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|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
415
|
A
|
Mashmokh and Lights
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Mashmokh works in a factory. At the end of each day he must turn off all of the lights.
The lights on the factory are indexed from 1 to *n*. There are *n* buttons in Mashmokh's room indexed from 1 to *n* as well. If Mashmokh pushes button with index *i*, then each light with index not less than *i* that is still turned on turns off.
Mashmokh is not very clever. So instead of pushing the first button he pushes some of the buttons randomly each night. He pushed *m* distinct buttons *b*1,<=*b*2,<=...,<=*b**m* (the buttons were pushed consecutively in the given order) this night. Now he wants to know for each light the index of the button that turned this light off. Please note that the index of button *b**i* is actually *b**i*, not *i*.
Please, help Mashmokh, print these indices.
|
The first line of the input contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100), the number of the factory lights and the pushed buttons respectively. The next line contains *m* distinct space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*).
It is guaranteed that all lights will be turned off after pushing all buttons.
|
Output *n* space-separated integers where the *i*-th number is index of the button that turns the *i*-th light off.
|
[
"5 4\n4 3 1 2\n",
"5 5\n5 4 3 2 1\n"
] |
[
"1 1 3 4 4 \n",
"1 2 3 4 5 \n"
] |
In the first sample, after pressing button number 4, lights 4 and 5 are turned off and lights 1, 2 and 3 are still on. Then after pressing button number 3, light number 3 is turned off as well. Pressing button number 1 turns off lights number 1 and 2 as well so pressing button number 2 in the end has no effect. Thus button number 4 turned lights 4 and 5 off, button number 3 turned light 3 off and button number 1 turned light 1 and 2 off.
| 500
|
[
{
"input": "5 4\n4 3 1 2",
"output": "1 1 3 4 4 "
},
{
"input": "5 5\n5 4 3 2 1",
"output": "1 2 3 4 5 "
},
{
"input": "16 11\n8 5 12 10 14 2 6 3 15 9 1",
"output": "1 2 2 2 5 5 5 8 8 8 8 8 8 8 8 8 "
},
{
"input": "79 22\n76 32 48 28 33 44 58 59 1 51 77 13 15 64 49 72 74 21 61 12 60 57",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 28 28 28 28 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 76 76 76 76 "
},
{
"input": "25 19\n3 12 21 11 19 6 5 15 4 16 20 8 9 1 22 23 25 18 13",
"output": "1 1 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 "
},
{
"input": "48 8\n42 27 40 1 18 3 19 2",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 27 27 27 27 27 27 27 27 27 27 27 27 27 27 27 42 42 42 42 42 42 42 "
},
{
"input": "44 19\n13 20 7 10 9 14 43 17 18 39 21 42 37 1 33 8 35 4 6",
"output": "1 1 1 1 1 1 7 7 7 7 7 7 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 "
},
{
"input": "80 29\n79 51 28 73 65 39 10 1 59 29 7 70 64 3 35 17 24 71 74 2 6 49 66 80 13 18 60 15 12",
"output": "1 1 1 1 1 1 1 1 1 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 28 28 28 28 28 28 28 28 28 28 28 28 28 28 28 28 28 28 28 28 28 28 28 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 79 79 "
},
{
"input": "31 4\n8 18 30 1",
"output": "1 1 1 1 1 1 1 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 "
},
{
"input": "62 29\n61 55 35 13 51 56 23 6 8 26 27 40 48 11 18 12 19 50 54 14 24 21 32 17 43 33 1 2 3",
"output": "1 1 1 1 1 6 6 6 6 6 6 6 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 35 35 35 35 35 35 35 35 35 35 35 35 35 35 35 35 35 35 35 35 55 55 55 55 55 55 61 61 "
},
{
"input": "5 4\n2 3 4 1",
"output": "1 2 2 2 2 "
},
{
"input": "39 37\n2 5 17 24 19 33 35 16 20 3 1 34 10 36 15 37 14 8 28 21 13 31 30 29 7 25 32 12 6 27 22 4 11 39 18 9 26",
"output": "1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 "
},
{
"input": "100 100\n100 99 98 97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 80 79 78 77 76 75 74 73 72 71 70 69 68 67 66 65 64 63 62 61 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1",
"output": "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 "
},
{
"input": "1 1\n1",
"output": "1 "
},
{
"input": "18 3\n18 1 11",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 18 "
},
{
"input": "67 20\n66 23 40 49 3 39 60 43 52 47 16 36 22 5 41 10 55 34 64 1",
"output": "1 1 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 23 66 66 "
},
{
"input": "92 52\n9 85 44 13 27 61 8 1 28 41 6 14 70 67 39 71 56 80 34 21 5 10 40 73 63 38 90 57 37 36 82 86 65 46 7 54 81 12 45 49 83 59 64 26 62 25 60 24 91 47 53 55",
"output": "1 1 1 1 1 1 1 8 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 "
},
{
"input": "66 36\n44 62 32 29 3 15 47 30 50 42 35 2 33 65 10 13 56 12 1 16 7 36 39 11 25 28 20 52 46 38 37 8 61 49 48 14",
"output": "1 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 29 29 29 32 32 32 32 32 32 32 32 32 32 32 32 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 "
},
{
"input": "32 8\n27 23 1 13 18 24 17 26",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 23 23 23 23 27 27 27 27 27 27 "
},
{
"input": "26 13\n1 14 13 2 4 24 21 22 16 3 10 12 6",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 "
},
{
"input": "31 20\n10 11 20 2 4 26 31 7 13 12 28 1 30 18 21 8 3 16 15 19",
"output": "1 2 2 2 2 2 2 2 2 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 "
},
{
"input": "86 25\n22 62 8 23 53 77 9 31 43 1 58 16 72 11 15 35 60 39 79 4 82 64 76 63 59",
"output": "1 1 1 1 1 1 1 8 8 8 8 8 8 8 8 8 8 8 8 8 8 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 "
},
{
"input": "62 54\n2 5 4 47 40 61 37 31 41 16 44 42 48 32 10 6 62 38 52 49 11 20 55 22 3 36 25 21 50 8 28 14 18 39 34 54 53 19 46 27 15 23 12 24 60 17 33 57 58 1 35 29 51 7",
"output": "1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 "
},
{
"input": "57 19\n43 45 37 40 42 55 16 33 47 32 34 35 9 41 1 6 8 15 5",
"output": "1 1 1 1 1 1 1 1 9 9 9 9 9 9 9 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 37 37 37 37 37 37 43 43 43 43 43 43 43 43 43 43 43 43 43 43 43 "
},
{
"input": "32 14\n4 7 13 1 25 22 9 27 6 28 30 2 14 21",
"output": "1 1 1 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 "
},
{
"input": "57 12\n8 53 51 38 1 6 16 33 13 46 28 35",
"output": "1 1 1 1 1 1 1 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 "
},
{
"input": "87 9\n57 34 78 1 52 67 56 6 54",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 34 34 34 34 34 34 34 34 34 34 34 34 34 34 34 34 34 34 34 34 34 34 34 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 "
},
{
"input": "88 42\n85 45 52 14 63 53 70 71 16 86 66 47 12 22 10 72 4 31 3 69 11 77 17 25 46 75 23 1 21 84 44 20 18 33 48 88 41 83 67 61 73 34",
"output": "1 1 3 4 4 4 4 4 4 10 10 12 12 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 85 85 85 85 "
},
{
"input": "27 25\n9 21 17 5 16 3 23 7 12 4 14 11 13 1 15 19 27 8 20 10 22 25 6 18 26",
"output": "1 1 3 3 5 5 5 5 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 "
},
{
"input": "89 28\n5 22 79 42 16 35 66 48 57 55 1 37 29 31 40 38 45 62 41 87 64 89 81 13 60 44 71 82",
"output": "1 1 1 1 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 "
},
{
"input": "17 4\n4 3 1 2",
"output": "1 1 3 4 4 4 4 4 4 4 4 4 4 4 4 4 4 "
}
] | 1,618,421,348
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 61
| 0
|
n,k=map(int,input().split())
l=list(map(int,input().split()))
m=[0]*n
for i in range (len(l)):
# print(i)
for j in range (l[i]-1,len(m)):
# print(j,end=" ")
if m[j]==0:
m[j]=l[i]
print(m)
|
Title: Mashmokh and Lights
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mashmokh works in a factory. At the end of each day he must turn off all of the lights.
The lights on the factory are indexed from 1 to *n*. There are *n* buttons in Mashmokh's room indexed from 1 to *n* as well. If Mashmokh pushes button with index *i*, then each light with index not less than *i* that is still turned on turns off.
Mashmokh is not very clever. So instead of pushing the first button he pushes some of the buttons randomly each night. He pushed *m* distinct buttons *b*1,<=*b*2,<=...,<=*b**m* (the buttons were pushed consecutively in the given order) this night. Now he wants to know for each light the index of the button that turned this light off. Please note that the index of button *b**i* is actually *b**i*, not *i*.
Please, help Mashmokh, print these indices.
Input Specification:
The first line of the input contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100), the number of the factory lights and the pushed buttons respectively. The next line contains *m* distinct space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*).
It is guaranteed that all lights will be turned off after pushing all buttons.
Output Specification:
Output *n* space-separated integers where the *i*-th number is index of the button that turns the *i*-th light off.
Demo Input:
['5 4\n4 3 1 2\n', '5 5\n5 4 3 2 1\n']
Demo Output:
['1 1 3 4 4 \n', '1 2 3 4 5 \n']
Note:
In the first sample, after pressing button number 4, lights 4 and 5 are turned off and lights 1, 2 and 3 are still on. Then after pressing button number 3, light number 3 is turned off as well. Pressing button number 1 turns off lights number 1 and 2 as well so pressing button number 2 in the end has no effect. Thus button number 4 turned lights 4 and 5 off, button number 3 turned light 3 off and button number 1 turned light 1 and 2 off.
|
```python
n,k=map(int,input().split())
l=list(map(int,input().split()))
m=[0]*n
for i in range (len(l)):
# print(i)
for j in range (l[i]-1,len(m)):
# print(j,end=" ")
if m[j]==0:
m[j]=l[i]
print(m)
```
| 0
|
|
136
|
A
|
Presents
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Little Petya very much likes gifts. Recently he has received a new laptop as a New Year gift from his mother. He immediately decided to give it to somebody else as what can be more pleasant than giving somebody gifts. And on this occasion he organized a New Year party at his place and invited *n* his friends there.
If there's one thing Petya likes more that receiving gifts, that's watching others giving gifts to somebody else. Thus, he safely hid the laptop until the next New Year and made up his mind to watch his friends exchanging gifts while he does not participate in the process. He numbered all his friends with integers from 1 to *n*. Petya remembered that a friend number *i* gave a gift to a friend number *p**i*. He also remembered that each of his friends received exactly one gift.
Now Petya wants to know for each friend *i* the number of a friend who has given him a gift.
|
The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the quantity of friends Petya invited to the party. The second line contains *n* space-separated integers: the *i*-th number is *p**i* — the number of a friend who gave a gift to friend number *i*. It is guaranteed that each friend received exactly one gift. It is possible that some friends do not share Petya's ideas of giving gifts to somebody else. Those friends gave the gifts to themselves.
|
Print *n* space-separated integers: the *i*-th number should equal the number of the friend who gave a gift to friend number *i*.
|
[
"4\n2 3 4 1\n",
"3\n1 3 2\n",
"2\n1 2\n"
] |
[
"4 1 2 3\n",
"1 3 2\n",
"1 2\n"
] |
none
| 500
|
[
{
"input": "4\n2 3 4 1",
"output": "4 1 2 3"
},
{
"input": "3\n1 3 2",
"output": "1 3 2"
},
{
"input": "2\n1 2",
"output": "1 2"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "10\n1 3 2 6 4 5 7 9 8 10",
"output": "1 3 2 5 6 4 7 9 8 10"
},
{
"input": "5\n5 4 3 2 1",
"output": "5 4 3 2 1"
},
{
"input": "20\n2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19"
},
{
"input": "21\n3 2 1 6 5 4 9 8 7 12 11 10 15 14 13 18 17 16 21 20 19",
"output": "3 2 1 6 5 4 9 8 7 12 11 10 15 14 13 18 17 16 21 20 19"
},
{
"input": "10\n3 4 5 6 7 8 9 10 1 2",
"output": "9 10 1 2 3 4 5 6 7 8"
},
{
"input": "8\n1 5 3 7 2 6 4 8",
"output": "1 5 3 7 2 6 4 8"
},
{
"input": "50\n49 22 4 2 20 46 7 32 5 19 48 24 26 15 45 21 44 11 50 43 39 17 31 1 42 34 3 27 36 25 12 30 13 33 28 35 18 6 8 37 38 14 10 9 29 16 40 23 41 47",
"output": "24 4 27 3 9 38 7 39 44 43 18 31 33 42 14 46 22 37 10 5 16 2 48 12 30 13 28 35 45 32 23 8 34 26 36 29 40 41 21 47 49 25 20 17 15 6 50 11 1 19"
},
{
"input": "34\n13 20 33 30 15 11 27 4 8 2 29 25 24 7 3 22 18 10 26 16 5 1 32 9 34 6 12 14 28 19 31 21 23 17",
"output": "22 10 15 8 21 26 14 9 24 18 6 27 1 28 5 20 34 17 30 2 32 16 33 13 12 19 7 29 11 4 31 23 3 25"
},
{
"input": "92\n23 1 6 4 84 54 44 76 63 34 61 20 48 13 28 78 26 46 90 72 24 55 91 89 53 38 82 5 79 92 29 32 15 64 11 88 60 70 7 66 18 59 8 57 19 16 42 21 80 71 62 27 75 86 36 9 83 73 74 50 43 31 56 30 17 33 40 81 49 12 10 41 22 77 25 68 51 2 47 3 58 69 87 67 39 37 35 65 14 45 52 85",
"output": "2 78 80 4 28 3 39 43 56 71 35 70 14 89 33 46 65 41 45 12 48 73 1 21 75 17 52 15 31 64 62 32 66 10 87 55 86 26 85 67 72 47 61 7 90 18 79 13 69 60 77 91 25 6 22 63 44 81 42 37 11 51 9 34 88 40 84 76 82 38 50 20 58 59 53 8 74 16 29 49 68 27 57 5 92 54 83 36 24 19 23 30"
},
{
"input": "49\n30 24 33 48 7 3 17 2 8 35 10 39 23 40 46 32 18 21 26 22 1 16 47 45 41 28 31 6 12 43 27 11 13 37 19 15 44 5 29 42 4 38 20 34 14 9 25 36 49",
"output": "21 8 6 41 38 28 5 9 46 11 32 29 33 45 36 22 7 17 35 43 18 20 13 2 47 19 31 26 39 1 27 16 3 44 10 48 34 42 12 14 25 40 30 37 24 15 23 4 49"
},
{
"input": "12\n3 8 7 4 6 5 2 1 11 9 10 12",
"output": "8 7 1 4 6 5 3 2 10 11 9 12"
},
{
"input": "78\n16 56 36 78 21 14 9 77 26 57 70 61 41 47 18 44 5 31 50 74 65 52 6 39 22 62 67 69 43 7 64 29 24 40 48 51 73 54 72 12 19 34 4 25 55 33 17 35 23 53 10 8 27 32 42 68 20 63 3 2 1 71 58 46 13 30 49 11 37 66 38 60 28 75 15 59 45 76",
"output": "61 60 59 43 17 23 30 52 7 51 68 40 65 6 75 1 47 15 41 57 5 25 49 33 44 9 53 73 32 66 18 54 46 42 48 3 69 71 24 34 13 55 29 16 77 64 14 35 67 19 36 22 50 38 45 2 10 63 76 72 12 26 58 31 21 70 27 56 28 11 62 39 37 20 74 78 8 4"
},
{
"input": "64\n64 57 40 3 15 8 62 18 33 59 51 19 22 13 4 37 47 45 50 35 63 11 58 42 46 21 7 2 41 48 32 23 28 38 17 12 24 27 49 31 60 6 30 25 61 52 26 54 9 14 29 20 44 39 55 10 34 16 5 56 1 36 53 43",
"output": "61 28 4 15 59 42 27 6 49 56 22 36 14 50 5 58 35 8 12 52 26 13 32 37 44 47 38 33 51 43 40 31 9 57 20 62 16 34 54 3 29 24 64 53 18 25 17 30 39 19 11 46 63 48 55 60 2 23 10 41 45 7 21 1"
},
{
"input": "49\n38 20 49 32 14 41 39 45 25 48 40 19 26 43 34 12 10 3 35 42 5 7 46 47 4 2 13 22 16 24 33 15 11 18 29 31 23 9 44 36 6 17 37 1 30 28 8 21 27",
"output": "44 26 18 25 21 41 22 47 38 17 33 16 27 5 32 29 42 34 12 2 48 28 37 30 9 13 49 46 35 45 36 4 31 15 19 40 43 1 7 11 6 20 14 39 8 23 24 10 3"
},
{
"input": "78\n17 50 30 48 33 12 42 4 18 53 76 67 38 3 20 72 51 55 60 63 46 10 57 45 54 32 24 62 8 11 35 44 65 74 58 28 2 6 56 52 39 23 47 49 61 1 66 41 15 77 7 27 78 13 14 34 5 31 37 21 40 16 29 69 59 43 64 36 70 19 25 73 71 75 9 68 26 22",
"output": "46 37 14 8 57 38 51 29 75 22 30 6 54 55 49 62 1 9 70 15 60 78 42 27 71 77 52 36 63 3 58 26 5 56 31 68 59 13 41 61 48 7 66 32 24 21 43 4 44 2 17 40 10 25 18 39 23 35 65 19 45 28 20 67 33 47 12 76 64 69 73 16 72 34 74 11 50 53"
},
{
"input": "29\n14 21 27 1 4 18 10 17 20 23 2 24 7 9 28 22 8 25 12 15 11 6 16 29 3 26 19 5 13",
"output": "4 11 25 5 28 22 13 17 14 7 21 19 29 1 20 23 8 6 27 9 2 16 10 12 18 26 3 15 24"
},
{
"input": "82\n6 1 10 75 28 66 61 81 78 63 17 19 58 34 49 12 67 50 41 44 3 15 59 38 51 72 36 11 46 29 18 64 27 23 13 53 56 68 2 25 47 40 69 54 42 5 60 55 4 16 24 79 57 20 7 73 32 80 76 52 82 37 26 31 65 8 39 62 33 71 30 9 77 43 48 74 70 22 14 45 35 21",
"output": "2 39 21 49 46 1 55 66 72 3 28 16 35 79 22 50 11 31 12 54 82 78 34 51 40 63 33 5 30 71 64 57 69 14 81 27 62 24 67 42 19 45 74 20 80 29 41 75 15 18 25 60 36 44 48 37 53 13 23 47 7 68 10 32 65 6 17 38 43 77 70 26 56 76 4 59 73 9 52 58 8 61"
},
{
"input": "82\n74 18 15 69 71 77 19 26 80 20 66 7 30 82 22 48 21 44 52 65 64 61 35 49 12 8 53 81 54 16 11 9 40 46 13 1 29 58 5 41 55 4 78 60 6 51 56 2 38 36 34 62 63 25 17 67 45 14 32 37 75 79 10 47 27 39 31 68 59 24 50 43 72 70 42 28 76 23 57 3 73 33",
"output": "36 48 80 42 39 45 12 26 32 63 31 25 35 58 3 30 55 2 7 10 17 15 78 70 54 8 65 76 37 13 67 59 82 51 23 50 60 49 66 33 40 75 72 18 57 34 64 16 24 71 46 19 27 29 41 47 79 38 69 44 22 52 53 21 20 11 56 68 4 74 5 73 81 1 61 77 6 43 62 9 28 14"
},
{
"input": "45\n2 32 34 13 3 15 16 33 22 12 31 38 42 14 27 7 36 8 4 19 45 41 5 35 10 11 39 20 29 44 17 9 6 40 37 28 25 21 1 30 24 18 43 26 23",
"output": "39 1 5 19 23 33 16 18 32 25 26 10 4 14 6 7 31 42 20 28 38 9 45 41 37 44 15 36 29 40 11 2 8 3 24 17 35 12 27 34 22 13 43 30 21"
},
{
"input": "45\n4 32 33 39 43 21 22 35 45 7 14 5 16 9 42 31 24 36 17 29 41 25 37 34 27 20 11 44 3 13 19 2 1 10 26 30 38 18 6 8 15 23 40 28 12",
"output": "33 32 29 1 12 39 10 40 14 34 27 45 30 11 41 13 19 38 31 26 6 7 42 17 22 35 25 44 20 36 16 2 3 24 8 18 23 37 4 43 21 15 5 28 9"
},
{
"input": "74\n48 72 40 67 17 4 27 53 11 32 25 9 74 2 41 24 56 22 14 21 33 5 18 55 20 7 29 36 69 13 52 19 38 30 68 59 66 34 63 6 47 45 54 44 62 12 50 71 16 10 8 64 57 73 46 26 49 42 3 23 35 1 61 39 70 60 65 43 15 28 37 51 58 31",
"output": "62 14 59 6 22 40 26 51 12 50 9 46 30 19 69 49 5 23 32 25 20 18 60 16 11 56 7 70 27 34 74 10 21 38 61 28 71 33 64 3 15 58 68 44 42 55 41 1 57 47 72 31 8 43 24 17 53 73 36 66 63 45 39 52 67 37 4 35 29 65 48 2 54 13"
},
{
"input": "47\n9 26 27 10 6 34 28 42 39 22 45 21 11 43 14 47 38 15 40 32 46 1 36 29 17 25 2 23 31 5 24 4 7 8 12 19 16 44 37 20 18 33 30 13 35 41 3",
"output": "22 27 47 32 30 5 33 34 1 4 13 35 44 15 18 37 25 41 36 40 12 10 28 31 26 2 3 7 24 43 29 20 42 6 45 23 39 17 9 19 46 8 14 38 11 21 16"
},
{
"input": "49\n14 38 6 29 9 49 36 43 47 3 44 20 34 15 7 11 1 28 12 40 16 37 31 10 42 41 33 21 18 30 5 27 17 35 25 26 45 19 2 13 23 32 4 22 46 48 24 39 8",
"output": "17 39 10 43 31 3 15 49 5 24 16 19 40 1 14 21 33 29 38 12 28 44 41 47 35 36 32 18 4 30 23 42 27 13 34 7 22 2 48 20 26 25 8 11 37 45 9 46 6"
},
{
"input": "100\n78 56 31 91 90 95 16 65 58 77 37 89 33 61 10 76 62 47 35 67 69 7 63 83 22 25 49 8 12 30 39 44 57 64 48 42 32 11 70 43 55 50 99 24 85 73 45 14 54 21 98 84 74 2 26 18 9 36 80 53 75 46 66 86 59 93 87 68 94 13 72 28 79 88 92 29 52 82 34 97 19 38 1 41 27 4 40 5 96 100 51 6 20 23 81 15 17 3 60 71",
"output": "83 54 98 86 88 92 22 28 57 15 38 29 70 48 96 7 97 56 81 93 50 25 94 44 26 55 85 72 76 30 3 37 13 79 19 58 11 82 31 87 84 36 40 32 47 62 18 35 27 42 91 77 60 49 41 2 33 9 65 99 14 17 23 34 8 63 20 68 21 39 100 71 46 53 61 16 10 1 73 59 95 78 24 52 45 64 67 74 12 5 4 75 66 69 6 89 80 51 43 90"
},
{
"input": "22\n12 8 11 2 16 7 13 6 22 21 20 10 4 14 18 1 5 15 3 19 17 9",
"output": "16 4 19 13 17 8 6 2 22 12 3 1 7 14 18 5 21 15 20 11 10 9"
},
{
"input": "72\n16 11 49 51 3 27 60 55 23 40 66 7 53 70 13 5 15 32 18 72 33 30 8 31 46 12 28 67 25 38 50 22 69 34 71 52 58 39 24 35 42 9 41 26 62 1 63 65 36 64 68 61 37 14 45 47 6 57 54 20 17 2 56 59 29 10 4 48 21 43 19 44",
"output": "46 62 5 67 16 57 12 23 42 66 2 26 15 54 17 1 61 19 71 60 69 32 9 39 29 44 6 27 65 22 24 18 21 34 40 49 53 30 38 10 43 41 70 72 55 25 56 68 3 31 4 36 13 59 8 63 58 37 64 7 52 45 47 50 48 11 28 51 33 14 35 20"
},
{
"input": "63\n21 56 11 10 62 24 20 42 28 52 38 2 37 43 48 22 7 8 40 14 13 46 53 1 23 4 60 63 51 36 25 12 39 32 49 16 58 44 31 61 33 50 55 54 45 6 47 41 9 57 30 29 26 18 19 27 15 34 3 35 59 5 17",
"output": "24 12 59 26 62 46 17 18 49 4 3 32 21 20 57 36 63 54 55 7 1 16 25 6 31 53 56 9 52 51 39 34 41 58 60 30 13 11 33 19 48 8 14 38 45 22 47 15 35 42 29 10 23 44 43 2 50 37 61 27 40 5 28"
},
{
"input": "18\n2 16 8 4 18 12 3 6 5 9 10 15 11 17 14 13 1 7",
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},
{
"input": "47\n6 9 10 41 25 3 4 37 20 1 36 22 29 27 11 24 43 31 12 17 34 42 38 39 13 2 7 21 18 5 15 35 44 26 33 46 19 40 30 14 28 23 47 32 45 8 16",
"output": "10 26 6 7 30 1 27 46 2 3 15 19 25 40 31 47 20 29 37 9 28 12 42 16 5 34 14 41 13 39 18 44 35 21 32 11 8 23 24 38 4 22 17 33 45 36 43"
},
{
"input": "96\n41 91 48 88 29 57 1 19 44 43 37 5 10 75 25 63 30 78 76 53 8 92 18 70 39 17 49 60 9 16 3 34 86 59 23 79 55 45 72 51 28 33 96 40 26 54 6 32 89 61 85 74 7 82 52 31 64 66 94 95 11 22 2 73 35 13 42 71 14 47 84 69 50 67 58 12 77 46 38 68 15 36 20 93 27 90 83 56 87 4 21 24 81 62 80 65",
"output": "7 63 31 90 12 47 53 21 29 13 61 76 66 69 81 30 26 23 8 83 91 62 35 92 15 45 85 41 5 17 56 48 42 32 65 82 11 79 25 44 1 67 10 9 38 78 70 3 27 73 40 55 20 46 37 88 6 75 34 28 50 94 16 57 96 58 74 80 72 24 68 39 64 52 14 19 77 18 36 95 93 54 87 71 51 33 89 4 49 86 2 22 84 59 60 43"
},
{
"input": "73\n67 24 39 22 23 20 48 34 42 40 19 70 65 69 64 21 53 11 59 15 26 10 30 33 72 29 55 25 56 71 8 9 57 49 41 61 13 12 6 27 66 36 47 50 73 60 2 37 7 4 51 17 1 46 14 62 35 3 45 63 43 58 54 32 31 5 28 44 18 52 68 38 16",
"output": "53 47 58 50 66 39 49 31 32 22 18 38 37 55 20 73 52 69 11 6 16 4 5 2 28 21 40 67 26 23 65 64 24 8 57 42 48 72 3 10 35 9 61 68 59 54 43 7 34 44 51 70 17 63 27 29 33 62 19 46 36 56 60 15 13 41 1 71 14 12 30 25 45"
},
{
"input": "81\n25 2 78 40 12 80 69 13 49 43 17 33 23 54 32 61 77 66 27 71 24 26 42 55 60 9 5 30 7 37 45 63 53 11 38 44 68 34 28 52 67 22 57 46 47 50 8 16 79 62 4 36 20 14 73 64 6 76 35 74 58 10 29 81 59 31 19 1 75 39 70 18 41 21 72 65 3 48 15 56 51",
"output": "68 2 77 51 27 57 29 47 26 62 34 5 8 54 79 48 11 72 67 53 74 42 13 21 1 22 19 39 63 28 66 15 12 38 59 52 30 35 70 4 73 23 10 36 31 44 45 78 9 46 81 40 33 14 24 80 43 61 65 25 16 50 32 56 76 18 41 37 7 71 20 75 55 60 69 58 17 3 49 6 64"
},
{
"input": "12\n12 3 1 5 11 6 7 10 2 8 9 4",
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},
{
"input": "47\n7 21 41 18 40 31 12 28 24 14 43 23 33 10 19 38 26 8 34 15 29 44 5 13 39 25 3 27 20 42 35 9 2 1 30 46 36 32 4 22 37 45 6 47 11 16 17",
"output": "34 33 27 39 23 43 1 18 32 14 45 7 24 10 20 46 47 4 15 29 2 40 12 9 26 17 28 8 21 35 6 38 13 19 31 37 41 16 25 5 3 30 11 22 42 36 44"
},
{
"input": "8\n1 3 5 2 4 8 6 7",
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},
{
"input": "38\n28 8 2 33 20 32 26 29 23 31 15 38 11 37 18 21 22 19 4 34 1 35 16 7 17 6 27 30 36 12 9 24 25 13 5 3 10 14",
"output": "21 3 36 19 35 26 24 2 31 37 13 30 34 38 11 23 25 15 18 5 16 17 9 32 33 7 27 1 8 28 10 6 4 20 22 29 14 12"
},
{
"input": "10\n2 9 4 6 10 1 7 5 3 8",
"output": "6 1 9 3 8 4 7 10 2 5"
},
{
"input": "23\n20 11 15 1 5 12 23 9 2 22 13 19 16 14 7 4 8 21 6 17 18 10 3",
"output": "4 9 23 16 5 19 15 17 8 22 2 6 11 14 3 13 20 21 12 1 18 10 7"
},
{
"input": "10\n2 4 9 3 6 8 10 5 1 7",
"output": "9 1 4 2 8 5 10 6 3 7"
},
{
"input": "55\n9 48 23 49 11 24 4 22 34 32 17 45 39 13 14 21 19 25 2 31 37 7 55 36 20 51 5 12 54 10 35 40 43 1 46 18 53 41 38 26 29 50 3 42 52 27 8 28 47 33 6 16 30 44 15",
"output": "34 19 43 7 27 51 22 47 1 30 5 28 14 15 55 52 11 36 17 25 16 8 3 6 18 40 46 48 41 53 20 10 50 9 31 24 21 39 13 32 38 44 33 54 12 35 49 2 4 42 26 45 37 29 23"
},
{
"input": "58\n49 13 12 54 2 38 56 11 33 25 26 19 28 8 23 41 20 36 46 55 15 35 9 7 32 37 58 6 3 14 47 31 40 30 53 44 4 50 29 34 10 43 39 57 5 22 27 45 51 42 24 16 18 21 52 17 48 1",
"output": "58 5 29 37 45 28 24 14 23 41 8 3 2 30 21 52 56 53 12 17 54 46 15 51 10 11 47 13 39 34 32 25 9 40 22 18 26 6 43 33 16 50 42 36 48 19 31 57 1 38 49 55 35 4 20 7 44 27"
},
{
"input": "34\n20 25 2 3 33 29 1 16 14 7 21 9 32 31 6 26 22 4 27 23 24 10 34 12 19 15 5 18 28 17 13 8 11 30",
"output": "7 3 4 18 27 15 10 32 12 22 33 24 31 9 26 8 30 28 25 1 11 17 20 21 2 16 19 29 6 34 14 13 5 23"
},
{
"input": "53\n47 29 46 25 23 13 7 31 33 4 38 11 35 16 42 14 15 43 34 39 28 18 6 45 30 1 40 20 2 37 5 32 24 12 44 26 27 3 19 51 36 21 22 9 10 50 41 48 49 53 8 17 52",
"output": "26 29 38 10 31 23 7 51 44 45 12 34 6 16 17 14 52 22 39 28 42 43 5 33 4 36 37 21 2 25 8 32 9 19 13 41 30 11 20 27 47 15 18 35 24 3 1 48 49 46 40 53 50"
},
{
"input": "99\n77 87 90 48 53 38 68 6 28 57 35 82 63 71 60 41 3 12 86 65 10 59 22 67 33 74 93 27 24 1 61 43 25 4 51 52 15 88 9 31 30 42 89 49 23 21 29 32 46 73 37 16 5 69 56 26 92 64 20 54 75 14 98 13 94 2 95 7 36 66 58 8 50 78 84 45 11 96 76 62 97 80 40 39 47 85 34 79 83 17 91 72 19 44 70 81 55 99 18",
"output": "30 66 17 34 53 8 68 72 39 21 77 18 64 62 37 52 90 99 93 59 46 23 45 29 33 56 28 9 47 41 40 48 25 87 11 69 51 6 84 83 16 42 32 94 76 49 85 4 44 73 35 36 5 60 97 55 10 71 22 15 31 80 13 58 20 70 24 7 54 95 14 92 50 26 61 79 1 74 88 82 96 12 89 75 86 19 2 38 43 3 91 57 27 65 67 78 81 63 98"
},
{
"input": "32\n17 29 2 6 30 8 26 7 1 27 10 9 13 24 31 21 15 19 22 18 4 11 25 28 32 3 23 12 5 14 20 16",
"output": "9 3 26 21 29 4 8 6 12 11 22 28 13 30 17 32 1 20 18 31 16 19 27 14 23 7 10 24 2 5 15 25"
},
{
"input": "65\n18 40 1 60 17 19 4 6 12 49 28 58 2 25 13 14 64 56 61 34 62 30 59 51 26 8 33 63 36 48 46 7 43 21 31 27 11 44 29 5 32 23 35 9 53 57 52 50 15 38 42 3 54 65 55 41 20 24 22 47 45 10 39 16 37",
"output": "3 13 52 7 40 8 32 26 44 62 37 9 15 16 49 64 5 1 6 57 34 59 42 58 14 25 36 11 39 22 35 41 27 20 43 29 65 50 63 2 56 51 33 38 61 31 60 30 10 48 24 47 45 53 55 18 46 12 23 4 19 21 28 17 54"
},
{
"input": "71\n35 50 55 58 25 32 26 40 63 34 44 53 24 18 37 7 64 27 56 65 1 19 2 43 42 14 57 47 22 13 59 61 39 67 30 45 54 38 33 48 6 5 3 69 36 21 41 4 16 46 20 17 15 12 10 70 68 23 60 31 52 29 66 28 51 49 62 11 8 9 71",
"output": "21 23 43 48 42 41 16 69 70 55 68 54 30 26 53 49 52 14 22 51 46 29 58 13 5 7 18 64 62 35 60 6 39 10 1 45 15 38 33 8 47 25 24 11 36 50 28 40 66 2 65 61 12 37 3 19 27 4 31 59 32 67 9 17 20 63 34 57 44 56 71"
},
{
"input": "74\n33 8 42 63 64 61 31 74 11 50 68 14 36 25 57 30 7 44 21 15 6 9 23 59 46 3 73 16 62 51 40 60 41 54 5 39 35 28 48 4 58 12 66 69 13 26 71 1 24 19 29 52 37 2 20 43 18 72 17 56 34 38 65 67 27 10 47 70 53 32 45 55 49 22",
"output": "48 54 26 40 35 21 17 2 22 66 9 42 45 12 20 28 59 57 50 55 19 74 23 49 14 46 65 38 51 16 7 70 1 61 37 13 53 62 36 31 33 3 56 18 71 25 67 39 73 10 30 52 69 34 72 60 15 41 24 32 6 29 4 5 63 43 64 11 44 68 47 58 27 8"
},
{
"input": "96\n78 10 82 46 38 91 77 69 2 27 58 80 79 44 59 41 6 31 76 11 42 48 51 37 19 87 43 25 52 32 1 39 63 29 21 65 53 74 92 16 15 95 90 83 30 73 71 5 50 17 96 33 86 60 67 64 20 26 61 40 55 88 94 93 9 72 47 57 14 45 22 3 54 68 13 24 4 7 56 81 89 70 49 8 84 28 18 62 35 36 75 23 66 85 34 12",
"output": "31 9 72 77 48 17 78 84 65 2 20 96 75 69 41 40 50 87 25 57 35 71 92 76 28 58 10 86 34 45 18 30 52 95 89 90 24 5 32 60 16 21 27 14 70 4 67 22 83 49 23 29 37 73 61 79 68 11 15 54 59 88 33 56 36 93 55 74 8 82 47 66 46 38 91 19 7 1 13 12 80 3 44 85 94 53 26 62 81 43 6 39 64 63 42 51"
},
{
"input": "7\n2 1 5 7 3 4 6",
"output": "2 1 5 6 3 7 4"
},
{
"input": "51\n8 33 37 2 16 22 24 30 4 9 5 15 27 3 18 39 31 26 10 17 46 41 25 14 6 1 29 48 36 20 51 49 21 43 19 13 38 50 47 34 11 23 28 12 42 7 32 40 44 45 35",
"output": "26 4 14 9 11 25 46 1 10 19 41 44 36 24 12 5 20 15 35 30 33 6 42 7 23 18 13 43 27 8 17 47 2 40 51 29 3 37 16 48 22 45 34 49 50 21 39 28 32 38 31"
},
{
"input": "27\n12 14 7 3 20 21 25 13 22 15 23 4 2 24 10 17 19 8 26 11 27 18 9 5 6 1 16",
"output": "26 13 4 12 24 25 3 18 23 15 20 1 8 2 10 27 16 22 17 5 6 9 11 14 7 19 21"
},
{
"input": "71\n51 13 20 48 54 23 24 64 14 62 71 67 57 53 3 30 55 43 33 25 39 40 66 6 46 18 5 19 61 16 32 68 70 41 60 44 29 49 27 69 50 38 10 17 45 56 9 21 26 63 28 35 7 59 1 65 2 15 8 11 12 34 37 47 58 22 31 4 36 42 52",
"output": "55 57 15 68 27 24 53 59 47 43 60 61 2 9 58 30 44 26 28 3 48 66 6 7 20 49 39 51 37 16 67 31 19 62 52 69 63 42 21 22 34 70 18 36 45 25 64 4 38 41 1 71 14 5 17 46 13 65 54 35 29 10 50 8 56 23 12 32 40 33 11"
},
{
"input": "9\n8 5 2 6 1 9 4 7 3",
"output": "5 3 9 7 2 4 8 1 6"
},
{
"input": "29\n10 24 11 5 26 25 2 9 22 15 8 14 29 21 4 1 23 17 3 12 13 16 18 28 19 20 7 6 27",
"output": "16 7 19 15 4 28 27 11 8 1 3 20 21 12 10 22 18 23 25 26 14 9 17 2 6 5 29 24 13"
},
{
"input": "60\n39 25 42 4 55 60 16 18 47 1 11 40 7 50 19 35 49 54 12 3 30 38 2 58 17 26 45 6 33 43 37 32 52 36 15 23 27 59 24 20 28 14 8 9 13 29 44 46 41 21 5 48 51 22 31 56 57 53 10 34",
"output": "10 23 20 4 51 28 13 43 44 59 11 19 45 42 35 7 25 8 15 40 50 54 36 39 2 26 37 41 46 21 55 32 29 60 16 34 31 22 1 12 49 3 30 47 27 48 9 52 17 14 53 33 58 18 5 56 57 24 38 6"
},
{
"input": "50\n37 45 22 5 12 21 28 24 18 47 20 25 8 50 14 2 34 43 11 16 49 41 48 1 19 31 39 46 32 23 15 42 3 35 38 30 44 26 10 9 40 36 7 17 33 4 27 6 13 29",
"output": "24 16 33 46 4 48 43 13 40 39 19 5 49 15 31 20 44 9 25 11 6 3 30 8 12 38 47 7 50 36 26 29 45 17 34 42 1 35 27 41 22 32 18 37 2 28 10 23 21 14"
},
{
"input": "30\n8 29 28 16 17 25 27 15 21 11 6 20 2 13 1 30 5 4 24 10 14 3 23 18 26 9 12 22 19 7",
"output": "15 13 22 18 17 11 30 1 26 20 10 27 14 21 8 4 5 24 29 12 9 28 23 19 6 25 7 3 2 16"
},
{
"input": "46\n15 2 44 43 38 19 31 42 4 37 29 30 24 45 27 41 8 20 33 7 35 3 18 46 36 26 1 28 21 40 16 22 32 11 14 13 12 9 25 39 10 6 23 17 5 34",
"output": "27 2 22 9 45 42 20 17 38 41 34 37 36 35 1 31 44 23 6 18 29 32 43 13 39 26 15 28 11 12 7 33 19 46 21 25 10 5 40 30 16 8 4 3 14 24"
},
{
"input": "9\n4 8 6 5 3 9 2 7 1",
"output": "9 7 5 1 4 3 8 2 6"
},
{
"input": "46\n31 30 33 23 45 7 36 8 11 3 32 39 41 20 1 28 6 27 18 24 17 5 16 37 26 13 22 14 2 38 15 46 9 4 19 21 12 44 10 35 25 34 42 43 40 29",
"output": "15 29 10 34 22 17 6 8 33 39 9 37 26 28 31 23 21 19 35 14 36 27 4 20 41 25 18 16 46 2 1 11 3 42 40 7 24 30 12 45 13 43 44 38 5 32"
},
{
"input": "66\n27 12 37 48 46 21 34 58 38 28 66 2 64 32 44 31 13 36 40 15 19 11 22 5 30 29 6 7 61 39 20 42 23 54 51 33 50 9 60 8 57 45 49 10 62 41 59 3 55 63 52 24 25 26 43 56 65 4 16 14 1 35 18 17 53 47",
"output": "61 12 48 58 24 27 28 40 38 44 22 2 17 60 20 59 64 63 21 31 6 23 33 52 53 54 1 10 26 25 16 14 36 7 62 18 3 9 30 19 46 32 55 15 42 5 66 4 43 37 35 51 65 34 49 56 41 8 47 39 29 45 50 13 57 11"
},
{
"input": "13\n3 12 9 2 8 5 13 4 11 1 10 7 6",
"output": "10 4 1 8 6 13 12 5 3 11 9 2 7"
},
{
"input": "80\n21 25 56 50 20 61 7 74 51 69 8 2 46 57 45 71 14 52 17 43 9 30 70 78 31 10 38 13 23 15 37 79 6 16 77 73 80 4 49 48 18 28 26 58 33 41 64 22 54 72 59 60 40 63 53 27 1 5 75 67 62 34 19 39 68 65 44 55 3 32 11 42 76 12 35 47 66 36 24 29",
"output": "57 12 69 38 58 33 7 11 21 26 71 74 28 17 30 34 19 41 63 5 1 48 29 79 2 43 56 42 80 22 25 70 45 62 75 78 31 27 64 53 46 72 20 67 15 13 76 40 39 4 9 18 55 49 68 3 14 44 51 52 6 61 54 47 66 77 60 65 10 23 16 50 36 8 59 73 35 24 32 37"
},
{
"input": "63\n9 49 53 25 40 46 43 51 54 22 58 16 23 26 10 47 5 27 2 8 61 59 19 35 63 56 28 20 34 4 62 38 6 55 36 31 57 15 29 33 1 48 50 37 7 30 18 42 32 52 12 41 14 21 45 11 24 17 39 13 44 60 3",
"output": "41 19 63 30 17 33 45 20 1 15 56 51 60 53 38 12 58 47 23 28 54 10 13 57 4 14 18 27 39 46 36 49 40 29 24 35 44 32 59 5 52 48 7 61 55 6 16 42 2 43 8 50 3 9 34 26 37 11 22 62 21 31 25"
},
{
"input": "26\n11 4 19 13 17 9 2 24 6 5 22 23 14 15 3 25 16 8 18 10 21 1 12 26 7 20",
"output": "22 7 15 2 10 9 25 18 6 20 1 23 4 13 14 17 5 19 3 26 21 11 12 8 16 24"
},
{
"input": "69\n40 22 11 66 4 27 31 29 64 53 37 55 51 2 7 36 18 52 6 1 30 21 17 20 14 9 59 62 49 68 3 50 65 57 44 5 67 46 33 13 34 15 24 48 63 58 38 25 41 35 16 54 32 10 60 61 39 12 69 8 23 45 26 47 56 43 28 19 42",
"output": "20 14 31 5 36 19 15 60 26 54 3 58 40 25 42 51 23 17 68 24 22 2 61 43 48 63 6 67 8 21 7 53 39 41 50 16 11 47 57 1 49 69 66 35 62 38 64 44 29 32 13 18 10 52 12 65 34 46 27 55 56 28 45 9 33 4 37 30 59"
},
{
"input": "6\n4 3 6 5 1 2",
"output": "5 6 2 1 4 3"
},
{
"input": "9\n7 8 5 3 1 4 2 9 6",
"output": "5 7 4 6 3 9 1 2 8"
},
{
"input": "41\n27 24 16 30 25 8 32 2 26 20 39 33 41 22 40 14 36 9 28 4 34 11 31 23 19 18 17 35 3 10 6 13 5 15 29 38 7 21 1 12 37",
"output": "39 8 29 20 33 31 37 6 18 30 22 40 32 16 34 3 27 26 25 10 38 14 24 2 5 9 1 19 35 4 23 7 12 21 28 17 41 36 11 15 13"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "20\n2 6 4 18 7 10 17 13 16 8 14 9 20 5 19 12 1 3 15 11",
"output": "17 1 18 3 14 2 5 10 12 6 20 16 8 11 19 9 7 4 15 13"
},
{
"input": "2\n2 1",
"output": "2 1"
},
{
"input": "60\n2 4 31 51 11 7 34 20 3 14 18 23 48 54 15 36 38 60 49 40 5 33 41 26 55 58 10 8 13 9 27 30 37 1 21 59 44 57 35 19 46 43 42 45 12 22 39 32 24 16 6 56 53 52 25 17 47 29 50 28",
"output": "34 1 9 2 21 51 6 28 30 27 5 45 29 10 15 50 56 11 40 8 35 46 12 49 55 24 31 60 58 32 3 48 22 7 39 16 33 17 47 20 23 43 42 37 44 41 57 13 19 59 4 54 53 14 25 52 38 26 36 18"
},
{
"input": "14\n14 6 3 12 11 2 7 1 10 9 8 5 4 13",
"output": "8 6 3 13 12 2 7 11 10 9 5 4 14 1"
},
{
"input": "81\n13 43 79 8 7 21 73 46 63 4 62 78 56 11 70 68 61 53 60 49 16 27 59 47 69 5 22 44 77 57 52 48 1 9 72 81 28 55 58 33 51 18 31 17 41 20 42 3 32 54 19 2 75 34 64 10 65 50 30 29 67 12 71 66 74 15 26 23 6 38 25 35 37 24 80 76 40 45 39 36 14",
"output": "33 52 48 10 26 69 5 4 34 56 14 62 1 81 66 21 44 42 51 46 6 27 68 74 71 67 22 37 60 59 43 49 40 54 72 80 73 70 79 77 45 47 2 28 78 8 24 32 20 58 41 31 18 50 38 13 30 39 23 19 17 11 9 55 57 64 61 16 25 15 63 35 7 65 53 76 29 12 3 75 36"
},
{
"input": "42\n41 11 10 8 21 37 32 19 31 25 1 15 36 5 6 27 4 3 13 7 16 17 2 23 34 24 38 28 12 20 30 42 18 26 39 35 33 40 9 14 22 29",
"output": "11 23 18 17 14 15 20 4 39 3 2 29 19 40 12 21 22 33 8 30 5 41 24 26 10 34 16 28 42 31 9 7 37 25 36 13 6 27 35 38 1 32"
},
{
"input": "97\n20 6 76 42 4 18 35 59 39 63 27 7 66 47 61 52 15 36 88 93 19 33 10 92 1 34 46 86 78 57 51 94 77 29 26 73 41 2 58 97 43 65 17 74 21 49 25 3 91 82 95 12 96 13 84 90 69 24 72 37 16 55 54 71 64 62 48 89 11 70 80 67 30 40 44 85 53 83 79 9 56 45 75 87 22 14 81 68 8 38 60 50 28 23 31 32 5",
"output": "25 38 48 5 97 2 12 89 80 23 69 52 54 86 17 61 43 6 21 1 45 85 94 58 47 35 11 93 34 73 95 96 22 26 7 18 60 90 9 74 37 4 41 75 82 27 14 67 46 92 31 16 77 63 62 81 30 39 8 91 15 66 10 65 42 13 72 88 57 70 64 59 36 44 83 3 33 29 79 71 87 50 78 55 76 28 84 19 68 56 49 24 20 32 51 53 40"
},
{
"input": "62\n15 27 46 6 8 51 14 56 23 48 42 49 52 22 20 31 29 12 47 3 62 34 37 35 32 57 19 25 5 60 61 38 18 10 11 55 45 53 17 30 9 36 4 50 41 16 44 28 40 59 24 1 13 39 26 7 33 58 2 43 21 54",
"output": "52 59 20 43 29 4 56 5 41 34 35 18 53 7 1 46 39 33 27 15 61 14 9 51 28 55 2 48 17 40 16 25 57 22 24 42 23 32 54 49 45 11 60 47 37 3 19 10 12 44 6 13 38 62 36 8 26 58 50 30 31 21"
},
{
"input": "61\n35 27 4 61 52 32 41 46 14 37 17 54 55 31 11 26 44 49 15 30 9 50 45 39 7 38 53 3 58 40 13 56 18 19 28 6 43 5 21 42 20 34 2 25 36 12 33 57 16 60 1 8 59 10 22 23 24 48 51 47 29",
"output": "51 43 28 3 38 36 25 52 21 54 15 46 31 9 19 49 11 33 34 41 39 55 56 57 44 16 2 35 61 20 14 6 47 42 1 45 10 26 24 30 7 40 37 17 23 8 60 58 18 22 59 5 27 12 13 32 48 29 53 50 4"
},
{
"input": "59\n31 26 36 15 17 19 10 53 11 34 13 46 55 9 44 7 8 37 32 52 47 25 51 22 35 39 41 4 43 24 5 27 20 57 6 38 3 28 21 40 50 18 14 56 33 45 12 2 49 59 54 29 16 48 42 58 1 30 23",
"output": "57 48 37 28 31 35 16 17 14 7 9 47 11 43 4 53 5 42 6 33 39 24 59 30 22 2 32 38 52 58 1 19 45 10 25 3 18 36 26 40 27 55 29 15 46 12 21 54 49 41 23 20 8 51 13 44 34 56 50"
},
{
"input": "10\n2 10 7 4 1 5 8 6 3 9",
"output": "5 1 9 4 6 8 3 7 10 2"
},
{
"input": "14\n14 2 1 8 6 12 11 10 9 7 3 4 5 13",
"output": "3 2 11 12 13 5 10 4 9 8 7 6 14 1"
},
{
"input": "43\n28 38 15 14 31 42 27 30 19 33 43 26 22 29 18 32 3 13 1 8 35 34 4 12 11 17 41 21 5 25 39 37 20 23 7 24 16 10 40 9 6 36 2",
"output": "19 43 17 23 29 41 35 20 40 38 25 24 18 4 3 37 26 15 9 33 28 13 34 36 30 12 7 1 14 8 5 16 10 22 21 42 32 2 31 39 27 6 11"
},
{
"input": "86\n39 11 20 31 28 76 29 64 35 21 41 71 12 82 5 37 80 73 38 26 79 75 23 15 59 45 47 6 3 62 50 49 51 22 2 65 86 60 70 42 74 17 1 30 55 44 8 66 81 27 57 77 43 13 54 32 72 46 48 56 14 34 78 52 36 85 24 19 69 83 25 61 7 4 84 33 63 58 18 40 68 10 67 9 16 53",
"output": "43 35 29 74 15 28 73 47 84 82 2 13 54 61 24 85 42 79 68 3 10 34 23 67 71 20 50 5 7 44 4 56 76 62 9 65 16 19 1 80 11 40 53 46 26 58 27 59 32 31 33 64 86 55 45 60 51 78 25 38 72 30 77 8 36 48 83 81 69 39 12 57 18 41 22 6 52 63 21 17 49 14 70 75 66 37"
},
{
"input": "99\n65 78 56 98 33 24 61 40 29 93 1 64 57 22 25 52 67 95 50 3 31 15 90 68 71 83 38 36 6 46 89 26 4 87 14 88 72 37 23 43 63 12 80 96 5 34 73 86 9 48 92 62 99 10 16 20 66 27 28 2 82 70 30 94 49 8 84 69 18 60 58 59 44 39 21 7 91 76 54 19 75 85 74 47 55 32 97 77 51 13 35 79 45 42 11 41 17 81 53",
"output": "11 60 20 33 45 29 76 66 49 54 95 42 90 35 22 55 97 69 80 56 75 14 39 6 15 32 58 59 9 63 21 86 5 46 91 28 38 27 74 8 96 94 40 73 93 30 84 50 65 19 89 16 99 79 85 3 13 71 72 70 7 52 41 12 1 57 17 24 68 62 25 37 47 83 81 78 88 2 92 43 98 61 26 67 82 48 34 36 31 23 77 51 10 64 18 44 87 4 53"
},
{
"input": "100\n42 23 48 88 36 6 18 70 96 1 34 40 46 22 39 55 85 93 45 67 71 75 59 9 21 3 86 63 65 68 20 38 73 31 84 90 50 51 56 95 72 33 49 19 83 76 54 74 100 30 17 98 15 94 4 97 5 99 81 27 92 32 89 12 13 91 87 29 60 11 52 43 35 58 10 25 16 80 28 2 44 61 8 82 66 69 41 24 57 62 78 37 79 77 53 7 14 47 26 64",
"output": "10 80 26 55 57 6 96 83 24 75 70 64 65 97 53 77 51 7 44 31 25 14 2 88 76 99 60 79 68 50 34 62 42 11 73 5 92 32 15 12 87 1 72 81 19 13 98 3 43 37 38 71 95 47 16 39 89 74 23 69 82 90 28 100 29 85 20 30 86 8 21 41 33 48 22 46 94 91 93 78 59 84 45 35 17 27 67 4 63 36 66 61 18 54 40 9 56 52 58 49"
},
{
"input": "99\n8 68 94 75 71 60 57 58 6 11 5 48 65 41 49 12 46 72 95 59 13 70 74 7 84 62 17 36 55 76 38 79 2 85 23 10 32 99 87 50 83 28 54 91 53 51 1 3 97 81 21 89 93 78 61 26 82 96 4 98 25 40 31 44 24 47 30 52 14 16 39 27 9 29 45 18 67 63 37 43 90 66 19 69 88 22 92 77 34 42 73 80 56 64 20 35 15 33 86",
"output": "47 33 48 59 11 9 24 1 73 36 10 16 21 69 97 70 27 76 83 95 51 86 35 65 61 56 72 42 74 67 63 37 98 89 96 28 79 31 71 62 14 90 80 64 75 17 66 12 15 40 46 68 45 43 29 93 7 8 20 6 55 26 78 94 13 82 77 2 84 22 5 18 91 23 4 30 88 54 32 92 50 57 41 25 34 99 39 85 52 81 44 87 53 3 19 58 49 60 38"
},
{
"input": "99\n12 99 88 13 7 19 74 47 23 90 16 29 26 11 58 60 64 98 37 18 82 67 72 46 51 85 17 92 87 20 77 36 78 71 57 35 80 54 73 15 14 62 97 45 31 79 94 56 76 96 28 63 8 44 38 86 49 2 52 66 61 59 10 43 55 50 22 34 83 53 95 40 81 21 30 42 27 3 5 41 1 70 69 25 93 48 65 6 24 89 91 33 39 68 9 4 32 84 75",
"output": "81 58 78 96 79 88 5 53 95 63 14 1 4 41 40 11 27 20 6 30 74 67 9 89 84 13 77 51 12 75 45 97 92 68 36 32 19 55 93 72 80 76 64 54 44 24 8 86 57 66 25 59 70 38 65 48 35 15 62 16 61 42 52 17 87 60 22 94 83 82 34 23 39 7 99 49 31 33 46 37 73 21 69 98 26 56 29 3 90 10 91 28 85 47 71 50 43 18 2"
},
{
"input": "99\n20 79 26 75 99 69 98 47 93 62 18 42 43 38 90 66 67 8 13 84 76 58 81 60 64 46 56 23 78 17 86 36 19 52 85 39 48 27 96 49 37 95 5 31 10 24 12 1 80 35 92 33 16 68 57 54 32 29 45 88 72 77 4 87 97 89 59 3 21 22 61 94 83 15 44 34 70 91 55 9 51 50 73 11 14 6 40 7 63 25 2 82 41 65 28 74 71 30 53",
"output": "48 91 68 63 43 86 88 18 80 45 84 47 19 85 74 53 30 11 33 1 69 70 28 46 90 3 38 95 58 98 44 57 52 76 50 32 41 14 36 87 93 12 13 75 59 26 8 37 40 82 81 34 99 56 79 27 55 22 67 24 71 10 89 25 94 16 17 54 6 77 97 61 83 96 4 21 62 29 2 49 23 92 73 20 35 31 64 60 66 15 78 51 9 72 42 39 65 7 5"
},
{
"input": "99\n74 20 9 1 60 85 65 13 4 25 40 99 5 53 64 3 36 31 73 44 55 50 45 63 98 51 68 6 47 37 71 82 88 34 84 18 19 12 93 58 86 7 11 46 90 17 33 27 81 69 42 59 56 32 95 52 76 61 96 62 78 43 66 21 49 97 75 14 41 72 89 16 30 79 22 23 15 83 91 38 48 2 87 26 28 80 94 70 54 92 57 10 8 35 67 77 29 24 39",
"output": "4 82 16 9 13 28 42 93 3 92 43 38 8 68 77 72 46 36 37 2 64 75 76 98 10 84 48 85 97 73 18 54 47 34 94 17 30 80 99 11 69 51 62 20 23 44 29 81 65 22 26 56 14 89 21 53 91 40 52 5 58 60 24 15 7 63 95 27 50 88 31 70 19 1 67 57 96 61 74 86 49 32 78 35 6 41 83 33 71 45 79 90 39 87 55 59 66 25 12"
},
{
"input": "99\n50 94 2 18 69 90 59 83 75 68 77 97 39 78 25 7 16 9 49 4 42 89 44 48 17 96 61 70 3 10 5 81 56 57 88 6 98 1 46 67 92 37 11 30 85 41 8 36 51 29 20 71 19 79 74 93 43 34 55 40 38 21 64 63 32 24 72 14 12 86 82 15 65 23 66 22 28 53 13 26 95 99 91 52 76 27 60 45 47 33 73 84 31 35 54 80 58 62 87",
"output": "38 3 29 20 31 36 16 47 18 30 43 69 79 68 72 17 25 4 53 51 62 76 74 66 15 80 86 77 50 44 93 65 90 58 94 48 42 61 13 60 46 21 57 23 88 39 89 24 19 1 49 84 78 95 59 33 34 97 7 87 27 98 64 63 73 75 40 10 5 28 52 67 91 55 9 85 11 14 54 96 32 71 8 92 45 70 99 35 22 6 83 41 56 2 81 26 12 37 82"
},
{
"input": "99\n19 93 14 34 39 37 33 15 52 88 7 43 69 27 9 77 94 31 48 22 63 70 79 17 50 6 81 8 76 58 23 74 86 11 57 62 41 87 75 51 12 18 68 56 95 3 80 83 84 29 24 61 71 78 59 96 20 85 90 28 45 36 38 97 1 49 40 98 44 67 13 73 72 91 47 10 30 54 35 42 4 2 92 26 64 60 53 21 5 82 46 32 55 66 16 89 99 65 25",
"output": "65 82 46 81 89 26 11 28 15 76 34 41 71 3 8 95 24 42 1 57 88 20 31 51 99 84 14 60 50 77 18 92 7 4 79 62 6 63 5 67 37 80 12 69 61 91 75 19 66 25 40 9 87 78 93 44 35 30 55 86 52 36 21 85 98 94 70 43 13 22 53 73 72 32 39 29 16 54 23 47 27 90 48 49 58 33 38 10 96 59 74 83 2 17 45 56 64 68 97"
},
{
"input": "99\n86 25 50 51 62 39 41 67 44 20 45 14 80 88 66 7 36 59 13 84 78 58 96 75 2 43 48 47 69 12 19 98 22 38 28 55 11 76 68 46 53 70 85 34 16 33 91 30 8 40 74 60 94 82 87 32 37 4 5 10 89 73 90 29 35 26 23 57 27 65 24 3 9 83 77 72 6 31 15 92 93 79 64 18 63 42 56 1 52 97 17 81 71 21 49 99 54 95 61",
"output": "88 25 72 58 59 77 16 49 73 60 37 30 19 12 79 45 91 84 31 10 94 33 67 71 2 66 69 35 64 48 78 56 46 44 65 17 57 34 6 50 7 86 26 9 11 40 28 27 95 3 4 89 41 97 36 87 68 22 18 52 99 5 85 83 70 15 8 39 29 42 93 76 62 51 24 38 75 21 82 13 92 54 74 20 43 1 55 14 61 63 47 80 81 53 98 23 90 32 96"
},
{
"input": "100\n66 44 99 15 43 79 28 33 88 90 49 68 82 38 9 74 4 58 29 81 31 94 10 42 89 21 63 40 62 61 18 6 84 72 48 25 67 69 71 85 98 34 83 70 65 78 91 77 93 41 23 24 87 11 55 12 59 73 36 97 7 14 26 39 30 27 45 20 50 17 53 2 57 47 95 56 75 19 37 96 16 35 8 3 76 60 13 86 5 32 64 80 46 51 54 100 1 22 52 92",
"output": "97 72 84 17 89 32 61 83 15 23 54 56 87 62 4 81 70 31 78 68 26 98 51 52 36 63 66 7 19 65 21 90 8 42 82 59 79 14 64 28 50 24 5 2 67 93 74 35 11 69 94 99 71 95 55 76 73 18 57 86 30 29 27 91 45 1 37 12 38 44 39 34 58 16 77 85 48 46 6 92 20 13 43 33 40 88 53 9 25 10 47 100 49 22 75 80 60 41 3 96"
},
{
"input": "99\n3 73 32 37 25 15 93 63 85 8 91 78 80 5 39 48 46 7 83 70 23 96 9 29 77 53 30 20 56 50 13 45 21 76 87 99 65 31 16 18 14 72 51 28 43 2 81 34 38 40 66 54 74 26 71 4 61 17 58 24 22 33 49 36 42 11 12 55 60 27 62 90 79 92 94 68 1 52 84 41 86 35 69 75 47 10 64 88 97 98 67 19 89 95 59 82 57 44 6",
"output": "77 46 1 56 14 99 18 10 23 86 66 67 31 41 6 39 58 40 92 28 33 61 21 60 5 54 70 44 24 27 38 3 62 48 82 64 4 49 15 50 80 65 45 98 32 17 85 16 63 30 43 78 26 52 68 29 97 59 95 69 57 71 8 87 37 51 91 76 83 20 55 42 2 53 84 34 25 12 73 13 47 96 19 79 9 81 35 88 93 72 11 74 7 75 94 22 89 90 36"
},
{
"input": "100\n100 99 98 97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 80 79 78 77 76 75 74 73 72 71 70 69 68 67 66 65 64 63 62 61 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1",
"output": "100 99 98 97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 80 79 78 77 76 75 74 73 72 71 70 69 68 67 66 65 64 63 62 61 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1"
}
] | 1,691,706,723
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 92
| 0
|
n = int(input())
lis = input()
lis = list(map(int,lis.split()))
res = []
for i in range(n):
for j in range(n):
if lis[j] == i+1:
res.append(j+1)
break
print(res)
|
Title: Presents
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Petya very much likes gifts. Recently he has received a new laptop as a New Year gift from his mother. He immediately decided to give it to somebody else as what can be more pleasant than giving somebody gifts. And on this occasion he organized a New Year party at his place and invited *n* his friends there.
