided that he'll first jump by only one unit, and each subsequent jump will be exactly one longer than the previous one. He can go either left or right with each jump. He wonders how many jumps he needs to reach *x*.
Input Specification:
The input data consists of only one integer *x* (<=-<=109<=≤<=*x*<=≤<=109).
Output Specification:
Output the minimal number of jumps that Jack requires to reach *x*.
Demo Input:
['2\n', '6\n', '0\n']
Demo Output:
['3\n', '3\n', '0\n']
Note:
none
|
```python
"""
Jack is working on his jumping skills recently. Currently he's located at point zero of the number line. He would like to get to the point x. In order to train, he has decided that he'll first jump by only one unit, and each subsequent jump will be exactly one longer than the previous one. He can go either left or right with each jump. He wonders how many jumps he needs to reach x.
Input
The input data consists of only one integer x ( - 109 ≤ x ≤ 109).
Output
Output the minimal number of jumps that Jack requires to reach x.
"""
from math import *
def is_even(a):
return False if a%2 else True
def calc_sum(n):
return n * (n + 1)/2
def check(a, x):
return (not is_even(a) and not is_even(x)) or (is_even(a) and is_even(x))
x = abs(int(input()))
if x == 0:
print(0)
else:
n = (-1+pow(1+4*2*x, 0.5))/2
ceil_n = ceil(n)
if check(calc_sum(ceil_n), x):
print(ceil_n)
else:
if check(calc_sum(ceil_n + 1), x):
print(ceil_n + 1)
else:
print(ceil_n + 2)
```
| 3.891
|
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"
}
] | 1,694,096,553
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 81
| 154
| 0
|
n = int(input())
a = 0
b = 0
c = 0
for _ in range(n):
m, n, k = map(int,input().split())
a = a + m
b = b + n
c = c + k
if a == 0 and b == 0 and c == 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())
a = 0
b = 0
c = 0
for _ in range(n):
m, n, k = map(int,input().split())
a = a + m
b = b + n
c = c + k
if a == 0 and b == 0 and c == 0:
print('YES')
else:
print('NO')
```
| 3.9615
|
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"
}
] | 1,624,352,295
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 45
| 124
| 0
|
n,m = map(int,input().split())
prime = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]
if m in prime:
if m == prime[prime.index(n)+1]:
print('YES')
else :
print('NO')
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]
if m in prime:
if m == prime[prime.index(n)+1]:
print('YES')
else :
print('NO')
else :
print('NO')
```
| 3.969
|
4
|
C
|
Registration System
|
PROGRAMMING
| 1,300
|
[
"data structures",
"hashing",
"implementation"
] |
C. Registration system
|
5
|
64
|
A new e-mail service "Berlandesk" is going to be opened in Berland in the near future. The site administration wants to launch their project as soon as possible, that's why they ask you to help. You're suggested to implement the prototype of site registration system. The system should work on the following principle.
Each time a new user wants to register, he sends to the system a request with his name. If such a name does not exist in the system database, it is inserted into the database, and the user gets the response OK, confirming the successful registration. If the name already exists in the system database, the system makes up a new user name, sends it to the user as a prompt and also inserts the prompt into the database. The new name is formed by the following rule. Numbers, starting with 1, are appended one after another to name (name1, name2, ...), among these numbers the least *i* is found so that name*i* does not yet exist in the database.
|
The first line contains number *n* (1<=≤<=*n*<=≤<=105). The following *n* lines contain the requests to the system. Each request is a non-empty line, and consists of not more than 32 characters, which are all lowercase Latin letters.
|
Print *n* lines, which are system responses to the requests: OK in case of successful registration, or a prompt with a new name, if the requested name is already taken.
|
[
"4\nabacaba\nacaba\nabacaba\nacab\n",
"6\nfirst\nfirst\nsecond\nsecond\nthird\nthird\n"
] |
[
"OK\nOK\nabacaba1\nOK\n",
"OK\nfirst1\nOK\nsecond1\nOK\nthird1\n"
] |
none
| 0
|
[
{
"input": "4\nabacaba\nacaba\nabacaba\nacab",
"output": "OK\nOK\nabacaba1\nOK"
},
{
"input": "6\nfirst\nfirst\nsecond\nsecond\nthird\nthird",
"output": "OK\nfirst1\nOK\nsecond1\nOK\nthird1"
},
{
"input": "1\nn",
"output": "OK"
},
{
"input": "2\nu\nu",
"output": "OK\nu1"
},
{
"input": "3\nb\nb\nb",
"output": "OK\nb1\nb2"
},
{
"input": "2\nc\ncn",
"output": "OK\nOK"
},
{
"input": "3\nvhn\nvhn\nh",
"output": "OK\nvhn1\nOK"
},
{
"input": "4\nd\nhd\nd\nh",
"output": "OK\nOK\nd1\nOK"
},
{
"input": "10\nbhnqaptmp\nbhnqaptmp\nbhnqaptmp\nbhnqaptmp\nbhnqaptmp\nbhnqaptmp\nbhnqaptmp\nbhnqaptmp\nbhnqaptmp\nbhnqaptmp",
"output": "OK\nbhnqaptmp1\nbhnqaptmp2\nbhnqaptmp3\nbhnqaptmp4\nbhnqaptmp5\nbhnqaptmp6\nbhnqaptmp7\nbhnqaptmp8\nbhnqaptmp9"
},
{
"input": "10\nfpqhfouqdldravpjttarh\nfpqhfouqdldravpjttarh\nfpqhfouqdldravpjttarh\nfpqhfouqdldravpjttarh\nfpqhfouqdldravpjttarh\nfpqhfouqdldravpjttarh\njmvlplnrmba\nfpqhfouqdldravpjttarh\njmvlplnrmba\nfpqhfouqdldravpjttarh",
"output": "OK\nfpqhfouqdldravpjttarh1\nfpqhfouqdldravpjttarh2\nfpqhfouqdldravpjttarh3\nfpqhfouqdldravpjttarh4\nfpqhfouqdldravpjttarh5\nOK\nfpqhfouqdldravpjttarh6\njmvlplnrmba1\nfpqhfouqdldravpjttarh7"
},
{
"input": "10\niwexcrupuubwzbooj\niwexcrupuubwzbooj\njzsyjnxttliyfpunxyhsouhunenzxedi\njzsyjnxttliyfpunxyhsouhunenzxedi\njzsyjnxttliyfpunxyhsouhunenzxedi\njzsyjnxttliyfpunxyhsouhunenzxedi\njzsyjnxttliyfpunxyhsouhunenzxedi\niwexcrupuubwzbooj\niwexcrupuubwzbooj\niwexcrupuubwzbooj",
"output": "OK\niwexcrupuubwzbooj1\nOK\njzsyjnxttliyfpunxyhsouhunenzxedi1\njzsyjnxttliyfpunxyhsouhunenzxedi2\njzsyjnxttliyfpunxyhsouhunenzxedi3\njzsyjnxttliyfpunxyhsouhunenzxedi4\niwexcrupuubwzbooj2\niwexcrupuubwzbooj3\niwexcrupuubwzbooj4"
},
{
"input": "10\nzzzzzzzzzzzzzzzzzzzzzzzzzzz\nzzzzzzzzzzzzzzzzzzzzzzzzzzz\nzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz\nzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz\nzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz\nzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz\nzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz\nzzzzzzzzzzzzzzzzzzzzzzzzzzz\nzzzzzzzzzzzzzzzzzzzzzzzzzzz\nzzzzzzzzzzzzzzzzzzzzzzzzzzz",
"output": "OK\nzzzzzzzzzzzzzzzzzzzzzzzzzzz1\nOK\nzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz1\nzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz2\nzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz3\nzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz4\nzzzzzzzzzzzzzzzzzzzzzzzzzzz2\nzzzzzzzzzzzzzzzzzzzzzzzzzzz3\nzzzzzzzzzzzzzzzzzzzzzzzzzzz4"
},
{
"input": "20\nzzzzzzzzz\nzzzzzzzzzzzzz\nz\nzzzzzzzzzzzzz\nzzzzzzzzz\nzzzzzzzzz\nzzzzzzzzzzzzz\nzzzzzzzzzzzzz\nzzzzzzzzzzzzzzzzzzzzzzzz\nzzzzzzzzzzzzz\nzzzzzzzzzzzzz\nzzzzzzzzzzzzzzzzzzzzzzzz\nzzzzzzzzzzzzzzzzzzzzzzzz\nzzzzzzzzzzzzz\nzzzzzzzzzzzzz\nz\nzzzzzzzzz\nzzzzzzzzz\nzzzzzzzzzzzzz\nzzzzzzzzzzzzzzzzzzzzzzzz",
"output": "OK\nOK\nOK\nzzzzzzzzzzzzz1\nzzzzzzzzz1\nzzzzzzzzz2\nzzzzzzzzzzzzz2\nzzzzzzzzzzzzz3\nOK\nzzzzzzzzzzzzz4\nzzzzzzzzzzzzz5\nzzzzzzzzzzzzzzzzzzzzzzzz1\nzzzzzzzzzzzzzzzzzzzzzzzz2\nzzzzzzzzzzzzz6\nzzzzzzzzzzzzz7\nz1\nzzzzzzzzz3\nzzzzzzzzz4\nzzzzzzzzzzzzz8\nzzzzzzzzzzzzzzzzzzzzzzzz3"
}
] | 1,694,559,627
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 25
| 1,432
| 1,228,800
|
# Asif Islam - asifislam510
num_entries = int(input())
system = {}
while num_entries > 0:
name = input()
if name in system:
system[name] += 1
print(name + str(system[name]))
else:
system[name] = 0
print("OK")
num_entries -= 1
|
Title: Registration System
Time Limit: 5 seconds
Memory Limit: 64 megabytes
Problem Description:
A new e-mail service "Berlandesk" is going to be opened in Berland in the near future. The site administration wants to launch their project as soon as possible, that's why they ask you to help. You're suggested to implement the prototype of site registration system. The system should work on the following principle.
Each time a new user wants to register, he sends to the system a request with his name. If such a name does not exist in the system database, it is inserted into the database, and the user gets the response OK, confirming the successful registration. If the name already exists in the system database, the system makes up a new user name, sends it to the user as a prompt and also inserts the prompt into the database. The new name is formed by the following rule. Numbers, starting with 1, are appended one after another to name (name1, name2, ...), among these numbers the least *i* is found so that name*i* does not yet exist in the database.
Input Specification:
The first line contains number *n* (1<=≤<=*n*<=≤<=105). The following *n* lines contain the requests to the system. Each request is a non-empty line, and consists of not more than 32 characters, which are all lowercase Latin letters.
Output Specification:
Print *n* lines, which are system responses to the requests: OK in case of successful registration, or a prompt with a new name, if the requested name is already taken.
Demo Input:
['4\nabacaba\nacaba\nabacaba\nacab\n', '6\nfirst\nfirst\nsecond\nsecond\nthird\nthird\n']
Demo Output:
['OK\nOK\nabacaba1\nOK\n', 'OK\nfirst1\nOK\nsecond1\nOK\nthird1\n']
Note:
none
|
```python
# Asif Islam - asifislam510
num_entries = int(input())
system = {}
while num_entries > 0:
name = input()
if name in system:
system[name] += 1
print(name + str(system[name]))
else:
system[name] = 0
print("OK")
num_entries -= 1
```
| 3.847645
|
285
|
B
|
Find Marble
|
PROGRAMMING
| 1,200
|
[
"implementation"
] | null | null |
Petya and Vasya are playing a game. Petya's got *n* non-transparent glasses, standing in a row. The glasses' positions are indexed with integers from 1 to *n* from left to right. Note that the positions are indexed but the glasses are not.
First Petya puts a marble under the glass in position *s*. Then he performs some (possibly zero) shuffling operations. One shuffling operation means moving the glass from the first position to position *p*1, the glass from the second position to position *p*2 and so on. That is, a glass goes from position *i* to position *p**i*. Consider all glasses are moving simultaneously during one shuffling operation. When the glasses are shuffled, the marble doesn't travel from one glass to another: it moves together with the glass it was initially been put in.
After all shuffling operations Petya shows Vasya that the ball has moved to position *t*. Vasya's task is to say what minimum number of shuffling operations Petya has performed or determine that Petya has made a mistake and the marble could not have got from position *s* to position *t*.
|
The first line contains three integers: *n*,<=*s*,<=*t* (1<=≤<=*n*<=≤<=105; 1<=≤<=*s*,<=*t*<=≤<=*n*) — the number of glasses, the ball's initial and final position. The second line contains *n* space-separated integers: *p*1,<=*p*2,<=...,<=*p**n* (1<=≤<=*p**i*<=≤<=*n*) — the shuffling operation parameters. It is guaranteed that all *p**i*'s are distinct.
Note that *s* can equal *t*.
|
If the marble can move from position *s* to position *t*, then print on a single line a non-negative integer — the minimum number of shuffling operations, needed to get the marble to position *t*. If it is impossible, print number -1.
|
[
"4 2 1\n2 3 4 1\n",
"4 3 3\n4 1 3 2\n",
"4 3 4\n1 2 3 4\n",
"3 1 3\n2 1 3\n"
] |
[
"3\n",
"0\n",
"-1\n",
"-1\n"
] |
none
| 1,000
|
[
{
"input": "4 2 1\n2 3 4 1",
"output": "3"
},
{
"input": "4 3 3\n4 1 3 2",
"output": "0"
},
{
"input": "4 3 4\n1 2 3 4",
"output": "-1"
},
{
"input": "3 1 3\n2 1 3",
"output": "-1"
},
{
"input": "1 1 1\n1",
"output": "0"
},
{
"input": "10 6 7\n10 7 8 1 5 6 2 9 4 3",
"output": "-1"
},
{
"input": "10 3 6\n5 6 7 3 8 4 2 1 10 9",
"output": "3"
},
{
"input": "10 10 4\n4 2 6 9 5 3 8 1 10 7",
"output": "4"
},
{
"input": "100 90 57\n19 55 91 50 31 23 60 84 38 1 22 51 27 76 28 98 11 44 61 63 15 93 52 3 66 16 53 36 18 62 35 85 78 37 73 64 87 74 46 26 82 69 49 33 83 89 56 67 71 25 39 94 96 17 21 6 47 68 34 42 57 81 13 10 54 2 48 80 20 77 4 5 59 30 90 95 45 75 8 88 24 41 40 14 97 32 7 9 65 70 100 99 72 58 92 29 79 12 86 43",
"output": "-1"
},
{
"input": "100 11 20\n80 25 49 55 22 98 35 59 88 14 91 20 68 66 53 50 77 45 82 63 96 93 85 46 37 74 84 9 7 95 41 86 23 36 33 27 81 39 18 13 12 92 24 71 3 48 83 61 31 87 28 79 75 38 11 21 29 69 44 100 72 62 32 43 30 16 47 56 89 60 42 17 26 70 94 99 4 6 2 73 8 52 65 1 15 90 67 51 78 10 5 76 57 54 34 58 19 64 40 97",
"output": "26"
},
{
"input": "100 84 83\n30 67 53 89 94 54 92 17 26 57 15 5 74 85 10 61 18 70 91 75 14 11 93 41 25 78 88 81 20 51 35 4 62 1 97 39 68 52 47 77 64 3 2 72 60 80 8 83 65 98 21 22 45 7 58 31 43 38 90 99 49 87 55 36 29 6 37 23 66 76 59 79 40 86 63 44 82 32 48 16 50 100 28 96 46 12 27 13 24 9 19 84 73 69 71 42 56 33 34 95",
"output": "71"
},
{
"input": "100 6 93\n74 62 67 81 40 85 35 42 59 72 80 28 79 41 16 19 33 63 13 10 69 76 70 93 49 84 89 94 8 37 11 90 26 52 47 7 36 95 86 75 56 15 61 99 88 12 83 21 20 3 100 17 32 82 6 5 43 25 66 68 73 78 18 77 92 27 23 2 4 39 60 48 22 24 14 97 29 34 54 64 71 57 87 38 9 50 30 53 51 45 44 31 58 91 98 65 55 1 46 96",
"output": "-1"
},
{
"input": "100 27 56\n58 18 50 41 33 37 14 87 77 73 61 53 15 8 70 68 45 96 54 78 39 67 51 60 80 12 93 99 20 92 17 79 4 13 62 91 69 29 49 36 98 34 90 35 84 64 38 83 28 89 97 94 9 16 26 48 10 57 23 75 27 88 44 21 72 76 30 43 32 2 71 24 100 1 31 81 42 40 47 55 86 85 66 5 52 22 95 74 11 19 7 82 6 25 56 63 65 59 46 3",
"output": "20"
},
{
"input": "87 42 49\n45 55 24 44 56 72 74 23 4 7 37 67 22 6 58 76 40 36 3 20 26 87 64 75 49 70 62 42 31 1 80 33 25 59 78 27 32 2 41 61 66 28 19 85 15 69 52 77 50 14 16 34 18 43 73 83 11 39 29 9 35 13 81 54 79 21 60 46 71 57 12 17 5 47 38 30 10 84 53 63 68 8 51 65 48 86 82",
"output": "-1"
},
{
"input": "2 1 2\n1 2",
"output": "-1"
},
{
"input": "2 1 2\n2 1",
"output": "1"
},
{
"input": "2 2 2\n1 2",
"output": "0"
},
{
"input": "2 2 2\n2 1",
"output": "0"
},
{
"input": "2 1 1\n2 1",
"output": "0"
}
] | 1,688,452,065
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 33
| 154
| 13,516,800
|
n,s,t=map(int,input().split())
p=[0]+list(map(int,input().split()))
ans=0
while ans<=n and s!=t:
s=p[s]
ans+=1
print([-1,ans][ans<=n])
|
Title: Find Marble
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya and Vasya are playing a game. Petya's got *n* non-transparent glasses, standing in a row. The glasses' positions are indexed with integers from 1 to *n* from left to right. Note that the positions are indexed but the glasses are not.
First Petya puts a marble under the glass in position *s*. Then he performs some (possibly zero) shuffling operations. One shuffling operation means moving the glass from the first position to position *p*1, the glass from the second position to position *p*2 and so on. That is, a glass goes from position *i* to position *p**i*. Consider all glasses are moving simultaneously during one shuffling operation. When the glasses are shuffled, the marble doesn't travel from one glass to another: it moves together with the glass it was initially been put in.
After all shuffling operations Petya shows Vasya that the ball has moved to position *t*. Vasya's task is to say what minimum number of shuffling operations Petya has performed or determine that Petya has made a mistake and the marble could not have got from position *s* to position *t*.
Input Specification:
The first line contains three integers: *n*,<=*s*,<=*t* (1<=≤<=*n*<=≤<=105; 1<=≤<=*s*,<=*t*<=≤<=*n*) — the number of glasses, the ball's initial and final position. The second line contains *n* space-separated integers: *p*1,<=*p*2,<=...,<=*p**n* (1<=≤<=*p**i*<=≤<=*n*) — the shuffling operation parameters. It is guaranteed that all *p**i*'s are distinct.
Note that *s* can equal *t*.
Output Specification:
If the marble can move from position *s* to position *t*, then print on a single line a non-negative integer — the minimum number of shuffling operations, needed to get the marble to position *t*. If it is impossible, print number -1.
Demo Input:
['4 2 1\n2 3 4 1\n', '4 3 3\n4 1 3 2\n', '4 3 4\n1 2 3 4\n', '3 1 3\n2 1 3\n']
Demo Output:
['3\n', '0\n', '-1\n', '-1\n']
Note:
none
|
```python
n,s,t=map(int,input().split())
p=[0]+list(map(int,input().split()))
ans=0
while ans<=n and s!=t:
s=p[s]
ans+=1
print([-1,ans][ans<=n])
```
| 3
|
|
443
|
A
|
Anton and Letters
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"implementation"
] | null | null |
Recently, Anton has found a set. The set consists of small English letters. Anton carefully wrote out all the letters from the set in one line, separated by a comma. He also added an opening curved bracket at the beginning of the line and a closing curved bracket at the end of the line.
Unfortunately, from time to time Anton would forget writing some letter and write it again. He asks you to count the total number of distinct letters in his set.
|
The first and the single line contains the set of letters. The length of the line doesn't exceed 1000. It is guaranteed that the line starts from an opening curved bracket and ends with a closing curved bracket. Between them, small English letters are listed, separated by a comma. Each comma is followed by a space.
|
Print a single number — the number of distinct letters in Anton's set.
|
[
"{a, b, c}\n",
"{b, a, b, a}\n",
"{}\n"
] |
[
"3\n",
"2\n",
"0\n"
] |
none
| 500
|
[
{
"input": "{a, b, c}",
"output": "3"
},
{
"input": "{b, a, b, a}",
"output": "2"
},
{
"input": "{}",
"output": "0"
},
{
"input": "{a, a, c, b, b, b, c, c, c, c}",
"output": "3"
},
{
"input": "{a, c, b, b}",
"output": "3"
},
{
"input": "{a, b}",
"output": "2"
},
{
"input": "{a}",
"output": "1"
},
{
"input": "{b, a, b, a, b, c, c, b, c, b}",
"output": "3"
},
{
"input": "{e, g, c, e}",
"output": "3"
},
{
"input": "{a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a}",
"output": "1"
},
{
"input": "{a, a, b}",
"output": "2"
},
{
"input": "{a, b, b, b, a, b, a, a, a, a, a, a, b, a, b, a, a, a, a, a, b, a, b, a}",
"output": "2"
},
{
"input": "{j, u, a, c, f, w, e, w, x, t, h, p, v, n, i, l, x, n, i, b, u, c, a, a}",
"output": "16"
},
{
"input": "{x, i, w, c, p, e, h, z, k, i}",
"output": "9"
},
{
"input": "{t, k, o, x, r, d, q, j, k, e, z, w, y, r, z, s, s, e, s, b, k, i}",
"output": "15"
},
{
"input": "{y}",
"output": "1"
},
{
"input": "{x}",
"output": "1"
},
{
"input": "{b, z, a, z}",
"output": "3"
},
{
"input": "{z}",
"output": "1"
},
{
"input": "{a, z}",
"output": "2"
},
{
"input": "{a, b, z}",
"output": "3"
},
{
"input": "{s, q, z, r, t, a, b, h, j, i, o, z, r, q}",
"output": "11"
}
] | 1,683,115,803
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 27
| 61
| 409,600
|
import re
X = str(input().split())
Y = re.sub(r'[^A-Za-z]+', '', X)
print(len(set(Y)))
|
Title: Anton and Letters
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently, Anton has found a set. The set consists of small English letters. Anton carefully wrote out all the letters from the set in one line, separated by a comma. He also added an opening curved bracket at the beginning of the line and a closing curved bracket at the end of the line.
Unfortunately, from time to time Anton would forget writing some letter and write it again. He asks you to count the total number of distinct letters in his set.
Input Specification:
The first and the single line contains the set of letters. The length of the line doesn't exceed 1000. It is guaranteed that the line starts from an opening curved bracket and ends with a closing curved bracket. Between them, small English letters are listed, separated by a comma. Each comma is followed by a space.
Output Specification:
Print a single number — the number of distinct letters in Anton's set.
Demo Input:
['{a, b, c}\n', '{b, a, b, a}\n', '{}\n']
Demo Output:
['3\n', '2\n', '0\n']
Note:
none
|
```python
import re
X = str(input().split())
Y = re.sub(r'[^A-Za-z]+', '', X)
print(len(set(Y)))
```
| 3
|
|
84
|
A
|
Toy Army
|
PROGRAMMING
| 900
|
[
"math",
"number theory"
] |
A. Toy Army
|
2
|
256
|
The hero of our story, Valera, and his best friend Arcady are still in school, and therefore they spend all the free time playing turn-based strategy "GAGA: Go And Go Again". The gameplay is as follows.
There are two armies on the playing field each of which consists of *n* men (*n* is always even). The current player specifies for each of her soldiers an enemy's soldier he will shoot (a target) and then all the player's soldiers shot simultaneously. This is a game world, and so each soldier shoots perfectly, that is he absolutely always hits the specified target. If an enemy soldier is hit, he will surely die. It may happen that several soldiers had been indicated the same target. Killed soldiers do not participate in the game anymore.
The game "GAGA" consists of three steps: first Valera makes a move, then Arcady, then Valera again and the game ends.
You are asked to calculate the maximum total number of soldiers that may be killed during the game.
|
The input data consist of a single integer *n* (2<=≤<=*n*<=≤<=108, *n* is even). Please note that before the game starts there are 2*n* soldiers on the fields.
|
Print a single number — a maximum total number of soldiers that could be killed in the course of the game in three turns.
|
[
"2\n",
"4\n"
] |
[
"3\n",
"6\n"
] |
The first sample test:
1) Valera's soldiers 1 and 2 shoot at Arcady's soldier 1.
2) Arcady's soldier 2 shoots at Valera's soldier 1.
3) Valera's soldier 1 shoots at Arcady's soldier 2.
There are 3 soldiers killed in total: Valera's soldier 1 and Arcady's soldiers 1 and 2.
| 500
|
[
{
"input": "2",
"output": "3"
},
{
"input": "4",
"output": "6"
},
{
"input": "6",
"output": "9"
},
{
"input": "8",
"output": "12"
},
{
"input": "10",
"output": "15"
},
{
"input": "140",
"output": "210"
},
{
"input": "500",
"output": "750"
},
{
"input": "1000",
"output": "1500"
},
{
"input": "2000",
"output": "3000"
},
{
"input": "50000",
"output": "75000"
},
{
"input": "10000",
"output": "15000"
},
{
"input": "25460",
"output": "38190"
},
{
"input": "54646",
"output": "81969"
},
{
"input": "59790",
"output": "89685"
},
{
"input": "578456",
"output": "867684"
},
{
"input": "56798056",
"output": "85197084"
},
{
"input": "8457980",
"output": "12686970"
},
{
"input": "5687986",
"output": "8531979"
},
{
"input": "10984932",
"output": "16477398"
},
{
"input": "99999994",
"output": "149999991"
},
{
"input": "99999996",
"output": "149999994"
},
{
"input": "99999998",
"output": "149999997"
},
{
"input": "100000000",
"output": "150000000"
}
] | 1,581,095,841
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 23
| 248
| 307,200
|
n = int(input())
k = n//2
#hello and welcome to this course on machine learning
print(n+k)
|
Title: Toy Army
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The hero of our story, Valera, and his best friend Arcady are still in school, and therefore they spend all the free time playing turn-based strategy "GAGA: Go And Go Again". The gameplay is as follows.
There are two armies on the playing field each of which consists of *n* men (*n* is always even). The current player specifies for each of her soldiers an enemy's soldier he will shoot (a target) and then all the player's soldiers shot simultaneously. This is a game world, and so each soldier shoots perfectly, that is he absolutely always hits the specified target. If an enemy soldier is hit, he will surely die. It may happen that several soldiers had been indicated the same target. Killed soldiers do not participate in the game anymore.
The game "GAGA" consists of three steps: first Valera makes a move, then Arcady, then Valera again and the game ends.
You are asked to calculate the maximum total number of soldiers that may be killed during the game.
Input Specification:
The input data consist of a single integer *n* (2<=≤<=*n*<=≤<=108, *n* is even). Please note that before the game starts there are 2*n* soldiers on the fields.
Output Specification:
Print a single number — a maximum total number of soldiers that could be killed in the course of the game in three turns.
Demo Input:
['2\n', '4\n']
Demo Output:
['3\n', '6\n']
Note:
The first sample test:
1) Valera's soldiers 1 and 2 shoot at Arcady's soldier 1.
2) Arcady's soldier 2 shoots at Valera's soldier 1.
3) Valera's soldier 1 shoots at Arcady's soldier 2.
There are 3 soldiers killed in total: Valera's soldier 1 and Arcady's soldiers 1 and 2.
|
```python
n = int(input())
k = n//2
#hello and welcome to this course on machine learning
print(n+k)
```
| 3.937428
|
855
|
B
|
Marvolo Gaunt's Ring
|
PROGRAMMING
| 1,500
|
[
"brute force",
"data structures",
"dp"
] | null | null |
Professor Dumbledore is helping Harry destroy the Horcruxes. He went to Gaunt Shack as he suspected a Horcrux to be present there. He saw Marvolo Gaunt's Ring and identified it as a Horcrux. Although he destroyed it, he is still affected by its curse. Professor Snape is helping Dumbledore remove the curse. For this, he wants to give Dumbledore exactly *x* drops of the potion he made.
Value of *x* is calculated as maximum of *p*·*a**i*<=+<=*q*·*a**j*<=+<=*r*·*a**k* for given *p*,<=*q*,<=*r* and array *a*1,<=*a*2,<=... *a**n* such that 1<=≤<=*i*<=≤<=*j*<=≤<=*k*<=≤<=*n*. Help Snape find the value of *x*. Do note that the value of *x* may be negative.
|
First line of input contains 4 integers *n*,<=*p*,<=*q*,<=*r* (<=-<=109<=≤<=*p*,<=*q*,<=*r*<=≤<=109,<=1<=≤<=*n*<=≤<=105).
Next line of input contains *n* space separated integers *a*1,<=*a*2,<=... *a**n* (<=-<=109<=≤<=*a**i*<=≤<=109).
|
Output a single integer the maximum value of *p*·*a**i*<=+<=*q*·*a**j*<=+<=*r*·*a**k* that can be obtained provided 1<=≤<=*i*<=≤<=*j*<=≤<=*k*<=≤<=*n*.
|
[
"5 1 2 3\n1 2 3 4 5\n",
"5 1 2 -3\n-1 -2 -3 -4 -5\n"
] |
[
"30\n",
"12\n"
] |
In the first sample case, we can take *i* = *j* = *k* = 5, thus making the answer as 1·5 + 2·5 + 3·5 = 30.
In second sample case, selecting *i* = *j* = 1 and *k* = 5 gives the answer 12.
| 1,000
|
[
{
"input": "5 1 2 3\n1 2 3 4 5",
"output": "30"
},
{
"input": "5 1 2 -3\n-1 -2 -3 -4 -5",
"output": "12"
},
{
"input": "5 886327859 82309257 -68295239\n-731225382 354766539 -48222231 -474691998 360965777",
"output": "376059240645059046"
},
{
"input": "4 -96405765 -495906217 625385006\n-509961652 392159235 -577128498 -744548876",
"output": "547306902373544674"
},
{
"input": "43 959134961 -868367850 142426380\n921743429 63959718 -797293233 122041422 -407576197 700139744 299598010 168207043 362252658 591926075 941946099 812263640 -76679927 -824267725 89529990 -73303355 83596189 -982699817 -235197848 654773327 125211479 -497091570 -2301804 203486596 -126652024 309810546 -581289415 -740125230 64425927 -501018049 304730559 34930193 -762964086 723645139 -826821494 495947907 816331024 9932423 -876541603 -782692568 322360800 841436938 40787162",
"output": "1876641179289775029"
},
{
"input": "1 0 0 0\n0",
"output": "0"
},
{
"input": "1 1000000000 1000000000 1000000000\n1000000000",
"output": "3000000000000000000"
},
{
"input": "1 -1000000000 -1000000000 1000000000\n1000000000",
"output": "-1000000000000000000"
},
{
"input": "1 -1000000000 -1000000000 -1000000000\n1000000000",
"output": "-3000000000000000000"
},
{
"input": "3 1000000000 1000000000 1000000000\n-1000000000 -1000000000 -1000000000",
"output": "-3000000000000000000"
},
{
"input": "1 1 1 1\n-1",
"output": "-3"
},
{
"input": "1 -1 -1 -1\n1",
"output": "-3"
},
{
"input": "1 1000000000 1000000000 1000000000\n-1000000000",
"output": "-3000000000000000000"
},
{
"input": "1 1 2 3\n-1",
"output": "-6"
},
{
"input": "3 -1000000000 -1000000000 -1000000000\n1000000000 1000000000 1000000000",
"output": "-3000000000000000000"
},
{
"input": "2 -1000000000 -1000000000 -1000000000\n1000000000 1000000000",
"output": "-3000000000000000000"
},
{
"input": "3 1 1 1\n-1 -1 -1",
"output": "-3"
},
{
"input": "1 -1000000000 0 0\n1000000000",
"output": "-1000000000000000000"
},
{
"input": "1 -100 -100 -100\n100",
"output": "-30000"
},
{
"input": "5 -1000000000 -1000000000 -1000000000\n1000000000 1000000000 1000000000 1000000000 1000000000",
"output": "-3000000000000000000"
},
{
"input": "1 999999999 999999999 999999999\n-999999999",
"output": "-2999999994000000003"
},
{
"input": "3 -1000000000 -1000000000 1\n1000000000 1000000000 1000000000",
"output": "-1999999999000000000"
},
{
"input": "3 -2 3 -2\n1 2 1",
"output": "2"
},
{
"input": "2 1 -1 1\n1 -1",
"output": "1"
},
{
"input": "1 -1000000000 1 -1000000000\n1000000000",
"output": "-1999999999000000000"
},
{
"input": "1 1000000000 1000000000 -1000000000\n-1000000000",
"output": "-1000000000000000000"
},
{
"input": "1 -1000000000 -1000000000 0\n1000000000",
"output": "-2000000000000000000"
}
] | 1,688,919,572
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 82
| 327
| 26,112,000
|
n,p,q,r = [int(_) for _ in input().split()]
arr = [int(_) for _ in input().split()]
dp = [[0]*3 for _ in range(100005)]
dp[0][0] = arr[0]*p
for i in range(1,n):
dp[i][0] = max(dp[i-1][0], (p*arr[i]))
dp[0][1] = dp[0][0]+ arr[0]*q
for i in range(1,n):
dp[i][1] = max(dp[i-1][1], dp[i][0]+ q*arr[i])
dp[0][2] = dp[0][1] + arr[0]*r
for i in range(1,n):
dp[i][2] = max(dp[i-1][2], dp[i][1] + arr[i]*r)
print(dp[n-1][2])
|
Title: Marvolo Gaunt's Ring
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Professor Dumbledore is helping Harry destroy the Horcruxes. He went to Gaunt Shack as he suspected a Horcrux to be present there. He saw Marvolo Gaunt's Ring and identified it as a Horcrux. Although he destroyed it, he is still affected by its curse. Professor Snape is helping Dumbledore remove the curse. For this, he wants to give Dumbledore exactly *x* drops of the potion he made.