If there's one thing Petya likes more that receiving gifts, that's watching others giving gifts to somebody else. Thus, he safely hid the laptop until the next New Year and made up his mind to watch his friends exchanging gifts while he does not participate in the process. He numbered all his friends with integers from 1 to *n*. Petya remembered that a friend number *i* gave a gift to a friend number *p**i*. He also remembered that each of his friends received exactly one gift.
Now Petya wants to know for each friend *i* the number of a friend who has given him a gift.
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the quantity of friends Petya invited to the party. The second line contains *n* space-separated integers: the *i*-th number is *p**i* — the number of a friend who gave a gift to friend number *i*. It is guaranteed that each friend received exactly one gift. It is possible that some friends do not share Petya's ideas of giving gifts to somebody else. Those friends gave the gifts to themselves.
Output Specification:
Print *n* space-separated integers: the *i*-th number should equal the number of the friend who gave a gift to friend number *i*.
Demo Input:
['4\n2 3 4 1\n', '3\n1 3 2\n', '2\n1 2\n']
Demo Output:
['4 1 2 3\n', '1 3 2\n', '1 2\n']
Note:
none
|
```python
n = int(input())
lis = input()
lis = list(map(int,lis.split()))
res = []
for i in range(n):
for j in range(n):
if lis[j] == i+1:
res.append(j+1)
break
print(res)
```
| 0
|
|
202
|
A
|
LLPS
|
PROGRAMMING
| 800
|
[
"binary search",
"bitmasks",
"brute force",
"greedy",
"implementation",
"strings"
] | null | null |
This problem's actual name, "Lexicographically Largest Palindromic Subsequence" is too long to fit into the page headline.
You are given string *s* consisting of lowercase English letters only. Find its lexicographically largest palindromic subsequence.
We'll call a non-empty string *s*[*p*1*p*2... *p**k*] = *s**p*1*s**p*2... *s**p**k* (1 <=≤<= *p*1<=<<=*p*2<=<<=...<=<<=*p**k* <=≤<= |*s*|) a subsequence of string *s* = *s*1*s*2... *s*|*s*|, where |*s*| is the length of string *s*. For example, strings "abcb", "b" and "abacaba" are subsequences of string "abacaba".
String *x* = *x*1*x*2... *x*|*x*| is lexicographically larger than string *y* = *y*1*y*2... *y*|*y*| if either |*x*| > |*y*| and *x*1<==<=*y*1, *x*2<==<=*y*2, ...,<=*x*|*y*|<==<=*y*|*y*|, or there exists such number *r* (*r*<=<<=|*x*|, *r*<=<<=|*y*|) that *x*1<==<=*y*1, *x*2<==<=*y*2, ..., *x**r*<==<=*y**r* and *x**r*<=<=+<=<=1<=><=*y**r*<=<=+<=<=1. Characters in the strings are compared according to their ASCII codes. For example, string "ranger" is lexicographically larger than string "racecar" and string "poster" is lexicographically larger than string "post".
String *s* = *s*1*s*2... *s*|*s*| is a palindrome if it matches string *rev*(*s*) = *s*|*s*|*s*|*s*|<=-<=1... *s*1. In other words, a string is a palindrome if it reads the same way from left to right and from right to left. For example, palindromic strings are "racecar", "refer" and "z".
|
The only input line contains a non-empty string *s* consisting of lowercase English letters only. Its length does not exceed 10.
|
Print the lexicographically largest palindromic subsequence of string *s*.
|
[
"radar\n",
"bowwowwow\n",
"codeforces\n",
"mississipp\n"
] |
[
"rr\n",
"wwwww\n",
"s\n",
"ssss\n"
] |
Among all distinct subsequences of string "radar" the following ones are palindromes: "a", "d", "r", "aa", "rr", "ada", "rar", "rdr", "raar" and "radar". The lexicographically largest of them is "rr".
| 500
|
[
{
"input": "radar",
"output": "rr"
},
{
"input": "bowwowwow",
"output": "wwwww"
},
{
"input": "codeforces",
"output": "s"
},
{
"input": "mississipp",
"output": "ssss"
},
{
"input": "tourist",
"output": "u"
},
{
"input": "romka",
"output": "r"
},
{
"input": "helloworld",
"output": "w"
},
{
"input": "zzzzzzzazz",
"output": "zzzzzzzzz"
},
{
"input": "testcase",
"output": "tt"
},
{
"input": "hahahahaha",
"output": "hhhhh"
},
{
"input": "abbbbbbbbb",
"output": "bbbbbbbbb"
},
{
"input": "zaz",
"output": "zz"
},
{
"input": "aza",
"output": "z"
},
{
"input": "dcbaedcba",
"output": "e"
},
{
"input": "abcdeabcd",
"output": "e"
},
{
"input": "edcbabcde",
"output": "ee"
},
{
"input": "aaaaaaaaab",
"output": "b"
},
{
"input": "testzzzzzz",
"output": "zzzzzz"
},
{
"input": "zzzzzzwait",
"output": "zzzzzz"
},
{
"input": "rrrrrqponm",
"output": "rrrrr"
},
{
"input": "zzyzyy",
"output": "zzz"
},
{
"input": "aababb",
"output": "bbb"
},
{
"input": "zanzibar",
"output": "zz"
},
{
"input": "hhgfedcbaa",
"output": "hh"
},
{
"input": "aabcdefghh",
"output": "hh"
},
{
"input": "aruaru",
"output": "uu"
},
{
"input": "uraura",
"output": "uu"
},
{
"input": "aru",
"output": "u"
},
{
"input": "aburvabur",
"output": "v"
},
{
"input": "ura",
"output": "u"
},
{
"input": "eurottat",
"output": "u"
},
{
"input": "referee",
"output": "rr"
},
{
"input": "joking",
"output": "o"
},
{
"input": "seriously",
"output": "y"
},
{
"input": "sets",
"output": "t"
},
{
"input": "test",
"output": "tt"
},
{
"input": "klmgameklm",
"output": "mmm"
},
{
"input": "dfkjafdkdd",
"output": "kk"
},
{
"input": "zzzzzzzzzz",
"output": "zzzzzzzzzz"
},
{
"input": "aaaaaaaaaa",
"output": "aaaaaaaaaa"
},
{
"input": "buzz",
"output": "zz"
},
{
"input": "b",
"output": "b"
},
{
"input": "y",
"output": "y"
},
{
"input": "yy",
"output": "yy"
},
{
"input": "zz",
"output": "zz"
},
{
"input": "bb",
"output": "bb"
},
{
"input": "aa",
"output": "aa"
},
{
"input": "a",
"output": "a"
},
{
"input": "z",
"output": "z"
},
{
"input": "zz",
"output": "zz"
},
{
"input": "zzazazzzaz",
"output": "zzzzzzz"
},
{
"input": "hellhellhe",
"output": "llll"
},
{
"input": "hellohello",
"output": "oo"
},
{
"input": "refer",
"output": "rr"
}
] | 1,659,096,459
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 54
| 92
| 0
|
x = input().strip()
a = max(x)
print(a * (x.count(a)))
|
Title: LLPS
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
This problem's actual name, "Lexicographically Largest Palindromic Subsequence" is too long to fit into the page headline.
You are given string *s* consisting of lowercase English letters only. Find its lexicographically largest palindromic subsequence.
We'll call a non-empty string *s*[*p*1*p*2... *p**k*] = *s**p*1*s**p*2... *s**p**k* (1 <=≤<= *p*1<=<<=*p*2<=<<=...<=<<=*p**k* <=≤<= |*s*|) a subsequence of string *s* = *s*1*s*2... *s*|*s*|, where |*s*| is the length of string *s*. For example, strings "abcb", "b" and "abacaba" are subsequences of string "abacaba".
String *x* = *x*1*x*2... *x*|*x*| is lexicographically larger than string *y* = *y*1*y*2... *y*|*y*| if either |*x*| > |*y*| and *x*1<==<=*y*1, *x*2<==<=*y*2, ...,<=*x*|*y*|<==<=*y*|*y*|, or there exists such number *r* (*r*<=<<=|*x*|, *r*<=<<=|*y*|) that *x*1<==<=*y*1, *x*2<==<=*y*2, ..., *x**r*<==<=*y**r* and *x**r*<=<=+<=<=1<=><=*y**r*<=<=+<=<=1. Characters in the strings are compared according to their ASCII codes. For example, string "ranger" is lexicographically larger than string "racecar" and string "poster" is lexicographically larger than string "post".
String *s* = *s*1*s*2... *s*|*s*| is a palindrome if it matches string *rev*(*s*) = *s*|*s*|*s*|*s*|<=-<=1... *s*1. In other words, a string is a palindrome if it reads the same way from left to right and from right to left. For example, palindromic strings are "racecar", "refer" and "z".
Input Specification:
The only input line contains a non-empty string *s* consisting of lowercase English letters only. Its length does not exceed 10.
Output Specification:
Print the lexicographically largest palindromic subsequence of string *s*.
Demo Input:
['radar\n', 'bowwowwow\n', 'codeforces\n', 'mississipp\n']
Demo Output:
['rr\n', 'wwwww\n', 's\n', 'ssss\n']
Note:
Among all distinct subsequences of string "radar" the following ones are palindromes: "a", "d", "r", "aa", "rr", "ada", "rar", "rdr", "raar" and "radar". The lexicographically largest of them is "rr".
|
```python
x = input().strip()
a = max(x)
print(a * (x.count(a)))
```
| 3
|
|
71
|
A
|
Way Too Long Words
|
PROGRAMMING
| 800
|
[
"strings"
] |
A. Way Too Long Words
|
1
|
256
|
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
|
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
|
[
"4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n"
] |
[
"word\nl10n\ni18n\np43s\n"
] |
none
| 500
|
[
{
"input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis",
"output": "word\nl10n\ni18n\np43s"
},
{
"input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm",
"output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m"
},
{
"input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd",
"output": "n20n\nj27b\nk15d"
},
{
"input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt",
"output": "t41t"
},
{
"input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings",
"output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz",
"output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz"
},
{
"input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij",
"output": "a98j"
},
{
"input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco",
"output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco"
},
{
"input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt",
"output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t"
},
{
"input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr",
"output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr"
}
] | 1,670,179,290
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 20
| 31
| 0
|
amount = int(input())
for x in range(amount):
x = input()
if len(x) > 10:
y = int(len(x)) - 2
print(x[0],y,x[-1], sep = '')
else:
if x.isdigit():
print()
else:
print(x)
|
Title: Way Too Long Words
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output Specification:
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
Demo Input:
['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n']
Demo Output:
['word\nl10n\ni18n\np43s\n']
Note:
none
|
```python
amount = int(input())
for x in range(amount):
x = input()
if len(x) > 10:
y = int(len(x)) - 2
print(x[0],y,x[-1], sep = '')
else:
if x.isdigit():
print()
else:
print(x)
```
| 3.9845
|
602
|
B
|
Approximating a Constant Range
|
PROGRAMMING
| 1,400
|
[
"dp",
"implementation",
"two pointers"
] | null | null |
When Xellos was doing a practice course in university, he once had to measure the intensity of an effect that slowly approached equilibrium. A good way to determine the equilibrium intensity would be choosing a sufficiently large number of consecutive data points that seems as constant as possible and taking their average. Of course, with the usual sizes of data, it's nothing challenging — but why not make a similar programming contest problem while we're at it?
You're given a sequence of *n* data points *a*1,<=...,<=*a**n*. There aren't any big jumps between consecutive data points — for each 1<=≤<=*i*<=<<=*n*, it's guaranteed that |*a**i*<=+<=1<=-<=*a**i*|<=≤<=1.
A range [*l*,<=*r*] of data points is said to be almost constant if the difference between the largest and the smallest value in that range is at most 1. Formally, let *M* be the maximum and *m* the minimum value of *a**i* for *l*<=≤<=*i*<=≤<=*r*; the range [*l*,<=*r*] is almost constant if *M*<=-<=*m*<=≤<=1.
Find the length of the longest almost constant range.
|
The first line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100<=000) — the number of data points.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100<=000).
|
Print a single number — the maximum length of an almost constant range of the given sequence.
|
[
"5\n1 2 3 3 2\n",
"11\n5 4 5 5 6 7 8 8 8 7 6\n"
] |
[
"4\n",
"5\n"
] |
In the first sample, the longest almost constant range is [2, 5]; its length (the number of data points in it) is 4.
In the second sample, there are three almost constant ranges of length 4: [1, 4], [6, 9] and [7, 10]; the only almost constant range of the maximum length 5 is [6, 10].
| 1,000
|
[
{
"input": "5\n1 2 3 3 2",
"output": "4"
},
{
"input": "11\n5 4 5 5 6 7 8 8 8 7 6",
"output": "5"
},
{
"input": "2\n3 2",
"output": "2"
},
{
"input": "4\n1001 1000 1000 1001",
"output": "4"
},
{
"input": "4\n1 1 2 3",
"output": "3"
},
{
"input": "3\n1 2 1",
"output": "3"
},
{
"input": "3\n1 2 3",
"output": "2"
},
{
"input": "18\n10 9 8 7 6 5 4 3 2 1 2 3 4 5 6 7 8 9",
"output": "3"
},
{
"input": "3\n1 2 2",
"output": "3"
},
{
"input": "4\n10 9 10 9",
"output": "4"
},
{
"input": "4\n4 3 2 3",
"output": "3"
},
{
"input": "4\n8 8 7 7",
"output": "4"
},
{
"input": "3\n99998 99999 100000",
"output": "2"
},
{
"input": "3\n100000 99999 99998",
"output": "2"
},
{
"input": "3\n1 1 1",
"output": "3"
},
{
"input": "2\n99999 100000",
"output": "2"
},
{
"input": "2\n100000 100000",
"output": "2"
},
{
"input": "2\n1 1",
"output": "2"
},
{
"input": "15\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000",
"output": "15"
}
] | 1,620,540,873
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 107
| 156
| 30,822,400
|
n = int(input())
arr = list(map(int, input().split()))
ans = 1
pdiff = e = g = 0
for i in range(1, n):
diff = arr[i] - arr[i - 1]
if diff:
if diff == pdiff:
g = e + 1
e = i - 1
pdiff = diff
ans = max(ans, i - g + 1)
print(ans)
|
Title: Approximating a Constant Range
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
When Xellos was doing a practice course in university, he once had to measure the intensity of an effect that slowly approached equilibrium. A good way to determine the equilibrium intensity would be choosing a sufficiently large number of consecutive data points that seems as constant as possible and taking their average. Of course, with the usual sizes of data, it's nothing challenging — but why not make a similar programming contest problem while we're at it?
You're given a sequence of *n* data points *a*1,<=...,<=*a**n*. There aren't any big jumps between consecutive data points — for each 1<=≤<=*i*<=<<=*n*, it's guaranteed that |*a**i*<=+<=1<=-<=*a**i*|<=≤<=1.
A range [*l*,<=*r*] of data points is said to be almost constant if the difference between the largest and the smallest value in that range is at most 1. Formally, let *M* be the maximum and *m* the minimum value of *a**i* for *l*<=≤<=*i*<=≤<=*r*; the range [*l*,<=*r*] is almost constant if *M*<=-<=*m*<=≤<=1.
Find the length of the longest almost constant range.
Input Specification:
The first line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100<=000) — the number of data points.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100<=000).
Output Specification:
Print a single number — the maximum length of an almost constant range of the given sequence.
Demo Input:
['5\n1 2 3 3 2\n', '11\n5 4 5 5 6 7 8 8 8 7 6\n']
Demo Output:
['4\n', '5\n']
Note:
In the first sample, the longest almost constant range is [2, 5]; its length (the number of data points in it) is 4.
In the second sample, there are three almost constant ranges of length 4: [1, 4], [6, 9] and [7, 10]; the only almost constant range of the maximum length 5 is [6, 10].
|
```python
n = int(input())
arr = list(map(int, input().split()))
ans = 1
pdiff = e = g = 0
for i in range(1, n):
diff = arr[i] - arr[i - 1]
if diff:
if diff == pdiff:
g = e + 1
e = i - 1
pdiff = diff
ans = max(ans, i - g + 1)
print(ans)
```
| 3
|
|
832
|
A
|
Sasha and Sticks
|
PROGRAMMING
| 800
|
[
"games",
"math"
] | null | null |
It's one more school day now. Sasha doesn't like classes and is always bored at them. So, each day he invents some game and plays in it alone or with friends.
Today he invented one simple game to play with Lena, with whom he shares a desk. The rules are simple. Sasha draws *n* sticks in a row. After that the players take turns crossing out exactly *k* sticks from left or right in each turn. Sasha moves first, because he is the inventor of the game. If there are less than *k* sticks on the paper before some turn, the game ends. Sasha wins if he makes strictly more moves than Lena. Sasha wants to know the result of the game before playing, you are to help him.
|
The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=1018, *k*<=≤<=*n*) — the number of sticks drawn by Sasha and the number *k* — the number of sticks to be crossed out on each turn.
|
If Sasha wins, print "YES" (without quotes), otherwise print "NO" (without quotes).
You can print each letter in arbitrary case (upper of lower).
|
[
"1 1\n",
"10 4\n"
] |
[
"YES\n",
"NO\n"
] |
In the first example Sasha crosses out 1 stick, and then there are no sticks. So Lena can't make a move, and Sasha wins.
In the second example Sasha crosses out 4 sticks, then Lena crosses out 4 sticks, and after that there are only 2 sticks left. Sasha can't make a move. The players make equal number of moves, so Sasha doesn't win.
| 500
|
[
{
"input": "1 1",
"output": "YES"
},
{
"input": "10 4",
"output": "NO"
},
{
"input": "251656215122324104 164397544865601257",
"output": "YES"
},
{
"input": "963577813436662285 206326039287271924",
"output": "NO"
},
{
"input": "1000000000000000000 1",
"output": "NO"
},
{
"input": "253308697183523656 25332878317796706",
"output": "YES"
},
{
"input": "669038685745448997 501718093668307460",
"output": "YES"
},
{
"input": "116453141993601660 87060381463547965",
"output": "YES"
},
{
"input": "766959657 370931668",
"output": "NO"
},
{
"input": "255787422422806632 146884995820359999",
"output": "YES"
},
{
"input": "502007866464507926 71266379084204128",
"output": "YES"
},
{
"input": "257439908778973480 64157133126869976",
"output": "NO"
},
{
"input": "232709385 91708542",
"output": "NO"
},
{
"input": "252482458300407528 89907711721009125",
"output": "NO"
},
{
"input": "6 2",
"output": "YES"
},
{
"input": "6 3",
"output": "NO"
},
{
"input": "6 4",
"output": "YES"
},
{
"input": "6 5",
"output": "YES"
},
{
"input": "6 6",
"output": "YES"
},
{
"input": "258266151957056904 30153168463725364",
"output": "NO"
},
{
"input": "83504367885565783 52285355047292458",
"output": "YES"
},
{
"input": "545668929424440387 508692735816921376",
"output": "YES"
},
{
"input": "547321411485639939 36665750286082900",
"output": "NO"
},
{
"input": "548973893546839491 183137237979822911",
"output": "NO"
},
{
"input": "544068082 193116851",
"output": "NO"
},
{
"input": "871412474 749817171",
"output": "YES"
},
{
"input": "999999999 1247",
"output": "NO"
},
{
"input": "851941088 712987048",
"output": "YES"
},
{
"input": "559922900 418944886",
"output": "YES"
},
{
"input": "293908937 37520518",
"output": "YES"
},
{
"input": "650075786 130049650",
"output": "NO"
},
{
"input": "1000000000 1000000000",
"output": "YES"
},
{
"input": "548147654663723363 107422751713800746",
"output": "YES"
},
{
"input": "828159210 131819483",
"output": "NO"
},
{
"input": "6242634 4110365",
"output": "YES"
},
{
"input": "458601973 245084155",
"output": "YES"
},
{
"input": "349593257 18089089",
"output": "YES"
},
{
"input": "814768821 312514745",
"output": "NO"
},
{
"input": "697884949 626323363",
"output": "YES"
},
{
"input": "667011589 54866795",
"output": "NO"
},
{
"input": "1000000000000000000 2",
"output": "NO"
},
{
"input": "1000000000000000000 3",
"output": "YES"
},
{
"input": "1000000000000000000 4",
"output": "NO"
},
{
"input": "999999999999999 1",
"output": "YES"
},
{
"input": "17 4",
"output": "NO"
},
{
"input": "2 2",
"output": "YES"
},
{
"input": "1000000000000000 2",
"output": "NO"
},
{
"input": "12 4",
"output": "YES"
},
{
"input": "6 1",
"output": "NO"
},
{
"input": "2 1",
"output": "NO"
},
{
"input": "10000000005 1",
"output": "YES"
},
{
"input": "10000000000000009 2",
"output": "NO"
},
{
"input": "12457895452123 1",
"output": "YES"
},
{
"input": "999999999999999999 9",
"output": "YES"
},
{
"input": "1000000000000 3",
"output": "YES"
},
{
"input": "13099714659575475 6549849616514894",
"output": "NO"
},
{
"input": "100000000000000001 1",
"output": "YES"
},
{
"input": "825175814723458 324",
"output": "YES"
},
{
"input": "20 4",
"output": "YES"
},
{
"input": "100000176877 4",
"output": "YES"
},
{
"input": "100000 3",
"output": "YES"
},
{
"input": "946744073709551614 10",
"output": "YES"
}
] | 1,632,033,967
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 2
| 62
| 20,172,800
|
a, b = list(map(int, input().split()))
if b // a % 2 == 0:
print('NO')
else:
print('YES')
|
Title: Sasha and Sticks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It's one more school day now. Sasha doesn't like classes and is always bored at them. So, each day he invents some game and plays in it alone or with friends.
Today he invented one simple game to play with Lena, with whom he shares a desk. The rules are simple. Sasha draws *n* sticks in a row. After that the players take turns crossing out exactly *k* sticks from left or right in each turn. Sasha moves first, because he is the inventor of the game. If there are less than *k* sticks on the paper before some turn, the game ends. Sasha wins if he makes strictly more moves than Lena. Sasha wants to know the result of the game before playing, you are to help him.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=1018, *k*<=≤<=*n*) — the number of sticks drawn by Sasha and the number *k* — the number of sticks to be crossed out on each turn.
Output Specification:
If Sasha wins, print "YES" (without quotes), otherwise print "NO" (without quotes).
You can print each letter in arbitrary case (upper of lower).
Demo Input:
['1 1\n', '10 4\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first example Sasha crosses out 1 stick, and then there are no sticks. So Lena can't make a move, and Sasha wins.
In the second example Sasha crosses out 4 sticks, then Lena crosses out 4 sticks, and after that there are only 2 sticks left. Sasha can't make a move. The players make equal number of moves, so Sasha doesn't win.
|
```python
a, b = list(map(int, input().split()))
if b // a % 2 == 0:
print('NO')
else:
print('YES')
```
| 0
|
|
821
|
A
|
Okabe and Future Gadget Laboratory
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Okabe needs to renovate the Future Gadget Laboratory after he tried doing some crazy experiments! The lab is represented as an *n* by *n* square grid of integers. A good lab is defined as a lab in which every number not equal to 1 can be expressed as the sum of a number in the same row and a number in the same column. In other words, for every *x*,<=*y* such that 1<=≤<=*x*,<=*y*<=≤<=*n* and *a**x*,<=*y*<=≠<=1, there should exist two indices *s* and *t* so that *a**x*,<=*y*<==<=*a**x*,<=*s*<=+<=*a**t*,<=*y*, where *a**i*,<=*j* denotes the integer in *i*-th row and *j*-th column.
Help Okabe determine whether a given lab is good!
|
The first line of input contains the integer *n* (1<=≤<=*n*<=≤<=50) — the size of the lab.
The next *n* lines contain *n* space-separated integers denoting a row of the grid. The *j*-th integer in the *i*-th row is *a**i*,<=*j* (1<=≤<=*a**i*,<=*j*<=≤<=105).
|
Print "Yes" if the given lab is good and "No" otherwise.
You can output each letter in upper or lower case.
|
[
"3\n1 1 2\n2 3 1\n6 4 1\n",
"3\n1 5 2\n1 1 1\n1 2 3\n"
] |
[
"Yes\n",
"No\n"
] |
In the first sample test, the 6 in the bottom left corner is valid because it is the sum of the 2 above it and the 4 on the right. The same holds for every number not equal to 1 in this table, so the answer is "Yes".
In the second sample test, the 5 cannot be formed as the sum of an integer in the same row and an integer in the same column. Thus the answer is "No".
| 500
|
[
{
"input": "3\n1 1 2\n2 3 1\n6 4 1",
"output": "Yes"
},
{
"input": "3\n1 5 2\n1 1 1\n1 2 3",
"output": "No"
},
{
"input": "1\n1",
"output": "Yes"
},
{
"input": "4\n1 1 1 1\n1 11 1 2\n2 5 1 4\n3 9 4 1",
"output": "Yes"
},
{
"input": "4\n1 1 1 1\n1 7 1 1\n1 3 1 2\n2 6 3 1",
"output": "Yes"
},
{
"input": "4\n1 1 1 1\n1 12 1 2\n4 4 1 3\n5 10 6 1",
"output": "Yes"
},
{
"input": "4\n1 1 1 1\n1 13 1 2\n4 5 1 3\n5 11 6 1",
"output": "Yes"
},
{
"input": "4\n1 1 1 1\n1 13 1 2\n4 5 1 3\n7 11 6 1",
"output": "No"
},
{
"input": "4\n1 1 1 1\n1 14 1 6\n4 5 1 5\n3 8 2 1",
"output": "Yes"
},
{
"input": "10\n3 2 6 5 1 2 1 1 2 1\n1 2 2 3 1 4 1 1 1 1\n1 1 3 1 2 2 2 1 1 1\n1 3 1 2 1 2 1 1 1 1\n12 1 4 3 3 1 1 1 2 1\n8 1 1 1 1 3 6 5 1 3\n1 1 1 2 4 1 3 2 1 1\n1 1 1 1 1 1 1 2 1 1\n1 2 1 1 1 1 1 1 1 1\n2 1 1 1 1 1 1 1 1 2",
"output": "Yes"
},
{
"input": "4\n1 2 3 5\n1 1 1 1\n1 1 1 1\n1 1 1 1",
"output": "No"
},
{
"input": "1\n2",
"output": "No"
},
{
"input": "3\n1 1 4\n1 1 2\n1 1 2",
"output": "No"
},
{
"input": "1\n5",
"output": "No"
},
{
"input": "3\n1 2 1\n2 2 3\n1 3 1",
"output": "No"
},
{
"input": "3\n2 1 1\n2 1 1\n2 1 1",
"output": "No"
},
{
"input": "1\n60000",
"output": "No"
}
] | 1,498,407,483
| 6,183
|
Python 3
|
WRONG_ANSWER
|
PRETESTS
| 2
| 46
| 5,529,600
|
n=int(input())
numArray=[[0 for i in range(n) ] for j in range(n)]
numDict={}
for i in range(n):
numList=input().split(" ")
for j in range(n):
iNum=int(numList[j])
numArray[i][j]=iNum
if iNum!=1:
numDict[(i,j)]=iNum
if len(numDict)==0:
print("No")
else:
iFlag=True
for key in numDict:
x,y=key
iNum=numDict[key]
for i in range(n):
iRight=False
for j in range(n):
if numArray[x][j]+numArray[i][y]==iNum:
iRight=True
break
elif i==n-1 and j==n-1:
print("No")
iFlag=False
if iRight:
break
if iFlag:
print("Yes")
|
Title: Okabe and Future Gadget Laboratory
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Okabe needs to renovate the Future Gadget Laboratory after he tried doing some crazy experiments! The lab is represented as an *n* by *n* square grid of integers. A good lab is defined as a lab in which every number not equal to 1 can be expressed as the sum of a number in the same row and a number in the same column. In other words, for every *x*,<=*y* such that 1<=≤<=*x*,<=*y*<=≤<=*n* and *a**x*,<=*y*<=≠<=1, there should exist two indices *s* and *t* so that *a**x*,<=*y*<==<=*a**x*,<=*s*<=+<=*a**t*,<=*y*, where *a**i*,<=*j* denotes the integer in *i*-th row and *j*-th column.
Help Okabe determine whether a given lab is good!
Input Specification:
The first line of input contains the integer *n* (1<=≤<=*n*<=≤<=50) — the size of the lab.
The next *n* lines contain *n* space-separated integers denoting a row of the grid. The *j*-th integer in the *i*-th row is *a**i*,<=*j* (1<=≤<=*a**i*,<=*j*<=≤<=105).
Output Specification:
Print "Yes" if the given lab is good and "No" otherwise.
You can output each letter in upper or lower case.
Demo Input:
['3\n1 1 2\n2 3 1\n6 4 1\n', '3\n1 5 2\n1 1 1\n1 2 3\n']
Demo Output:
['Yes\n', 'No\n']
Note:
In the first sample test, the 6 in the bottom left corner is valid because it is the sum of the 2 above it and the 4 on the right. The same holds for every number not equal to 1 in this table, so the answer is "Yes".
In the second sample test, the 5 cannot be formed as the sum of an integer in the same row and an integer in the same column. Thus the answer is "No".
|
```python
n=int(input())
numArray=[[0 for i in range(n) ] for j in range(n)]
numDict={}
for i in range(n):
numList=input().split(" ")
for j in range(n):
iNum=int(numList[j])
numArray[i][j]=iNum
if iNum!=1:
numDict[(i,j)]=iNum
if len(numDict)==0:
print("No")
else:
iFlag=True
for key in numDict:
x,y=key
iNum=numDict[key]
for i in range(n):
iRight=False
for j in range(n):
if numArray[x][j]+numArray[i][y]==iNum:
iRight=True
break
elif i==n-1 and j==n-1:
print("No")
iFlag=False
if iRight:
break
if iFlag:
print("Yes")
```
| 0
|
|
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"
}
] | 1,696,590,636
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 15
| 0
|
n = int(input())
segments = []
# Read and store the segments for each student
for _ in range(n):
li, ri = map(int, input().split())
segments.append((li, ri))
# Sort the segments based on li values in ascending order
segments.sort(key=lambda x: x[0])
left = 0
right = 100
total_length = 0
# Iterate through the sorted segments
for li, ri in segments:
if li > right:
# There's a gap between current segment and Alexey's available segment
total_length += li - right
# Update left and right based on the current segment
left = max(left, li)
right = min(right, ri)
# Add the remaining available segment length to total_length
if left < right:
total_length += right - left
print(total_length)
|
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 = []
# Read and store the segments for each student
for _ in range(n):
li, ri = map(int, input().split())
segments.append((li, ri))
# Sort the segments based on li values in ascending order
segments.sort(key=lambda x: x[0])
left = 0
right = 100
total_length = 0
# Iterate through the sorted segments
for li, ri in segments:
if li > right:
# There's a gap between current segment and Alexey's available segment
total_length += li - right
# Update left and right based on the current segment
left = max(left, li)
right = min(right, ri)
# Add the remaining available segment length to total_length
if left < right:
total_length += right - left
print(total_length)
```
| 0
|
|
569
|
A
|
Music
|
PROGRAMMING
| 1,500
|
[
"implementation",
"math"
] | null | null |
Little Lesha loves listening to music via his smartphone. But the smartphone doesn't have much memory, so Lesha listens to his favorite songs in a well-known social network InTalk.
Unfortunately, internet is not that fast in the city of Ekaterinozavodsk and the song takes a lot of time to download. But Lesha is quite impatient. The song's duration is *T* seconds. Lesha downloads the first *S* seconds of the song and plays it. When the playback reaches the point that has not yet been downloaded, Lesha immediately plays the song from the start (the loaded part of the song stays in his phone, and the download is continued from the same place), and it happens until the song is downloaded completely and Lesha listens to it to the end. For *q* seconds of real time the Internet allows you to download *q*<=-<=1 seconds of the track.
Tell Lesha, for how many times he will start the song, including the very first start.
|
The single line contains three integers *T*,<=*S*,<=*q* (2<=≤<=*q*<=≤<=104, 1<=≤<=*S*<=<<=*T*<=≤<=105).
|
Print a single integer — the number of times the song will be restarted.
|
[
"5 2 2\n",
"5 4 7\n",
"6 2 3\n"
] |
[
"2\n",
"1\n",
"1\n"
] |
In the first test, the song is played twice faster than it is downloaded, which means that during four first seconds Lesha reaches the moment that has not been downloaded, and starts the song again. After another two seconds, the song is downloaded completely, and thus, Lesha starts the song twice.
In the second test, the song is almost downloaded, and Lesha will start it only once.
In the third sample test the download finishes and Lesha finishes listening at the same moment. Note that song isn't restarted in this case.
| 500
|
[
{
"input": "5 2 2",
"output": "2"
},
{
"input": "5 4 7",
"output": "1"
},
{
"input": "6 2 3",
"output": "1"
},
{
"input": "2 1 2",
"output": "1"
},
{
"input": "2 1 3",
"output": "1"
},
{
"input": "2 1 10000",
"output": "1"
},
{
"input": "12326 6163 2",
"output": "1"
},
{
"input": "10000 2500 4",
"output": "1"
},
{
"input": "100000 99999 4",
"output": "1"
},
{
"input": "12351 1223 6",
"output": "2"
},
{
"input": "100000 1 10000",
"output": "2"
},
{
"input": "10028 13 10000",
"output": "1"
},
{
"input": "100000 99999 2",
"output": "1"
},
{
"input": "100000 99999 3",
"output": "1"
},
{
"input": "100000 1 2",
"output": "17"
},
{
"input": "100000 1 3",
"output": "11"
},
{
"input": "100000 1 4",
"output": "9"
},
{
"input": "100000 1 5",
"output": "8"
},
{
"input": "100000 3125 2",
"output": "5"
},
{
"input": "12628 1804 7",
"output": "1"
},
{
"input": "100000 45 13",
"output": "4"
},
{
"input": "100000 500 3",
"output": "5"
},
{
"input": "356 2 3",
"output": "5"
},
{
"input": "50 2 2",
"output": "5"
},
{
"input": "65465 12 3",
"output": "8"
},
{
"input": "10033 3 8",
"output": "4"
},
{
"input": "100000 3 2",
"output": "16"
},
{
"input": "64 1 8",
"output": "2"
},
{
"input": "10000 9 2",
"output": "11"
},
{
"input": "25 2 2",
"output": "4"
},
{
"input": "129 2 2",
"output": "7"
},
{
"input": "6562 1 3",
"output": "9"
},
{
"input": "100000 1 10",
"output": "5"
}
] | 1,439,768,279
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 33
| 62
| 0
|
t, s, q = map(int, input().split())
r = 0
while s < t:
s *= q
r += 1
print(r)
|
Title: Music
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Lesha loves listening to music via his smartphone. But the smartphone doesn't have much memory, so Lesha listens to his favorite songs in a well-known social network InTalk.
Unfortunately, internet is not that fast in the city of Ekaterinozavodsk and the song takes a lot of time to download. But Lesha is quite impatient. The song's duration is *T* seconds. Lesha downloads the first *S* seconds of the song and plays it. When the playback reaches the point that has not yet been downloaded, Lesha immediately plays the song from the start (the loaded part of the song stays in his phone, and the download is continued from the same place), and it happens until the song is downloaded completely and Lesha listens to it to the end. For *q* seconds of real time the Internet allows you to download *q*<=-<=1 seconds of the track.
Tell Lesha, for how many times he will start the song, including the very first start.
Input Specification:
The single line contains three integers *T*,<=*S*,<=*q* (2<=≤<=*q*<=≤<=104, 1<=≤<=*S*<=<<=*T*<=≤<=105).
Output Specification:
Print a single integer — the number of times the song will be restarted.
Demo Input:
['5 2 2\n', '5 4 7\n', '6 2 3\n']
Demo Output:
['2\n', '1\n', '1\n']
Note:
In the first test, the song is played twice faster than it is downloaded, which means that during four first seconds Lesha reaches the moment that has not been downloaded, and starts the song again. After another two seconds, the song is downloaded completely, and thus, Lesha starts the song twice.
In the second test, the song is almost downloaded, and Lesha will start it only once.
In the third sample test the download finishes and Lesha finishes listening at the same moment. Note that song isn't restarted in this case.
|
```python
t, s, q = map(int, input().split())
r = 0
while s < t:
s *= q
r += 1
print(r)
```
| 3
|
|
1,007
|
A
|
Reorder the Array
|
PROGRAMMING
| 1,300
|
[
"combinatorics",
"data structures",
"math",
"sortings",
"two pointers"
] | null | null |
You are given an array of integers. Vasya can permute (change order) its integers. He wants to do it so that as many as possible integers will become on a place where a smaller integer used to stand. Help Vasya find the maximal number of such integers.
For instance, if we are given an array $[10, 20, 30, 40]$, we can permute it so that it becomes $[20, 40, 10, 30]$. Then on the first and the second positions the integers became larger ($20>10$, $40>20$) and did not on the third and the fourth, so for this permutation, the number that Vasya wants to maximize equals $2$. Read the note for the first example, there is one more demonstrative test case.
Help Vasya to permute integers in such way that the number of positions in a new array, where integers are greater than in the original one, is maximal.
|
The first line contains a single integer $n$ ($1 \leq n \leq 10^5$) — the length of the array.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^9$) — the elements of the array.
|
Print a single integer — the maximal number of the array's elements which after a permutation will stand on the position where a smaller element stood in the initial array.
|
[
"7\n10 1 1 1 5 5 3\n",
"5\n1 1 1 1 1\n"
] |
[
"4\n",
"0\n"
] |
In the first sample, one of the best permutations is $[1, 5, 5, 3, 10, 1, 1]$. On the positions from second to fifth the elements became larger, so the answer for this permutation is 4.
In the second sample, there is no way to increase any element with a permutation, so the answer is 0.
| 500
|
[
{
"input": "7\n10 1 1 1 5 5 3",
"output": "4"
},
{
"input": "5\n1 1 1 1 1",
"output": "0"
},
{
"input": "6\n300000000 200000000 300000000 200000000 1000000000 300000000",
"output": "3"
},
{
"input": "10\n1 2 3 4 5 6 7 8 9 10",
"output": "9"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "7\n3 5 2 2 5 2 4",
"output": "4"
},
{
"input": "5\n1 5 4 2 3",
"output": "4"
}
] | 1,650,868,032
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 31
| 0
|
a= int(input())
b= int(input())
c= int(input())
d= int(input())
difference=((a * b) - (c * d))
print("DIFERENCA = %d" %difference)
|
Title: Reorder the Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given an array of integers. Vasya can permute (change order) its integers. He wants to do it so that as many as possible integers will become on a place where a smaller integer used to stand. Help Vasya find the maximal number of such integers.
For instance, if we are given an array $[10, 20, 30, 40]$, we can permute it so that it becomes $[20, 40, 10, 30]$. Then on the first and the second positions the integers became larger ($20>10$, $40>20$) and did not on the third and the fourth, so for this permutation, the number that Vasya wants to maximize equals $2$. Read the note for the first example, there is one more demonstrative test case.
Help Vasya to permute integers in such way that the number of positions in a new array, where integers are greater than in the original one, is maximal.
Input Specification:
The first line contains a single integer $n$ ($1 \leq n \leq 10^5$) — the length of the array.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^9$) — the elements of the array.
Output Specification:
Print a single integer — the maximal number of the array's elements which after a permutation will stand on the position where a smaller element stood in the initial array.
Demo Input:
['7\n10 1 1 1 5 5 3\n', '5\n1 1 1 1 1\n']
Demo Output:
['4\n', '0\n']
Note:
In the first sample, one of the best permutations is $[1, 5, 5, 3, 10, 1, 1]$. On the positions from second to fifth the elements became larger, so the answer for this permutation is 4.
In the second sample, there is no way to increase any element with a permutation, so the answer is 0.
|
```python
a= int(input())
b= int(input())
c= int(input())
d= int(input())
difference=((a * b) - (c * d))
print("DIFERENCA = %d" %difference)
```
| -1
|
|
71
|
A
|
Way Too Long Words
|
PROGRAMMING
| 800
|
[
"strings"
] |
A. Way Too Long Words
|
1
|
256
|
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
|
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
|
[
"4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n"
] |
[
"word\nl10n\ni18n\np43s\n"
] |
none
| 500
|
[
{
"input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis",
"output": "word\nl10n\ni18n\np43s"
},
{
"input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm",
"output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m"
},
{
"input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd",
"output": "n20n\nj27b\nk15d"
},
{
"input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt",
"output": "t41t"
},
{
"input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings",
"output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz",
"output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz"
},
{
"input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij",
"output": "a98j"
},
{
"input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco",
"output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco"
},
{
"input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt",
"output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t"
},
{
"input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr",
"output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr"
}
] | 1,693,597,536
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 31
| 0
|
n = int(input())
ans = [] * n
for i in range(n):
s = input()
a = len(s)
if a > 4:
if a - 2 >= 10:
s = s[0] + str(a - 2) + s[-1]
else:
s = s[0] + "0" + str(a - 2) + s[-1]
ans.append(s)
for i in range(n):
print(ans[i])
|
Title: Way Too Long Words
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output Specification:
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
Demo Input:
['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n']
Demo Output:
['word\nl10n\ni18n\np43s\n']
Note:
none
|
```python
n = int(input())
ans = [] * n
for i in range(n):
s = input()
a = len(s)
if a > 4:
if a - 2 >= 10:
s = s[0] + str(a - 2) + s[-1]
else:
s = s[0] + "0" + str(a - 2) + s[-1]
ans.append(s)
for i in range(n):
print(ans[i])
```
| 0
|
854
|
A
|
Fraction
|
PROGRAMMING
| 800
|
[
"brute force",
"constructive algorithms",
"math"
] | null | null |
Petya is a big fan of mathematics, especially its part related to fractions. Recently he learned that a fraction is called proper iff its numerator is smaller than its denominator (*a*<=<<=*b*) and that the fraction is called irreducible if its numerator and its denominator are coprime (they do not have positive common divisors except 1).
During his free time, Petya thinks about proper irreducible fractions and converts them to decimals using the calculator. One day he mistakenly pressed addition button (<=+<=) instead of division button (÷) and got sum of numerator and denominator that was equal to *n* instead of the expected decimal notation.
Petya wanted to restore the original fraction, but soon he realized that it might not be done uniquely. That's why he decided to determine maximum possible proper irreducible fraction such that sum of its numerator and denominator equals *n*. Help Petya deal with this problem.
|
In the only line of input there is an integer *n* (3<=≤<=*n*<=≤<=1000), the sum of numerator and denominator of the fraction.
|
Output two space-separated positive integers *a* and *b*, numerator and denominator of the maximum possible proper irreducible fraction satisfying the given sum.
|
[
"3\n",
"4\n",
"12\n"
] |
[
"1 2\n",
"1 3\n",
"5 7\n"
] |
none
| 500
|
[
{
"input": "3",
"output": "1 2"
},
{
"input": "4",
"output": "1 3"
},
{
"input": "12",
"output": "5 7"
},
{
"input": "34",
"output": "15 19"
},
{
"input": "13",
"output": "6 7"
},
{
"input": "11",
"output": "5 6"
},
{
"input": "24",
"output": "11 13"
},
{
"input": "17",
"output": "8 9"
},
{
"input": "10",
"output": "3 7"
},
{
"input": "69",
"output": "34 35"
},
{
"input": "100",
"output": "49 51"
},
{
"input": "57",
"output": "28 29"
},
{
"input": "1000",
"output": "499 501"
},
{
"input": "999",
"output": "499 500"
},
{
"input": "998",
"output": "497 501"
},
{
"input": "997",
"output": "498 499"
},
{
"input": "996",
"output": "497 499"
},
{
"input": "995",
"output": "497 498"
},
{
"input": "994",
"output": "495 499"
},
{
"input": "5",
"output": "2 3"
},
{
"input": "6",
"output": "1 5"
},
{
"input": "8",
"output": "3 5"
},
{
"input": "9",
"output": "4 5"
},
{
"input": "423",
"output": "211 212"
},
{
"input": "876",
"output": "437 439"
},
{
"input": "29",
"output": "14 15"
}
] | 1,505,297,676
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 26
| 46
| 0
|
def nod(a, b):
while (a!=0) and (b!=0):
if a>b: a%=b;
else: b%=a;
return a+b
n = int(input())
a = n//2 if n%2==1 else n//2-1
b = n - a
while nod(a, b)!=1:
a-=1
b+=1
print (a, b)
|
Title: Fraction
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya is a big fan of mathematics, especially its part related to fractions. Recently he learned that a fraction is called proper iff its numerator is smaller than its denominator (*a*<=<<=*b*) and that the fraction is called irreducible if its numerator and its denominator are coprime (they do not have positive common divisors except 1).
During his free time, Petya thinks about proper irreducible fractions and converts them to decimals using the calculator. One day he mistakenly pressed addition button (<=+<=) instead of division button (÷) and got sum of numerator and denominator that was equal to *n* instead of the expected decimal notation.
Petya wanted to restore the original fraction, but soon he realized that it might not be done uniquely. That's why he decided to determine maximum possible proper irreducible fraction such that sum of its numerator and denominator equals *n*. Help Petya deal with this problem.
Input Specification:
In the only line of input there is an integer *n* (3<=≤<=*n*<=≤<=1000), the sum of numerator and denominator of the fraction.
Output Specification:
Output two space-separated positive integers *a* and *b*, numerator and denominator of the maximum possible proper irreducible fraction satisfying the given sum.
Demo Input:
['3\n', '4\n', '12\n']
Demo Output:
['1 2\n', '1 3\n', '5 7\n']
Note:
none
|
```python
def nod(a, b):
while (a!=0) and (b!=0):
if a>b: a%=b;
else: b%=a;
return a+b
n = int(input())
a = n//2 if n%2==1 else n//2-1
b = n - a
while nod(a, b)!=1:
a-=1
b+=1
print (a, b)
```
| 3
|
|
595
|
A
|
Vitaly and Night
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"implementation"
] | null | null |
One day Vitaly was going home late at night and wondering: how many people aren't sleeping at that moment? To estimate, Vitaly decided to look which windows are lit in the house he was passing by at that moment.
Vitaly sees a building of *n* floors and 2·*m* windows on each floor. On each floor there are *m* flats numbered from 1 to *m*, and two consecutive windows correspond to each flat. If we number the windows from 1 to 2·*m* from left to right, then the *j*-th flat of the *i*-th floor has windows 2·*j*<=-<=1 and 2·*j* in the corresponding row of windows (as usual, floors are enumerated from the bottom). Vitaly thinks that people in the flat aren't sleeping at that moment if at least one of the windows corresponding to this flat has lights on.
Given the information about the windows of the given house, your task is to calculate the number of flats where, according to Vitaly, people aren't sleeping.
|
The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of floors in the house and the number of flats on each floor respectively.
Next *n* lines describe the floors from top to bottom and contain 2·*m* characters each. If the *i*-th window of the given floor has lights on, then the *i*-th character of this line is '1', otherwise it is '0'.
|
Print a single integer — the number of flats that have lights on in at least one window, that is, the flats where, according to Vitaly, people aren't sleeping.
|
[
"2 2\n0 0 0 1\n1 0 1 1\n",
"1 3\n1 1 0 1 0 0\n"
] |
[
"3\n",
"2\n"
] |
In the first test case the house has two floors, two flats on each floor. That is, in total there are 4 flats. The light isn't on only on the second floor in the left flat. That is, in both rooms of the flat the light is off.
In the second test case the house has one floor and the first floor has three flats. The light is on in the leftmost flat (in both windows) and in the middle flat (in one window). In the right flat the light is off.
| 500
|
[
{
"input": "2 2\n0 0 0 1\n1 0 1 1",
"output": "3"
},
{
"input": "1 3\n1 1 0 1 0 0",
"output": "2"
},
{
"input": "3 3\n1 1 1 1 1 1\n1 1 0 1 1 0\n1 0 0 0 1 1",
"output": "8"
},
{
"input": "1 5\n1 0 1 1 1 0 1 1 1 1",
"output": "5"
},
{
"input": "1 100\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 0 1 1 1 1 0 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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 0 0 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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",
"output": "99"
},
{
"input": "1 100\n0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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 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 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 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 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "6"
},
{
"input": "1 100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 0\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n0 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 0\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 0\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n0 1\n1 1\n1 1\n1 0\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1",
"output": "100"
},
{
"input": "100 1\n0 0\n0 0\n0 1\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n1 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 1\n0 0\n0 0\n0 0\n1 0\n0 0\n0 0\n0 0\n1 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 1\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 1\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n1 0",
"output": "8"
},
{
"input": "100 1\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "1 1\n0 0",
"output": "0"
},
{
"input": "1 1\n0 1",
"output": "1"
},
{
"input": "1 1\n1 0",
"output": "1"
},
{
"input": "1 1\n1 1",
"output": "1"
}
] | 1,447,091,775
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 0
| 78
| 307,200
|
n,m = map(int,input().split())
b = 0
for i in range(n):
l = list(map(int,input().split()))
for i in range(m):
if(l[i*2]==1 or l[2*i+1]==1):
print(n,i)
b += 1
print(str(b))
|
Title: Vitaly and Night
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Vitaly was going home late at night and wondering: how many people aren't sleeping at that moment? To estimate, Vitaly decided to look which windows are lit in the house he was passing by at that moment.
Vitaly sees a building of *n* floors and 2·*m* windows on each floor. On each floor there are *m* flats numbered from 1 to *m*, and two consecutive windows correspond to each flat. If we number the windows from 1 to 2·*m* from left to right, then the *j*-th flat of the *i*-th floor has windows 2·*j*<=-<=1 and 2·*j* in the corresponding row of windows (as usual, floors are enumerated from the bottom). Vitaly thinks that people in the flat aren't sleeping at that moment if at least one of the windows corresponding to this flat has lights on.
Given the information about the windows of the given house, your task is to calculate the number of flats where, according to Vitaly, people aren't sleeping.
Input Specification:
The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of floors in the house and the number of flats on each floor respectively.
Next *n* lines describe the floors from top to bottom and contain 2·*m* characters each. If the *i*-th window of the given floor has lights on, then the *i*-th character of this line is '1', otherwise it is '0'.
Output Specification:
Print a single integer — the number of flats that have lights on in at least one window, that is, the flats where, according to Vitaly, people aren't sleeping.
Demo Input:
['2 2\n0 0 0 1\n1 0 1 1\n', '1 3\n1 1 0 1 0 0\n']
Demo Output:
['3\n', '2\n']
Note:
In the first test case the house has two floors, two flats on each floor. That is, in total there are 4 flats. The light isn't on only on the second floor in the left flat. That is, in both rooms of the flat the light is off.
In the second test case the house has one floor and the first floor has three flats. The light is on in the leftmost flat (in both windows) and in the middle flat (in one window). In the right flat the light is off.
|
```python
n,m = map(int,input().split())
b = 0
for i in range(n):
l = list(map(int,input().split()))
for i in range(m):
if(l[i*2]==1 or l[2*i+1]==1):
print(n,i)
b += 1
print(str(b))
```
| 0
|
|
842
|
C
|
Ilya And The Tree
|
PROGRAMMING
| 2,000
|
[
"dfs and similar",
"graphs",
"math",
"number theory",
"trees"
] | null | null |
Ilya is very fond of graphs, especially trees. During his last trip to the forest Ilya found a very interesting tree rooted at vertex 1. There is an integer number written on each vertex of the tree; the number written on vertex *i* is equal to *a**i*.
Ilya believes that the beauty of the vertex *x* is the greatest common divisor of all numbers written on the vertices on the path from the root to *x*, including this vertex itself. In addition, Ilya can change the number in one arbitrary vertex to 0 or leave all vertices unchanged. Now for each vertex Ilya wants to know the maximum possible beauty it can have.
For each vertex the answer must be considered independently.
The beauty of the root equals to number written on it.
|
First line contains one integer number *n* — the number of vertices in tree (1<=≤<=*n*<=≤<=2·105).
Next line contains *n* integer numbers *a**i* (1<=≤<=*i*<=≤<=*n*, 1<=≤<=*a**i*<=≤<=2·105).
Each of next *n*<=-<=1 lines contains two integer numbers *x* and *y* (1<=≤<=*x*,<=*y*<=≤<=*n*, *x*<=≠<=*y*), which means that there is an edge (*x*,<=*y*) in the tree.
|
Output *n* numbers separated by spaces, where *i*-th number equals to maximum possible beauty of vertex *i*.
|
[
"2\n6 2\n1 2\n",
"3\n6 2 3\n1 2\n1 3\n",
"1\n10\n"
] |
[
"6 6 \n",
"6 6 6 \n",
"10 \n"
] |
none
| 1,500
|
[
{
"input": "2\n6 2\n1 2",
"output": "6 6 "
},
{
"input": "3\n6 2 3\n1 2\n1 3",
"output": "6 6 6 "
},
{
"input": "1\n10",
"output": "10 "
},
{
"input": "10\n2 3 4 5 6 7 8 9 10 11\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n4 8\n8 9\n9 10",
"output": "2 3 2 1 1 1 1 1 1 1 "
},
{
"input": "4\n6 2 3 2\n1 2\n2 3\n3 4",
"output": "6 6 3 2 "
}
] | 1,504,098,952
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 46
| 0
|
def read(f = int): return f(input())
def readlist(f = int): return list(map(f, input().split()))
n = read()
a = readlist()
g = {x:[] for x in range(n)}
for _ in range(n-1):
x, y = readlist()
g[x-1].append(y-1)
g[y-1].append(x-1)
def gcd(a, b): return a if b == 0 else gcd(b, a%b)
# vis = [0 for _ in range(n)]
# par = [-1 for _ in range(n)]
bty = {-1:(0,0)}
def bfs(v, p):
# vis[v] = 1
# par[v] = p
bty[v] = (max(gcd(bty[p][0], a[v]), bty[p][1]), gcd(bty[p][1], a[v]))
for u in g[v]:
# if vis[u] == 0:
if u not in bty:
bfs(u, v)
bfs(0, -1)
print(' '.join([str(max(bty[i])) for i in range(n)]))
|
Title: Ilya And The Tree
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ilya is very fond of graphs, especially trees. During his last trip to the forest Ilya found a very interesting tree rooted at vertex 1. There is an integer number written on each vertex of the tree; the number written on vertex *i* is equal to *a**i*.
Ilya believes that the beauty of the vertex *x* is the greatest common divisor of all numbers written on the vertices on the path from the root to *x*, including this vertex itself. In addition, Ilya can change the number in one arbitrary vertex to 0 or leave all vertices unchanged. Now for each vertex Ilya wants to know the maximum possible beauty it can have.
For each vertex the answer must be considered independently.
The beauty of the root equals to number written on it.
Input Specification:
First line contains one integer number *n* — the number of vertices in tree (1<=≤<=*n*<=≤<=2·105).
Next line contains *n* integer numbers *a**i* (1<=≤<=*i*<=≤<=*n*, 1<=≤<=*a**i*<=≤<=2·105).
Each of next *n*<=-<=1 lines contains two integer numbers *x* and *y* (1<=≤<=*x*,<=*y*<=≤<=*n*, *x*<=≠<=*y*), which means that there is an edge (*x*,<=*y*) in the tree.
Output Specification:
Output *n* numbers separated by spaces, where *i*-th number equals to maximum possible beauty of vertex *i*.
Demo Input:
['2\n6 2\n1 2\n', '3\n6 2 3\n1 2\n1 3\n', '1\n10\n']
Demo Output:
['6 6 \n', '6 6 6 \n', '10 \n']
Note:
none
|
```python
def read(f = int): return f(input())
def readlist(f = int): return list(map(f, input().split()))
n = read()
a = readlist()
g = {x:[] for x in range(n)}
for _ in range(n-1):
x, y = readlist()
g[x-1].append(y-1)
g[y-1].append(x-1)
def gcd(a, b): return a if b == 0 else gcd(b, a%b)
# vis = [0 for _ in range(n)]
# par = [-1 for _ in range(n)]
bty = {-1:(0,0)}
def bfs(v, p):
# vis[v] = 1
# par[v] = p
bty[v] = (max(gcd(bty[p][0], a[v]), bty[p][1]), gcd(bty[p][1], a[v]))
for u in g[v]:
# if vis[u] == 0:
if u not in bty:
bfs(u, v)
bfs(0, -1)
print(' '.join([str(max(bty[i])) for i in range(n)]))
```
| 0
|
|
50
|
A
|
Domino piling
|
PROGRAMMING
| 800
|
[
"greedy",
"math"
] |
A. Domino piling
|
2
|
256
|
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
|
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
|
Output one number — the maximal number of dominoes, which can be placed.
|
[
"2 4\n",
"3 3\n"
] |
[
"4\n",
"4\n"
] |
none
| 500
|
[
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "2 6",
"output": "6"
},
{
"input": "2 7",
"output": "7"
},
{
"input": "2 14",
"output": "14"
},
{
"input": "2 15",
"output": "15"
},
{
"input": "1 4",
"output": "2"
},
{
"input": "2 16",
"output": "16"
},
{
"input": "3 5",
"output": "7"
},
{
"input": "3 6",
"output": "9"
},
{
"input": "3 10",
"output": "15"
},
{
"input": "3 14",
"output": "21"
},
{
"input": "3 15",
"output": "22"
},
{
"input": "3 16",
"output": "24"
},
{
"input": "5 7",
"output": "17"
},
{
"input": "16 16",
"output": "128"
},
{
"input": "15 16",
"output": "120"
},
{
"input": "2 3",
"output": "3"
},
{
"input": "15 15",
"output": "112"
},
{
"input": "14 16",
"output": "112"
},
{
"input": "11 13",
"output": "71"
},
{
"input": "5 16",
"output": "40"
},
{
"input": "8 15",
"output": "60"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "3 4",
"output": "6"
},
{
"input": "4 4",
"output": "8"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "1 3",
"output": "1"
},
{
"input": "14 15",
"output": "105"
}
] | 1,617,528,502
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 60
| 0
|
n, m = list(input().split())
result = (n*m) // 2
print(result)
|
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
n, m = list(input().split())
result = (n*m) // 2
print(result)
```
| -1
|
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"
}
] | 1,632,297,654
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 110
| 109
| 20,172,800
|
c, v0, v1, a, l = map(int, input().split())
page, day, add = v0, 1, v0
while page < c:
add = min(v1, add + a)
page += add - l
day += 1
print(day)
|
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, v0, v1, a, l = map(int, input().split())
page, day, add = v0, 1, v0
while page < c:
add = min(v1, add + a)
page += add - l
day += 1
print(day)
```
| 3
|
|
102
|
B
|
Sum of Digits
|
PROGRAMMING
| 1,000
|
[
"implementation"
] |
B. Sum of Digits
|
2
|
265
|
Having watched the last Harry Potter film, little Gerald also decided to practice magic. He found in his father's magical book a spell that turns any number in the sum of its digits. At the moment Gerald learned that, he came across a number *n*. How many times can Gerald put a spell on it until the number becomes one-digit?
|
The first line contains the only integer *n* (0<=≤<=*n*<=≤<=10100000). It is guaranteed that *n* doesn't contain any leading zeroes.
|
Print the number of times a number can be replaced by the sum of its digits until it only contains one digit.
|
[
"0\n",
"10\n",
"991\n"
] |
[
"0\n",
"1\n",
"3\n"
] |
In the first sample the number already is one-digit — Herald can't cast a spell.
The second test contains number 10. After one casting of a spell it becomes 1, and here the process is completed. Thus, Gerald can only cast the spell once.
The third test contains number 991. As one casts a spell the following transformations take place: 991 → 19 → 10 → 1. After three transformations the number becomes one-digit.
| 1,000
|
[
{
"input": "0",
"output": "0"
},
{
"input": "10",
"output": "1"
},
{
"input": "991",
"output": "3"
},
{
"input": "99",
"output": "2"
},
{
"input": "100",
"output": "1"
},
{
"input": "123456789",
"output": "2"
},
{
"input": "32",
"output": "1"
},
{
"input": "86",
"output": "2"
},
{
"input": "2",
"output": "0"
},
{
"input": "8",
"output": "0"
},
{
"input": "34",
"output": "1"
},
{
"input": "13",
"output": "1"
},
{
"input": "28",
"output": "2"
},
{
"input": "23",
"output": "1"
},
{
"input": "57",
"output": "2"
},
{
"input": "29",
"output": "2"
},
{
"input": "353869285",
"output": "3"
},
{
"input": "549548646",
"output": "2"
},
{
"input": "858893424",
"output": "2"
},
{
"input": "644818852",
"output": "3"
},
{
"input": "360322525",
"output": "3"
},
{
"input": "122937520",
"output": "2"
},
{
"input": "288403032",
"output": "2"
},
{
"input": "677257481",
"output": "3"
},
{
"input": "58059561",
"output": "3"
},
{
"input": "211288847",
"output": "2"
},
{
"input": "339900034079539584",
"output": "2"
},
{
"input": "784084029310269952",
"output": "3"
},
{
"input": "774730557264864000",
"output": "3"
},
{
"input": "478233807148352256",
"output": "3"
},
{
"input": "165646874056812544",
"output": "3"
},
{
"input": "477533739511673792",
"output": "3"
},
{
"input": "660119944377002240",
"output": "3"
},
{
"input": "448375218506078080",
"output": "3"
},
{
"input": "77213334785795616",
"output": "3"
},
{
"input": "165767221702271872",
"output": "3"
},
{
"input": "1",
"output": "0"
},
{
"input": "5",
"output": "0"
},
{
"input": "8",
"output": "0"
},
{
"input": "156161456146164104103460452045416165146141414651641564105461465156445621465146514562146145465165145145614561465145614561",
"output": "3"
},
{
"input": "9",
"output": "0"
}
] | 1,676,027,418
| 2,147,483,647
|
PyPy 3-64
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 6
| 2,000
| 9,420,800
|
def sum_digits(num):
tot = 0
while num:
tot += (num % 10)
num //= 10
return tot
if __name__ == "__main__":
n = int(input())
ans = 0
while n >= 10:
ans += 1
n = sum_digits(n)
print(ans)
|
Title: Sum of Digits
Time Limit: 2 seconds
Memory Limit: 265 megabytes
Problem Description:
Having watched the last Harry Potter film, little Gerald also decided to practice magic. He found in his father's magical book a spell that turns any number in the sum of its digits. At the moment Gerald learned that, he came across a number *n*. How many times can Gerald put a spell on it until the number becomes one-digit?
Input Specification:
The first line contains the only integer *n* (0<=≤<=*n*<=≤<=10100000). It is guaranteed that *n* doesn't contain any leading zeroes.
Output Specification:
Print the number of times a number can be replaced by the sum of its digits until it only contains one digit.
Demo Input:
['0\n', '10\n', '991\n']
Demo Output:
['0\n', '1\n', '3\n']
Note:
In the first sample the number already is one-digit — Herald can't cast a spell.
The second test contains number 10. After one casting of a spell it becomes 1, and here the process is completed. Thus, Gerald can only cast the spell once.
The third test contains number 991. As one casts a spell the following transformations take place: 991 → 19 → 10 → 1. After three transformations the number becomes one-digit.
|
```python
def sum_digits(num):
tot = 0
while num:
tot += (num % 10)
num //= 10
return tot
if __name__ == "__main__":
n = int(input())
ans = 0
while n >= 10:
ans += 1
n = sum_digits(n)
print(ans)
```
| 0
|
340
|
B
|
Maximal Area Quadrilateral
|
PROGRAMMING
| 2,100
|
[
"brute force",
"geometry"
] | null | null |
Iahub has drawn a set of *n* points in the cartesian plane which he calls "special points". A quadrilateral is a simple polygon without self-intersections with four sides (also called edges) and four vertices (also called corners). Please note that a quadrilateral doesn't have to be convex. A special quadrilateral is one which has all four vertices in the set of special points. Given the set of special points, please calculate the maximal area of a special quadrilateral.
|
The first line contains integer *n* (4<=≤<=*n*<=≤<=300). Each of the next *n* lines contains two integers: *x**i*, *y**i* (<=-<=1000<=≤<=*x**i*,<=*y**i*<=≤<=1000) — the cartesian coordinates of *i*th special point. It is guaranteed that no three points are on the same line. It is guaranteed that no two points coincide.
|
Output a single real number — the maximal area of a special quadrilateral. The answer will be considered correct if its absolute or relative error does't exceed 10<=-<=9.
|
[
"5\n0 0\n0 4\n4 0\n4 4\n2 3\n"
] |
[
"16.000000"
] |
In the test example we can choose first 4 points to be the vertices of the quadrilateral. They form a square by side 4, so the area is 4·4 = 16.
| 3,000
|
[
{
"input": "5\n0 0\n0 4\n4 0\n4 4\n2 3",
"output": "16.000000"
},
{
"input": "10\n-6 -4\n-7 5\n-7 -7\n5 -7\n4 -9\n-6 7\n2 9\n-4 -6\n2 10\n-10 -4",
"output": "166.000000"
},
{
"input": "4\n-3 3\n0 3\n-2 -1\n2 2",
"output": "11.000000"
},
{
"input": "5\n-4 -3\n-3 -2\n3 3\n-1 2\n3 -3",
"output": "29.500000"
},
{
"input": "6\n-4 -3\n-1 3\n0 0\n2 2\n2 1\n-3 1",
"output": "15.000000"
},
{
"input": "7\n-2 -1\n4 3\n2 2\n-4 0\n-2 4\n0 0\n1 -3",
"output": "32.500000"
},
{
"input": "4\n-874 606\n-996 -207\n897 847\n775 191",
"output": "1261820.500000"
},
{
"input": "10\n156 -415\n879 198\n-250 -676\n-594 -433\n-207 368\n296 -641\n-387 -795\n143 -304\n-468 390\n-873 226",
"output": "1129219.500000"
},
{
"input": "50\n-768 -243\n-741 -984\n-370 213\n-808 571\n-726 442\n234 452\n-105 -990\n-876 -278\n987 473\n-968 -531\n-274 -842\n259 -655\n-59 -555\n976 -396\n878 -85\n551 213\n675 599\n-990 -507\n1 48\n-147 919\n-218 798\n-191 928\n916 263\n-975 169\n567 -967\n394 16\n-224 915\n280 -613\n804 -877\n988 -576\n-256 -708\n757 546\n777 99\n-579 -608\n-102 1\n-309 636\n-24 -718\n644 -84\n111 -822\n-722 544\n78 595\n-194 716\n-409 -845\n-291 441\n388 379\n-950 277\n-718 359\n881 198\n198 670\n828 -820",
"output": "2425414.000000"
},
{
"input": "4\n0 0\n0 5\n5 0\n1 1",
"output": "10.000000"
}
] | 1,688,918,232
| 2,147,483,647
|
PyPy 3-64
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 6
| 1,000
| 15,052,800
|
from dataclasses import dataclass
def gauss_s(i, j, h, k):
point1 = i
point2 = j
point3 = h
point4 = k
s = (point1[0] * point2[1] + point2[0]*point3[1] + point3[0]*point4[1] + point4[0]*point1[1] - point1[1] * point2[0] - point2[1]*point3[0] - point3[1]*point4[0] - point4[1]*point1[0])/2
return s
def solve():
points = []
max_ = 0
n = int(input())
for _ in range(n):
points.append(list(map(int, input().split())))
for i in points:
for j in points:
for h in points:
for k in points:
if i != j and i != h and i != k and j != h and j !=k and h != k and k != i:
max_ = max(max_, gauss_s(i, j, h, k))
print(max_)
def main():
solve()
main()
|
Title: Maximal Area Quadrilateral
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Iahub has drawn a set of *n* points in the cartesian plane which he calls "special points". A quadrilateral is a simple polygon without self-intersections with four sides (also called edges) and four vertices (also called corners). Please note that a quadrilateral doesn't have to be convex. A special quadrilateral is one which has all four vertices in the set of special points. Given the set of special points, please calculate the maximal area of a special quadrilateral.
Input Specification:
The first line contains integer *n* (4<=≤<=*n*<=≤<=300). Each of the next *n* lines contains two integers: *x**i*, *y**i* (<=-<=1000<=≤<=*x**i*,<=*y**i*<=≤<=1000) — the cartesian coordinates of *i*th special point. It is guaranteed that no three points are on the same line. It is guaranteed that no two points coincide.
Output Specification:
Output a single real number — the maximal area of a special quadrilateral. The answer will be considered correct if its absolute or relative error does't exceed 10<=-<=9.
Demo Input:
['5\n0 0\n0 4\n4 0\n4 4\n2 3\n']
Demo Output:
['16.000000']
Note:
In the test example we can choose first 4 points to be the vertices of the quadrilateral. They form a square by side 4, so the area is 4·4 = 16.
|
```python
from dataclasses import dataclass
def gauss_s(i, j, h, k):
point1 = i
point2 = j
point3 = h
point4 = k
s = (point1[0] * point2[1] + point2[0]*point3[1] + point3[0]*point4[1] + point4[0]*point1[1] - point1[1] * point2[0] - point2[1]*point3[0] - point3[1]*point4[0] - point4[1]*point1[0])/2
return s
def solve():
points = []
max_ = 0
n = int(input())
for _ in range(n):
points.append(list(map(int, input().split())))
for i in points:
for j in points:
for h in points:
for k in points:
if i != j and i != h and i != k and j != h and j !=k and h != k and k != i:
max_ = max(max_, gauss_s(i, j, h, k))
print(max_)
def main():
solve()
main()
```
| 0
|
|
463
|
B
|
Caisa and Pylons
|
PROGRAMMING
| 1,100
|
[
"brute force",
"implementation",
"math"
] | null | null |
Caisa solved the problem with the sugar and now he is on the way back to home.
Caisa is playing a mobile game during his path. There are (*n*<=+<=1) pylons numbered from 0 to *n* in this game. The pylon with number 0 has zero height, the pylon with number *i* (*i*<=><=0) has height *h**i*. The goal of the game is to reach *n*-th pylon, and the only move the player can do is to jump from the current pylon (let's denote its number as *k*) to the next one (its number will be *k*<=+<=1). When the player have made such a move, its energy increases by *h**k*<=-<=*h**k*<=+<=1 (if this value is negative the player loses energy). The player must have non-negative amount of energy at any moment of the time.
Initially Caisa stand at 0 pylon and has 0 energy. The game provides a special opportunity: one can pay a single dollar and increase the height of anyone pylon by one. Caisa may use that opportunity several times, but he doesn't want to spend too much money. What is the minimal amount of money he must paid to reach the goal of the game?
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=105). The next line contains *n* integers *h*1, *h*2,<=..., *h**n* (1<=<=≤<=<=*h**i*<=<=≤<=<=105) representing the heights of the pylons.
|
Print a single number representing the minimum number of dollars paid by Caisa.
|
[
"5\n3 4 3 2 4\n",
"3\n4 4 4\n"
] |
[
"4\n",
"4\n"
] |
In the first sample he can pay 4 dollars and increase the height of pylon with number 0 by 4 units. Then he can safely pass to the last pylon.
| 1,000
|
[
{
"input": "5\n3 4 3 2 4",
"output": "4"
},
{
"input": "3\n4 4 4",
"output": "4"
},
{
"input": "99\n1401 2019 1748 3785 3236 3177 3443 3772 2138 1049 353 908 310 2388 1322 88 2160 2783 435 2248 1471 706 2468 2319 3156 3506 2794 1999 1983 2519 2597 3735 537 344 3519 3772 3872 2961 3895 2010 10 247 3269 671 2986 942 758 1146 77 1545 3745 1547 2250 2565 217 1406 2070 3010 3404 404 1528 2352 138 2065 3047 3656 2188 2919 2616 2083 1280 2977 2681 548 4000 1667 1489 1109 3164 1565 2653 3260 3463 903 1824 3679 2308 245 2689 2063 648 568 766 785 2984 3812 440 1172 2730",
"output": "4000"
},
{
"input": "68\n477 1931 3738 3921 2306 1823 3328 2057 661 3993 2967 3520 171 1739 1525 1817 209 3475 1902 2666 518 3283 3412 3040 3383 2331 1147 1460 1452 1800 1327 2280 82 1416 2200 2388 3238 1879 796 250 1872 114 121 2042 1853 1645 211 2061 1472 2464 726 1989 1746 489 1380 1128 2819 2527 2939 622 678 265 2902 1111 2032 1453 3850 1621",
"output": "3993"
},
{
"input": "30\n30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1",
"output": "30"
},
{
"input": "3\n3 2 1",
"output": "3"
},
{
"input": "1\n69",
"output": "69"
}
] | 1,663,607,638
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 31
| 0
|
n = int(input())
heights = list(map(int, input().split()))
sum = 0
for i in range(0, n - 1):
if (i == 0):
sum += -heights[i]
sum += heights[i] - heights[i + 1]
print(abs(sum))
|
Title: Caisa and Pylons
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Caisa solved the problem with the sugar and now he is on the way back to home.
Caisa is playing a mobile game during his path. There are (*n*<=+<=1) pylons numbered from 0 to *n* in this game. The pylon with number 0 has zero height, the pylon with number *i* (*i*<=><=0) has height *h**i*. The goal of the game is to reach *n*-th pylon, and the only move the player can do is to jump from the current pylon (let's denote its number as *k*) to the next one (its number will be *k*<=+<=1). When the player have made such a move, its energy increases by *h**k*<=-<=*h**k*<=+<=1 (if this value is negative the player loses energy). The player must have non-negative amount of energy at any moment of the time.
Initially Caisa stand at 0 pylon and has 0 energy. The game provides a special opportunity: one can pay a single dollar and increase the height of anyone pylon by one. Caisa may use that opportunity several times, but he doesn't want to spend too much money. What is the minimal amount of money he must paid to reach the goal of the game?
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105). The next line contains *n* integers *h*1, *h*2,<=..., *h**n* (1<=<=≤<=<=*h**i*<=<=≤<=<=105) representing the heights of the pylons.
Output Specification:
Print a single number representing the minimum number of dollars paid by Caisa.
Demo Input:
['5\n3 4 3 2 4\n', '3\n4 4 4\n']
Demo Output:
['4\n', '4\n']
Note:
In the first sample he can pay 4 dollars and increase the height of pylon with number 0 by 4 units. Then he can safely pass to the last pylon.
|
```python
n = int(input())
heights = list(map(int, input().split()))
sum = 0
for i in range(0, n - 1):
if (i == 0):
sum += -heights[i]
sum += heights[i] - heights[i + 1]
print(abs(sum))
```
| 0
|
|
298
|
A
|
Snow Footprints
|
PROGRAMMING
| 1,300
|
[
"greedy",
"implementation"
] | null | null |
There is a straight snowy road, divided into *n* blocks. The blocks are numbered from 1 to *n* from left to right. If one moves from the *i*-th block to the (*i*<=+<=1)-th block, he will leave a right footprint on the *i*-th block. Similarly, if one moves from the *i*-th block to the (*i*<=-<=1)-th block, he will leave a left footprint on the *i*-th block. If there already is a footprint on the *i*-th block, the new footprint will cover the old one.
At the beginning, there were no footprints. Then polar bear Alice starts from the *s*-th block, makes a sequence of moves and ends in the *t*-th block. It is known that Alice never moves outside of the road.
You are given the description of Alice's footprints. Your task is to find a pair of possible values of *s*,<=*t* by looking at the footprints.
|
The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=1000).
The second line contains the description of the road — the string that consists of *n* characters. Each character will be either "." (a block without footprint), or "L" (a block with a left footprint), "R" (a block with a right footprint).
It's guaranteed that the given string contains at least one character not equal to ".". Also, the first and the last character will always be ".". It's guaranteed that a solution exists.
|
Print two space-separated integers — the values of *s* and *t*. If there are several possible solutions you can print any of them.
|
[
"9\n..RRLL...\n",
"11\n.RRRLLLLL..\n"
] |
[
"3 4\n",
"7 5\n"
] |
The first test sample is the one in the picture.
| 500
|
[
{
"input": "11\n.RRRLLLLL..",
"output": "7 5"
},
{
"input": "4\n.RL.",
"output": "3 2"
},
{
"input": "3\n.L.",
"output": "2 1"
},
{
"input": "3\n.R.",
"output": "2 3"
}
] | 1,610,041,871
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 23
| 218
| 0
|
n = int(input())
word = input()
if word.count("R")==0:
print(word.rfind("L")+1,word.find("L"))
elif word.count("L")==0:
print(word.find("R")+1,word.rfind("R")+2)
else:
print(word.find("R")+1,word.rfind("R")+1)
|
Title: Snow Footprints
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There is a straight snowy road, divided into *n* blocks. The blocks are numbered from 1 to *n* from left to right. If one moves from the *i*-th block to the (*i*<=+<=1)-th block, he will leave a right footprint on the *i*-th block. Similarly, if one moves from the *i*-th block to the (*i*<=-<=1)-th block, he will leave a left footprint on the *i*-th block. If there already is a footprint on the *i*-th block, the new footprint will cover the old one.
At the beginning, there were no footprints. Then polar bear Alice starts from the *s*-th block, makes a sequence of moves and ends in the *t*-th block. It is known that Alice never moves outside of the road.
You are given the description of Alice's footprints. Your task is to find a pair of possible values of *s*,<=*t* by looking at the footprints.
Input Specification:
The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=1000).
The second line contains the description of the road — the string that consists of *n* characters. Each character will be either "." (a block without footprint), or "L" (a block with a left footprint), "R" (a block with a right footprint).
It's guaranteed that the given string contains at least one character not equal to ".". Also, the first and the last character will always be ".". It's guaranteed that a solution exists.
Output Specification:
Print two space-separated integers — the values of *s* and *t*. If there are several possible solutions you can print any of them.
Demo Input:
['9\n..RRLL...\n', '11\n.RRRLLLLL..\n']
Demo Output:
['3 4\n', '7 5\n']
Note:
The first test sample is the one in the picture.
|
```python
n = int(input())
word = input()
if word.count("R")==0:
print(word.rfind("L")+1,word.find("L"))
elif word.count("L")==0:
print(word.find("R")+1,word.rfind("R")+2)
else:
print(word.find("R")+1,word.rfind("R")+1)
```
| 3
|
|
959
|
A
|
Mahmoud and Ehab and the even-odd game
|
PROGRAMMING
| 800
|
[
"games",
"math"
] | null | null |
Mahmoud and Ehab play a game called the even-odd game. Ehab chooses his favorite integer *n* and then they take turns, starting from Mahmoud. In each player's turn, he has to choose an integer *a* and subtract it from *n* such that:
- 1<=≤<=*a*<=≤<=*n*. - If it's Mahmoud's turn, *a* has to be even, but if it's Ehab's turn, *a* has to be odd.
If the current player can't choose any number satisfying the conditions, he loses. Can you determine the winner if they both play optimally?
|
The only line contains an integer *n* (1<=≤<=*n*<=≤<=109), the number at the beginning of the game.
|
Output "Mahmoud" (without quotes) if Mahmoud wins and "Ehab" (without quotes) otherwise.
|
[
"1\n",
"2\n"
] |
[
"Ehab",
"Mahmoud"
] |
In the first sample, Mahmoud can't choose any integer *a* initially because there is no positive even integer less than or equal to 1 so Ehab wins.
In the second sample, Mahmoud has to choose *a* = 2 and subtract it from *n*. It's Ehab's turn and *n* = 0. There is no positive odd integer less than or equal to 0 so Mahmoud wins.
| 500
|
[
{
"input": "1",
"output": "Ehab"
},
{
"input": "2",
"output": "Mahmoud"
},
{
"input": "10000",
"output": "Mahmoud"
},
{
"input": "33333",
"output": "Ehab"
},
{
"input": "5",
"output": "Ehab"
},
{
"input": "1000000000",
"output": "Mahmoud"
},
{
"input": "999999999",
"output": "Ehab"
},
{
"input": "123123123",
"output": "Ehab"
},
{
"input": "22222221",
"output": "Ehab"
},
{
"input": "22222220",
"output": "Mahmoud"
},
{
"input": "3",
"output": "Ehab"
},
{
"input": "4",
"output": "Mahmoud"
},
{
"input": "6",
"output": "Mahmoud"
},
{
"input": "7",
"output": "Ehab"
},
{
"input": "8",
"output": "Mahmoud"
},
{
"input": "9",
"output": "Ehab"
},
{
"input": "10",
"output": "Mahmoud"
},
{
"input": "536870912",
"output": "Mahmoud"
},
{
"input": "536870913",
"output": "Ehab"
},
{
"input": "536870911",
"output": "Ehab"
}
] | 1,616,330,758
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 20
| 124
| 0
|
j=int(input())
k=0
i=0
if j%2==0:
print("Mahmoud")
else:
print("Ehab")
|
Title: Mahmoud and Ehab and the even-odd game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mahmoud and Ehab play a game called the even-odd game. Ehab chooses his favorite integer *n* and then they take turns, starting from Mahmoud. In each player's turn, he has to choose an integer *a* and subtract it from *n* such that:
- 1<=≤<=*a*<=≤<=*n*. - If it's Mahmoud's turn, *a* has to be even, but if it's Ehab's turn, *a* has to be odd.
If the current player can't choose any number satisfying the conditions, he loses. Can you determine the winner if they both play optimally?
Input Specification:
The only line contains an integer *n* (1<=≤<=*n*<=≤<=109), the number at the beginning of the game.
Output Specification:
Output "Mahmoud" (without quotes) if Mahmoud wins and "Ehab" (without quotes) otherwise.
Demo Input:
['1\n', '2\n']
Demo Output:
['Ehab', 'Mahmoud']
Note:
In the first sample, Mahmoud can't choose any integer *a* initially because there is no positive even integer less than or equal to 1 so Ehab wins.
In the second sample, Mahmoud has to choose *a* = 2 and subtract it from *n*. It's Ehab's turn and *n* = 0. There is no positive odd integer less than or equal to 0 so Mahmoud wins.
|
```python
j=int(input())
k=0
i=0
if j%2==0:
print("Mahmoud")
else:
print("Ehab")
```
| 3
|
|
126
|
B
|
Password
|
PROGRAMMING
| 1,700
|
[
"binary search",
"dp",
"hashing",
"string suffix structures",
"strings"
] | null | null |
Asterix, Obelix and their temporary buddies Suffix and Prefix has finally found the Harmony temple. However, its doors were firmly locked and even Obelix had no luck opening them.
A little later they found a string *s*, carved on a rock below the temple's gates. Asterix supposed that that's the password that opens the temple and read the string aloud. However, nothing happened. Then Asterix supposed that a password is some substring *t* of the string *s*.
Prefix supposed that the substring *t* is the beginning of the string *s*; Suffix supposed that the substring *t* should be the end of the string *s*; and Obelix supposed that *t* should be located somewhere inside the string *s*, that is, *t* is neither its beginning, nor its end.
Asterix chose the substring *t* so as to please all his companions. Besides, from all acceptable variants Asterix chose the longest one (as Asterix loves long strings). When Asterix read the substring *t* aloud, the temple doors opened.
You know the string *s*. Find the substring *t* or determine that such substring does not exist and all that's been written above is just a nice legend.
|
You are given the string *s* whose length can vary from 1 to 106 (inclusive), consisting of small Latin letters.
|
Print the string *t*. If a suitable *t* string does not exist, then print "Just a legend" without the quotes.
|
[
"fixprefixsuffix\n",
"abcdabc\n"
] |
[
"fix",
"Just a legend"
] |
none
| 1,000
|
[
{
"input": "fixprefixsuffix",
"output": "fix"
},
{
"input": "abcdabc",
"output": "Just a legend"
},
{
"input": "qwertyqwertyqwerty",
"output": "qwerty"
},
{
"input": "papapapap",
"output": "papap"
},
{
"input": "aaaaaaaaaa",
"output": "aaaaaaaa"
},
{
"input": "ghbdtn",
"output": "Just a legend"
},
{
"input": "a",
"output": "Just a legend"
},
{
"input": "aa",
"output": "Just a legend"
},
{
"input": "ab",
"output": "Just a legend"
},
{
"input": "aaa",
"output": "a"
},
{
"input": "aba",
"output": "Just a legend"
},
{
"input": "aab",
"output": "Just a legend"
},
{
"input": "abb",
"output": "Just a legend"
},
{
"input": "abc",
"output": "Just a legend"
},
{
"input": "aaabaabaaaaab",
"output": "Just a legend"
},
{
"input": "aabaaabaaaaab",
"output": "aab"
},
{
"input": "aaabaaaabab",
"output": "Just a legend"
},
{
"input": "abcabcabcabcabc",
"output": "abcabcabc"
},
{
"input": "aaaaabaaaa",
"output": "aaaa"
},
{
"input": "aaaabaaaaaaa",
"output": "aaaa"
},
{
"input": "ghghghgxghghghg",
"output": "ghghg"
},
{
"input": "kincenvizh",
"output": "Just a legend"
},
{
"input": "amcksgurlgqzqizdauqminfzshiweejkevbazyzylrrghumnvqeqqdedyopgtvxakqwpvxntxgrkrcxabhrgoxngrwrxrvcguuyw",
"output": "Just a legend"
},
{
"input": "kwuaizneqxfflhmyruotjlkqksinoanvkyvqptkkntnpjdyzicceelgooajdgpkneuhyvhdtmasiglplajxolxovlhkwuaizneqx",
"output": "Just a legend"
},
{
"input": "nfbdzgdlbjhrlvfryyjbvtsmzacxglcvukmyexdgpuiwvqbnfbdzgdlbjhrlvfryyjbtuomcwbwvlhefnfbdzgdlbjhrlvfryyjb",
"output": "nfbdzgdlbjhrlvfryyjb"
},
{
"input": "btbdpnzdenxueteteytvkwnegodyhmdwhmrmbftrifytzudumzlacwytsuzefqfpjesdblxhpcsidgksisjdmhyxmmugrjmtfcgxmcgpmvbxbyszdugvmoyjeeqsywcvuqskcjotyhvdijyslgfvyujxthngnonasgnbzzygyikheelirdfsvuicbiazfdeqecbtbdpnzdenxueteteytvkwnegodyhmdwhmrmbftrifytzudumzlacwytsuzefqfpjesdblxhpcsidgksisjdmhyxmmugrjmtfcgxmcgpccdtdteohtjjuicfjfcbzpynlqhecyqqrjpmexgfqywnfmusldgzvgnzdmuuujtkdfnlytzgstquwqpuccxbtbdpnzdenxueteteytvkwnegodyhmdwhmrmbftrifytzudumzlacwytsuzefqfpjesdblxhpcsidgksisjdmhyxmmugrjmtfcgxmcgp",
"output": "btbdpnzdenxueteteytvkwnegodyhmdwhmrmbftrifytzudumzlacwytsuzefqfpjesdblxhpcsidgksisjdmhyxmmugrjmtfcgxmcgp"
},
{
"input": "abcazabca",
"output": "a"
}
] | 1,679,054,958
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 62
| 0
|
def kmp_preprocess(pattern):
n = len(pattern)
pi = [0] * n
j = 0
for i in range(1, n):
while j > 0 and pattern[i] != pattern[j]:
j = pi[j-1]
if pattern[i] == pattern[j]:
j += 1
pi[i] = j
return pi
def solve(s):
n = len(s)
pi = kmp_preprocess(s)
i = pi[-1]
while i > 0 and 2*i > n:
i = pi[i-1]
if i == 0:
print("Just a legend")
else:
print(s[:i])
# Example usage
s = input()
solve(s)
|
Title: Password
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Asterix, Obelix and their temporary buddies Suffix and Prefix has finally found the Harmony temple. However, its doors were firmly locked and even Obelix had no luck opening them.
A little later they found a string *s*, carved on a rock below the temple's gates. Asterix supposed that that's the password that opens the temple and read the string aloud. However, nothing happened. Then Asterix supposed that a password is some substring *t* of the string *s*.
Prefix supposed that the substring *t* is the beginning of the string *s*; Suffix supposed that the substring *t* should be the end of the string *s*; and Obelix supposed that *t* should be located somewhere inside the string *s*, that is, *t* is neither its beginning, nor its end.
Asterix chose the substring *t* so as to please all his companions. Besides, from all acceptable variants Asterix chose the longest one (as Asterix loves long strings). When Asterix read the substring *t* aloud, the temple doors opened.
You know the string *s*. Find the substring *t* or determine that such substring does not exist and all that's been written above is just a nice legend.
Input Specification:
You are given the string *s* whose length can vary from 1 to 106 (inclusive), consisting of small Latin letters.
Output Specification:
Print the string *t*. If a suitable *t* string does not exist, then print "Just a legend" without the quotes.
Demo Input:
['fixprefixsuffix\n', 'abcdabc\n']
Demo Output:
['fix', 'Just a legend']
Note:
none
|
```python
def kmp_preprocess(pattern):
n = len(pattern)
pi = [0] * n
j = 0
for i in range(1, n):
while j > 0 and pattern[i] != pattern[j]:
j = pi[j-1]
if pattern[i] == pattern[j]:
j += 1
pi[i] = j
return pi
def solve(s):
n = len(s)
pi = kmp_preprocess(s)
i = pi[-1]
while i > 0 and 2*i > n:
i = pi[i-1]
if i == 0:
print("Just a legend")
else:
print(s[:i])
# Example usage
s = input()
solve(s)
```
| 0
|
|
451
|
A
|
Game With Sticks
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
After winning gold and silver in IOI 2014, Akshat and Malvika want to have some fun. Now they are playing a game on a grid made of *n* horizontal and *m* vertical sticks.
An intersection point is any point on the grid which is formed by the intersection of one horizontal stick and one vertical stick.
In the grid shown below, *n*<==<=3 and *m*<==<=3. There are *n*<=+<=*m*<==<=6 sticks in total (horizontal sticks are shown in red and vertical sticks are shown in green). There are *n*·*m*<==<=9 intersection points, numbered from 1 to 9.
The rules of the game are very simple. The players move in turns. Akshat won gold, so he makes the first move. During his/her move, a player must choose any remaining intersection point and remove from the grid all sticks which pass through this point. A player will lose the game if he/she cannot make a move (i.e. there are no intersection points remaining on the grid at his/her move).
Assume that both players play optimally. Who will win the game?
|
The first line of input contains two space-separated integers, *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
|
Print a single line containing "Akshat" or "Malvika" (without the quotes), depending on the winner of the game.
|
[
"2 2\n",
"2 3\n",
"3 3\n"
] |
[
"Malvika\n",
"Malvika\n",
"Akshat\n"
] |
Explanation of the first sample:
The grid has four intersection points, numbered from 1 to 4.
If Akshat chooses intersection point 1, then he will remove two sticks (1 - 2 and 1 - 3). The resulting grid will look like this.
Now there is only one remaining intersection point (i.e. 4). Malvika must choose it and remove both remaining sticks. After her move the grid will be empty.
In the empty grid, Akshat cannot make any move, hence he will lose.
Since all 4 intersection points of the grid are equivalent, Akshat will lose no matter which one he picks.
| 500
|
[
{
"input": "2 2",
"output": "Malvika"
},
{
"input": "2 3",
"output": "Malvika"
},
{
"input": "3 3",
"output": "Akshat"
},
{
"input": "20 68",
"output": "Malvika"
},
{
"input": "1 1",
"output": "Akshat"
},
{
"input": "1 2",
"output": "Akshat"
},
{
"input": "1 3",
"output": "Akshat"
},
{
"input": "2 1",
"output": "Akshat"
},
{
"input": "2 2",
"output": "Malvika"
},
{
"input": "3 1",
"output": "Akshat"
},
{
"input": "3 2",
"output": "Malvika"
},
{
"input": "68 42",
"output": "Malvika"
},
{
"input": "1 35",
"output": "Akshat"
},
{
"input": "25 70",
"output": "Akshat"
},
{
"input": "59 79",
"output": "Akshat"
},
{
"input": "65 63",
"output": "Akshat"
},
{
"input": "46 6",
"output": "Malvika"
},
{
"input": "28 82",
"output": "Malvika"
},
{
"input": "98 98",
"output": "Malvika"
},
{
"input": "98 99",
"output": "Malvika"
},
{
"input": "98 100",
"output": "Malvika"
},
{
"input": "99 98",
"output": "Malvika"
},
{
"input": "99 99",
"output": "Akshat"
},
{
"input": "99 100",
"output": "Akshat"
},
{
"input": "100 98",
"output": "Malvika"
},
{
"input": "100 99",
"output": "Akshat"
},
{
"input": "100 100",
"output": "Malvika"
},
{
"input": "3 4",
"output": "Akshat"
}
] | 1,682,370,592
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 5
| 108
| 0
|
sticks = [int(i) for i in input().split()]
if sticks[0] % 2 == 0 or sticks[1] % 2 == 0:
print('Malvika')
else:
print('Akshat')
|
Title: Game With Sticks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After winning gold and silver in IOI 2014, Akshat and Malvika want to have some fun. Now they are playing a game on a grid made of *n* horizontal and *m* vertical sticks.
An intersection point is any point on the grid which is formed by the intersection of one horizontal stick and one vertical stick.
In the grid shown below, *n*<==<=3 and *m*<==<=3. There are *n*<=+<=*m*<==<=6 sticks in total (horizontal sticks are shown in red and vertical sticks are shown in green). There are *n*·*m*<==<=9 intersection points, numbered from 1 to 9.
The rules of the game are very simple. The players move in turns. Akshat won gold, so he makes the first move. During his/her move, a player must choose any remaining intersection point and remove from the grid all sticks which pass through this point. A player will lose the game if he/she cannot make a move (i.e. there are no intersection points remaining on the grid at his/her move).
Assume that both players play optimally. Who will win the game?
Input Specification:
The first line of input contains two space-separated integers, *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
Output Specification:
Print a single line containing "Akshat" or "Malvika" (without the quotes), depending on the winner of the game.
Demo Input:
['2 2\n', '2 3\n', '3 3\n']
Demo Output:
['Malvika\n', 'Malvika\n', 'Akshat\n']
Note:
Explanation of the first sample:
The grid has four intersection points, numbered from 1 to 4.
If Akshat chooses intersection point 1, then he will remove two sticks (1 - 2 and 1 - 3). The resulting grid will look like this.
Now there is only one remaining intersection point (i.e. 4). Malvika must choose it and remove both remaining sticks. After her move the grid will be empty.
In the empty grid, Akshat cannot make any move, hence he will lose.
Since all 4 intersection points of the grid are equivalent, Akshat will lose no matter which one he picks.
|
```python
sticks = [int(i) for i in input().split()]
if sticks[0] % 2 == 0 or sticks[1] % 2 == 0:
print('Malvika')
else:
print('Akshat')
```
| 0
|
|
116
|
A
|
Tram
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Linear Kingdom has exactly one tram line. It has *n* stops, numbered from 1 to *n* in the order of tram's movement. At the *i*-th stop *a**i* passengers exit the tram, while *b**i* passengers enter it. The tram is empty before it arrives at the first stop. Also, when the tram arrives at the last stop, all passengers exit so that it becomes empty.
Your task is to calculate the tram's minimum capacity such that the number of people inside the tram at any time never exceeds this capacity. Note that at each stop all exiting passengers exit before any entering passenger enters the tram.
|
The first line contains a single number *n* (2<=≤<=*n*<=≤<=1000) — the number of the tram's stops.
Then *n* lines follow, each contains two integers *a**i* and *b**i* (0<=≤<=*a**i*,<=*b**i*<=≤<=1000) — the number of passengers that exits the tram at the *i*-th stop, and the number of passengers that enter the tram at the *i*-th stop. The stops are given from the first to the last stop in the order of tram's movement.
- The number of people who exit at a given stop does not exceed the total number of people in the tram immediately before it arrives at the stop. More formally, . This particularly means that *a*1<==<=0. - At the last stop, all the passengers exit the tram and it becomes empty. More formally, . - No passenger will enter the train at the last stop. That is, *b**n*<==<=0.
|
Print a single integer denoting the minimum possible capacity of the tram (0 is allowed).
|
[
"4\n0 3\n2 5\n4 2\n4 0\n"
] |
[
"6\n"
] |
For the first example, a capacity of 6 is sufficient:
- At the first stop, the number of passengers inside the tram before arriving is 0. Then, 3 passengers enter the tram, and the number of passengers inside the tram becomes 3. - At the second stop, 2 passengers exit the tram (1 passenger remains inside). Then, 5 passengers enter the tram. There are 6 passengers inside the tram now. - At the third stop, 4 passengers exit the tram (2 passengers remain inside). Then, 2 passengers enter the tram. There are 4 passengers inside the tram now. - Finally, all the remaining passengers inside the tram exit the tram at the last stop. There are no passenger inside the tram now, which is in line with the constraints.