Value of *x* is calculated as maximum of *p*·*a**i*<=+<=*q*·*a**j*<=+<=*r*·*a**k* for given *p*,<=*q*,<=*r* and array *a*1,<=*a*2,<=... *a**n* such that 1<=≤<=*i*<=≤<=*j*<=≤<=*k*<=≤<=*n*. Help Snape find the value of *x*. Do note that the value of *x* may be negative.
Input Specification:
First line of input contains 4 integers *n*,<=*p*,<=*q*,<=*r* (<=-<=109<=≤<=*p*,<=*q*,<=*r*<=≤<=109,<=1<=≤<=*n*<=≤<=105).
Next line of input contains *n* space separated integers *a*1,<=*a*2,<=... *a**n* (<=-<=109<=≤<=*a**i*<=≤<=109).
Output Specification:
Output a single integer the maximum value of *p*·*a**i*<=+<=*q*·*a**j*<=+<=*r*·*a**k* that can be obtained provided 1<=≤<=*i*<=≤<=*j*<=≤<=*k*<=≤<=*n*.
Demo Input:
['5 1 2 3\n1 2 3 4 5\n', '5 1 2 -3\n-1 -2 -3 -4 -5\n']
Demo Output:
['30\n', '12\n']
Note:
In the first sample case, we can take *i* = *j* = *k* = 5, thus making the answer as 1·5 + 2·5 + 3·5 = 30.
In second sample case, selecting *i* = *j* = 1 and *k* = 5 gives the answer 12.
|
```python
n,p,q,r = [int(_) for _ in input().split()]
arr = [int(_) for _ in input().split()]
dp = [[0]*3 for _ in range(100005)]
dp[0][0] = arr[0]*p
for i in range(1,n):
dp[i][0] = max(dp[i-1][0], (p*arr[i]))
dp[0][1] = dp[0][0]+ arr[0]*q
for i in range(1,n):
dp[i][1] = max(dp[i-1][1], dp[i][0]+ q*arr[i])
dp[0][2] = dp[0][1] + arr[0]*r
for i in range(1,n):
dp[i][2] = max(dp[i-1][2], dp[i][1] + arr[i]*r)
print(dp[n-1][2])
```
| 3
|
|
644
|
A
|
Parliament of Berland
|
PROGRAMMING
| 1,000
|
[
"*special",
"constructive algorithms"
] | null | null |
There are *n* parliamentarians in Berland. They are numbered with integers from 1 to *n*. It happened that all parliamentarians with odd indices are Democrats and all parliamentarians with even indices are Republicans.
New parliament assembly hall is a rectangle consisting of *a*<=×<=*b* chairs — *a* rows of *b* chairs each. Two chairs are considered neighbouring if they share as side. For example, chair number 5 in row number 2 is neighbouring to chairs number 4 and 6 in this row and chairs with number 5 in rows 1 and 3. Thus, chairs have four neighbours in general, except for the chairs on the border of the hall
We know that if two parliamentarians from one political party (that is two Democrats or two Republicans) seat nearby they spent all time discussing internal party issues.
Write the program that given the number of parliamentarians and the sizes of the hall determine if there is a way to find a seat for any parliamentarian, such that no two members of the same party share neighbouring seats.
|
The first line of the input contains three integers *n*, *a* and *b* (1<=≤<=*n*<=≤<=10<=000, 1<=≤<=*a*,<=*b*<=≤<=100) — the number of parliamentarians, the number of rows in the assembly hall and the number of seats in each row, respectively.
|
If there is no way to assigns seats to parliamentarians in a proper way print -1.
Otherwise print the solution in *a* lines, each containing *b* integers. The *j*-th integer of the *i*-th line should be equal to the index of parliamentarian occupying this seat, or 0 if this seat should remain empty. If there are multiple possible solution, you may print any of them.
|
[
"3 2 2\n",
"8 4 3\n",
"10 2 2\n"
] |
[
"0 3\n1 2\n",
"7 8 3\n0 1 4\n6 0 5\n0 2 0\n",
"-1\n"
] |
In the first sample there are many other possible solutions. For example,
and
The following assignment
is incorrect, because parliamentarians 1 and 3 are both from Democrats party but will occupy neighbouring seats.
| 500
|
[
{
"input": "3 2 2",
"output": "1 2 \n0 3 "
},
{
"input": "8 4 3",
"output": "1 2 3 \n4 5 6 \n7 8 0 \n0 0 0 "
},
{
"input": "10 2 2",
"output": "-1"
},
{
"input": "1 1 1",
"output": "1 "
},
{
"input": "8 3 3",
"output": "1 2 3 \n4 5 6 \n7 8 0 "
},
{
"input": "1 1 100",
"output": "1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 "
},
{
"input": "1 100 1",
"output": "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 \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 "
},
{
"input": "12 4 3",
"output": "1 2 3 \n4 5 6 \n7 8 9 \n10 11 12 "
},
{
"input": "64 8 9",
"output": "1 2 3 4 5 6 7 8 9 \n10 11 12 13 14 15 16 17 18 \n19 20 21 22 23 24 25 26 27 \n28 29 30 31 32 33 34 35 36 \n37 38 39 40 41 42 43 44 45 \n46 47 48 49 50 51 52 53 54 \n55 56 57 58 59 60 61 62 63 \n64 0 0 0 0 0 0 0 0 "
},
{
"input": "13 2 6",
"output": "-1"
},
{
"input": "41 6 7",
"output": "1 2 3 4 5 6 7 \n8 9 10 11 12 13 14 \n15 16 17 18 19 20 21 \n22 23 24 25 26 27 28 \n29 30 31 32 33 34 35 \n36 37 38 39 40 41 0 "
},
{
"input": "9999 100 100",
"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 \n102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 138 137 140 139 142 141 144 143 146 145 148 147 150 149 152 151 154 153 1..."
},
{
"input": "10000 100 100",
"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 \n102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 138 137 140 139 142 141 144 143 146 145 148 147 150 149 152 151 154 153 1..."
},
{
"input": "2099 70 30",
"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 \n32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 54 53 56 55 58 57 60 59 \n61 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 \n92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 \n121 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 \n152 151 1..."
},
{
"input": "2098 30 70",
"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 \n72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 138 137 140 139 \n141 142 143 144 145 146 147 148 149 150 151 152 153 154..."
},
{
"input": "10000 1 1",
"output": "-1"
},
{
"input": "1583 49 36",
"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 \n38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 54 53 56 55 58 57 60 59 62 61 64 63 66 65 68 67 70 69 72 71 \n73 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 \n110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 138 137 140 139 142 141 144 143 \n145 146 147 148 149 150 151 152 153..."
},
{
"input": "4825 77 88",
"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 \n90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 138 137 140 139 142 141 144 143 146 145 148 147 150 149 152 151 154 153 1..."
},
{
"input": "26 1 33",
"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 0 0 0 0 0 0 0 "
},
{
"input": "274 25 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 \n78 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 \n..."
},
{
"input": "694 49 22",
"output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 \n24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 \n45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 \n68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 \n89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 \n112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 \n133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152..."
},
{
"input": "3585 77 62",
"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 \n64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 \n125 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": "3 1 6",
"output": "1 2 3 0 0 0 "
},
{
"input": "352 25 59",
"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 \n60 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 \n119 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": "150 53 3",
"output": "1 2 3 \n4 5 6 \n7 8 9 \n10 11 12 \n13 14 15 \n16 17 18 \n19 20 21 \n22 23 24 \n25 26 27 \n28 29 30 \n31 32 33 \n34 35 36 \n37 38 39 \n40 41 42 \n43 44 45 \n46 47 48 \n49 50 51 \n52 53 54 \n55 56 57 \n58 59 60 \n61 62 63 \n64 65 66 \n67 68 69 \n70 71 72 \n73 74 75 \n76 77 78 \n79 80 81 \n82 83 84 \n85 86 87 \n88 89 90 \n91 92 93 \n94 95 96 \n97 98 99 \n100 101 102 \n103 104 105 \n106 107 108 \n109 110 111 \n112 113 114 \n115 116 117 \n118 119 120 \n121 122 123 \n124 125 126 \n127 128 129 \n130 131 132 \n133..."
},
{
"input": "4227 91 80",
"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 \n82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 138 137 140 139 142 141 144 143 146 145 148 147 150 149 152 151 154 153 1..."
},
{
"input": "378 19 25",
"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 \n26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 \n51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 \n76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 \n101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 \n126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 \n151 152..."
},
{
"input": "2357 43 65",
"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 \n66 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 \n131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..."
},
{
"input": "232 71 9",
"output": "1 2 3 4 5 6 7 8 9 \n10 11 12 13 14 15 16 17 18 \n19 20 21 22 23 24 25 26 27 \n28 29 30 31 32 33 34 35 36 \n37 38 39 40 41 42 43 44 45 \n46 47 48 49 50 51 52 53 54 \n55 56 57 58 59 60 61 62 63 \n64 65 66 67 68 69 70 71 72 \n73 74 75 76 77 78 79 80 81 \n82 83 84 85 86 87 88 89 90 \n91 92 93 94 95 96 97 98 99 \n100 101 102 103 104 105 106 107 108 \n109 110 111 112 113 114 115 116 117 \n118 119 120 121 122 123 124 125 126 \n127 128 129 130 131 132 133 134 135 \n136 137 138 139 140 141 142 143 144 \n145 146 147..."
},
{
"input": "2362 91 62",
"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 \n64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 \n125 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": "4601 59 78",
"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 \n80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 138 137 140 139 142 141 144 143 146 145 148 147 150 149 152 151 154 153 1..."
},
{
"input": "4439 74 60",
"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 \n62 61 64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 \n121 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": "3733 89 42",
"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 \n44 43 46 45 48 47 50 49 52 51 54 53 56 55 58 57 60 59 62 61 64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 \n85 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 \n128 127 130 129 132 131 134 133 136 135 138 137 140 139 142 141 144 143 146 145 148 147 150 149 152 151 154 1..."
},
{
"input": "335 12 28",
"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 \n30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 54 53 56 55 \n57 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 \n86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 \n113 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 \n142 141 144 143 146 145 148 147 150 149 152 151 1..."
},
{
"input": "440 26 17",
"output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 \n18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 \n35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 \n52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 \n69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 \n86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 \n103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 \n120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 \n137 138 139 140 141 142 143 144 145 146 147 148 149 150 151..."
},
{
"input": "109 37 3",
"output": "1 2 3 \n4 5 6 \n7 8 9 \n10 11 12 \n13 14 15 \n16 17 18 \n19 20 21 \n22 23 24 \n25 26 27 \n28 29 30 \n31 32 33 \n34 35 36 \n37 38 39 \n40 41 42 \n43 44 45 \n46 47 48 \n49 50 51 \n52 53 54 \n55 56 57 \n58 59 60 \n61 62 63 \n64 65 66 \n67 68 69 \n70 71 72 \n73 74 75 \n76 77 78 \n79 80 81 \n82 83 84 \n85 86 87 \n88 89 90 \n91 92 93 \n94 95 96 \n97 98 99 \n100 101 102 \n103 104 105 \n106 107 108 \n109 0 0 "
},
{
"input": "4416 52 85",
"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 \n86 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 1..."
},
{
"input": "5025 75 67",
"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 \n68 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 \n135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..."
},
{
"input": "4983 89 56",
"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 \n58 57 60 59 62 61 64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 \n113 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": "950 17 56",
"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 \n58 57 60 59 62 61 64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 \n113 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": "1637 40 41",
"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 \n42 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 \n83 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 \n124 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 1..."
},
{
"input": "1142 52 22",
"output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 \n24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 \n45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 \n68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 \n89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 \n112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 \n133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152..."
},
{
"input": "907 70 13",
"output": "1 2 3 4 5 6 7 8 9 10 11 12 13 \n14 15 16 17 18 19 20 21 22 23 24 25 26 \n27 28 29 30 31 32 33 34 35 36 37 38 39 \n40 41 42 43 44 45 46 47 48 49 50 51 52 \n53 54 55 56 57 58 59 60 61 62 63 64 65 \n66 67 68 69 70 71 72 73 74 75 76 77 78 \n79 80 81 82 83 84 85 86 87 88 89 90 91 \n92 93 94 95 96 97 98 99 100 101 102 103 104 \n105 106 107 108 109 110 111 112 113 114 115 116 117 \n118 119 120 121 122 123 124 125 126 127 128 129 130 \n131 132 133 134 135 136 137 138 139 140 141 142 143 \n144 145 146 147 148 149 1..."
},
{
"input": "7279 80 91",
"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 \n92 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 1..."
},
{
"input": "1653 87 19",
"output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 \n20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 \n39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 \n58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 \n77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 \n96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 \n115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 \n134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 1..."
},
{
"input": "15 2 8",
"output": "1 2 3 4 5 6 7 8 \n10 9 12 11 14 13 0 15 "
},
{
"input": "1459 17 86",
"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 \n88 87 90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 138 137 140 139 142 141 144 143 146 145 148 147 150 149 152 151 154 153 1..."
},
{
"input": "3035 40 76",
"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 \n78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 138 137 140 139 142 141 144 143 146 145 148 147 150 149 152 151 \n153 154..."
},
{
"input": "3095 50 62",
"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 \n64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 \n125 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": "3055 65 47",
"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 \n48 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 \n95 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 \n142 143 144 145 146 147 148 149 150 151 152 153 1..."
},
{
"input": "2638 80 33",
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},
{
"input": "29 3 11",
"output": "1 2 3 4 5 6 7 8 9 10 11 \n12 13 14 15 16 17 18 19 20 21 22 \n23 24 25 26 27 28 29 0 0 0 0 "
},
{
"input": "16 18 1",
"output": "1 \n2 \n3 \n4 \n5 \n6 \n7 \n8 \n9 \n10 \n11 \n12 \n13 \n14 \n15 \n16 \n0 \n0 "
},
{
"input": "2240 27 83",
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},
{
"input": "1264 55 23",
"output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 \n24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 \n47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 \n70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 \n93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 \n116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 \n139 140 141 142 143 144 145 146 147 148 149 150 151 152..."
},
{
"input": "5400 75 72",
"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 \n74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 138 137 140 139 142 141 144 143 \n145 146 147 148 149 150 151 152 153 154..."
},
{
"input": "46 3 16",
"output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 \n18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 \n33 34 35 36 37 38 39 40 41 42 43 44 45 46 0 0 "
},
{
"input": "1512 27 56",
"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 \n58 57 60 59 62 61 64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 \n113 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": "4206 86 12",
"output": "-1"
},
{
"input": "2358 14 56",
"output": "-1"
},
{
"input": "5420 35 96",
"output": "-1"
},
{
"input": "7758 63 41",
"output": "-1"
},
{
"input": "9806 87 93",
"output": "-1"
},
{
"input": "99 1 97",
"output": "-1"
},
{
"input": "1053 25 42",
"output": "-1"
},
{
"input": "4217 49 86",
"output": "-1"
},
{
"input": "2312 77 30",
"output": "-1"
},
{
"input": "74 1 71",
"output": "-1"
},
{
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},
{
"input": "1 57 99",
"output": "1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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 0 0 0 0 0 0..."
},
{
"input": "4785 55 87",
"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 \n88 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 1..."
},
{
"input": "4611 53 87",
"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 \n88 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 1..."
},
{
"input": "9801 99 99",
"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 \n100 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 1..."
},
{
"input": "8096 92 88",
"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 \n90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 138 137 140 139 142 141 144 143 146 145 148 147 150 149 152 151 154 153 1..."
},
{
"input": "5916 68 87",
"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 \n88 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 1..."
},
{
"input": "8928 93 96",
"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 \n98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 138 137 140 139 142 141 144 143 146 145 148 147 150 149 152 151 154 153 1..."
},
{
"input": "7743 89 87",
"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 \n88 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 1..."
},
{
"input": "3128 46 68",
"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 \n70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 \n137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..."
},
{
"input": "3800 40 95",
"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 \n96 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 1..."
},
{
"input": "5368 61 88",
"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 \n90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 138 137 140 139 142 141 144 143 146 145 148 147 150 149 152 151 154 153 1..."
},
{
"input": "2583 41 63",
"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 \n64 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 \n127 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": "4056 52 78",
"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 \n80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 138 137 140 139 142 141 144 143 146 145 148 147 150 149 152 151 154 153 1..."
},
{
"input": "3822 42 91",
"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 \n92 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 1..."
},
{
"input": "5358 57 94",
"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 \n96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 138 137 140 139 142 141 144 143 146 145 148 147 150 149 152 151 154 153 1..."
},
{
"input": "2961 47 63",
"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 \n64 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 \n127 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..."
}
] | 1,458,229,134
| 110,334
|
Python 3
|
OK
|
TESTS
| 85
| 78
| 716,800
|
n, a, b = map(int, (input().split()))
if n > (a * b):
print(-1)
else:
for i in range(a):
if (b % 2 == 0) and ((i + 1) % 2 == 0):
for c in range(b * (i + 1), b * (i + 1) - b, -1):
if c <= n:
print(c, end='\t')
else:
print(0, end='\t')
else:
for c in range((b * (i + 1) - b) + 1, b * (i + 1) + 1):
if c <= n:
print(c, end='\t')
else:
print(0, end='\t')
print()
# n, a, b = map(int, (input().split()))
# if n > (a * b):
# print(-1)
# else:
# c = 0
# for i in range(a):
# if (b % 2 == 0) and (i % 2 == 1):
# c += b + 1
# for _ in range(b):
# if b % 2 == 0 and (i % 2 == 1):
# c -= 1
# else:
# c += 1
# if c <= n:
# print(c, end='\t')
# else:
# print(0, end='\t')
# print()
# if (b % 2 == 0) and (i % 2 == 1):
# c += b - 1
|
Title: Parliament of Berland
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* parliamentarians in Berland. They are numbered with integers from 1 to *n*. It happened that all parliamentarians with odd indices are Democrats and all parliamentarians with even indices are Republicans.
New parliament assembly hall is a rectangle consisting of *a*<=×<=*b* chairs — *a* rows of *b* chairs each. Two chairs are considered neighbouring if they share as side. For example, chair number 5 in row number 2 is neighbouring to chairs number 4 and 6 in this row and chairs with number 5 in rows 1 and 3. Thus, chairs have four neighbours in general, except for the chairs on the border of the hall
We know that if two parliamentarians from one political party (that is two Democrats or two Republicans) seat nearby they spent all time discussing internal party issues.
Write the program that given the number of parliamentarians and the sizes of the hall determine if there is a way to find a seat for any parliamentarian, such that no two members of the same party share neighbouring seats.
Input Specification:
The first line of the input contains three integers *n*, *a* and *b* (1<=≤<=*n*<=≤<=10<=000, 1<=≤<=*a*,<=*b*<=≤<=100) — the number of parliamentarians, the number of rows in the assembly hall and the number of seats in each row, respectively.
Output Specification:
If there is no way to assigns seats to parliamentarians in a proper way print -1.
Otherwise print the solution in *a* lines, each containing *b* integers. The *j*-th integer of the *i*-th line should be equal to the index of parliamentarian occupying this seat, or 0 if this seat should remain empty. If there are multiple possible solution, you may print any of them.
Demo Input:
['3 2 2\n', '8 4 3\n', '10 2 2\n']
Demo Output:
['0 3\n1 2\n', '7 8 3\n0 1 4\n6 0 5\n0 2 0\n', '-1\n']
Note:
In the first sample there are many other possible solutions. For example,
and
The following assignment
is incorrect, because parliamentarians 1 and 3 are both from Democrats party but will occupy neighbouring seats.
|
```python
n, a, b = map(int, (input().split()))
if n > (a * b):
print(-1)
else:
for i in range(a):
if (b % 2 == 0) and ((i + 1) % 2 == 0):
for c in range(b * (i + 1), b * (i + 1) - b, -1):
if c <= n:
print(c, end='\t')
else:
print(0, end='\t')
else:
for c in range((b * (i + 1) - b) + 1, b * (i + 1) + 1):
if c <= n:
print(c, end='\t')
else:
print(0, end='\t')
print()
# n, a, b = map(int, (input().split()))
# if n > (a * b):
# print(-1)
# else:
# c = 0
# for i in range(a):
# if (b % 2 == 0) and (i % 2 == 1):
# c += b + 1
# for _ in range(b):
# if b % 2 == 0 and (i % 2 == 1):
# c -= 1
# else:
# c += 1
# if c <= n:
# print(c, end='\t')
# else:
# print(0, end='\t')
# print()
# if (b % 2 == 0) and (i % 2 == 1):
# c += b - 1
```
| 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"
}
] | 1,670,193,021
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 35
| 92
| 0
|
a,b = map(int,input().split())
if a*b < 2:
print(0)
elif (a*b)%2 == 0:
print(int(a*b/2))
else:
print(int((a*b - 1)/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())
if a*b < 2:
print(0)
elif (a*b)%2 == 0:
print(int(a*b/2))
else:
print(int((a*b - 1)/2))
```
| 3.977
|
158
|
A
|
Next Round
|
PROGRAMMING
| 800
|
[
"*special",
"implementation"
] | null | null |
"Contestant who earns a score equal to or greater than the *k*-th place finisher's score will advance to the next round, as long as the contestant earns a positive score..." — an excerpt from contest rules.
A total of *n* participants took part in the contest (*n*<=≥<=*k*), and you already know their scores. Calculate how many participants will advance to the next round.
|
The first line of the input contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=50) separated by a single space.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100), where *a**i* is the score earned by the participant who got the *i*-th place. The given sequence is non-increasing (that is, for all *i* from 1 to *n*<=-<=1 the following condition is fulfilled: *a**i*<=≥<=*a**i*<=+<=1).
|
Output the number of participants who advance to the next round.
|
[
"8 5\n10 9 8 7 7 7 5 5\n",
"4 2\n0 0 0 0\n"
] |
[
"6\n",
"0\n"
] |
In the first example the participant on the 5th place earned 7 points. As the participant on the 6th place also earned 7 points, there are 6 advancers.
In the second example nobody got a positive score.
| 500
|
[
{
"input": "8 5\n10 9 8 7 7 7 5 5",
"output": "6"
},
{
"input": "4 2\n0 0 0 0",
"output": "0"
},
{
"input": "5 1\n1 1 1 1 1",
"output": "5"
},
{
"input": "5 5\n1 1 1 1 1",
"output": "5"
},
{
"input": "1 1\n10",
"output": "1"
},
{
"input": "17 14\n16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0",
"output": "14"
},
{
"input": "5 5\n3 2 1 0 0",
"output": "3"
},
{
"input": "8 6\n10 9 8 7 7 7 5 5",
"output": "6"
},
{
"input": "8 7\n10 9 8 7 7 7 5 5",
"output": "8"
},
{
"input": "8 4\n10 9 8 7 7 7 5 5",
"output": "6"
},
{
"input": "8 3\n10 9 8 7 7 7 5 5",
"output": "3"
},
{
"input": "8 1\n10 9 8 7 7 7 5 5",
"output": "1"
},
{
"input": "8 2\n10 9 8 7 7 7 5 5",
"output": "2"
},
{
"input": "1 1\n100",
"output": "1"
},
{
"input": "1 1\n0",
"output": "0"
},
{
"input": "50 25\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": "50"
},
{
"input": "50 25\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 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": "25"
},
{
"input": "50 25\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 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": "26"
},
{
"input": "50 25\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 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": "50"
},
{
"input": "11 5\n100 99 98 97 96 95 94 93 92 91 90",
"output": "5"
},
{
"input": "10 4\n100 81 70 69 64 43 34 29 15 3",
"output": "4"
},
{
"input": "11 6\n87 71 62 52 46 46 43 35 32 25 12",
"output": "6"
},
{
"input": "17 12\n99 88 86 82 75 75 74 65 58 52 45 30 21 16 7 2 2",
"output": "12"
},
{
"input": "20 3\n98 98 96 89 87 82 82 80 76 74 74 68 61 60 43 32 30 22 4 2",
"output": "3"
},
{
"input": "36 12\n90 87 86 85 83 80 79 78 76 70 69 69 61 61 59 58 56 48 45 44 42 41 33 31 27 25 23 21 20 19 15 14 12 7 5 5",
"output": "12"
},
{
"input": "49 8\n99 98 98 96 92 92 90 89 89 86 86 85 83 80 79 76 74 69 67 67 58 56 55 51 49 47 47 46 45 41 41 40 39 34 34 33 25 23 18 15 13 13 11 9 5 4 3 3 1",
"output": "9"
},
{
"input": "49 29\n100 98 98 96 96 96 95 87 85 84 81 76 74 70 63 63 63 62 57 57 56 54 53 52 50 47 45 41 41 39 38 31 30 28 27 26 23 22 20 15 15 11 7 6 6 4 2 1 0",
"output": "29"
},
{
"input": "49 34\n99 98 96 96 93 92 90 89 88 86 85 85 82 76 73 69 66 64 63 63 60 59 57 57 56 55 54 54 51 48 47 44 42 42 40 39 38 36 33 26 24 23 19 17 17 14 12 7 4",
"output": "34"
},
{
"input": "50 44\n100 100 99 97 95 91 91 84 83 83 79 71 70 69 69 62 61 60 59 59 58 58 58 55 55 54 52 48 47 45 44 44 38 36 32 31 28 28 25 25 24 24 24 22 17 15 14 13 12 4",
"output": "44"
},
{
"input": "50 13\n99 95 94 94 88 87 81 79 78 76 74 72 72 69 68 67 67 67 66 63 62 61 58 57 55 55 54 51 50 50 48 48 42 41 38 35 34 32 31 30 26 24 13 13 12 6 5 4 3 3",
"output": "13"
},
{
"input": "50 30\n100 98 96 94 91 89 88 81 81 81 81 81 76 73 72 71 70 69 66 64 61 59 59 56 52 50 49 48 43 39 36 35 34 34 31 29 27 26 24 22 16 16 15 14 14 14 9 7 4 3",
"output": "30"
},
{
"input": "2 1\n10 10",
"output": "2"
},
{
"input": "2 2\n10 10",
"output": "2"
},
{
"input": "2 2\n10 0",
"output": "1"
},
{
"input": "2 2\n10 1",
"output": "2"
},
{
"input": "2 1\n10 0",
"output": "1"
},
{
"input": "2 1\n10 2",
"output": "1"
},
{
"input": "50 13\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": "0"
},
{
"input": "50 1\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": "0"
},
{
"input": "50 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": "0"
},
{
"input": "10 1\n5 5 5 3 3 3 0 0 0 0",
"output": "3"
},
{
"input": "10 2\n5 5 5 3 3 3 0 0 0 0",
"output": "3"
},
{
"input": "10 3\n5 5 5 3 3 3 0 0 0 0",
"output": "3"
},
{
"input": "10 4\n5 5 5 3 3 3 0 0 0 0",
"output": "6"
},
{
"input": "10 5\n5 5 5 3 3 3 0 0 0 0",
"output": "6"
},
{
"input": "10 6\n5 5 5 3 3 3 0 0 0 0",
"output": "6"
},
{
"input": "10 7\n5 5 5 3 3 3 0 0 0 0",
"output": "6"
},
{
"input": "10 8\n5 5 5 3 3 3 0 0 0 0",
"output": "6"
},
{
"input": "10 9\n5 5 5 3 3 3 0 0 0 0",
"output": "6"
},
{
"input": "10 10\n5 5 5 3 3 3 0 0 0 0",
"output": "6"
}
] | 1,696,671,251
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 50
| 92
| 0
|
n_k = input().split()
n_k = [int(n_k[i]) for i in range(len(n_k))]
points = input().split()
points = [int(points[i]) for i in range(len(points))]
for winner in range(n_k[1]):
max_ = max(points)
if max_ == 0:break
for i in range(points.count(max_)):
points.remove(max_)
if n_k[0] - len(points) >= n_k[1]:
break
print(n_k[0] - len(points))
|
Title: Next Round
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
"Contestant who earns a score equal to or greater than the *k*-th place finisher's score will advance to the next round, as long as the contestant earns a positive score..." — an excerpt from contest rules.
A total of *n* participants took part in the contest (*n*<=≥<=*k*), and you already know their scores. Calculate how many participants will advance to the next round.
Input Specification:
The first line of the input contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=50) separated by a single space.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100), where *a**i* is the score earned by the participant who got the *i*-th place. The given sequence is non-increasing (that is, for all *i* from 1 to *n*<=-<=1 the following condition is fulfilled: *a**i*<=≥<=*a**i*<=+<=1).
Output Specification:
Output the number of participants who advance to the next round.
Demo Input:
['8 5\n10 9 8 7 7 7 5 5\n', '4 2\n0 0 0 0\n']
Demo Output:
['6\n', '0\n']
Note:
In the first example the participant on the 5th place earned 7 points. As the participant on the 6th place also earned 7 points, there are 6 advancers.
In the second example nobody got a positive score.
|
```python
n_k = input().split()
n_k = [int(n_k[i]) for i in range(len(n_k))]
points = input().split()
points = [int(points[i]) for i in range(len(points))]
for winner in range(n_k[1]):
max_ = max(points)
if max_ == 0:break
for i in range(points.count(max_)):
points.remove(max_)
if n_k[0] - len(points) >= n_k[1]:
break
print(n_k[0] - len(points))
```
| 3
|
|
999
|
C
|
Alphabetic Removals
|
PROGRAMMING
| 1,200
|
[
"implementation"
] | null | null |
You are given a string $s$ consisting of $n$ lowercase Latin letters. Polycarp wants to remove exactly $k$ characters ($k \le n$) from the string $s$. Polycarp uses the following algorithm $k$ times:
- if there is at least one letter 'a', remove the leftmost occurrence and stop the algorithm, otherwise go to next item; - if there is at least one letter 'b', remove the leftmost occurrence and stop the algorithm, otherwise go to next item; - ... - remove the leftmost occurrence of the letter 'z' and stop the algorithm.
This algorithm removes a single letter from the string. Polycarp performs this algorithm exactly $k$ times, thus removing exactly $k$ characters.
Help Polycarp find the resulting string.
|
The first line of input contains two integers $n$ and $k$ ($1 \le k \le n \le 4 \cdot 10^5$) — the length of the string and the number of letters Polycarp will remove.
The second line contains the string $s$ consisting of $n$ lowercase Latin letters.
|
Print the string that will be obtained from $s$ after Polycarp removes exactly $k$ letters using the above algorithm $k$ times.
If the resulting string is empty, print nothing. It is allowed to print nothing or an empty line (line break).
|
[
"15 3\ncccaabababaccbc\n",
"15 9\ncccaabababaccbc\n",
"1 1\nu\n"
] |
[
"cccbbabaccbc\n",
"cccccc\n",
""
] |
none
| 0
|
[
{
"input": "15 3\ncccaabababaccbc",
"output": "cccbbabaccbc"
},
{
"input": "15 9\ncccaabababaccbc",
"output": "cccccc"
},
{
"input": "5 2\nzyzyx",
"output": "zzy"
},
{
"input": "4 3\nhack",
"output": "k"
},
{
"input": "4 3\nzzzz",
"output": "z"
},
{
"input": "6 5\naaccdd",
"output": "d"
},
{
"input": "2 1\nzz",
"output": "z"
},
{
"input": "14 5\nhxehmvkybeklnj",
"output": "xmvkyklnj"
}
] | 1,622,708,508
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 44
| 124
| 1,228,800
|
al="abcdefghijklmnopqrstuvwxyz"
n,k=[int(x)for x in input().rstrip().split()]
a=input()
for i in al:
c=a.count(i)
a=a.replace(i,"",min(c,k))
k-=c
if k<1:
break
print(a)
|
Title: Alphabetic Removals
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a string $s$ consisting of $n$ lowercase Latin letters. Polycarp wants to remove exactly $k$ characters ($k \le n$) from the string $s$. Polycarp uses the following algorithm $k$ times:
- if there is at least one letter 'a', remove the leftmost occurrence and stop the algorithm, otherwise go to next item; - if there is at least one letter 'b', remove the leftmost occurrence and stop the algorithm, otherwise go to next item; - ... - remove the leftmost occurrence of the letter 'z' and stop the algorithm.