Since the number of passengers inside the tram never exceeds 6, a capacity of 6 is sufficient. Furthermore it is not possible for the tram to have a capacity less than 6. Hence, 6 is the correct answer.
| 500
|
[
{
"input": "4\n0 3\n2 5\n4 2\n4 0",
"output": "6"
},
{
"input": "5\n0 4\n4 6\n6 5\n5 4\n4 0",
"output": "6"
},
{
"input": "10\n0 5\n1 7\n10 8\n5 3\n0 5\n3 3\n8 8\n0 6\n10 1\n9 0",
"output": "18"
},
{
"input": "3\n0 1\n1 1\n1 0",
"output": "1"
},
{
"input": "4\n0 1\n0 1\n1 0\n1 0",
"output": "2"
},
{
"input": "3\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "3\n0 1000\n1000 1000\n1000 0",
"output": "1000"
},
{
"input": "5\n0 73\n73 189\n189 766\n766 0\n0 0",
"output": "766"
},
{
"input": "5\n0 0\n0 0\n0 0\n0 1\n1 0",
"output": "1"
},
{
"input": "5\n0 917\n917 923\n904 992\n1000 0\n11 0",
"output": "1011"
},
{
"input": "5\n0 1\n1 2\n2 1\n1 2\n2 0",
"output": "2"
},
{
"input": "5\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "20\n0 7\n2 1\n2 2\n5 7\n2 6\n6 10\n2 4\n0 4\n7 4\n8 0\n10 6\n2 1\n6 1\n1 7\n0 3\n8 7\n6 3\n6 3\n1 1\n3 0",
"output": "22"
},
{
"input": "5\n0 1000\n1000 1000\n1000 1000\n1000 1000\n1000 0",
"output": "1000"
},
{
"input": "10\n0 592\n258 598\n389 203\n249 836\n196 635\n478 482\n994 987\n1000 0\n769 0\n0 0",
"output": "1776"
},
{
"input": "10\n0 1\n1 0\n0 0\n0 0\n0 0\n0 1\n1 1\n0 1\n1 0\n1 0",
"output": "2"
},
{
"input": "10\n0 926\n926 938\n938 931\n931 964\n937 989\n983 936\n908 949\n997 932\n945 988\n988 0",
"output": "1016"
},
{
"input": "10\n0 1\n1 2\n1 2\n2 2\n2 2\n2 2\n1 1\n1 1\n2 1\n2 0",
"output": "3"
},
{
"input": "10\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "10\n0 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 0",
"output": "1000"
},
{
"input": "50\n0 332\n332 268\n268 56\n56 711\n420 180\n160 834\n149 341\n373 777\n763 93\n994 407\n86 803\n700 132\n471 608\n429 467\n75 5\n638 305\n405 853\n316 478\n643 163\n18 131\n648 241\n241 766\n316 847\n640 380\n923 759\n789 41\n125 421\n421 9\n9 388\n388 829\n408 108\n462 856\n816 411\n518 688\n290 7\n405 912\n397 772\n396 652\n394 146\n27 648\n462 617\n514 433\n780 35\n710 705\n460 390\n194 508\n643 56\n172 469\n1000 0\n194 0",
"output": "2071"
},
{
"input": "50\n0 0\n0 1\n1 1\n0 1\n0 0\n1 0\n0 0\n1 0\n0 0\n0 0\n0 0\n0 0\n0 1\n0 0\n0 0\n0 1\n1 0\n0 1\n0 0\n1 1\n1 0\n0 1\n0 0\n1 1\n0 1\n1 0\n1 1\n1 0\n0 0\n1 1\n1 0\n0 1\n0 0\n0 1\n1 1\n1 1\n1 1\n1 0\n1 1\n1 0\n0 1\n1 0\n0 0\n0 1\n1 1\n1 1\n0 1\n0 0\n1 0\n1 0",
"output": "3"
},
{
"input": "50\n0 926\n926 971\n915 980\n920 965\n954 944\n928 952\n955 980\n916 980\n906 935\n944 913\n905 923\n912 922\n965 934\n912 900\n946 930\n931 983\n979 905\n925 969\n924 926\n910 914\n921 977\n934 979\n962 986\n942 909\n976 903\n982 982\n991 941\n954 929\n902 980\n947 983\n919 924\n917 943\n916 905\n907 913\n964 977\n984 904\n905 999\n950 970\n986 906\n993 970\n960 994\n963 983\n918 986\n980 900\n931 986\n993 997\n941 909\n907 909\n1000 0\n278 0",
"output": "1329"
},
{
"input": "2\n0 863\n863 0",
"output": "863"
},
{
"input": "50\n0 1\n1 2\n2 2\n1 1\n1 1\n1 2\n1 2\n1 1\n1 2\n1 1\n1 1\n1 2\n1 2\n1 1\n2 1\n2 2\n1 2\n2 2\n1 2\n2 1\n2 1\n2 2\n2 1\n1 2\n1 2\n2 1\n1 1\n2 2\n1 1\n2 1\n2 2\n2 1\n1 2\n2 2\n1 2\n1 1\n1 1\n2 1\n2 1\n2 2\n2 1\n2 1\n1 2\n1 2\n1 2\n1 2\n2 0\n2 0\n2 0\n0 0",
"output": "8"
},
{
"input": "50\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "100\n0 1\n0 0\n0 0\n1 0\n0 0\n0 1\n0 1\n1 1\n0 0\n0 0\n1 1\n0 0\n1 1\n0 1\n1 1\n0 1\n1 1\n1 0\n1 0\n0 0\n1 0\n0 1\n1 0\n0 0\n0 0\n1 1\n1 1\n0 1\n0 0\n1 0\n1 1\n0 1\n1 0\n1 1\n0 1\n1 1\n1 0\n0 0\n0 0\n0 1\n0 0\n0 1\n1 1\n0 0\n1 1\n1 1\n0 0\n0 1\n1 0\n0 1\n0 0\n0 1\n0 1\n1 1\n1 1\n1 1\n0 0\n0 0\n1 1\n0 1\n0 1\n1 0\n0 0\n0 0\n1 1\n0 1\n0 1\n1 1\n1 1\n0 1\n1 1\n1 1\n0 0\n1 0\n0 1\n0 0\n0 0\n1 1\n1 1\n1 1\n1 1\n0 1\n1 0\n1 0\n1 0\n1 0\n1 0\n0 0\n1 0\n1 0\n0 0\n1 0\n0 0\n0 1\n1 0\n0 1\n1 0\n1 0\n1 0\n1 0",
"output": "11"
},
{
"input": "100\n0 2\n1 2\n2 1\n1 2\n1 2\n2 1\n2 2\n1 1\n1 1\n2 1\n1 2\n2 1\n1 2\n2 2\n2 2\n2 2\n1 2\n2 2\n2 1\n1 1\n1 1\n1 1\n2 2\n1 2\n2 2\n1 1\n1 1\n1 1\n1 1\n2 2\n1 2\n2 1\n1 1\n2 2\n1 1\n2 1\n1 1\n2 2\n2 1\n1 2\n1 1\n1 2\n2 1\n2 2\n1 1\n2 1\n1 1\n2 1\n1 1\n1 2\n2 2\n2 2\n1 1\n2 2\n1 2\n2 1\n2 1\n1 1\n1 1\n1 2\n1 2\n1 1\n1 1\n2 1\n1 2\n1 2\n2 1\n2 2\n2 2\n2 2\n2 1\n2 2\n1 1\n1 2\n1 2\n1 1\n2 2\n2 2\n1 1\n2 1\n1 1\n1 2\n1 2\n1 2\n1 1\n1 1\n2 2\n1 2\n2 1\n2 1\n2 1\n1 2\n1 2\n1 1\n2 2\n1 2\n2 0\n2 0\n2 0\n1 0",
"output": "7"
},
{
"input": "100\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "2\n0 1\n1 0",
"output": "1"
},
{
"input": "2\n0 900\n900 0",
"output": "900"
},
{
"input": "2\n0 1\n1 0",
"output": "1"
},
{
"input": "2\n0 0\n0 0",
"output": "0"
},
{
"input": "2\n0 1000\n1000 0",
"output": "1000"
},
{
"input": "3\n0 802\n175 188\n815 0",
"output": "815"
},
{
"input": "3\n0 910\n910 976\n976 0",
"output": "976"
},
{
"input": "3\n0 2\n2 1\n1 0",
"output": "2"
}
] | 1,699,694,179
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 71
| 186
| 2,048,000
|
n = int(input())
curr = 0
ans = 0
for rep1 in range(0,n):
a, b = (int(x) for x in input().split())
curr -= a
curr += b
if curr>ans:
ans=curr
print(ans)
|
Title: Tram
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Linear Kingdom has exactly one tram line. It has *n* stops, numbered from 1 to *n* in the order of tram's movement. At the *i*-th stop *a**i* passengers exit the tram, while *b**i* passengers enter it. The tram is empty before it arrives at the first stop. Also, when the tram arrives at the last stop, all passengers exit so that it becomes empty.
Your task is to calculate the tram's minimum capacity such that the number of people inside the tram at any time never exceeds this capacity. Note that at each stop all exiting passengers exit before any entering passenger enters the tram.
Input Specification:
The first line contains a single number *n* (2<=≤<=*n*<=≤<=1000) — the number of the tram's stops.
Then *n* lines follow, each contains two integers *a**i* and *b**i* (0<=≤<=*a**i*,<=*b**i*<=≤<=1000) — the number of passengers that exits the tram at the *i*-th stop, and the number of passengers that enter the tram at the *i*-th stop. The stops are given from the first to the last stop in the order of tram's movement.
- The number of people who exit at a given stop does not exceed the total number of people in the tram immediately before it arrives at the stop. More formally, . This particularly means that *a*1<==<=0. - At the last stop, all the passengers exit the tram and it becomes empty. More formally, . - No passenger will enter the train at the last stop. That is, *b**n*<==<=0.
Output Specification:
Print a single integer denoting the minimum possible capacity of the tram (0 is allowed).
Demo Input:
['4\n0 3\n2 5\n4 2\n4 0\n']
Demo Output:
['6\n']
Note:
For the first example, a capacity of 6 is sufficient:
- At the first stop, the number of passengers inside the tram before arriving is 0. Then, 3 passengers enter the tram, and the number of passengers inside the tram becomes 3. - At the second stop, 2 passengers exit the tram (1 passenger remains inside). Then, 5 passengers enter the tram. There are 6 passengers inside the tram now. - At the third stop, 4 passengers exit the tram (2 passengers remain inside). Then, 2 passengers enter the tram. There are 4 passengers inside the tram now. - Finally, all the remaining passengers inside the tram exit the tram at the last stop. There are no passenger inside the tram now, which is in line with the constraints.
Since the number of passengers inside the tram never exceeds 6, a capacity of 6 is sufficient. Furthermore it is not possible for the tram to have a capacity less than 6. Hence, 6 is the correct answer.
|
```python
n = int(input())
curr = 0
ans = 0
for rep1 in range(0,n):
a, b = (int(x) for x in input().split())
curr -= a
curr += b
if curr>ans:
ans=curr
print(ans)
```
| 3
|
|
373
|
A
|
Collecting Beats is Fun
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Cucumber boy is fan of Kyubeat, a famous music game.
Kyubeat has 16 panels for playing arranged in 4<=×<=4 table. When a panel lights up, he has to press that panel.
Each panel has a timing to press (the preffered time when a player should press it), and Cucumber boy is able to press at most *k* panels in a time with his one hand. Cucumber boy is trying to press all panels in perfect timing, that is he wants to press each panel exactly in its preffered time. If he cannot press the panels with his two hands in perfect timing, his challenge to press all the panels in perfect timing will fail.
You are given one scene of Kyubeat's panel from the music Cucumber boy is trying. Tell him is he able to press all the panels in perfect timing.
|
The first line contains a single integer *k* (1<=≤<=*k*<=≤<=5) — the number of panels Cucumber boy can press with his one hand.
Next 4 lines contain 4 characters each (digits from 1 to 9, or period) — table of panels. If a digit *i* was written on the panel, it means the boy has to press that panel in time *i*. If period was written on the panel, he doesn't have to press that panel.
|
Output "YES" (without quotes), if he is able to press all the panels in perfect timing. If not, output "NO" (without quotes).
|
[
"1\n.135\n1247\n3468\n5789\n",
"5\n..1.\n1111\n..1.\n..1.\n",
"1\n....\n12.1\n.2..\n.2..\n"
] |
[
"YES\n",
"YES\n",
"NO\n"
] |
In the third sample boy cannot press all panels in perfect timing. He can press all the panels in timing in time 1, but he cannot press the panels in time 2 in timing with his two hands.
| 500
|
[
{
"input": "1\n.135\n1247\n3468\n5789",
"output": "YES"
},
{
"input": "5\n..1.\n1111\n..1.\n..1.",
"output": "YES"
},
{
"input": "1\n....\n12.1\n.2..\n.2..",
"output": "NO"
},
{
"input": "1\n6981\n.527\n4163\n2345",
"output": "YES"
},
{
"input": "5\n9999\n9999\n9999\n9999",
"output": "NO"
},
{
"input": "2\n4444\n3333\n2222\n1111",
"output": "YES"
},
{
"input": "3\n2123\n1232\n2321\n3213",
"output": "NO"
},
{
"input": "2\n1...\n.1..\n..1.\n...1",
"output": "YES"
},
{
"input": "2\n1.1.\n.1.1\n2.2.\n.222",
"output": "NO"
},
{
"input": "1\n1..2\n.3.4\n567.\n.89.",
"output": "YES"
},
{
"input": "1\n1122\n3344\n5588\n6699",
"output": "YES"
},
{
"input": "4\n1111\n1221\n1221\n1111",
"output": "NO"
},
{
"input": "5\n3141\n5926\n5358\n9793",
"output": "YES"
},
{
"input": "5\n5454\n4343\n3232\n2121",
"output": "YES"
},
{
"input": "5\n1222\n2221\n2221\n1122",
"output": "NO"
},
{
"input": "1\n....\n5..5\n6..6\n7..7",
"output": "YES"
},
{
"input": "1\n9875\n8643\n7421\n531.",
"output": "YES"
},
{
"input": "1\n..1.\n..1.\n..1.\n..1.",
"output": "NO"
},
{
"input": "4\n4.4.\n4.4.\n4444\n..4.",
"output": "NO"
},
{
"input": "5\n4.4.\n4.4.\n4444\n..4.",
"output": "YES"
},
{
"input": "3\n1.1.\n.1.1\n1.1.\n.1.1",
"output": "NO"
},
{
"input": "2\n1131\n4412\n2569\n3478",
"output": "YES"
},
{
"input": "2\n8888\n8888\n8888\n8888",
"output": "NO"
}
] | 1,554,327,111
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 3
| 109
| 0
|
k = int(input())
tab = []
for i in range(4):
tab.append(input())
tab = ''.join(tab)
maxx = int(max(list(tab)))
maxkeys = k * 2;
for i in range(1, maxx+1):
if tab.count(str(i)) > maxkeys:
print("NO")
exit(0)
print("YES")
|
Title: Collecting Beats is Fun
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Cucumber boy is fan of Kyubeat, a famous music game.
Kyubeat has 16 panels for playing arranged in 4<=×<=4 table. When a panel lights up, he has to press that panel.
Each panel has a timing to press (the preffered time when a player should press it), and Cucumber boy is able to press at most *k* panels in a time with his one hand. Cucumber boy is trying to press all panels in perfect timing, that is he wants to press each panel exactly in its preffered time. If he cannot press the panels with his two hands in perfect timing, his challenge to press all the panels in perfect timing will fail.
You are given one scene of Kyubeat's panel from the music Cucumber boy is trying. Tell him is he able to press all the panels in perfect timing.
Input Specification:
The first line contains a single integer *k* (1<=≤<=*k*<=≤<=5) — the number of panels Cucumber boy can press with his one hand.
Next 4 lines contain 4 characters each (digits from 1 to 9, or period) — table of panels. If a digit *i* was written on the panel, it means the boy has to press that panel in time *i*. If period was written on the panel, he doesn't have to press that panel.
Output Specification:
Output "YES" (without quotes), if he is able to press all the panels in perfect timing. If not, output "NO" (without quotes).
Demo Input:
['1\n.135\n1247\n3468\n5789\n', '5\n..1.\n1111\n..1.\n..1.\n', '1\n....\n12.1\n.2..\n.2..\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
In the third sample boy cannot press all panels in perfect timing. He can press all the panels in timing in time 1, but he cannot press the panels in time 2 in timing with his two hands.
|
```python
k = int(input())
tab = []
for i in range(4):
tab.append(input())
tab = ''.join(tab)
maxx = int(max(list(tab)))
maxkeys = k * 2;
for i in range(1, maxx+1):
if tab.count(str(i)) > maxkeys:
print("NO")
exit(0)
print("YES")
```
| -1
|
|
841
|
A
|
Generous Kefa
|
PROGRAMMING
| 900
|
[
"brute force",
"implementation"
] | null | null |
One day Kefa found *n* baloons. For convenience, we denote color of *i*-th baloon as *s**i* — lowercase letter of the Latin alphabet. Also Kefa has *k* friends. Friend will be upset, If he get two baloons of the same color. Kefa want to give out all baloons to his friends. Help Kefa to find out, can he give out all his baloons, such that no one of his friens will be upset — print «YES», if he can, and «NO», otherwise. Note, that Kefa's friend will not upset, if he doesn't get baloons at all.
|
The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100) — the number of baloons and friends.
Next line contains string *s* — colors of baloons.
|
Answer to the task — «YES» or «NO» in a single line.
You can choose the case (lower or upper) for each letter arbitrary.
|
[
"4 2\naabb\n",
"6 3\naacaab\n"
] |
[
"YES\n",
"NO\n"
] |
In the first sample Kefa can give 1-st and 3-rd baloon to the first friend, and 2-nd and 4-th to the second.
In the second sample Kefa needs to give to all his friends baloons of color a, but one baloon will stay, thats why answer is «NO».
| 500
|
[
{
"input": "4 2\naabb",
"output": "YES"
},
{
"input": "6 3\naacaab",
"output": "NO"
},
{
"input": "2 2\nlu",
"output": "YES"
},
{
"input": "5 3\novvoo",
"output": "YES"
},
{
"input": "36 13\nbzbzcffczzcbcbzzfzbbfzfzzbfbbcbfccbf",
"output": "YES"
},
{
"input": "81 3\nooycgmvvrophvcvpoupepqllqttwcocuilvyxbyumdmmfapvpnxhjhxfuagpnntonibicaqjvwfhwxhbv",
"output": "NO"
},
{
"input": "100 100\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx",
"output": "YES"
},
{
"input": "100 1\nnubcvvjvbjgnjsdkajimdcxvewbcytvfkihunycdrlconddlwgzjasjlsrttlrzsumzpyumpveglfqzmaofbshbojmwuwoxxvrod",
"output": "NO"
},
{
"input": "100 13\nvyldolgryldqrvoldvzvrdrgorlorszddtgqvrlisxxrxdxlqtvtgsrqlzixoyrozxzogqxlsgzdddzqrgitxxritoolzolgrtvl",
"output": "YES"
},
{
"input": "18 6\njzwtnkvmscqhmdlsxy",
"output": "YES"
},
{
"input": "21 2\nfscegcqgzesefghhwcexs",
"output": "NO"
},
{
"input": "32 22\ncduamsptaklqtxlyoutlzepxgyfkvngc",
"output": "YES"
},
{
"input": "49 27\noxyorfnkzwsfllnyvdhdanppuzrnbxehugvmlkgeymqjlmfxd",
"output": "YES"
},
{
"input": "50 24\nxxutzjwbggcwvxztttkmzovtmuwttzcbwoztttohzzxghuuthv",
"output": "YES"
},
{
"input": "57 35\nglxshztrqqfyxthqamagvtmrdparhelnzrqvcwqxjytkbuitovkdxueul",
"output": "YES"
},
{
"input": "75 23\nittttiiuitutuiiuuututiuttiuiuutuuuiuiuuuuttuuttuutuiiuiuiiuiitttuututuiuuii",
"output": "NO"
},
{
"input": "81 66\nfeqevfqfebhvubhuuvfuqheuqhbeeuebehuvhffvbqvqvfbqqvvhevqffbqqhvvqhfeehuhqeqhueuqqq",
"output": "YES"
},
{
"input": "93 42\npqeiafraiavfcteumflpcbpozcomlvpovlzdbldvoopnhdoeqaopzthiuzbzmeieiatthdeqovaqfipqlddllmfcrrnhb",
"output": "YES"
},
{
"input": "100 53\nizszyqyndzwzyzgsdagdwdazadiawizinagqqgczaqqnawgijziziawzszdjdcqjdjqiwgadydcnqisaayjiqqsscwwzjzaycwwc",
"output": "YES"
},
{
"input": "100 14\nvkrdcqbvkwuckpmnbydmczdxoagdsgtqxvhaxntdcxhjcrjyvukhugoglbmyoaqexgtcfdgemmizoniwtmisqqwcwfusmygollab",
"output": "YES"
},
{
"input": "100 42\naaaaaiiiiaiiiaaiaiiaaiiiiiaaaaaiaiiiaiiiiaiiiaaaaaiiiaaaiiaaiiiaiiiaiaaaiaiiiiaaiiiaiiaiaiiaiiiaaaia",
"output": "NO"
},
{
"input": "100 89\ntjbkmydejporbqhcbztkcumxjjgsrvxpuulbhzeeckkbchpbxwhedrlhjsabcexcohgdzouvsgphjdthpuqrlkgzxvqbuhqxdsmf",
"output": "YES"
},
{
"input": "100 100\njhpyiuuzizhubhhpxbbhpyxzhbpjphzppuhiahihiappbhuypyauhizpbibzixjbzxzpbphuiaypyujappuxiyuyaajaxjupbahb",
"output": "YES"
},
{
"input": "100 3\nsszoovvzysavsvzsozzvoozvysozsaszayaszasaysszzzysosyayyvzozovavzoyavsooaoyvoozvvozsaosvayyovazzszzssa",
"output": "NO"
},
{
"input": "100 44\ndluthkxwnorabqsukgnxnvhmsmzilyulpursnxkdsavgemiuizbyzebhyjejgqrvuckhaqtuvdmpziesmpmewpvozdanjyvwcdgo",
"output": "YES"
},
{
"input": "100 90\ntljonbnwnqounictqqctgonktiqoqlocgoblngijqokuquoolciqwnctgoggcbojtwjlculoikbggquqncittwnjbkgkgubnioib",
"output": "YES"
},
{
"input": "100 79\nykxptzgvbqxlregvkvucewtydvnhqhuggdsyqlvcfiuaiddnrrnstityyehiamrggftsqyduwxpuldztyzgmfkehprrneyvtknmf",
"output": "YES"
},
{
"input": "100 79\naagwekyovbviiqeuakbqbqifwavkfkutoriovgfmittulhwojaptacekdirgqoovlleeoqkkdukpadygfwavppohgdrmymmulgci",
"output": "YES"
},
{
"input": "100 93\nearrehrehenaddhdnrdddhdahnadndheeennrearrhraharddreaeraddhehhhrdnredanndneheddrraaneerreedhnadnerhdn",
"output": "YES"
},
{
"input": "100 48\nbmmaebaebmmmbbmxvmammbvvebvaemvbbaxvbvmaxvvmveaxmbbxaaemxmxvxxxvxbmmxaaaevvaxmvamvvmaxaxavexbmmbmmev",
"output": "YES"
},
{
"input": "100 55\nhsavbkehaaesffaeeffakhkhfehbbvbeasahbbbvkesbfvkefeesesevbsvfkbffakvshsbkahfkfakebsvafkbvsskfhfvaasss",
"output": "YES"
},
{
"input": "100 2\ncscffcffsccffsfsfffccssfsscfsfsssffcffsscfccssfffcfscfsscsccccfsssffffcfcfsfffcsfsccffscffcfccccfffs",
"output": "NO"
},
{
"input": "100 3\nzrgznxgdpgfoiifrrrsjfuhvtqxjlgochhyemismjnanfvvpzzvsgajcbsulxyeoepjfwvhkqogiiwqxjkrpsyaqdlwffoockxnc",
"output": "NO"
},
{
"input": "100 5\njbltyyfjakrjeodqepxpkjideulofbhqzxjwlarufwzwsoxhaexpydpqjvhybmvjvntuvhvflokhshpicbnfgsqsmrkrfzcrswwi",
"output": "NO"
},
{
"input": "100 1\nfnslnqktlbmxqpvcvnemxcutebdwepoxikifkzaaixzzydffpdxodmsxjribmxuqhueifdlwzytxkklwhljswqvlejedyrgguvah",
"output": "NO"
},
{
"input": "100 21\nddjenetwgwmdtjbpzssyoqrtirvoygkjlqhhdcjgeurqpunxpupwaepcqkbjjfhnvgpyqnozhhrmhfwararmlcvpgtnopvjqsrka",
"output": "YES"
},
{
"input": "100 100\nnjrhiauqlgkkpkuvciwzivjbbplipvhslqgdkfnmqrxuxnycmpheenmnrglotzuyxycosfediqcuadklsnzjqzfxnbjwvfljnlvq",
"output": "YES"
},
{
"input": "100 100\nbbbbbbbtbbttbtbbbttbttbtbbttttbbbtbttbbbtbttbtbbttttbbbbbtbbttbtbbtbttbbbtbtbtbtbtbtbbbttbbtbtbtbbtb",
"output": "YES"
},
{
"input": "14 5\nfssmmsfffmfmmm",
"output": "NO"
},
{
"input": "2 1\nff",
"output": "NO"
},
{
"input": "2 1\nhw",
"output": "YES"
},
{
"input": "2 2\nss",
"output": "YES"
},
{
"input": "1 1\nl",
"output": "YES"
},
{
"input": "100 50\nfffffttttttjjjuuuvvvvvdddxxxxwwwwgggbsssncccczzyyyyyhhhhhkrreeeeeeaaaaaiiillllllllooooqqqqqqmmpppppp",
"output": "YES"
},
{
"input": "100 50\nbbbbbbbbgggggggggggaaaaaaaahhhhhhhhhhpppppppppsssssssrrrrrrrrllzzzzzzzeeeeeeekkkkkkkwwwwwwwwjjjjjjjj",
"output": "YES"
},
{
"input": "100 50\nwwwwwwwwwwwwwwxxxxxxxxxxxxxxxxxxxxxxxxzzzzzzzzzzzzzzzzzzbbbbbbbbbbbbbbbbbbbbjjjjjjjjjjjjjjjjjjjjjjjj",
"output": "YES"
},
{
"input": "100 80\nbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm",
"output": "YES"
},
{
"input": "100 10\nbbttthhhhiiiiiiijjjjjvvvvpppssssseeeeeeewwwwgggkkkkkkkkmmmddddduuuzzzzllllnnnnnxxyyyffffccraaaaooooq",
"output": "YES"
},
{
"input": "100 20\nssssssssssbbbbbbbhhhhhhhyyyyyyyzzzzzzzzzzzzcccccxxxxxxxxxxddddmmmmmmmeeeeeeejjjjjjjjjwwwwwwwtttttttt",
"output": "YES"
},
{
"input": "1 2\na",
"output": "YES"
},
{
"input": "3 1\nabb",
"output": "NO"
},
{
"input": "2 1\naa",
"output": "NO"
},
{
"input": "2 1\nab",
"output": "YES"
},
{
"input": "6 2\naaaaaa",
"output": "NO"
},
{
"input": "8 4\naaaaaaaa",
"output": "NO"
},
{
"input": "4 2\naaaa",
"output": "NO"
},
{
"input": "4 3\naaaa",
"output": "NO"
},
{
"input": "1 3\na",
"output": "YES"
},
{
"input": "4 3\nzzzz",
"output": "NO"
},
{
"input": "4 1\naaaa",
"output": "NO"
},
{
"input": "3 4\nabc",
"output": "YES"
},
{
"input": "2 5\nab",
"output": "YES"
},
{
"input": "2 4\nab",
"output": "YES"
},
{
"input": "1 10\na",
"output": "YES"
},
{
"input": "5 2\nzzzzz",
"output": "NO"
},
{
"input": "53 26\naaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbbbbbbbbbbbbbbb",
"output": "NO"
},
{
"input": "4 1\nabab",
"output": "NO"
},
{
"input": "4 1\nabcb",
"output": "NO"
},
{
"input": "4 2\nabbb",
"output": "NO"
},
{
"input": "5 2\nabccc",
"output": "NO"
},
{
"input": "2 3\nab",
"output": "YES"
},
{
"input": "4 3\nbbbs",
"output": "YES"
},
{
"input": "10 2\nazzzzzzzzz",
"output": "NO"
},
{
"input": "1 2\nb",
"output": "YES"
},
{
"input": "1 3\nb",
"output": "YES"
},
{
"input": "4 5\nabcd",
"output": "YES"
},
{
"input": "4 6\naabb",
"output": "YES"
},
{
"input": "5 2\naaaab",
"output": "NO"
},
{
"input": "3 5\naaa",
"output": "YES"
},
{
"input": "5 3\nazzzz",
"output": "NO"
},
{
"input": "4 100\naabb",
"output": "YES"
},
{
"input": "3 10\naaa",
"output": "YES"
},
{
"input": "3 4\naaa",
"output": "YES"
},
{
"input": "12 5\naaaaabbbbbbb",
"output": "NO"
},
{
"input": "5 2\naabbb",
"output": "NO"
},
{
"input": "10 5\nzzzzzzzzzz",
"output": "NO"
},
{
"input": "2 4\naa",
"output": "YES"
},
{
"input": "1 5\na",
"output": "YES"
},
{
"input": "10 5\naaaaaaaaaa",
"output": "NO"
},
{
"input": "6 3\naaaaaa",
"output": "NO"
},
{
"input": "7 1\nabcdeee",
"output": "NO"
},
{
"input": "18 3\naaaaaabbbbbbcccccc",
"output": "NO"
},
{
"input": "8 2\naabbccdd",
"output": "YES"
},
{
"input": "4 2\nzzzz",
"output": "NO"
},
{
"input": "4 2\nabaa",
"output": "NO"
},
{
"input": "3 2\naaa",
"output": "NO"
},
{
"input": "3 1\nzzz",
"output": "NO"
},
{
"input": "5 4\nzzzzz",
"output": "NO"
},
{
"input": "6 2\naabbbc",
"output": "NO"
},
{
"input": "3 6\naaa",
"output": "YES"
},
{
"input": "2 1\nzz",
"output": "NO"
},
{
"input": "10 3\naaaeeeeeee",
"output": "NO"
},
{
"input": "4 5\naabb",
"output": "YES"
},
{
"input": "3 1\naaa",
"output": "NO"
},
{
"input": "5 2\naazzz",
"output": "NO"
},
{
"input": "6 2\nabbbbc",
"output": "NO"
},
{
"input": "4 2\nxxxx",
"output": "NO"
},
{
"input": "6 3\nzzzzzz",
"output": "NO"
},
{
"input": "3 2\nabb",
"output": "YES"
},
{
"input": "3 2\nzzz",
"output": "NO"
},
{
"input": "6 5\nzzzzzz",
"output": "NO"
},
{
"input": "6 3\nbcaaaa",
"output": "NO"
},
{
"input": "100 100\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "YES"
},
{
"input": "3 6\nabc",
"output": "YES"
}
] | 1,612,569,198
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 114
| 77
| 0
|
num = input().split()
balloons_original = input()
balloons = []
result = ""
for j in balloons_original:
balloons.append(j)
for i in range(int(num[0])):
if balloons.count(balloons[i]) > int(num[1]):
result = "NO"
break
if result == "":
result = "YES"
print(result)
|
Title: Generous Kefa
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Kefa found *n* baloons. For convenience, we denote color of *i*-th baloon as *s**i* — lowercase letter of the Latin alphabet. Also Kefa has *k* friends. Friend will be upset, If he get two baloons of the same color. Kefa want to give out all baloons to his friends. Help Kefa to find out, can he give out all his baloons, such that no one of his friens will be upset — print «YES», if he can, and «NO», otherwise. Note, that Kefa's friend will not upset, if he doesn't get baloons at all.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100) — the number of baloons and friends.
Next line contains string *s* — colors of baloons.
Output Specification:
Answer to the task — «YES» or «NO» in a single line.
You can choose the case (lower or upper) for each letter arbitrary.
Demo Input:
['4 2\naabb\n', '6 3\naacaab\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample Kefa can give 1-st and 3-rd baloon to the first friend, and 2-nd and 4-th to the second.
In the second sample Kefa needs to give to all his friends baloons of color a, but one baloon will stay, thats why answer is «NO».
|
```python
num = input().split()
balloons_original = input()
balloons = []
result = ""
for j in balloons_original:
balloons.append(j)
for i in range(int(num[0])):
if balloons.count(balloons[i]) > int(num[1]):
result = "NO"
break
if result == "":
result = "YES"
print(result)
```
| 3
|
|
652
|
B
|
z-sort
|
PROGRAMMING
| 1,000
|
[
"sortings"
] | null | null |
A student of *z*-school found a kind of sorting called *z*-sort. The array *a* with *n* elements are *z*-sorted if two conditions hold:
1. *a**i*<=≥<=*a**i*<=-<=1 for all even *i*, 1. *a**i*<=≤<=*a**i*<=-<=1 for all odd *i*<=><=1.
For example the arrays [1,2,1,2] and [1,1,1,1] are *z*-sorted while the array [1,2,3,4] isn’t *z*-sorted.
Can you make the array *z*-sorted?
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of elements in the array *a*.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=109) — the elements of the array *a*.
|
If it's possible to make the array *a* *z*-sorted print *n* space separated integers *a**i* — the elements after *z*-sort. Otherwise print the only word "Impossible".
|
[
"4\n1 2 2 1\n",
"5\n1 3 2 2 5\n"
] |
[
"1 2 1 2\n",
"1 5 2 3 2\n"
] |
none
| 0
|
[
{
"input": "4\n1 2 2 1",
"output": "1 2 1 2"
},
{
"input": "5\n1 3 2 2 5",
"output": "1 5 2 3 2"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1 1",
"output": "1 1 1 1 1 1 1 1 1 1"
},
{
"input": "10\n1 9 7 6 2 4 7 8 1 3",
"output": "1 9 1 8 2 7 3 7 4 6"
},
{
"input": "100\n82 51 81 14 37 17 78 92 64 15 8 86 89 8 87 77 66 10 15 12 100 25 92 47 21 78 20 63 13 49 41 36 41 79 16 87 87 69 3 76 80 60 100 49 70 59 72 8 38 71 45 97 71 14 76 54 81 4 59 46 39 29 92 3 49 22 53 99 59 52 74 31 92 43 42 23 44 9 82 47 7 40 12 9 3 55 37 85 46 22 84 52 98 41 21 77 63 17 62 91",
"output": "3 100 3 100 3 99 4 98 7 97 8 92 8 92 8 92 9 92 9 91 10 89 12 87 12 87 13 87 14 86 14 85 15 84 15 82 16 82 17 81 17 81 20 80 21 79 21 78 22 78 22 77 23 77 25 76 29 76 31 74 36 72 37 71 37 71 38 70 39 69 40 66 41 64 41 63 41 63 42 62 43 60 44 59 45 59 46 59 46 55 47 54 47 53 49 52 49 52 49 51"
},
{
"input": "3\n1 2 6",
"output": "1 6 2"
},
{
"input": "136\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1"
},
{
"input": "3\n1 2 3",
"output": "1 3 2"
},
{
"input": "7\n999999998 999999999 999999999 999999999 999999999 999999999 1000000000",
"output": "999999998 1000000000 999999999 999999999 999999999 999999999 999999999"
},
{
"input": "3\n100 1 2",
"output": "1 100 2"
}
] | 1,596,367,432
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 2
| 108
| 6,963,200
|
n=int(input())
x=list(map(int,input().split()))
x.sort()
lengtho=n-1
if n%2==0:
point=n//2-1
else:
point=n//2
for i in range(point):
print(x[i],x[lengtho-i],end=' ')
if n%2==0:
print(x[i],x[lengtho-i])
else:print(x[i])
|
Title: z-sort
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A student of *z*-school found a kind of sorting called *z*-sort. The array *a* with *n* elements are *z*-sorted if two conditions hold:
1. *a**i*<=≥<=*a**i*<=-<=1 for all even *i*, 1. *a**i*<=≤<=*a**i*<=-<=1 for all odd *i*<=><=1.
For example the arrays [1,2,1,2] and [1,1,1,1] are *z*-sorted while the array [1,2,3,4] isn’t *z*-sorted.
Can you make the array *z*-sorted?
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of elements in the array *a*.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=109) — the elements of the array *a*.
Output Specification:
If it's possible to make the array *a* *z*-sorted print *n* space separated integers *a**i* — the elements after *z*-sort. Otherwise print the only word "Impossible".
Demo Input:
['4\n1 2 2 1\n', '5\n1 3 2 2 5\n']
Demo Output:
['1 2 1 2\n', '1 5 2 3 2\n']
Note:
none
|
```python
n=int(input())
x=list(map(int,input().split()))
x.sort()
lengtho=n-1
if n%2==0:
point=n//2-1
else:
point=n//2
for i in range(point):
print(x[i],x[lengtho-i],end=' ')
if n%2==0:
print(x[i],x[lengtho-i])
else:print(x[i])
```
| -1
|
|
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"
}
] | 1,630,026,966
| 2,147,483,647
|
PyPy 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 248
| 22,220,800
|
from typing_extensions import final
num_v = int(input())
vec_lst = []
final_vec = 0
for i in range(num_v):
l = list(map(int, input().split()))
vec_lst.append(l)
for dim in range(3):
final_vec = 0
for vec in vec_lst:
final_vec += vec[dim]
if final_vec != 0:
print("NO")
break
if final_vec == 0: print("YES")
|
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
from typing_extensions import final
num_v = int(input())
vec_lst = []
final_vec = 0
for i in range(num_v):
l = list(map(int, input().split()))
vec_lst.append(l)
for dim in range(3):
final_vec = 0
for vec in vec_lst:
final_vec += vec[dim]
if final_vec != 0:
print("NO")
break
if final_vec == 0: print("YES")
```
| -1
|
626
|
A
|
Robot Sequence
|
PROGRAMMING
| 1,000
|
[
"brute force",
"implementation"
] | null | null |
Calvin the robot lies in an infinite rectangular grid. Calvin's source code contains a list of *n* commands, each either 'U', 'R', 'D', or 'L' — instructions to move a single square up, right, down, or left, respectively. How many ways can Calvin execute a non-empty contiguous substrings of commands and return to the same square he starts in? Two substrings are considered different if they have different starting or ending indices.
|
The first line of the input contains a single positive integer, *n* (1<=≤<=*n*<=≤<=200) — the number of commands.
The next line contains *n* characters, each either 'U', 'R', 'D', or 'L' — Calvin's source code.
|
Print a single integer — the number of contiguous substrings that Calvin can execute and return to his starting square.
|
[
"6\nURLLDR\n",
"4\nDLUU\n",
"7\nRLRLRLR\n"
] |
[
"2\n",
"0\n",
"12\n"
] |
In the first case, the entire source code works, as well as the "RL" substring in the second and third characters.
Note that, in the third case, the substring "LR" appears three times, and is therefore counted three times to the total result.
| 500
|
[
{
"input": "6\nURLLDR",
"output": "2"
},
{
"input": "4\nDLUU",
"output": "0"
},
{
"input": "7\nRLRLRLR",
"output": "12"
},
{
"input": "1\nR",
"output": "0"
},
{
"input": "100\nURDLURDLURDLURDLURDLURDLURDLURDLURDLURDLURDLURDLURDLURDLURDLURDLURDLURDLURDLURDLURDLURDLURDLURDLURDL",
"output": "1225"
},
{
"input": "200\nLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "100"
},
{
"input": "20\nLDURLDURRLRUDLRRUDLU",
"output": "29"
},
{
"input": "140\nDLDLULULDRDDDLLUDRRDLLUULLDDLDLUURLDLDRDUDDLRRDURUUUUURLDUDDLLRRLLDRRRDDDDDUDUULLURRDLDULUDLLUUDRRLUDULUDUDULULUURURRDUURRDLULLURUDDDDRDRDRD",
"output": "125"
},
{
"input": "194\nULLLDLLDRUUDURRULLRLUUURDRLLURDUDDUDLULRLDRUDURLDLRDLLLLUDDRRRULULULUDDULRURURLLDLDLDRUDUUDULRULDDRRLRDRULLDRULLLLRRDDLLLLULDRLUULRUUULDUUDLDLDUUUDDLDDRULDRRLUURRULLDULRRDLLRDURDLUUDUDLLUDDULDDD",
"output": "282"
},
{
"input": "200\nDDDURLLUUULUDDURRDLLDDLLRLUULUULDDDLRRDLRRDUDURDUDRRLLDRDUDDLDDRDLURRRLLRDRRLLLRDDDRDRRLLRRLULRUULRLDLUDRRRDDUUURLLUDRLDUDRLLRLRRLUDLRULDUDDRRLLRLURDLRUDDDURLRDUDUUURLLULULRDRLDLDRURDDDLLRUDDRDUDDDLRU",
"output": "408"
},
{
"input": "197\nDUUDUDUDUDUUDUUDUUUDDDDUUUDUUUDUUUUUDUUUDDUDDDUUDUDDDUUDDUUUUUUUDUDDDDDUUUUUDDDDDDUUUUDDUDDUDDDUDUUUDUUDUDUDUUUDUDDDDUUDDUDDDDUDDDUDUUUDUUDUUUDDDDUUUDUUDDUUUUUDDDDUUDUUDDDDUDDUUDUUUDDDDUDUUUDDDUUDU",
"output": "1995"
},
{
"input": "200\nLLLLRLLRLLRRRRLLRRLRRLRRRLLLRRLRRRRLLRRLLRRRLRLRLRRLLRLLRRLLLRRRRLRLLRLLLRLLLRRLLLRLRLRRRRRRRLRRRLRLRLLLLRLRRRRRLRRLRLLLLRLLLRRLRRLLRLRLLLRRLLRRLRRRRRLRLRRLRLLRLLLLRLRRRLRRLRLLRLRRLRRRRRLRRLLLRRRRRLLR",
"output": "1368"
},
{
"input": "184\nUUUDDUDDDDDUDDDDUDDUUUUUDDDUUDDUDUUDUUUDDUDDDDDDDDDDUDUDDUUDDDUUDDUDUDDDUUDUDUUUUDDUDUUUDDUDUUUUDUUDDUUDUUUDUDUDDUDUDDDUUDDDDUUUUUDDDUDUDUDUDUDUUUDUDDUUDDUDUUDUDUUUDUUDDDDUDDDDUDUUDUUD",
"output": "1243"
},
{
"input": "187\nRLLRLRRLLRRLRRRRLLRLLRLLLLRRRLLLRLLLLRRLRLRRRRRRLLRRLRLLRRRLLRRLLLRRLRRLRLLLLRRRRLRRLLRRLRRRRLLLLRRLRLRLRRRRRLLRLRLRLRLRLRLLLRLLLLLRRRLLRLRRRLLLRRLLLLLRLLRLLLRRRLLLRRLRRRLLLRRLRLLRRLRLRLR",
"output": "1501"
},
{
"input": "190\nUULLLUUULLLULLUULUUUUULUUULLULLULUULLUULLUUULULUULLUULLUUULULLLLLLULLLLLULUULLULLULLLUULUULLLUUUULLLLUUULLUUULLLULULUULULLUULULULUUULLUUUULLUUULULUULLLLULLLLLUULLUULULLULUUUUUULULLLULLUULUUU",
"output": "0"
},
{
"input": "46\nULUURRRRLDRDRDDDURRRLLLDDULLRRRRRLUDDLRDRULLLL",
"output": "23"
},
{
"input": "70\nUUDRLDRDRUDLLURURULRDULRRDULDUDDRUULLDDDDDRLLRDURRDULRDLRUUUDDLRUURRLD",
"output": "86"
},
{
"input": "198\nURLLUDRDUUDRDLLRURULLRRLRRUULRLULUUDRRURLRUURRDRUUDRLRURLLULRDDDDDRDDRRRLRUDULLDDLLLUDRLDRUDRDLDUULLUUUULULLRLDDRDURDRURLULDRURLLDDULURULDLUUUUULDLURRLLDLULLDULRUURRLDLLUUURDLDDUDUULRLUDULLULDRDRLRL",
"output": "160"
},
{
"input": "22\nDUDDDURURUDURRUDRDULUL",
"output": "10"
},
{
"input": "200\nUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUD",
"output": "10000"
},
{
"input": "4\nRRDR",
"output": "0"
},
{
"input": "6\nUULLLL",
"output": "0"
},
{
"input": "2\nDU",
"output": "1"
},
{
"input": "6\nUURRRR",
"output": "0"
},
{
"input": "101\nRDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD",
"output": "0"
}
] | 1,455,408,522
| 8,322
|
Python 3
|
OK
|
TESTS
| 25
| 62
| 0
|
n=eval(input())
s=input()
i=0
ans=0
while(i<n):
r,l,u,d=0,0,0,0
g=i
while(g<n):
if(s[g]=='R'):
r+=1
elif(s[g]=='L'):
l+=1
elif(s[g]=='U'):
u+=1
elif(s[g]=='D'):
d+=1
if(d==u and l ==r):
ans+=1
g+=1
i+=1
print(ans)
|
Title: Robot Sequence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Calvin the robot lies in an infinite rectangular grid. Calvin's source code contains a list of *n* commands, each either 'U', 'R', 'D', or 'L' — instructions to move a single square up, right, down, or left, respectively. How many ways can Calvin execute a non-empty contiguous substrings of commands and return to the same square he starts in? Two substrings are considered different if they have different starting or ending indices.
Input Specification:
The first line of the input contains a single positive integer, *n* (1<=≤<=*n*<=≤<=200) — the number of commands.
The next line contains *n* characters, each either 'U', 'R', 'D', or 'L' — Calvin's source code.
Output Specification:
Print a single integer — the number of contiguous substrings that Calvin can execute and return to his starting square.
Demo Input:
['6\nURLLDR\n', '4\nDLUU\n', '7\nRLRLRLR\n']
Demo Output:
['2\n', '0\n', '12\n']
Note:
In the first case, the entire source code works, as well as the "RL" substring in the second and third characters.
Note that, in the third case, the substring "LR" appears three times, and is therefore counted three times to the total result.
|
```python
n=eval(input())
s=input()
i=0
ans=0
while(i<n):
r,l,u,d=0,0,0,0
g=i
while(g<n):
if(s[g]=='R'):
r+=1
elif(s[g]=='L'):
l+=1
elif(s[g]=='U'):
u+=1
elif(s[g]=='D'):
d+=1
if(d==u and l ==r):
ans+=1
g+=1
i+=1
print(ans)
```
| 3
|
|
75
|
B
|
Facetook Priority Wall
|
PROGRAMMING
| 1,500
|
[
"expression parsing",
"implementation",
"strings"
] |
B. Facetook Priority Wall
|
2
|
256
|
Facetook is a well known social network website, and it will launch a new feature called Facetook Priority Wall. This feature will sort all posts from your friends according to the priority factor (it will be described).
This priority factor will be affected by three types of actions:
- 1. "*X* posted on *Y*'s wall" (15 points), - 2. "*X* commented on *Y*'s post" (10 points), - 3. "*X* likes *Y*'s post" (5 points).
*X* and *Y* will be two distinct names. And each action will increase the priority factor between *X* and *Y* (and vice versa) by the above value of points (the priority factor between *X* and *Y* is the same as the priority factor between *Y* and *X*).
You will be given *n* actions with the above format (without the action number and the number of points), and you have to print all the distinct names in these actions sorted according to the priority factor with you.
|
The first line contains your name. The second line contains an integer *n*, which is the number of actions (1<=≤<=*n*<=≤<=100). Then *n* lines follow, it is guaranteed that each one contains exactly 1 action in the format given above. There is exactly one space between each two words in a line, and there are no extra spaces. All the letters are lowercase. All names in the input will consist of at least 1 letter and at most 10 small Latin letters.
|
Print *m* lines, where *m* is the number of distinct names in the input (excluding yourself). Each line should contain just 1 name. The names should be sorted according to the priority factor with you in the descending order (the highest priority factor should come first). If two or more names have the same priority factor, print them in the alphabetical (lexicographical) order.
Note, that you should output all the names that are present in the input data (excluding yourself), even if that person has a zero priority factor.
The lexicographical comparison is performed by the standard "<" operator in modern programming languages. The line *a* is lexicographically smaller than the line *b*, if either *a* is the prefix of *b*, or if exists such an *i* (1<=≤<=*i*<=≤<=*min*(|*a*|,<=|*b*|)), that *a**i*<=<<=*b**i*, and for any *j* (1<=≤<=*j*<=<<=*i*) *a**j*<==<=*b**j*, where |*a*| and |*b*| stand for the lengths of strings *a* and *b* correspondently.
|
[
"ahmed\n3\nahmed posted on fatma's wall\nfatma commented on ahmed's post\nmona likes ahmed's post\n",
"aba\n1\nlikes likes posted's post\n"
] |
[
"fatma\nmona\n",
"likes\nposted\n"
] |
none
| 1,000
|
[
{
"input": "ahmed\n3\nahmed posted on fatma's wall\nfatma commented on ahmed's post\nmona likes ahmed's post",
"output": "fatma\nmona"
},
{
"input": "aba\n1\nlikes likes posted's post",
"output": "likes\nposted"
},
{
"input": "nu\n5\ng commented on pwyndmh's post\nqv posted on g's wall\ng likes nu's post\ng posted on nu's wall\nqv commented on pwyndmh's post",
"output": "g\npwyndmh\nqv"
},
{
"input": "szfwtzfp\n5\nzqx posted on szfwtzfp's wall\nr commented on scguem's post\nr posted on civ's wall\nr likes scguem's post\nr likes scguem's post",
"output": "zqx\nciv\nr\nscguem"
},
{
"input": "oaquudhavr\n3\ni posted on cwfwujpc's wall\ni likes oaquudhavr's post\noaquudhavr commented on cwfwujpc's post",
"output": "cwfwujpc\ni"
},
{
"input": "eo\n4\neo commented on xkgjgwxtrx's post\neo posted on iqquh's wall\nn commented on xkgjgwxtrx's post\niqquh commented on n's post",
"output": "iqquh\nxkgjgwxtrx\nn"
},
{
"input": "plwun\n3\neusjuq commented on plwun's post\nagktgdar likes eusjuq's post\nagppcoil likes agktgdar's post",
"output": "eusjuq\nagktgdar\nagppcoil"
},
{
"input": "fgzrn\n3\nzhl likes fgzrn's post\nxryet likes fgzrn's post\nzhl commented on fgzrn's post",
"output": "zhl\nxryet"
},
{
"input": "qatugmdjwg\n3\nb posted on cf's wall\nyjxkat posted on b's wall\nko commented on qatugmdjwg's post",
"output": "ko\nb\ncf\nyjxkat"
},
{
"input": "dagwdwxsuf\n5\nesrvncb commented on dagwdwxsuf's post\nzcepigpbz posted on dagwdwxsuf's wall\nesrvncb commented on zcepigpbz's post\nesrvncb commented on dagwdwxsuf's post\ndagwdwxsuf commented on esrvncb's post",
"output": "esrvncb\nzcepigpbz"
},
{
"input": "a\n1\nb likes c's post",
"output": "b\nc"
},
{
"input": "a\n1\nc likes b's post",
"output": "b\nc"
},
{
"input": "wuaiz\n10\nmnbggnud posted on xttaqvel's wall\ns posted on xopffmspf's wall\nkysxb likes qnrtpzkh's post\ngptks likes quebtsup's post\nkgmd commented on kmtnhsiue's post\newqjtxtiyn commented on a's post\nol posted on iglplaj's wall\nif posted on yuo's wall\nfs posted on dwjtuhgrq's wall\nygmdprun likes tzfneuly's post",
"output": "a\ndwjtuhgrq\newqjtxtiyn\nfs\ngptks\nif\niglplaj\nkgmd\nkmtnhsiue\nkysxb\nmnbggnud\nol\nqnrtpzkh\nquebtsup\ns\ntzfneuly\nxopffmspf\nxttaqvel\nygmdprun\nyuo"
},
{
"input": "fzhzg\n11\nv likes xyf's post\nktqtpzhlh commented on ffsxarrn's post\nktqtpzhlh commented on lbt's post\njcdwpcycj commented on qbuigcgflm's post\nl likes pmg's post\nracszbmsk posted on ojr's wall\nojr commented on n's post\nnzqx commented on lkj's post\nv posted on lzoca's wall\nnwqnoham commented on gyivezpu's post\nfzhzg likes uqvzgzrpac's post",
"output": "uqvzgzrpac\nffsxarrn\ngyivezpu\njcdwpcycj\nktqtpzhlh\nl\nlbt\nlkj\nlzoca\nn\nnwqnoham\nnzqx\nojr\npmg\nqbuigcgflm\nracszbmsk\nv\nxyf"
},
{
"input": "qdrnpb\n12\nymklhj commented on dkcbo's post\nhcucrenckl posted on mut's wall\nnvkyta commented on eo's post\npvgow likes mut's post\nob likes wlwcxtf's post\npvgow commented on advpu's post\nkfflyfbr commented on igozjnrxw's post\nsq commented on qdrnpb's post\nmrvn posted on lahduc's wall\ngsnlicy likes u's post\ndltqujf commented on qgzk's post\nr posted on bey's wall",
"output": "sq\nadvpu\nbey\ndkcbo\ndltqujf\neo\ngsnlicy\nhcucrenckl\nigozjnrxw\nkfflyfbr\nlahduc\nmrvn\nmut\nnvkyta\nob\npvgow\nqgzk\nr\nu\nwlwcxtf\nymklhj"
},
{
"input": "biycvwb\n13\nhp likes cigobksf's post\nmcoqt commented on gaswzwat's post\nnz posted on xyvetbokl's wall\nqbnwy commented on ylkfbwjy's post\nqdwktrro likes rxgujnzecs's post\nbbsw commented on hwtatkfnps's post\ngspx posted on ugjxfnahuc's wall\nxlmut likes plle's post\numbwlleag commented on xfwlhen's post\nrlwxqksbwi commented on rypqtrgf's post\nbj posted on vovq's wall\nozpdpb commented on zti's post\nhqj posted on rxgujnzecs's wall",
"output": "bbsw\nbj\ncigobksf\ngaswzwat\ngspx\nhp\nhqj\nhwtatkfnps\nmcoqt\nnz\nozpdpb\nplle\nqbnwy\nqdwktrro\nrlwxqksbwi\nrxgujnzecs\nrypqtrgf\nugjxfnahuc\numbwlleag\nvovq\nxfwlhen\nxlmut\nxyvetbokl\nylkfbwjy\nzti"
},
{
"input": "kmircqsffq\n14\nfrnf likes xgmmp's post\nfnfdpupayp commented on syz's post\nxefshpn commented on xgmmp's post\nm posted on gdwydzktok's wall\neskm likes pqmbnuc's post\npnqiapduhz likes zzqvjdz's post\nx likes nouuurc's post\nvnyxhoukuo posted on uhblapjab's wall\nblpjpxn likes zvwbger's post\nj posted on vuknetvl's wall\nscsw commented on xaggwxlxe's post\npqmbnuc commented on ojwaibie's post\niaazdlqdew commented on kmircqsffq's post\nqznqshxdi commented on umdqztoqun's post",
"output": "iaazdlqdew\nblpjpxn\neskm\nfnfdpupayp\nfrnf\ngdwydzktok\nj\nm\nnouuurc\nojwaibie\npnqiapduhz\npqmbnuc\nqznqshxdi\nscsw\nsyz\nuhblapjab\numdqztoqun\nvnyxhoukuo\nvuknetvl\nx\nxaggwxlxe\nxefshpn\nxgmmp\nzvwbger\nzzqvjdz"
},
{
"input": "posted\n3\nposted posted on fatma's wall\nfatma commented on posted's post\nmona likes posted's post",
"output": "fatma\nmona"
},
{
"input": "posted\n3\nposted posted on wall's wall\nwall commented on posted's post\nmona likes posted's post",
"output": "wall\nmona"
},
{
"input": "posted\n3\nposted posted on wall's wall\nwall commented on posted's post\npost likes posted's post",
"output": "wall\npost"
},
{
"input": "wall\n5\nwall posted on posted's wall\nwall posted on on's wall\nwall posted on commented's wall\nwall posted on likes's wall\nwall posted on post's wall",
"output": "commented\nlikes\non\npost\nposted"
},
{
"input": "commented\n5\non commented on commented's post\npos commented on commented's post\nlikes commented on commented's post\nposted commented on commented's post\nwall commented on commented's post",
"output": "likes\non\npos\nposted\nwall"
},
{
"input": "likes\n3\nlikes posted on post's wall\nlikes commented on on's post\nlikes likes commented's post",
"output": "post\non\ncommented"
},
{
"input": "on\n4\non posted on posted's wall\non commented on commented's post\non posted on wall's wall\non commented on post's post",
"output": "posted\nwall\ncommented\npost"
},
{
"input": "wall\n9\nwall posted on posted's wall\non commented on wall's post\nwall likes post's post\nposted posted on wall's wall\nwall commented on post's post\nlikes likes wall's post\nwall posted on on's wall\npost commented on wall's post\nwall likes likes's post",
"output": "posted\non\npost\nlikes"
},
{
"input": "post\n9\npost posted on wall's wall\non commented on post's post\npost likes likes's post\ncommented posted on post's wall\npost commented on likes's post\nlikes likes post's post\npost posted on posted's wall\non commented on post's post\npost likes commented's post",
"output": "commented\nlikes\non\nposted\nwall"
},
{
"input": "ahmed\n9\npost posted on ahmeds's wall\nahmeds commented on post's post\npost likes ahmeds's post\nahmeds posted on post's wall\npost commented on ahmeds's post\nahmeds likes post's post\npost posted on ahmeds's wall\nahmeds commented on post's post\npost likes ahmeds's post",
"output": "ahmeds\npost"
}
] | 1,639,417,994
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 60
| 248
| 0
|
n = 0
name = ""
s = ""
name = input()
n = int(input())
score = {}
points = {
"posted" : 15 ,
"commented" : 10 ,
"likes" : 5
}
for i in range(n) :
s = input().split(" ")
j = 3
if(s[1] == "likes") :
j = 2
s[j] = s[j][:len(s[j])-2]
if s[0] != name and s[0] not in score :
score[s[0]] = 0
if s[j] != name and s[j] not in score :
score[s[j]] = 0
if(name == s[0]) :
score[s[j]] += points[s[1]]
if name == s[j]:
score[s[0]] += points[s[1]]
for e in sorted( [(-value,key) for (key,value) in score.items()]) :
print(e[1])
|
Title: Facetook Priority Wall
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Facetook is a well known social network website, and it will launch a new feature called Facetook Priority Wall. This feature will sort all posts from your friends according to the priority factor (it will be described).