This algorithm removes a single letter from the string. Polycarp performs this algorithm exactly $k$ times, thus removing exactly $k$ characters.
Help Polycarp find the resulting string.
Input Specification:
The first line of input contains two integers $n$ and $k$ ($1 \le k \le n \le 4 \cdot 10^5$) — the length of the string and the number of letters Polycarp will remove.
The second line contains the string $s$ consisting of $n$ lowercase Latin letters.
Output Specification:
Print the string that will be obtained from $s$ after Polycarp removes exactly $k$ letters using the above algorithm $k$ times.
If the resulting string is empty, print nothing. It is allowed to print nothing or an empty line (line break).
Demo Input:
['15 3\ncccaabababaccbc\n', '15 9\ncccaabababaccbc\n', '1 1\nu\n']
Demo Output:
['cccbbabaccbc\n', 'cccccc\n', '']
Note:
none
|
```python
al="abcdefghijklmnopqrstuvwxyz"
n,k=[int(x)for x in input().rstrip().split()]
a=input()
for i in al:
c=a.count(i)
a=a.replace(i,"",min(c,k))
k-=c
if k<1:
break
print(a)
```
| 3
|
|
898
|
A
|
Rounding
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
Vasya has a non-negative integer *n*. He wants to round it to nearest integer, which ends up with 0. If *n* already ends up with 0, Vasya considers it already rounded.
For example, if *n*<==<=4722 answer is 4720. If *n*<==<=5 Vasya can round it to 0 or to 10. Both ways are correct.
For given *n* find out to which integer will Vasya round it.
|
The first line contains single integer *n* (0<=≤<=*n*<=≤<=109) — number that Vasya has.
|
Print result of rounding *n*. Pay attention that in some cases answer isn't unique. In that case print any correct answer.
|
[
"5\n",
"113\n",
"1000000000\n",
"5432359\n"
] |
[
"0\n",
"110\n",
"1000000000\n",
"5432360\n"
] |
In the first example *n* = 5. Nearest integers, that ends up with zero are 0 and 10. Any of these answers is correct, so you can print 0 or 10.
| 500
|
[
{
"input": "5",
"output": "0"
},
{
"input": "113",
"output": "110"
},
{
"input": "1000000000",
"output": "1000000000"
},
{
"input": "5432359",
"output": "5432360"
},
{
"input": "999999994",
"output": "999999990"
},
{
"input": "10",
"output": "10"
},
{
"input": "9",
"output": "10"
},
{
"input": "1",
"output": "0"
},
{
"input": "0",
"output": "0"
},
{
"input": "3",
"output": "0"
},
{
"input": "4",
"output": "0"
},
{
"input": "6",
"output": "10"
},
{
"input": "7",
"output": "10"
},
{
"input": "8",
"output": "10"
},
{
"input": "19",
"output": "20"
},
{
"input": "100",
"output": "100"
},
{
"input": "997",
"output": "1000"
},
{
"input": "9994",
"output": "9990"
},
{
"input": "10002",
"output": "10000"
},
{
"input": "100000",
"output": "100000"
},
{
"input": "99999",
"output": "100000"
},
{
"input": "999999999",
"output": "1000000000"
},
{
"input": "999999998",
"output": "1000000000"
},
{
"input": "999999995",
"output": "999999990"
},
{
"input": "999999990",
"output": "999999990"
},
{
"input": "1000000",
"output": "1000000"
},
{
"input": "1000010",
"output": "1000010"
},
{
"input": "10000010",
"output": "10000010"
},
{
"input": "100000011",
"output": "100000010"
},
{
"input": "400000003",
"output": "400000000"
},
{
"input": "234234",
"output": "234230"
},
{
"input": "675621",
"output": "675620"
},
{
"input": "43532",
"output": "43530"
},
{
"input": "4576453",
"output": "4576450"
},
{
"input": "65754674",
"output": "65754670"
},
{
"input": "3245526",
"output": "3245530"
},
{
"input": "123445",
"output": "123440"
},
{
"input": "234217",
"output": "234220"
},
{
"input": "23451218",
"output": "23451220"
},
{
"input": "1231239",
"output": "1231240"
},
{
"input": "1923140",
"output": "1923140"
},
{
"input": "307910310",
"output": "307910310"
},
{
"input": "780961030",
"output": "780961030"
},
{
"input": "103509421",
"output": "103509420"
},
{
"input": "576560141",
"output": "576560140"
},
{
"input": "48851642",
"output": "48851640"
},
{
"input": "226935072",
"output": "226935070"
},
{
"input": "844450763",
"output": "844450760"
},
{
"input": "22534183",
"output": "22534180"
},
{
"input": "640049874",
"output": "640049870"
},
{
"input": "818133304",
"output": "818133300"
},
{
"input": "730616285",
"output": "730616280"
},
{
"input": "613732415",
"output": "613732410"
},
{
"input": "380991216",
"output": "380991220"
},
{
"input": "559074636",
"output": "559074640"
},
{
"input": "176590327",
"output": "176590330"
},
{
"input": "354673757",
"output": "354673760"
},
{
"input": "267156738",
"output": "267156740"
},
{
"input": "150272868",
"output": "150272870"
},
{
"input": "62755859",
"output": "62755860"
},
{
"input": "945871979",
"output": "945871980"
},
{
"input": "46",
"output": "50"
},
{
"input": "999",
"output": "1000"
},
{
"input": "1397",
"output": "1400"
}
] | 1,611,873,164
| 164
|
PyPy 3
|
OK
|
TESTS
| 64
| 109
| 0
|
n = int(input())
q,r = divmod(n, 10)
a = q*10
b = (q+1)*10
if n-a <= b-n:
print(a)
else:
print(b)
|
Title: Rounding
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has a non-negative integer *n*. He wants to round it to nearest integer, which ends up with 0. If *n* already ends up with 0, Vasya considers it already rounded.
For example, if *n*<==<=4722 answer is 4720. If *n*<==<=5 Vasya can round it to 0 or to 10. Both ways are correct.
For given *n* find out to which integer will Vasya round it.
Input Specification:
The first line contains single integer *n* (0<=≤<=*n*<=≤<=109) — number that Vasya has.
Output Specification:
Print result of rounding *n*. Pay attention that in some cases answer isn't unique. In that case print any correct answer.
Demo Input:
['5\n', '113\n', '1000000000\n', '5432359\n']
Demo Output:
['0\n', '110\n', '1000000000\n', '5432360\n']
Note:
In the first example *n* = 5. Nearest integers, that ends up with zero are 0 and 10. Any of these answers is correct, so you can print 0 or 10.
|
```python
n = int(input())
q,r = divmod(n, 10)
a = q*10
b = (q+1)*10
if n-a <= b-n:
print(a)
else:
print(b)
```
| 3
|
|
637
|
B
|
Chat Order
|
PROGRAMMING
| 1,200
|
[
"*special",
"binary search",
"constructive algorithms",
"data structures",
"sortings"
] | null | null |
Polycarp is a big lover of killing time in social networks. A page with a chatlist in his favourite network is made so that when a message is sent to some friend, his friend's chat rises to the very top of the page. The relative order of the other chats doesn't change. If there was no chat with this friend before, then a new chat is simply inserted to the top of the list.
Assuming that the chat list is initially empty, given the sequence of Polycaprus' messages make a list of chats after all of his messages are processed. Assume that no friend wrote any message to Polycarpus.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of Polycarpus' messages. Next *n* lines enlist the message recipients in the order in which the messages were sent. The name of each participant is a non-empty sequence of lowercase English letters of length at most 10.
|
Print all the recipients to who Polycarp talked to in the order of chats with them, from top to bottom.
|
[
"4\nalex\nivan\nroman\nivan\n",
"8\nalina\nmaria\nekaterina\ndarya\ndarya\nekaterina\nmaria\nalina\n"
] |
[
"ivan\nroman\nalex\n",
"alina\nmaria\nekaterina\ndarya\n"
] |
In the first test case Polycarpus first writes to friend by name "alex", and the list looks as follows:
1. alex
Then Polycarpus writes to friend by name "ivan" and the list looks as follows:
1. ivan 1. alex
Polycarpus writes the third message to friend by name "roman" and the list looks as follows:
1. roman 1. ivan 1. alex
Polycarpus writes the fourth message to friend by name "ivan", to who he has already sent a message, so the list of chats changes as follows:
1. ivan 1. roman 1. alex
| 1,000
|
[
{
"input": "4\nalex\nivan\nroman\nivan",
"output": "ivan\nroman\nalex"
},
{
"input": "8\nalina\nmaria\nekaterina\ndarya\ndarya\nekaterina\nmaria\nalina",
"output": "alina\nmaria\nekaterina\ndarya"
},
{
"input": "1\nwdi",
"output": "wdi"
},
{
"input": "2\nypg\nypg",
"output": "ypg"
},
{
"input": "3\nexhll\nexhll\narruapexj",
"output": "arruapexj\nexhll"
},
{
"input": "3\nfv\nle\nle",
"output": "le\nfv"
},
{
"input": "8\nm\nm\nm\nm\nm\nm\nm\nm",
"output": "m"
},
{
"input": "10\nr\nr\ni\nw\nk\nr\nb\nu\nu\nr",
"output": "r\nu\nb\nk\nw\ni"
},
{
"input": "7\ne\nfau\ncmk\nnzs\nby\nwx\ntjmok",
"output": "tjmok\nwx\nby\nnzs\ncmk\nfau\ne"
},
{
"input": "6\nklrj\nwe\nklrj\nwe\nwe\nwe",
"output": "we\nklrj"
},
{
"input": "8\nzncybqmh\naeebef\nzncybqmh\nn\naeebef\nzncybqmh\nzncybqmh\nzncybqmh",
"output": "zncybqmh\naeebef\nn"
},
{
"input": "30\nkqqcbs\nvap\nkymomn\nj\nkqqcbs\nfuzlzoum\nkymomn\ndbh\nfuzlzoum\nkymomn\nvap\nvlgzs\ndbh\nvlgzs\nbvy\ndbh\nkymomn\nkymomn\neoqql\nkymomn\nkymomn\nkqqcbs\nvlgzs\nkqqcbs\nkqqcbs\nfuzlzoum\nvlgzs\nrylgdoo\nvlgzs\nrylgdoo",
"output": "rylgdoo\nvlgzs\nfuzlzoum\nkqqcbs\nkymomn\neoqql\ndbh\nbvy\nvap\nj"
},
{
"input": "40\nji\nv\nv\nns\nji\nn\nji\nv\nfvy\nvje\nns\nvje\nv\nhas\nv\nusm\nhas\nfvy\nvje\nkdb\nn\nv\nji\nji\nn\nhas\nv\nji\nkdb\nr\nvje\nns\nv\nusm\nn\nvje\nhas\nns\nhas\nn",
"output": "n\nhas\nns\nvje\nusm\nv\nr\nkdb\nji\nfvy"
},
{
"input": "50\njcg\nvle\njopb\nepdb\nnkef\nfv\nxj\nufe\nfuy\noqta\ngbc\nyuz\nec\nyji\nkuux\ncwm\ntq\nnno\nhp\nzry\nxxpp\ntjvo\ngyz\nkwo\nvwqz\nyaqc\njnj\nwoav\nqcv\ndcu\ngc\nhovn\nop\nevy\ndc\ntrpu\nyb\nuzfa\npca\noq\nnhxy\nsiqu\nde\nhphy\nc\nwovu\nf\nbvv\ndsik\nlwyg",
"output": "lwyg\ndsik\nbvv\nf\nwovu\nc\nhphy\nde\nsiqu\nnhxy\noq\npca\nuzfa\nyb\ntrpu\ndc\nevy\nop\nhovn\ngc\ndcu\nqcv\nwoav\njnj\nyaqc\nvwqz\nkwo\ngyz\ntjvo\nxxpp\nzry\nhp\nnno\ntq\ncwm\nkuux\nyji\nec\nyuz\ngbc\noqta\nfuy\nufe\nxj\nfv\nnkef\nepdb\njopb\nvle\njcg"
},
{
"input": "100\nvhh\nvhh\nvhh\nfa\nfa\nvhh\nvhh\nvhh\nfa\nfa\nfa\nvhh\nfa\nvhh\nvhh\nvhh\nfa\nvhh\nvhh\nfa\nfa\nfa\nfa\nfa\nfa\nvhh\nfa\nfa\nvhh\nvhh\nvhh\nfa\nfa\nfa\nvhh\nfa\nvhh\nfa\nvhh\nvhh\nfa\nvhh\nfa\nvhh\nvhh\nvhh\nfa\nvhh\nfa\nfa\nvhh\nfa\nvhh\nvhh\nvhh\nvhh\nfa\nvhh\nvhh\nvhh\nvhh\nfa\nvhh\nvhh\nvhh\nvhh\nvhh\nfa\nvhh\nvhh\nfa\nfa\nfa\nvhh\nfa\nfa\nvhh\nfa\nvhh\nfa\nfa\nfa\nfa\nfa\nfa\nvhh\nvhh\nfa\nvhh\nfa\nfa\nvhh\nfa\nfa\nvhh\nfa\nvhh\nvhh\nfa\nvhh",
"output": "vhh\nfa"
},
{
"input": "2\naa\nbb",
"output": "bb\naa"
},
{
"input": "2\naa\na",
"output": "a\naa"
},
{
"input": "3\naa\naa\naa",
"output": "aa"
},
{
"input": "5\naa\na\naa\na\naa",
"output": "aa\na"
},
{
"input": "7\naaaa\naaaa\naaa\na\naa\naaaaaaa\naaa",
"output": "aaa\naaaaaaa\naa\na\naaaa"
},
{
"input": "5\na\naa\naaa\naaaa\na",
"output": "a\naaaa\naaa\naa"
},
{
"input": "12\naaaaa\naaaaaa\naaaa\naaaaaa\naa\naaaa\naaaa\naaaaaa\na\naaa\naaaaaaaa\naa",
"output": "aa\naaaaaaaa\naaa\na\naaaaaa\naaaa\naaaaa"
},
{
"input": "3\na\naa\naaa",
"output": "aaa\naa\na"
},
{
"input": "9\nzzz\nzzzzz\nzzz\nzzzz\nzz\nzzzz\nzzzzz\nzzzz\nzzzzzzz",
"output": "zzzzzzz\nzzzz\nzzzzz\nzz\nzzz"
}
] | 1,694,280,754
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 95
| 1,357
| 31,846,400
|
x = []
s = set()
n = int(input())
for g in range(0,n):
i = input()
x.append(i)
for d in x[::-1]:
if d not in s:
print(d)
s.add(d)
|
Title: Chat Order
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp is a big lover of killing time in social networks. A page with a chatlist in his favourite network is made so that when a message is sent to some friend, his friend's chat rises to the very top of the page. The relative order of the other chats doesn't change. If there was no chat with this friend before, then a new chat is simply inserted to the top of the list.
Assuming that the chat list is initially empty, given the sequence of Polycaprus' messages make a list of chats after all of his messages are processed. Assume that no friend wrote any message to Polycarpus.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of Polycarpus' messages. Next *n* lines enlist the message recipients in the order in which the messages were sent. The name of each participant is a non-empty sequence of lowercase English letters of length at most 10.
Output Specification:
Print all the recipients to who Polycarp talked to in the order of chats with them, from top to bottom.
Demo Input:
['4\nalex\nivan\nroman\nivan\n', '8\nalina\nmaria\nekaterina\ndarya\ndarya\nekaterina\nmaria\nalina\n']
Demo Output:
['ivan\nroman\nalex\n', 'alina\nmaria\nekaterina\ndarya\n']
Note:
In the first test case Polycarpus first writes to friend by name "alex", and the list looks as follows:
1. alex
Then Polycarpus writes to friend by name "ivan" and the list looks as follows:
1. ivan 1. alex
Polycarpus writes the third message to friend by name "roman" and the list looks as follows:
1. roman 1. ivan 1. alex
Polycarpus writes the fourth message to friend by name "ivan", to who he has already sent a message, so the list of chats changes as follows:
1. ivan 1. roman 1. alex
|
```python
x = []
s = set()
n = int(input())
for g in range(0,n):
i = input()
x.append(i)
for d in x[::-1]:
if d not in s:
print(d)
s.add(d)
```
| 3
|
|
1,009
|
A
|
Game Shopping
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Maxim wants to buy some games at the local game shop. There are $n$ games in the shop, the $i$-th game costs $c_i$.
Maxim has a wallet which can be represented as an array of integers. His wallet contains $m$ bills, the $j$-th bill has value $a_j$.
Games in the shop are ordered from left to right, Maxim tries to buy every game in that order.
When Maxim stands at the position $i$ in the shop, he takes the first bill from his wallet (if his wallet is empty then he proceeds to the next position immediately) and tries to buy the $i$-th game using this bill. After Maxim tried to buy the $n$-th game, he leaves the shop.
Maxim buys the $i$-th game if and only if the value of the first bill (which he takes) from his wallet is greater or equal to the cost of the $i$-th game. If he successfully buys the $i$-th game, the first bill from his wallet disappears and the next bill becomes first. Otherwise Maxim leaves the first bill in his wallet (this bill still remains the first one) and proceeds to the next game.
For example, for array $c = [2, 4, 5, 2, 4]$ and array $a = [5, 3, 4, 6]$ the following process takes place: Maxim buys the first game using the first bill (its value is $5$), the bill disappears, after that the second bill (with value $3$) becomes the first one in Maxim's wallet, then Maxim doesn't buy the second game because $c_2 > a_2$, the same with the third game, then he buys the fourth game using the bill of value $a_2$ (the third bill becomes the first one in Maxim's wallet) and buys the fifth game using the bill of value $a_3$.
Your task is to get the number of games Maxim will buy.
|
The first line of the input contains two integers $n$ and $m$ ($1 \le n, m \le 1000$) — the number of games and the number of bills in Maxim's wallet.
The second line of the input contains $n$ integers $c_1, c_2, \dots, c_n$ ($1 \le c_i \le 1000$), where $c_i$ is the cost of the $i$-th game.
The third line of the input contains $m$ integers $a_1, a_2, \dots, a_m$ ($1 \le a_j \le 1000$), where $a_j$ is the value of the $j$-th bill from the Maxim's wallet.
|
Print a single integer — the number of games Maxim will buy.
|
[
"5 4\n2 4 5 2 4\n5 3 4 6\n",
"5 2\n20 40 50 20 40\n19 20\n",
"6 4\n4 8 15 16 23 42\n1000 1000 1000 1000\n"
] |
[
"3\n",
"0\n",
"4\n"
] |
The first example is described in the problem statement.
In the second example Maxim cannot buy any game because the value of the first bill in his wallet is smaller than the cost of any game in the shop.
In the third example the values of the bills in Maxim's wallet are large enough to buy any game he encounter until he runs out of bills in his wallet.
| 0
|
[
{
"input": "5 4\n2 4 5 2 4\n5 3 4 6",
"output": "3"
},
{
"input": "5 2\n20 40 50 20 40\n19 20",
"output": "0"
},
{
"input": "6 4\n4 8 15 16 23 42\n1000 1000 1000 1000",
"output": "4"
},
{
"input": "5 1\n1 1 1 1 1\n5",
"output": "1"
},
{
"input": "5 1\n10 1 1 1 1\n1000",
"output": "1"
},
{
"input": "5 1\n100 100 100 100 100\n100",
"output": "1"
},
{
"input": "2 1\n2 1\n1",
"output": "1"
},
{
"input": "2 3\n3 1\n2 4 2",
"output": "1"
},
{
"input": "1 5\n4\n1 4 3 3 2",
"output": "0"
},
{
"input": "5 3\n4 2 3 1 1\n2 1 3",
"output": "3"
},
{
"input": "3 5\n5 2 5\n1 4 1 4 2",
"output": "0"
},
{
"input": "7 3\n9 7 10 2 1 1 1\n8 9 6",
"output": "3"
},
{
"input": "5 3\n2 5 3 3 2\n2 5 3",
"output": "3"
}
] | 1,531,662,583
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 19
| 109
| 0
|
a=[int(q) for q in input().strip().split()]
b=[int(q) for q in input().strip().split()]
c=[int(q) for q in input().strip().split()]
ct=0
while len(b)>0 and len(c)>0:
if b[0]<=c[0]:
c.remove(c[0])
ct+=1
b.remove(b[0])
print(ct)
|
Title: Game Shopping
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Maxim wants to buy some games at the local game shop. There are $n$ games in the shop, the $i$-th game costs $c_i$.
Maxim has a wallet which can be represented as an array of integers. His wallet contains $m$ bills, the $j$-th bill has value $a_j$.
Games in the shop are ordered from left to right, Maxim tries to buy every game in that order.
When Maxim stands at the position $i$ in the shop, he takes the first bill from his wallet (if his wallet is empty then he proceeds to the next position immediately) and tries to buy the $i$-th game using this bill. After Maxim tried to buy the $n$-th game, he leaves the shop.
Maxim buys the $i$-th game if and only if the value of the first bill (which he takes) from his wallet is greater or equal to the cost of the $i$-th game. If he successfully buys the $i$-th game, the first bill from his wallet disappears and the next bill becomes first. Otherwise Maxim leaves the first bill in his wallet (this bill still remains the first one) and proceeds to the next game.
For example, for array $c = [2, 4, 5, 2, 4]$ and array $a = [5, 3, 4, 6]$ the following process takes place: Maxim buys the first game using the first bill (its value is $5$), the bill disappears, after that the second bill (with value $3$) becomes the first one in Maxim's wallet, then Maxim doesn't buy the second game because $c_2 > a_2$, the same with the third game, then he buys the fourth game using the bill of value $a_2$ (the third bill becomes the first one in Maxim's wallet) and buys the fifth game using the bill of value $a_3$.
Your task is to get the number of games Maxim will buy.
Input Specification:
The first line of the input contains two integers $n$ and $m$ ($1 \le n, m \le 1000$) — the number of games and the number of bills in Maxim's wallet.
The second line of the input contains $n$ integers $c_1, c_2, \dots, c_n$ ($1 \le c_i \le 1000$), where $c_i$ is the cost of the $i$-th game.
The third line of the input contains $m$ integers $a_1, a_2, \dots, a_m$ ($1 \le a_j \le 1000$), where $a_j$ is the value of the $j$-th bill from the Maxim's wallet.
Output Specification:
Print a single integer — the number of games Maxim will buy.
Demo Input:
['5 4\n2 4 5 2 4\n5 3 4 6\n', '5 2\n20 40 50 20 40\n19 20\n', '6 4\n4 8 15 16 23 42\n1000 1000 1000 1000\n']
Demo Output:
['3\n', '0\n', '4\n']
Note:
The first example is described in the problem statement.
In the second example Maxim cannot buy any game because the value of the first bill in his wallet is smaller than the cost of any game in the shop.
In the third example the values of the bills in Maxim's wallet are large enough to buy any game he encounter until he runs out of bills in his wallet.
|
```python
a=[int(q) for q in input().strip().split()]
b=[int(q) for q in input().strip().split()]
c=[int(q) for q in input().strip().split()]
ct=0
while len(b)>0 and len(c)>0:
if b[0]<=c[0]:
c.remove(c[0])
ct+=1
b.remove(b[0])
print(ct)
```
| 3
|
|
898
|
A
|
Rounding
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
Vasya has a non-negative integer *n*. He wants to round it to nearest integer, which ends up with 0. If *n* already ends up with 0, Vasya considers it already rounded.
For example, if *n*<==<=4722 answer is 4720. If *n*<==<=5 Vasya can round it to 0 or to 10. Both ways are correct.
For given *n* find out to which integer will Vasya round it.
|
The first line contains single integer *n* (0<=≤<=*n*<=≤<=109) — number that Vasya has.
|
Print result of rounding *n*. Pay attention that in some cases answer isn't unique. In that case print any correct answer.
|
[
"5\n",
"113\n",
"1000000000\n",
"5432359\n"
] |
[
"0\n",
"110\n",
"1000000000\n",
"5432360\n"
] |
In the first example *n* = 5. Nearest integers, that ends up with zero are 0 and 10. Any of these answers is correct, so you can print 0 or 10.
| 500
|
[
{
"input": "5",
"output": "0"
},
{
"input": "113",
"output": "110"
},
{
"input": "1000000000",
"output": "1000000000"
},
{
"input": "5432359",
"output": "5432360"
},
{
"input": "999999994",
"output": "999999990"
},
{
"input": "10",
"output": "10"
},
{
"input": "9",
"output": "10"
},
{
"input": "1",
"output": "0"
},
{
"input": "0",
"output": "0"
},
{
"input": "3",
"output": "0"
},
{
"input": "4",
"output": "0"
},
{
"input": "6",
"output": "10"
},
{
"input": "7",
"output": "10"
},
{
"input": "8",
"output": "10"
},
{
"input": "19",
"output": "20"
},
{
"input": "100",
"output": "100"
},
{
"input": "997",
"output": "1000"
},
{
"input": "9994",
"output": "9990"
},
{
"input": "10002",
"output": "10000"
},
{
"input": "100000",
"output": "100000"
},
{
"input": "99999",
"output": "100000"
},
{
"input": "999999999",
"output": "1000000000"
},
{
"input": "999999998",
"output": "1000000000"
},
{
"input": "999999995",
"output": "999999990"
},
{
"input": "999999990",
"output": "999999990"
},
{
"input": "1000000",
"output": "1000000"
},
{
"input": "1000010",
"output": "1000010"
},
{
"input": "10000010",
"output": "10000010"
},
{
"input": "100000011",
"output": "100000010"
},
{
"input": "400000003",
"output": "400000000"
},
{
"input": "234234",
"output": "234230"
},
{
"input": "675621",
"output": "675620"
},
{
"input": "43532",
"output": "43530"
},
{
"input": "4576453",
"output": "4576450"
},
{
"input": "65754674",
"output": "65754670"
},
{
"input": "3245526",
"output": "3245530"
},
{
"input": "123445",
"output": "123440"
},
{
"input": "234217",
"output": "234220"
},
{
"input": "23451218",
"output": "23451220"
},
{
"input": "1231239",
"output": "1231240"
},
{
"input": "1923140",
"output": "1923140"
},
{
"input": "307910310",
"output": "307910310"
},
{
"input": "780961030",
"output": "780961030"
},
{
"input": "103509421",
"output": "103509420"
},
{
"input": "576560141",
"output": "576560140"
},
{
"input": "48851642",
"output": "48851640"
},
{
"input": "226935072",
"output": "226935070"
},
{
"input": "844450763",
"output": "844450760"
},
{
"input": "22534183",
"output": "22534180"
},
{
"input": "640049874",
"output": "640049870"
},
{
"input": "818133304",
"output": "818133300"
},
{
"input": "730616285",
"output": "730616280"
},
{
"input": "613732415",
"output": "613732410"
},
{
"input": "380991216",
"output": "380991220"
},
{
"input": "559074636",
"output": "559074640"
},
{
"input": "176590327",
"output": "176590330"
},
{
"input": "354673757",
"output": "354673760"
},
{
"input": "267156738",
"output": "267156740"
},
{
"input": "150272868",
"output": "150272870"
},
{
"input": "62755859",
"output": "62755860"
},
{
"input": "945871979",
"output": "945871980"
},
{
"input": "46",
"output": "50"
},
{
"input": "999",
"output": "1000"
},
{
"input": "1397",
"output": "1400"
}
] | 1,700,516,745
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 64
| 46
| 0
|
n = int(input())
r = (n%10)
if r <= 5:
print (n-r)
else:
print (n+(10-r))
|
Title: Rounding
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has a non-negative integer *n*. He wants to round it to nearest integer, which ends up with 0. If *n* already ends up with 0, Vasya considers it already rounded.
For example, if *n*<==<=4722 answer is 4720. If *n*<==<=5 Vasya can round it to 0 or to 10. Both ways are correct.
For given *n* find out to which integer will Vasya round it.
Input Specification:
The first line contains single integer *n* (0<=≤<=*n*<=≤<=109) — number that Vasya has.
Output Specification:
Print result of rounding *n*. Pay attention that in some cases answer isn't unique. In that case print any correct answer.
Demo Input:
['5\n', '113\n', '1000000000\n', '5432359\n']
Demo Output:
['0\n', '110\n', '1000000000\n', '5432360\n']
Note:
In the first example *n* = 5. Nearest integers, that ends up with zero are 0 and 10. Any of these answers is correct, so you can print 0 or 10.
|
```python
n = int(input())
r = (n%10)
if r <= 5:
print (n-r)
else:
print (n+(10-r))
```
| 3
|
|
300
|
A
|
Array
|
PROGRAMMING
| 1,100
|
[
"brute force",
"constructive algorithms",
"implementation"
] | null | null |
Vitaly has an array of *n* distinct integers. Vitaly wants to divide this array into three non-empty sets so as the following conditions hold:
1. The product of all numbers in the first set is less than zero (<=<<=0). 1. The product of all numbers in the second set is greater than zero (<=><=0). 1. The product of all numbers in the third set is equal to zero. 1. Each number from the initial array must occur in exactly one set.
Help Vitaly. Divide the given array.
|
The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=100). The second line contains *n* space-separated distinct integers *a*1,<=*a*2,<=...,<=*a**n* (|*a**i*|<=≤<=103) — the array elements.
|
In the first line print integer *n*1 (*n*1<=><=0) — the number of elements in the first set. Then print *n*1 numbers — the elements that got to the first set.
In the next line print integer *n*2 (*n*2<=><=0) — the number of elements in the second set. Then print *n*2 numbers — the elements that got to the second set.
In the next line print integer *n*3 (*n*3<=><=0) — the number of elements in the third set. Then print *n*3 numbers — the elements that got to the third set.
The printed sets must meet the described conditions. It is guaranteed that the solution exists. If there are several solutions, you are allowed to print any of them.
|
[
"3\n-1 2 0\n",
"4\n-1 -2 -3 0\n"
] |
[
"1 -1\n1 2\n1 0\n",
"1 -1\n2 -3 -2\n1 0\n"
] |
none
| 500
|
[
{
"input": "3\n-1 2 0",
"output": "1 -1\n1 2\n1 0"
},
{
"input": "4\n-1 -2 -3 0",
"output": "1 -1\n2 -3 -2\n1 0"
},
{
"input": "5\n-1 -2 1 2 0",
"output": "1 -1\n2 1 2\n2 0 -2"
},
{
"input": "100\n-64 -51 -75 -98 74 -26 -1 -8 -99 -76 -53 -80 -43 -22 -100 -62 -34 -5 -65 -81 -18 -91 -92 -16 -23 -95 -9 -19 -44 -46 -79 52 -35 4 -87 -7 -90 -20 -71 -61 -67 -50 -66 -68 -49 -27 -32 -57 -85 -59 -30 -36 -3 -77 86 -25 -94 -56 60 -24 -37 -72 -41 -31 11 -48 28 -38 -42 -39 -33 -70 -84 0 -93 -73 -14 -69 -40 -97 -6 -55 -45 -54 -10 -29 -96 -12 -83 -15 -21 -47 17 -2 -63 -89 88 13 -58 -82",
"output": "89 -64 -51 -75 -98 -26 -1 -8 -99 -76 -53 -80 -43 -22 -100 -62 -34 -5 -65 -81 -18 -91 -92 -16 -23 -95 -9 -19 -44 -46 -79 -35 -87 -7 -90 -20 -71 -61 -67 -50 -66 -68 -49 -27 -32 -57 -85 -59 -30 -36 -3 -77 -25 -94 -56 -24 -37 -72 -41 -31 -48 -38 -42 -39 -33 -70 -84 -93 -73 -14 -69 -40 -97 -6 -55 -45 -54 -10 -29 -96 -12 -83 -15 -21 -47 -2 -63 -89 -58 -82\n10 74 52 4 86 60 11 28 17 88 13\n1 0"
},
{
"input": "100\n3 -66 -17 54 24 -29 76 89 32 -37 93 -16 99 -25 51 78 23 68 -95 59 18 34 -45 77 9 39 -10 19 8 73 -5 60 12 31 0 2 26 40 48 30 52 49 27 4 87 57 85 58 -61 50 83 80 69 67 91 97 -96 11 100 56 82 53 13 -92 -72 70 1 -94 -63 47 21 14 74 7 6 33 55 65 64 -41 81 42 36 28 38 20 43 71 90 -88 22 84 -86 15 75 62 44 35 98 46",
"output": "19 -66 -17 -29 -37 -16 -25 -95 -45 -10 -5 -61 -96 -92 -72 -94 -63 -41 -88 -86\n80 3 54 24 76 89 32 93 99 51 78 23 68 59 18 34 77 9 39 19 8 73 60 12 31 2 26 40 48 30 52 49 27 4 87 57 85 58 50 83 80 69 67 91 97 11 100 56 82 53 13 70 1 47 21 14 74 7 6 33 55 65 64 81 42 36 28 38 20 43 71 90 22 84 15 75 62 44 35 98 46\n1 0"
},
{
"input": "100\n-17 16 -70 32 -60 75 -100 -9 -68 -30 -42 86 -88 -98 -47 -5 58 -14 -94 -73 -80 -51 -66 -85 -53 49 -25 -3 -45 -69 -11 -64 83 74 -65 67 13 -91 81 6 -90 -54 -12 -39 0 -24 -71 -41 -44 57 -93 -20 -92 18 -43 -52 -55 -84 -89 -19 40 -4 -99 -26 -87 -36 -56 -61 -62 37 -95 -28 63 23 35 -82 1 -2 -78 -96 -21 -77 -76 -27 -10 -97 -8 46 -15 -48 -34 -59 -7 -29 50 -33 -72 -79 22 38",
"output": "75 -17 -70 -60 -100 -9 -68 -30 -42 -88 -98 -47 -5 -14 -94 -73 -80 -51 -66 -85 -53 -25 -3 -45 -69 -11 -64 -65 -91 -90 -54 -12 -39 -24 -71 -41 -44 -93 -20 -92 -43 -52 -55 -84 -89 -19 -4 -99 -26 -87 -36 -56 -61 -62 -95 -28 -82 -2 -78 -96 -21 -77 -76 -27 -10 -97 -8 -15 -48 -34 -59 -7 -29 -33 -72 -79\n24 16 32 75 86 58 49 83 74 67 13 81 6 57 18 40 37 63 23 35 1 46 50 22 38\n1 0"
},
{
"input": "100\n-97 -90 61 78 87 -52 -3 65 83 38 30 -60 35 -50 -73 -77 44 -32 -81 17 -67 58 -6 -34 47 -28 71 -45 69 -80 -4 -7 -57 -79 43 -27 -31 29 16 -89 -21 -93 95 -82 74 -5 -70 -20 -18 36 -64 -66 72 53 62 -68 26 15 76 -40 -99 8 59 88 49 -23 9 10 56 -48 -98 0 100 -54 25 94 13 -63 42 39 -1 55 24 -12 75 51 41 84 -96 -85 -2 -92 14 -46 -91 -19 -11 -86 22 -37",
"output": "51 -97 -90 -52 -3 -60 -50 -73 -77 -32 -81 -67 -6 -34 -28 -45 -80 -4 -7 -57 -79 -27 -31 -89 -21 -93 -82 -5 -70 -20 -18 -64 -66 -68 -40 -99 -23 -48 -98 -54 -63 -1 -12 -96 -85 -2 -92 -46 -91 -19 -11 -86\n47 61 78 87 65 83 38 30 35 44 17 58 47 71 69 43 29 16 95 74 36 72 53 62 26 15 76 8 59 88 49 9 10 56 100 25 94 13 42 39 55 24 75 51 41 84 14 22\n2 0 -37"
},
{
"input": "100\n-75 -60 -18 -92 -71 -9 -37 -34 -82 28 -54 93 -83 -76 -58 -88 -17 -97 64 -39 -96 -81 -10 -98 -47 -100 -22 27 14 -33 -19 -99 87 -66 57 -21 -90 -70 -32 -26 24 -77 -74 13 -44 16 -5 -55 -2 -6 -7 -73 -1 -68 -30 -95 -42 69 0 -20 -79 59 -48 -4 -72 -67 -46 62 51 -52 -86 -40 56 -53 85 -35 -8 49 50 65 29 11 -43 -15 -41 -12 -3 -80 -31 -38 -91 -45 -25 78 94 -23 -63 84 89 -61",
"output": "73 -75 -60 -18 -92 -71 -9 -37 -34 -82 -54 -83 -76 -58 -88 -17 -97 -39 -96 -81 -10 -98 -47 -100 -22 -33 -19 -99 -66 -21 -90 -70 -32 -26 -77 -74 -44 -5 -55 -2 -6 -7 -73 -1 -68 -30 -95 -42 -20 -79 -48 -4 -72 -67 -46 -52 -86 -40 -53 -35 -8 -43 -15 -41 -12 -3 -80 -31 -38 -91 -45 -25 -23 -63\n25 28 93 64 27 14 87 57 24 13 16 69 59 62 51 56 85 49 50 65 29 11 78 94 84 89\n2 0 -61"
},
{
"input": "100\n-87 -48 -76 -1 -10 -17 -22 -19 -27 -99 -43 49 38 -20 -45 -64 44 -96 -35 -74 -65 -41 -21 -75 37 -12 -67 0 -3 5 -80 -93 -81 -97 -47 -63 53 -100 95 -79 -83 -90 -32 88 -77 -16 -23 -54 -28 -4 -73 -98 -25 -39 60 -56 -34 -2 -11 -55 -52 -69 -68 -29 -82 -62 -36 -13 -6 -89 8 -72 18 -15 -50 -71 -70 -92 -42 -78 -61 -9 -30 -85 -91 -94 84 -86 -7 -57 -14 40 -33 51 -26 46 59 -31 -58 -66",
"output": "83 -87 -48 -76 -1 -10 -17 -22 -19 -27 -99 -43 -20 -45 -64 -96 -35 -74 -65 -41 -21 -75 -12 -67 -3 -80 -93 -81 -97 -47 -63 -100 -79 -83 -90 -32 -77 -16 -23 -54 -28 -4 -73 -98 -25 -39 -56 -34 -2 -11 -55 -52 -69 -68 -29 -82 -62 -36 -13 -6 -89 -72 -15 -50 -71 -70 -92 -42 -78 -61 -9 -30 -85 -91 -94 -86 -7 -57 -14 -33 -26 -31 -58 -66\n16 49 38 44 37 5 53 95 88 60 8 18 84 40 51 46 59\n1 0"
},
{
"input": "100\n-95 -28 -43 -72 -11 -24 -37 -35 -44 -66 -45 -62 -96 -51 -55 -23 -31 -26 -59 -17 77 -69 -10 -12 -78 -14 -52 -57 -40 -75 4 -98 -6 7 -53 -3 -90 -63 -8 -20 88 -91 -32 -76 -80 -97 -34 -27 -19 0 70 -38 -9 -49 -67 73 -36 2 81 -39 -65 -83 -64 -18 -94 -79 -58 -16 87 -22 -74 -25 -13 -46 -89 -47 5 -15 -54 -99 56 -30 -60 -21 -86 33 -1 -50 -68 -100 -85 -29 92 -48 -61 42 -84 -93 -41 -82",
"output": "85 -95 -28 -43 -72 -11 -24 -37 -35 -44 -66 -45 -62 -96 -51 -55 -23 -31 -26 -59 -17 -69 -10 -12 -78 -14 -52 -57 -40 -75 -98 -6 -53 -3 -90 -63 -8 -20 -91 -32 -76 -80 -97 -34 -27 -19 -38 -9 -49 -67 -36 -39 -65 -83 -64 -18 -94 -79 -58 -16 -22 -74 -25 -13 -46 -89 -47 -15 -54 -99 -30 -60 -21 -86 -1 -50 -68 -100 -85 -29 -48 -61 -84 -93 -41 -82\n14 77 4 7 88 70 73 2 81 87 5 56 33 92 42\n1 0"
},
{
"input": "100\n-12 -41 57 13 83 -36 53 69 -6 86 -75 87 11 -5 -4 -14 -37 -84 70 2 -73 16 31 34 -45 94 -9 26 27 52 -42 46 96 21 32 7 -18 61 66 -51 95 -48 -76 90 80 -40 89 77 78 54 -30 8 88 33 -24 82 -15 19 1 59 44 64 -97 -60 43 56 35 47 39 50 29 28 -17 -67 74 23 85 -68 79 0 65 55 -3 92 -99 72 93 -71 38 -10 -100 -98 81 62 91 -63 -58 49 -20 22",
"output": "35 -12 -41 -36 -6 -75 -5 -4 -14 -37 -84 -73 -45 -9 -42 -18 -51 -48 -76 -40 -30 -24 -15 -97 -60 -17 -67 -68 -3 -99 -71 -10 -100 -98 -63 -58\n63 57 13 83 53 69 86 87 11 70 2 16 31 34 94 26 27 52 46 96 21 32 7 61 66 95 90 80 89 77 78 54 8 88 33 82 19 1 59 44 64 43 56 35 47 39 50 29 28 74 23 85 79 65 55 92 72 93 38 81 62 91 49 22\n2 0 -20"
},
{
"input": "100\n-34 81 85 -96 50 20 54 86 22 10 -19 52 65 44 30 53 63 71 17 98 -92 4 5 -99 89 -23 48 9 7 33 75 2 47 -56 42 70 -68 57 51 83 82 94 91 45 46 25 95 11 -12 62 -31 -87 58 38 67 97 -60 66 73 -28 13 93 29 59 -49 77 37 -43 -27 0 -16 72 15 79 61 78 35 21 3 8 84 1 -32 36 74 -88 26 100 6 14 40 76 18 90 24 69 80 64 55 41",
"output": "19 -34 -96 -19 -92 -99 -23 -56 -68 -12 -31 -87 -60 -28 -49 -43 -27 -16 -32 -88\n80 81 85 50 20 54 86 22 10 52 65 44 30 53 63 71 17 98 4 5 89 48 9 7 33 75 2 47 42 70 57 51 83 82 94 91 45 46 25 95 11 62 58 38 67 97 66 73 13 93 29 59 77 37 72 15 79 61 78 35 21 3 8 84 1 36 74 26 100 6 14 40 76 18 90 24 69 80 64 55 41\n1 0"
},
{
"input": "100\n-1000 -986 -979 -955 -966 -963 -973 -959 -972 -906 -924 -927 -929 -918 -977 -967 -921 -989 -911 -995 -945 -919 -971 -913 -912 -933 -969 -975 -920 -988 -997 -994 -953 -962 -940 -905 -978 -948 -957 -996 0 -976 -949 -931 -903 -985 -923 -993 -944 -909 -938 -946 -934 -992 -904 -980 -954 -943 -917 -968 -991 -956 -902 -942 -999 -998 -908 -928 -930 -914 -922 -936 -960 -937 -939 -926 -965 -925 -951 -910 -907 -970 -990 -984 -964 -987 -916 -947 -982 -950 -974 -915 -932 -958 -981 -941 -961 -983 -952 -935",
"output": "97 -1000 -986 -979 -955 -966 -963 -973 -959 -972 -906 -924 -927 -929 -918 -977 -967 -921 -989 -911 -995 -945 -919 -971 -913 -912 -933 -969 -975 -920 -988 -997 -994 -953 -962 -940 -905 -978 -948 -957 -996 -976 -949 -931 -903 -985 -923 -993 -944 -909 -938 -946 -934 -992 -904 -980 -954 -943 -917 -968 -991 -956 -902 -942 -999 -998 -908 -928 -930 -914 -922 -936 -960 -937 -939 -926 -965 -925 -951 -910 -907 -970 -990 -984 -964 -987 -916 -947 -982 -950 -974 -915 -932 -958 -981 -941 -961 -983\n2 -935 -952\n1 0"
},
{
"input": "99\n-1000 -986 -979 -955 -966 -963 -973 -959 -972 -906 -924 -927 -929 -918 -977 -967 -921 -989 -911 -995 -945 -919 -971 -913 -912 -933 -969 -975 -920 -988 -997 -994 -953 -962 -940 -905 -978 -948 -957 -996 0 -976 -949 -931 -903 -985 -923 -993 -944 -909 -938 -946 -934 -992 -904 -980 -954 -943 -917 -968 -991 -956 -902 -942 -999 -998 -908 -928 -930 -914 -922 -936 -960 -937 -939 -926 -965 -925 -951 -910 -907 -970 -990 -984 -964 -987 -916 -947 -982 -950 -974 -915 -932 -958 -981 -941 -961 -983 -952",
"output": "95 -1000 -986 -979 -955 -966 -963 -973 -959 -972 -906 -924 -927 -929 -918 -977 -967 -921 -989 -911 -995 -945 -919 -971 -913 -912 -933 -969 -975 -920 -988 -997 -994 -953 -962 -940 -905 -978 -948 -957 -996 -976 -949 -931 -903 -985 -923 -993 -944 -909 -938 -946 -934 -992 -904 -980 -954 -943 -917 -968 -991 -956 -902 -942 -999 -998 -908 -928 -930 -914 -922 -936 -960 -937 -939 -926 -965 -925 -951 -910 -907 -970 -990 -984 -964 -987 -916 -947 -982 -950 -974 -915 -932 -958 -981 -941\n2 -952 -983\n2 0 -961"
},
{
"input": "59\n-990 -876 -641 -726 718 -53 803 -954 894 -265 -587 -665 904 349 754 -978 441 794 -768 -428 -569 -476 188 -620 -290 -333 45 705 -201 109 165 446 13 122 714 -562 -15 -86 -960 43 329 578 287 -776 -14 -71 915 886 -259 337 -495 913 -498 -669 -673 818 225 647 0",
"output": "29 -990 -876 -641 -726 -53 -954 -265 -587 -665 -978 -768 -428 -569 -476 -620 -290 -333 -201 -562 -15 -86 -960 -776 -14 -71 -259 -495 -498 -669\n28 718 803 894 904 349 754 441 794 188 45 705 109 165 446 13 122 714 43 329 578 287 915 886 337 913 818 225 647\n2 0 -673"
},
{
"input": "64\n502 885 -631 -906 735 687 642 -29 -696 -165 -524 15 -129 -663 -846 -501 -651 895 -341 -833 -142 33 -847 688 945 -192 -587 -930 603 849 736 676 788 256 863 -509 319 -49 -807 -158 218 -886 -143 -639 118 -156 -291 325 892 -916 -622 -960 -959 -731 -943 436 -535 861 745 589 -159 376 -182 0",
"output": "35 -631 -906 -29 -696 -165 -524 -129 -663 -846 -501 -651 -341 -833 -142 -847 -192 -587 -930 -509 -49 -807 -158 -886 -143 -639 -156 -291 -916 -622 -960 -959 -731 -943 -535 -159\n27 502 885 735 687 642 15 895 33 688 945 603 849 736 676 788 256 863 319 218 118 325 892 436 861 745 589 376\n2 0 -182"
},
{
"input": "5\n-1 -2 -3 -4 0",
"output": "1 -1\n2 -4 -3\n2 0 -2"
},
{
"input": "3\n-101 101 0",
"output": "1 -101\n1 101\n1 0"
},
{
"input": "21\n-100 -200 -300 -400 -500 -600 -700 -800 -900 -1000 0 100 200 300 400 500 600 700 800 900 1000",
"output": "9 -100 -200 -300 -400 -500 -600 -700 -800 -900\n10 100 200 300 400 500 600 700 800 900 1000\n2 0 -1000"
},
{
"input": "4\n0 -1 -2 -3",
"output": "1 -1\n2 -3 -2\n1 0"
}
] | 1,665,778,209
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 20
| 186
| 0
|
n=int(input())
a=list(map(int,input().split()))
aa=[]
bb=[]
cc=[]
for i in a:
if i<0:
aa.append(i)
elif i>0:
bb.append(i)
else:
cc.append(i)
l=[]
g=[]
z=[]
l.append(aa[0])
minus=0
if len(bb)==0:
if (len(aa)-1)%2==0:
g=aa[1:]
minus=len(g)
else:
g=aa[1:len(aa)-1]
minus=len(g)
else:
g=bb
if len(aa)-(1+minus)==0:
z=[0]
else:
z=cc+aa[(1+minus):]
ans=""
for i in range(len(l)):
if i==0:
ans+=str(len(l))+" "
ans+=str(l[i])+" "
ans+="\n"
for i in range(len(g)):
if i==0:
ans+=str(len(g))+" "
ans+=str(g[i])+" "
ans+="\n"
for i in range(len(z)):
if i==0:
ans+=str(len(z))+" "
ans+=str(z[i])+" "
print(ans)
|
Title: Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vitaly has an array of *n* distinct integers. Vitaly wants to divide this array into three non-empty sets so as the following conditions hold:
1. The product of all numbers in the first set is less than zero (<=<<=0). 1. The product of all numbers in the second set is greater than zero (<=><=0). 1. The product of all numbers in the third set is equal to zero. 1. Each number from the initial array must occur in exactly one set.
Help Vitaly. Divide the given array.
Input Specification:
The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=100). The second line contains *n* space-separated distinct integers *a*1,<=*a*2,<=...,<=*a**n* (|*a**i*|<=≤<=103) — the array elements.
Output Specification:
In the first line print integer *n*1 (*n*1<=><=0) — the number of elements in the first set. Then print *n*1 numbers — the elements that got to the first set.
In the next line print integer *n*2 (*n*2<=><=0) — the number of elements in the second set. Then print *n*2 numbers — the elements that got to the second set.
In the next line print integer *n*3 (*n*3<=><=0) — the number of elements in the third set. Then print *n*3 numbers — the elements that got to the third set.
The printed sets must meet the described conditions. It is guaranteed that the solution exists. If there are several solutions, you are allowed to print any of them.
Demo Input:
['3\n-1 2 0\n', '4\n-1 -2 -3 0\n']
Demo Output:
['1 -1\n1 2\n1 0\n', '1 -1\n2 -3 -2\n1 0\n']
Note:
none
|
```python
n=int(input())
a=list(map(int,input().split()))
aa=[]
bb=[]
cc=[]
for i in a:
if i<0:
aa.append(i)
elif i>0:
bb.append(i)
else:
cc.append(i)
l=[]
g=[]
z=[]
l.append(aa[0])
minus=0
if len(bb)==0:
if (len(aa)-1)%2==0:
g=aa[1:]
minus=len(g)
else:
g=aa[1:len(aa)-1]
minus=len(g)
else:
g=bb
if len(aa)-(1+minus)==0:
z=[0]
else:
z=cc+aa[(1+minus):]
ans=""
for i in range(len(l)):
if i==0:
ans+=str(len(l))+" "
ans+=str(l[i])+" "
ans+="\n"
for i in range(len(g)):
if i==0:
ans+=str(len(g))+" "
ans+=str(g[i])+" "
ans+="\n"
for i in range(len(z)):
if i==0:
ans+=str(len(z))+" "
ans+=str(z[i])+" "
print(ans)
```
| 3
|
|
141
|
A
|
Amusing Joke
|
PROGRAMMING
| 800
|
[
"implementation",
"sortings",
"strings"
] | null | null |
So, the New Year holidays are over. Santa Claus and his colleagues can take a rest and have guests at last. When two "New Year and Christmas Men" meet, thear assistants cut out of cardboard the letters from the guest's name and the host's name in honor of this event. Then the hung the letters above the main entrance. One night, when everyone went to bed, someone took all the letters of our characters' names. Then he may have shuffled the letters and put them in one pile in front of the door.
The next morning it was impossible to find the culprit who had made the disorder. But everybody wondered whether it is possible to restore the names of the host and his guests from the letters lying at the door? That is, we need to verify that there are no extra letters, and that nobody will need to cut more letters.
Help the "New Year and Christmas Men" and their friends to cope with this problem. You are given both inscriptions that hung over the front door the previous night, and a pile of letters that were found at the front door next morning.
|
The input file consists of three lines: the first line contains the guest's name, the second line contains the name of the residence host and the third line contains letters in a pile that were found at the door in the morning. All lines are not empty and contain only uppercase Latin letters. The length of each line does not exceed 100.
|
Print "YES" without the quotes, if the letters in the pile could be permuted to make the names of the "New Year and Christmas Men". Otherwise, print "NO" without the quotes.
|
[
"SANTACLAUS\nDEDMOROZ\nSANTAMOROZDEDCLAUS\n",
"PAPAINOEL\nJOULUPUKKI\nJOULNAPAOILELUPUKKI\n",
"BABBONATALE\nFATHERCHRISTMAS\nBABCHRISTMASBONATALLEFATHER\n"
] |
[
"YES\n",
"NO\n",
"NO\n"
] |
In the first sample the letters written in the last line can be used to write the names and there won't be any extra letters left.
In the second sample letter "P" is missing from the pile and there's an extra letter "L".
In the third sample there's an extra letter "L".
| 500
|
[
{
"input": "SANTACLAUS\nDEDMOROZ\nSANTAMOROZDEDCLAUS",
"output": "YES"
},
{
"input": "PAPAINOEL\nJOULUPUKKI\nJOULNAPAOILELUPUKKI",
"output": "NO"
},
{
"input": "BABBONATALE\nFATHERCHRISTMAS\nBABCHRISTMASBONATALLEFATHER",
"output": "NO"
},
{
"input": "B\nA\nAB",
"output": "YES"
},
{
"input": "ONDOL\nJNPB\nONLNJBODP",
"output": "YES"
},
{
"input": "Y\nW\nYW",
"output": "YES"
},
{
"input": "OI\nM\nIMO",
"output": "YES"
},
{
"input": "VFQRWWWACX\nGHZJPOQUSXRAQDGOGMR\nOPAWDOUSGWWCGQXXQAZJRQRGHRMVF",
"output": "YES"
},
{
"input": "JUTCN\nPIGMZOPMEUFADQBW\nNWQGZMAIPUPOMCDUB",
"output": "NO"
},
{
"input": "Z\nO\nZOCNDOLTBZKQLTBOLDEGXRHZGTTPBJBLSJCVSVXISQZCSFDEBXRCSGBGTHWOVIXYHACAGBRYBKBJAEPIQZHVEGLYH",
"output": "NO"
},
{
"input": "IQ\nOQ\nQOQIGGKFNHJSGCGM",
"output": "NO"
},
{
"input": "ROUWANOPNIGTVMIITVMZ\nOQTUPZMTKUGY\nVTVNGZITGPUNPMQOOATUUIYIWMMKZOTR",
"output": "YES"
},
{
"input": "OVQELLOGFIOLEHXMEMBJDIGBPGEYFG\nJNKFPFFIJOFHRIFHXEWYZOPDJBZTJZKBWQTECNHRFSJPJOAPQT\nYAIPFFFEXJJNEJPLREIGODEGQZVMCOBDFKWTMWJSBEBTOFFQOHIQJLHFNXIGOHEZRZLFOKJBJPTPHPGY",
"output": "YES"
},
{
"input": "NBJGVNGUISUXQTBOBKYHQCOOVQWUXWPXBUDLXPKX\nNSFQDFUMQDQWQ\nWXKKVNTDQQFXCUQBIMQGQHSLVGWSBFYBUPOWPBDUUJUXQNOQDNXOX",
"output": "YES"
},
{
"input": "IJHHGKCXWDBRWJUPRDBZJLNTTNWKXLUGJSBWBOAUKWRAQWGFNL\nNJMWRMBCNPHXTDQQNZ\nWDNJRCLILNQRHWBANLTXWMJBPKUPGKJDJZAQWKTZFBRCTXHHBNXRGUQUNBNMWODGSJWW",
"output": "YES"
},
{
"input": "SRROWANGUGZHCIEFYMQVTWVOMDWPUZJFRDUMVFHYNHNTTGNXCJ\nDJYWGLBFCCECXFHOLORDGDCNRHPWXNHXFCXQCEZUHRRNAEKUIX\nWCUJDNYHNHYOPWMHLDCDYRWBVOGHFFUKOZTXJRXJHRGWICCMRNEVNEGQWTZPNFCSHDRFCFQDCXMHTLUGZAXOFNXNVGUEXIACRERU",
"output": "YES"
},
{
"input": "H\nJKFGHMIAHNDBMFXWYQLZRSVNOTEGCQSVUBYUOZBTNKTXPFQDCMKAGFITEUGOYDFIYQIORMFJEOJDNTFVIQEBICSNGKOSNLNXJWC\nBQSVDOGIHCHXSYNYTQFCHNJGYFIXTSOQINZOKSVQJMTKNTGFNXAVTUYEONMBQMGJLEWJOFGEARIOPKFUFCEMUBRBDNIIDFZDCLWK",
"output": "YES"
},
{
"input": "DSWNZRFVXQ\nPVULCZGOOU\nUOLVZXNUPOQRZGWFVDSCANQTCLEIE",
"output": "NO"
},
{
"input": "EUHTSCENIPXLTSBMLFHD\nIZAVSZPDLXOAGESUSE\nLXAELAZ",
"output": "NO"
},
{
"input": "WYSJFEREGELSKRQRXDXCGBODEFZVSI\nPEJKMGFLBFFDWRCRFSHVEFLEBTJCVCHRJTLDTISHPOGFWPLEWNYJLMXWIAOTYOXMV\nHXERTZWLEXTPIOTFRVMEJVYFFJLRPFMXDEBNSGCEOFFCWTKIDDGCFYSJKGLHBORWEPLDRXRSJYBGASSVCMHEEJFLVI",
"output": "NO"
},
{
"input": "EPBMDIUQAAUGLBIETKOKFLMTCVEPETWJRHHYKCKU\nHGMAETVPCFZYNNKDQXVXUALHYLOTCHM\nECGXACVKEYMCEDOTMKAUFHLHOMT",
"output": "NO"
},
{
"input": "NUBKQEJHALANSHEIFUZHYEZKKDRFHQKAJHLAOWTZIMOCWOVVDW\nEFVOBIGAUAUSQGVSNBKNOBDMINODMFSHDL\nKLAMKNTHBFFOHVKWICHBKNDDQNEISODUSDNLUSIOAVWY",
"output": "NO"
},
{
"input": "VXINHOMEQCATZUGAJEIUIZZLPYFGUTVLNBNWCUVMEENUXKBWBGZTMRJJVJDLVSLBABVCEUDDSQFHOYPYQTWVAGTWOLKYISAGHBMC\nZMRGXPZSHOGCSAECAPGVOIGCWEOWWOJXLGYRDMPXBLOKZVRACPYQLEQGFQCVYXAGBEBELUTDAYEAGPFKXRULZCKFHZCHVCWIRGPK\nRCVUXGQVNWFGRUDLLENNDQEJHYYVWMKTLOVIPELKPWCLSQPTAXAYEMGWCBXEVAIZGGDDRBRT",
"output": "NO"
},
{
"input": "PHBDHHWUUTZAHELGSGGOPOQXSXEZIXHZTOKYFBQLBDYWPVCNQSXHEAXRRPVHFJBVBYCJIFOTQTWSUOWXLKMVJJBNLGTVITWTCZZ\nFUPDLNVIHRWTEEEHOOEC\nLOUSUUSZCHJBPEWIILUOXEXRQNCJEGTOBRVZLTTZAHTKVEJSNGHFTAYGY",
"output": "NO"
},
{
"input": "GDSLNIIKTO\nJF\nPDQYFKDTNOLI",
"output": "NO"
},
{
"input": "AHOKHEKKPJLJIIWJRCGY\nORELJCSIX\nZVWPXVFWFSWOXXLIHJKPXIOKRELYE",
"output": "NO"
},
{
"input": "ZWCOJFORBPHXCOVJIDPKVECMHVHCOC\nTEV\nJVGTBFTLFVIEPCCHODOFOMCVZHWXVCPEH",
"output": "NO"
},
{
"input": "AGFIGYWJLVMYZGNQHEHWKJIAWBPUAQFERMCDROFN\nPMJNHMVNRGCYZAVRWNDSMLSZHFNYIUWFPUSKKIGU\nMCDVPPRXGUAYLSDRHRURZASXUWZSIIEZCPXUVEONKNGNWRYGOSFMCKESMVJZHWWUCHWDQMLASLNNMHAU",
"output": "NO"
},
{
"input": "XLOWVFCZSSXCSYQTIIDKHNTKNKEEDFMDZKXSPVLBIDIREDUAIN\nZKIWNDGBISDB\nSLPKLYFYSRNRMOSWYLJJDGFFENPOXYLPZFTQDANKBDNZDIIEWSUTTKYBKVICLG",
"output": "NO"
},
{
"input": "PMUKBTRKFIAYVGBKHZHUSJYSSEPEOEWPOSPJLWLOCTUYZODLTUAFCMVKGQKRRUSOMPAYOTBTFPXYAZXLOADDEJBDLYOTXJCJYTHA\nTWRRAJLCQJTKOKWCGUH\nEWDPNXVCXWCDQCOYKKSOYTFSZTOOPKPRDKFJDETKSRAJRVCPDOBWUGPYRJPUWJYWCBLKOOTUPBESTOFXZHTYLLMCAXDYAEBUTAHM",
"output": "NO"
},
{
"input": "QMIMGQRQDMJDPNFEFXSXQMCHEJKTWCTCVZPUAYICOIRYOWKUSIWXJLHDYWSBOITHTMINXFKBKAWZTXXBJIVYCRWKXNKIYKLDDXL\nV\nFWACCXBVDOJFIUAVYRALBYJKXXWIIFORRUHKHCXLDBZMXIYJWISFEAWTIQFIZSBXMKNOCQKVKRWDNDAMQSTKYLDNYVTUCGOJXJTW",
"output": "NO"
},
{
"input": "XJXPVOOQODELPPWUISSYVVXRJTYBPDHJNENQEVQNVFIXSESKXVYPVVHPMOSX\nLEXOPFPVPSZK\nZVXVPYEYOYXVOISVLXPOVHEQVXPNQJIOPFDTXEUNMPEPPHELNXKKWSVSOXSBPSJDPVJVSRFQ",
"output": "YES"
},
{
"input": "OSKFHGYNQLSRFSAHPXKGPXUHXTRBJNAQRBSSWJVEENLJCDDHFXVCUNPZAIVVO\nFNUOCXAGRRHNDJAHVVLGGEZQHWARYHENBKHP\nUOEFNWVXCUNERLKVTHAGPSHKHDYFPYWZHJKHQLSNFBJHVJANRXCNSDUGVDABGHVAOVHBJZXGRACHRXEGNRPQEAPORQSILNXFS",
"output": "YES"
},
{
"input": "VYXYVVACMLPDHONBUTQFZTRREERBLKUJYKAHZRCTRLRCLOZYWVPBRGDQPFPQIF\nFE\nRNRPEVDRLYUQFYRZBCQLCYZEABKLRXCJLKVZBVFUEYRATOMDRTHFPGOWQVTIFPPH",
"output": "YES"
},
{
"input": "WYXUZQJQNLASEGLHPMSARWMTTQMQLVAZLGHPIZTRVTCXDXBOLNXZPOFCTEHCXBZ\nBLQZRRWP\nGIQZXPLTTMNHQVWPPEAPLOCDMBSTHRCFLCQRRZXLVAOQEGZBRUZJXXZTMAWLZHSLWNQTYXB",
"output": "YES"
},
{
"input": "MKVJTSSTDGKPVVDPYSRJJYEVGKBMSIOKHLZQAEWLRIBINVRDAJIBCEITKDHUCCVY\nPUJJQFHOGZKTAVNUGKQUHMKTNHCCTI\nQVJKUSIGTSVYUMOMLEGHWYKSKQTGATTKBNTKCJKJPCAIRJIRMHKBIZISEGFHVUVQZBDERJCVAKDLNTHUDCHONDCVVJIYPP",
"output": "YES"
},
{
"input": "OKNJOEYVMZXJMLVJHCSPLUCNYGTDASKSGKKCRVIDGEIBEWRVBVRVZZTLMCJLXHJIA\nDJBFVRTARTFZOWN\nAGHNVUNJVCPLWSVYBJKZSVTFGLELZASLWTIXDDJXCZDICTVIJOTMVEYOVRNMJGRKKHRMEBORAKFCZJBR",
"output": "YES"
},
{
"input": "OQZACLPSAGYDWHFXDFYFRRXWGIEJGSXWUONAFWNFXDTGVNDEWNQPHUXUJNZWWLBPYL\nOHBKWRFDRQUAFRCMT\nWIQRYXRJQWWRUWCYXNXALKFZGXFTLOODWRDPGURFUFUQOHPWBASZNVWXNCAGHWEHFYESJNFBMNFDDAPLDGT",
"output": "YES"
},
{
"input": "OVIRQRFQOOWVDEPLCJETWQSINIOPLTLXHSQWUYUJNFBMKDNOSHNJQQCDHZOJVPRYVSV\nMYYDQKOOYPOOUELCRIT\nNZSOTVLJTTVQLFHDQEJONEOUOFOLYVSOIYUDNOSIQVIRMVOERCLMYSHPCQKIDRDOQPCUPQBWWRYYOXJWJQPNKH",
"output": "YES"
},
{
"input": "WGMBZWNMSJXNGDUQUJTCNXDSJJLYRDOPEGPQXYUGBESDLFTJRZDDCAAFGCOCYCQMDBWK\nYOBMOVYTUATTFGJLYUQD\nDYXVTLQCYFJUNJTUXPUYOPCBCLBWNSDUJRJGWDOJDSQAAMUOJWSYERDYDXYTMTOTMQCGQZDCGNFBALGGDFKZMEBG",
"output": "YES"
},
{
"input": "CWLRBPMEZCXAPUUQFXCUHAQTLPBTXUUKWVXKBHKNSSJFEXLZMXGVFHHVTPYAQYTIKXJJE\nMUFOSEUEXEQTOVLGDSCWM\nJUKEQCXOXWEHCGKFPBIGMWVJLXUONFXBYTUAXERYTXKCESKLXAEHVPZMMUFTHLXTTZSDMBJLQPEUWCVUHSQQVUASPF",
"output": "YES"
},
{
"input": "IDQRX\nWETHO\nODPDGBHVUVSSISROHQJTUKPUCLXABIZQQPPBPKOSEWGEHRSRRNBAVLYEMZISMWWGKHVTXKUGUXEFBSWOIWUHRJGMWBMHQLDZHBWA",
"output": "NO"
},
{
"input": "IXFDY\nJRMOU\nDF",
"output": "NO"
},
{
"input": "JPSPZ\nUGCUB\nJMZZZZZZZZ",
"output": "NO"
},
{
"input": "AC\nA\nBBA",
"output": "NO"
},
{
"input": "UIKWWKXLSHTOOZOVGXKYSOJEHAUEEG\nKZXQDWJJWRXFHKJDQHJK\nXMZHTFOGEXAUJXXJUYVJIFOTKLZHDKELJWERHMGAWGKWAQKEKHIDWGGZVYOHKXRPWSJDPESFJUMKQYWBYUTHQYEFZUGKQOBHYDWB",
"output": "NO"
},
{
"input": "PXWRXRPFLR\nPJRWWXIVHODV\nXW",
"output": "NO"
},
{
"input": "CHTAZVHGSHCVIBK\nEQINEBKXEPYJSAZIMLDF\nZCZZZZDZMCZZEZDZZEZZZZQZZBZZZOZZCZE",
"output": "NO"
},
{
"input": "GXPZFSELJJNDAXYRV\nUYBKPMVBSOVOJWMONLTJOJCNQKMTAHEWLHOWIIBH\nHCWNFWJPEJIWOVPTBMVCRJLSISSVNOHCKLBFMIUAIMASQWPXEYXBOXQGFEMYJLBKDCZIMJNHOJEDGGANIVYKQTUOSOVOPWHVJGXH",
"output": "NO"
},
{
"input": "LFGJCJJDUTUP\nOVSBILTIYCJCRHKCIXCETJQJJ\nGIJJTJCLTJJJ",
"output": "NO"
},
{
"input": "GIO\nPRL\nPRL",
"output": "NO"
},
{
"input": "A\nB\nABC",
"output": "NO"
},
{
"input": "KKK\nKKK\nZZZZZ",
"output": "NO"
},
{
"input": "ZMYGQLDBLAPN\nZFJBKWHROVNPSJQUDFTHOCGREUFLYIWYICD\nZMJZZEDAZANKZZZZZZEZZBZDZZZZZZKHZZFZZZDZNZMDZZA",
"output": "NO"
}
] | 1,688,806,194
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 54
| 92
| 0
|
h=input()
u=input()
i=input()
funny=h+u
slojna=sorted(funny)
legko=sorted(i)
if slojna==legko:
print("YES")
else:
print("NO")
# Sat Jul 08 2023 11:49:53 GMT+0300 (Moscow Standard Time)
|
Title: Amusing Joke
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
So, the New Year holidays are over. Santa Claus and his colleagues can take a rest and have guests at last. When two "New Year and Christmas Men" meet, thear assistants cut out of cardboard the letters from the guest's name and the host's name in honor of this event. Then the hung the letters above the main entrance. One night, when everyone went to bed, someone took all the letters of our characters' names. Then he may have shuffled the letters and put them in one pile in front of the door.