This priority factor will be affected by three types of actions:
- 1. "*X* posted on *Y*'s wall" (15 points), - 2. "*X* commented on *Y*'s post" (10 points), - 3. "*X* likes *Y*'s post" (5 points).
*X* and *Y* will be two distinct names. And each action will increase the priority factor between *X* and *Y* (and vice versa) by the above value of points (the priority factor between *X* and *Y* is the same as the priority factor between *Y* and *X*).
You will be given *n* actions with the above format (without the action number and the number of points), and you have to print all the distinct names in these actions sorted according to the priority factor with you.
Input Specification:
The first line contains your name. The second line contains an integer *n*, which is the number of actions (1<=≤<=*n*<=≤<=100). Then *n* lines follow, it is guaranteed that each one contains exactly 1 action in the format given above. There is exactly one space between each two words in a line, and there are no extra spaces. All the letters are lowercase. All names in the input will consist of at least 1 letter and at most 10 small Latin letters.
Output Specification:
Print *m* lines, where *m* is the number of distinct names in the input (excluding yourself). Each line should contain just 1 name. The names should be sorted according to the priority factor with you in the descending order (the highest priority factor should come first). If two or more names have the same priority factor, print them in the alphabetical (lexicographical) order.
Note, that you should output all the names that are present in the input data (excluding yourself), even if that person has a zero priority factor.
The lexicographical comparison is performed by the standard "<" operator in modern programming languages. The line *a* is lexicographically smaller than the line *b*, if either *a* is the prefix of *b*, or if exists such an *i* (1<=≤<=*i*<=≤<=*min*(|*a*|,<=|*b*|)), that *a**i*<=<<=*b**i*, and for any *j* (1<=≤<=*j*<=<<=*i*) *a**j*<==<=*b**j*, where |*a*| and |*b*| stand for the lengths of strings *a* and *b* correspondently.
Demo Input:
["ahmed\n3\nahmed posted on fatma's wall\nfatma commented on ahmed's post\nmona likes ahmed's post\n", "aba\n1\nlikes likes posted's post\n"]
Demo Output:
['fatma\nmona\n', 'likes\nposted\n']
Note:
none
|
```python
n = 0
name = ""
s = ""
name = input()
n = int(input())
score = {}
points = {
"posted" : 15 ,
"commented" : 10 ,
"likes" : 5
}
for i in range(n) :
s = input().split(" ")
j = 3
if(s[1] == "likes") :
j = 2
s[j] = s[j][:len(s[j])-2]
if s[0] != name and s[0] not in score :
score[s[0]] = 0
if s[j] != name and s[j] not in score :
score[s[j]] = 0
if(name == s[0]) :
score[s[j]] += points[s[1]]
if name == s[j]:
score[s[0]] += points[s[1]]
for e in sorted( [(-value,key) for (key,value) in score.items()]) :
print(e[1])
```
| 3.938
|
873
|
D
|
Merge Sort
|
PROGRAMMING
| 1,800
|
[
"constructive algorithms",
"divide and conquer"
] | null | null |
Merge sort is a well-known sorting algorithm. The main function that sorts the elements of array *a* with indices from [*l*,<=*r*) can be implemented as follows:
1. If the segment [*l*,<=*r*) is already sorted in non-descending order (that is, for any *i* such that *l*<=≤<=*i*<=<<=*r*<=-<=1 *a*[*i*]<=≤<=*a*[*i*<=+<=1]), then end the function call; 1. Let ; 1. Call *mergesort*(*a*,<=*l*,<=*mid*); 1. Call *mergesort*(*a*,<=*mid*,<=*r*); 1. Merge segments [*l*,<=*mid*) and [*mid*,<=*r*), making the segment [*l*,<=*r*) sorted in non-descending order. The merge algorithm doesn't call any other functions.
The array in this problem is 0-indexed, so to sort the whole array, you need to call *mergesort*(*a*,<=0,<=*n*).
The number of calls of function *mergesort* is very important, so Ivan has decided to calculate it while sorting the array. For example, if *a*<==<={1,<=2,<=3,<=4}, then there will be 1 call of *mergesort* — *mergesort*(0,<=4), which will check that the array is sorted and then end. If *a*<==<={2,<=1,<=3}, then the number of calls is 3: first of all, you call *mergesort*(0,<=3), which then sets *mid*<==<=1 and calls *mergesort*(0,<=1) and *mergesort*(1,<=3), which do not perform any recursive calls because segments (0,<=1) and (1,<=3) are sorted.
Ivan has implemented the program that counts the number of *mergesort* calls, but now he needs to test it. To do this, he needs to find an array *a* such that *a* is a permutation of size *n* (that is, the number of elements in *a* is *n*, and every integer number from [1,<=*n*] can be found in this array), and the number of *mergesort* calls when sorting the array is exactly *k*.
Help Ivan to find an array he wants!
|
The first line contains two numbers *n* and *k* (1<=≤<=*n*<=≤<=100000, 1<=≤<=*k*<=≤<=200000) — the size of a desired permutation and the number of *mergesort* calls required to sort it.
|
If a permutation of size *n* such that there will be exactly *k* calls of *mergesort* while sorting it doesn't exist, output <=-<=1. Otherwise output *n* integer numbers *a*[0],<=*a*[1],<=...,<=*a*[*n*<=-<=1] — the elements of a permutation that would meet the required conditions. If there are multiple answers, print any of them.
|
[
"3 3\n",
"4 1\n",
"5 6\n"
] |
[
"2 1 3 ",
"1 2 3 4 ",
"-1\n"
] |
none
| 0
|
[
{
"input": "3 3",
"output": "2 1 3 "
},
{
"input": "4 1",
"output": "1 2 3 4 "
},
{
"input": "5 6",
"output": "-1"
},
{
"input": "100 100",
"output": "-1"
},
{
"input": "10000 10001",
"output": "3 1 5 2 7 4 8 10 6 12 9 13 15 11 17 14 18 20 16 22 19 23 25 21 27 24 28 30 26 32 29 33 35 31 37 34 38 40 36 42 39 44 41 46 43 47 49 45 51 48 52 54 50 56 53 57 59 55 61 58 62 64 60 66 63 67 69 65 71 68 72 74 70 76 73 77 79 75 81 78 83 80 85 82 86 88 84 90 87 91 93 89 95 92 96 98 94 100 97 101 103 99 105 102 106 108 104 110 107 111 113 109 115 112 116 118 114 120 117 122 119 124 121 125 127 123 129 126 130 132 128 134 131 135 137 133 139 136 140 142 138 144 141 145 147 143 149 146 150 152 148 154 151 155 157..."
},
{
"input": "10000 20001",
"output": "-1"
},
{
"input": "10000 30001",
"output": "-1"
},
{
"input": "20000 10001",
"output": "3 1 5 2 7 4 8 10 6 12 9 13 15 11 17 14 18 20 16 22 19 23 25 21 27 24 28 30 26 32 29 33 35 31 37 34 38 40 36 42 39 44 41 46 43 47 49 45 51 48 52 54 50 56 53 57 59 55 61 58 62 64 60 66 63 67 69 65 71 68 72 74 70 76 73 77 79 75 81 78 83 80 85 82 86 88 84 90 87 91 93 89 95 92 96 98 94 100 97 101 103 99 105 102 106 108 104 110 107 111 113 109 115 112 116 118 114 120 117 122 119 124 121 125 127 123 129 126 130 132 128 134 131 135 137 133 139 136 140 142 138 144 141 145 147 143 149 146 150 152 148 154 151 155 157..."
},
{
"input": "20000 20001",
"output": "3 1 5 2 7 4 8 10 6 12 9 13 15 11 17 14 18 20 16 22 19 23 25 21 27 24 28 30 26 32 29 33 35 31 37 34 38 40 36 42 39 44 41 46 43 47 49 45 51 48 52 54 50 56 53 57 59 55 61 58 62 64 60 66 63 67 69 65 71 68 72 74 70 76 73 77 79 75 81 78 83 80 85 82 86 88 84 90 87 91 93 89 95 92 96 98 94 100 97 101 103 99 105 102 106 108 104 110 107 111 113 109 115 112 116 118 114 120 117 122 119 124 121 125 127 123 129 126 130 132 128 134 131 135 137 133 139 136 140 142 138 144 141 145 147 143 149 146 150 152 148 154 151 155 157..."
},
{
"input": "20000 30001",
"output": "3 1 5 2 7 4 8 10 6 12 9 13 15 11 17 14 18 20 16 22 19 23 25 21 27 24 28 30 26 32 29 33 35 31 37 34 38 40 36 42 39 44 41 46 43 47 49 45 51 48 52 54 50 56 53 57 59 55 61 58 62 64 60 66 63 67 69 65 71 68 72 74 70 76 73 77 79 75 81 78 83 80 85 82 86 88 84 90 87 91 93 89 95 92 96 98 94 100 97 101 103 99 105 102 106 108 104 110 107 111 113 109 115 112 116 118 114 120 117 122 119 124 121 125 127 123 129 126 130 132 128 134 131 135 137 133 139 136 140 142 138 144 141 145 147 143 149 146 150 152 148 154 151 155 157..."
},
{
"input": "30000 10001",
"output": "2 4 1 6 3 8 5 9 11 7 13 10 15 12 16 18 14 20 17 22 19 24 21 26 23 28 25 30 27 31 33 29 35 32 37 34 38 40 36 42 39 44 41 45 47 43 49 46 51 48 53 50 55 52 57 54 59 56 60 62 58 64 61 66 63 67 69 65 71 68 73 70 74 76 72 78 75 80 77 82 79 84 81 86 83 88 85 89 91 87 93 90 95 92 97 94 99 96 101 98 103 100 104 106 102 108 105 110 107 112 109 114 111 116 113 118 115 119 121 117 123 120 125 122 126 128 124 130 127 132 129 133 135 131 137 134 139 136 141 138 143 140 145 142 147 144 148 150 146 152 149 154 151 155 157..."
},
{
"input": "30000 20001",
"output": "2 4 1 6 3 8 5 9 11 7 13 10 15 12 16 18 14 20 17 22 19 24 21 26 23 28 25 30 27 31 33 29 35 32 37 34 38 40 36 42 39 44 41 45 47 43 49 46 51 48 53 50 55 52 57 54 59 56 60 62 58 64 61 66 63 67 69 65 71 68 73 70 74 76 72 78 75 80 77 82 79 84 81 86 83 88 85 89 91 87 93 90 95 92 97 94 99 96 101 98 103 100 104 106 102 108 105 110 107 112 109 114 111 116 113 118 115 119 121 117 123 120 125 122 126 128 124 130 127 132 129 133 135 131 137 134 139 136 141 138 143 140 145 142 147 144 148 150 146 152 149 154 151 155 157..."
},
{
"input": "30000 30001",
"output": "2 4 1 6 3 8 5 9 11 7 13 10 15 12 16 18 14 20 17 22 19 24 21 26 23 28 25 30 27 31 33 29 35 32 37 34 38 40 36 42 39 44 41 45 47 43 49 46 51 48 53 50 55 52 57 54 59 56 60 62 58 64 61 66 63 67 69 65 71 68 73 70 74 76 72 78 75 80 77 82 79 84 81 86 83 88 85 89 91 87 93 90 95 92 97 94 99 96 101 98 103 100 104 106 102 108 105 110 107 112 109 114 111 116 113 118 115 119 121 117 123 120 125 122 126 128 124 130 127 132 129 133 135 131 137 134 139 136 141 138 143 140 145 142 147 144 148 150 146 152 149 154 151 155 157..."
},
{
"input": "40000 10001",
"output": "3 1 5 2 7 4 8 10 6 12 9 13 15 11 17 14 18 20 16 22 19 23 25 21 27 24 28 30 26 32 29 33 35 31 37 34 38 40 36 42 39 44 41 46 43 47 49 45 51 48 52 54 50 56 53 57 59 55 61 58 62 64 60 66 63 67 69 65 71 68 72 74 70 76 73 77 79 75 81 78 83 80 85 82 86 88 84 90 87 91 93 89 95 92 96 98 94 100 97 101 103 99 105 102 106 108 104 110 107 111 113 109 115 112 116 118 114 120 117 122 119 124 121 125 127 123 129 126 130 132 128 134 131 135 137 133 139 136 140 142 138 144 141 145 147 143 149 146 150 152 148 154 151 155 157..."
},
{
"input": "40000 20001",
"output": "3 1 5 2 7 4 8 10 6 12 9 13 15 11 17 14 18 20 16 22 19 23 25 21 27 24 28 30 26 32 29 33 35 31 37 34 38 40 36 42 39 44 41 46 43 47 49 45 51 48 52 54 50 56 53 57 59 55 61 58 62 64 60 66 63 67 69 65 71 68 72 74 70 76 73 77 79 75 81 78 83 80 85 82 86 88 84 90 87 91 93 89 95 92 96 98 94 100 97 101 103 99 105 102 106 108 104 110 107 111 113 109 115 112 116 118 114 120 117 122 119 124 121 125 127 123 129 126 130 132 128 134 131 135 137 133 139 136 140 142 138 144 141 145 147 143 149 146 150 152 148 154 151 155 157..."
},
{
"input": "40000 30001",
"output": "3 1 5 2 7 4 8 10 6 12 9 13 15 11 17 14 18 20 16 22 19 23 25 21 27 24 28 30 26 32 29 33 35 31 37 34 38 40 36 42 39 44 41 46 43 47 49 45 51 48 52 54 50 56 53 57 59 55 61 58 62 64 60 66 63 67 69 65 71 68 72 74 70 76 73 77 79 75 81 78 83 80 85 82 86 88 84 90 87 91 93 89 95 92 96 98 94 100 97 101 103 99 105 102 106 108 104 110 107 111 113 109 115 112 116 118 114 120 117 122 119 124 121 125 127 123 129 126 130 132 128 134 131 135 137 133 139 136 140 142 138 144 141 145 147 143 149 146 150 152 148 154 151 155 157..."
},
{
"input": "50000 10001",
"output": "2 4 1 5 7 3 8 10 6 11 13 9 14 16 12 17 19 15 20 22 18 23 25 21 26 28 24 29 31 27 32 34 30 35 37 33 38 40 36 41 43 39 44 46 42 47 49 45 50 52 48 53 55 51 56 58 54 59 61 57 62 64 60 65 67 63 68 70 66 71 73 69 74 76 72 77 79 75 80 82 78 83 85 81 86 88 84 89 91 87 92 94 90 96 93 98 95 99 101 97 102 104 100 105 107 103 108 110 106 111 113 109 114 116 112 117 119 115 120 122 118 123 125 121 126 128 124 129 131 127 132 134 130 135 137 133 138 140 136 141 143 139 145 142 147 144 148 150 146 151 153 149 154 156 152..."
},
{
"input": "50000 20001",
"output": "2 4 1 5 7 3 8 10 6 11 13 9 14 16 12 17 19 15 20 22 18 23 25 21 26 28 24 29 31 27 32 34 30 35 37 33 38 40 36 41 43 39 44 46 42 47 49 45 50 52 48 53 55 51 56 58 54 59 61 57 62 64 60 65 67 63 68 70 66 71 73 69 74 76 72 77 79 75 80 82 78 83 85 81 86 88 84 89 91 87 92 94 90 96 93 98 95 99 101 97 102 104 100 105 107 103 108 110 106 111 113 109 114 116 112 117 119 115 120 122 118 123 125 121 126 128 124 129 131 127 132 134 130 135 137 133 138 140 136 141 143 139 145 142 147 144 148 150 146 151 153 149 154 156 152..."
},
{
"input": "50000 30001",
"output": "2 4 1 5 7 3 8 10 6 11 13 9 14 16 12 17 19 15 20 22 18 23 25 21 26 28 24 29 31 27 32 34 30 35 37 33 38 40 36 41 43 39 44 46 42 47 49 45 50 52 48 53 55 51 56 58 54 59 61 57 62 64 60 65 67 63 68 70 66 71 73 69 74 76 72 77 79 75 80 82 78 83 85 81 86 88 84 89 91 87 92 94 90 96 93 98 95 99 101 97 102 104 100 105 107 103 108 110 106 111 113 109 114 116 112 117 119 115 120 122 118 123 125 121 126 128 124 129 131 127 132 134 130 135 137 133 138 140 136 141 143 139 145 142 147 144 148 150 146 151 153 149 154 156 152..."
},
{
"input": "60000 10001",
"output": "2 4 1 6 3 8 5 9 11 7 13 10 15 12 16 18 14 20 17 22 19 24 21 26 23 28 25 30 27 31 33 29 35 32 37 34 38 40 36 42 39 44 41 45 47 43 49 46 51 48 53 50 55 52 57 54 59 56 60 62 58 64 61 66 63 67 69 65 71 68 73 70 74 76 72 78 75 80 77 82 79 84 81 86 83 88 85 89 91 87 93 90 95 92 97 94 99 96 101 98 103 100 104 106 102 108 105 110 107 112 109 114 111 116 113 118 115 119 121 117 123 120 125 122 126 128 124 130 127 132 129 133 135 131 137 134 139 136 141 138 143 140 145 142 147 144 148 150 146 152 149 154 151 155 157..."
},
{
"input": "60000 20001",
"output": "2 4 1 6 3 8 5 9 11 7 13 10 15 12 16 18 14 20 17 22 19 24 21 26 23 28 25 30 27 31 33 29 35 32 37 34 38 40 36 42 39 44 41 45 47 43 49 46 51 48 53 50 55 52 57 54 59 56 60 62 58 64 61 66 63 67 69 65 71 68 73 70 74 76 72 78 75 80 77 82 79 84 81 86 83 88 85 89 91 87 93 90 95 92 97 94 99 96 101 98 103 100 104 106 102 108 105 110 107 112 109 114 111 116 113 118 115 119 121 117 123 120 125 122 126 128 124 130 127 132 129 133 135 131 137 134 139 136 141 138 143 140 145 142 147 144 148 150 146 152 149 154 151 155 157..."
},
{
"input": "60000 30001",
"output": "2 4 1 6 3 8 5 9 11 7 13 10 15 12 16 18 14 20 17 22 19 24 21 26 23 28 25 30 27 31 33 29 35 32 37 34 38 40 36 42 39 44 41 45 47 43 49 46 51 48 53 50 55 52 57 54 59 56 60 62 58 64 61 66 63 67 69 65 71 68 73 70 74 76 72 78 75 80 77 82 79 84 81 86 83 88 85 89 91 87 93 90 95 92 97 94 99 96 101 98 103 100 104 106 102 108 105 110 107 112 109 114 111 116 113 118 115 119 121 117 123 120 125 122 126 128 124 130 127 132 129 133 135 131 137 134 139 136 141 138 143 140 145 142 147 144 148 150 146 152 149 154 151 155 157..."
},
{
"input": "70000 10001",
"output": "3 1 5 2 7 4 9 6 11 8 13 10 15 12 16 18 14 20 17 22 19 24 21 26 23 28 25 30 27 32 29 33 35 31 37 34 39 36 41 38 43 40 45 42 47 44 49 46 50 52 48 54 51 56 53 58 55 60 57 62 59 64 61 66 63 67 69 65 71 68 73 70 75 72 77 74 79 76 81 78 83 80 84 86 82 88 85 90 87 92 89 94 91 96 93 98 95 100 97 101 103 99 105 102 107 104 109 106 111 108 113 110 115 112 117 114 118 120 116 122 119 124 121 126 123 128 125 130 127 132 129 134 131 135 137 133 139 136 141 138 143 140 145 142 147 144 149 146 151 148 152 154 150 156 153..."
},
{
"input": "70000 20001",
"output": "3 1 5 2 7 4 9 6 11 8 13 10 15 12 16 18 14 20 17 22 19 24 21 26 23 28 25 30 27 32 29 33 35 31 37 34 39 36 41 38 43 40 45 42 47 44 49 46 50 52 48 54 51 56 53 58 55 60 57 62 59 64 61 66 63 67 69 65 71 68 73 70 75 72 77 74 79 76 81 78 83 80 84 86 82 88 85 90 87 92 89 94 91 96 93 98 95 100 97 101 103 99 105 102 107 104 109 106 111 108 113 110 115 112 117 114 118 120 116 122 119 124 121 126 123 128 125 130 127 132 129 134 131 135 137 133 139 136 141 138 143 140 145 142 147 144 149 146 151 148 152 154 150 156 153..."
},
{
"input": "70000 30001",
"output": "3 1 5 2 7 4 9 6 11 8 13 10 15 12 16 18 14 20 17 22 19 24 21 26 23 28 25 30 27 32 29 33 35 31 37 34 39 36 41 38 43 40 45 42 47 44 49 46 50 52 48 54 51 56 53 58 55 60 57 62 59 64 61 66 63 67 69 65 71 68 73 70 75 72 77 74 79 76 81 78 83 80 84 86 82 88 85 90 87 92 89 94 91 96 93 98 95 100 97 101 103 99 105 102 107 104 109 106 111 108 113 110 115 112 117 114 118 120 116 122 119 124 121 126 123 128 125 130 127 132 129 134 131 135 137 133 139 136 141 138 143 140 145 142 147 144 149 146 151 148 152 154 150 156 153..."
},
{
"input": "80000 10001",
"output": "3 1 5 2 7 4 8 10 6 12 9 13 15 11 17 14 18 20 16 22 19 23 25 21 27 24 28 30 26 32 29 33 35 31 37 34 38 40 36 42 39 44 41 46 43 47 49 45 51 48 52 54 50 56 53 57 59 55 61 58 62 64 60 66 63 67 69 65 71 68 72 74 70 76 73 77 79 75 81 78 83 80 85 82 86 88 84 90 87 91 93 89 95 92 96 98 94 100 97 101 103 99 105 102 106 108 104 110 107 111 113 109 115 112 116 118 114 120 117 122 119 124 121 125 127 123 129 126 130 132 128 134 131 135 137 133 139 136 140 142 138 144 141 145 147 143 149 146 150 152 148 154 151 155 157..."
},
{
"input": "80000 20001",
"output": "3 1 5 2 7 4 8 10 6 12 9 13 15 11 17 14 18 20 16 22 19 23 25 21 27 24 28 30 26 32 29 33 35 31 37 34 38 40 36 42 39 44 41 46 43 47 49 45 51 48 52 54 50 56 53 57 59 55 61 58 62 64 60 66 63 67 69 65 71 68 72 74 70 76 73 77 79 75 81 78 83 80 85 82 86 88 84 90 87 91 93 89 95 92 96 98 94 100 97 101 103 99 105 102 106 108 104 110 107 111 113 109 115 112 116 118 114 120 117 122 119 124 121 125 127 123 129 126 130 132 128 134 131 135 137 133 139 136 140 142 138 144 141 145 147 143 149 146 150 152 148 154 151 155 157..."
},
{
"input": "80000 30001",
"output": "3 1 5 2 7 4 8 10 6 12 9 13 15 11 17 14 18 20 16 22 19 23 25 21 27 24 28 30 26 32 29 33 35 31 37 34 38 40 36 42 39 44 41 46 43 47 49 45 51 48 52 54 50 56 53 57 59 55 61 58 62 64 60 66 63 67 69 65 71 68 72 74 70 76 73 77 79 75 81 78 83 80 85 82 86 88 84 90 87 91 93 89 95 92 96 98 94 100 97 101 103 99 105 102 106 108 104 110 107 111 113 109 115 112 116 118 114 120 117 122 119 124 121 125 127 123 129 126 130 132 128 134 131 135 137 133 139 136 140 142 138 144 141 145 147 143 149 146 150 152 148 154 151 155 157..."
},
{
"input": "90000 10001",
"output": "3 1 4 6 2 8 5 9 11 7 13 10 14 16 12 17 19 15 20 22 18 24 21 25 27 23 28 30 26 31 33 29 35 32 36 38 34 39 41 37 42 44 40 46 43 47 49 45 50 52 48 53 55 51 57 54 58 60 56 61 63 59 64 66 62 68 65 69 71 67 72 74 70 75 77 73 79 76 80 82 78 83 85 81 86 88 84 90 87 91 93 89 94 96 92 97 99 95 101 98 102 104 100 105 107 103 108 110 106 112 109 113 115 111 116 118 114 119 121 117 123 120 124 126 122 127 129 125 130 132 128 134 131 135 137 133 138 140 136 141 143 139 145 142 146 148 144 149 151 147 152 154 150 156 153..."
},
{
"input": "90000 20001",
"output": "3 1 4 6 2 8 5 9 11 7 13 10 14 16 12 17 19 15 20 22 18 24 21 25 27 23 28 30 26 31 33 29 35 32 36 38 34 39 41 37 42 44 40 46 43 47 49 45 50 52 48 53 55 51 57 54 58 60 56 61 63 59 64 66 62 68 65 69 71 67 72 74 70 75 77 73 79 76 80 82 78 83 85 81 86 88 84 90 87 91 93 89 94 96 92 97 99 95 101 98 102 104 100 105 107 103 108 110 106 112 109 113 115 111 116 118 114 119 121 117 123 120 124 126 122 127 129 125 130 132 128 134 131 135 137 133 138 140 136 141 143 139 145 142 146 148 144 149 151 147 152 154 150 156 153..."
},
{
"input": "90000 30001",
"output": "3 1 4 6 2 8 5 9 11 7 13 10 14 16 12 17 19 15 20 22 18 24 21 25 27 23 28 30 26 31 33 29 35 32 36 38 34 39 41 37 42 44 40 46 43 47 49 45 50 52 48 53 55 51 57 54 58 60 56 61 63 59 64 66 62 68 65 69 71 67 72 74 70 75 77 73 79 76 80 82 78 83 85 81 86 88 84 90 87 91 93 89 94 96 92 97 99 95 101 98 102 104 100 105 107 103 108 110 106 112 109 113 115 111 116 118 114 119 121 117 123 120 124 126 122 127 129 125 130 132 128 134 131 135 137 133 138 140 136 141 143 139 145 142 146 148 144 149 151 147 152 154 150 156 153..."
},
{
"input": "100000 10001",
"output": "2 4 1 5 7 3 8 10 6 11 13 9 14 16 12 17 19 15 20 22 18 23 25 21 26 28 24 29 31 27 32 34 30 35 37 33 38 40 36 41 43 39 44 46 42 47 49 45 50 52 48 53 55 51 56 58 54 59 61 57 62 64 60 65 67 63 68 70 66 71 73 69 74 76 72 77 79 75 80 82 78 83 85 81 86 88 84 89 91 87 92 94 90 96 93 98 95 99 101 97 102 104 100 105 107 103 108 110 106 111 113 109 114 116 112 117 119 115 120 122 118 123 125 121 126 128 124 129 131 127 132 134 130 135 137 133 138 140 136 141 143 139 145 142 147 144 148 150 146 151 153 149 154 156 152..."
},
{
"input": "100000 20001",
"output": "2 4 1 5 7 3 8 10 6 11 13 9 14 16 12 17 19 15 20 22 18 23 25 21 26 28 24 29 31 27 32 34 30 35 37 33 38 40 36 41 43 39 44 46 42 47 49 45 50 52 48 53 55 51 56 58 54 59 61 57 62 64 60 65 67 63 68 70 66 71 73 69 74 76 72 77 79 75 80 82 78 83 85 81 86 88 84 89 91 87 92 94 90 96 93 98 95 99 101 97 102 104 100 105 107 103 108 110 106 111 113 109 114 116 112 117 119 115 120 122 118 123 125 121 126 128 124 129 131 127 132 134 130 135 137 133 138 140 136 141 143 139 145 142 147 144 148 150 146 151 153 149 154 156 152..."
},
{
"input": "100000 30001",
"output": "2 4 1 5 7 3 8 10 6 11 13 9 14 16 12 17 19 15 20 22 18 23 25 21 26 28 24 29 31 27 32 34 30 35 37 33 38 40 36 41 43 39 44 46 42 47 49 45 50 52 48 53 55 51 56 58 54 59 61 57 62 64 60 65 67 63 68 70 66 71 73 69 74 76 72 77 79 75 80 82 78 83 85 81 86 88 84 89 91 87 92 94 90 96 93 98 95 99 101 97 102 104 100 105 107 103 108 110 106 111 113 109 114 116 112 117 119 115 120 122 118 123 125 121 126 128 124 129 131 127 132 134 130 135 137 133 138 140 136 141 143 139 145 142 147 144 148 150 146 151 153 149 154 156 152..."
},
{
"input": "100000 199999",
"output": "2 4 1 5 7 3 8 10 6 11 13 9 14 16 12 17 19 15 20 22 18 23 25 21 26 28 24 29 31 27 32 34 30 35 37 33 38 40 36 41 43 39 44 46 42 47 49 45 50 52 48 53 55 51 56 58 54 59 61 57 62 64 60 65 67 63 68 70 66 71 73 69 74 76 72 77 79 75 80 82 78 83 85 81 86 88 84 89 91 87 92 94 90 96 93 98 95 99 101 97 102 104 100 105 107 103 108 110 106 111 113 109 114 116 112 117 119 115 120 122 118 123 125 121 126 128 124 129 131 127 132 134 130 135 137 133 138 140 136 141 143 139 145 142 147 144 148 150 146 151 153 149 154 156 152..."
},
{
"input": "10 17",
"output": "3 1 4 6 2 8 5 9 7 10 "
}
] | 1,648,399,119
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 36
| 124
| 6,553,600
|
n, k = map(int,input().split())
a = []
def merge(start, end, call):
if call < 2:
a.extend(range(start, end))
return call
if end - start == 1:
a.append(start)
return call
call -= 2
mid = (start + end + 1) // 2
call = merge(mid, end, call)
call = merge(start, mid, call)
return call
call = merge(1, n + 1, k - 1)
if call == 0:
print(' '.join(map(str, a)))
else:
print(-1)
|
Title: Merge Sort
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Merge sort is a well-known sorting algorithm. The main function that sorts the elements of array *a* with indices from [*l*,<=*r*) can be implemented as follows:
1. If the segment [*l*,<=*r*) is already sorted in non-descending order (that is, for any *i* such that *l*<=≤<=*i*<=<<=*r*<=-<=1 *a*[*i*]<=≤<=*a*[*i*<=+<=1]), then end the function call; 1. Let ; 1. Call *mergesort*(*a*,<=*l*,<=*mid*); 1. Call *mergesort*(*a*,<=*mid*,<=*r*); 1. Merge segments [*l*,<=*mid*) and [*mid*,<=*r*), making the segment [*l*,<=*r*) sorted in non-descending order. The merge algorithm doesn't call any other functions.
The array in this problem is 0-indexed, so to sort the whole array, you need to call *mergesort*(*a*,<=0,<=*n*).
The number of calls of function *mergesort* is very important, so Ivan has decided to calculate it while sorting the array. For example, if *a*<==<={1,<=2,<=3,<=4}, then there will be 1 call of *mergesort* — *mergesort*(0,<=4), which will check that the array is sorted and then end. If *a*<==<={2,<=1,<=3}, then the number of calls is 3: first of all, you call *mergesort*(0,<=3), which then sets *mid*<==<=1 and calls *mergesort*(0,<=1) and *mergesort*(1,<=3), which do not perform any recursive calls because segments (0,<=1) and (1,<=3) are sorted.
Ivan has implemented the program that counts the number of *mergesort* calls, but now he needs to test it. To do this, he needs to find an array *a* such that *a* is a permutation of size *n* (that is, the number of elements in *a* is *n*, and every integer number from [1,<=*n*] can be found in this array), and the number of *mergesort* calls when sorting the array is exactly *k*.
Help Ivan to find an array he wants!
Input Specification:
The first line contains two numbers *n* and *k* (1<=≤<=*n*<=≤<=100000, 1<=≤<=*k*<=≤<=200000) — the size of a desired permutation and the number of *mergesort* calls required to sort it.
Output Specification:
If a permutation of size *n* such that there will be exactly *k* calls of *mergesort* while sorting it doesn't exist, output <=-<=1. Otherwise output *n* integer numbers *a*[0],<=*a*[1],<=...,<=*a*[*n*<=-<=1] — the elements of a permutation that would meet the required conditions. If there are multiple answers, print any of them.
Demo Input:
['3 3\n', '4 1\n', '5 6\n']
Demo Output:
['2 1 3 ', '1 2 3 4 ', '-1\n']
Note:
none
|
```python
n, k = map(int,input().split())
a = []
def merge(start, end, call):
if call < 2:
a.extend(range(start, end))
return call
if end - start == 1:
a.append(start)
return call
call -= 2
mid = (start + end + 1) // 2
call = merge(mid, end, call)
call = merge(start, mid, call)
return call
call = merge(1, n + 1, k - 1)
if call == 0:
print(' '.join(map(str, a)))
else:
print(-1)
```
| 3
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings *a* and *b* of equal length are called equivalent in one of the two cases:
1. They are equal. 1. If we split string *a* into two halves of the same size *a*1 and *a*2, and string *b* into two halves of the same size *b*1 and *b*2, then one of the following is correct: *a*1 is equivalent to *b*1, and *a*2 is equivalent to *b*2 1. *a*1 is equivalent to *b*2, and *a*2 is equivalent to *b*1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn!
|
The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200<=000 and consists of lowercase English letters. The strings have the same length.
|
Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.
|
[
"aaba\nabaa\n",
"aabb\nabab\n"
] |
[
"YES\n",
"NO\n"
] |
In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa".
| 0
|
[
{
"input": "aaba\nabaa",
"output": "YES"
},
{
"input": "aabb\nabab",
"output": "NO"
},
{
"input": "a\na",
"output": "YES"
},
{
"input": "a\nb",
"output": "NO"
},
{
"input": "ab\nab",
"output": "YES"
},
{
"input": "ab\nba",
"output": "YES"
},
{
"input": "ab\nbb",
"output": "NO"
},
{
"input": "zzaa\naazz",
"output": "YES"
},
{
"input": "azza\nzaaz",
"output": "YES"
},
{
"input": "abc\nabc",
"output": "YES"
},
{
"input": "abc\nacb",
"output": "NO"
},
{
"input": "azzz\nzzaz",
"output": "YES"
},
{
"input": "abcd\ndcab",
"output": "YES"
},
{
"input": "abcd\ncdab",
"output": "YES"
},
{
"input": "abcd\ndcba",
"output": "YES"
},
{
"input": "abcd\nacbd",
"output": "NO"
},
{
"input": "oloaxgddgujq\noloaxgujqddg",
"output": "YES"
},
{
"input": "uwzwdxfmosmqatyv\ndxfmzwwusomqvyta",
"output": "YES"
},
{
"input": "hagnzomowtledfdotnll\nledfdotnllomowthagnz",
"output": "YES"
},
{
"input": "snyaydaeobufdg\nsnyaydaeobufdg",
"output": "YES"
},
{
"input": "baaaaa\nabaaaa",
"output": "NO"
},
{
"input": "hhiisug\nmzdjwju",
"output": "NO"
},
{
"input": "bbbabbabaaab\naaaabbabbbbb",
"output": "NO"
},
{
"input": "bbaaab\naababb",
"output": "NO"
},
{
"input": "aabbaaaa\naaaaabab",
"output": "NO"
},
{
"input": "aab\naba",
"output": "NO"
},
{
"input": "abcddd\nbacddd",
"output": "NO"
},
{
"input": "qgiufelsfhanx\naaaaaaaaaaaaa",
"output": "NO"
},
{
"input": "aabaababaaba\naababaaababa",
"output": "NO"
},
{
"input": "nocdqzdriyyil\naaaaaaaaaaaaa",
"output": "NO"
},
{
"input": "zdmctxl\nkojqhgw",
"output": "NO"
},
{
"input": "yhwepqwyhwepqwyhwepqweahnqtueahnqtueahnqtuyhwepqwyhwepqwyhwepqwyhwepqweahnqtueahnqtuyhwepqweahnqtueahnqtueahnqtueahnqtueahnqtueahnqtu\neahnqtueahnqtueahnqtuyhwepqweahnqtuyhwepqwyhwepqweahnqtuyhwepqweahnqtuyhwepqweahnqtueahnqtuyhwepqweahnqtueahnqtuyhwepqwyhwepqwyhwepqw",
"output": "NO"
},
{
"input": "abc\nbac",
"output": "NO"
},
{
"input": "ottceez\npcstdvz",
"output": "NO"
}
] | 1,513,583,821
| 2,147,483,647
|
PyPy 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
#include <bits/stdc++.h>
#define ll long long int
#define mod 1000000007
#define pii pair<int, int>
#define fr(n) for (int i = 0; i < n; i++)
#define fr1(n) for (int i = 1; i <= n; i++)
using namespace std;
string s, t;
bool cmp(int a, int b, int l) {
if (s.substr(a, l) == t.substr(b, l)) return true;
if (l & 1) return false;
l >>= 1;
return (cmp(a, b, l) && cmp(a + l, b + l, l)) || (cmp(a, b + l, l) && cmp(a + l, b, l));
}
int main() {
ios::sync_with_stdio(false);
ofstream out (".out");
ifstream in (".in");
cin >> s >> t;
if (cmp(0, 0, (int)s.length())) cout << "YES";
else cout << "NO";
}
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings *a* and *b* of equal length are called equivalent in one of the two cases:
1. They are equal. 1. If we split string *a* into two halves of the same size *a*1 and *a*2, and string *b* into two halves of the same size *b*1 and *b*2, then one of the following is correct: *a*1 is equivalent to *b*1, and *a*2 is equivalent to *b*2 1. *a*1 is equivalent to *b*2, and *a*2 is equivalent to *b*1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn!
Input Specification:
The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200<=000 and consists of lowercase English letters. The strings have the same length.
Output Specification:
Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.
Demo Input:
['aaba\nabaa\n', 'aabb\nabab\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa".
|
```python
#include <bits/stdc++.h>
#define ll long long int
#define mod 1000000007
#define pii pair<int, int>
#define fr(n) for (int i = 0; i < n; i++)
#define fr1(n) for (int i = 1; i <= n; i++)
using namespace std;
string s, t;
bool cmp(int a, int b, int l) {
if (s.substr(a, l) == t.substr(b, l)) return true;
if (l & 1) return false;
l >>= 1;
return (cmp(a, b, l) && cmp(a + l, b + l, l)) || (cmp(a, b + l, l) && cmp(a + l, b, l));
}
int main() {
ios::sync_with_stdio(false);
ofstream out (".out");
ifstream in (".in");
cin >> s >> t;
if (cmp(0, 0, (int)s.length())) cout << "YES";
else cout << "NO";
}
```
| -1
|
|
453
|
A
|
Little Pony and Expected Maximum
|
PROGRAMMING
| 1,600
|
[
"probabilities"
] | null | null |
Twilight Sparkle was playing Ludo with her friends Rainbow Dash, Apple Jack and Flutter Shy. But she kept losing. Having returned to the castle, Twilight Sparkle became interested in the dice that were used in the game.
The dice has *m* faces: the first face of the dice contains a dot, the second one contains two dots, and so on, the *m*-th face contains *m* dots. Twilight Sparkle is sure that when the dice is tossed, each face appears with probability . Also she knows that each toss is independent from others. Help her to calculate the expected maximum number of dots she could get after tossing the dice *n* times.
|
A single line contains two integers *m* and *n* (1<=≤<=*m*,<=*n*<=≤<=105).
|
Output a single real number corresponding to the expected maximum. The answer will be considered correct if its relative or absolute error doesn't exceed 10<=<=-<=4.
|
[
"6 1\n",
"6 3\n",
"2 2\n"
] |
[
"3.500000000000\n",
"4.958333333333\n",
"1.750000000000\n"
] |
Consider the third test example. If you've made two tosses:
1. You can get 1 in the first toss, and 2 in the second. Maximum equals to 2. 1. You can get 1 in the first toss, and 1 in the second. Maximum equals to 1. 1. You can get 2 in the first toss, and 1 in the second. Maximum equals to 2. 1. You can get 2 in the first toss, and 2 in the second. Maximum equals to 2.
The probability of each outcome is 0.25, that is expectation equals to:
You can read about expectation using the following link: http://en.wikipedia.org/wiki/Expected_value
| 500
|
[
{
"input": "6 1",
"output": "3.500000000000"
},
{
"input": "6 3",
"output": "4.958333333333"
},
{
"input": "2 2",
"output": "1.750000000000"
},
{
"input": "5 4",
"output": "4.433600000000"
},
{
"input": "5 8",
"output": "4.814773760000"
},
{
"input": "3 10",
"output": "2.982641534996"
},
{
"input": "3 6",
"output": "2.910836762689"
},
{
"input": "1 8",
"output": "1.000000000000"
},
{
"input": "24438 9",
"output": "21994.699969310015"
},
{
"input": "94444 9",
"output": "85000.099992058866"
},
{
"input": "8 66716",
"output": "8.000000000000"
},
{
"input": "4 25132",
"output": "4.000000000000"
},
{
"input": "51520 73331",
"output": "51519.682650242677"
},
{
"input": "54230 31747",
"output": "54228.743352775018"
},
{
"input": "24236 90163",
"output": "24235.975171545670"
},
{
"input": "26946 99523",
"output": "26945.974480086279"
},
{
"input": "50323 7",
"output": "44033.124988408454"
},
{
"input": "53033 3",
"output": "39775.249995286234"
},
{
"input": "55743 5",
"output": "46452.999992525307"
},
{
"input": "59964 79",
"output": "59214.949890211828"
},
{
"input": "1 1",
"output": "1.000000000000"
},
{
"input": "1 1",
"output": "1.000000000000"
},
{
"input": "3 1",
"output": "2.000000000000"
},
{
"input": "1 2",
"output": "1.000000000000"
},
{
"input": "53513 34040",
"output": "53511.875329020870"
},
{
"input": "100000 100000",
"output": "99999.418033254507"
},
{
"input": "1 100000",
"output": "1.000000000000"
},
{
"input": "100000 1",
"output": "50000.499999999935"
},
{
"input": "2 100000",
"output": "2.000000000000"
},
{
"input": "100000 2",
"output": "66667.166665000332"
},
{
"input": "50000 100000",
"output": "49999.843487110789"
},
{
"input": "99999 1111",
"output": "99909.571915885972"
},
{
"input": "99999 99999",
"output": "99998.418033254609"
},
{
"input": "1000 1000",
"output": "999.419018443269"
},
{
"input": "50000 50000",
"output": "49999.418043215679"
},
{
"input": "88888 88888",
"output": "88887.418034499773"
},
{
"input": "99999 100000",
"output": "99998.418042461126"
},
{
"input": "100000 555",
"output": "99820.643422392372"
},
{
"input": "10000 10000",
"output": "9999.418122897887"
},
{
"input": "100000 5001",
"output": "99980.503829474910"
},
{
"input": "100000 1000",
"output": "99900.599066768002"
}
] | 1,408,884,846
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 12
| 1,000
| 0
|
def power(a, n):
if n == 0:
return 1
if not n % 2:
return power(a, n // 2) ** 2
else:
return power(a, n - 1) * a
m, n = tuple(map(int, input().split()))
answer = 0.0
for i in range(1, m + 1):
answer += (power(i / m, n) - power((i - 1) / m, n)) * i
print(answer)
|
Title: Little Pony and Expected Maximum
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Twilight Sparkle was playing Ludo with her friends Rainbow Dash, Apple Jack and Flutter Shy. But she kept losing. Having returned to the castle, Twilight Sparkle became interested in the dice that were used in the game.
The dice has *m* faces: the first face of the dice contains a dot, the second one contains two dots, and so on, the *m*-th face contains *m* dots. Twilight Sparkle is sure that when the dice is tossed, each face appears with probability . Also she knows that each toss is independent from others. Help her to calculate the expected maximum number of dots she could get after tossing the dice *n* times.
Input Specification:
A single line contains two integers *m* and *n* (1<=≤<=*m*,<=*n*<=≤<=105).
Output Specification:
Output a single real number corresponding to the expected maximum. The answer will be considered correct if its relative or absolute error doesn't exceed 10<=<=-<=4.
Demo Input:
['6 1\n', '6 3\n', '2 2\n']
Demo Output:
['3.500000000000\n', '4.958333333333\n', '1.750000000000\n']
Note:
Consider the third test example. If you've made two tosses:
1. You can get 1 in the first toss, and 2 in the second. Maximum equals to 2. 1. You can get 1 in the first toss, and 1 in the second. Maximum equals to 1. 1. You can get 2 in the first toss, and 1 in the second. Maximum equals to 2. 1. You can get 2 in the first toss, and 2 in the second. Maximum equals to 2.
The probability of each outcome is 0.25, that is expectation equals to:
You can read about expectation using the following link: http://en.wikipedia.org/wiki/Expected_value
|
```python
def power(a, n):
if n == 0:
return 1
if not n % 2:
return power(a, n // 2) ** 2
else:
return power(a, n - 1) * a
m, n = tuple(map(int, input().split()))
answer = 0.0
for i in range(1, m + 1):
answer += (power(i / m, n) - power((i - 1) / m, n)) * i
print(answer)
```
| 0
|
|
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 "
}
] | 1,565,623,899
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 6
| 2,000
| 1,536,000
|
n,m = map(int,input().split())
x_arr = {}
y_arr = {}
for i in range(m) :
x,y = map(int,input().split())
if x not in x_arr.keys() :
x_arr[x] = 1
if y not in y_arr.keys() :
y_arr[y] = 1
count = (n-len(list(x_arr.keys())))*(n-len(list(y_arr.keys())))
print(count,end=" ")
|
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
n,m = map(int,input().split())
x_arr = {}
y_arr = {}
for i in range(m) :
x,y = map(int,input().split())
if x not in x_arr.keys() :
x_arr[x] = 1
if y not in y_arr.keys() :
y_arr[y] = 1
count = (n-len(list(x_arr.keys())))*(n-len(list(y_arr.keys())))
print(count,end=" ")
```
| 0
|
|
834
|
B
|
The Festive Evening
|
PROGRAMMING
| 1,100
|
[
"data structures",
"implementation"
] | null | null |
It's the end of July – the time when a festive evening is held at Jelly Castle! Guests from all over the kingdom gather here to discuss new trends in the world of confectionery. Yet some of the things discussed here are not supposed to be disclosed to the general public: the information can cause discord in the kingdom of Sweetland in case it turns out to reach the wrong hands. So it's a necessity to not let any uninvited guests in.
There are 26 entrances in Jelly Castle, enumerated with uppercase English letters from A to Z. Because of security measures, each guest is known to be assigned an entrance he should enter the castle through. The door of each entrance is opened right before the first guest's arrival and closed right after the arrival of the last guest that should enter the castle through this entrance. No two guests can enter the castle simultaneously.
For an entrance to be protected from possible intrusion, a candy guard should be assigned to it. There are *k* such guards in the castle, so if there are more than *k* opened doors, one of them is going to be left unguarded! Notice that a guard can't leave his post until the door he is assigned to is closed.
Slastyona had a suspicion that there could be uninvited guests at the evening. She knows the order in which the invited guests entered the castle, and wants you to help her check whether there was a moment when more than *k* doors were opened.
|
Two integers are given in the first string: the number of guests *n* and the number of guards *k* (1<=≤<=*n*<=≤<=106, 1<=≤<=*k*<=≤<=26).
In the second string, *n* uppercase English letters *s*1*s*2... *s**n* are given, where *s**i* is the entrance used by the *i*-th guest.
|
Output «YES» if at least one door was unguarded during some time, and «NO» otherwise.
You can output each letter in arbitrary case (upper or lower).
|
[
"5 1\nAABBB\n",
"5 1\nABABB\n"
] |
[
"NO\n",
"YES\n"
] |
In the first sample case, the door A is opened right before the first guest's arrival and closed when the second guest enters the castle. The door B is opened right before the arrival of the third guest, and closed after the fifth one arrives. One guard can handle both doors, as the first one is closed before the second one is opened.
In the second sample case, the door B is opened before the second guest's arrival, but the only guard can't leave the door A unattended, as there is still one more guest that should enter the castle through this door.
| 1,000
|
[
{
"input": "5 1\nAABBB",
"output": "NO"
},
{
"input": "5 1\nABABB",
"output": "YES"
},
{
"input": "26 1\nABCDEFGHIJKLMNOPQRSTUVWXYZ",
"output": "NO"
},
{
"input": "27 1\nABCDEFGHIJKLMNOPQRSTUVWXYZA",
"output": "YES"
},
{
"input": "5 2\nABACA",
"output": "NO"
},
{
"input": "6 2\nABCABC",
"output": "YES"
},
{
"input": "8 3\nABCBCDCA",
"output": "NO"
},
{
"input": "73 2\nDEBECECBBADAADEAABEAEEEAEBEAEBCDDBABBAEBACCBEEBBAEADEECACEDEEDABACDCDBBBD",
"output": "YES"
},
{
"input": "44 15\nHGJIFCGGCDGIJDHBIBGAEABCIABIGBDEADBBBAGDFDHA",
"output": "NO"
},
{
"input": "41 19\nTMEYYIIELFDCMBDKWWKYNRNDUPRONYROXQCLVQALP",
"output": "NO"
},
{
"input": "377 3\nEADADBBBBDEAABBAEBABACDBDBBCACAADBEAEACDEAABACADEEDEACACDADABBBBDDEECBDABACACBAECBADAEBDEEBDBCDAEADBCDDACACDCCEEDBCCBBCEDBECBABCDDBBDEADEDAEACDECECBEBACBCCDCDBDAECDECADBCBEDBBDAAEBCAAECCDCCDBDDEBADEEBDCAEABBDEDBBDDEAECCBDDCDEACDAECCBDDABABEAEDCDEDBAECBDEACEBCECEACDCBABCBAAEAADACADBBBBABEADBCADEBCBECCABBDDDEEBCDEBADEBDAAABBEABADEDEAEABCEEBEEDEAEBEABCEDDBACBCCADEBAAAAAEABABBCE",
"output": "YES"
},
{
"input": "433 3\nFZDDHMJGBZCHFUXBBPIEBBEFDWOMXXEPOMDGSMPIUZOMRZQNSJAVNATGIWPDFISKFQXJNVFXPHOZDAEZFDAHDXXQKZMGNSGKQNWGNGJGJZVVITKNFLVCPMZSDMCHBTVAWYVZLIXXIADXNYILEYNIQHKMOGMVOCWGHCWIYMPEPADSJAAKEGTUSEDWAHMNYJDIHBKHVUHLYGNGZDBULRXLSAJHPCMNWCEAAPYMHDTYWPADOTJTXTXUKLCHWKUSZRHEKQEFPVJEJJHRWCKYOIWALRTIBUMNOCRXLSIKQCJVQXEPGOHRUDJDKMUUUDORURWXJNVRVMNOUNRFKSVMTMZGOIJLXEPAMVGESOADYIGZXRBJDIWKNOWTCSROAQTBECHTOZVSQUOOJRZIBAUHMKAXDCIMDZJFMABGRNTGPUJAUNFPFWCJG",
"output": "YES"
},
{
"input": "5 2\nABCAB",
"output": "YES"
},
{
"input": "5 1\nAZAZA",
"output": "YES"
},
{
"input": "7 2\nABCDBCD",
"output": "YES"
},
{
"input": "3 26\nAAB",
"output": "NO"
}
] | 1,631,264,022
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 34
| 171
| 26,419,200
|
n, k = map(int, input().split())
l = input()
first = [-1]*26
last = [-1]*26
for i in range(n):
if first[ord(l[i])-ord('A')] == -1:
first[ord(l[i])-ord('A')] = i
for i in range(n-1,-1,-1):
if last[ord(l[i])-ord('A')] == -1:
last[ord(l[i])-ord('A')] = i
crt = 0
check = True
for i in range(n):
if first[ord(l[i])-ord('A')] == i:
crt += 1
if crt > k:
check = False
break
if last[ord(l[i])-ord('A')] == i:
crt -= 1
if check:
print("NO")
else:
print("YES")
|
Title: The Festive Evening
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It's the end of July – the time when a festive evening is held at Jelly Castle! Guests from all over the kingdom gather here to discuss new trends in the world of confectionery. Yet some of the things discussed here are not supposed to be disclosed to the general public: the information can cause discord in the kingdom of Sweetland in case it turns out to reach the wrong hands. So it's a necessity to not let any uninvited guests in.
There are 26 entrances in Jelly Castle, enumerated with uppercase English letters from A to Z. Because of security measures, each guest is known to be assigned an entrance he should enter the castle through. The door of each entrance is opened right before the first guest's arrival and closed right after the arrival of the last guest that should enter the castle through this entrance. No two guests can enter the castle simultaneously.
For an entrance to be protected from possible intrusion, a candy guard should be assigned to it. There are *k* such guards in the castle, so if there are more than *k* opened doors, one of them is going to be left unguarded! Notice that a guard can't leave his post until the door he is assigned to is closed.
Slastyona had a suspicion that there could be uninvited guests at the evening. She knows the order in which the invited guests entered the castle, and wants you to help her check whether there was a moment when more than *k* doors were opened.
Input Specification:
Two integers are given in the first string: the number of guests *n* and the number of guards *k* (1<=≤<=*n*<=≤<=106, 1<=≤<=*k*<=≤<=26).
In the second string, *n* uppercase English letters *s*1*s*2... *s**n* are given, where *s**i* is the entrance used by the *i*-th guest.
Output Specification:
Output «YES» if at least one door was unguarded during some time, and «NO» otherwise.
You can output each letter in arbitrary case (upper or lower).
Demo Input:
['5 1\nAABBB\n', '5 1\nABABB\n']
Demo Output:
['NO\n', 'YES\n']
Note:
In the first sample case, the door A is opened right before the first guest's arrival and closed when the second guest enters the castle. The door B is opened right before the arrival of the third guest, and closed after the fifth one arrives. One guard can handle both doors, as the first one is closed before the second one is opened.