The next morning it was impossible to find the culprit who had made the disorder. But everybody wondered whether it is possible to restore the names of the host and his guests from the letters lying at the door? That is, we need to verify that there are no extra letters, and that nobody will need to cut more letters.
Help the "New Year and Christmas Men" and their friends to cope with this problem. You are given both inscriptions that hung over the front door the previous night, and a pile of letters that were found at the front door next morning.
Input Specification:
The input file consists of three lines: the first line contains the guest's name, the second line contains the name of the residence host and the third line contains letters in a pile that were found at the door in the morning. All lines are not empty and contain only uppercase Latin letters. The length of each line does not exceed 100.
Output Specification:
Print "YES" without the quotes, if the letters in the pile could be permuted to make the names of the "New Year and Christmas Men". Otherwise, print "NO" without the quotes.
Demo Input:
['SANTACLAUS\nDEDMOROZ\nSANTAMOROZDEDCLAUS\n', 'PAPAINOEL\nJOULUPUKKI\nJOULNAPAOILELUPUKKI\n', 'BABBONATALE\nFATHERCHRISTMAS\nBABCHRISTMASBONATALLEFATHER\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
In the first sample the letters written in the last line can be used to write the names and there won't be any extra letters left.
In the second sample letter "P" is missing from the pile and there's an extra letter "L".
In the third sample there's an extra letter "L".
|
```python
h=input()
u=input()
i=input()
funny=h+u
slojna=sorted(funny)
legko=sorted(i)
if slojna==legko:
print("YES")
else:
print("NO")
# Sat Jul 08 2023 11:49:53 GMT+0300 (Moscow Standard Time)
```
| 3
|
|
397
|
A
|
On Segment's Own Points
|
PROGRAMMING
| 0
|
[
"implementation"
] | null | null |
Our old friend Alexey has finally entered the University of City N — the Berland capital. Alexey expected his father to get him a place to live in but his father said it was high time for Alexey to practice some financial independence. So, Alexey is living in a dorm.
The dorm has exactly one straight dryer — a 100 centimeter long rope to hang clothes on. The dryer has got a coordinate system installed: the leftmost end of the dryer has coordinate 0, and the opposite end has coordinate 100. Overall, the university has *n* students. Dean's office allows *i*-th student to use the segment (*l**i*,<=*r**i*) of the dryer. However, the dean's office actions are contradictory and now one part of the dryer can belong to multiple students!
Alexey don't like when someone touch his clothes. That's why he want make it impossible to someone clothes touch his ones. So Alexey wonders: what is the total length of the parts of the dryer that he may use in a such way that clothes of the others (*n*<=-<=1) students aren't drying there. Help him! Note that Alexey, as the most respected student, has number 1.
|
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100). The (*i*<=+<=1)-th line contains integers *l**i* and *r**i* (0<=≤<=*l**i*<=<<=*r**i*<=≤<=100) — the endpoints of the corresponding segment for the *i*-th student.
|
On a single line print a single number *k*, equal to the sum of lengths of the parts of the dryer which are inside Alexey's segment and are outside all other segments.
|
[
"3\n0 5\n2 8\n1 6\n",
"3\n0 10\n1 5\n7 15\n"
] |
[
"1\n",
"3\n"
] |
Note that it's not important are clothes drying on the touching segments (e.g. (0, 1) and (1, 2)) considered to be touching or not because you need to find the length of segments.
In the first test sample Alexey may use the only segment (0, 1). In such case his clothes will not touch clothes on the segments (1, 6) and (2, 8). The length of segment (0, 1) is 1.
In the second test sample Alexey may dry his clothes on segments (0, 1) and (5, 7). Overall length of these segments is 3.
| 500
|
[
{
"input": "3\n0 5\n2 8\n1 6",
"output": "1"
},
{
"input": "3\n0 10\n1 5\n7 15",
"output": "3"
},
{
"input": "1\n0 100",
"output": "100"
},
{
"input": "2\n1 9\n1 9",
"output": "0"
},
{
"input": "2\n1 9\n5 10",
"output": "4"
},
{
"input": "2\n1 9\n3 5",
"output": "6"
},
{
"input": "2\n3 5\n1 9",
"output": "0"
},
{
"input": "10\n43 80\n39 75\n26 71\n4 17\n11 57\n31 42\n1 62\n9 19\n27 76\n34 53",
"output": "4"
},
{
"input": "50\n33 35\n98 99\n1 2\n4 6\n17 18\n63 66\n29 30\n35 37\n44 45\n73 75\n4 5\n39 40\n92 93\n96 97\n23 27\n49 50\n2 3\n60 61\n43 44\n69 70\n7 8\n45 46\n21 22\n85 86\n48 49\n41 43\n70 71\n10 11\n27 28\n71 72\n6 7\n15 16\n46 47\n89 91\n54 55\n19 21\n86 87\n37 38\n77 82\n84 85\n54 55\n93 94\n45 46\n37 38\n75 76\n22 23\n50 52\n38 39\n1 2\n66 67",
"output": "2"
},
{
"input": "2\n1 5\n7 9",
"output": "4"
},
{
"input": "2\n1 5\n3 5",
"output": "2"
},
{
"input": "2\n1 5\n1 2",
"output": "3"
},
{
"input": "5\n5 10\n5 10\n5 10\n5 10\n5 10",
"output": "0"
},
{
"input": "6\n1 99\n33 94\n68 69\n3 35\n93 94\n5 98",
"output": "3"
},
{
"input": "11\n2 98\n63 97\n4 33\n12 34\n34 65\n23 31\n43 54\n82 99\n15 84\n23 52\n4 50",
"output": "2"
},
{
"input": "10\n95 96\n19 20\n72 73\n1 2\n25 26\n48 49\n90 91\n22 23\n16 17\n16 17",
"output": "1"
},
{
"input": "11\n1 100\n63 97\n4 33\n12 34\n34 65\n23 31\n43 54\n82 99\n15 84\n23 52\n4 50",
"output": "4"
},
{
"input": "21\n0 100\n81 90\n11 68\n18 23\n75 78\n45 86\n37 58\n15 21\n40 98\n53 100\n10 70\n14 75\n1 92\n23 81\n13 66\n93 100\n6 34\n22 87\n27 84\n15 63\n54 91",
"output": "1"
},
{
"input": "10\n60 66\n5 14\n1 3\n55 56\n70 87\n34 35\n16 21\n23 24\n30 31\n25 27",
"output": "6"
},
{
"input": "40\n29 31\n22 23\n59 60\n70 71\n42 43\n13 15\n11 12\n64 65\n1 2\n62 63\n54 56\n8 9\n2 3\n53 54\n27 28\n48 49\n72 73\n17 18\n46 47\n18 19\n43 44\n39 40\n83 84\n63 64\n52 53\n33 34\n3 4\n24 25\n74 75\n0 1\n61 62\n68 69\n80 81\n5 6\n36 37\n81 82\n50 51\n66 67\n69 70\n20 21",
"output": "2"
},
{
"input": "15\n22 31\n0 4\n31 40\n77 80\n81 83\n11 13\n59 61\n53 59\n51 53\n87 88\n14 22\n43 45\n8 10\n45 47\n68 71",
"output": "9"
},
{
"input": "31\n0 100\n2 97\n8 94\n9 94\n14 94\n15 93\n15 90\n17 88\n19 88\n19 87\n20 86\n25 86\n30 85\n32 85\n35 82\n35 81\n36 80\n37 78\n38 74\n38 74\n39 71\n40 69\n40 68\n41 65\n43 62\n44 62\n45 61\n45 59\n46 57\n49 54\n50 52",
"output": "5"
},
{
"input": "21\n0 97\n46 59\n64 95\n3 16\n86 95\n55 71\n51 77\n26 28\n47 88\n30 40\n26 34\n2 12\n9 10\n4 19\n35 36\n41 92\n1 16\n41 78\n56 81\n23 35\n40 68",
"output": "7"
},
{
"input": "27\n0 97\n7 9\n6 9\n12 33\n12 26\n15 27\n10 46\n33 50\n31 47\n15 38\n12 44\n21 35\n24 37\n51 52\n65 67\n58 63\n53 60\n63 68\n57 63\n60 68\n55 58\n74 80\n70 75\n89 90\n81 85\n93 99\n93 98",
"output": "19"
},
{
"input": "20\n23 24\n22 23\n84 86\n6 10\n40 45\n11 13\n24 27\n81 82\n53 58\n87 90\n14 15\n49 50\n70 75\n75 78\n98 100\n66 68\n18 21\n1 2\n92 93\n34 37",
"output": "1"
},
{
"input": "11\n2 100\n34 65\n4 50\n63 97\n82 99\n43 54\n23 52\n4 33\n15 84\n23 31\n12 34",
"output": "3"
},
{
"input": "60\n73 75\n6 7\n69 70\n15 16\n54 55\n66 67\n7 8\n39 40\n38 39\n37 38\n1 2\n46 47\n7 8\n21 22\n23 27\n15 16\n45 46\n37 38\n60 61\n4 6\n63 66\n10 11\n33 35\n43 44\n2 3\n4 6\n10 11\n93 94\n45 46\n7 8\n44 45\n41 43\n35 37\n17 18\n48 49\n89 91\n27 28\n46 47\n71 72\n1 2\n75 76\n49 50\n84 85\n17 18\n98 99\n54 55\n46 47\n19 21\n77 82\n29 30\n4 5\n70 71\n85 86\n96 97\n86 87\n92 93\n22 23\n50 52\n44 45\n63 66",
"output": "2"
},
{
"input": "40\n47 48\n42 44\n92 94\n15 17\n20 22\n11 13\n37 39\n6 8\n39 40\n35 37\n21 22\n41 42\n77 78\n76 78\n69 71\n17 19\n18 19\n17 18\n84 85\n9 10\n11 12\n51 52\n99 100\n7 8\n97 99\n22 23\n60 62\n7 8\n67 69\n20 22\n13 14\n89 91\n15 17\n12 13\n56 57\n37 39\n29 30\n24 26\n37 38\n25 27",
"output": "1"
},
{
"input": "10\n28 36\n18 26\n28 35\n95 100\n68 72\n41 42\n76 84\n99 100\n6 8\n58 60",
"output": "1"
},
{
"input": "20\n69 72\n88 92\n77 80\n64 69\n66 67\n79 81\n91 96\n78 83\n81 86\n11 12\n48 53\n22 23\n81 84\n89 92\n56 60\n1 4\n1 5\n60 62\n20 23\n63 66",
"output": "3"
},
{
"input": "71\n1 99\n11 69\n86 92\n7 49\n31 70\n42 53\n48 81\n86 96\n36 91\n19 38\n39 91\n41 64\n87 93\n83 97\n40 41\n3 32\n15 18\n58 65\n22 32\n1 71\n58 86\n64 77\n15 69\n4 34\n42 89\n9 66\n15 18\n58 65\n59 96\n39 89\n19 38\n6 63\n26 73\n29 47\n55 88\n5 78\n41 74\n48 81\n20 71\n59 96\n42 49\n4 69\n41 74\n87 93\n0 65\n2 34\n15 18\n10 56\n55 88\n33 56\n42 89\n86 92\n42 81\n65 82\n5 78\n13 52\n32 85\n7 65\n59 96\n4 65\n46 69\n10 56\n42 89\n4 69\n0 65\n32 35\n5 78\n32 75\n42 53\n55 59\n64 77",
"output": "2"
},
{
"input": "1\n1 2",
"output": "1"
}
] | 1,583,057,505
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 37
| 109
| 512,000
|
n = int(input())
segments = []
for _ in range(n):
segments.append(list(map(int, input().split())))
l = segments[0][0]
r = segments[0][1]
list_of_tuples = []
for i in range(l,r):
list_of_tuples.append((i,i+1))
i=1
while i<n:
for j in range(segments[i][0],segments[i][1]):
if (j,j+1) in list_of_tuples:
list_of_tuples.remove((j,j+1))
i += 1
print(len(list_of_tuples))
|
Title: On Segment's Own Points
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Our old friend Alexey has finally entered the University of City N — the Berland capital. Alexey expected his father to get him a place to live in but his father said it was high time for Alexey to practice some financial independence. So, Alexey is living in a dorm.
The dorm has exactly one straight dryer — a 100 centimeter long rope to hang clothes on. The dryer has got a coordinate system installed: the leftmost end of the dryer has coordinate 0, and the opposite end has coordinate 100. Overall, the university has *n* students. Dean's office allows *i*-th student to use the segment (*l**i*,<=*r**i*) of the dryer. However, the dean's office actions are contradictory and now one part of the dryer can belong to multiple students!
Alexey don't like when someone touch his clothes. That's why he want make it impossible to someone clothes touch his ones. So Alexey wonders: what is the total length of the parts of the dryer that he may use in a such way that clothes of the others (*n*<=-<=1) students aren't drying there. Help him! Note that Alexey, as the most respected student, has number 1.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100). The (*i*<=+<=1)-th line contains integers *l**i* and *r**i* (0<=≤<=*l**i*<=<<=*r**i*<=≤<=100) — the endpoints of the corresponding segment for the *i*-th student.
Output Specification:
On a single line print a single number *k*, equal to the sum of lengths of the parts of the dryer which are inside Alexey's segment and are outside all other segments.
Demo Input:
['3\n0 5\n2 8\n1 6\n', '3\n0 10\n1 5\n7 15\n']
Demo Output:
['1\n', '3\n']
Note:
Note that it's not important are clothes drying on the touching segments (e.g. (0, 1) and (1, 2)) considered to be touching or not because you need to find the length of segments.
In the first test sample Alexey may use the only segment (0, 1). In such case his clothes will not touch clothes on the segments (1, 6) and (2, 8). The length of segment (0, 1) is 1.
In the second test sample Alexey may dry his clothes on segments (0, 1) and (5, 7). Overall length of these segments is 3.
|
```python
n = int(input())
segments = []
for _ in range(n):
segments.append(list(map(int, input().split())))
l = segments[0][0]
r = segments[0][1]
list_of_tuples = []
for i in range(l,r):
list_of_tuples.append((i,i+1))
i=1
while i<n:
for j in range(segments[i][0],segments[i][1]):
if (j,j+1) in list_of_tuples:
list_of_tuples.remove((j,j+1))
i += 1
print(len(list_of_tuples))
```
| 3
|
|
911
|
C
|
Three Garlands
|
PROGRAMMING
| 1,400
|
[
"brute force",
"constructive algorithms"
] | null | null |
Mishka is decorating the Christmas tree. He has got three garlands, and all of them will be put on the tree. After that Mishka will switch these garlands on.
When a garland is switched on, it periodically changes its state — sometimes it is lit, sometimes not. Formally, if *i*-th garland is switched on during *x*-th second, then it is lit only during seconds *x*, *x*<=+<=*k**i*, *x*<=+<=2*k**i*, *x*<=+<=3*k**i* and so on.
Mishka wants to switch on the garlands in such a way that during each second after switching the garlands on there would be at least one lit garland. Formally, Mishka wants to choose three integers *x*1, *x*2 and *x*3 (not necessarily distinct) so that he will switch on the first garland during *x*1-th second, the second one — during *x*2-th second, and the third one — during *x*3-th second, respectively, and during each second starting from *max*(*x*1,<=*x*2,<=*x*3) at least one garland will be lit.
Help Mishka by telling him if it is possible to do this!
|
The first line contains three integers *k*1, *k*2 and *k*3 (1<=≤<=*k**i*<=≤<=1500) — time intervals of the garlands.
|
If Mishka can choose moments of time to switch on the garlands in such a way that each second after switching the garlands on at least one garland will be lit, print YES.
Otherwise, print NO.
|
[
"2 2 3\n",
"4 2 3\n"
] |
[
"YES\n",
"NO\n"
] |
In the first example Mishka can choose *x*<sub class="lower-index">1</sub> = 1, *x*<sub class="lower-index">2</sub> = 2, *x*<sub class="lower-index">3</sub> = 1. The first garland will be lit during seconds 1, 3, 5, 7, ..., the second — 2, 4, 6, 8, ..., which already cover all the seconds after the 2-nd one. It doesn't even matter what *x*<sub class="lower-index">3</sub> is chosen. Our choice will lead third to be lit during seconds 1, 4, 7, 10, ..., though.
In the second example there is no way to choose such moments of time, there always be some seconds when no garland is lit.
| 0
|
[
{
"input": "2 2 3",
"output": "YES"
},
{
"input": "4 2 3",
"output": "NO"
},
{
"input": "1499 1498 1500",
"output": "NO"
},
{
"input": "1500 1500 1500",
"output": "NO"
},
{
"input": "100 4 1",
"output": "YES"
},
{
"input": "4 2 4",
"output": "YES"
},
{
"input": "3 3 3",
"output": "YES"
},
{
"input": "2 3 6",
"output": "NO"
},
{
"input": "2 3 3",
"output": "NO"
},
{
"input": "4 4 2",
"output": "YES"
},
{
"input": "1 1 1",
"output": "YES"
},
{
"input": "2 11 2",
"output": "YES"
},
{
"input": "4 4 4",
"output": "NO"
},
{
"input": "4 4 5",
"output": "NO"
},
{
"input": "3 3 2",
"output": "NO"
},
{
"input": "3 6 6",
"output": "NO"
},
{
"input": "2 3 2",
"output": "YES"
},
{
"input": "1 1 3",
"output": "YES"
},
{
"input": "3 3 4",
"output": "NO"
},
{
"input": "2 4 4",
"output": "YES"
},
{
"input": "2 2 2",
"output": "YES"
},
{
"input": "2 10 10",
"output": "NO"
},
{
"input": "3 4 4",
"output": "NO"
},
{
"input": "2 5 5",
"output": "NO"
},
{
"input": "2 4 5",
"output": "NO"
},
{
"input": "228 2 2",
"output": "YES"
},
{
"input": "2 998 1000",
"output": "NO"
},
{
"input": "2 6 6",
"output": "NO"
},
{
"input": "6 4 7",
"output": "NO"
},
{
"input": "2 5 2",
"output": "YES"
},
{
"input": "2 100 100",
"output": "NO"
},
{
"input": "7 7 2",
"output": "NO"
},
{
"input": "3 3 6",
"output": "NO"
},
{
"input": "82 3 82",
"output": "NO"
},
{
"input": "2 3 5",
"output": "NO"
},
{
"input": "1 218 924",
"output": "YES"
},
{
"input": "4 4 123",
"output": "NO"
},
{
"input": "4 4 3",
"output": "NO"
},
{
"input": "3 4 2",
"output": "NO"
},
{
"input": "2 2 5",
"output": "YES"
},
{
"input": "2 10 2",
"output": "YES"
},
{
"input": "5 2 2",
"output": "YES"
},
{
"input": "3 3 9",
"output": "NO"
},
{
"input": "1 5 5",
"output": "YES"
},
{
"input": "2 4 6",
"output": "NO"
},
{
"input": "15 3 3",
"output": "NO"
},
{
"input": "1 5 10",
"output": "YES"
},
{
"input": "2 3 14",
"output": "NO"
},
{
"input": "1265 2 593",
"output": "NO"
},
{
"input": "2 2 567",
"output": "YES"
},
{
"input": "1 6 5",
"output": "YES"
},
{
"input": "2 2 7",
"output": "YES"
},
{
"input": "2 2 1500",
"output": "YES"
},
{
"input": "3 6 9",
"output": "NO"
},
{
"input": "1 46 79",
"output": "YES"
},
{
"input": "4 3 3",
"output": "NO"
},
{
"input": "2 4 8",
"output": "NO"
},
{
"input": "1493 1489 1487",
"output": "NO"
},
{
"input": "1 2 3",
"output": "YES"
},
{
"input": "1 2 5",
"output": "YES"
},
{
"input": "1 2 8",
"output": "YES"
},
{
"input": "3 4 5",
"output": "NO"
},
{
"input": "2 2 4",
"output": "YES"
},
{
"input": "3 2 3",
"output": "NO"
},
{
"input": "7 2 2",
"output": "YES"
},
{
"input": "3 2 2",
"output": "YES"
},
{
"input": "6 7 4",
"output": "NO"
}
] | 1,590,648,209
| 2,609
|
PyPy 3
|
OK
|
TESTS
| 67
| 1,933
| 8,089,600
|
import sys
#import math
#import random
#sys.setrecursionlimit(100000000)
input = sys.stdin.readline
############ ---- USER DEFINED INPUT FUNCTIONS ---- ############
def inp():
return(int(input()))
def inara():
return(list(map(int,input().split())))
def insr():
s = input()
return(list(s[:len(s) - 1]))
def invr():
return(map(int,input().split()))
################################################################
############ ---- THE ACTUAL CODE STARTS BELOW ---- ############
k1,k2,k3=invr()
for x1 in range(1,1200):
for x2 in range(1,1200):
ache=[0]*105
now=x1
while now<105:
ache[now]=1
now+=k1
now=x2
while now<105:
ache[now]=1
now+=k2
baki=[]
for i in range(1,105):
if ache[i]==0:
baki.append(i)
if len(baki)<2:
#print(str(x1)+" "+str(x2),end=" ")
print("YES")
exit(0)
else:
flag=True
for i in range(1,len(baki)):
flag&=((baki[i]-baki[i-1])%k3==0)
if not flag:
break
if flag:
print("YES")
exit(0)
print("NO")
|
Title: Three Garlands
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mishka is decorating the Christmas tree. He has got three garlands, and all of them will be put on the tree. After that Mishka will switch these garlands on.
When a garland is switched on, it periodically changes its state — sometimes it is lit, sometimes not. Formally, if *i*-th garland is switched on during *x*-th second, then it is lit only during seconds *x*, *x*<=+<=*k**i*, *x*<=+<=2*k**i*, *x*<=+<=3*k**i* and so on.
Mishka wants to switch on the garlands in such a way that during each second after switching the garlands on there would be at least one lit garland. Formally, Mishka wants to choose three integers *x*1, *x*2 and *x*3 (not necessarily distinct) so that he will switch on the first garland during *x*1-th second, the second one — during *x*2-th second, and the third one — during *x*3-th second, respectively, and during each second starting from *max*(*x*1,<=*x*2,<=*x*3) at least one garland will be lit.
Help Mishka by telling him if it is possible to do this!
Input Specification:
The first line contains three integers *k*1, *k*2 and *k*3 (1<=≤<=*k**i*<=≤<=1500) — time intervals of the garlands.
Output Specification:
If Mishka can choose moments of time to switch on the garlands in such a way that each second after switching the garlands on at least one garland will be lit, print YES.
Otherwise, print NO.
Demo Input:
['2 2 3\n', '4 2 3\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first example Mishka can choose *x*<sub class="lower-index">1</sub> = 1, *x*<sub class="lower-index">2</sub> = 2, *x*<sub class="lower-index">3</sub> = 1. The first garland will be lit during seconds 1, 3, 5, 7, ..., the second — 2, 4, 6, 8, ..., which already cover all the seconds after the 2-nd one. It doesn't even matter what *x*<sub class="lower-index">3</sub> is chosen. Our choice will lead third to be lit during seconds 1, 4, 7, 10, ..., though.
In the second example there is no way to choose such moments of time, there always be some seconds when no garland is lit.
|
```python
import sys
#import math
#import random
#sys.setrecursionlimit(100000000)
input = sys.stdin.readline
############ ---- USER DEFINED INPUT FUNCTIONS ---- ############
def inp():
return(int(input()))
def inara():
return(list(map(int,input().split())))
def insr():
s = input()
return(list(s[:len(s) - 1]))
def invr():
return(map(int,input().split()))
################################################################
############ ---- THE ACTUAL CODE STARTS BELOW ---- ############
k1,k2,k3=invr()
for x1 in range(1,1200):
for x2 in range(1,1200):
ache=[0]*105
now=x1
while now<105:
ache[now]=1
now+=k1
now=x2
while now<105:
ache[now]=1
now+=k2
baki=[]
for i in range(1,105):
if ache[i]==0:
baki.append(i)
if len(baki)<2:
#print(str(x1)+" "+str(x2),end=" ")
print("YES")
exit(0)
else:
flag=True
for i in range(1,len(baki)):
flag&=((baki[i]-baki[i-1])%k3==0)
if not flag:
break
if flag:
print("YES")
exit(0)
print("NO")
```
| 3
|
|
792
|
A
|
New Bus Route
|
PROGRAMMING
| 1,100
|
[
"implementation",
"sortings"
] | null | null |
There are *n* cities situated along the main road of Berland. Cities are represented by their coordinates — integer numbers *a*1,<=*a*2,<=...,<=*a**n*. All coordinates are pairwise distinct.
It is possible to get from one city to another only by bus. But all buses and roads are very old, so the Minister of Transport decided to build a new bus route. The Minister doesn't want to spend large amounts of money — he wants to choose two cities in such a way that the distance between them is minimal possible. The distance between two cities is equal to the absolute value of the difference between their coordinates.
It is possible that there are multiple pairs of cities with minimal possible distance, so the Minister wants to know the quantity of such pairs.
Your task is to write a program that will calculate the minimal possible distance between two pairs of cities and the quantity of pairs which have this distance.
|
The first line contains one integer number *n* (2<=≤<=*n*<=≤<=2·105).
The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=109<=≤<=*a**i*<=≤<=109). All numbers *a**i* are pairwise distinct.
|
Print two integer numbers — the minimal distance and the quantity of pairs with this distance.
|
[
"4\n6 -3 0 4\n",
"3\n-2 0 2\n"
] |
[
"2 1\n",
"2 2\n"
] |
In the first example the distance between the first city and the fourth city is |4 - 6| = 2, and it is the only pair with this distance.
| 0
|
[
{
"input": "4\n6 -3 0 4",
"output": "2 1"
},
{
"input": "3\n-2 0 2",
"output": "2 2"
},
{
"input": "2\n1 2",
"output": "1 1"
},
{
"input": "2\n1000000000 -1000000000",
"output": "2000000000 1"
},
{
"input": "5\n-979619606 -979619602 -979619604 -979619605 -979619603",
"output": "1 4"
},
{
"input": "5\n-799147771 -799147773 -799147764 -799147774 -799147770",
"output": "1 2"
},
{
"input": "20\n553280626 553280623 553280627 553280624 553280625 553280618 553280620 553280629 553280637 553280631 553280628 553280636 553280635 553280632 553280634 553280622 553280633 553280621 553280630 553280619",
"output": "1 19"
},
{
"input": "20\n105619866 106083760 106090730 105809555 106115212 105155938 105979518 106075627 106145216 105637844 105925719 105498536 105927000 106155938 106134226 106125969 106130588 105464813 106145509 106114971",
"output": "241 1"
},
{
"input": "10\n570685866 570685854 570685858 570685850 570685856 570685864 570685860 570685852 570685862 570685868",
"output": "2 9"
},
{
"input": "2\n1 1000000000",
"output": "999999999 1"
},
{
"input": "6\n1 2 3 4 5 6",
"output": "1 5"
},
{
"input": "3\n7 10 12",
"output": "2 1"
},
{
"input": "5\n-7 -5 -4 -3 -1",
"output": "1 2"
},
{
"input": "4\n-6 -4 -2 1",
"output": "2 2"
},
{
"input": "4\n3 5 7 8",
"output": "1 1"
},
{
"input": "9\n-9 -8 -7 -6 -5 -4 -3 -2 -1",
"output": "1 8"
},
{
"input": "2\n15 13",
"output": "2 1"
},
{
"input": "2\n14 13",
"output": "1 1"
},
{
"input": "2\n12 13",
"output": "1 1"
},
{
"input": "2\n-1000000000 13265920",
"output": "1013265920 1"
},
{
"input": "2\n8 10",
"output": "2 1"
}
] | 1,637,303,192
| 92
|
PyPy 3-64
|
OK
|
TESTS
| 45
| 249
| 22,323,200
|
import sys
input = sys.stdin.readline
n = int(input())
l = sorted([int(i) for i in input().split()])
res = 10**18
ans = 0
for i in range(n-1):
t = l[i+1]-l[i]
if t == res:
ans += 1
elif t < res:
res = t
ans = 1
print(res, ans)
|
Title: New Bus Route
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* cities situated along the main road of Berland. Cities are represented by their coordinates — integer numbers *a*1,<=*a*2,<=...,<=*a**n*. All coordinates are pairwise distinct.
It is possible to get from one city to another only by bus. But all buses and roads are very old, so the Minister of Transport decided to build a new bus route. The Minister doesn't want to spend large amounts of money — he wants to choose two cities in such a way that the distance between them is minimal possible. The distance between two cities is equal to the absolute value of the difference between their coordinates.
It is possible that there are multiple pairs of cities with minimal possible distance, so the Minister wants to know the quantity of such pairs.
Your task is to write a program that will calculate the minimal possible distance between two pairs of cities and the quantity of pairs which have this distance.
Input Specification:
The first line contains one integer number *n* (2<=≤<=*n*<=≤<=2·105).
The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=109<=≤<=*a**i*<=≤<=109). All numbers *a**i* are pairwise distinct.
Output Specification:
Print two integer numbers — the minimal distance and the quantity of pairs with this distance.
Demo Input:
['4\n6 -3 0 4\n', '3\n-2 0 2\n']
Demo Output:
['2 1\n', '2 2\n']
Note:
In the first example the distance between the first city and the fourth city is |4 - 6| = 2, and it is the only pair with this distance.
|
```python
import sys
input = sys.stdin.readline
n = int(input())
l = sorted([int(i) for i in input().split()])
res = 10**18
ans = 0
for i in range(n-1):
t = l[i+1]-l[i]
if t == res:
ans += 1
elif t < res:
res = t
ans = 1
print(res, ans)
```
| 3
|
|
56
|
A
|
Bar
|
PROGRAMMING
| 1,000
|
[
"implementation"
] |
A. Bar
|
2
|
256
|
According to Berland laws it is only allowed to sell alcohol to people not younger than 18 years. Vasya's job is to monitor the law's enforcement. Tonight he entered a bar and saw *n* people sitting there. For every one of them Vasya happened to determine either the age or the drink the person is having. Vasya can check any person, i.e. learn his age and the drink he is having at the same time. What minimal number of people should Vasya check additionally to make sure that there are no clients under 18 having alcohol drinks?
The list of all alcohol drinks in Berland is: ABSINTH, BEER, BRANDY, CHAMPAGNE, GIN, RUM, SAKE, TEQUILA, VODKA, WHISKEY, WINE
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) which is the number of the bar's clients. Then follow *n* lines, each describing one visitor. A line either contains his age (an integer from 0 to 1000) or his drink (a string of capital Latin letters from 1 to 100 in length). It is guaranteed that the input data does not contain spaces and other unnecessary separators.
Only the drinks from the list given above should be considered alcohol.
|
Print a single number which is the number of people Vasya should check to guarantee the law enforcement.
|
[
"5\n18\nVODKA\nCOKE\n19\n17\n"
] |
[
"2\n"
] |
In the sample test the second and fifth clients should be checked.
| 500
|
[
{
"input": "5\n18\nVODKA\nCOKE\n19\n17",
"output": "2"
},
{
"input": "2\n2\nGIN",
"output": "2"
},
{
"input": "3\nWHISKEY\n3\nGIN",
"output": "3"
},
{
"input": "4\n813\nIORBQITQXMPTFAEMEQDQIKFGKGOTNKTOSZCBRPXJLUKVLVHJYNRUJXK\nRUM\nRHVRWGODYWWTYZFLFYKCVUFFRTQDINKNWPKFHZBFWBHWINWJW",
"output": "1"
},
{
"input": "4\nSAKE\nSAKE\n13\n2",
"output": "4"
},
{
"input": "2\n0\n17",
"output": "2"
},
{
"input": "1\n0",
"output": "1"
}
] | 1,580,374,511
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 28
| 218
| 0
|
lis = ["ABSINTH","BEER","BRANDY","CHAMPAGNE","GIN","RUM","SAKE","TEQUILA","VODKA","WHISKEY","WINE"]
n = int(input())
k = 0
for i in range(n):
s = input()
if s.isnumeric():
s = int(s)
if s < 18:
k = k + 1
else:
if s in lis:
k = k + 1
print(k)
|
Title: Bar
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
According to Berland laws it is only allowed to sell alcohol to people not younger than 18 years. Vasya's job is to monitor the law's enforcement. Tonight he entered a bar and saw *n* people sitting there. For every one of them Vasya happened to determine either the age or the drink the person is having. Vasya can check any person, i.e. learn his age and the drink he is having at the same time. What minimal number of people should Vasya check additionally to make sure that there are no clients under 18 having alcohol drinks?
The list of all alcohol drinks in Berland is: ABSINTH, BEER, BRANDY, CHAMPAGNE, GIN, RUM, SAKE, TEQUILA, VODKA, WHISKEY, WINE
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) which is the number of the bar's clients. Then follow *n* lines, each describing one visitor. A line either contains his age (an integer from 0 to 1000) or his drink (a string of capital Latin letters from 1 to 100 in length). It is guaranteed that the input data does not contain spaces and other unnecessary separators.
Only the drinks from the list given above should be considered alcohol.
Output Specification:
Print a single number which is the number of people Vasya should check to guarantee the law enforcement.
Demo Input:
['5\n18\nVODKA\nCOKE\n19\n17\n']
Demo Output:
['2\n']
Note:
In the sample test the second and fifth clients should be checked.
|
```python
lis = ["ABSINTH","BEER","BRANDY","CHAMPAGNE","GIN","RUM","SAKE","TEQUILA","VODKA","WHISKEY","WINE"]
n = int(input())
k = 0
for i in range(n):
s = input()
if s.isnumeric():
s = int(s)
if s < 18:
k = k + 1
else:
if s in lis:
k = k + 1
print(k)
```
| 3.9455
|
676
|
A
|
Nicholas and Permutation
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"implementation"
] | null | null |
Nicholas has an array *a* that contains *n* distinct integers from 1 to *n*. In other words, Nicholas has a permutation of size *n*.
Nicholas want the minimum element (integer 1) and the maximum element (integer *n*) to be as far as possible from each other. He wants to perform exactly one swap in order to maximize the distance between the minimum and the maximum elements. The distance between two elements is considered to be equal to the absolute difference between their positions.
|
The first line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100) — the size of the permutation.
The second line of the input contains *n* distinct integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*), where *a**i* is equal to the element at the *i*-th position.
|
Print a single integer — the maximum possible distance between the minimum and the maximum elements Nicholas can achieve by performing exactly one swap.
|
[
"5\n4 5 1 3 2\n",
"7\n1 6 5 3 4 7 2\n",
"6\n6 5 4 3 2 1\n"
] |
[
"3\n",
"6\n",
"5\n"
] |
In the first sample, one may obtain the optimal answer by swapping elements 1 and 2.
In the second sample, the minimum and the maximum elements will be located in the opposite ends of the array if we swap 7 and 2.
In the third sample, the distance between the minimum and the maximum elements is already maximum possible, so we just perform some unnecessary swap, for example, one can swap 5 and 2.
| 500
|
[
{
"input": "5\n4 5 1 3 2",
"output": "3"
},
{
"input": "7\n1 6 5 3 4 7 2",
"output": "6"
},
{
"input": "6\n6 5 4 3 2 1",
"output": "5"
},
{
"input": "2\n1 2",
"output": "1"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "3\n2 3 1",
"output": "2"
},
{
"input": "4\n4 1 3 2",
"output": "3"
},
{
"input": "5\n1 4 5 2 3",
"output": "4"
},
{
"input": "6\n4 6 3 5 2 1",
"output": "5"
},
{
"input": "7\n1 5 3 6 2 4 7",
"output": "6"
},
{
"input": "100\n76 70 67 54 40 1 48 63 64 36 42 90 99 27 47 17 93 7 13 84 16 57 74 5 83 61 19 56 52 92 38 91 82 79 34 66 71 28 37 98 35 94 77 53 73 10 26 80 15 32 8 81 3 95 44 46 72 6 33 11 21 85 4 30 24 51 49 96 87 55 14 31 12 60 45 9 29 22 58 18 88 2 50 59 20 86 23 41 100 39 62 68 69 97 78 43 25 89 65 75",
"output": "94"
},
{
"input": "8\n4 5 3 8 6 7 1 2",
"output": "6"
},
{
"input": "9\n6 8 5 3 4 7 9 2 1",
"output": "8"
},
{
"input": "10\n8 7 10 1 2 3 4 6 5 9",
"output": "7"
},
{
"input": "11\n5 4 6 9 10 11 7 3 1 2 8",
"output": "8"
},
{
"input": "12\n3 6 7 8 9 10 12 5 4 2 11 1",
"output": "11"
},
{
"input": "13\n8 4 3 7 5 11 9 1 10 2 13 12 6",
"output": "10"
},
{
"input": "14\n6 10 13 9 7 1 12 14 3 2 5 4 11 8",
"output": "8"
},
{
"input": "15\n3 14 13 12 7 2 4 11 15 1 8 6 5 10 9",
"output": "9"
},
{
"input": "16\n11 6 9 8 7 14 12 13 10 15 2 5 3 1 4 16",
"output": "15"
},
{
"input": "17\n13 12 5 3 9 16 8 14 2 4 10 1 6 11 7 15 17",
"output": "16"
},
{
"input": "18\n8 6 14 17 9 11 15 13 5 3 18 1 2 7 12 16 4 10",
"output": "11"
},
{
"input": "19\n12 19 3 11 15 6 18 14 5 10 2 13 9 7 4 8 17 16 1",
"output": "18"
},
{
"input": "20\n15 17 10 20 7 2 16 9 13 6 18 5 19 8 11 14 4 12 3 1",
"output": "19"
},
{
"input": "21\n1 9 14 18 13 12 11 20 16 2 4 19 15 7 6 17 8 5 3 10 21",
"output": "20"
},
{
"input": "22\n8 3 17 4 16 21 14 11 10 15 6 18 13 12 22 20 5 2 9 7 19 1",
"output": "21"
},
{
"input": "23\n1 23 11 20 9 3 12 4 7 17 5 15 2 10 18 16 8 22 14 13 19 21 6",
"output": "22"
},
{
"input": "24\n2 10 23 22 20 19 18 16 11 12 15 17 21 8 24 13 1 5 6 7 14 3 9 4",
"output": "16"
},
{
"input": "25\n12 13 22 17 1 18 14 5 21 2 10 4 3 23 11 6 20 8 24 16 15 19 9 7 25",
"output": "24"
},
{
"input": "26\n6 21 20 16 26 17 11 2 24 4 1 12 14 8 25 7 15 10 22 5 13 18 9 23 19 3",
"output": "21"
},
{
"input": "27\n20 14 18 10 5 3 9 4 24 22 21 27 17 15 26 2 23 7 12 11 6 8 19 25 16 13 1",
"output": "26"
},
{
"input": "28\n28 13 16 6 1 12 4 27 22 7 18 3 21 26 25 11 5 10 20 24 19 15 14 8 23 17 9 2",
"output": "27"
},
{
"input": "29\n21 11 10 25 2 5 9 16 29 8 17 4 15 13 6 22 7 24 19 12 18 20 1 3 23 28 27 14 26",
"output": "22"
},
{
"input": "30\n6 19 14 22 26 17 27 8 25 3 24 30 4 18 23 16 9 13 29 20 15 2 5 11 28 12 1 10 21 7",
"output": "26"
},
{
"input": "31\n29 13 26 27 9 28 2 16 30 21 12 11 3 31 23 6 22 20 1 5 14 24 19 18 8 4 10 17 15 25 7",
"output": "18"
},
{
"input": "32\n15 32 11 3 18 23 19 14 5 8 6 21 13 24 25 4 16 9 27 20 17 31 2 22 7 12 30 1 26 10 29 28",
"output": "30"
},
{
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"output": "29"
},
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"output": "33"
},
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"output": "21"
},
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"output": "35"
},
{
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"output": "35"
},
{
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"output": "34"
},
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"output": "38"
},
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"output": "39"
},
{
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"output": "34"
},
{
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"output": "41"
},
{
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"output": "42"
},
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"output": "37"
},
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"input": "45\n45 29 24 2 31 5 34 41 26 44 33 43 15 3 4 11 21 37 27 12 14 39 23 42 16 6 13 19 8 38 20 9 25 22 40 17 32 35 18 10 28 7 30 36 1",
"output": "44"
},
{
"input": "46\n29 3 12 33 45 40 19 17 25 27 28 1 16 23 24 46 31 8 44 15 5 32 22 11 4 36 34 10 35 26 21 7 14 2 18 9 20 41 6 43 42 37 38 13 39 30",
"output": "34"
},
{
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"output": "41"
},
{
"input": "48\n29 26 14 18 34 33 13 39 32 1 37 20 35 19 28 48 30 23 46 27 5 22 24 38 12 15 8 36 43 45 16 47 6 9 31 40 44 17 2 41 11 42 25 4 21 3 10 7",
"output": "38"
},
{
"input": "49\n16 7 42 32 11 35 15 8 23 41 6 20 47 24 9 45 49 2 37 48 25 28 5 18 3 19 12 4 22 33 13 14 10 36 44 17 40 38 30 26 1 43 29 46 21 34 27 39 31",
"output": "40"
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{
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"output": "38"
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"output": "43"
},
{
"input": "52\n16 45 23 7 15 19 43 20 4 32 35 36 9 50 5 26 38 46 13 33 12 2 48 37 41 31 10 28 8 42 3 21 11 1 17 27 34 30 44 40 6 51 49 47 25 22 18 24 52 29 14 39",
"output": "48"
},
{
"input": "53\n53 30 50 22 51 31 32 38 12 7 39 43 1 23 6 8 24 52 2 21 34 13 3 35 5 15 19 11 47 18 9 20 29 4 36 45 27 41 25 48 16 46 44 17 10 14 42 26 40 28 33 37 49",
"output": "52"
},
{
"input": "54\n6 39 17 3 45 52 16 21 23 48 42 36 13 37 46 10 43 27 49 7 38 32 31 30 15 25 2 29 8 51 54 19 41 44 24 34 22 5 20 14 12 1 33 40 4 26 9 35 18 28 47 50 11 53",
"output": "41"
},
{
"input": "55\n26 15 31 21 32 43 34 51 7 12 5 44 17 54 18 25 48 47 20 3 41 24 45 2 11 22 29 39 37 53 35 28 36 9 50 10 30 38 19 13 4 8 27 1 42 6 49 23 55 40 33 16 46 14 52",
"output": "48"
},
{
"input": "56\n6 20 38 46 10 11 40 19 5 1 47 33 4 18 32 36 37 45 56 49 48 52 12 26 31 14 2 9 24 3 16 51 41 43 23 17 34 7 29 50 55 25 39 44 22 27 54 8 28 35 30 42 13 53 21 15",
"output": "46"
},
{
"input": "57\n39 28 53 36 3 6 12 56 55 20 50 19 43 42 18 40 24 52 38 17 33 23 22 41 14 7 26 44 45 16 35 1 8 47 31 5 30 51 32 4 37 25 13 34 54 21 46 10 15 11 2 27 29 48 49 9 57",
"output": "56"
},
{
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"output": "57"
},
{
"input": "59\n1 27 10 37 53 9 14 49 46 26 50 42 59 11 47 15 24 56 43 45 44 38 5 8 58 30 52 12 23 32 22 3 31 41 2 25 29 6 54 16 35 33 18 55 4 51 57 28 40 19 13 21 7 39 36 48 34 17 20",
"output": "58"
},
{
"input": "60\n60 27 34 32 54 55 33 12 40 3 47 44 50 39 38 59 11 25 17 15 16 30 21 31 10 52 5 23 4 48 6 26 36 57 14 22 8 56 58 9 24 7 37 53 42 43 20 49 51 19 2 46 28 18 35 13 29 45 41 1",
"output": "59"
},
{
"input": "61\n61 11 26 29 31 40 32 30 35 3 18 52 9 53 42 4 50 54 20 58 28 49 22 12 2 19 16 15 57 34 51 43 7 17 25 41 56 47 55 60 46 14 44 45 24 27 33 1 48 13 59 23 38 39 6 5 36 10 8 37 21",
"output": "60"
},
{
"input": "62\n21 23 34 38 11 61 55 30 37 48 54 51 46 47 6 56 36 49 1 35 12 28 29 20 43 42 5 8 22 57 44 4 53 10 58 33 27 25 16 45 50 40 18 15 3 41 39 2 7 60 59 13 32 24 52 31 14 9 19 26 17 62",
"output": "61"
},
{
"input": "63\n2 5 29 48 31 26 21 16 47 24 43 22 61 28 6 39 60 27 14 52 37 7 53 8 62 56 63 10 50 18 44 13 4 9 25 11 23 42 45 41 59 12 32 36 40 51 1 35 49 54 57 20 19 34 38 46 33 3 55 15 30 58 17",
"output": "46"
},
{
"input": "64\n23 5 51 40 12 46 44 8 64 31 58 55 45 24 54 39 21 19 52 61 30 42 16 18 15 32 53 22 28 26 11 25 48 56 27 9 29 41 35 49 59 38 62 7 34 1 20 33 60 17 2 3 43 37 57 14 6 36 13 10 50 4 63 47",
"output": "55"
},
{
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"output": "62"
},
{
"input": "66\n66 39 3 2 55 53 60 54 12 49 10 30 59 26 32 46 50 56 7 13 43 36 24 28 11 8 6 21 35 25 42 57 23 45 64 5 34 61 27 51 52 9 15 1 38 17 63 48 37 20 58 14 47 19 22 41 31 44 33 65 4 62 40 18 16 29",
"output": "65"
},
{
"input": "67\n66 16 2 53 35 38 49 28 18 6 36 58 21 47 27 5 50 62 44 12 52 37 11 56 15 31 25 65 17 29 59 41 7 42 4 43 39 10 1 40 24 13 20 54 19 67 46 60 51 45 64 30 8 33 26 9 3 22 34 23 57 48 55 14 63 61 32",
"output": "45"
},
{
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"output": "64"
},
{
"input": "69\n29 49 25 51 21 35 11 61 39 54 40 37 60 42 27 33 59 53 34 10 46 2 23 69 8 47 58 36 1 38 19 12 7 48 13 3 6 22 18 5 65 24 50 41 66 44 67 57 4 56 62 43 9 30 14 15 28 31 64 26 16 55 68 17 32 20 45 52 63",
"output": "45"
},
{
"input": "70\n19 12 15 18 36 16 61 69 24 7 11 13 3 48 55 21 37 17 43 31 41 22 28 32 27 63 38 49 59 56 30 25 67 51 52 45 50 44 66 57 26 60 5 46 33 6 23 34 8 40 2 68 14 39 65 64 62 42 47 54 10 53 9 1 70 58 20 4 29 35",
"output": "64"
},
{
"input": "71\n40 6 62 3 41 52 31 66 27 16 35 5 17 60 2 15 51 22 67 61 71 53 1 64 8 45 28 18 50 30 12 69 20 26 10 37 36 49 70 32 33 11 57 14 9 55 4 58 29 25 44 65 39 48 24 47 19 46 56 38 34 42 59 63 54 23 7 68 43 13 21",
"output": "50"
},
{
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"output": "57"
},
{
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"output": "45"
},
{
"input": "74\n19 51 59 34 8 40 42 55 65 16 74 26 49 63 64 70 35 72 7 12 43 18 61 27 47 31 13 32 71 22 25 67 9 1 48 50 33 10 21 46 11 45 17 37 28 60 69 66 38 2 30 3 39 15 53 68 57 41 6 36 24 73 4 23 5 62 44 14 20 29 52 54 56 58",
"output": "63"
},
{
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"output": "74"
},
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"output": "75"
},
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"output": "62"
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"output": "70"
},
{
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"output": "77"
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{
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"output": "52"
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{
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"output": "69"
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"output": "53"
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"output": "66"
},
{
"input": "84\n75 8 66 21 61 63 72 51 52 13 59 25 28 58 64 53 79 41 34 7 67 11 39 56 44 24 50 9 49 55 1 80 26 6 73 74 27 69 65 37 18 43 36 17 30 3 47 29 76 78 32 22 12 68 46 5 42 81 57 31 33 83 54 48 14 62 10 16 4 20 71 70 35 15 45 19 60 77 2 23 84 40 82 38",
"output": "80"
},
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"output": "84"
},
{
"input": "86\n64 56 41 10 31 69 47 39 37 36 27 19 9 42 15 6 78 59 52 17 71 45 72 14 2 54 38 79 4 18 16 8 46 75 50 82 44 24 20 55 58 86 61 43 35 32 33 40 63 30 28 60 13 53 12 57 77 81 76 66 73 84 85 62 68 22 51 5 49 7 1 70 80 65 34 48 23 21 83 11 74 26 29 67 25 3",
"output": "70"
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{
"input": "87\n14 20 82 47 39 75 71 45 3 37 63 19 32 68 7 41 48 76 27 46 84 49 4 44 26 69 17 64 1 18 58 33 11 23 21 86 67 52 70 16 77 78 6 74 15 87 10 59 13 34 22 2 65 38 66 61 51 57 35 60 81 40 36 80 31 43 83 56 79 55 29 5 12 8 50 30 53 72 54 9 24 25 42 62 73 28 85",
"output": "58"
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"output": "87"
},
{
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"output": "88"
},
{
"input": "90\n2 67 26 58 9 49 76 22 60 30 77 20 13 7 37 81 47 16 19 12 14 45 41 68 85 54 28 24 46 1 27 43 32 89 53 35 59 75 18 51 17 64 66 80 31 88 87 90 38 72 55 71 42 11 73 69 62 78 23 74 65 79 84 4 86 52 10 6 3 82 56 5 48 33 21 57 40 29 61 63 34 36 83 8 15 44 50 70 39 25",
"output": "60"
},
{
"input": "91\n91 69 56 16 73 55 14 82 80 46 57 81 22 71 63 76 43 37 77 75 70 3 26 2 28 17 51 38 30 67 41 47 54 62 34 25 84 11 87 39 32 52 31 36 50 19 21 53 29 24 79 8 74 64 44 7 6 18 10 42 13 9 83 58 4 88 65 60 20 90 66 49 86 89 78 48 5 27 23 59 61 15 72 45 40 33 68 85 35 12 1",
"output": "90"
},
{
"input": "92\n67 57 76 78 25 89 6 82 11 16 26 17 59 48 73 10 21 31 27 80 4 5 22 13 92 55 45 85 63 28 75 60 54 88 91 47 29 35 7 87 1 39 43 51 71 84 83 81 46 9 38 56 90 24 37 41 19 86 50 61 79 20 18 14 69 23 62 65 49 52 58 53 36 2 68 64 15 42 30 34 66 32 44 40 8 33 3 77 74 12 70 72",
"output": "67"
},
{
"input": "93\n76 35 5 87 7 21 59 71 24 37 2 73 31 74 4 52 28 20 56 27 65 86 16 45 85 67 68 70 47 72 91 88 14 32 62 69 78 41 15 22 57 18 50 13 39 58 17 83 64 51 25 11 38 77 82 90 8 26 29 61 10 43 79 53 48 6 23 55 63 49 81 92 80 44 89 60 66 30 1 9 36 33 19 46 75 93 3 12 42 84 40 54 34",
"output": "85"
},
{
"input": "94\n29 85 82 78 61 83 80 63 11 38 50 43 9 24 4 87 79 45 3 17 90 7 34 27 1 76 26 39 84 47 22 41 81 19 44 23 56 92 35 31 72 62 70 53 40 88 13 14 73 2 59 86 46 94 15 12 77 57 89 42 75 48 18 51 32 55 71 30 49 91 20 60 5 93 33 64 21 36 10 28 8 65 66 69 74 58 6 52 25 67 16 37 54 68",
"output": "69"
},
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"output": "94"
},
{
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"output": "86"
},
{
"input": "97\n40 63 44 64 84 92 38 41 28 91 3 70 76 67 94 96 35 79 29 22 78 88 85 8 21 1 93 54 71 80 37 17 13 26 62 59 75 87 69 33 89 49 77 61 12 39 6 36 58 18 73 50 82 45 74 52 11 34 95 7 23 30 15 32 31 16 55 19 20 83 60 72 10 53 51 14 27 9 68 47 5 2 81 46 57 86 56 43 48 66 24 25 4 42 65 97 90",
"output": "95"
},
{
"input": "98\n85 94 69 86 22 52 27 79 53 91 35 55 33 88 8 75 76 95 64 54 67 30 70 49 6 16 2 48 80 32 25 90 98 46 9 96 36 81 10 92 28 11 37 97 15 41 38 40 83 44 29 47 23 3 31 61 87 39 78 20 68 12 17 73 59 18 77 72 43 51 84 24 89 65 26 7 74 93 21 19 5 14 50 42 82 71 60 56 34 62 58 57 45 66 13 63 4 1",
"output": "97"
},
{
"input": "99\n33 48 19 41 59 64 16 12 17 13 7 1 9 6 4 92 61 49 60 25 74 65 22 97 30 32 10 62 14 55 80 66 82 78 31 23 87 93 27 98 20 29 88 84 77 34 83 96 79 90 56 89 58 72 52 47 21 76 24 70 44 94 5 39 8 18 57 36 40 68 43 75 3 2 35 99 63 26 67 73 15 11 53 28 42 46 69 50 51 95 38 37 54 85 81 91 45 86 71",
"output": "87"
},
{
"input": "100\n28 30 77 4 81 67 31 25 66 56 88 73 83 51 57 34 21 90 38 76 22 99 53 70 91 3 64 54 6 94 8 5 97 80 50 45 61 40 16 95 36 98 9 2 17 44 72 55 18 58 47 12 87 24 7 32 14 23 65 41 63 48 62 39 92 27 43 19 46 13 42 52 96 84 26 69 100 79 93 49 35 60 71 59 68 15 10 29 20 1 78 33 75 86 11 85 74 82 89 37",
"output": "89"
},
{
"input": "100\n100 97 35 55 45 3 46 98 77 64 94 85 73 43 49 79 72 9 70 62 80 88 29 58 61 20 89 83 66 86 82 15 6 87 42 96 90 75 63 38 81 40 5 23 4 18 41 19 99 60 8 12 76 51 39 93 53 26 21 50 47 28 13 30 68 59 34 54 24 56 31 27 65 16 32 10 36 52 44 91 22 14 33 25 7 78 67 17 57 37 92 11 2 69 84 95 74 71 48 1",
"output": "99"
},
{
"input": "100\n83 96 73 70 30 25 7 77 58 89 76 85 49 82 45 51 14 62 50 9 31 32 16 15 97 64 4 37 20 93 24 10 80 71 100 39 75 72 78 74 8 29 53 86 79 48 3 68 90 99 56 87 63 94 36 1 40 65 6 44 43 84 17 52 34 95 38 47 60 57 98 59 33 41 46 81 23 27 19 2 54 91 55 35 26 12 92 18 28 66 69 21 5 67 13 11 22 88 61 42",
"output": "65"
},
{
"input": "100\n96 80 47 60 56 9 78 20 37 72 68 15 100 94 51 26 65 38 50 19 4 70 25 63 22 30 13 58 43 69 18 33 5 66 39 73 12 55 95 92 97 1 14 83 10 28 64 31 46 91 32 86 74 54 29 52 89 53 90 44 62 40 16 24 67 81 36 34 7 23 79 87 75 98 84 3 41 77 76 42 71 35 49 61 2 27 59 82 99 85 21 11 45 6 88 48 17 57 8 93",
"output": "87"
},
{
"input": "100\n5 6 88 37 97 51 25 81 54 17 57 98 99 44 67 24 30 93 100 36 8 38 84 42 21 4 75 31 85 48 70 77 43 50 65 94 29 32 68 86 56 39 69 47 20 60 52 53 10 34 79 2 95 40 89 64 71 26 22 46 1 62 91 76 83 41 9 78 16 63 13 3 28 92 27 49 7 12 96 72 80 23 14 19 18 66 59 87 90 45 73 82 33 74 35 61 55 15 58 11",
"output": "81"
},
{
"input": "100\n100 97 92 12 62 17 19 58 37 26 30 95 31 35 87 10 13 43 98 61 28 89 76 1 23 21 11 22 50 56 91 74 3 24 96 55 64 67 14 4 71 16 18 9 77 68 51 81 32 82 46 88 86 60 29 66 72 85 70 7 53 63 33 45 83 2 25 94 52 93 5 69 20 47 49 54 57 39 34 27 90 80 78 59 40 42 79 6 38 8 48 15 65 73 99 44 41 84 36 75",
"output": "99"
},
{
"input": "100\n22 47 34 65 69 5 68 78 53 54 41 23 80 51 11 8 2 85 81 75 25 58 29 73 30 49 10 71 17 96 76 89 79 20 12 15 55 7 46 32 19 3 82 35 74 44 38 40 92 14 6 50 97 63 45 93 37 18 62 77 87 36 83 9 90 61 57 28 39 43 52 42 24 56 21 84 26 99 88 59 33 70 4 60 98 95 94 100 13 48 66 72 16 31 64 91 1 86 27 67",
"output": "96"
},
{
"input": "100\n41 67 94 18 14 83 59 12 19 54 13 68 75 26 15 65 80 40 23 30 34 78 47 21 63 79 4 70 3 31 86 69 92 10 61 74 97 100 9 99 32 27 91 55 85 52 16 17 28 1 64 29 58 76 98 25 84 7 2 96 20 72 36 46 49 82 93 44 45 6 38 87 57 50 53 35 60 33 8 89 39 42 37 48 62 81 73 43 95 11 66 88 90 22 24 77 71 51 5 56",
"output": "62"
},
{
"input": "100\n1 88 38 56 62 99 39 80 12 33 57 24 28 84 37 42 10 95 83 58 8 40 20 2 30 78 60 79 36 71 51 31 27 65 22 47 6 19 61 94 75 4 74 35 15 23 92 9 70 13 11 59 90 18 66 81 64 72 16 32 34 67 46 91 21 87 77 97 82 41 7 86 26 43 45 3 93 17 52 96 50 63 48 5 53 44 29 25 98 54 49 14 73 69 89 55 76 85 68 100",
"output": "99"
},
{
"input": "100\n22 59 25 77 68 79 32 45 20 28 61 60 38 86 33 10 100 15 53 75 78 39 67 13 66 34 96 4 63 23 73 29 31 35 71 55 16 14 72 56 94 97 17 93 47 84 57 8 21 51 54 85 26 76 49 81 2 92 62 44 91 87 11 24 95 69 5 7 99 6 65 48 70 12 41 18 74 27 42 3 80 30 50 98 58 37 82 89 83 36 40 52 19 9 88 46 43 1 90 64",
"output": "97"
},
{
"input": "100\n12 1 76 78 97 82 59 80 48 8 91 51 54 74 16 10 89 99 83 63 93 90 55 25 30 33 29 6 9 65 92 79 44 39 15 58 37 46 32 19 27 3 75 49 62 71 98 42 69 50 26 81 96 5 7 61 60 21 20 36 18 34 40 4 47 85 64 38 22 84 2 68 11 56 31 66 17 14 95 43 53 35 23 52 70 13 72 45 41 77 73 87 88 94 28 86 24 67 100 57",
"output": "98"
},
{
"input": "100\n66 100 53 88 7 73 54 41 31 42 8 46 65 90 78 14 94 30 79 39 89 5 83 50 38 61 37 86 22 95 60 98 34 57 91 10 75 25 15 43 23 17 96 35 93 48 87 47 56 13 19 9 82 62 67 80 11 55 99 70 18 26 58 85 12 44 16 45 4 49 20 71 92 24 81 2 76 32 6 21 84 36 52 97 59 63 40 51 27 64 68 3 77 72 28 33 29 1 74 69",
"output": "98"
},
{
"input": "100\n56 64 1 95 72 39 9 49 87 29 94 7 32 6 30 48 50 25 31 78 90 45 60 44 80 68 17 20 73 15 75 98 83 13 71 22 36 26 96 88 35 3 85 54 16 41 92 99 69 86 93 33 43 62 77 46 47 37 12 10 18 40 27 4 63 55 28 59 23 34 61 53 76 42 51 91 21 70 8 58 38 19 5 66 84 11 52 24 81 82 79 67 97 65 57 74 2 89 100 14",
"output": "98"
},
{
"input": "3\n1 2 3",
"output": "2"
},
{
"input": "3\n1 3 2",
"output": "2"
},
{
"input": "3\n2 1 3",
"output": "2"
},
{
"input": "3\n2 3 1",
"output": "2"
},
{
"input": "3\n3 1 2",
"output": "2"
},
{
"input": "3\n3 2 1",
"output": "2"
},
{
"input": "4\n1 2 3 4",
"output": "3"
},
{
"input": "4\n1 2 4 3",
"output": "3"
},
{
"input": "4\n1 3 2 4",
"output": "3"
},
{
"input": "4\n1 3 4 2",
"output": "3"
},
{
"input": "4\n1 4 2 3",
"output": "3"
},
{
"input": "4\n1 4 3 2",
"output": "3"
},
{
"input": "4\n2 1 3 4",
"output": "3"
},
{
"input": "4\n2 1 4 3",
"output": "2"
},
{
"input": "4\n2 4 1 3",
"output": "2"
},
{
"input": "4\n2 4 3 1",
"output": "3"
},
{
"input": "4\n3 1 2 4",
"output": "3"
},
{
"input": "4\n3 1 4 2",
"output": "2"
},
{
"input": "4\n3 2 1 4",
"output": "3"
},
{
"input": "4\n3 2 4 1",
"output": "3"
},
{
"input": "4\n3 4 1 2",
"output": "2"
},
{
"input": "4\n3 4 2 1",
"output": "3"
},
{
"input": "4\n4 1 2 3",
"output": "3"
},
{
"input": "4\n4 1 3 2",
"output": "3"
},
{
"input": "4\n4 2 1 3",
"output": "3"
},
{
"input": "4\n4 2 3 1",
"output": "3"
},
{
"input": "4\n4 3 1 2",
"output": "3"
},
{
"input": "4\n4 3 2 1",
"output": "3"
},
{
"input": "8\n2 5 6 4 8 3 1 7",
"output": "6"
},
{
"input": "5\n2 3 1 5 4",
"output": "3"
},
{
"input": "6\n2 5 3 6 4 1",
"output": "5"
},
{
"input": "6\n5 4 2 6 1 3",
"output": "4"
},
{
"input": "6\n4 2 3 1 6 5",
"output": "4"
},
{
"input": "6\n5 4 2 1 6 3",
"output": "4"
},
{
"input": "9\n7 2 3 4 5 6 1 9 8",
"output": "7"
},
{
"input": "6\n3 2 1 4 6 5",
"output": "4"
},
{
"input": "6\n2 3 4 1 6 5",
"output": "4"
},
{
"input": "10\n5 2 3 4 1 6 7 8 10 9",
"output": "8"
},
{
"input": "6\n5 2 3 1 6 4",
"output": "4"
},
{
"input": "10\n2 9 3 4 1 10 5 6 7 8",
"output": "5"
},
{
"input": "10\n2 3 4 5 6 7 1 8 10 9",
"output": "8"
},
{
"input": "8\n2 3 4 5 1 6 8 7",
"output": "6"
},
{
"input": "6\n2 1 3 4 5 6",
"output": "5"
}
] | 1,685,950,052
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 159
| 62
| 0
|
n = int(input())
x = [int(x) for x in input().split()]
minpos = x.index(min(x))
maxpos = x.index(max(x))
if(minpos>maxpos):
d1 = (n-1)-(minpos)
d2 = maxpos-0
print((minpos-maxpos)+max(d1,d2))
else:
d1 = (n-1)-(maxpos)
d2 = minpos-0
print((maxpos-minpos)+max(d1,d2))
|
Title: Nicholas and Permutation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Nicholas has an array *a* that contains *n* distinct integers from 1 to *n*. In other words, Nicholas has a permutation of size *n*.