In the second sample case, the door B is opened before the second guest's arrival, but the only guard can't leave the door A unattended, as there is still one more guest that should enter the castle through this door.
|
```python
n, k = map(int, input().split())
l = input()
first = [-1]*26
last = [-1]*26
for i in range(n):
if first[ord(l[i])-ord('A')] == -1:
first[ord(l[i])-ord('A')] = i
for i in range(n-1,-1,-1):
if last[ord(l[i])-ord('A')] == -1:
last[ord(l[i])-ord('A')] = i
crt = 0
check = True
for i in range(n):
if first[ord(l[i])-ord('A')] == i:
crt += 1
if crt > k:
check = False
break
if last[ord(l[i])-ord('A')] == i:
crt -= 1
if check:
print("NO")
else:
print("YES")
```
| 3
|
|
149
|
A
|
Business trip
|
PROGRAMMING
| 900
|
[
"greedy",
"implementation",
"sortings"
] | null | null |
What joy! Petya's parents went on a business trip for the whole year and the playful kid is left all by himself. Petya got absolutely happy. He jumped on the bed and threw pillows all day long, until...
Today Petya opened the cupboard and found a scary note there. His parents had left him with duties: he should water their favourite flower all year, each day, in the morning, in the afternoon and in the evening. "Wait a second!" — thought Petya. He know for a fact that if he fulfills the parents' task in the *i*-th (1<=≤<=*i*<=≤<=12) month of the year, then the flower will grow by *a**i* centimeters, and if he doesn't water the flower in the *i*-th month, then the flower won't grow this month. Petya also knows that try as he might, his parents won't believe that he has been watering the flower if it grows strictly less than by *k* centimeters.
Help Petya choose the minimum number of months when he will water the flower, given that the flower should grow no less than by *k* centimeters.
|
The first line contains exactly one integer *k* (0<=≤<=*k*<=≤<=100). The next line contains twelve space-separated integers: the *i*-th (1<=≤<=*i*<=≤<=12) number in the line represents *a**i* (0<=≤<=*a**i*<=≤<=100).
|
Print the only integer — the minimum number of months when Petya has to water the flower so that the flower grows no less than by *k* centimeters. If the flower can't grow by *k* centimeters in a year, print -1.
|
[
"5\n1 1 1 1 2 2 3 2 2 1 1 1\n",
"0\n0 0 0 0 0 0 0 1 1 2 3 0\n",
"11\n1 1 4 1 1 5 1 1 4 1 1 1\n"
] |
[
"2\n",
"0\n",
"3\n"
] |
Let's consider the first sample test. There it is enough to water the flower during the seventh and the ninth month. Then the flower grows by exactly five centimeters.
In the second sample Petya's parents will believe him even if the flower doesn't grow at all (*k* = 0). So, it is possible for Petya not to water the flower at all.
| 500
|
[
{
"input": "5\n1 1 1 1 2 2 3 2 2 1 1 1",
"output": "2"
},
{
"input": "0\n0 0 0 0 0 0 0 1 1 2 3 0",
"output": "0"
},
{
"input": "11\n1 1 4 1 1 5 1 1 4 1 1 1",
"output": "3"
},
{
"input": "15\n20 1 1 1 1 2 2 1 2 2 1 1",
"output": "1"
},
{
"input": "7\n8 9 100 12 14 17 21 10 11 100 23 10",
"output": "1"
},
{
"input": "52\n1 12 3 11 4 5 10 6 9 7 8 2",
"output": "6"
},
{
"input": "50\n2 2 3 4 5 4 4 5 7 3 2 7",
"output": "-1"
},
{
"input": "0\n55 81 28 48 99 20 67 95 6 19 10 93",
"output": "0"
},
{
"input": "93\n85 40 93 66 92 43 61 3 64 51 90 21",
"output": "1"
},
{
"input": "99\n36 34 22 0 0 0 52 12 0 0 33 47",
"output": "2"
},
{
"input": "99\n28 32 31 0 10 35 11 18 0 0 32 28",
"output": "3"
},
{
"input": "99\n19 17 0 1 18 11 29 9 29 22 0 8",
"output": "4"
},
{
"input": "76\n2 16 11 10 12 0 20 4 4 14 11 14",
"output": "5"
},
{
"input": "41\n2 1 7 7 4 2 4 4 9 3 10 0",
"output": "6"
},
{
"input": "47\n8 2 2 4 3 1 9 4 2 7 7 8",
"output": "7"
},
{
"input": "58\n6 11 7 0 5 6 3 9 4 9 5 1",
"output": "8"
},
{
"input": "32\n5 2 4 1 5 0 5 1 4 3 0 3",
"output": "9"
},
{
"input": "31\n6 1 0 4 4 5 1 0 5 3 2 0",
"output": "9"
},
{
"input": "35\n2 3 0 0 6 3 3 4 3 5 0 6",
"output": "9"
},
{
"input": "41\n3 1 3 4 3 6 6 1 4 4 0 6",
"output": "11"
},
{
"input": "97\n0 5 3 12 10 16 22 8 21 17 21 10",
"output": "5"
},
{
"input": "100\n21 21 0 0 4 13 0 26 0 0 0 15",
"output": "6"
},
{
"input": "100\n0 0 16 5 22 0 5 0 25 0 14 13",
"output": "7"
},
{
"input": "97\n17 0 10 0 0 0 18 0 14 23 15 0",
"output": "6"
},
{
"input": "100\n0 9 0 18 7 0 0 14 33 3 0 16",
"output": "7"
},
{
"input": "95\n5 2 13 0 15 18 17 0 6 11 0 8",
"output": "9"
},
{
"input": "94\n11 13 0 9 15 8 8 16 3 7 1 3",
"output": "11"
},
{
"input": "96\n8 4 12 15 8 0 4 10 6 6 12 11",
"output": "11"
},
{
"input": "100\n5 5 3 8 6 5 0 3 3 8 1 3",
"output": "-1"
},
{
"input": "100\n1 0 0 1 1 0 1 1 1 1 2 1",
"output": "-1"
},
{
"input": "100\n6 3 2 0 4 1 2 2 2 2 1 1",
"output": "-1"
},
{
"input": "0\n0 0 0 0 0 0 0 0 0 0 0 0",
"output": "0"
},
{
"input": "100\n0 0 0 0 0 0 0 0 0 0 0 0",
"output": "-1"
},
{
"input": "0\n100 100 100 100 100 100 100 100 100 100 100 100",
"output": "0"
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 100 100",
"output": "1"
},
{
"input": "12\n1 1 1 1 1 1 1 1 1 1 1 1",
"output": "12"
},
{
"input": "13\n1 1 1 1 1 1 1 1 1 1 1 2",
"output": "12"
},
{
"input": "15\n10 1 1 1 1 1 1 1 1 1 1 1",
"output": "6"
},
{
"input": "1\n0 0 0 0 0 0 0 0 0 0 0 0",
"output": "-1"
}
] | 1,696,116,565
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 39
| 124
| 0
|
k = int(input())
total = 0
for i, n in enumerate(sorted(map(int, input().split()), reverse=True)):
if total >= k:
print(i)
quit()
total += n
if total >= k:
print(i+1)
else:
print(-1)
|
Title: Business trip
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
What joy! Petya's parents went on a business trip for the whole year and the playful kid is left all by himself. Petya got absolutely happy. He jumped on the bed and threw pillows all day long, until...
Today Petya opened the cupboard and found a scary note there. His parents had left him with duties: he should water their favourite flower all year, each day, in the morning, in the afternoon and in the evening. "Wait a second!" — thought Petya. He know for a fact that if he fulfills the parents' task in the *i*-th (1<=≤<=*i*<=≤<=12) month of the year, then the flower will grow by *a**i* centimeters, and if he doesn't water the flower in the *i*-th month, then the flower won't grow this month. Petya also knows that try as he might, his parents won't believe that he has been watering the flower if it grows strictly less than by *k* centimeters.
Help Petya choose the minimum number of months when he will water the flower, given that the flower should grow no less than by *k* centimeters.
Input Specification:
The first line contains exactly one integer *k* (0<=≤<=*k*<=≤<=100). The next line contains twelve space-separated integers: the *i*-th (1<=≤<=*i*<=≤<=12) number in the line represents *a**i* (0<=≤<=*a**i*<=≤<=100).
Output Specification:
Print the only integer — the minimum number of months when Petya has to water the flower so that the flower grows no less than by *k* centimeters. If the flower can't grow by *k* centimeters in a year, print -1.
Demo Input:
['5\n1 1 1 1 2 2 3 2 2 1 1 1\n', '0\n0 0 0 0 0 0 0 1 1 2 3 0\n', '11\n1 1 4 1 1 5 1 1 4 1 1 1\n']
Demo Output:
['2\n', '0\n', '3\n']
Note:
Let's consider the first sample test. There it is enough to water the flower during the seventh and the ninth month. Then the flower grows by exactly five centimeters.
In the second sample Petya's parents will believe him even if the flower doesn't grow at all (*k* = 0). So, it is possible for Petya not to water the flower at all.
|
```python
k = int(input())
total = 0
for i, n in enumerate(sorted(map(int, input().split()), reverse=True)):
if total >= k:
print(i)
quit()
total += n
if total >= k:
print(i+1)
else:
print(-1)
```
| 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"
}
] | 1,603,963,620
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 5
| 93
| 0
|
s = input()
a = 'heidi'
if s.index("h")<s.index("e")<s.index("i")<s.index("d")<len(s)-s[::-1].index("i")-1:
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()
a = 'heidi'
if s.index("h")<s.index("e")<s.index("i")<s.index("d")<len(s)-s[::-1].index("i")-1:
print("YES")
else:
print("NO")
```
| -1
|
|
757
|
C
|
Felicity is Coming!
|
PROGRAMMING
| 1,900
|
[
"data structures",
"hashing",
"sortings",
"strings"
] | null | null |
It's that time of the year, Felicity is around the corner and you can see people celebrating all around the Himalayan region. The Himalayan region has *n* gyms. The *i*-th gym has *g**i* Pokemon in it. There are *m* distinct Pokemon types in the Himalayan region numbered from 1 to *m*. There is a special evolution camp set up in the fest which claims to evolve any Pokemon. The type of a Pokemon could change after evolving, subject to the constraint that if two Pokemon have the same type before evolving, they will have the same type after evolving. Also, if two Pokemon have different types before evolving, they will have different types after evolving. It is also possible that a Pokemon has the same type before and after evolving.
Formally, an evolution plan is a permutation *f* of {1,<=2,<=...,<=*m*}, such that *f*(*x*)<==<=*y* means that a Pokemon of type *x* evolves into a Pokemon of type *y*.
The gym leaders are intrigued by the special evolution camp and all of them plan to evolve their Pokemons. The protocol of the mountain states that in each gym, for every type of Pokemon, the number of Pokemon of that type before evolving any Pokemon should be equal the number of Pokemon of that type after evolving all the Pokemons according to the evolution plan. They now want to find out how many distinct evolution plans exist which satisfy the protocol.
Two evolution plans *f*1 and *f*2 are distinct, if they have at least one Pokemon type evolving into a different Pokemon type in the two plans, i. e. there exists an *i* such that *f*1(*i*)<=≠<=*f*2(*i*).
Your task is to find how many distinct evolution plans are possible such that if all Pokemon in all the gyms are evolved, the number of Pokemon of each type in each of the gyms remains the same. As the answer can be large, output it modulo 109<=+<=7.
|
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=105, 1<=≤<=*m*<=≤<=106) — the number of gyms and the number of Pokemon types.
The next *n* lines contain the description of Pokemons in the gyms. The *i*-th of these lines begins with the integer *g**i* (1<=≤<=*g**i*<=≤<=105) — the number of Pokemon in the *i*-th gym. After that *g**i* integers follow, denoting types of the Pokemons in the *i*-th gym. Each of these integers is between 1 and *m*.
The total number of Pokemons (the sum of all *g**i*) does not exceed 5·105.
|
Output the number of valid evolution plans modulo 109<=+<=7.
|
[
"2 3\n2 1 2\n2 2 3\n",
"1 3\n3 1 2 3\n",
"2 4\n2 1 2\n3 2 3 4\n",
"2 2\n3 2 2 1\n2 1 2\n",
"3 7\n2 1 2\n2 3 4\n3 5 6 7\n"
] |
[
"1\n",
"6\n",
"2\n",
"1\n",
"24\n"
] |
In the first case, the only possible evolution plan is:
In the second case, any permutation of (1, 2, 3) is valid.
In the third case, there are two possible plans:
In the fourth case, the only possible evolution plan is:
| 1,500
|
[
{
"input": "2 3\n2 1 2\n2 2 3",
"output": "1"
},
{
"input": "1 3\n3 1 2 3",
"output": "6"
},
{
"input": "2 4\n2 1 2\n3 2 3 4",
"output": "2"
},
{
"input": "2 2\n3 2 2 1\n2 1 2",
"output": "1"
},
{
"input": "3 7\n2 1 2\n2 3 4\n3 5 6 7",
"output": "24"
},
{
"input": "10 100\n16 25 7 48 43 16 23 66 3 17 31 64 27 7 17 11 60\n62 76 82 99 77 19 26 66 46 9 54 77 8 34 76 70 48 53 35 69 29 84 22 16 53 36 27 24 81 2 86 67 45 22 54 96 37 8 3 22 9 30 63 61 86 19 16 47 3 72 39 36 1 50 1 18 7 44 52 66 90 3 63\n3 22 61 39\n9 28 69 91 62 98 23 45 9 10\n2 42 20\n3 90 46 55\n2 71 9\n1 7\n1 44\n1 94",
"output": "732842622"
},
{
"input": "10 100\n26 69 60 30 8 89 7 54 66 100 75 4 17 48 40 20 78 56 94 23 48 55 40 9 23 55 30\n3 94 78 64\n50 57 81 62 43 95 4 22 29 9 67 17 82 13 69 13 30 85 3 44 5 85 70 4 50 9 30 85 67 64 7 59 98 78 68 61 63 35 35 94 87 37 18 12 83 26 77 48 67 72 82\n7 59 52 92 41 37 11 17\n1 65\n2 75 82\n4 28 66 33 70\n1 81\n2 4 31\n1 12",
"output": "510562296"
},
{
"input": "10 100\n53 9 10 7 62 66 82 38 22 82 14 48 7 77 51 37 5 10 12 68 88 36 49 80 80 71 48 72 6 49 87 21 48 17 75 43 25 75 55 36 10 82 2 28 14 53 25 66 7 70 58 53 74 86\n32 84 95 55 32 79 75 12 94 80 13 29 49 87 26 69 51 73 52 30 87 17 75 60 1 82 15 34 26 83 95 60 13\n8 61 39 91 78 19 32 91 26\n1 22\n1 87\n1 55\n1 87\n1 39\n1 70\n1 40",
"output": "51603121"
},
{
"input": "10 100\n46 62 64 81 19 35 65 30 81 64 54 95 98 18 78 54 19 68 34 16 37 22 55 63 41 87 65 33 22 15 5 99 35 49 79 47 54 50 97 54 3 100 86 91 3 24 36\n36 25 29 71 1 64 18 92 22 86 76 91 87 79 29 33 61 36 87 22 10 25 7 96 56 67 38 66 43 35 55 54 90 65 83 56 11\n4 36 73 34 11\n2 28 94\n2 97 100\n5 52 69 13 11 78\n1 78\n2 71 8\n1 33\n1 11",
"output": "166939681"
},
{
"input": "10 100\n73 10 13 55 73 7 41 18 37 47 97 43 96 52 97 75 42 23 52 61 89 100 64 43 98 95 86 86 39 85 31 74 30 82 84 51 84 21 35 61 3 15 71 45 99 12 48 54 39 96 85 57 45 35 92 57 65 97 42 91 86 47 64 35 67 52 11 34 24 41 45 42 87 50\n9 77 91 42 99 98 20 43 82 35\n10 96 48 77 64 81 66 3 38 58 9\n1 61\n2 47 35\n1 7\n1 61\n1 70\n1 88\n1 83",
"output": "8656282"
},
{
"input": "100 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1",
"output": "1"
},
{
"input": "100 2\n1 1\n1 2\n1 2\n1 1\n1 2\n1 1\n1 2\n1 1\n1 1\n1 1\n1 2\n1 1\n1 1\n1 1\n1 2\n1 1\n1 2\n1 1\n1 2\n1 2\n1 1\n1 1\n1 1\n1 2\n1 2\n1 2\n1 2\n1 1\n1 2\n1 1\n1 1\n1 2\n1 1\n1 1\n1 1\n1 2\n1 2\n1 1\n1 1\n1 1\n1 1\n1 2\n1 1\n1 2\n1 1\n1 2\n1 1\n1 2\n1 1\n1 2\n1 1\n1 2\n1 2\n1 2\n1 1\n1 1\n1 1\n1 1\n1 1\n1 2\n1 1\n1 1\n1 1\n1 1\n1 2\n1 2\n1 2\n1 1\n1 1\n1 1\n1 1\n1 1\n1 2\n1 1\n1 2\n1 1\n1 2\n1 2\n1 2\n1 1\n1 2\n1 2\n1 2\n1 2\n1 1\n1 2\n1 2\n1 1\n1 2\n1 2\n1 2\n1 2\n1 2\n1 1\n1 2\n1 2\n1 1\n1 2\n1 1\n1 2",
"output": "1"
},
{
"input": "2 1000000\n1 1\n1 2",
"output": "44455173"
},
{
"input": "5 262143\n1 1\n1 2\n1 3\n1 4\n1 5",
"output": "943283753"
},
{
"input": "65 3\n1 1\n1 1\n1 1\n1 1\n2 1 2\n1 1\n2 1 2\n2 1 2\n2 1 2\n1 1\n1 1\n2 1 2\n1 1\n2 1 2\n2 1 2\n2 1 2\n2 1 2\n2 1 2\n2 1 2\n1 1\n1 1\n2 1 2\n1 1\n1 1\n2 1 2\n2 1 2\n2 1 2\n1 1\n2 1 2\n2 1 2\n1 1\n2 1 2\n1 1\n2 1 2\n1 1\n1 1\n1 1\n1 1\n2 1 2\n1 1\n1 1\n2 1 2\n1 1\n1 1\n1 1\n2 1 2\n1 1\n2 1 2\n2 1 2\n1 1\n2 1 2\n2 1 2\n1 1\n2 1 2\n1 1\n2 1 2\n2 1 2\n1 1\n1 1\n2 1 2\n1 1\n1 1\n1 1\n1 1\n2 1 2",
"output": "1"
},
{
"input": "1 1000000\n1 1",
"output": "128233642"
},
{
"input": "20 3\n4 1 3 3 3\n6 1 3 3 3 3 3\n1 1\n2 1 3\n2 1 2\n1 1\n8 1 2 2 2 2 2 2 2\n3 1 3 3\n3 1 3 3\n5 1 2 2 2 2\n10 1 3 3 3 3 3 3 3 3 3\n15 1 3 3 3 3 3 3 3 3 3 3 3 3 3 3\n10 1 2 2 2 2 2 2 2 2 2\n3 1 2 2\n1 1\n1 1\n7 1 3 3 3 3 3 3\n1 1\n1 1\n1 1",
"output": "1"
},
{
"input": "20 3\n1 1\n5 1 2 2 2 2\n6 1 3 3 3 3 3\n2 1 2\n3 1 3 3\n3 1 3 3\n4 1 3 3 3\n2 1 3\n3 1 3 3\n5 1 2 2 2 2\n3 1 3 3\n7 1 2 2 2 2 2 2\n3 1 2 2\n6 1 3 3 3 3 3\n3 1 3 3\n2 1 2\n3 1 3 3\n2 1 2\n1 1\n1 1",
"output": "1"
},
{
"input": "65 3\n1 1\n1 1\n2 1 2\n2 1 2\n2 1 2\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n2 1 2\n2 1 2\n1 1\n1 1\n2 1 2\n1 1\n1 1\n1 1\n2 1 2\n2 1 2\n2 1 2\n2 1 2\n2 1 2\n2 1 2\n2 1 2\n2 1 2\n1 1\n2 1 2\n2 1 2\n1 1\n2 1 2\n2 1 2\n2 1 2\n2 1 2\n2 1 2\n1 1\n2 1 2\n1 1\n1 1\n2 1 2\n2 1 2\n2 1 2\n1 1\n2 1 2\n2 1 2\n1 1\n1 1\n2 1 2\n2 1 2\n1 1\n1 1\n2 1 2\n2 1 2\n2 1 2\n2 1 2\n2 1 2\n1 1\n1 1\n2 1 2\n1 1\n2 1 2",
"output": "1"
},
{
"input": "20 3\n2 1 2\n8 1 3 3 3 3 3 3 3\n4 1 3 3 3\n2 1 2\n3 1 2 2\n9 1 3 3 3 3 3 3 3 3\n2 1 2\n3 1 2 2\n2 1 2\n3 1 3 3\n9 1 3 3 3 3 3 3 3 3\n2 1 2\n15 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2\n2 1 3\n4 1 3 3 3\n2 1 2\n1 1\n1 1\n1 1\n1 1",
"output": "1"
},
{
"input": "20 3\n2 1 3\n3 1 2 2\n5 1 3 3 3 3\n3 1 2 2\n1 1\n5 1 3 3 3 3\n4 1 3 3 3\n5 1 3 3 3 3\n4 1 3 3 3\n3 1 2 2\n2 1 3\n5 1 3 3 3 3\n5 1 2 2 2 2\n6 1 2 2 2 2 2\n3 1 2 2\n5 1 3 3 3 3\n5 1 2 2 2 2\n3 1 3 3\n4 1 2 2 2\n2 1 2",
"output": "1"
},
{
"input": "3 3\n6 1 1 1 1 1 1\n6 2 2 2 2 2 2\n2 1 1",
"output": "1"
}
] | 1,484,618,322
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 20
| 2,000
| 12,697,600
|
from collections import defaultdict, Counter
from itertools import combinations
from functools import reduce
from operator import __add__, __and__, __mul__
n, m = map(int, input().split())
eq = defaultdict(lambda: set(range(1, m + 1)))
for i in range(n):
_, *gs = map(int, input().split())
counter = Counter(gs)
for i in range(m):
counter[i + 1] += 0
ds = defaultdict(set)
for item, count in counter.items():
ds[count].add(item)
for k, vs in ds.items():
for v in vs:
eq[v] &= set(vs)
ps = set(map(tuple, eq.values()))
MODULUS = 10**9 + 7
total = reduce(lambda x, y: (x * reduce(__mul__, range(1, len(y) + 1), 1)) % MODULUS, ps, 1)
print(total)
|
Title: Felicity is Coming!
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It's that time of the year, Felicity is around the corner and you can see people celebrating all around the Himalayan region. The Himalayan region has *n* gyms. The *i*-th gym has *g**i* Pokemon in it. There are *m* distinct Pokemon types in the Himalayan region numbered from 1 to *m*. There is a special evolution camp set up in the fest which claims to evolve any Pokemon. The type of a Pokemon could change after evolving, subject to the constraint that if two Pokemon have the same type before evolving, they will have the same type after evolving. Also, if two Pokemon have different types before evolving, they will have different types after evolving. It is also possible that a Pokemon has the same type before and after evolving.
Formally, an evolution plan is a permutation *f* of {1,<=2,<=...,<=*m*}, such that *f*(*x*)<==<=*y* means that a Pokemon of type *x* evolves into a Pokemon of type *y*.
The gym leaders are intrigued by the special evolution camp and all of them plan to evolve their Pokemons. The protocol of the mountain states that in each gym, for every type of Pokemon, the number of Pokemon of that type before evolving any Pokemon should be equal the number of Pokemon of that type after evolving all the Pokemons according to the evolution plan. They now want to find out how many distinct evolution plans exist which satisfy the protocol.
Two evolution plans *f*1 and *f*2 are distinct, if they have at least one Pokemon type evolving into a different Pokemon type in the two plans, i. e. there exists an *i* such that *f*1(*i*)<=≠<=*f*2(*i*).
Your task is to find how many distinct evolution plans are possible such that if all Pokemon in all the gyms are evolved, the number of Pokemon of each type in each of the gyms remains the same. As the answer can be large, output it modulo 109<=+<=7.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=105, 1<=≤<=*m*<=≤<=106) — the number of gyms and the number of Pokemon types.
The next *n* lines contain the description of Pokemons in the gyms. The *i*-th of these lines begins with the integer *g**i* (1<=≤<=*g**i*<=≤<=105) — the number of Pokemon in the *i*-th gym. After that *g**i* integers follow, denoting types of the Pokemons in the *i*-th gym. Each of these integers is between 1 and *m*.
The total number of Pokemons (the sum of all *g**i*) does not exceed 5·105.
Output Specification:
Output the number of valid evolution plans modulo 109<=+<=7.
Demo Input:
['2 3\n2 1 2\n2 2 3\n', '1 3\n3 1 2 3\n', '2 4\n2 1 2\n3 2 3 4\n', '2 2\n3 2 2 1\n2 1 2\n', '3 7\n2 1 2\n2 3 4\n3 5 6 7\n']
Demo Output:
['1\n', '6\n', '2\n', '1\n', '24\n']
Note:
In the first case, the only possible evolution plan is:
In the second case, any permutation of (1, 2, 3) is valid.
In the third case, there are two possible plans:
In the fourth case, the only possible evolution plan is:
|
```python
from collections import defaultdict, Counter
from itertools import combinations
from functools import reduce
from operator import __add__, __and__, __mul__
n, m = map(int, input().split())
eq = defaultdict(lambda: set(range(1, m + 1)))
for i in range(n):
_, *gs = map(int, input().split())
counter = Counter(gs)
for i in range(m):
counter[i + 1] += 0
ds = defaultdict(set)
for item, count in counter.items():
ds[count].add(item)
for k, vs in ds.items():
for v in vs:
eq[v] &= set(vs)
ps = set(map(tuple, eq.values()))
MODULUS = 10**9 + 7
total = reduce(lambda x, y: (x * reduce(__mul__, range(1, len(y) + 1), 1)) % MODULUS, ps, 1)
print(total)
```
| 0
|
|
527
|
C
|
Glass Carving
|
PROGRAMMING
| 1,500
|
[
"binary search",
"data structures",
"implementation"
] | null | null |
Leonid wants to become a glass carver (the person who creates beautiful artworks by cutting the glass). He already has a rectangular *w* mm <=×<= *h* mm sheet of glass, a diamond glass cutter and lots of enthusiasm. What he lacks is understanding of what to carve and how.
In order not to waste time, he decided to practice the technique of carving. To do this, he makes vertical and horizontal cuts through the entire sheet. This process results in making smaller rectangular fragments of glass. Leonid does not move the newly made glass fragments. In particular, a cut divides each fragment of glass that it goes through into smaller fragments.
After each cut Leonid tries to determine what area the largest of the currently available glass fragments has. Since there appear more and more fragments, this question takes him more and more time and distracts him from the fascinating process.
Leonid offers to divide the labor — he will cut glass, and you will calculate the area of the maximum fragment after each cut. Do you agree?
|
The first line contains three integers *w*,<=*h*,<=*n* (2<=≤<=*w*,<=*h*<=≤<=200<=000, 1<=≤<=*n*<=≤<=200<=000).
Next *n* lines contain the descriptions of the cuts. Each description has the form *H* *y* or *V* *x*. In the first case Leonid makes the horizontal cut at the distance *y* millimeters (1<=≤<=*y*<=≤<=*h*<=-<=1) from the lower edge of the original sheet of glass. In the second case Leonid makes a vertical cut at distance *x* (1<=≤<=*x*<=≤<=*w*<=-<=1) millimeters from the left edge of the original sheet of glass. It is guaranteed that Leonid won't make two identical cuts.
|
After each cut print on a single line the area of the maximum available glass fragment in mm2.
|
[
"4 3 4\nH 2\nV 2\nV 3\nV 1\n",
"7 6 5\nH 4\nV 3\nV 5\nH 2\nV 1\n"
] |
[
"8\n4\n4\n2\n",
"28\n16\n12\n6\n4\n"
] |
Picture for the first sample test:
| 1,500
|
[
{
"input": "4 3 4\nH 2\nV 2\nV 3\nV 1",
"output": "8\n4\n4\n2"
},
{
"input": "7 6 5\nH 4\nV 3\nV 5\nH 2\nV 1",
"output": "28\n16\n12\n6\n4"
},
{
"input": "2 2 1\nV 1",
"output": "2"
},
{
"input": "2 2 1\nH 1",
"output": "2"
},
{
"input": "2 2 2\nV 1\nH 1",
"output": "2\n1"
},
{
"input": "2 2 2\nH 1\nV 1",
"output": "2\n1"
},
{
"input": "10 10 10\nV 6\nH 8\nV 4\nV 8\nH 2\nH 5\nV 9\nH 7\nH 3\nV 7",
"output": "60\n48\n32\n32\n24\n12\n12\n12\n8\n8"
},
{
"input": "5 15 10\nH 8\nH 9\nV 1\nH 2\nH 6\nH 4\nH 1\nV 2\nH 13\nV 3",
"output": "40\n40\n32\n24\n24\n24\n24\n18\n12\n8"
},
{
"input": "15 5 10\nV 13\nV 10\nV 3\nH 2\nV 9\nV 7\nV 2\nH 1\nV 4\nH 3",
"output": "65\n50\n35\n21\n18\n12\n12\n12\n9\n6"
},
{
"input": "2 3 1\nH 1",
"output": "4"
},
{
"input": "200000 200000 1\nH 1",
"output": "39999800000"
},
{
"input": "2 4 1\nH 2",
"output": "4"
}
] | 1,609,928,711
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 22
| 1,840
| 24,473,600
|
w, h, n = map(int, input().split())
l, r = [-1] * (w+1), [-1] * (w+1)
t, b = [-1] * (h+1), [-1] * (h+1)
l[0], b[0], t[h], r[w] = 0, 0, h, w
V, H = [0] * n, [0] * n
for i in range(n):
line, idx = input().split()
idx = int(idx)
if line == 'V':
r[idx] = w
V[i] = idx
else:
t[idx] = h
H[i] = idx
left, max_w = 0, 0
for i in range(1, w+1):
if r[i] != -1:
l[i] = left
r[left] = i
max_w = max(max_w, i - left)
left = i
bottom, max_h = 0, 0
for i in range(1 ,h+1):
if t[i] != -1:
b[i] = bottom
t[bottom] = i
max_h = max(max_h, i - bottom)
bottom = i
res = [0] * n
res[n-1] = max_h * max_w
for i in range(n-1, 0, -1):
if V[i] != 0:
max_w = max(max_w, r[V[i]] - l[V[i]])
r[l[V[i]]] = r[V[i]]
l[r[V[i]]] = l[V[i]]
else:
max_h = max(max_h, t[H[i]] - b[H[i]])
b[t[H[i]]] = b[H[i]]
t[b[H[i]]] = t[H[i]]
res[i-1] = max_h * max_w
for i in range(n):
print(res[i])
|
Title: Glass Carving
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Leonid wants to become a glass carver (the person who creates beautiful artworks by cutting the glass). He already has a rectangular *w* mm <=×<= *h* mm sheet of glass, a diamond glass cutter and lots of enthusiasm. What he lacks is understanding of what to carve and how.
In order not to waste time, he decided to practice the technique of carving. To do this, he makes vertical and horizontal cuts through the entire sheet. This process results in making smaller rectangular fragments of glass. Leonid does not move the newly made glass fragments. In particular, a cut divides each fragment of glass that it goes through into smaller fragments.
After each cut Leonid tries to determine what area the largest of the currently available glass fragments has. Since there appear more and more fragments, this question takes him more and more time and distracts him from the fascinating process.
Leonid offers to divide the labor — he will cut glass, and you will calculate the area of the maximum fragment after each cut. Do you agree?
Input Specification:
The first line contains three integers *w*,<=*h*,<=*n* (2<=≤<=*w*,<=*h*<=≤<=200<=000, 1<=≤<=*n*<=≤<=200<=000).
Next *n* lines contain the descriptions of the cuts. Each description has the form *H* *y* or *V* *x*. In the first case Leonid makes the horizontal cut at the distance *y* millimeters (1<=≤<=*y*<=≤<=*h*<=-<=1) from the lower edge of the original sheet of glass. In the second case Leonid makes a vertical cut at distance *x* (1<=≤<=*x*<=≤<=*w*<=-<=1) millimeters from the left edge of the original sheet of glass. It is guaranteed that Leonid won't make two identical cuts.
Output Specification:
After each cut print on a single line the area of the maximum available glass fragment in mm2.
Demo Input:
['4 3 4\nH 2\nV 2\nV 3\nV 1\n', '7 6 5\nH 4\nV 3\nV 5\nH 2\nV 1\n']
Demo Output:
['8\n4\n4\n2\n', '28\n16\n12\n6\n4\n']
Note:
Picture for the first sample test:
|
```python
w, h, n = map(int, input().split())
l, r = [-1] * (w+1), [-1] * (w+1)
t, b = [-1] * (h+1), [-1] * (h+1)
l[0], b[0], t[h], r[w] = 0, 0, h, w
V, H = [0] * n, [0] * n
for i in range(n):
line, idx = input().split()
idx = int(idx)
if line == 'V':
r[idx] = w
V[i] = idx
else:
t[idx] = h
H[i] = idx
left, max_w = 0, 0
for i in range(1, w+1):
if r[i] != -1:
l[i] = left
r[left] = i
max_w = max(max_w, i - left)
left = i
bottom, max_h = 0, 0
for i in range(1 ,h+1):
if t[i] != -1:
b[i] = bottom
t[bottom] = i
max_h = max(max_h, i - bottom)
bottom = i
res = [0] * n
res[n-1] = max_h * max_w
for i in range(n-1, 0, -1):
if V[i] != 0:
max_w = max(max_w, r[V[i]] - l[V[i]])
r[l[V[i]]] = r[V[i]]
l[r[V[i]]] = l[V[i]]
else:
max_h = max(max_h, t[H[i]] - b[H[i]])
b[t[H[i]]] = b[H[i]]
t[b[H[i]]] = t[H[i]]
res[i-1] = max_h * max_w
for i in range(n):
print(res[i])
```
| 3
|
|
296
|
A
|
Yaroslav and Permutations
|
PROGRAMMING
| 1,100
|
[
"greedy",
"math"
] | null | null |
Yaroslav has an array that consists of *n* integers. In one second Yaroslav can swap two neighboring array elements. Now Yaroslav is wondering if he can obtain an array where any two neighboring elements would be distinct in a finite time.
Help Yaroslav.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of elements in the array. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000) — the array elements.
|
In the single line print "YES" (without the quotes) if Yaroslav can obtain the array he needs, and "NO" (without the quotes) otherwise.
|
[
"1\n1\n",
"3\n1 1 2\n",
"4\n7 7 7 7\n"
] |
[
"YES\n",
"YES\n",
"NO\n"
] |
In the first sample the initial array fits well.
In the second sample Yaroslav can get array: 1, 2, 1. He can swap the last and the second last elements to obtain it.
In the third sample Yarosav can't get the array he needs.
| 500
|
[
{
"input": "1\n1",
"output": "YES"
},
{
"input": "3\n1 1 2",
"output": "YES"
},
{
"input": "4\n7 7 7 7",
"output": "NO"
},
{
"input": "4\n479 170 465 146",
"output": "YES"
},
{
"input": "5\n996 437 605 996 293",
"output": "YES"
},
{
"input": "6\n727 539 896 668 36 896",
"output": "YES"
},
{
"input": "7\n674 712 674 674 674 674 674",
"output": "NO"
},
{
"input": "8\n742 742 742 742 742 289 742 742",
"output": "NO"
},
{
"input": "9\n730 351 806 806 806 630 85 757 967",
"output": "YES"
},
{
"input": "10\n324 539 83 440 834 640 440 440 440 440",
"output": "YES"
},
{
"input": "7\n925 830 925 98 987 162 356",
"output": "YES"
},
{
"input": "68\n575 32 53 351 151 942 725 967 431 108 192 8 338 458 288 754 384 946 910 210 759 222 589 423 947 507 31 414 169 901 592 763 656 411 360 625 538 549 484 596 42 603 351 292 837 375 21 597 22 349 200 669 485 282 735 54 1000 419 939 901 789 128 468 729 894 649 484 808",
"output": "YES"
},
{
"input": "22\n618 814 515 310 617 936 452 601 250 520 557 799 304 225 9 845 610 990 703 196 486 94",
"output": "YES"
},
{
"input": "44\n459 581 449 449 449 449 449 449 449 623 449 449 449 449 449 449 449 449 889 449 203 273 329 449 449 449 449 449 449 845 882 323 22 449 449 893 449 449 449 449 449 870 449 402",
"output": "NO"
},
{
"input": "90\n424 3 586 183 286 89 427 618 758 833 933 170 155 722 190 977 330 369 693 426 556 435 550 442 513 146 61 719 754 140 424 280 997 688 530 550 438 867 950 194 196 298 417 287 106 489 283 456 735 115 702 317 672 787 264 314 356 186 54 913 809 833 946 314 757 322 559 647 983 482 145 197 223 130 162 536 451 174 467 45 660 293 440 254 25 155 511 746 650 187",
"output": "YES"
},
{
"input": "14\n959 203 478 315 788 788 373 834 488 519 774 764 193 103",
"output": "YES"
},
{
"input": "81\n544 528 528 528 528 4 506 528 32 528 528 528 528 528 528 528 528 975 528 528 528 528 528 528 528 528 528 528 528 528 528 20 528 528 528 528 528 528 528 528 852 528 528 120 528 528 61 11 528 528 528 228 528 165 883 528 488 475 628 528 528 528 528 528 528 597 528 528 528 528 528 528 528 528 528 528 528 412 528 521 925",
"output": "NO"
},
{
"input": "89\n354 356 352 355 355 355 352 354 354 352 355 356 355 352 354 356 354 355 355 354 353 352 352 355 355 356 352 352 353 356 352 353 354 352 355 352 353 353 353 354 353 354 354 353 356 353 353 354 354 354 354 353 352 353 355 356 356 352 356 354 353 352 355 354 356 356 356 354 354 356 354 355 354 355 353 352 354 355 352 355 355 354 356 353 353 352 356 352 353",
"output": "YES"
},
{
"input": "71\n284 284 285 285 285 284 285 284 284 285 284 285 284 284 285 284 285 285 285 285 284 284 285 285 284 284 284 285 284 285 284 285 285 284 284 284 285 284 284 285 285 285 284 284 285 284 285 285 284 285 285 284 285 284 284 284 285 285 284 285 284 285 285 285 285 284 284 285 285 284 285",
"output": "NO"
},
{
"input": "28\n602 216 214 825 814 760 814 28 76 814 814 288 814 814 222 707 11 490 814 543 914 705 814 751 976 814 814 99",
"output": "YES"
},
{
"input": "48\n546 547 914 263 986 945 914 914 509 871 324 914 153 571 914 914 914 528 970 566 544 914 914 914 410 914 914 589 609 222 914 889 691 844 621 68 914 36 914 39 630 749 914 258 945 914 727 26",
"output": "YES"
},
{
"input": "56\n516 76 516 197 516 427 174 516 706 813 94 37 516 815 516 516 937 483 16 516 842 516 638 691 516 635 516 516 453 263 516 516 635 257 125 214 29 81 516 51 362 516 677 516 903 516 949 654 221 924 516 879 516 516 972 516",
"output": "YES"
},
{
"input": "46\n314 723 314 314 314 235 314 314 314 314 270 314 59 972 314 216 816 40 314 314 314 314 314 314 314 381 314 314 314 314 314 314 314 789 314 957 114 942 314 314 29 314 314 72 314 314",
"output": "NO"
},
{
"input": "72\n169 169 169 599 694 81 250 529 865 406 817 169 667 169 965 169 169 663 65 169 903 169 942 763 169 807 169 603 169 169 13 169 169 810 169 291 169 169 169 169 169 169 169 713 169 440 169 169 169 169 169 480 169 169 867 169 169 169 169 169 169 169 169 393 169 169 459 169 99 169 601 800",
"output": "NO"
},
{
"input": "100\n317 316 317 316 317 316 317 316 317 316 316 317 317 316 317 316 316 316 317 316 317 317 316 317 316 316 316 316 316 316 317 316 317 317 317 317 317 317 316 316 316 317 316 317 316 317 316 317 317 316 317 316 317 317 316 317 316 317 316 317 316 316 316 317 317 317 317 317 316 317 317 316 316 316 316 317 317 316 317 316 316 316 316 316 316 317 316 316 317 317 317 317 317 317 317 317 317 316 316 317",
"output": "NO"
},
{
"input": "100\n510 510 510 162 969 32 510 511 510 510 911 183 496 875 903 461 510 510 123 578 510 510 510 510 510 755 510 673 510 510 763 510 510 909 510 435 487 959 807 510 368 788 557 448 284 332 510 949 510 510 777 112 857 926 487 510 510 510 678 510 510 197 829 427 698 704 409 509 510 238 314 851 510 651 510 455 682 510 714 635 973 510 443 878 510 510 510 591 510 24 596 510 43 183 510 510 671 652 214 784",
"output": "YES"
},
{
"input": "100\n476 477 474 476 476 475 473 476 474 475 473 477 476 476 474 476 474 475 476 477 473 473 473 474 474 476 473 473 476 476 475 476 473 474 473 473 477 475 475 475 476 475 477 477 477 476 475 475 475 473 476 477 475 476 477 473 474 477 473 475 476 476 474 477 476 474 473 477 473 475 477 473 476 474 477 473 475 477 473 476 476 475 476 475 474 473 477 473 475 473 477 473 473 474 475 473 477 476 477 474",
"output": "YES"
},
{
"input": "100\n498 498 498 498 498 499 498 499 499 499 498 498 498 498 499 498 499 499 498 499 498 498 498 499 499 499 498 498 499 499 498 498 498 499 498 499 498 498 498 499 498 499 498 498 498 498 499 498 498 499 498 498 499 498 499 499 498 499 499 499 498 498 498 498 499 498 499 498 499 499 499 499 498 498 499 499 498 499 499 498 498 499 499 498 498 499 499 499 498 498 499 498 498 498 499 499 499 498 498 499",
"output": "NO"
},
{
"input": "100\n858 53 816 816 816 816 816 816 816 181 816 816 816 816 579 879 816 948 171 816 816 150 866 816 816 816 897 816 816 816 816 816 816 706 816 539 816 816 816 816 816 816 423 487 816 615 254 816 816 816 816 83 816 816 816 816 816 816 816 816 816 816 816 136 775 999 816 816 816 644 816 816 816 816 927 816 802 816 856 816 816 816 816 816 816 816 816 816 816 700 816 816 816 816 982 477 816 891 806 816",
"output": "NO"
},
{
"input": "100\n167 169 169 167 169 169 167 167 167 167 168 166 170 170 169 170 170 170 169 168 166 167 170 169 167 169 168 169 166 170 166 167 170 166 166 167 169 166 166 169 166 167 168 168 170 167 168 166 168 170 167 168 167 169 169 166 168 167 170 168 167 169 168 169 166 168 168 169 169 166 170 168 167 169 170 168 167 169 168 167 168 168 166 169 170 170 166 166 167 170 167 168 167 167 169 169 166 166 169 167",
"output": "YES"
},
{
"input": "100\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000",
"output": "NO"
},
{
"input": "99\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000",
"output": "NO"
},
{
"input": "100\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "YES"
},
{
"input": "99\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "YES"
},
{
"input": "2\n1 1",
"output": "NO"
},
{
"input": "1\n1000",
"output": "YES"
},
{
"input": "12\n2 2 4 4 4 4 6 6 6 6 6 6",
"output": "YES"
}
] | 1,679,435,722
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 37
| 122
| 0
|
#61
n = int(input())
ins = [int(x) for x in input().split()]
countDict = {}
maxCount = 1
for elemnt in ins:
try:
countDict[elemnt] += 1
if countDict[elemnt] > maxCount:
maxCount = countDict[elemnt]
except:
countDict[elemnt] = 1
if n - maxCount >= maxCount - 1:
print("YES")
else:
print("NO")
|
Title: Yaroslav and Permutations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Yaroslav has an array that consists of *n* integers. In one second Yaroslav can swap two neighboring array elements. Now Yaroslav is wondering if he can obtain an array where any two neighboring elements would be distinct in a finite time.
Help Yaroslav.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of elements in the array. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000) — the array elements.
Output Specification:
In the single line print "YES" (without the quotes) if Yaroslav can obtain the array he needs, and "NO" (without the quotes) otherwise.
Demo Input:
['1\n1\n', '3\n1 1 2\n', '4\n7 7 7 7\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
In the first sample the initial array fits well.
In the second sample Yaroslav can get array: 1, 2, 1. He can swap the last and the second last elements to obtain it.
In the third sample Yarosav can't get the array he needs.
|
```python
#61
n = int(input())
ins = [int(x) for x in input().split()]
countDict = {}
maxCount = 1
for elemnt in ins:
try:
countDict[elemnt] += 1
if countDict[elemnt] > maxCount:
maxCount = countDict[elemnt]
except:
countDict[elemnt] = 1
if n - maxCount >= maxCount - 1:
print("YES")
else:
print("NO")
```
| 3
|
|
62
|
A
|
A Student's Dream
|
PROGRAMMING
| 1,300
|
[
"greedy",
"math"
] |
A. A Student's Dream
|
2
|
256
|
Statistics claims that students sleep no more than three hours a day. But even in the world of their dreams, while they are snoring peacefully, the sense of impending doom is still upon them.
A poor student is dreaming that he is sitting the mathematical analysis exam. And he is examined by the most formidable professor of all times, a three times Soviet Union Hero, a Noble Prize laureate in student expulsion, venerable Petr Palych.
The poor student couldn't answer a single question. Thus, instead of a large spacious office he is going to apply for a job to thorium mines. But wait a minute! Petr Palych decided to give the student the last chance! Yes, that is possible only in dreams.
So the professor began: "Once a Venusian girl and a Marsian boy met on the Earth and decided to take a walk holding hands. But the problem is the girl has *a**l* fingers on her left hand and *a**r* fingers on the right one. The boy correspondingly has *b**l* and *b**r* fingers. They can only feel comfortable when holding hands, when no pair of the girl's fingers will touch each other. That is, they are comfortable when between any two girl's fingers there is a boy's finger. And in addition, no three fingers of the boy should touch each other. Determine if they can hold hands so that the both were comfortable."
The boy any the girl don't care who goes to the left and who goes to the right. The difference is only that if the boy goes to the left of the girl, he will take her left hand with his right one, and if he goes to the right of the girl, then it is vice versa.
|
The first line contains two positive integers not exceeding 100. They are the number of fingers on the Venusian girl's left and right hand correspondingly. The second line contains two integers not exceeding 100. They are the number of fingers on the Marsian boy's left and right hands correspondingly.
|
Print YES or NO, that is, the answer to Petr Palych's question.
|
[
"5 1\n10 5\n",
"4 5\n3 3\n",
"1 2\n11 6\n"
] |
[
"YES",
"YES",
"NO"
] |
The boy and the girl don't really care who goes to the left.
| 500
|
[
{
"input": "5 1\n10 5",
"output": "YES"
},
{
"input": "4 5\n3 3",
"output": "YES"
},
{
"input": "1 2\n11 6",
"output": "NO"
},
{
"input": "1 1\n1 1",
"output": "YES"
},
{
"input": "2 2\n1 1",
"output": "YES"
},
{
"input": "3 3\n1 1",
"output": "NO"
},
{
"input": "4 4\n1 1",
"output": "NO"
},
{
"input": "100 100\n50 50",
"output": "NO"
},
{
"input": "100 3\n4 1",
"output": "YES"
},
{
"input": "100 5\n1 1",
"output": "NO"
},
{
"input": "100 4\n1 1",
"output": "NO"
},
{
"input": "100 1\n4 1",
"output": "YES"
},
{
"input": "1 100\n1 4",
"output": "YES"
},
{
"input": "1 100\n5 4",
"output": "YES"
},
{
"input": "1 100\n1 5",
"output": "NO"
},
{
"input": "43 100\n65 24",
"output": "NO"
},
{
"input": "4 2\n12 1",
"output": "NO"
},
{
"input": "6 11\n13 11",
"output": "YES"
},
{
"input": "2 6\n12 12",
"output": "YES"
},
{
"input": "14 7\n2 9",
"output": "NO"
},
{
"input": "1 14\n7 14",
"output": "NO"
},
{
"input": "6 11\n2 10",
"output": "YES"
},
{
"input": "5 12\n13 11",
"output": "YES"
},
{
"input": "15 1\n11 9",
"output": "NO"
},
{
"input": "7 12\n10 6",
"output": "YES"
},
{
"input": "15 7\n15 15",
"output": "YES"
},
{
"input": "1 5\n14 1",
"output": "YES"
},
{
"input": "2 4\n6 6",
"output": "YES"
},
{
"input": "12 8\n4 12",
"output": "YES"
},
{
"input": "6 14\n5 5",
"output": "YES"
},
{
"input": "19 17\n5 8",
"output": "NO"
},
{
"input": "9 21\n13 16",
"output": "YES"
},
{
"input": "11 2\n11 22",
"output": "YES"
},
{
"input": "15 3\n12 16",
"output": "YES"
},
{
"input": "13 2\n13 5",
"output": "NO"
},
{
"input": "21 1\n5 19",
"output": "NO"
},
{
"input": "9 15\n16 2",
"output": "YES"
},
{
"input": "7 18\n23 19",
"output": "YES"
},
{
"input": "13 17\n19 1",
"output": "YES"
},
{
"input": "5 15\n13 9",
"output": "YES"
},
{
"input": "11 17\n6 4",
"output": "NO"
},
{
"input": "18 3\n16 15",
"output": "NO"
},
{
"input": "5 23\n12 17",
"output": "NO"
},
{
"input": "25 8\n14 24",
"output": "YES"
},
{
"input": "18 22\n22 19",
"output": "YES"
},
{
"input": "2 25\n8 24",
"output": "NO"
},
{
"input": "7 25\n18 15",
"output": "YES"
},
{
"input": "8 22\n2 3",
"output": "NO"
},
{
"input": "25 9\n16 12",
"output": "YES"
},
{
"input": "19 4\n25 17",
"output": "NO"
},
{
"input": "24 43\n96 39",
"output": "YES"
},
{
"input": "13 23\n19 63",
"output": "NO"
},
{
"input": "93 12\n87 54",
"output": "NO"
},
{
"input": "94 35\n53 79",
"output": "YES"
},
{
"input": "65 8\n73 25",
"output": "NO"
},
{
"input": "25 14\n19 91",
"output": "YES"
},
{
"input": "58 86\n40 46",
"output": "NO"
},
{
"input": "82 60\n100 38",
"output": "YES"
},
{
"input": "36 62\n81 12",
"output": "YES"
},
{
"input": "30 38\n12 100",
"output": "NO"
}
] | 1,590,843,982
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 5
| 278
| 0
|
import math
a, b = map(int, input().split())
c, d = map(int, input().split())
if a >= ((2/3)*d) or b >= ((2/3)*c) :
print("YES", '\n')
else :
print('NO', '\n')
|
Title: A Student's Dream
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Statistics claims that students sleep no more than three hours a day. But even in the world of their dreams, while they are snoring peacefully, the sense of impending doom is still upon them.
A poor student is dreaming that he is sitting the mathematical analysis exam. And he is examined by the most formidable professor of all times, a three times Soviet Union Hero, a Noble Prize laureate in student expulsion, venerable Petr Palych.
The poor student couldn't answer a single question. Thus, instead of a large spacious office he is going to apply for a job to thorium mines. But wait a minute! Petr Palych decided to give the student the last chance! Yes, that is possible only in dreams.
So the professor began: "Once a Venusian girl and a Marsian boy met on the Earth and decided to take a walk holding hands. But the problem is the girl has *a**l* fingers on her left hand and *a**r* fingers on the right one. The boy correspondingly has *b**l* and *b**r* fingers. They can only feel comfortable when holding hands, when no pair of the girl's fingers will touch each other. That is, they are comfortable when between any two girl's fingers there is a boy's finger. And in addition, no three fingers of the boy should touch each other. Determine if they can hold hands so that the both were comfortable."
The boy any the girl don't care who goes to the left and who goes to the right. The difference is only that if the boy goes to the left of the girl, he will take her left hand with his right one, and if he goes to the right of the girl, then it is vice versa.
Input Specification:
The first line contains two positive integers not exceeding 100. They are the number of fingers on the Venusian girl's left and right hand correspondingly. The second line contains two integers not exceeding 100. They are the number of fingers on the Marsian boy's left and right hands correspondingly.
Output Specification:
Print YES or NO, that is, the answer to Petr Palych's question.