Nicholas want the minimum element (integer 1) and the maximum element (integer *n*) to be as far as possible from each other. He wants to perform exactly one swap in order to maximize the distance between the minimum and the maximum elements. The distance between two elements is considered to be equal to the absolute difference between their positions.
Input Specification:
The first line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100) — the size of the permutation.
The second line of the input contains *n* distinct integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*), where *a**i* is equal to the element at the *i*-th position.
Output Specification:
Print a single integer — the maximum possible distance between the minimum and the maximum elements Nicholas can achieve by performing exactly one swap.
Demo Input:
['5\n4 5 1 3 2\n', '7\n1 6 5 3 4 7 2\n', '6\n6 5 4 3 2 1\n']
Demo Output:
['3\n', '6\n', '5\n']
Note:
In the first sample, one may obtain the optimal answer by swapping elements 1 and 2.
In the second sample, the minimum and the maximum elements will be located in the opposite ends of the array if we swap 7 and 2.
In the third sample, the distance between the minimum and the maximum elements is already maximum possible, so we just perform some unnecessary swap, for example, one can swap 5 and 2.
|
```python
n = int(input())
x = [int(x) for x in input().split()]
minpos = x.index(min(x))
maxpos = x.index(max(x))
if(minpos>maxpos):
d1 = (n-1)-(minpos)
d2 = maxpos-0
print((minpos-maxpos)+max(d1,d2))
else:
d1 = (n-1)-(maxpos)
d2 = minpos-0
print((maxpos-minpos)+max(d1,d2))
```
| 3
|
|
550
|
B
|
Preparing Olympiad
|
PROGRAMMING
| 1,400
|
[
"bitmasks",
"brute force"
] | null | null |
You have *n* problems. You have estimated the difficulty of the *i*-th one as integer *c**i*. Now you want to prepare a problemset for a contest, using some of the problems you've made.
A problemset for the contest must consist of at least two problems. You think that the total difficulty of the problems of the contest must be at least *l* and at most *r*. Also, you think that the difference between difficulties of the easiest and the hardest of the chosen problems must be at least *x*.
Find the number of ways to choose a problemset for the contest.
|
The first line contains four integers *n*, *l*, *r*, *x* (1<=≤<=*n*<=≤<=15, 1<=≤<=*l*<=≤<=*r*<=≤<=109, 1<=≤<=*x*<=≤<=106) — the number of problems you have, the minimum and maximum value of total difficulty of the problemset and the minimum difference in difficulty between the hardest problem in the pack and the easiest one, respectively.
The second line contains *n* integers *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=106) — the difficulty of each problem.
|
Print the number of ways to choose a suitable problemset for the contest.
|
[
"3 5 6 1\n1 2 3\n",
"4 40 50 10\n10 20 30 25\n",
"5 25 35 10\n10 10 20 10 20\n"
] |
[
"2\n",
"2\n",
"6\n"
] |
In the first example two sets are suitable, one consisting of the second and third problem, another one consisting of all three problems.
In the second example, two sets of problems are suitable — the set of problems with difficulties 10 and 30 as well as the set of problems with difficulties 20 and 30.
In the third example any set consisting of one problem of difficulty 10 and one problem of difficulty 20 is suitable.
| 750
|
[
{
"input": "3 5 6 1\n1 2 3",
"output": "2"
},
{
"input": "4 40 50 10\n10 20 30 25",
"output": "2"
},
{
"input": "5 25 35 10\n10 10 20 10 20",
"output": "6"
},
{
"input": "4 15 60 10\n10 20 30 25",
"output": "6"
},
{
"input": "1 10 20 1\n15",
"output": "0"
},
{
"input": "10 626451 11471247 246428\n369649 684428 303821 287098 422756 301599 720377 177567 515216 750602",
"output": "914"
},
{
"input": "15 1415849 15540979 356865\n8352 960238 276753 259695 712845 945369 60023 920446 181269 392011 318488 857649 30681 740872 115749",
"output": "31485"
},
{
"input": "7 1000 2000 1\n10 20 30 40 50 60 70",
"output": "0"
},
{
"input": "4 10 20 1\n4 6 4 6",
"output": "9"
},
{
"input": "4 10 20 1\n5 15 13 7",
"output": "4"
},
{
"input": "2 10 20 5\n5 10",
"output": "1"
},
{
"input": "5 1098816 3969849 167639\n85627 615007 794045 530104 7091",
"output": "15"
},
{
"input": "13 700147 8713522 390093\n996812 94040 954140 545670 369698 423872 365802 784830 700267 960664 949252 84637 257447",
"output": "8026"
},
{
"input": "15 4531977 20754263 137419\n637830 85299 755530 64382 896833 879525 331501 148182 741013 192101 112217 52165 702790 988594 587499",
"output": "6759"
},
{
"input": "15 2572491 5084070 823435\n570344 78552 775918 501843 844935 71141 331498 636557 435494 715447 992666 831188 28969 171046 989614",
"output": "15078"
},
{
"input": "15 4789415 23152928 233992\n502422 273992 449428 947379 700461 681985 857134 243310 478052 77769 936151 642380 464695 281772 964693",
"output": "10875"
},
{
"input": "3 390224 390224 1\n264237 125987 288891",
"output": "1"
},
{
"input": "7 1652707 1652707 1\n492387 684636 235422 332532 924898 499872 192988",
"output": "1"
},
{
"input": "10 501107 501107 1\n843967 30518 196518 619138 204862 690754 274071 550121 173607 359971",
"output": "1"
},
{
"input": "15 6627289 6627289 1\n683844 183950 184972 764255 211665 842336 790234 815301 914823 513046 93547 713159 554415 200951 388028",
"output": "1"
},
{
"input": "15 5083470 5083470 1\n978510 643688 591921 723137 573784 346171 920030 352119 528857 365128 627302 308557 716247 263519 654230",
"output": "1"
},
{
"input": "15 6558665 6558665 1\n572491 435494 916457 775918 823435 78552 501843 331498 71141 844935 636557 992666 570344 831188 715447",
"output": "1"
},
{
"input": "10 159699 10967276 3542\n998862 999751 995306 992648 992661 991407 997503 998809 999740 997669",
"output": "942"
},
{
"input": "5 2815840 8479687 4082\n991137 992161 997887 998891 994990",
"output": "14"
},
{
"input": "15 2898377 6694755 721\n992733 999159 990076 996808 990975 993338 993234 994757 997873 993303 994409 993801 998027 990495 999287",
"output": "9819"
},
{
"input": "6 20 70 1\n10 10 20 20 30 30",
"output": "35"
},
{
"input": "6 20 70 1\n10 10 10 10 10 10",
"output": "0"
},
{
"input": "15 1 1000000000 1\n10 20 30 40 50 60 70 80 90 100 110 120 130 140 150",
"output": "32752"
},
{
"input": "6 30 40 1\n19 20 21 14 15 16",
"output": "13"
},
{
"input": "4 5 234 2\n10 9 12 11",
"output": "8"
}
] | 1,692,088,409
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 233
| 0
|
n, l, r, x = map(int, input().split())
a = list(map(int, input().split()))
ans = 0
#find all subsets using bitmasking and find values and compare with restrictions
for i in range(1 << n):
mx = -1
mn = 1000000
sum_val = 0
for j in range(n):
if i & (1 << j):
sum_val += a[j]
mx = max(mx, a[j])
mn = min(mn, a[j])
if l <= sum_val <= r and (mx - mn) >= x:
ans += 1
print(ans)
|
Title: Preparing Olympiad
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have *n* problems. You have estimated the difficulty of the *i*-th one as integer *c**i*. Now you want to prepare a problemset for a contest, using some of the problems you've made.
A problemset for the contest must consist of at least two problems. You think that the total difficulty of the problems of the contest must be at least *l* and at most *r*. Also, you think that the difference between difficulties of the easiest and the hardest of the chosen problems must be at least *x*.
Find the number of ways to choose a problemset for the contest.
Input Specification:
The first line contains four integers *n*, *l*, *r*, *x* (1<=≤<=*n*<=≤<=15, 1<=≤<=*l*<=≤<=*r*<=≤<=109, 1<=≤<=*x*<=≤<=106) — the number of problems you have, the minimum and maximum value of total difficulty of the problemset and the minimum difference in difficulty between the hardest problem in the pack and the easiest one, respectively.
The second line contains *n* integers *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=106) — the difficulty of each problem.
Output Specification:
Print the number of ways to choose a suitable problemset for the contest.
Demo Input:
['3 5 6 1\n1 2 3\n', '4 40 50 10\n10 20 30 25\n', '5 25 35 10\n10 10 20 10 20\n']
Demo Output:
['2\n', '2\n', '6\n']
Note:
In the first example two sets are suitable, one consisting of the second and third problem, another one consisting of all three problems.
In the second example, two sets of problems are suitable — the set of problems with difficulties 10 and 30 as well as the set of problems with difficulties 20 and 30.
In the third example any set consisting of one problem of difficulty 10 and one problem of difficulty 20 is suitable.
|
```python
n, l, r, x = map(int, input().split())
a = list(map(int, input().split()))
ans = 0
#find all subsets using bitmasking and find values and compare with restrictions
for i in range(1 << n):
mx = -1
mn = 1000000
sum_val = 0
for j in range(n):
if i & (1 << j):
sum_val += a[j]
mx = max(mx, a[j])
mn = min(mn, a[j])
if l <= sum_val <= r and (mx - mn) >= x:
ans += 1
print(ans)
```
| 3
|
|
701
|
B
|
Cells Not Under Attack
|
PROGRAMMING
| 1,200
|
[
"data structures",
"math"
] | null | null |
Vasya has the square chessboard of size *n*<=×<=*n* and *m* rooks. Initially the chessboard is empty. Vasya will consequently put the rooks on the board one after another.
The cell of the field is under rook's attack, if there is at least one rook located in the same row or in the same column with this cell. If there is a rook located in the cell, this cell is also under attack.
You are given the positions of the board where Vasya will put rooks. For each rook you have to determine the number of cells which are not under attack after Vasya puts it on the board.
|
The first line of the input contains two integers *n* and *m* (1<=≤<=*n*<=≤<=100<=000, 1<=≤<=*m*<=≤<=*min*(100<=000,<=*n*2)) — the size of the board and the number of rooks.
Each of the next *m* lines contains integers *x**i* and *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*) — the number of the row and the number of the column where Vasya will put the *i*-th rook. Vasya puts rooks on the board in the order they appear in the input. It is guaranteed that any cell will contain no more than one rook.
|
Print *m* integer, the *i*-th of them should be equal to the number of cells that are not under attack after first *i* rooks are put.
|
[
"3 3\n1 1\n3 1\n2 2\n",
"5 2\n1 5\n5 1\n",
"100000 1\n300 400\n"
] |
[
"4 2 0 \n",
"16 9 \n",
"9999800001 \n"
] |
On the picture below show the state of the board after put each of the three rooks. The cells which painted with grey color is not under the attack.
| 750
|
[
{
"input": "3 3\n1 1\n3 1\n2 2",
"output": "4 2 0 "
},
{
"input": "5 2\n1 5\n5 1",
"output": "16 9 "
},
{
"input": "100000 1\n300 400",
"output": "9999800001 "
},
{
"input": "10 4\n2 8\n1 8\n9 8\n6 9",
"output": "81 72 63 48 "
},
{
"input": "30 30\n3 13\n27 23\n18 24\n18 19\n14 20\n7 10\n27 13\n20 27\n11 1\n21 10\n2 9\n28 12\n29 19\n28 27\n27 29\n30 12\n27 2\n4 5\n8 19\n21 2\n24 27\n14 22\n20 3\n18 3\n23 9\n28 6\n15 12\n2 2\n16 27\n1 14",
"output": "841 784 729 702 650 600 600 552 506 484 441 400 380 380 361 342 324 289 272 272 255 240 225 225 210 196 182 182 168 143 "
},
{
"input": "70 31\n22 39\n33 43\n50 27\n70 9\n20 67\n61 24\n60 4\n60 28\n4 25\n30 29\n46 47\n51 48\n37 5\n14 29\n45 44\n68 35\n52 21\n7 37\n18 43\n44 22\n26 12\n39 37\n51 55\n50 23\n51 16\n16 49\n22 62\n35 45\n56 2\n20 51\n3 37",
"output": "4761 4624 4489 4356 4225 4096 3969 3906 3782 3660 3540 3422 3306 3249 3136 3025 2916 2809 2756 2652 2550 2499 2450 2401 2352 2256 2208 2115 2024 1978 1935 "
},
{
"input": "330 17\n259 262\n146 20\n235 69\n84 74\n131 267\n153 101\n32 232\n214 212\n239 157\n121 156\n10 45\n266 78\n52 258\n109 279\n193 276\n239 142\n321 89",
"output": "108241 107584 106929 106276 105625 104976 104329 103684 103041 102400 101761 101124 100489 99856 99225 98910 98282 "
},
{
"input": "500 43\n176 85\n460 171\n233 260\n73 397\n474 35\n290 422\n309 318\n280 415\n485 169\n106 22\n355 129\n180 301\n205 347\n197 93\n263 318\n336 382\n314 350\n476 214\n367 277\n333 166\n500 376\n236 17\n94 73\n116 204\n166 50\n168 218\n144 369\n340 91\n274 360\n171 360\n41 251\n262 478\n27 163\n151 491\n208 415\n448 386\n293 486\n371 479\n330 435\n220 374\n163 316\n155 158\n26 126",
"output": "249001 248004 247009 246016 245025 244036 243049 242064 241081 240100 239121 238144 237169 236196 235710 234740 233772 232806 231842 230880 229920 228962 228006 227052 226100 225150 224202 223256 222312 221840 220899 219960 219023 218088 217620 216688 215758 214830 213904 212980 212058 211138 210220 "
},
{
"input": "99999 1\n54016 16192",
"output": "9999600004 "
},
{
"input": "99991 9\n80814 65974\n12100 98787\n9390 76191\n5628 47659\n80075 25361\n75330 1630\n38758 99962\n33848 40352\n43732 52281",
"output": "9998000100 9997800121 9997600144 9997400169 9997200196 9997000225 9996800256 9996600289 9996400324 "
},
{
"input": "1 1\n1 1",
"output": "0 "
}
] | 1,589,790,356
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 40
| 1,543
| 30,208,000
|
m,n=map(int,input().split())
s,q=m,m
row=[0]*(m+1)
co=[0]*(m+1)
for i in range(n):
a,b=map(int,input().split())
if row[a]==0:
row[a]=1
s=s-1
if co[b]==0:
co[b]=1
q=q-1
if i!=n-1:
print(s*q,end=" ")
else:
print(s*q)
|
Title: Cells Not Under Attack
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has the square chessboard of size *n*<=×<=*n* and *m* rooks. Initially the chessboard is empty. Vasya will consequently put the rooks on the board one after another.
The cell of the field is under rook's attack, if there is at least one rook located in the same row or in the same column with this cell. If there is a rook located in the cell, this cell is also under attack.
You are given the positions of the board where Vasya will put rooks. For each rook you have to determine the number of cells which are not under attack after Vasya puts it on the board.
Input Specification:
The first line of the input contains two integers *n* and *m* (1<=≤<=*n*<=≤<=100<=000, 1<=≤<=*m*<=≤<=*min*(100<=000,<=*n*2)) — the size of the board and the number of rooks.
Each of the next *m* lines contains integers *x**i* and *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*) — the number of the row and the number of the column where Vasya will put the *i*-th rook. Vasya puts rooks on the board in the order they appear in the input. It is guaranteed that any cell will contain no more than one rook.
Output Specification:
Print *m* integer, the *i*-th of them should be equal to the number of cells that are not under attack after first *i* rooks are put.
Demo Input:
['3 3\n1 1\n3 1\n2 2\n', '5 2\n1 5\n5 1\n', '100000 1\n300 400\n']
Demo Output:
['4 2 0 \n', '16 9 \n', '9999800001 \n']
Note:
On the picture below show the state of the board after put each of the three rooks. The cells which painted with grey color is not under the attack.
|
```python
m,n=map(int,input().split())
s,q=m,m
row=[0]*(m+1)
co=[0]*(m+1)
for i in range(n):
a,b=map(int,input().split())
if row[a]==0:
row[a]=1
s=s-1
if co[b]==0:
co[b]=1
q=q-1
if i!=n-1:
print(s*q,end=" ")
else:
print(s*q)
```
| 3
|
|
399
|
B
|
Red and Blue Balls
|
PROGRAMMING
| 0
|
[] | null | null |
User ainta has a stack of *n* red and blue balls. He can apply a certain operation which changes the colors of the balls inside the stack.
- While the top ball inside the stack is red, pop the ball from the top of the stack. - Then replace the blue ball on the top with a red ball. - And finally push some blue balls to the stack until the stack has total of *n* balls inside.
If there are no blue balls inside the stack, ainta can't apply this operation. Given the initial state of the stack, ainta wants to know the maximum number of operations he can repeatedly apply.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=50) — the number of balls inside the stack.
The second line contains a string *s* (|*s*|<==<=*n*) describing the initial state of the stack. The *i*-th character of the string *s* denotes the color of the *i*-th ball (we'll number the balls from top to bottom of the stack). If the character is "R", the color is red. If the character is "B", the color is blue.
|
Print the maximum number of operations ainta can repeatedly apply.
Please, do not write 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.
|
[
"3\nRBR\n",
"4\nRBBR\n",
"5\nRBBRR\n"
] |
[
"2\n",
"6\n",
"6\n"
] |
The first example is depicted below.
The explanation how user ainta applies the first operation. He pops out one red ball, changes the color of the ball in the middle from blue to red, and pushes one blue ball.
The explanation how user ainta applies the second operation. He will not pop out red balls, he simply changes the color of the ball on the top from blue to red.
From now on, ainta can't apply any operation because there are no blue balls inside the stack. ainta applied two operations, so the answer is 2.
The second example is depicted below. The blue arrow denotes a single operation.
| 1,000
|
[
{
"input": "3\nRBR",
"output": "2"
},
{
"input": "4\nRBBR",
"output": "6"
},
{
"input": "5\nRBBRR",
"output": "6"
},
{
"input": "5\nRBRBR",
"output": "10"
},
{
"input": "10\nRRBRRBBRRR",
"output": "100"
},
{
"input": "10\nBRBRRRRRRR",
"output": "5"
},
{
"input": "10\nBRRRRRRRRR",
"output": "1"
},
{
"input": "20\nBRBRRRRRRRRRRRRRRRRR",
"output": "5"
},
{
"input": "30\nRRBBBBBBBBBBBBBBBBBBBBBBBBBBBB",
"output": "1073741820"
},
{
"input": "50\nBRRRBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB",
"output": "1125899906842609"
},
{
"input": "20\nRRRBRBBBBBRRRRRRRRRR",
"output": "1000"
},
{
"input": "20\nRRBRBBBBBRRRRRRRRRRR",
"output": "500"
},
{
"input": "1\nR",
"output": "0"
},
{
"input": "1\nB",
"output": "1"
},
{
"input": "2\nRR",
"output": "0"
},
{
"input": "2\nBR",
"output": "1"
},
{
"input": "50\nRRRRRRRRRRBBBBBBRRBBRRRBRRBBBRRRRRRRRRRRRRRRRRRRRR",
"output": "479001600"
},
{
"input": "50\nRBRRRRRBRBRRBBBBBBRRRBRRRRRBBBRRBRRRRRBBBRRRRRRRRR",
"output": "1929382195842"
},
{
"input": "48\nRBRBRRRRBRBRRBRRRRRRRBBBRRBRBRRRBBRRRRRRRRRRRRRR",
"output": "13235135754"
},
{
"input": "30\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "0"
},
{
"input": "50\nRRBBBBBBBBBBBBBBBBRBRRBBBRBBRBBBRRBRBBBBBRBBRBBRBR",
"output": "402373705727996"
},
{
"input": "50\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB",
"output": "1125899906842623"
},
{
"input": "19\nRRRRRBRRBRRRRBRBBBB",
"output": "500000"
},
{
"input": "32\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBR",
"output": "2147483647"
},
{
"input": "3\nBBB",
"output": "7"
},
{
"input": "3\nBBR",
"output": "3"
},
{
"input": "3\nBRB",
"output": "5"
},
{
"input": "3\nBRR",
"output": "1"
},
{
"input": "3\nRBB",
"output": "6"
},
{
"input": "3\nRBR",
"output": "2"
},
{
"input": "3\nRRB",
"output": "4"
},
{
"input": "3\nRRR",
"output": "0"
},
{
"input": "2\nRB",
"output": "2"
},
{
"input": "2\nBB",
"output": "3"
}
] | 1,501,423,535
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 34
| 62
| 4,608,000
|
n = int(input())
s = input()
ans = 0
for i in range(len(s)):
ch = s[i]
if ch == 'B':
ans |= 1 << i
print(ans)
|
Title: Red and Blue Balls
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
User ainta has a stack of *n* red and blue balls. He can apply a certain operation which changes the colors of the balls inside the stack.
- While the top ball inside the stack is red, pop the ball from the top of the stack. - Then replace the blue ball on the top with a red ball. - And finally push some blue balls to the stack until the stack has total of *n* balls inside.
If there are no blue balls inside the stack, ainta can't apply this operation. Given the initial state of the stack, ainta wants to know the maximum number of operations he can repeatedly apply.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=50) — the number of balls inside the stack.
The second line contains a string *s* (|*s*|<==<=*n*) describing the initial state of the stack. The *i*-th character of the string *s* denotes the color of the *i*-th ball (we'll number the balls from top to bottom of the stack). If the character is "R", the color is red. If the character is "B", the color is blue.
Output Specification:
Print the maximum number of operations ainta can repeatedly apply.
Please, do not write 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.
Demo Input:
['3\nRBR\n', '4\nRBBR\n', '5\nRBBRR\n']
Demo Output:
['2\n', '6\n', '6\n']
Note:
The first example is depicted below.
The explanation how user ainta applies the first operation. He pops out one red ball, changes the color of the ball in the middle from blue to red, and pushes one blue ball.
The explanation how user ainta applies the second operation. He will not pop out red balls, he simply changes the color of the ball on the top from blue to red.
From now on, ainta can't apply any operation because there are no blue balls inside the stack. ainta applied two operations, so the answer is 2.
The second example is depicted below. The blue arrow denotes a single operation.
|
```python
n = int(input())
s = input()
ans = 0
for i in range(len(s)):
ch = s[i]
if ch == 'B':
ans |= 1 << i
print(ans)
```
| 3
|
|
774
|
D
|
Lie or Truth
|
PROGRAMMING
| 1,500
|
[
"*special",
"constructive algorithms",
"implementation",
"sortings"
] | null | null |
Vasya has a sequence of cubes and exactly one integer is written on each cube. Vasya exhibited all his cubes in a row. So the sequence of numbers written on the cubes in the order from the left to the right equals to *a*1,<=*a*2,<=...,<=*a**n*.
While Vasya was walking, his little brother Stepan played with Vasya's cubes and changed their order, so now the sequence of numbers written on the cubes became equal to *b*1,<=*b*2,<=...,<=*b**n*.
Stepan said that he swapped only cubes which where on the positions between *l* and *r*, inclusive, and did not remove or add any other cubes (i. e. he said that he reordered cubes between positions *l* and *r*, inclusive, in some way).
Your task is to determine if it is possible that Stepan said the truth, or it is guaranteed that Stepan deceived his brother.
|
The first line contains three integers *n*, *l*, *r* (1<=≤<=*n*<=≤<=105, 1<=≤<=*l*<=≤<=*r*<=≤<=*n*) — the number of Vasya's cubes and the positions told by Stepan.
The second line contains the sequence *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the sequence of integers written on cubes in the Vasya's order.
The third line contains the sequence *b*1,<=*b*2,<=...,<=*b**n* (1<=≤<=*b**i*<=≤<=*n*) — the sequence of integers written on cubes after Stepan rearranged their order.
It is guaranteed that Stepan did not remove or add other cubes, he only rearranged Vasya's cubes.
|
Print "LIE" (without quotes) if it is guaranteed that Stepan deceived his brother. In the other case, print "TRUTH" (without quotes).
|
[
"5 2 4\n3 4 2 3 1\n3 2 3 4 1\n",
"3 1 2\n1 2 3\n3 1 2\n",
"4 2 4\n1 1 1 1\n1 1 1 1\n"
] |
[
"TRUTH\n",
"LIE\n",
"TRUTH\n"
] |
In the first example there is a situation when Stepan said the truth. Initially the sequence of integers on the cubes was equal to [3, 4, 2, 3, 1]. Stepan could at first swap cubes on positions 2 and 3 (after that the sequence of integers on cubes became equal to [3, 2, 4, 3, 1]), and then swap cubes in positions 3 and 4 (after that the sequence of integers on cubes became equal to [3, 2, 3, 4, 1]).
In the second example it is not possible that Stepan said truth because he said that he swapped cubes only between positions 1 and 2, but we can see that it is guaranteed that he changed the position of the cube which was on the position 3 at first. So it is guaranteed that Stepan deceived his brother.
In the third example for any values *l* and *r* there is a situation when Stepan said the truth.
| 0
|
[
{
"input": "5 2 4\n3 4 2 3 1\n3 2 3 4 1",
"output": "TRUTH"
},
{
"input": "3 1 2\n1 2 3\n3 1 2",
"output": "LIE"
},
{
"input": "4 2 4\n1 1 1 1\n1 1 1 1",
"output": "TRUTH"
},
{
"input": "5 1 3\n2 2 2 1 2\n2 2 2 1 2",
"output": "TRUTH"
},
{
"input": "7 1 4\n2 5 5 5 4 3 4\n2 5 5 5 4 3 4",
"output": "TRUTH"
},
{
"input": "10 1 10\n6 7 6 1 10 10 9 5 3 9\n7 10 9 6 1 5 9 3 10 6",
"output": "TRUTH"
},
{
"input": "1 1 1\n1\n1",
"output": "TRUTH"
},
{
"input": "4 3 4\n1 2 3 4\n2 1 3 4",
"output": "LIE"
},
{
"input": "7 2 4\n1 2 3 4 5 7 6\n1 2 3 4 5 6 7",
"output": "LIE"
},
{
"input": "5 1 2\n1 2 3 4 5\n1 2 3 5 4",
"output": "LIE"
},
{
"input": "8 3 6\n5 3 1 1 1 1 3 5\n3 3 1 1 1 1 5 5",
"output": "LIE"
},
{
"input": "4 2 2\n2 1 2 2\n1 2 2 2",
"output": "LIE"
}
] | 1,491,409,368
| 2,868
|
Python 3
|
OK
|
TESTS
| 52
| 140
| 14,950,400
|
n, l, r = map(int, input().split())
l -= 1
r -= 1
a = list(map(int, input().split()))
b = list(map(int, input().split()))
for i in range(n):
if (b[i] != a[i]) and (i < l or i > r):
print("LIE")
break
else:
print("TRUTH")
|
Title: Lie or Truth
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has a sequence of cubes and exactly one integer is written on each cube. Vasya exhibited all his cubes in a row. So the sequence of numbers written on the cubes in the order from the left to the right equals to *a*1,<=*a*2,<=...,<=*a**n*.
While Vasya was walking, his little brother Stepan played with Vasya's cubes and changed their order, so now the sequence of numbers written on the cubes became equal to *b*1,<=*b*2,<=...,<=*b**n*.
Stepan said that he swapped only cubes which where on the positions between *l* and *r*, inclusive, and did not remove or add any other cubes (i. e. he said that he reordered cubes between positions *l* and *r*, inclusive, in some way).
Your task is to determine if it is possible that Stepan said the truth, or it is guaranteed that Stepan deceived his brother.
Input Specification:
The first line contains three integers *n*, *l*, *r* (1<=≤<=*n*<=≤<=105, 1<=≤<=*l*<=≤<=*r*<=≤<=*n*) — the number of Vasya's cubes and the positions told by Stepan.
The second line contains the sequence *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the sequence of integers written on cubes in the Vasya's order.
The third line contains the sequence *b*1,<=*b*2,<=...,<=*b**n* (1<=≤<=*b**i*<=≤<=*n*) — the sequence of integers written on cubes after Stepan rearranged their order.
It is guaranteed that Stepan did not remove or add other cubes, he only rearranged Vasya's cubes.
Output Specification:
Print "LIE" (without quotes) if it is guaranteed that Stepan deceived his brother. In the other case, print "TRUTH" (without quotes).
Demo Input:
['5 2 4\n3 4 2 3 1\n3 2 3 4 1\n', '3 1 2\n1 2 3\n3 1 2\n', '4 2 4\n1 1 1 1\n1 1 1 1\n']
Demo Output:
['TRUTH\n', 'LIE\n', 'TRUTH\n']
Note:
In the first example there is a situation when Stepan said the truth. Initially the sequence of integers on the cubes was equal to [3, 4, 2, 3, 1]. Stepan could at first swap cubes on positions 2 and 3 (after that the sequence of integers on cubes became equal to [3, 2, 4, 3, 1]), and then swap cubes in positions 3 and 4 (after that the sequence of integers on cubes became equal to [3, 2, 3, 4, 1]).
In the second example it is not possible that Stepan said truth because he said that he swapped cubes only between positions 1 and 2, but we can see that it is guaranteed that he changed the position of the cube which was on the position 3 at first. So it is guaranteed that Stepan deceived his brother.
In the third example for any values *l* and *r* there is a situation when Stepan said the truth.
|
```python
n, l, r = map(int, input().split())
l -= 1
r -= 1
a = list(map(int, input().split()))
b = list(map(int, input().split()))
for i in range(n):
if (b[i] != a[i]) and (i < l or i > r):
print("LIE")
break
else:
print("TRUTH")
```
| 3
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
The GCD table *G* of size *n*<=×<=*n* for an array of positive integers *a* of length *n* is defined by formula
Let us remind you that the greatest common divisor (GCD) of two positive integers *x* and *y* is the greatest integer that is divisor of both *x* and *y*, it is denoted as . For example, for array *a*<==<={4,<=3,<=6,<=2} of length 4 the GCD table will look as follows:
Given all the numbers of the GCD table *G*, restore array *a*.
|
The first line contains number *n* (1<=≤<=*n*<=≤<=500) — the length of array *a*. The second line contains *n*2 space-separated numbers — the elements of the GCD table of *G* for array *a*.