Demo Input:
['5 1\n10 5\n', '4 5\n3 3\n', '1 2\n11 6\n']
Demo Output:
['YES', 'YES', 'NO']
Note:
The boy and the girl don't really care who goes to the left.
|
```python
import math
a, b = map(int, input().split())
c, d = map(int, input().split())
if a >= ((2/3)*d) or b >= ((2/3)*c) :
print("YES", '\n')
else :
print('NO', '\n')
```
| 0
|
58
|
A
|
Chat room
|
PROGRAMMING
| 1,000
|
[
"greedy",
"strings"
] |
A. Chat room
|
1
|
256
|
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
|
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
|
If Vasya managed to say hello, print "YES", otherwise print "NO".
|
[
"ahhellllloou\n",
"hlelo\n"
] |
[
"YES\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "ahhellllloou",
"output": "YES"
},
{
"input": "hlelo",
"output": "NO"
},
{
"input": "helhcludoo",
"output": "YES"
},
{
"input": "hehwelloho",
"output": "YES"
},
{
"input": "pnnepelqomhhheollvlo",
"output": "YES"
},
{
"input": "tymbzjyqhymedasloqbq",
"output": "NO"
},
{
"input": "yehluhlkwo",
"output": "NO"
},
{
"input": "hatlevhhalrohairnolsvocafgueelrqmlqlleello",
"output": "YES"
},
{
"input": "hhhtehdbllnhwmbyhvelqqyoulretpbfokflhlhreeflxeftelziclrwllrpflflbdtotvlqgoaoqldlroovbfsq",
"output": "YES"
},
{
"input": "rzlvihhghnelqtwlexmvdjjrliqllolhyewgozkuovaiezgcilelqapuoeglnwmnlftxxiigzczlouooi",
"output": "YES"
},
{
"input": "pfhhwctyqdlkrwhebfqfelhyebwllhemtrmeblgrynmvyhioesqklclocxmlffuormljszllpoo",
"output": "YES"
},
{
"input": "lqllcolohwflhfhlnaow",
"output": "NO"
},
{
"input": "heheeellollvoo",
"output": "YES"
},
{
"input": "hellooo",
"output": "YES"
},
{
"input": "o",
"output": "NO"
},
{
"input": "hhqhzeclohlehljlhtesllylrolmomvuhcxsobtsckogdv",
"output": "YES"
},
{
"input": "yoegfuzhqsihygnhpnukluutocvvwuldiighpogsifealtgkfzqbwtmgghmythcxflebrkctlldlkzlagovwlstsghbouk",
"output": "YES"
},
{
"input": "uatqtgbvrnywfacwursctpagasnhydvmlinrcnqrry",
"output": "NO"
},
{
"input": "tndtbldbllnrwmbyhvqaqqyoudrstpbfokfoclnraefuxtftmgzicorwisrpfnfpbdtatvwqgyalqtdtrjqvbfsq",
"output": "NO"
},
{
"input": "rzlvirhgemelnzdawzpaoqtxmqucnahvqnwldklrmjiiyageraijfivigvozgwngiulttxxgzczptusoi",
"output": "YES"
},
{
"input": "kgyelmchocojsnaqdsyeqgnllytbqietpdlgknwwumqkxrexgdcnwoldicwzwofpmuesjuxzrasscvyuqwspm",
"output": "YES"
},
{
"input": "pnyvrcotjvgynbeldnxieghfltmexttuxzyac",
"output": "NO"
},
{
"input": "dtwhbqoumejligbenxvzhjlhosqojetcqsynlzyhfaevbdpekgbtjrbhlltbceobcok",
"output": "YES"
},
{
"input": "crrfpfftjwhhikwzeedrlwzblckkteseofjuxjrktcjfsylmlsvogvrcxbxtffujqshslemnixoeezivksouefeqlhhokwbqjz",
"output": "YES"
},
{
"input": "jhfbndhyzdvhbvhmhmefqllujdflwdpjbehedlsqfdsqlyelwjtyloxwsvasrbqosblzbowlqjmyeilcvotdlaouxhdpoeloaovb",
"output": "YES"
},
{
"input": "hwlghueoemiqtjhhpashjsouyegdlvoyzeunlroypoprnhlyiwiuxrghekaylndhrhllllwhbebezoglydcvykllotrlaqtvmlla",
"output": "YES"
},
{
"input": "wshiaunnqnqxodholbipwhhjmyeblhgpeleblklpzwhdunmpqkbuzloetmwwxmeltkrcomulxauzlwmlklldjodozxryghsnwgcz",
"output": "YES"
},
{
"input": "shvksednttggehroewuiptvvxtrzgidravtnjwuqrlnnkxbplctzkckinpkgjopjfoxdbojtcvsuvablcbkrzajrlhgobkcxeqti",
"output": "YES"
},
{
"input": "hyyhddqhxhekehkwfhlnlsihzefwchzerevcjtokefplholrbvxlltdlafjxrfhleglrvlolojoqaolagtbeyogxlbgfolllslli",
"output": "YES"
},
{
"input": "iaagrdhhelxpdegueiulflquvlzidoprzkehthkldaytrjwhyhprwjxlltinxvuilxohqgjqcvkkdcuoliekcnlilwgqlnlzouoo",
"output": "YES"
},
{
"input": "wfluaeseldgxyvxpwuhkptdmlflnlhktwxiabsvkolsquymrmhzczzoybvlilhmvsuunggvgxzgyyffk",
"output": "NO"
},
{
"input": "loee",
"output": "NO"
},
{
"input": "izdipwylefqmjbuoznfglgcfvedeouflkeehxbasaqmiooppfsqlhprospqgxvzrcpwlfdddngoqvpwvggsnvvxhmjoslwjudjlo",
"output": "NO"
},
{
"input": "pplynflbjlseflaegzqpdxswgunhnlglqhycrbcwdfhclmcrwekkzsoosmfblljeenwfjyifcvxihhhbyhmhlkvtskeglaovjnsi",
"output": "YES"
},
{
"input": "xqjqmenkodmlhzyzmmvofdngktygbbxbzpluzcohohmalkoeuwfikblltaaigv",
"output": "NO"
},
{
"input": "hello",
"output": "YES"
},
{
"input": "oohell",
"output": "NO"
},
{
"input": "hell",
"output": "NO"
},
{
"input": "eloellohoelo",
"output": "NO"
},
{
"input": "helllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllo",
"output": "YES"
}
] | 1,664,709,936
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 46
| 0
|
a = input()
s = ""
for i in a:
if i not in s:
s+=i
print("YES" if "helo" in s and a.count("l") > 1 else "NO")
|
Title: Chat room
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
Input Specification:
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
Output Specification:
If Vasya managed to say hello, print "YES", otherwise print "NO".
Demo Input:
['ahhellllloou\n', 'hlelo\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none
|
```python
a = input()
s = ""
for i in a:
if i not in s:
s+=i
print("YES" if "helo" in s and a.count("l") > 1 else "NO")
```
| 0
|
499
|
B
|
Lecture
|
PROGRAMMING
| 1,000
|
[
"implementation",
"strings"
] | null | null |
You have a new professor of graph theory and he speaks very quickly. You come up with the following plan to keep up with his lecture and make notes.
You know two languages, and the professor is giving the lecture in the first one. The words in both languages consist of lowercase English characters, each language consists of several words. For each language, all words are distinct, i.e. they are spelled differently. Moreover, the words of these languages have a one-to-one correspondence, that is, for each word in each language, there exists exactly one word in the other language having has the same meaning.
You can write down every word the professor says in either the first language or the second language. Of course, during the lecture you write down each word in the language in which the word is shorter. In case of equal lengths of the corresponding words you prefer the word of the first language.
You are given the text of the lecture the professor is going to read. Find out how the lecture will be recorded in your notes.
|
The first line contains two integers, *n* and *m* (1<=≤<=*n*<=≤<=3000, 1<=≤<=*m*<=≤<=3000) — the number of words in the professor's lecture and the number of words in each of these languages.
The following *m* lines contain the words. The *i*-th line contains two strings *a**i*, *b**i* meaning that the word *a**i* belongs to the first language, the word *b**i* belongs to the second language, and these two words have the same meaning. It is guaranteed that no word occurs in both languages, and each word occurs in its language exactly once.
The next line contains *n* space-separated strings *c*1,<=*c*2,<=...,<=*c**n* — the text of the lecture. It is guaranteed that each of the strings *c**i* belongs to the set of strings {*a*1,<=*a*2,<=... *a**m*}.
All the strings in the input are non-empty, each consisting of no more than 10 lowercase English letters.
|
Output exactly *n* words: how you will record the lecture in your notebook. Output the words of the lecture in the same order as in the input.
|
[
"4 3\ncodeforces codesecrof\ncontest round\nletter message\ncodeforces contest letter contest\n",
"5 3\njoll wuqrd\neuzf un\nhbnyiyc rsoqqveh\nhbnyiyc joll joll euzf joll\n"
] |
[
"codeforces round letter round\n",
"hbnyiyc joll joll un joll\n"
] |
none
| 500
|
[
{
"input": "4 3\ncodeforces codesecrof\ncontest round\nletter message\ncodeforces contest letter contest",
"output": "codeforces round letter round"
},
{
"input": "5 3\njoll wuqrd\neuzf un\nhbnyiyc rsoqqveh\nhbnyiyc joll joll euzf joll",
"output": "hbnyiyc joll joll un joll"
},
{
"input": "5 5\nqueyqj f\nb vn\ntabzvk qpfzoqx\nytnyonoc hnxsd\njpggvr lchinjmt\nqueyqj jpggvr b ytnyonoc b",
"output": "f jpggvr b hnxsd b"
},
{
"input": "10 22\nazbrll oen\ngh vdyayei\njphveblohx vfglv\nmfyxib jepnvhcuwo\nrpikazqj uam\nl rx\nokjenof qpnyi\nj tixqrno\nod itozmfct\nikkdxmirx ev\nqexftojc p\nkdazb zjs\nmbk ykvqjrxaxu\nhbcwhouzq pwt\nmirpsz zfaegpl\nuhkkvcj rlvwj\nef iqnnwtolrc\npjzfcpmeag ecdayth\nsa qcthz\ncbfhfxi qrnbvdryz\nwqel tj\natx smkbid\nef hbcwhouzq cbfhfxi hbcwhouzq mirpsz cbfhfxi cbfhfxi okjenof pjzfcpmeag kdazb",
"output": "ef pwt cbfhfxi pwt mirpsz cbfhfxi cbfhfxi qpnyi ecdayth zjs"
},
{
"input": "1 1\namit am\namit",
"output": "am"
},
{
"input": "1 1\na c\na",
"output": "a"
}
] | 1,618,792,414
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 0
| 61
| 0
|
q = [int(i) for i in input().split()]
num_words = q[0]
nil = q[1]
list_words = []
finale = []
for x in range(nil):
info = [str(i) for i in input().split()]
info0 = info[0]
info1 = info[1]
list_words.append(info0)
list_words.append(info1)
print(list_words)
lecture = [str(i) for i in input().split()]
for x in range(len(lecture)):
word = lecture[x]
number = list_words.index(word)
if len(list_words[number]) <= len(list_words[number+1]):
number = list_words.index(word)
else:
number = list_words.index(word)+1
finale.append(list_words[number])
print(' '.join(finale))
|
Title: Lecture
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have a new professor of graph theory and he speaks very quickly. You come up with the following plan to keep up with his lecture and make notes.
You know two languages, and the professor is giving the lecture in the first one. The words in both languages consist of lowercase English characters, each language consists of several words. For each language, all words are distinct, i.e. they are spelled differently. Moreover, the words of these languages have a one-to-one correspondence, that is, for each word in each language, there exists exactly one word in the other language having has the same meaning.
You can write down every word the professor says in either the first language or the second language. Of course, during the lecture you write down each word in the language in which the word is shorter. In case of equal lengths of the corresponding words you prefer the word of the first language.
You are given the text of the lecture the professor is going to read. Find out how the lecture will be recorded in your notes.
Input Specification:
The first line contains two integers, *n* and *m* (1<=≤<=*n*<=≤<=3000, 1<=≤<=*m*<=≤<=3000) — the number of words in the professor's lecture and the number of words in each of these languages.
The following *m* lines contain the words. The *i*-th line contains two strings *a**i*, *b**i* meaning that the word *a**i* belongs to the first language, the word *b**i* belongs to the second language, and these two words have the same meaning. It is guaranteed that no word occurs in both languages, and each word occurs in its language exactly once.
The next line contains *n* space-separated strings *c*1,<=*c*2,<=...,<=*c**n* — the text of the lecture. It is guaranteed that each of the strings *c**i* belongs to the set of strings {*a*1,<=*a*2,<=... *a**m*}.
All the strings in the input are non-empty, each consisting of no more than 10 lowercase English letters.
Output Specification:
Output exactly *n* words: how you will record the lecture in your notebook. Output the words of the lecture in the same order as in the input.
Demo Input:
['4 3\ncodeforces codesecrof\ncontest round\nletter message\ncodeforces contest letter contest\n', '5 3\njoll wuqrd\neuzf un\nhbnyiyc rsoqqveh\nhbnyiyc joll joll euzf joll\n']
Demo Output:
['codeforces round letter round\n', 'hbnyiyc joll joll un joll\n']
Note:
none
|
```python
q = [int(i) for i in input().split()]
num_words = q[0]
nil = q[1]
list_words = []
finale = []
for x in range(nil):
info = [str(i) for i in input().split()]
info0 = info[0]
info1 = info[1]
list_words.append(info0)
list_words.append(info1)
print(list_words)
lecture = [str(i) for i in input().split()]
for x in range(len(lecture)):
word = lecture[x]
number = list_words.index(word)
if len(list_words[number]) <= len(list_words[number+1]):
number = list_words.index(word)
else:
number = list_words.index(word)+1
finale.append(list_words[number])
print(' '.join(finale))
```
| 0
|
|
858
|
A
|
k-rounding
|
PROGRAMMING
| 1,100
|
[
"brute force",
"math",
"number theory"
] | null | null |
For a given positive integer *n* denote its *k*-rounding as the minimum positive integer *x*, such that *x* ends with *k* or more zeros in base 10 and is divisible by *n*.
For example, 4-rounding of 375 is 375·80<==<=30000. 30000 is the minimum integer such that it ends with 4 or more zeros and is divisible by 375.
Write a program that will perform the *k*-rounding of *n*.
|
The only line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=109, 0<=≤<=*k*<=≤<=8).
|
Print the *k*-rounding of *n*.
|
[
"375 4\n",
"10000 1\n",
"38101 0\n",
"123456789 8\n"
] |
[
"30000\n",
"10000\n",
"38101\n",
"12345678900000000\n"
] |
none
| 750
|
[
{
"input": "375 4",
"output": "30000"
},
{
"input": "10000 1",
"output": "10000"
},
{
"input": "38101 0",
"output": "38101"
},
{
"input": "123456789 8",
"output": "12345678900000000"
},
{
"input": "1 0",
"output": "1"
},
{
"input": "2 0",
"output": "2"
},
{
"input": "100 0",
"output": "100"
},
{
"input": "1000000000 0",
"output": "1000000000"
},
{
"input": "160 2",
"output": "800"
},
{
"input": "3 0",
"output": "3"
},
{
"input": "10 0",
"output": "10"
},
{
"input": "1 1",
"output": "10"
},
{
"input": "2 1",
"output": "10"
},
{
"input": "3 1",
"output": "30"
},
{
"input": "4 1",
"output": "20"
},
{
"input": "5 1",
"output": "10"
},
{
"input": "6 1",
"output": "30"
},
{
"input": "7 1",
"output": "70"
},
{
"input": "8 1",
"output": "40"
},
{
"input": "9 1",
"output": "90"
},
{
"input": "10 1",
"output": "10"
},
{
"input": "11 1",
"output": "110"
},
{
"input": "12 1",
"output": "60"
},
{
"input": "16 2",
"output": "400"
},
{
"input": "2 2",
"output": "100"
},
{
"input": "1 2",
"output": "100"
},
{
"input": "5 2",
"output": "100"
},
{
"input": "15 2",
"output": "300"
},
{
"input": "36 2",
"output": "900"
},
{
"input": "1 8",
"output": "100000000"
},
{
"input": "8 8",
"output": "100000000"
},
{
"input": "96 8",
"output": "300000000"
},
{
"input": "175 8",
"output": "700000000"
},
{
"input": "9999995 8",
"output": "199999900000000"
},
{
"input": "999999999 8",
"output": "99999999900000000"
},
{
"input": "12345678 8",
"output": "617283900000000"
},
{
"input": "78125 8",
"output": "100000000"
},
{
"input": "390625 8",
"output": "100000000"
},
{
"input": "1953125 8",
"output": "500000000"
},
{
"input": "9765625 8",
"output": "2500000000"
},
{
"input": "68359375 8",
"output": "17500000000"
},
{
"input": "268435456 8",
"output": "104857600000000"
},
{
"input": "125829120 8",
"output": "9830400000000"
},
{
"input": "128000 8",
"output": "400000000"
},
{
"input": "300000 8",
"output": "300000000"
},
{
"input": "3711871 8",
"output": "371187100000000"
},
{
"input": "55555 8",
"output": "1111100000000"
},
{
"input": "222222222 8",
"output": "11111111100000000"
},
{
"input": "479001600 8",
"output": "7484400000000"
},
{
"input": "655360001 7",
"output": "6553600010000000"
},
{
"input": "655360001 8",
"output": "65536000100000000"
},
{
"input": "1000000000 1",
"output": "1000000000"
},
{
"input": "1000000000 7",
"output": "1000000000"
},
{
"input": "1000000000 8",
"output": "1000000000"
},
{
"input": "100000000 8",
"output": "100000000"
},
{
"input": "10000000 8",
"output": "100000000"
},
{
"input": "1000000 8",
"output": "100000000"
},
{
"input": "10000009 8",
"output": "1000000900000000"
},
{
"input": "10000005 8",
"output": "200000100000000"
},
{
"input": "10000002 8",
"output": "500000100000000"
},
{
"input": "999999997 8",
"output": "99999999700000000"
},
{
"input": "999999997 7",
"output": "9999999970000000"
},
{
"input": "999999995 8",
"output": "19999999900000000"
},
{
"input": "123 8",
"output": "12300000000"
},
{
"input": "24 2",
"output": "600"
},
{
"input": "16 4",
"output": "10000"
},
{
"input": "123456787 8",
"output": "12345678700000000"
},
{
"input": "100000000 8",
"output": "100000000"
},
{
"input": "7 1",
"output": "70"
},
{
"input": "101 1",
"output": "1010"
},
{
"input": "50 2",
"output": "100"
},
{
"input": "999999818 1",
"output": "4999999090"
},
{
"input": "2 1",
"output": "10"
},
{
"input": "123 1",
"output": "1230"
},
{
"input": "16 1",
"output": "80"
},
{
"input": "1 1",
"output": "10"
},
{
"input": "1000000000 8",
"output": "1000000000"
},
{
"input": "15304 6",
"output": "1913000000"
},
{
"input": "3 8",
"output": "300000000"
},
{
"input": "4 2",
"output": "100"
},
{
"input": "100000 7",
"output": "10000000"
},
{
"input": "5 8",
"output": "100000000"
},
{
"input": "16724 6",
"output": "4181000000"
},
{
"input": "999999999 1",
"output": "9999999990"
},
{
"input": "999999990 8",
"output": "9999999900000000"
},
{
"input": "999999999 8",
"output": "99999999900000000"
},
{
"input": "100000 1",
"output": "100000"
},
{
"input": "8 3",
"output": "1000"
},
{
"input": "16768 6",
"output": "262000000"
},
{
"input": "123456789 1",
"output": "1234567890"
},
{
"input": "2 8",
"output": "100000000"
},
{
"input": "999999937 8",
"output": "99999993700000000"
},
{
"input": "5 1",
"output": "10"
},
{
"input": "2000000 7",
"output": "10000000"
},
{
"input": "1999998 2",
"output": "99999900"
},
{
"input": "125 3",
"output": "1000"
}
] | 1,505,744,075
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
b = input()
n = int(b[:b.find(" ")])
k = int(b[b.find(" ") + 1:])
a = 0
i = 0
while n % 10 == 0:
n = n // 10
i += 1
if i < k:
a = (int(str(n) + ("0" * (k - i))))
else:
a = int(str(n) + ("0" * i))
print (a)
|
Title: k-rounding
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
For a given positive integer *n* denote its *k*-rounding as the minimum positive integer *x*, such that *x* ends with *k* or more zeros in base 10 and is divisible by *n*.
For example, 4-rounding of 375 is 375·80<==<=30000. 30000 is the minimum integer such that it ends with 4 or more zeros and is divisible by 375.
Write a program that will perform the *k*-rounding of *n*.
Input Specification:
The only line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=109, 0<=≤<=*k*<=≤<=8).
Output Specification:
Print the *k*-rounding of *n*.
Demo Input:
['375 4\n', '10000 1\n', '38101 0\n', '123456789 8\n']
Demo Output:
['30000\n', '10000\n', '38101\n', '12345678900000000\n']
Note:
none
|
```python
b = input()
n = int(b[:b.find(" ")])
k = int(b[b.find(" ") + 1:])
a = 0
i = 0
while n % 10 == 0:
n = n // 10
i += 1
if i < k:
a = (int(str(n) + ("0" * (k - i))))
else:
a = int(str(n) + ("0" * i))
print (a)
```
| 0
|
|
454
|
B
|
Little Pony and Sort by Shift
|
PROGRAMMING
| 1,200
|
[
"implementation"
] | null | null |
One day, Twilight Sparkle is interested in how to sort a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* in non-decreasing order. Being a young unicorn, the only operation she can perform is a unit shift. That is, she can move the last element of the sequence to its beginning:
Help Twilight Sparkle to calculate: what is the minimum number of operations that she needs to sort the sequence?
|
The first line contains an integer *n* (2<=≤<=*n*<=≤<=105). The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105).
|
If it's impossible to sort the sequence output -1. Otherwise output the minimum number of operations Twilight Sparkle needs to sort it.
|
[
"2\n2 1\n",
"3\n1 3 2\n",
"2\n1 2\n"
] |
[
"1\n",
"-1\n",
"0\n"
] |
none
| 1,000
|
[
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "3\n1 3 2",
"output": "-1"
},
{
"input": "2\n1 2",
"output": "0"
},
{
"input": "6\n3 4 5 6 3 2",
"output": "-1"
},
{
"input": "3\n1 2 1",
"output": "1"
},
{
"input": "5\n1 1 2 1 1",
"output": "2"
},
{
"input": "4\n5 4 5 4",
"output": "-1"
},
{
"input": "7\n3 4 5 5 5 1 2",
"output": "2"
},
{
"input": "5\n2 2 1 2 2",
"output": "3"
},
{
"input": "5\n5 4 1 2 3",
"output": "-1"
},
{
"input": "4\n6 1 2 7",
"output": "-1"
},
{
"input": "5\n4 5 6 2 3",
"output": "2"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "4\n1 2 2 1",
"output": "1"
},
{
"input": "9\n4 5 6 7 1 2 3 4 10",
"output": "-1"
},
{
"input": "7\n2 3 4 1 2 3 4",
"output": "-1"
},
{
"input": "6\n1 2 1 2 1 2",
"output": "-1"
},
{
"input": "3\n3 2 1",
"output": "-1"
},
{
"input": "4\n1 4 4 1",
"output": "1"
},
{
"input": "5\n1 2 1 1 1",
"output": "3"
},
{
"input": "5\n4 6 7 3 5",
"output": "-1"
},
{
"input": "4\n2 3 1 4",
"output": "-1"
},
{
"input": "5\n5 4 3 2 1",
"output": "-1"
},
{
"input": "4\n2 4 1 4",
"output": "-1"
},
{
"input": "6\n4 5 6 1 2 7",
"output": "-1"
},
{
"input": "6\n1 2 3 1 1 1",
"output": "3"
},
{
"input": "5\n1 3 3 3 1",
"output": "1"
},
{
"input": "6\n5 6 7 5 5 5",
"output": "3"
},
{
"input": "5\n3 4 2 1 2",
"output": "-1"
},
{
"input": "3\n3 4 2",
"output": "1"
},
{
"input": "6\n1 1 2 2 1 1",
"output": "2"
},
{
"input": "4\n2 3 4 2",
"output": "1"
},
{
"input": "5\n3 5 7 7 3",
"output": "1"
},
{
"input": "4\n1 1 4 1",
"output": "1"
},
{
"input": "7\n1 5 6 1 1 1 1",
"output": "4"
},
{
"input": "5\n7 8 6 7 8",
"output": "-1"
},
{
"input": "4\n2 4 1 3",
"output": "-1"
}
] | 1,567,119,539
| 2,147,483,647
|
PyPy 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 3
| 1,000
| 11,776,000
|
n=int(input())
arr=list(map(int,input().strip().split()))
b=arr
c=0
while(arr!=sorted(arr)):
if c==n+1 and b==arr:
c=-1
break
x = arr[n - 1]
for i in range(n - 1, 0, -1):
arr[i] = arr[i - 1];
arr[0] = x;
c=c+1
print(c)
|
Title: Little Pony and Sort by Shift
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day, Twilight Sparkle is interested in how to sort a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* in non-decreasing order. Being a young unicorn, the only operation she can perform is a unit shift. That is, she can move the last element of the sequence to its beginning:
Help Twilight Sparkle to calculate: what is the minimum number of operations that she needs to sort the sequence?
Input Specification:
The first line contains an integer *n* (2<=≤<=*n*<=≤<=105). The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105).
Output Specification:
If it's impossible to sort the sequence output -1. Otherwise output the minimum number of operations Twilight Sparkle needs to sort it.
Demo Input:
['2\n2 1\n', '3\n1 3 2\n', '2\n1 2\n']
Demo Output:
['1\n', '-1\n', '0\n']
Note:
none
|
```python
n=int(input())
arr=list(map(int,input().strip().split()))
b=arr
c=0
while(arr!=sorted(arr)):
if c==n+1 and b==arr:
c=-1
break
x = arr[n - 1]
for i in range(n - 1, 0, -1):
arr[i] = arr[i - 1];
arr[0] = x;
c=c+1
print(c)
```
| 0
|
|
378
|
A
|
Playing with Dice
|
PROGRAMMING
| 800
|
[
"brute force"
] | null | null |
Two players are playing a game. First each of them writes an integer from 1 to 6, and then a dice is thrown. The player whose written number got closer to the number on the dice wins. If both payers have the same difference, it's a draw.
The first player wrote number *a*, the second player wrote number *b*. How many ways to throw a dice are there, at which the first player wins, or there is a draw, or the second player wins?
|
The single line contains two integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=6) — the numbers written on the paper by the first and second player, correspondingly.
|
Print three integers: the number of ways to throw the dice at which the first player wins, the game ends with a draw or the second player wins, correspondingly.
|
[
"2 5\n",
"2 4\n"
] |
[
"3 0 3\n",
"2 1 3\n"
] |
The dice is a standard cube-shaped six-sided object with each side containing a number from 1 to 6, and where all numbers on all sides are distinct.
You can assume that number *a* is closer to number *x* than number *b*, if |*a* - *x*| < |*b* - *x*|.
| 500
|
[
{
"input": "2 5",
"output": "3 0 3"
},
{
"input": "2 4",
"output": "2 1 3"
},
{
"input": "5 3",
"output": "2 1 3"
},
{
"input": "1 6",
"output": "3 0 3"
},
{
"input": "5 1",
"output": "3 1 2"
},
{
"input": "6 3",
"output": "2 0 4"
},
{
"input": "2 3",
"output": "2 0 4"
},
{
"input": "5 6",
"output": "5 0 1"
},
{
"input": "4 4",
"output": "0 6 0"
},
{
"input": "1 1",
"output": "0 6 0"
},
{
"input": "6 4",
"output": "1 1 4"
},
{
"input": "1 4",
"output": "2 0 4"
},
{
"input": "5 5",
"output": "0 6 0"
},
{
"input": "4 5",
"output": "4 0 2"
},
{
"input": "4 3",
"output": "3 0 3"
},
{
"input": "1 5",
"output": "2 1 3"
},
{
"input": "6 5",
"output": "1 0 5"
},
{
"input": "2 2",
"output": "0 6 0"
},
{
"input": "1 3",
"output": "1 1 4"
},
{
"input": "3 6",
"output": "4 0 2"
},
{
"input": "3 1",
"output": "4 1 1"
},
{
"input": "3 2",
"output": "4 0 2"
},
{
"input": "3 5",
"output": "3 1 2"
},
{
"input": "3 3",
"output": "0 6 0"
},
{
"input": "6 2",
"output": "2 1 3"
},
{
"input": "4 1",
"output": "4 0 2"
},
{
"input": "5 2",
"output": "3 0 3"
},
{
"input": "4 2",
"output": "3 1 2"
},
{
"input": "2 1",
"output": "5 0 1"
},
{
"input": "6 1",
"output": "3 0 3"
},
{
"input": "4 6",
"output": "4 1 1"
},
{
"input": "2 6",
"output": "3 1 2"
},
{
"input": "3 4",
"output": "3 0 3"
},
{
"input": "1 2",
"output": "1 0 5"
},
{
"input": "6 6",
"output": "0 6 0"
},
{
"input": "5 4",
"output": "2 0 4"
},
{
"input": "3 3",
"output": "0 6 0"
},
{
"input": "1 1",
"output": "0 6 0"
}
] | 1,594,514,197
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 38
| 140
| 20,172,800
|
a,b = map(int,input().split())
awin=0
bwin=0
draw=0
for i in range(1,7):
if abs(a-i)<abs(b-i):
awin+=1
elif abs(a-i)>abs(b-i):
bwin+=1
else:
draw+=1
print(awin,draw,bwin)
|
Title: Playing with Dice
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Two players are playing a game. First each of them writes an integer from 1 to 6, and then a dice is thrown. The player whose written number got closer to the number on the dice wins. If both payers have the same difference, it's a draw.
The first player wrote number *a*, the second player wrote number *b*. How many ways to throw a dice are there, at which the first player wins, or there is a draw, or the second player wins?
Input Specification:
The single line contains two integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=6) — the numbers written on the paper by the first and second player, correspondingly.
Output Specification:
Print three integers: the number of ways to throw the dice at which the first player wins, the game ends with a draw or the second player wins, correspondingly.
Demo Input:
['2 5\n', '2 4\n']
Demo Output:
['3 0 3\n', '2 1 3\n']
Note:
The dice is a standard cube-shaped six-sided object with each side containing a number from 1 to 6, and where all numbers on all sides are distinct.
You can assume that number *a* is closer to number *x* than number *b*, if |*a* - *x*| < |*b* - *x*|.
|
```python
a,b = map(int,input().split())
awin=0
bwin=0
draw=0
for i in range(1,7):
if abs(a-i)<abs(b-i):
awin+=1
elif abs(a-i)>abs(b-i):
bwin+=1
else:
draw+=1
print(awin,draw,bwin)
```
| 3
|
|
177
|
C1
|
Party
|
PROGRAMMING
| 1,500
|
[
"dfs and similar",
"dsu",
"graphs"
] | null | null |
To celebrate the second ABBYY Cup tournament, the Smart Beaver decided to throw a party. The Beaver has a lot of acquaintances, some of them are friends with each other, and some of them dislike each other. To make party successful, the Smart Beaver wants to invite only those of his friends who are connected by friendship relations, and not to invite those who dislike each other. Both friendship and dislike are mutual feelings.
More formally, for each invited person the following conditions should be fulfilled:
- all his friends should also be invited to the party; - the party shouldn't have any people he dislikes; - all people who are invited to the party should be connected with him by friendship either directly or through a chain of common friends of arbitrary length. We'll say that people *a*1 and *a**p* are connected through a chain of common friends if there exists a sequence of people *a*2,<=*a*3,<=...,<=*a**p*<=-<=1 such that all pairs of people *a**i* and *a**i*<=+<=1 (1<=≤<=*i*<=<<=*p*) are friends.
Help the Beaver find the maximum number of acquaintances he can invite.
|
The first line of input contains an integer *n* — the number of the Beaver's acquaintances.
The second line contains an integer *k* — the number of pairs of friends. Next *k* lines contain space-separated pairs of integers *u**i*,<=*v**i* — indices of people who form the *i*-th pair of friends.
The next line contains an integer *m* — the number of pairs of people who dislike each other. Next *m* lines describe pairs of people who dislike each other in the same format as the pairs of friends were described.
Each pair of people is mentioned in the input at most once . In particular, two persons cannot be friends and dislike each other at the same time.
The input limitations for getting 30 points are:
- 2<=≤<=*n*<=≤<=14
The input limitations for getting 100 points are:
- 2<=≤<=*n*<=≤<=2000
|
Output a single number — the maximum number of people that can be invited to the party. If a group of people that meets all the requirements is impossible to select, output 0.
|
[
"9\n8\n1 2\n1 3\n2 3\n4 5\n6 7\n7 8\n8 9\n9 6\n2\n1 6\n7 9\n"
] |
[
"3"
] |
Let's have a look at the example.
Two groups of people can be invited: {1, 2, 3} and {4, 5}, thus the answer will be the size of the largest of these groups. Group {6, 7, 8, 9} doesn't fit, since it includes people 7 and 9 who dislike each other. Group {1, 2, 3, 4, 5} also doesn't fit, because not all of its members are connected by a chain of common friends (for example, people 2 and 5 aren't connected).
| 30
|
[
{
"input": "9\n8\n1 2\n1 3\n2 3\n4 5\n6 7\n7 8\n8 9\n9 6\n2\n1 6\n7 9",
"output": "3"
},
{
"input": "2\n1\n1 2\n0",
"output": "2"
},
{
"input": "2\n0\n1\n1 2",
"output": "1"
},
{
"input": "3\n2\n1 2\n1 3\n1\n2 3",
"output": "0"
},
{
"input": "3\n3\n1 3\n2 1\n2 3\n0",
"output": "3"
},
{
"input": "4\n3\n1 2\n2 3\n3 1\n3\n1 4\n4 2\n3 4",
"output": "3"
},
{
"input": "7\n8\n1 2\n1 3\n1 4\n1 5\n2 4\n2 5\n3 4\n5 6\n3\n2 6\n5 7\n6 7",
"output": "1"
},
{
"input": "14\n20\n1 2\n4 5\n4 6\n4 11\n5 7\n5 8\n5 13\n5 14\n7 8\n7 14\n8 9\n8 11\n8 12\n8 14\n10 11\n10 12\n10 14\n11 13\n11 14\n12 14\n5\n1 8\n1 13\n2 10\n7 12\n8 10",
"output": "2"
},
{
"input": "2\n0\n0",
"output": "1"
},
{
"input": "14\n0\n0",
"output": "1"
},
{
"input": "14\n6\n1 2\n2 3\n3 4\n4 5\n8 9\n9 10\n3\n5 6\n6 7\n7 8",
"output": "5"
},
{
"input": "14\n10\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n2 3\n2 4\n2 5\n2 6\n1\n2 7",
"output": "1"
},
{
"input": "2\n0\n1\n1 2",
"output": "1"
},
{
"input": "13\n78\n11 1\n10 6\n6 2\n10 1\n11 6\n11 3\n5 3\n8 1\n12 11\n4 2\n10 3\n13 8\n9 8\n11 7\n7 5\n11 2\n7 1\n4 1\n11 10\n8 3\n13 11\n9 6\n13 9\n12 7\n12 8\n12 9\n10 2\n5 2\n12 10\n9 2\n9 7\n3 2\n7 4\n11 4\n13 1\n10 5\n11 5\n8 5\n10 4\n8 2\n10 9\n4 3\n9 5\n13 12\n13 5\n7 2\n12 4\n9 1\n10 8\n6 3\n6 4\n7 6\n7 3\n12 3\n5 4\n6 1\n12 5\n8 4\n13 3\n12 1\n9 3\n8 6\n11 9\n9 4\n8 7\n12 6\n5 1\n13 10\n13 6\n10 7\n13 4\n13 7\n13 2\n2 1\n6 5\n12 2\n11 8\n3 1\n0",
"output": "13"
},
{
"input": "13\n0\n78\n11 8\n8 4\n13 9\n6 1\n10 5\n5 1\n9 8\n11 3\n13 12\n6 2\n10 9\n9 1\n10 3\n13 6\n8 1\n11 10\n11 1\n11 9\n12 4\n12 11\n11 4\n8 6\n9 6\n13 10\n13 8\n7 2\n8 3\n10 1\n12 10\n6 5\n8 2\n5 4\n9 2\n13 1\n4 1\n13 2\n12 5\n10 7\n7 4\n8 5\n12 6\n4 3\n13 3\n12 2\n9 3\n11 7\n7 3\n2 1\n10 2\n13 7\n7 5\n13 4\n12 7\n4 2\n12 9\n11 5\n10 8\n11 2\n12 3\n3 1\n7 6\n10 6\n12 1\n10 4\n5 2\n9 4\n11 6\n9 7\n5 3\n7 1\n8 7\n6 3\n13 5\n12 8\n6 4\n13 11\n9 5\n3 2",
"output": "1"
}
] | 1,686,823,553
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS1
| 17
| 124
| 2,662,400
|
from collections import deque
n = int(input())
m = int(input())
f = dict()
for _ in range(m):
a, b = map(int,input().split())
if a not in f:
f[a] = set()
if b not in f:
f[b] = set()
f[a].add(b)
f[b].add(a)
k = int(input())
h = dict()
for _ in range(k):
a, b = map(int,input().split())
if a not in h:
h[a] = set()
if b not in h:
h[b] = set()
h[a].add(b)
h[b].add(a)
best = 0
check = set([i+1 for i in range(n)])
while len(check) > 0:
legit = True
banned = set()
temp = check.pop()
q = deque()
used = set()
q.append(temp)
while len(q) > 0:
temp = q.popleft()
used.add(temp)
if temp in h:
for i in h[temp]:
banned.add(i)
if temp in f:
for i in f[temp]:
if i not in used:
q.append(i)
for i in used:
if i in banned:
legit = False
break
if legit:
best = max(best, len(used))
print(best)
|
Title: Party
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
To celebrate the second ABBYY Cup tournament, the Smart Beaver decided to throw a party. The Beaver has a lot of acquaintances, some of them are friends with each other, and some of them dislike each other. To make party successful, the Smart Beaver wants to invite only those of his friends who are connected by friendship relations, and not to invite those who dislike each other. Both friendship and dislike are mutual feelings.
More formally, for each invited person the following conditions should be fulfilled:
- all his friends should also be invited to the party; - the party shouldn't have any people he dislikes; - all people who are invited to the party should be connected with him by friendship either directly or through a chain of common friends of arbitrary length. We'll say that people *a*1 and *a**p* are connected through a chain of common friends if there exists a sequence of people *a*2,<=*a*3,<=...,<=*a**p*<=-<=1 such that all pairs of people *a**i* and *a**i*<=+<=1 (1<=≤<=*i*<=<<=*p*) are friends.
Help the Beaver find the maximum number of acquaintances he can invite.
Input Specification:
The first line of input contains an integer *n* — the number of the Beaver's acquaintances.
The second line contains an integer *k* — the number of pairs of friends. Next *k* lines contain space-separated pairs of integers *u**i*,<=*v**i* — indices of people who form the *i*-th pair of friends.
The next line contains an integer *m* — the number of pairs of people who dislike each other. Next *m* lines describe pairs of people who dislike each other in the same format as the pairs of friends were described.
Each pair of people is mentioned in the input at most once . In particular, two persons cannot be friends and dislike each other at the same time.
The input limitations for getting 30 points are:
- 2<=≤<=*n*<=≤<=14
The input limitations for getting 100 points are:
- 2<=≤<=*n*<=≤<=2000
Output Specification:
Output a single number — the maximum number of people that can be invited to the party. If a group of people that meets all the requirements is impossible to select, output 0.
Demo Input:
['9\n8\n1 2\n1 3\n2 3\n4 5\n6 7\n7 8\n8 9\n9 6\n2\n1 6\n7 9\n']
Demo Output:
['3']
Note:
Let's have a look at the example.
Two groups of people can be invited: {1, 2, 3} and {4, 5}, thus the answer will be the size of the largest of these groups. Group {6, 7, 8, 9} doesn't fit, since it includes people 7 and 9 who dislike each other. Group {1, 2, 3, 4, 5} also doesn't fit, because not all of its members are connected by a chain of common friends (for example, people 2 and 5 aren't connected).
|
```python
from collections import deque
n = int(input())
m = int(input())
f = dict()
for _ in range(m):
a, b = map(int,input().split())
if a not in f:
f[a] = set()
if b not in f:
f[b] = set()
f[a].add(b)
f[b].add(a)
k = int(input())
h = dict()
for _ in range(k):
a, b = map(int,input().split())
if a not in h:
h[a] = set()
if b not in h:
h[b] = set()
h[a].add(b)
h[b].add(a)
best = 0
check = set([i+1 for i in range(n)])
while len(check) > 0:
legit = True
banned = set()
temp = check.pop()
q = deque()
used = set()
q.append(temp)
while len(q) > 0:
temp = q.popleft()
used.add(temp)
if temp in h:
for i in h[temp]:
banned.add(i)
if temp in f:
for i in f[temp]:
if i not in used:
q.append(i)
for i in used:
if i in banned:
legit = False
break
if legit:
best = max(best, len(used))
print(best)
```
| 3
|
|
802
|
A
|
Heidi and Library (easy)
|
PROGRAMMING
| 1,800
|
[
"greedy"
] | null | null |
Your search for Heidi is over – you finally found her at a library, dressed up as a human. In fact, she has spent so much time there that she now runs the place! Her job is to buy books and keep them at the library so that people can borrow and read them. There are *n* different books, numbered 1 through *n*.
We will look at the library's operation during *n* consecutive days. Heidi knows in advance that on the *i*-th day (1<=≤<=*i*<=≤<=*n*) precisely one person will come to the library, request to borrow the book *a**i*, read it in a few hours, and return the book later on the same day.
Heidi desperately wants to please all her guests, so she will make sure to always have the book *a**i* available in the library on the *i*-th day. During the night before the *i*-th day, she has the option of going to the bookstore (which operates at nights to avoid competition with the library) and buying any book for the price of 1 CHF. Of course, if she already has a book at the library, she does not need to buy it again. Initially, the library contains no books.
There is a problem, though. The capacity of the library is *k* – this means that at any time, there can be at most *k* books at the library. If buying a new book would cause Heidi to have more than *k* books, she must first get rid of some book that she already has, in order to make room for the new book. If she later needs a book that she got rid of, she will need to buy that book again.
You are given *k* and the sequence of requests for books *a*1,<=*a*2,<=...,<=*a**n*. What is the minimum cost (in CHF) of buying new books to satisfy all the requests?
|
The first line of input will contain two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=80). The second line will contain *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) – the sequence of book requests.
|
On a single line print the minimum cost of buying books at the store so as to satisfy all requests.
|
[
"4 80\n1 2 2 1\n",
"4 1\n1 2 2 1\n",
"4 2\n1 2 3 1\n"
] |
[
"2\n",
"3\n",
"3\n"
] |
In the first test case, Heidi is able to keep all books forever. Therefore, she only needs to buy the book 1 before the first day and the book 2 before the second day.
In the second test case, she can only keep one book at a time. Therefore she will need to buy new books on the first, second and fourth day.
In the third test case, before buying book 3 on the third day, she must decide which of the books 1 and 2 she should get rid of. Of course, she should keep the book 1, which will be requested on the fourth day.
| 0
|
[
{
"input": "4 80\n1 2 2 1",
"output": "2"
},
{
"input": "4 1\n1 2 2 1",
"output": "3"
},
{
"input": "4 2\n1 2 3 1",
"output": "3"
},
{
"input": "11 1\n1 2 3 5 1 10 10 1 1 3 5",
"output": "9"
},
{
"input": "5 2\n1 2 3 1 2",
"output": "4"
},
{
"input": "4 2\n1 2 3 2",
"output": "3"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "80 4\n9 9 2 6 3 10 2 5 4 9 6 7 5 5 3 8 5 3 2 10 7 8 5 3 4 9 4 3 9 5 2 10 8 4 7 3 8 3 5 2 3 7 8 4 2 4 4 7 2 2 5 7 5 8 10 10 5 1 1 3 5 2 10 8 7 9 7 4 8 3 2 8 7 9 10 9 7 1 5 5",
"output": "34"
},
{
"input": "80 4\n10 19 20 18 16 7 13 18 15 5 7 13 16 8 14 8 3 15 19 19 7 13 17 9 18 16 4 14 10 18 1 3 5 3 20 18 9 4 17 19 13 20 16 12 15 5 5 18 17 16 4 5 20 10 18 4 7 19 10 15 8 15 17 3 10 16 19 2 6 6 3 12 10 7 15 3 17 15 6 8",
"output": "49"
},
{
"input": "80 4\n28 34 9 3 29 12 19 17 22 10 21 2 26 18 14 7 7 10 37 39 10 1 9 37 33 4 25 21 23 2 4 2 35 1 11 19 33 31 18 10 23 1 26 20 17 31 18 27 31 22 33 7 2 5 30 24 18 32 1 14 2 33 7 26 2 10 1 10 5 19 37 33 33 34 28 20 1 22 11 14",
"output": "58"
},
{
"input": "80 4\n71 49 41 21 72 71 37 14 51 59 73 11 70 15 36 46 32 57 58 15 72 67 16 75 70 11 67 3 40 36 2 9 63 68 32 22 63 52 67 55 35 19 72 59 22 19 44 55 59 74 4 34 53 3 22 57 32 27 78 12 71 4 26 15 43 21 79 10 67 39 34 74 38 26 31 78 2 78 69 42",
"output": "62"
},
{
"input": "80 8\n16 13 11 16 3 4 1 4 4 16 6 6 1 12 19 18 12 15 2 10 2 18 18 13 3 17 16 15 7 6 19 8 2 14 17 13 1 14 4 2 3 16 2 15 13 15 9 10 7 14 7 2 1 18 19 15 7 3 19 8 9 4 12 4 3 4 9 10 6 5 4 4 9 4 20 8 17 7 1 14",
"output": "32"
},
{
"input": "80 8\n5 17 39 25 40 34 11 23 7 16 20 35 31 14 18 17 32 10 40 9 17 23 5 33 2 9 21 22 8 11 22 7 28 36 3 10 12 21 20 29 25 5 12 30 8 21 18 19 1 29 9 4 19 5 15 36 38 37 10 27 15 13 6 22 31 5 40 30 21 39 23 21 39 32 37 28 29 11 34 16",
"output": "51"
},
{
"input": "80 8\n8 72 32 27 27 20 69 28 77 25 8 4 75 11 41 71 57 17 45 65 79 8 61 15 24 80 39 36 34 13 76 37 16 71 64 77 11 58 30 26 61 23 18 30 68 65 12 47 69 65 3 55 71 3 32 4 20 39 47 25 75 49 34 60 48 56 77 70 59 59 75 6 5 23 55 30 62 66 4 4",
"output": "57"
},
{
"input": "80 12\n9 5 8 1 12 2 6 19 8 20 6 12 9 6 16 1 2 5 11 6 8 4 13 7 2 17 18 12 15 17 13 2 9 8 1 17 10 2 9 12 18 3 5 11 10 16 7 16 8 11 3 18 13 19 8 13 13 2 20 13 11 14 20 3 2 1 17 18 17 8 4 3 12 3 19 18 4 16 6 6",
"output": "25"
},
{
"input": "80 12\n27 12 25 30 13 27 12 17 35 25 1 28 35 16 23 20 38 1 37 2 35 29 16 26 37 4 23 39 24 2 16 21 39 21 23 38 33 9 38 22 40 36 23 39 1 2 4 14 22 26 32 4 31 38 4 5 4 15 35 12 5 32 37 38 11 14 16 26 36 38 2 40 10 15 33 38 36 20 35 12",
"output": "37"
},
{
"input": "80 12\n30 19 34 24 56 38 31 63 57 50 53 69 79 5 6 74 47 47 73 17 18 70 72 49 35 20 65 21 18 4 54 12 67 8 28 25 64 6 31 36 35 54 61 7 45 54 55 49 50 6 3 7 10 29 76 62 50 50 32 66 25 19 17 3 67 17 37 67 58 18 54 25 8 78 35 16 61 19 45 40",
"output": "55"
},
{
"input": "80 16\n4 27 31 28 8 17 28 31 20 7 39 5 40 13 28 6 23 1 16 4 34 2 13 6 6 9 18 1 25 19 33 26 33 16 24 5 13 23 25 9 10 16 25 34 39 8 4 6 33 25 7 40 32 23 13 17 32 20 28 25 33 20 29 2 40 34 23 6 28 2 12 12 9 36 18 39 32 8 11 15",
"output": "36"
},
{
"input": "80 16\n31 26 40 46 75 35 63 29 2 49 51 14 4 65 10 4 8 72 44 67 57 60 69 21 52 40 37 54 27 12 31 24 21 59 61 80 11 76 58 7 77 10 55 9 11 36 7 41 61 13 2 28 28 77 22 57 54 62 65 80 78 32 72 64 41 69 36 46 50 5 48 53 6 76 76 65 57 7 29 67",
"output": "53"
},
{
"input": "80 40\n34 71 32 39 65 8 13 4 7 4 18 66 20 12 57 74 58 50 30 27 31 48 1 6 63 63 7 32 56 48 42 35 45 55 52 76 52 26 40 15 8 38 73 47 55 75 17 22 36 59 28 19 6 79 58 7 40 66 48 39 71 67 55 61 71 24 60 39 63 6 47 70 8 10 68 49 49 67 53 15",
"output": "54"
},
{
"input": "80 60\n70 59 1 24 50 77 11 71 36 26 75 56 30 31 76 51 73 29 49 2 80 55 32 52 51 56 2 28 37 11 37 1 63 33 74 60 59 63 55 28 73 45 5 69 77 8 35 76 47 48 49 77 25 75 38 70 77 80 27 9 13 45 14 30 33 34 32 78 52 4 7 57 55 43 13 1 45 36 79 45",
"output": "50"
},
{
"input": "80 80\n73 32 65 26 35 2 10 73 80 64 67 14 40 34 15 59 8 71 5 9 49 63 32 51 23 1 13 9 34 6 80 31 32 60 31 27 65 20 21 74 26 51 17 11 19 37 4 50 73 22 71 55 76 7 51 21 50 63 6 75 66 55 37 62 10 45 52 21 42 18 80 12 7 28 54 50 9 5 25 42",
"output": "53"
},
{
"input": "80 4\n1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5",
"output": "23"
},
{
"input": "80 8\n1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8",
"output": "17"
},
{
"input": "80 16\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 1 2 3 4 5 6 7 8 9 10 11 12",
"output": "20"
},
{
"input": "80 32\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 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 1 2 3 4 5 6 7 8 9 10 11 12 13 14",
"output": "34"
},
{
"input": "80 80\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80",
"output": "80"
}
] | 1,683,831,959
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 46
| 0
|
library = [80,4]
requests = '9 9 2 6 3 10 2 5 4 9 6 7 5 5 3 8 5 3 2 10 7 8 5 3 4 9 4 3 9 5 2 10 8 4 7 3 8 3 5 2 3 7 8 4 2 4 4 7 2 2 5 7 5 8 10 10 5 1 1 3 5 2 10 8 7 9 7 4 8 3 2 8 7 9 10 9 7 1 5 5'.split(' ')
books = []
price = 0
for c in range(library[0]):
if requests[0] in books:
requests.pop(0)
else:
if len(books)+1 > library[1]:
bookwant = []
for c in books:
bookwant.append(requests.count(c))
nobook = books[bookwant.index(min(bookwant))]
if bookwant.count(nobook) > 1:
bad = []
for c in range(len(books)):
if bookwant[c] == nobook:
bad.append(books[c])
count = 0
while len(bad) > 1:
if requests[count] in bad:
bad.remove(requests[count])
count = count + 1
nobook = bad[0]
books.remove(nobook)
price = price + 1
books.append(requests[0])
requests.pop(0)
else:
price = price + 1
books.append(requests[0])
requests.pop(0)
print(price)
|
Title: Heidi and Library (easy)
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Your search for Heidi is over – you finally found her at a library, dressed up as a human. In fact, she has spent so much time there that she now runs the place! Her job is to buy books and keep them at the library so that people can borrow and read them. There are *n* different books, numbered 1 through *n*.
We will look at the library's operation during *n* consecutive days. Heidi knows in advance that on the *i*-th day (1<=≤<=*i*<=≤<=*n*) precisely one person will come to the library, request to borrow the book *a**i*, read it in a few hours, and return the book later on the same day.
Heidi desperately wants to please all her guests, so she will make sure to always have the book *a**i* available in the library on the *i*-th day. During the night before the *i*-th day, she has the option of going to the bookstore (which operates at nights to avoid competition with the library) and buying any book for the price of 1 CHF. Of course, if she already has a book at the library, she does not need to buy it again. Initially, the library contains no books.
There is a problem, though. The capacity of the library is *k* – this means that at any time, there can be at most *k* books at the library. If buying a new book would cause Heidi to have more than *k* books, she must first get rid of some book that she already has, in order to make room for the new book. If she later needs a book that she got rid of, she will need to buy that book again.
You are given *k* and the sequence of requests for books *a*1,<=*a*2,<=...,<=*a**n*. What is the minimum cost (in CHF) of buying new books to satisfy all the requests?
Input Specification:
The first line of input will contain two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=80). The second line will contain *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) – the sequence of book requests.
Output Specification:
On a single line print the minimum cost of buying books at the store so as to satisfy all requests.
Demo Input:
['4 80\n1 2 2 1\n', '4 1\n1 2 2 1\n', '4 2\n1 2 3 1\n']
Demo Output:
['2\n', '3\n', '3\n']
Note:
In the first test case, Heidi is able to keep all books forever. Therefore, she only needs to buy the book 1 before the first day and the book 2 before the second day.
In the second test case, she can only keep one book at a time. Therefore she will need to buy new books on the first, second and fourth day.
In the third test case, before buying book 3 on the third day, she must decide which of the books 1 and 2 she should get rid of. Of course, she should keep the book 1, which will be requested on the fourth day.
|
```python
library = [80,4]
requests = '9 9 2 6 3 10 2 5 4 9 6 7 5 5 3 8 5 3 2 10 7 8 5 3 4 9 4 3 9 5 2 10 8 4 7 3 8 3 5 2 3 7 8 4 2 4 4 7 2 2 5 7 5 8 10 10 5 1 1 3 5 2 10 8 7 9 7 4 8 3 2 8 7 9 10 9 7 1 5 5'.split(' ')
books = []
price = 0
for c in range(library[0]):
if requests[0] in books:
requests.pop(0)
else:
if len(books)+1 > library[1]:
bookwant = []
for c in books:
bookwant.append(requests.count(c))
nobook = books[bookwant.index(min(bookwant))]
if bookwant.count(nobook) > 1:
bad = []
for c in range(len(books)):
if bookwant[c] == nobook:
bad.append(books[c])
count = 0
while len(bad) > 1:
if requests[count] in bad:
bad.remove(requests[count])
count = count + 1
nobook = bad[0]
books.remove(nobook)
price = price + 1
books.append(requests[0])
requests.pop(0)
else:
price = price + 1
books.append(requests[0])
requests.pop(0)
print(price)
```
| 0
|
|
108
|
A
|
Palindromic Times
|
PROGRAMMING
| 1,000
|
[
"implementation",
"strings"
] |
A. Palindromic Times
|
2
|
256
|
Tattah is asleep if and only if Tattah is attending a lecture. This is a well-known formula among Tattah's colleagues.
On a Wednesday afternoon, Tattah was attending Professor HH's lecture. At 12:21, right before falling asleep, he was staring at the digital watch around Saher's wrist. He noticed that the digits on the clock were the same when read from both directions i.e. a palindrome.
In his sleep, he started dreaming about such rare moments of the day when the time displayed on a digital clock is a palindrome. As soon as he woke up, he felt destined to write a program that finds the next such moment.