All the numbers in the table are positive integers, not exceeding 109. Note that the elements are given in an arbitrary order. It is guaranteed that the set of the input data corresponds to some array *a*.
|
In the single line print *n* positive integers — the elements of array *a*. If there are multiple possible solutions, you are allowed to print any of them.
|
[
"4\n2 1 2 3 4 3 2 6 1 1 2 2 1 2 3 2\n",
"1\n42\n",
"2\n1 1 1 1\n"
] |
[
"4 3 6 2",
"42 ",
"1 1 "
] |
none
| 0
|
[
{
"input": "4\n2 1 2 3 4 3 2 6 1 1 2 2 1 2 3 2",
"output": "2 3 4 6 "
},
{
"input": "1\n42",
"output": "42 "
},
{
"input": "2\n1 1 1 1",
"output": "1 1 "
},
{
"input": "2\n54748096 1 641009859 1",
"output": "54748096 641009859 "
},
{
"input": "3\n1 7 923264237 374288891 7 524125987 1 1 1",
"output": "374288891 524125987 923264237 "
},
{
"input": "4\n1 1 1 1 1 702209411 496813081 673102149 1 1 561219907 1 1 1 1 1",
"output": "496813081 561219907 673102149 702209411 "
},
{
"input": "5\n1 1 1 1 1 9 564718673 585325539 1 1 3 1 9 1 1 365329221 3 291882089 3 1 412106895 1 1 1 3",
"output": "291882089 365329221 412106895 564718673 585325539 "
},
{
"input": "5\n1 161 1 534447872 161 233427865 1 7 7 73701396 1 401939237 4 1 1 1 1 1 7 115704211 1 4 1 7 1",
"output": "73701396 115704211 233427865 401939237 534447872 "
},
{
"input": "5\n2 11 1 1 2 4 2 1 181951 4 345484316 2 4 4 4 2 1 140772746 1 634524 4 521302304 1 2 11",
"output": "181951 634524 140772746 345484316 521302304 "
},
{
"input": "5\n27 675 1 1 347621274 5 2 13 189 738040275 5 1 189 13 1 959752125 770516962 769220855 5 5 2 675 1 1 27",
"output": "347621274 738040275 769220855 770516962 959752125 "
},
{
"input": "5\n2029 6087 2029 2029 6087 2029 527243766 4058 2029 2029 2029 2029 2029 2029 2029 2029 165353355 4058 2029 731472761 739767313 2029 2029 2029 585281282",
"output": "165353355 527243766 585281282 731472761 739767313 "
},
{
"input": "5\n537163 537163 537163 537163 537163 537163 1074326 537163 537163 537163 515139317 1074326 537163 537163 537163 539311652 321760637 170817834 537163 537163 537163 537163 537163 537163 392666153",
"output": "170817834 321760637 392666153 515139317 539311652 "
},
{
"input": "4\n1 188110 607844 2 1 1 695147 1 1 1 143380513 1 1 1 1 2",
"output": "188110 607844 695147 143380513 "
},
{
"input": "4\n3 1 96256522 120 360284388 3 3 2 2 2 3 12 12 2 1 198192381",
"output": "120 96256522 198192381 360284388 "
},
{
"input": "4\n67025 13405 1915 1915 1915 1915 5745 676469920 53620 5745 660330300 67025 53620 380098775 533084295 13405",
"output": "380098775 533084295 660330300 676469920 "
},
{
"input": "4\n700521 233507 759364764 467014 468181535 233507 233507 890362191 233507 700521 467014 233507 946637378 233507 233507 233507",
"output": "468181535 759364764 890362191 946637378 "
},
{
"input": "3\n484799 1 1 744137 1 1 909312183 1 1",
"output": "484799 744137 909312183 "
},
{
"input": "3\n1 716963379 1 1 205 1 1 964 1",
"output": "205 964 716963379 "
},
{
"input": "3\n5993 781145599 54740062 5993 5993 267030101 5993 5993 5993",
"output": "54740062 267030101 781145599 "
},
{
"input": "3\n121339 121339 121339 55451923 531222142 121339 121339 435485671 121339",
"output": "55451923 435485671 531222142 "
},
{
"input": "5\n4 4 4 4 4 4 4 4 4 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1",
"output": "1 2 4 4 4 "
},
{
"input": "4\n1 1 1 1 1 1 1 1 1 1 1 1 3 3 3 3",
"output": "1 1 3 3 "
},
{
"input": "6\n1 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 6 6 6 6 6 6 6 6 6",
"output": "1 3 3 6 6 6 "
}
] | 1,464,898,753
| 3,373
|
Python 3
|
OK
|
TESTS
| 54
| 811
| 23,859,200
|
import fractions
import collections
n = int(input())
l = sorted(map(int, input().split(" ")), reverse = True)
aux = []
cont = collections.Counter(l)
r = []
for i in l:
if not cont[i]:
continue
r.append(i)
cont[i] -= 1
for j in aux:
'''print(str(i) + " " + str(j))'''
cont[fractions.gcd(i, j)] -= 2
aux.append(i)
print(' '.join(map(str, r)))
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The GCD table *G* of size *n*<=×<=*n* for an array of positive integers *a* of length *n* is defined by formula
Let us remind you that the greatest common divisor (GCD) of two positive integers *x* and *y* is the greatest integer that is divisor of both *x* and *y*, it is denoted as . For example, for array *a*<==<={4,<=3,<=6,<=2} of length 4 the GCD table will look as follows:
Given all the numbers of the GCD table *G*, restore array *a*.
Input Specification:
The first line contains number *n* (1<=≤<=*n*<=≤<=500) — the length of array *a*. The second line contains *n*2 space-separated numbers — the elements of the GCD table of *G* for array *a*.
All the numbers in the table are positive integers, not exceeding 109. Note that the elements are given in an arbitrary order. It is guaranteed that the set of the input data corresponds to some array *a*.
Output Specification:
In the single line print *n* positive integers — the elements of array *a*. If there are multiple possible solutions, you are allowed to print any of them.
Demo Input:
['4\n2 1 2 3 4 3 2 6 1 1 2 2 1 2 3 2\n', '1\n42\n', '2\n1 1 1 1\n']
Demo Output:
['4 3 6 2', '42 ', '1 1 ']
Note:
none
|
```python
import fractions
import collections
n = int(input())
l = sorted(map(int, input().split(" ")), reverse = True)
aux = []
cont = collections.Counter(l)
r = []
for i in l:
if not cont[i]:
continue
r.append(i)
cont[i] -= 1
for j in aux:
'''print(str(i) + " " + str(j))'''
cont[fractions.gcd(i, j)] -= 2
aux.append(i)
print(' '.join(map(str, r)))
```
| 3
|
|
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"
}
] | 1,610,286,670
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 33
| 108
| 0
|
a = int(input())
b = str(a)
h = 0
t = len(b)
for x in range(t):
if b[x] == "4":
h += (2**(t-1-x))*1
else:
h += (2**(t-1-x))*2
print(h)
|
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
a = int(input())
b = str(a)
h = 0
t = len(b)
for x in range(t):
if b[x] == "4":
h += (2**(t-1-x))*1
else:
h += (2**(t-1-x))*2
print(h)
```
| 3
|
|
554
|
A
|
Kyoya and Photobooks
|
PROGRAMMING
| 900
|
[
"brute force",
"math",
"strings"
] | null | null |
Kyoya Ootori is selling photobooks of the Ouran High School Host Club. He has 26 photos, labeled "a" to "z", and he has compiled them into a photo booklet with some photos in some order (possibly with some photos being duplicated). A photo booklet can be described as a string of lowercase letters, consisting of the photos in the booklet in order. He now wants to sell some "special edition" photobooks, each with one extra photo inserted anywhere in the book. He wants to make as many distinct photobooks as possible, so he can make more money. He asks Haruhi, how many distinct photobooks can he make by inserting one extra photo into the photobook he already has?
Please help Haruhi solve this problem.
|
The first line of input will be a single string *s* (1<=≤<=|*s*|<=≤<=20). String *s* consists only of lowercase English letters.
|
Output a single integer equal to the number of distinct photobooks Kyoya Ootori can make.
|
[
"a\n",
"hi\n"
] |
[
"51\n",
"76\n"
] |
In the first case, we can make 'ab','ac',...,'az','ba','ca',...,'za', and 'aa', producing a total of 51 distinct photo booklets.
| 250
|
[
{
"input": "a",
"output": "51"
},
{
"input": "hi",
"output": "76"
},
{
"input": "y",
"output": "51"
},
{
"input": "kgan",
"output": "126"
},
{
"input": "zoabkyuvus",
"output": "276"
},
{
"input": "spyemhyznjieyhhbk",
"output": "451"
},
{
"input": "xulsyfkuizjauadjjopu",
"output": "526"
},
{
"input": "e",
"output": "51"
},
{
"input": "zv",
"output": "76"
},
{
"input": "jgv",
"output": "101"
},
{
"input": "zsfo",
"output": "126"
},
{
"input": "jselr",
"output": "151"
},
{
"input": "dwemig",
"output": "176"
},
{
"input": "mddoxsf",
"output": "201"
},
{
"input": "jgirkrmi",
"output": "226"
},
{
"input": "spkxurcum",
"output": "251"
},
{
"input": "fykkiubdkt",
"output": "276"
},
{
"input": "fznbcxsxygs",
"output": "301"
},
{
"input": "qcrvrdqcbtou",
"output": "326"
},
{
"input": "qktrbjzrqgmlr",
"output": "351"
},
{
"input": "foamodbvptlxxg",
"output": "376"
},
{
"input": "ydzpjhsidipricw",
"output": "401"
},
{
"input": "lpfpndmjfvqejdgf",
"output": "426"
},
{
"input": "ofkvparuvjtggnmab",
"output": "451"
},
{
"input": "xxncfutrtxcwdzwbgs",
"output": "476"
},
{
"input": "zovhffccflkgqncsdte",
"output": "501"
},
{
"input": "cskgsxywlvfeicoueglr",
"output": "526"
},
{
"input": "gggggggggggggggggggg",
"output": "526"
},
{
"input": "qdqdddqddqqddqddqdqd",
"output": "526"
},
{
"input": "takttttaakaaktakttkt",
"output": "526"
},
{
"input": "coccoooogogcgocccmcg",
"output": "526"
},
{
"input": "kskkskkkssksssk",
"output": "401"
},
{
"input": "lllllllllllllll",
"output": "401"
}
] | 1,669,886,266
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 33
| 46
| 0
|
s = (input())
print((len(s)+1)*25+1)
|
Title: Kyoya and Photobooks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kyoya Ootori is selling photobooks of the Ouran High School Host Club. He has 26 photos, labeled "a" to "z", and he has compiled them into a photo booklet with some photos in some order (possibly with some photos being duplicated). A photo booklet can be described as a string of lowercase letters, consisting of the photos in the booklet in order. He now wants to sell some "special edition" photobooks, each with one extra photo inserted anywhere in the book. He wants to make as many distinct photobooks as possible, so he can make more money. He asks Haruhi, how many distinct photobooks can he make by inserting one extra photo into the photobook he already has?
Please help Haruhi solve this problem.
Input Specification:
The first line of input will be a single string *s* (1<=≤<=|*s*|<=≤<=20). String *s* consists only of lowercase English letters.
Output Specification:
Output a single integer equal to the number of distinct photobooks Kyoya Ootori can make.
Demo Input:
['a\n', 'hi\n']
Demo Output:
['51\n', '76\n']
Note:
In the first case, we can make 'ab','ac',...,'az','ba','ca',...,'za', and 'aa', producing a total of 51 distinct photo booklets.
|
```python
s = (input())
print((len(s)+1)*25+1)
```
| 3
|
|
69
|
A
|
Young Physicist
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] |
A. Young Physicist
|
2
|
256
|
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — 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"
}
] | 1,645,697,927
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 81
| 92
| 0
|
n = int(input())
d = 0 ; e = 0; f = 0
for i in range(n):
a,b,c = map(int,input().split())
d += a ; e += b ; f += c
if d == 0 and e == 0 and f == 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())
d = 0 ; e = 0; f = 0
for i in range(n):
a,b,c = map(int,input().split())
d += a ; e += b ; f += c
if d == 0 and e == 0 and f == 0:
print("YES")
else:
print("NO")
```
| 3.977
|
61
|
A
|
Ultra-Fast Mathematician
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Ultra-Fast Mathematician
|
2
|
256
|
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
|
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
|
Write one line — the corresponding answer. Do not omit the leading 0s.
|
[
"1010100\n0100101\n",
"000\n111\n",
"1110\n1010\n",
"01110\n01100\n"
] |
[
"1110001\n",
"111\n",
"0100\n",
"00010\n"
] |
none
| 500
|
[
{
"input": "1010100\n0100101",
"output": "1110001"
},
{
"input": "000\n111",
"output": "111"
},
{
"input": "1110\n1010",
"output": "0100"
},
{
"input": "01110\n01100",
"output": "00010"
},
{
"input": "011101\n000001",
"output": "011100"
},
{
"input": "10\n01",
"output": "11"
},
{
"input": "00111111\n11011101",
"output": "11100010"
},
{
"input": "011001100\n101001010",
"output": "110000110"
},
{
"input": "1100100001\n0110101100",
"output": "1010001101"
},
{
"input": "00011101010\n10010100101",
"output": "10001001111"
},
{
"input": "100000101101\n111010100011",
"output": "011010001110"
},
{
"input": "1000001111010\n1101100110001",
"output": "0101101001011"
},
{
"input": "01011111010111\n10001110111010",
"output": "11010001101101"
},
{
"input": "110010000111100\n001100101011010",
"output": "111110101100110"
},
{
"input": "0010010111110000\n0000000011010110",
"output": "0010010100100110"
},
{
"input": "00111110111110000\n01111100001100000",
"output": "01000010110010000"
},
{
"input": "101010101111010001\n001001111101111101",
"output": "100011010010101100"
},
{
"input": "0110010101111100000\n0011000101000000110",
"output": "0101010000111100110"
},
{
"input": "11110100011101010111\n00001000011011000000",
"output": "11111100000110010111"
},
{
"input": "101010101111101101001\n111010010010000011111",
"output": "010000111101101110110"
},
{
"input": "0000111111100011000010\n1110110110110000001010",
"output": "1110001001010011001000"
},
{
"input": "10010010101000110111000\n00101110100110111000111",
"output": "10111100001110001111111"
},
{
"input": "010010010010111100000111\n100100111111100011001110",
"output": "110110101101011111001001"
},
{
"input": "0101110100100111011010010\n0101100011010111001010001",
"output": "0000010111110000010000011"
},
{
"input": "10010010100011110111111011\n10000110101100000001000100",
"output": "00010100001111110110111111"
},
{
"input": "000001111000000100001000000\n011100111101111001110110001",
"output": "011101000101111101111110001"
},
{
"input": "0011110010001001011001011100\n0000101101000011101011001010",
"output": "0011011111001010110010010110"
},
{
"input": "11111000000000010011001101111\n11101110011001010100010000000",
"output": "00010110011001000111011101111"
},
{
"input": "011001110000110100001100101100\n001010000011110000001000101001",
"output": "010011110011000100000100000101"
},
{
"input": "1011111010001100011010110101111\n1011001110010000000101100010101",
"output": "0000110100011100011111010111010"
},
{
"input": "10111000100001000001010110000001\n10111000001100101011011001011000",
"output": "00000000101101101010001111011001"
},
{
"input": "000001010000100001000000011011100\n111111111001010100100001100000111",
"output": "111110101001110101100001111011011"
},
{
"input": "1101000000000010011011101100000110\n1110000001100010011010000011011110",
"output": "0011000001100000000001101111011000"
},
{
"input": "01011011000010100001100100011110001\n01011010111000001010010100001110000",
"output": "00000001111010101011110000010000001"
},
{
"input": "000011111000011001000110111100000100\n011011000110000111101011100111000111",
"output": "011000111110011110101101011011000011"
},
{
"input": "1001000010101110001000000011111110010\n0010001011010111000011101001010110000",
"output": "1011001001111001001011101010101000010"
},
{
"input": "00011101011001100101111111000000010101\n10010011011011001011111000000011101011",
"output": "10001110000010101110000111000011111110"
},
{
"input": "111011100110001001101111110010111001010\n111111101101111001110010000101101000100",
"output": "000100001011110000011101110111010001110"
},
{
"input": "1111001001101000001000000010010101001010\n0010111100111110001011000010111110111001",
"output": "1101110101010110000011000000101011110011"
},
{
"input": "00100101111000000101011111110010100011010\n11101110001010010101001000111110101010100",
"output": "11001011110010010000010111001100001001110"
},
{
"input": "101011001110110100101001000111010101101111\n100111100110101011010100111100111111010110",
"output": "001100101000011111111101111011101010111001"
},
{
"input": "1111100001100101000111101001001010011100001\n1000110011000011110010001011001110001000001",
"output": "0111010010100110110101100010000100010100000"
},
{
"input": "01100111011111010101000001101110000001110101\n10011001011111110000000101011001001101101100",
"output": "11111110000000100101000100110111001100011001"
},
{
"input": "110010100111000100100101100000011100000011001\n011001111011100110000110111001110110100111011",
"output": "101011011100100010100011011001101010100100010"
},
{
"input": "0001100111111011010110100100111000000111000110\n1100101011000000000001010010010111001100110001",
"output": "1101001100111011010111110110101111001011110111"
},
{
"input": "00000101110110110001110010100001110100000100000\n10010000110011110001101000111111101010011010001",
"output": "10010101000101000000011010011110011110011110001"
},
{
"input": "110000100101011100100011001111110011111110010001\n101011111001011100110110111101110011010110101100",
"output": "011011011100000000010101110010000000101000111101"
},
{
"input": "0101111101011111010101011101000011101100000000111\n0000101010110110001110101011011110111001010100100",
"output": "0101010111101001011011110110011101010101010100011"
},
{
"input": "11000100010101110011101000011111001010110111111100\n00001111000111001011111110000010101110111001000011",
"output": "11001011010010111000010110011101100100001110111111"
},
{
"input": "101000001101111101101111111000001110110010101101010\n010011100111100001100000010001100101000000111011011",
"output": "111011101010011100001111101001101011110010010110001"
},
{
"input": "0011111110010001010100010110111000110011001101010100\n0111000000100010101010000100101000000100101000111001",
"output": "0100111110110011111110010010010000110111100101101101"
},
{
"input": "11101010000110000011011010000001111101000111011111100\n10110011110001010100010110010010101001010111100100100",
"output": "01011001110111010111001100010011010100010000111011000"
},
{
"input": "011000100001000001101000010110100110011110100111111011\n111011001000001001110011001111011110111110110011011111",
"output": "100011101001001000011011011001111000100000010100100100"
},
{
"input": "0111010110010100000110111011010110100000000111110110000\n1011100100010001101100000100111111101001110010000100110",
"output": "1100110010000101101010111111101001001001110101110010110"
},
{
"input": "10101000100111000111010001011011011011110100110101100011\n11101111000000001100100011111000100100000110011001101110",
"output": "01000111100111001011110010100011111111110010101100001101"
},
{
"input": "000000111001010001000000110001001011100010011101010011011\n110001101000010010000101000100001111101001100100001010010",
"output": "110001010001000011000101110101000100001011111001011001001"
},
{
"input": "0101011100111010000111110010101101111111000000111100011100\n1011111110000010101110111001000011100000100111111111000111",
"output": "1110100010111000101001001011101110011111100111000011011011"
},
{
"input": "11001000001100100111100111100100101011000101001111001001101\n10111110100010000011010100110100100011101001100000001110110",
"output": "01110110101110100100110011010000001000101100101111000111011"
},
{
"input": "010111011011101000000110000110100110001110100001110110111011\n101011110011101011101101011111010100100001100111100100111011",
"output": "111100101000000011101011011001110010101111000110010010000000"
},
{
"input": "1001011110110110000100011001010110000100011010010111010101110\n1101111100001000010111110011010101111010010100000001000010111",
"output": "0100100010111110010011101010000011111110001110010110010111001"
},
{
"input": "10000010101111100111110101111000010100110111101101111111111010\n10110110101100101010011001011010100110111011101100011001100111",
"output": "00110100000011001101101100100010110010001100000001100110011101"
},
{
"input": "011111010011111000001010101001101001000010100010111110010100001\n011111001011000011111001000001111001010110001010111101000010011",
"output": "000000011000111011110011101000010000010100101000000011010110010"
},
{
"input": "1111000000110001011101000100100100001111011100001111001100011111\n1101100110000101100001100000001001011011111011010101000101001010",
"output": "0010100110110100111100100100101101010100100111011010001001010101"
},
{
"input": "01100000101010010011001110100110110010000110010011011001100100011\n10110110010110111100100111000111000110010000000101101110000010111",
"output": "11010110111100101111101001100001110100010110010110110111100110100"
},
{
"input": "001111111010000100001100001010011001111110011110010111110001100111\n110000101001011000100010101100100110000111100000001101001110010111",
"output": "111111010011011100101110100110111111111001111110011010111111110000"
},
{
"input": "1011101011101101011110101101011101011000010011100101010101000100110\n0001000001001111010111100100111101100000000001110001000110000000110",
"output": "1010101010100010001001001001100000111000010010010100010011000100000"
},
{
"input": "01000001011001010011011100010000100100110101111011011011110000001110\n01011110000110011011000000000011000111100001010000000011111001110000",
"output": "00011111011111001000011100010011100011010100101011011000001001111110"
},
{
"input": "110101010100110101000001111110110100010010000100111110010100110011100\n111010010111111011100110101011001011001110110111110100000110110100111",
"output": "001111000011001110100111010101111111011100110011001010010010000111011"
},
{
"input": "1001101011000001011111100110010010000011010001001111011100010100110001\n1111100111110101001111010001010000011001001001010110001111000000100101",
"output": "0110001100110100010000110111000010011010011000011001010011010100010100"
},
{
"input": "00000111110010110001110110001010010101000111011001111111100110011110010\n00010111110100000100110101000010010001100001100011100000001100010100010",
"output": "00010000000110110101000011001000000100100110111010011111101010001010000"
},
{
"input": "100101011100101101000011010001011001101110101110001100010001010111001110\n100001111100101011011111110000001111000111001011111110000010101110111001",
"output": "000100100000000110011100100001010110101001100101110010010011111001110111"
},
{
"input": "1101100001000111001101001011101000111000011110000001001101101001111011010\n0101011101010100011011010110101000010010110010011110101100000110110001000",
"output": "1000111100010011010110011101000000101010101100011111100001101111001010010"
},
{
"input": "01101101010011110101100001110101111011100010000010001101111000011110111111\n00101111001101001100111010000101110000100101101111100111101110010100011011",
"output": "01000010011110111001011011110000001011000111101101101010010110001010100100"
},
{
"input": "101100101100011001101111110110110010100110110010100001110010110011001101011\n000001011010101011110011111101001110000111000010001101000010010000010001101",
"output": "101101110110110010011100001011111100100001110000101100110000100011011100110"
},
{
"input": "0010001011001010001100000010010011110110011000100000000100110000101111001110\n1100110100111000110100001110111001011101001100001010100001010011100110110001",
"output": "1110111111110010111000001100101010101011010100101010100101100011001001111111"
},
{
"input": "00101101010000000101011001101011001100010001100000101011101110000001111001000\n10010110010111000000101101000011101011001010000011011101101011010000000011111",
"output": "10111011000111000101110100101000100111011011100011110110000101010001111010111"
},
{
"input": "111100000100100000101001100001001111001010001000001000000111010000010101101011\n001000100010100101111011111011010110101100001111011000010011011011100010010110",
"output": "110100100110000101010010011010011001100110000111010000010100001011110111111101"
},
{
"input": "0110001101100100001111110101101000100101010010101010011001101001001101110000000\n0111011000000010010111011110010000000001000110001000011001101000000001110100111",
"output": "0001010101100110011000101011111000100100010100100010000000000001001100000100111"
},
{
"input": "10001111111001000101001011110101111010100001011010101100111001010001010010001000\n10000111010010011110111000111010101100000011110001101111001000111010100000000001",
"output": "00001000101011011011110011001111010110100010101011000011110001101011110010001001"
},
{
"input": "100110001110110000100101001110000011110110000110000000100011110100110110011001101\n110001110101110000000100101001101011111100100100001001000110000001111100011110110",
"output": "010111111011000000100001100111101000001010100010001001100101110101001010000111011"
},
{
"input": "0000010100100000010110111100011111111010011101000000100000011001001101101100111010\n0100111110011101010110101011110110010111001111000110101100101110111100101000111111",
"output": "0100101010111101000000010111101001101101010010000110001100110111110001000100000101"
},
{
"input": "11000111001010100001110000001001011010010010110000001110100101000001010101100110111\n11001100100100100001101010110100000111100011101110011010110100001001000011011011010",
"output": "00001011101110000000011010111101011101110001011110010100010001001000010110111101101"
},
{
"input": "010110100010001000100010101001101010011010111110100001000100101000111011100010100001\n110000011111101101010011111000101010111010100001001100001001100101000000111000000000",
"output": "100110111101100101110001010001000000100000011111101101001101001101111011011010100001"
},
{
"input": "0000011110101110010101110110110101100001011001101010101001000010000010000000101001101\n1100111111011100000110000111101110011111100111110001011001000010011111100001001100011",
"output": "1100100001110010010011110001011011111110111110011011110000000000011101100001100101110"
},
{
"input": "10100000101101110001100010010010100101100011010010101000110011100000101010110010000000\n10001110011011010010111011011101101111000111110000111000011010010101001100000001010011",
"output": "00101110110110100011011001001111001010100100100010010000101001110101100110110011010011"
},
{
"input": "001110000011111101101010011111000101010111010100001001100001001100101000000111000000000\n111010000000000000101001110011001000111011001100101010011001000011101001001011110000011",
"output": "110100000011111101000011101100001101101100011000100011111000001111000001001100110000011"
},
{
"input": "1110111100111011010101011011001110001010010010110011110010011111000010011111010101100001\n1001010101011001001010100010101100000110111101011000100010101111111010111100001110010010",
"output": "0111101001100010011111111001100010001100101111101011010000110000111000100011011011110011"
},
{
"input": "11100010001100010011001100001100010011010001101110011110100101110010101101011101000111111\n01110000000110111010110100001010000101011110100101010011000110101110101101110111011110001",
"output": "10010010001010101001111000000110010110001111001011001101100011011100000000101010011001110"
},
{
"input": "001101011001100101101100110000111000101011001001100100000100101000100000110100010111111101\n101001111110000010111101111110001001111001111101111010000110111000100100110010010001011111",
"output": "100100100111100111010001001110110001010010110100011110000010010000000100000110000110100010"
},
{
"input": "1010110110010101000110010010110101011101010100011001101011000110000000100011100100011000000\n0011011111100010001111101101000111001011101110100000110111100100101111010110101111011100011",
"output": "1001101001110111001001111111110010010110111010111001011100100010101111110101001011000100011"
},
{
"input": "10010010000111010111011111110010100101100000001100011100111011100010000010010001011100001100\n00111010100010110010000100010111010001111110100100100011101000101111111111001101101100100100",
"output": "10101000100101100101011011100101110100011110101000111111010011001101111101011100110000101000"
},
{
"input": "010101110001010101100000010111010000000111110011001101100011001000000011001111110000000010100\n010010111011100101010101111110110000000111000100001101101001001000001100101110001010000100001",
"output": "000111001010110000110101101001100000000000110111000000001010000000001111100001111010000110101"
},
{
"input": "1100111110011001000111101001001011000110011010111111100010111111001100111111011101100111101011\n1100000011001000110100110111000001011001010111101000010010100011000001100100111101101000010110",
"output": "0000111101010001110011011110001010011111001101010111110000011100001101011011100000001111111101"
},
{
"input": "00011000100100110111100101100100000000010011110111110010101110110011100001010111010011110100101\n00011011111011111011100101100111100101001110010111000010000111000100100100000001110101111011011",
"output": "00000011011111001100000000000011100101011101100000110000101001110111000101010110100110001111110"
},
{
"input": "000101011001001100000111100010110101111011110101111101000110001101011010111110110011100100000001\n011000101010011111011000111000100000000011011000000001111110001000001111101010110000011100001111",
"output": "011101110011010011011111011010010101111000101101111100111000000101010101010100000011111000001110"
},
{
"input": "1000101001011010000100100100010010011101011001110101111011101111111110010101001101010001010101001\n0110110010011100011111011111110111000000010001110100001010111110101011010011111011111110000110000",
"output": "1110011011000110011011111011100101011101001000000001110001010001010101000110110110101111010011001"
},
{
"input": "01111010010101100111110111111011011010100001011101010000111100101101101110111011001100101011100111\n00001100110110100001111011000010001001001100000010111101000001111011100000010111010010000011000010",
"output": "01110110100011000110001100111001010011101101011111101101111101010110001110101100011110101000100101"
},
{
"input": "111101011101111000001011001101111010110111001101110100100011111011011101001101010101011010111000110\n101111100101101101001100110011000001111010011101110111110110101110011011110011111100001001110101101",
"output": "010010111000010101000111111110111011001101010000000011010101010101000110111110101001010011001101011"
},
{
"input": "1010000011010110011101001101110001110010000111011101110010110110111111001001110100101100010101010001\n0111101000111100101100000101111010100100001000011101010100110011100011010011010101000100101011100011",
"output": "1101101011101010110001001000001011010110001111000000100110000101011100011010100001101000111110110010"
},
{
"input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001\n1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111010",
"output": "1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111011"
},
{
"input": "0\n0",
"output": "0"
},
{
"input": "0\n1",
"output": "1"
}
] | 1,662,443,013
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 102
| 61
| 0
|
n=str(input())
m=str(input())
o=' '
for i in range(len(m)):
if n[i]==m[i]:
o+='0'
else:
o+='1'
print(o)
|
Title: Ultra-Fast Mathematician
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
Input Specification:
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
Output Specification:
Write one line — the corresponding answer. Do not omit the leading 0s.
Demo Input:
['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n']
Demo Output:
['1110001\n', '111\n', '0100\n', '00010\n']
Note:
none
|
```python
n=str(input())
m=str(input())
o=' '
for i in range(len(m)):
if n[i]==m[i]:
o+='0'
else:
o+='1'
print(o)
```
| 3.98475
|
52
|
A
|
123-sequence
|
PROGRAMMING
| 900
|
[
"implementation"
] |
A. 123-sequence
|
2
|
256
|
There is a given sequence of integers *a*1,<=*a*2,<=...,<=*a**n*, where every number is from 1 to 3 inclusively. You have to replace the minimum number of numbers in it so that all the numbers in the sequence are equal to each other.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=106). The second line contains a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=3).
|
Print the minimum number of replacements needed to be performed to make all the numbers in the sequence equal.
|
[
"9\n1 3 2 2 2 1 1 2 3\n"
] |
[
"5\n"
] |
In the example all the numbers equal to 1 and 3 should be replaced by 2.
| 500
|
[
{
"input": "9\n1 3 2 2 2 1 1 2 3",
"output": "5"
},
{
"input": "6\n3 3 2 2 1 3",
"output": "3"
},
{
"input": "12\n3 1 3 1 2 1 3 2 2 1 2 1",
"output": "7"
},
{
"input": "15\n3 2 1 1 1 1 3 2 2 3 3 1 2 3 2",
"output": "10"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "2\n3 2",
"output": "1"
},
{
"input": "2\n3 1",
"output": "1"
},
{
"input": "18\n2 3 2 1 2 3 2 1 2 3 3 3 1 2 3 3 3 2",
"output": "10"
},
{
"input": "30\n2 1 3 2 3 2 2 2 2 3 2 2 3 2 1 1 3 1 3 2 1 2 3 1 1 3 3 1 3 1",
"output": "19"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "1\n2",
"output": "0"
},
{
"input": "1\n3",
"output": "0"
}
] | 1,676,667,309
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 23
| 342
| 11,059,200
|
print(int(input()) - max((map(input().split().count, map(str, range(1, 4))))))
|
Title: 123-sequence
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
There is a given sequence of integers *a*1,<=*a*2,<=...,<=*a**n*, where every number is from 1 to 3 inclusively. You have to replace the minimum number of numbers in it so that all the numbers in the sequence are equal to each other.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=106). The second line contains a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=3).
Output Specification:
Print the minimum number of replacements needed to be performed to make all the numbers in the sequence equal.
Demo Input:
['9\n1 3 2 2 2 1 1 2 3\n']
Demo Output:
['5\n']
Note:
In the example all the numbers equal to 1 and 3 should be replaced by 2.
|
```python
print(int(input()) - max((map(input().split().count, map(str, range(1, 4))))))
```
| 3.893901
|
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