However, he still hasn't mastered the skill of programming while sleeping, so your task is to help him.
|
The first and only line of the input starts with a string with the format "HH:MM" where "HH" is from "00" to "23" and "MM" is from "00" to "59". Both "HH" and "MM" have exactly two digits.
|
Print the palindromic time of day that comes soonest after the time given in the input. If the input time is palindromic, output the soonest palindromic time after the input time.
|
[
"12:21\n",
"23:59\n"
] |
[
"13:31\n",
"00:00\n"
] |
none
| 500
|
[
{
"input": "12:21",
"output": "13:31"
},
{
"input": "23:59",
"output": "00:00"
},
{
"input": "15:51",
"output": "20:02"
},
{
"input": "10:44",
"output": "11:11"
},
{
"input": "04:02",
"output": "04:40"
},
{
"input": "02:11",
"output": "02:20"
},
{
"input": "12:15",
"output": "12:21"
},
{
"input": "07:07",
"output": "10:01"
},
{
"input": "00:17",
"output": "01:10"
},
{
"input": "04:55",
"output": "05:50"
},
{
"input": "02:17",
"output": "02:20"
},
{
"input": "07:56",
"output": "10:01"
},
{
"input": "00:29",
"output": "01:10"
},
{
"input": "23:31",
"output": "23:32"
},
{
"input": "19:30",
"output": "20:02"
},
{
"input": "12:14",
"output": "12:21"
},
{
"input": "17:32",
"output": "20:02"
},
{
"input": "03:44",
"output": "04:40"
},
{
"input": "07:15",
"output": "10:01"
},
{
"input": "18:42",
"output": "20:02"
},
{
"input": "08:56",
"output": "10:01"
},
{
"input": "04:50",
"output": "05:50"
},
{
"input": "14:32",
"output": "14:41"
},
{
"input": "23:23",
"output": "23:32"
},
{
"input": "08:35",
"output": "10:01"
},
{
"input": "03:32",
"output": "04:40"
},
{
"input": "07:59",
"output": "10:01"
},
{
"input": "14:12",
"output": "14:41"
},
{
"input": "23:52",
"output": "00:00"
},
{
"input": "16:36",
"output": "20:02"
},
{
"input": "17:50",
"output": "20:02"
},
{
"input": "06:59",
"output": "10:01"
},
{
"input": "16:50",
"output": "20:02"
},
{
"input": "00:00",
"output": "01:10"
},
{
"input": "23:59",
"output": "00:00"
},
{
"input": "23:33",
"output": "00:00"
}
] | 1,594,825,735
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 36
| 280
| 20,172,800
|
h,m=map(int,input().split(':'))
while True:
m+=1
if m==60:
h+=1
m=0
if h==24:
h=0
a=str(h)
a='0'*(2-len(a))+a
b=str(m)
b='0'*(2-len(b))+b
a=a+":"+b
if(a==a[::-1]):
print(a)
exit()
|
Title: Palindromic Times
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Tattah is asleep if and only if Tattah is attending a lecture. This is a well-known formula among Tattah's colleagues.
On a Wednesday afternoon, Tattah was attending Professor HH's lecture. At 12:21, right before falling asleep, he was staring at the digital watch around Saher's wrist. He noticed that the digits on the clock were the same when read from both directions i.e. a palindrome.
In his sleep, he started dreaming about such rare moments of the day when the time displayed on a digital clock is a palindrome. As soon as he woke up, he felt destined to write a program that finds the next such moment.
However, he still hasn't mastered the skill of programming while sleeping, so your task is to help him.
Input Specification:
The first and only line of the input starts with a string with the format "HH:MM" where "HH" is from "00" to "23" and "MM" is from "00" to "59". Both "HH" and "MM" have exactly two digits.
Output Specification:
Print the palindromic time of day that comes soonest after the time given in the input. If the input time is palindromic, output the soonest palindromic time after the input time.
Demo Input:
['12:21\n', '23:59\n']
Demo Output:
['13:31\n', '00:00\n']
Note:
none
|
```python
h,m=map(int,input().split(':'))
while True:
m+=1
if m==60:
h+=1
m=0
if h==24:
h=0
a=str(h)
a='0'*(2-len(a))+a
b=str(m)
b='0'*(2-len(b))+b
a=a+":"+b
if(a==a[::-1]):
print(a)
exit()
```
| 3.892425
|
363
|
B
|
Fence
|
PROGRAMMING
| 1,100
|
[
"brute force",
"dp"
] | null | null |
There is a fence in front of Polycarpus's home. The fence consists of *n* planks of the same width which go one after another from left to right. The height of the *i*-th plank is *h**i* meters, distinct planks can have distinct heights.
Polycarpus has bought a posh piano and is thinking about how to get it into the house. In order to carry out his plan, he needs to take exactly *k* consecutive planks from the fence. Higher planks are harder to tear off the fence, so Polycarpus wants to find such *k* consecutive planks that the sum of their heights is minimal possible.
Write the program that finds the indexes of *k* consecutive planks with minimal total height. Pay attention, the fence is not around Polycarpus's home, it is in front of home (in other words, the fence isn't cyclic).
|
The first line of the input contains integers *n* and *k* (1<=≤<=*n*<=≤<=1.5·105,<=1<=≤<=*k*<=≤<=*n*) — the number of planks in the fence and the width of the hole for the piano. The second line contains the sequence of integers *h*1,<=*h*2,<=...,<=*h**n* (1<=≤<=*h**i*<=≤<=100), where *h**i* is the height of the *i*-th plank of the fence.
|
Print such integer *j* that the sum of the heights of planks *j*, *j*<=+<=1, ..., *j*<=+<=*k*<=-<=1 is the minimum possible. If there are multiple such *j*'s, print any of them.
|
[
"7 3\n1 2 6 1 1 7 1\n"
] |
[
"3\n"
] |
In the sample, your task is to find three consecutive planks with the minimum sum of heights. In the given case three planks with indexes 3, 4 and 5 have the required attribute, their total height is 8.
| 1,000
|
[
{
"input": "7 3\n1 2 6 1 1 7 1",
"output": "3"
},
{
"input": "1 1\n100",
"output": "1"
},
{
"input": "2 1\n10 20",
"output": "1"
},
{
"input": "10 5\n1 2 3 1 2 2 3 1 4 5",
"output": "1"
},
{
"input": "10 2\n3 1 4 1 4 6 2 1 4 6",
"output": "7"
},
{
"input": "2 2\n20 10",
"output": "1"
},
{
"input": "2 1\n20 1",
"output": "2"
},
{
"input": "3 1\n1 2 3",
"output": "1"
},
{
"input": "3 1\n2 1 3",
"output": "2"
},
{
"input": "3 1\n3 2 1",
"output": "3"
},
{
"input": "3 2\n1 2 3",
"output": "1"
},
{
"input": "3 2\n3 2 1",
"output": "2"
},
{
"input": "3 3\n1 2 3",
"output": "1"
},
{
"input": "4 2\n9 8 11 7",
"output": "1"
},
{
"input": "4 2\n10 1 2 3",
"output": "2"
},
{
"input": "6 3\n56 56 56 2 1 2",
"output": "4"
},
{
"input": "8 3\n1 1 1 1 2 60 90 1",
"output": "1"
},
{
"input": "4 1\n1 5 2 2",
"output": "1"
},
{
"input": "4 2\n4 6 7 4",
"output": "1"
},
{
"input": "10 4\n1 1 1 4 4 4 4 4 4 3",
"output": "1"
},
{
"input": "6 3\n1 2 1 3 1 1",
"output": "1"
},
{
"input": "5 2\n100 100 100 1 1",
"output": "4"
}
] | 1,686,824,001
| 2,147,483,647
|
PyPy 3-64
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 6
| 1,000
| 18,534,400
|
n,k=map(int,input().split())
l=list(map(int,input().split()))
s=[]
su=0
for i in range(0,n-k+1):
for j in range(k):
su+=l[i+j]
s.append(su)
su=0
mi=min(s)
print(s.index(mi)+1)
|
Title: Fence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There is a fence in front of Polycarpus's home. The fence consists of *n* planks of the same width which go one after another from left to right. The height of the *i*-th plank is *h**i* meters, distinct planks can have distinct heights.
Polycarpus has bought a posh piano and is thinking about how to get it into the house. In order to carry out his plan, he needs to take exactly *k* consecutive planks from the fence. Higher planks are harder to tear off the fence, so Polycarpus wants to find such *k* consecutive planks that the sum of their heights is minimal possible.
Write the program that finds the indexes of *k* consecutive planks with minimal total height. Pay attention, the fence is not around Polycarpus's home, it is in front of home (in other words, the fence isn't cyclic).
Input Specification:
The first line of the input contains integers *n* and *k* (1<=≤<=*n*<=≤<=1.5·105,<=1<=≤<=*k*<=≤<=*n*) — the number of planks in the fence and the width of the hole for the piano. The second line contains the sequence of integers *h*1,<=*h*2,<=...,<=*h**n* (1<=≤<=*h**i*<=≤<=100), where *h**i* is the height of the *i*-th plank of the fence.
Output Specification:
Print such integer *j* that the sum of the heights of planks *j*, *j*<=+<=1, ..., *j*<=+<=*k*<=-<=1 is the minimum possible. If there are multiple such *j*'s, print any of them.
Demo Input:
['7 3\n1 2 6 1 1 7 1\n']
Demo Output:
['3\n']
Note:
In the sample, your task is to find three consecutive planks with the minimum sum of heights. In the given case three planks with indexes 3, 4 and 5 have the required attribute, their total height is 8.
|
```python
n,k=map(int,input().split())
l=list(map(int,input().split()))
s=[]
su=0
for i in range(0,n-k+1):
for j in range(k):
su+=l[i+j]
s.append(su)
su=0
mi=min(s)
print(s.index(mi)+1)
```
| 0
|
|
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"
}
] | 1,596,737,239
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 139
| 7,475,200
|
import string
import random
n, k = map(int, input().split())
alpha = list(string.ascii_lowercase)
random_items = list(set(random.choices(population=alpha, k=k)))
if k > n:
print("".join(random_items + list((random_items[0])*(n-k))))
if k == n:
print("".join(random.choices(population=alpha, k=k)))
else:
print("".join(random.choices(population=alpha, k=n)))
|
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
import string
import random
n, k = map(int, input().split())
alpha = list(string.ascii_lowercase)
random_items = list(set(random.choices(population=alpha, k=k)))
if k > n:
print("".join(random_items + list((random_items[0])*(n-k))))
if k == n:
print("".join(random.choices(population=alpha, k=k)))
else:
print("".join(random.choices(population=alpha, k=n)))
```
| 0
|
|
603
|
A
|
Alternative Thinking
|
PROGRAMMING
| 1,600
|
[
"dp",
"greedy",
"math"
] | null | null |
Kevin has just recevied his disappointing results on the USA Identification of Cows Olympiad (USAICO) in the form of a binary string of length *n*. Each character of Kevin's string represents Kevin's score on one of the *n* questions of the olympiad—'1' for a correctly identified cow and '0' otherwise.
However, all is not lost. Kevin is a big proponent of alternative thinking and believes that his score, instead of being the sum of his points, should be the length of the longest alternating subsequence of his string. Here, we define an alternating subsequence of a string as a not-necessarily contiguous subsequence where no two consecutive elements are equal. For example, {0,<=1,<=0,<=1}, {1,<=0,<=1}, and {1,<=0,<=1,<=0} are alternating sequences, while {1,<=0,<=0} and {0,<=1,<=0,<=1,<=1} are not.
Kevin, being the sneaky little puffball that he is, is willing to hack into the USAICO databases to improve his score. In order to be subtle, he decides that he will flip exactly one substring—that is, take a contiguous non-empty substring of his score and change all '0's in that substring to '1's and vice versa. After such an operation, Kevin wants to know the length of the longest possible alternating subsequence that his string could have.
|
The first line contains the number of questions on the olympiad *n* (1<=≤<=*n*<=≤<=100<=000).
The following line contains a binary string of length *n* representing Kevin's results on the USAICO.
|
Output a single integer, the length of the longest possible alternating subsequence that Kevin can create in his string after flipping a single substring.
|
[
"8\n10000011\n",
"2\n01\n"
] |
[
"5\n",
"2\n"
] |
In the first sample, Kevin can flip the bolded substring '10000011' and turn his string into '10011011', which has an alternating subsequence of length 5: '10011011'.
In the second sample, Kevin can flip the entire string and still have the same score.
| 500
|
[
{
"input": "8\n10000011",
"output": "5"
},
{
"input": "2\n01",
"output": "2"
},
{
"input": "5\n10101",
"output": "5"
},
{
"input": "75\n010101010101010101010101010101010101010101010101010101010101010101010101010",
"output": "75"
},
{
"input": "11\n00000000000",
"output": "3"
},
{
"input": "56\n10101011010101010101010101010101010101011010101010101010",
"output": "56"
},
{
"input": "50\n01011010110101010101010101010101010101010101010100",
"output": "49"
},
{
"input": "7\n0110100",
"output": "7"
},
{
"input": "8\n11011111",
"output": "5"
},
{
"input": "6\n000000",
"output": "3"
},
{
"input": "5\n01000",
"output": "5"
},
{
"input": "59\n10101010101010101010101010101010101010101010101010101010101",
"output": "59"
},
{
"input": "88\n1010101010101010101010101010101010101010101010101010101010101010101010101010101010101010",
"output": "88"
},
{
"input": "93\n010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010",
"output": "93"
},
{
"input": "70\n0101010101010101010101010101010101010101010101010101010101010101010101",
"output": "70"
},
{
"input": "78\n010101010101010101010101010101101010101010101010101010101010101010101010101010",
"output": "78"
},
{
"input": "83\n10101010101010101010101010101010101010101010101010110101010101010101010101010101010",
"output": "83"
},
{
"input": "87\n101010101010101010101010101010101010101010101010101010101010101010101010101010010101010",
"output": "87"
},
{
"input": "65\n01010101101010101010101010101010101010101010101010101010101010101",
"output": "65"
},
{
"input": "69\n010101010101010101101010101010101010101010101010101010101010101010101",
"output": "69"
},
{
"input": "74\n01010101010101010101010101010101010101010101010101010101010101000101010101",
"output": "74"
},
{
"input": "77\n01010101010101001010101010101010100101010101010101010101010101010101010101010",
"output": "77"
},
{
"input": "60\n101010110101010101010101010110101010101010101010101010101010",
"output": "60"
},
{
"input": "89\n01010101010101010101010101010101010101010101010101010101101010101010101010100101010101010",
"output": "89"
},
{
"input": "68\n01010101010101010101010101010101010100101010100101010101010100101010",
"output": "67"
},
{
"input": "73\n0101010101010101010101010101010101010101010111011010101010101010101010101",
"output": "72"
},
{
"input": "55\n1010101010101010010101010101101010101010101010100101010",
"output": "54"
},
{
"input": "85\n1010101010101010101010101010010101010101010101101010101010101010101011010101010101010",
"output": "84"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "10\n1111111111",
"output": "3"
},
{
"input": "2\n10",
"output": "2"
},
{
"input": "2\n11",
"output": "2"
},
{
"input": "2\n00",
"output": "2"
},
{
"input": "3\n000",
"output": "3"
},
{
"input": "3\n001",
"output": "3"
},
{
"input": "3\n010",
"output": "3"
},
{
"input": "3\n011",
"output": "3"
},
{
"input": "3\n100",
"output": "3"
},
{
"input": "3\n101",
"output": "3"
},
{
"input": "3\n110",
"output": "3"
},
{
"input": "3\n111",
"output": "3"
},
{
"input": "4\n0000",
"output": "3"
},
{
"input": "4\n0001",
"output": "4"
},
{
"input": "4\n0010",
"output": "4"
},
{
"input": "4\n0011",
"output": "4"
},
{
"input": "4\n0100",
"output": "4"
},
{
"input": "4\n0101",
"output": "4"
},
{
"input": "4\n0110",
"output": "4"
},
{
"input": "4\n0111",
"output": "4"
},
{
"input": "4\n1000",
"output": "4"
},
{
"input": "4\n1001",
"output": "4"
},
{
"input": "4\n1010",
"output": "4"
},
{
"input": "4\n1011",
"output": "4"
},
{
"input": "4\n1100",
"output": "4"
},
{
"input": "4\n1101",
"output": "4"
},
{
"input": "4\n1110",
"output": "4"
},
{
"input": "4\n1111",
"output": "3"
},
{
"input": "5\n00000",
"output": "3"
},
{
"input": "5\n00001",
"output": "4"
},
{
"input": "5\n00010",
"output": "5"
},
{
"input": "5\n00011",
"output": "4"
},
{
"input": "5\n00100",
"output": "5"
},
{
"input": "5\n00101",
"output": "5"
},
{
"input": "5\n00110",
"output": "5"
},
{
"input": "5\n00111",
"output": "4"
},
{
"input": "5\n01000",
"output": "5"
},
{
"input": "5\n01001",
"output": "5"
},
{
"input": "5\n01010",
"output": "5"
},
{
"input": "5\n01011",
"output": "5"
},
{
"input": "5\n01100",
"output": "5"
},
{
"input": "5\n01101",
"output": "5"
},
{
"input": "5\n01110",
"output": "5"
},
{
"input": "5\n01111",
"output": "4"
},
{
"input": "5\n10000",
"output": "4"
},
{
"input": "5\n10001",
"output": "5"
},
{
"input": "5\n10010",
"output": "5"
},
{
"input": "5\n10100",
"output": "5"
},
{
"input": "5\n10101",
"output": "5"
},
{
"input": "5\n10110",
"output": "5"
},
{
"input": "5\n10111",
"output": "5"
},
{
"input": "5\n11000",
"output": "4"
},
{
"input": "5\n11001",
"output": "5"
},
{
"input": "5\n11010",
"output": "5"
},
{
"input": "5\n11011",
"output": "5"
},
{
"input": "5\n11100",
"output": "4"
},
{
"input": "5\n11101",
"output": "5"
},
{
"input": "5\n11110",
"output": "4"
},
{
"input": "5\n11111",
"output": "3"
}
] | 1,666,774,427
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 5
| 46
| 0
|
def inp():
return int(input())
def inlt():
return list(map(int, input().split()))
def insr():
s = input()
return list(s[:len(s)])
def invr():
return map(int, input().split())
n = inp()
s = input()
i = 0
flip = True
ans = 0
if n == 1:
print(0)
exit(0)
while i < n:
j = i+1
while j < n and s[i] == s[j]:
j += 1
if j - i >= 3 and flip:
flip = False
ans += 1
i = j
for i in range(2,n-2):
if (s[i] != s[i-1]) and (s[i] != s[i+1]) and (s[i-2] == s[i-1]) and (s[i+1] == s[i+2]):
print(ans + 2)
exit(0)
if not flip:
print(ans+2)
exit(0)
if flip:
for i in range(1,n-2):
if s[i] != s[i+1]:
if s[i-1] == s[i] and s[i+1] == s[i+2]:
print(ans+2)
exit(0)
if flip:
if n >= 2:
if s[0] == s[1]:
flip = False
ans += 1
if flip:
if n >= 2:
if s[-1] == s[-2]:
flip = False
ans += 1
print(ans)
|
Title: Alternative Thinking
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kevin has just recevied his disappointing results on the USA Identification of Cows Olympiad (USAICO) in the form of a binary string of length *n*. Each character of Kevin's string represents Kevin's score on one of the *n* questions of the olympiad—'1' for a correctly identified cow and '0' otherwise.
However, all is not lost. Kevin is a big proponent of alternative thinking and believes that his score, instead of being the sum of his points, should be the length of the longest alternating subsequence of his string. Here, we define an alternating subsequence of a string as a not-necessarily contiguous subsequence where no two consecutive elements are equal. For example, {0,<=1,<=0,<=1}, {1,<=0,<=1}, and {1,<=0,<=1,<=0} are alternating sequences, while {1,<=0,<=0} and {0,<=1,<=0,<=1,<=1} are not.
Kevin, being the sneaky little puffball that he is, is willing to hack into the USAICO databases to improve his score. In order to be subtle, he decides that he will flip exactly one substring—that is, take a contiguous non-empty substring of his score and change all '0's in that substring to '1's and vice versa. After such an operation, Kevin wants to know the length of the longest possible alternating subsequence that his string could have.
Input Specification:
The first line contains the number of questions on the olympiad *n* (1<=≤<=*n*<=≤<=100<=000).
The following line contains a binary string of length *n* representing Kevin's results on the USAICO.
Output Specification:
Output a single integer, the length of the longest possible alternating subsequence that Kevin can create in his string after flipping a single substring.
Demo Input:
['8\n10000011\n', '2\n01\n']
Demo Output:
['5\n', '2\n']
Note:
In the first sample, Kevin can flip the bolded substring '10000011' and turn his string into '10011011', which has an alternating subsequence of length 5: '10011011'.
In the second sample, Kevin can flip the entire string and still have the same score.
|
```python
def inp():
return int(input())
def inlt():
return list(map(int, input().split()))
def insr():
s = input()
return list(s[:len(s)])
def invr():
return map(int, input().split())
n = inp()
s = input()
i = 0
flip = True
ans = 0
if n == 1:
print(0)
exit(0)
while i < n:
j = i+1
while j < n and s[i] == s[j]:
j += 1
if j - i >= 3 and flip:
flip = False
ans += 1
i = j
for i in range(2,n-2):
if (s[i] != s[i-1]) and (s[i] != s[i+1]) and (s[i-2] == s[i-1]) and (s[i+1] == s[i+2]):
print(ans + 2)
exit(0)
if not flip:
print(ans+2)
exit(0)
if flip:
for i in range(1,n-2):
if s[i] != s[i+1]:
if s[i-1] == s[i] and s[i+1] == s[i+2]:
print(ans+2)
exit(0)
if flip:
if n >= 2:
if s[0] == s[1]:
flip = False
ans += 1
if flip:
if n >= 2:
if s[-1] == s[-2]:
flip = False
ans += 1
print(ans)
```
| 0
|
|
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"
}
] | 1,694,412,637
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 2
| 62
| 0
|
beautiful_year = []
for year in range(1000,9000):
year_list = list(str(year))
g = year_list[0]
s = year_list[1]
b = year_list[2]
q = year_list[3]
if g==s or g==b or g==q or s==b or s==q or b==q:
True
else:
beautiful_year.append(year)
year1 = int(input())
index1 = beautiful_year.index(year1)
index2 = index1 + 1
print(beautiful_year[index2])
|
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
beautiful_year = []
for year in range(1000,9000):
year_list = list(str(year))
g = year_list[0]
s = year_list[1]
b = year_list[2]
q = year_list[3]
if g==s or g==b or g==q or s==b or s==q or b==q:
True
else:
beautiful_year.append(year)
year1 = int(input())
index1 = beautiful_year.index(year1)
index2 = index1 + 1
print(beautiful_year[index2])
```
| -1
|
|
984
|
A
|
Game
|
PROGRAMMING
| 800
|
[
"sortings"
] | null | null |
Two players play a game.
Initially there are $n$ integers $a_1, a_2, \ldots, a_n$ written on the board. Each turn a player selects one number and erases it from the board. This continues until there is only one number left on the board, i. e. $n - 1$ turns are made. The first player makes the first move, then players alternate turns.
The first player wants to minimize the last number that would be left on the board, while the second player wants to maximize it.
You want to know what number will be left on the board after $n - 1$ turns if both players make optimal moves.
|
The first line contains one integer $n$ ($1 \le n \le 1000$) — the number of numbers on the board.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$).
|
Print one number that will be left on the board.
|
[
"3\n2 1 3\n",
"3\n2 2 2\n"
] |
[
"2",
"2"
] |
In the first sample, the first player erases $3$ and the second erases $1$. $2$ is left on the board.
In the second sample, $2$ is left on the board regardless of the actions of the players.
| 500
|
[
{
"input": "3\n2 1 3",
"output": "2"
},
{
"input": "3\n2 2 2",
"output": "2"
},
{
"input": "9\n44 53 51 80 5 27 74 79 94",
"output": "53"
},
{
"input": "10\n38 82 23 37 96 4 81 60 67 86",
"output": "60"
},
{
"input": "10\n58 26 77 15 53 81 68 48 22 65",
"output": "53"
},
{
"input": "1\n124",
"output": "124"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "3\n1 1 1000",
"output": "1"
},
{
"input": "2\n322 322",
"output": "322"
},
{
"input": "3\n9 92 12",
"output": "12"
},
{
"input": "3\n1 2 2",
"output": "2"
}
] | 1,670,322,481
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 35
| 93
| 2,867,200
|
import math
import string
import sys
input = sys.stdin.readline
# take a list input in a single line
def invr():
return list(map(int, input().split()))
def solve():
n = int(input())
string = input()
lst = list(map(int, string.split()))
lst.sort()
ans = 0
if( n%2 == 0):
ans = lst[int((n/2)) - 1]
else:
ans = lst[int((n-1)/2)]
print(ans)
# t = int(input())
# while t > 0:
# solve()
# t -= 1
solve()
|
Title: Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Two players play a game.
Initially there are $n$ integers $a_1, a_2, \ldots, a_n$ written on the board. Each turn a player selects one number and erases it from the board. This continues until there is only one number left on the board, i. e. $n - 1$ turns are made. The first player makes the first move, then players alternate turns.
The first player wants to minimize the last number that would be left on the board, while the second player wants to maximize it.
You want to know what number will be left on the board after $n - 1$ turns if both players make optimal moves.
Input Specification:
The first line contains one integer $n$ ($1 \le n \le 1000$) — the number of numbers on the board.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$).
Output Specification:
Print one number that will be left on the board.
Demo Input:
['3\n2 1 3\n', '3\n2 2 2\n']
Demo Output:
['2', '2']
Note:
In the first sample, the first player erases $3$ and the second erases $1$. $2$ is left on the board.
In the second sample, $2$ is left on the board regardless of the actions of the players.
|
```python
import math
import string
import sys
input = sys.stdin.readline
# take a list input in a single line
def invr():
return list(map(int, input().split()))
def solve():
n = int(input())
string = input()
lst = list(map(int, string.split()))
lst.sort()
ans = 0
if( n%2 == 0):
ans = lst[int((n/2)) - 1]
else:
ans = lst[int((n-1)/2)]
print(ans)
# t = int(input())
# while t > 0:
# solve()
# t -= 1
solve()
```
| 3
|
|
672
|
B
|
Different is Good
|
PROGRAMMING
| 1,000
|
[
"constructive algorithms",
"implementation",
"strings"
] | null | null |
A wise man told Kerem "Different is good" once, so Kerem wants all things in his life to be different.
Kerem recently got a string *s* consisting of lowercase English letters. Since Kerem likes it when things are different, he wants all substrings of his string *s* to be distinct. Substring is a string formed by some number of consecutive characters of the string. For example, string "aba" has substrings "" (empty substring), "a", "b", "a", "ab", "ba", "aba".
If string *s* has at least two equal substrings then Kerem will change characters at some positions to some other lowercase English letters. Changing characters is a very tiring job, so Kerem want to perform as few changes as possible.
Your task is to find the minimum number of changes needed to make all the substrings of the given string distinct, or determine that it is impossible.
|
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100<=000) — the length of the string *s*.
The second line contains the string *s* of length *n* consisting of only lowercase English letters.
|
If it's impossible to change the string *s* such that all its substring are distinct print -1. Otherwise print the minimum required number of changes.
|
[
"2\naa\n",
"4\nkoko\n",
"5\nmurat\n"
] |
[
"1\n",
"2\n",
"0\n"
] |
In the first sample one of the possible solutions is to change the first character to 'b'.
In the second sample, one may change the first character to 'a' and second character to 'b', so the string becomes "abko".
| 1,000
|
[
{
"input": "2\naa",
"output": "1"
},
{
"input": "4\nkoko",
"output": "2"
},
{
"input": "5\nmurat",
"output": "0"
},
{
"input": "6\nacbead",
"output": "1"
},
{
"input": "7\ncdaadad",
"output": "4"
},
{
"input": "25\npeoaicnbisdocqofsqdpgobpn",
"output": "12"
},
{
"input": "25\ntcqpchnqskqjacruoaqilgebu",
"output": "7"
},
{
"input": "13\naebaecedabbee",
"output": "8"
},
{
"input": "27\naaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "-1"
},
{
"input": "10\nbababbdaee",
"output": "6"
},
{
"input": "11\ndbadcdbdbca",
"output": "7"
},
{
"input": "12\nacceaabddaaa",
"output": "7"
},
{
"input": "13\nabddfbfaeecfa",
"output": "7"
},
{
"input": "14\neeceecacdbcbbb",
"output": "9"
},
{
"input": "15\ndcbceaaggabaheb",
"output": "8"
},
{
"input": "16\nhgiegfbadgcicbhd",
"output": "7"
},
{
"input": "17\nabhfibbdddfghgfdi",
"output": "10"
},
{
"input": "26\nbbbbbabbaababaaabaaababbaa",
"output": "24"
},
{
"input": "26\nahnxdnbfbcrirerssyzydihuee",
"output": "11"
},
{
"input": "26\nhwqeqhkpxwulbsiwmnlfyhgknc",
"output": "8"
},
{
"input": "26\nrvxmulriorilidecqwmfaemifj",
"output": "10"
},
{
"input": "26\naowpmreooavnmamogdoopuisge",
"output": "12"
},
{
"input": "26\ninimevtuefhvuefirdehmmfudh",
"output": "15"
},
{
"input": "26\naaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "25"
},
{
"input": "27\nqdcfjtblgglnilgassirrjekcjt",
"output": "-1"
},
{
"input": "27\nabcdefghijklmnopqrstuvwxyza",
"output": "-1"
},
{
"input": "26\nqwertyuiopasdfghjklzxcvbnm",
"output": "0"
},
{
"input": "5\nzzzzz",
"output": "4"
},
{
"input": "27\naaaaaaaaaaaaaaaaabaaaaaaaaa",
"output": "-1"
},
{
"input": "1\nq",
"output": "0"
},
{
"input": "27\nqwertyuioplkjhgfdsazxcvbnmm",
"output": "-1"
},
{
"input": "9\nxxxyyyzzz",
"output": "6"
},
{
"input": "45\naaabbbcccdddeeefffgghhiijjkkkkkkkkkkkkkkkkkkk",
"output": "-1"
},
{
"input": "27\nqwertyuiopasdfghjklzxcvbnmm",
"output": "-1"
},
{
"input": "26\nabcdefghijklmnopqrstuvwxyz",
"output": "0"
},
{
"input": "26\nabcdefghijklmnopqrstuvwxya",
"output": "1"
},
{
"input": "27\nabcdefghijklmnopqrstuvwxyzz",
"output": "-1"
},
{
"input": "26\naaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "25"
},
{
"input": "26\nqwertyuioplkjhgfdsazxcvbnm",
"output": "0"
},
{
"input": "10\nzzzzzzzzzz",
"output": "9"
},
{
"input": "1\na",
"output": "0"
},
{
"input": "30\nabcdefghtyabcdefghtyabcdefghty",
"output": "-1"
}
] | 1,652,178,124
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 31
| 102,400
|
from collections import Counter
n = int(input())
s = input()
res = 0
x = Counter(s)
for k in x:
if x[k] > 1:
res += 1
print(res)
|
Title: Different is Good
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A wise man told Kerem "Different is good" once, so Kerem wants all things in his life to be different.
Kerem recently got a string *s* consisting of lowercase English letters. Since Kerem likes it when things are different, he wants all substrings of his string *s* to be distinct. Substring is a string formed by some number of consecutive characters of the string. For example, string "aba" has substrings "" (empty substring), "a", "b", "a", "ab", "ba", "aba".
If string *s* has at least two equal substrings then Kerem will change characters at some positions to some other lowercase English letters. Changing characters is a very tiring job, so Kerem want to perform as few changes as possible.
Your task is to find the minimum number of changes needed to make all the substrings of the given string distinct, or determine that it is impossible.
Input Specification:
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100<=000) — the length of the string *s*.
The second line contains the string *s* of length *n* consisting of only lowercase English letters.
Output Specification:
If it's impossible to change the string *s* such that all its substring are distinct print -1. Otherwise print the minimum required number of changes.
Demo Input:
['2\naa\n', '4\nkoko\n', '5\nmurat\n']
Demo Output:
['1\n', '2\n', '0\n']
Note:
In the first sample one of the possible solutions is to change the first character to 'b'.
In the second sample, one may change the first character to 'a' and second character to 'b', so the string becomes "abko".
|
```python
from collections import Counter
n = int(input())
s = input()
res = 0
x = Counter(s)
for k in x:
if x[k] > 1:
res += 1
print(res)
```
| 0
|
|
924
|
C
|
Riverside Curio
|
PROGRAMMING
| 1,700
|
[
"data structures",
"dp",
"greedy"
] | 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.
| 1,250
|
[
{
"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"
}
] | 1,678,764,477
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 42
| 93
| 13,516,800
|
n=int(input())
arr=list(map(int,input().split()))
i=n-1
curr=0
t=[0]*(n)
while i>=0:
curr=max(curr-1,0)
curr=max(curr,arr[i]+1)
t[i]=curr
i-=1
curr=0
ans=0
for i in t:
curr=max(curr,i)
ans+=curr
for el in arr:
ans-=el+1
print(ans)
|
Title: Riverside Curio
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
n=int(input())
arr=list(map(int,input().split()))
i=n-1
curr=0
t=[0]*(n)
while i>=0:
curr=max(curr-1,0)
curr=max(curr,arr[i]+1)
t[i]=curr
i-=1
curr=0
ans=0
for i in t:
curr=max(curr,i)
ans+=curr
for el in arr:
ans-=el+1
print(ans)
```
| 3
|
|
192
|
A
|
Funky Numbers
|
PROGRAMMING
| 1,300
|
[
"binary search",
"brute force",
"implementation"
] | null | null |
As you very well know, this year's funkiest numbers are so called triangular numbers (that is, integers that are representable as , where *k* is some positive integer), and the coolest numbers are those that are representable as a sum of two triangular numbers.
A well-known hipster Andrew adores everything funky and cool but unfortunately, he isn't good at maths. Given number *n*, help him define whether this number can be represented by a sum of two triangular numbers (not necessarily different)!
|
The first input line contains an integer *n* (1<=≤<=*n*<=≤<=109).
|
Print "YES" (without the quotes), if *n* can be represented as a sum of two triangular numbers, otherwise print "NO" (without the quotes).
|
[
"256\n",
"512\n"
] |
[
"YES\n",
"NO\n"
] |
In the first sample number <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/92095692c6ea93e9e3b837a0408ba7543549d5b2.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample number 512 can not be represented as a sum of two triangular numbers.
| 500
|
[
{
"input": "256",
"output": "YES"
},
{
"input": "512",
"output": "NO"
},
{
"input": "80",
"output": "NO"
},
{
"input": "828",
"output": "YES"
},
{
"input": "6035",
"output": "NO"
},
{
"input": "39210",
"output": "YES"
},
{
"input": "79712",
"output": "NO"
},
{
"input": "190492",
"output": "YES"
},
{
"input": "5722367",
"output": "NO"
},
{
"input": "816761542",
"output": "YES"
},
{
"input": "1",
"output": "NO"
},
{
"input": "2",
"output": "YES"
},
{
"input": "3",
"output": "NO"
},
{
"input": "4",
"output": "YES"
},
{
"input": "5",
"output": "NO"
},
{
"input": "6",
"output": "YES"
},
{
"input": "7",
"output": "YES"
},
{
"input": "8",
"output": "NO"
},
{
"input": "9",
"output": "YES"
},
{
"input": "10",
"output": "NO"
},
{
"input": "12",
"output": "YES"
},
{
"input": "13",
"output": "YES"
},
{
"input": "14",
"output": "NO"
},
{
"input": "15",
"output": "NO"
},
{
"input": "16",
"output": "YES"
},
{
"input": "17",
"output": "NO"
},
{
"input": "18",
"output": "YES"
},
{
"input": "19",
"output": "NO"
},
{
"input": "20",
"output": "YES"
},
{
"input": "41",
"output": "NO"
},
{
"input": "11",
"output": "YES"
},
{
"input": "69",
"output": "YES"
},
{
"input": "82",
"output": "NO"
},
{
"input": "85",
"output": "NO"
},
{
"input": "736",
"output": "NO"
},
{
"input": "895",
"output": "YES"
},
{
"input": "934",
"output": "YES"
},
{
"input": "6213",
"output": "YES"
},
{
"input": "7405",
"output": "NO"
},
{
"input": "9919",
"output": "NO"
},
{
"input": "40942",
"output": "YES"
},
{
"input": "41992",
"output": "NO"
},
{
"input": "68535",
"output": "NO"
},
{
"input": "405718",
"output": "NO"
},
{
"input": "1046146",
"output": "YES"
},
{
"input": "3761248",
"output": "YES"
},
{
"input": "6195181",
"output": "YES"
},
{
"input": "35354345",
"output": "NO"
},
{
"input": "81282830",
"output": "NO"
},
{
"input": "187719774",
"output": "NO"
},
{
"input": "296798673",
"output": "NO"
},
{
"input": "938938476",
"output": "NO"
},
{
"input": "1000000000",
"output": "NO"
},
{
"input": "999887464",
"output": "YES"
},
{
"input": "999111944",
"output": "NO"
},
{
"input": "999966520",
"output": "YES"
},
{
"input": "999912080",
"output": "NO"
},
{
"input": "999992017",
"output": "YES"
},
{
"input": "999990474",
"output": "NO"
},
{
"input": "999999190",
"output": "YES"
},
{
"input": "999999125",
"output": "NO"
},
{
"input": "999999940",
"output": "YES"
},
{
"input": "999999995",
"output": "NO"
},
{
"input": "1000000000",
"output": "NO"
},
{
"input": "1",
"output": "NO"
},
{
"input": "999999999",
"output": "YES"
},
{
"input": "83495494",
"output": "NO"
},
{
"input": "968022000",
"output": "YES"
},
{
"input": "399980000",
"output": "YES"
},
{
"input": "4",
"output": "YES"
},
{
"input": "999999998",
"output": "NO"
}
] | 1,615,440,013
| 2,147,483,647
|
PyPy 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 6
| 2,000
| 5,632,000
|
def check(k):
l=[]
for i in range(1,k):
l.append(i*(i+1)//2)
for i in l:
for j in l:
if i+j==k:
return True
return False
if check(int(input())):
print('YES')
else:
print('NO')
|
Title: Funky Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As you very well know, this year's funkiest numbers are so called triangular numbers (that is, integers that are representable as , where *k* is some positive integer), and the coolest numbers are those that are representable as a sum of two triangular numbers.
A well-known hipster Andrew adores everything funky and cool but unfortunately, he isn't good at maths. Given number *n*, help him define whether this number can be represented by a sum of two triangular numbers (not necessarily different)!
Input Specification:
The first input line contains an integer *n* (1<=≤<=*n*<=≤<=109).
Output Specification:
Print "YES" (without the quotes), if *n* can be represented as a sum of two triangular numbers, otherwise print "NO" (without the quotes).
Demo Input:
['256\n', '512\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample number <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/92095692c6ea93e9e3b837a0408ba7543549d5b2.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample number 512 can not be represented as a sum of two triangular numbers.
|
```python
def check(k):
l=[]
for i in range(1,k):
l.append(i*(i+1)//2)
for i in l:
for j in l:
if i+j==k:
return True
return False
if check(int(input())):
print('YES')
else:
print('NO')
```
| 0
|
|
769
|
A
|
Year of University Entrance
|
PROGRAMMING
| 800
|
[
"*special",
"implementation",
"sortings"
] | null | null |
There is the faculty of Computer Science in Berland. In the social net "TheContact!" for each course of this faculty there is the special group whose name equals the year of university entrance of corresponding course of students at the university.
Each of students joins the group of his course and joins all groups for which the year of student's university entrance differs by no more than *x* from the year of university entrance of this student, where *x* — some non-negative integer. A value *x* is not given, but it can be uniquely determined from the available data. Note that students don't join other groups.
You are given the list of groups which the student Igor joined. According to this information you need to determine the year of Igor's university entrance.
|
The first line contains the positive odd integer *n* (1<=≤<=*n*<=≤<=5) — the number of groups which Igor joined.
The next line contains *n* distinct integers *a*1,<=*a*2,<=...,<=*a**n* (2010<=≤<=*a**i*<=≤<=2100) — years of student's university entrance for each group in which Igor is the member.
It is guaranteed that the input data is correct and the answer always exists. Groups are given randomly.
|
Print the year of Igor's university entrance.
|
[
"3\n2014 2016 2015\n",
"1\n2050\n"
] |
[
"2015\n",
"2050\n"
] |
In the first test the value *x* = 1. Igor entered the university in 2015. So he joined groups members of which are students who entered the university in 2014, 2015 and 2016.
In the second test the value *x* = 0. Igor entered only the group which corresponds to the year of his university entrance.
| 500
|
[
{
"input": "3\n2014 2016 2015",
"output": "2015"
},
{
"input": "1\n2050",
"output": "2050"
},
{
"input": "1\n2010",
"output": "2010"
},
{
"input": "1\n2011",
"output": "2011"
},
{
"input": "3\n2010 2011 2012",
"output": "2011"
},
{
"input": "3\n2049 2047 2048",
"output": "2048"
},
{
"input": "5\n2043 2042 2041 2044 2040",
"output": "2042"
},
{
"input": "5\n2012 2013 2014 2015 2016",
"output": "2014"
},
{
"input": "1\n2045",
"output": "2045"
},
{
"input": "1\n2046",
"output": "2046"
},
{
"input": "1\n2099",
"output": "2099"
},
{
"input": "1\n2100",
"output": "2100"
},
{
"input": "3\n2011 2010 2012",
"output": "2011"
},
{
"input": "3\n2011 2012 2010",
"output": "2011"
},
{
"input": "3\n2012 2011 2010",
"output": "2011"
},
{
"input": "3\n2010 2012 2011",
"output": "2011"
},
{
"input": "3\n2012 2010 2011",
"output": "2011"
},
{
"input": "3\n2047 2048 2049",
"output": "2048"
},
{
"input": "3\n2047 2049 2048",
"output": "2048"
},
{
"input": "3\n2048 2047 2049",
"output": "2048"
},
{
"input": "3\n2048 2049 2047",
"output": "2048"
},
{
"input": "3\n2049 2048 2047",
"output": "2048"
},
{
"input": "5\n2011 2014 2012 2013 2010",
"output": "2012"
},
{
"input": "5\n2014 2013 2011 2012 2015",
"output": "2013"
},
{
"input": "5\n2021 2023 2024 2020 2022",
"output": "2022"
},
{
"input": "5\n2081 2079 2078 2080 2077",
"output": "2079"
},
{
"input": "5\n2095 2099 2097 2096 2098",
"output": "2097"
},
{
"input": "5\n2097 2099 2100 2098 2096",
"output": "2098"
},
{
"input": "5\n2012 2010 2014 2011 2013",
"output": "2012"
},
{
"input": "5\n2012 2011 2013 2015 2014",
"output": "2013"
},
{
"input": "5\n2023 2024 2022 2021 2020",
"output": "2022"
},
{
"input": "5\n2077 2078 2080 2079 2081",
"output": "2079"
},
{
"input": "5\n2099 2096 2095 2097 2098",
"output": "2097"
},
{
"input": "5\n2097 2100 2098 2096 2099",
"output": "2098"
},
{
"input": "5\n2011 2014 2013 2010 2012",
"output": "2012"
},
{
"input": "5\n2013 2011 2015 2012 2014",
"output": "2013"
},
{
"input": "5\n2024 2020 2021 2023 2022",
"output": "2022"
},
{
"input": "5\n2079 2080 2077 2081 2078",
"output": "2079"
},
{
"input": "5\n2095 2097 2096 2098 2099",
"output": "2097"
},
{
"input": "5\n2099 2096 2100 2097 2098",
"output": "2098"
},
{
"input": "5\n2034 2033 2036 2032 2035",
"output": "2034"
},
{
"input": "5\n2030 2031 2033 2032 2029",
"output": "2031"
},
{
"input": "5\n2093 2092 2094 2096 2095",
"output": "2094"
},
{
"input": "5\n2012 2015 2014 2013 2011",
"output": "2013"
},
{
"input": "5\n2056 2057 2058 2059 2060",
"output": "2058"
}
] | 1,488,629,927
| 1,127
|
Python 3
|
OK
|
TESTS
| 45
| 61
| 4,608,000
|
'''input
1
2050
'''
# import sys
# from pprint import pprint
n = int(input())
arr = sorted([int(i) for i in input().split()])
print(arr[len(arr) // 2])
|
Title: Year of University Entrance
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There is the faculty of Computer Science in Berland. In the social net "TheContact!" for each course of this faculty there is the special group whose name equals the year of university entrance of corresponding course of students at the university.
Each of students joins the group of his course and joins all groups for which the year of student's university entrance differs by no more than *x* from the year of university entrance of this student, where *x* — some non-negative integer. A value *x* is not given, but it can be uniquely determined from the available data. Note that students don't join other groups.
You are given the list of groups which the student Igor joined. According to this information you need to determine the year of Igor's university entrance.
Input Specification:
The first line contains the positive odd integer *n* (1<=≤<=*n*<=≤<=5) — the number of groups which Igor joined.
The next line contains *n* distinct integers *a*1,<=*a*2,<=...,<=*a**n* (2010<=≤<=*a**i*<=≤<=2100) — years of student's university entrance for each group in which Igor is the member.
It is guaranteed that the input data is correct and the answer always exists. Groups are given randomly.
Output Specification:
Print the year of Igor's university entrance.
Demo Input:
['3\n2014 2016 2015\n', '1\n2050\n']
Demo Output:
['2015\n', '2050\n']
Note:
In the first test the value *x* = 1. Igor entered the university in 2015. So he joined groups members of which are students who entered the university in 2014, 2015 and 2016.
In the second test the value *x* = 0. Igor entered only the group which corresponds to the year of his university entrance.
|
```python
'''input
1
2050
'''
# import sys
# from pprint import pprint
n = int(input())
arr = sorted([int(i) for i in input().split()])
print(arr[len(arr) // 2])
```
| 3
|
|
160
|
A
|
Twins
|
PROGRAMMING
| 900
|
[
"greedy",
"sortings"
] | null | null |
Imagine that you have a twin brother or sister. Having another person that looks exactly like you seems very unusual. It's hard to say if having something of an alter ego is good or bad. And if you do have a twin, then you very well know what it's like.
Now let's imagine a typical morning in your family. You haven't woken up yet, and Mom is already going to work. She has been so hasty that she has nearly forgotten to leave the two of her darling children some money to buy lunches in the school cafeteria. She fished in the purse and found some number of coins, or to be exact, *n* coins of arbitrary values *a*1,<=*a*2,<=...,<=*a**n*. But as Mom was running out of time, she didn't split the coins for you two. So she scribbled a note asking you to split the money equally.
As you woke up, you found Mom's coins and read her note. "But why split the money equally?" — you thought. After all, your twin is sleeping and he won't know anything. So you decided to act like that: pick for yourself some subset of coins so that the sum of values of your coins is strictly larger than the sum of values of the remaining coins that your twin will have. However, you correctly thought that if you take too many coins, the twin will suspect the deception. So, you've decided to stick to the following strategy to avoid suspicions: you take the minimum number of coins, whose sum of values is strictly more than the sum of values of the remaining coins. On this basis, determine what minimum number of coins you need to take to divide them in the described manner.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of coins. The second line contains a sequence of *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=100) — the coins' values. All numbers are separated with spaces.
|
In the single line print the single number — the minimum needed number of coins.
|
[
"2\n3 3\n",
"3\n2 1 2\n"
] |
[
"2\n",
"2\n"
] |
In the first sample you will have to take 2 coins (you and your twin have sums equal to 6, 0 correspondingly). If you take 1 coin, you get sums 3, 3. If you take 0 coins, you get sums 0, 6. Those variants do not satisfy you as your sum should be strictly more that your twins' sum.
In the second sample one coin isn't enough for us, too. You can pick coins with values 1, 2 or 2, 2. In any case, the minimum number of coins equals 2.
| 500
|
[
{
"input": "2\n3 3",
"output": "2"
},
{
"input": "3\n2 1 2",
"output": "2"
},
{
"input": "1\n5",
"output": "1"
},
{
"input": "5\n4 2 2 2 2",
"output": "3"
},
{
"input": "7\n1 10 1 2 1 1 1",
"output": "1"
},
{
"input": "5\n3 2 3 3 1",
"output": "3"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "3\n2 1 3",
"output": "2"
},
{
"input": "6\n1 1 1 1 1 1",
"output": "4"
},
{
"input": "7\n10 10 5 5 5 5 1",
"output": "3"
},
{
"input": "20\n2 1 2 2 2 1 1 2 1 2 2 1 1 1 1 2 1 1 1 1",
"output": "8"
},
{
"input": "20\n4 2 4 4 3 4 2 2 4 2 3 1 1 2 2 3 3 3 1 4",
"output": "8"
},
{
"input": "20\n35 26 41 40 45 46 22 26 39 23 11 15 47 42 18 15 27 10 45 40",
"output": "8"
},
{
"input": "20\n7 84 100 10 31 35 41 2 63 44 57 4 63 11 23 49 98 71 16 90",
"output": "6"
},
{
"input": "50\n19 2 12 26 17 27 10 26 17 17 5 24 11 15 3 9 16 18 19 1 25 23 18 6 2 7 25 7 21 25 13 29 16 9 25 3 14 30 18 4 10 28 6 10 8 2 2 4 8 28",
"output": "14"
},
{
"input": "70\n2 18 18 47 25 5 14 9 19 46 36 49 33 32 38 23 32 39 8 29 31 17 24 21 10 15 33 37 46 21 22 11 20 35 39 13 11 30 28 40 39 47 1 17 24 24 21 46 12 2 20 43 8 16 44 11 45 10 13 44 31 45 45 46 11 10 33 35 23 42",
"output": "22"
},
{
"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": "51"
},
{
"input": "100\n1 2 2 1 2 1 1 2 1 1 1 2 2 1 1 1 2 2 2 1 2 1 1 1 1 1 2 1 2 1 2 1 2 1 2 1 1 1 2 1 1 1 1 1 2 2 1 2 1 2 1 2 2 2 1 2 1 2 2 1 1 2 2 1 1 2 2 2 1 1 2 1 1 2 2 1 2 1 1 2 2 1 2 1 1 2 2 1 1 1 1 2 1 1 1 1 2 2 2 2",
"output": "37"
},
{
"input": "100\n1 2 3 2 1 2 2 3 1 3 3 2 2 1 1 2 2 1 1 1 1 2 3 3 2 1 1 2 2 2 3 3 3 2 1 3 1 3 3 2 3 1 2 2 2 3 2 1 1 3 3 3 3 2 1 1 2 3 2 2 3 2 3 2 2 3 2 2 2 2 3 3 3 1 3 3 1 1 2 3 2 2 2 2 3 3 3 2 1 2 3 1 1 2 3 3 1 3 3 2",
"output": "36"
},
{
"input": "100\n5 5 4 3 5 1 2 5 1 1 3 5 4 4 1 1 1 1 5 4 4 5 1 5 5 1 2 1 3 1 5 1 3 3 3 2 2 2 1 1 5 1 3 4 1 1 3 2 5 2 2 5 5 4 4 1 3 4 3 3 4 5 3 3 3 1 2 1 4 2 4 4 1 5 1 3 5 5 5 5 3 4 4 3 1 2 5 2 3 5 4 2 4 5 3 2 4 2 4 3",
"output": "33"
},
{
"input": "100\n3 4 8 10 8 6 4 3 7 7 6 2 3 1 3 10 1 7 9 3 5 5 2 6 2 9 1 7 4 2 4 1 6 1 7 10 2 5 3 7 6 4 6 2 8 8 8 6 6 10 3 7 4 3 4 1 7 9 3 6 3 6 1 4 9 3 8 1 10 1 4 10 7 7 9 5 3 8 10 2 1 10 8 7 10 8 5 3 1 2 1 10 6 1 5 3 3 5 7 2",
"output": "30"
},
{
"input": "100\n16 9 11 8 11 4 9 17 4 8 4 10 9 10 6 3 3 15 1 6 1 15 12 18 6 14 13 18 1 7 18 4 10 7 10 12 3 16 14 4 10 8 10 7 19 13 15 1 4 8 16 10 6 4 3 16 11 10 7 3 4 16 1 20 1 11 4 16 10 7 7 12 18 19 3 17 19 3 4 19 2 12 11 3 18 20 2 2 14 4 20 13 13 11 16 20 19 14 7 2",
"output": "29"
},
{
"input": "100\n2 46 4 6 38 19 15 34 10 35 37 30 3 25 5 45 40 45 33 31 6 20 10 44 11 9 2 14 35 5 9 23 20 2 48 22 25 35 38 31 24 33 35 16 4 30 27 10 12 22 6 24 12 30 23 21 14 12 32 21 7 12 25 43 18 34 34 28 47 13 28 43 18 39 44 42 35 26 35 14 8 29 32 20 29 3 20 6 20 9 9 27 8 42 10 37 42 27 8 1",
"output": "30"
},
{
"input": "100\n85 50 17 89 65 89 5 20 86 26 16 21 85 14 44 31 87 31 6 2 48 67 8 80 79 1 48 36 97 1 5 30 79 50 78 12 2 55 76 100 54 40 26 81 97 96 68 56 87 14 51 17 54 37 52 33 69 62 38 63 74 15 62 78 9 19 67 2 60 58 93 60 18 96 55 48 34 7 79 82 32 58 90 67 20 50 27 15 7 89 98 10 11 15 99 49 4 51 77 52",
"output": "29"
},
{
"input": "100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100",
"output": "30"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "4\n4 1 2 3",
"output": "2"
},
{
"input": "5\n2 2 2 4 4",
"output": "2"
},
{
"input": "4\n100 99 99 100",
"output": "2"
}
] | 1,686,851,276
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 29
| 124
| 1,638,400
|
a = int(input())
b = list(map(int, input().split()))
count = 0
result = 0
for i in range(len(b)):
count += max(b)
b.pop(b.index(max(b)))
result += 1
if count > sum(b):
break
print(result)
|
Title: Twins
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Imagine that you have a twin brother or sister. Having another person that looks exactly like you seems very unusual. It's hard to say if having something of an alter ego is good or bad. And if you do have a twin, then you very well know what it's like.
Now let's imagine a typical morning in your family. You haven't woken up yet, and Mom is already going to work. She has been so hasty that she has nearly forgotten to leave the two of her darling children some money to buy lunches in the school cafeteria. She fished in the purse and found some number of coins, or to be exact, *n* coins of arbitrary values *a*1,<=*a*2,<=...,<=*a**n*. But as Mom was running out of time, she didn't split the coins for you two. So she scribbled a note asking you to split the money equally.
As you woke up, you found Mom's coins and read her note. "But why split the money equally?" — you thought. After all, your twin is sleeping and he won't know anything. So you decided to act like that: pick for yourself some subset of coins so that the sum of values of your coins is strictly larger than the sum of values of the remaining coins that your twin will have. However, you correctly thought that if you take too many coins, the twin will suspect the deception. So, you've decided to stick to the following strategy to avoid suspicions: you take the minimum number of coins, whose sum of values is strictly more than the sum of values of the remaining coins. On this basis, determine what minimum number of coins you need to take to divide them in the described manner.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of coins. The second line contains a sequence of *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=100) — the coins' values. All numbers are separated with spaces.
Output Specification:
In the single line print the single number — the minimum needed number of coins.
Demo Input:
['2\n3 3\n', '3\n2 1 2\n']
Demo Output:
['2\n', '2\n']
Note:
In the first sample you will have to take 2 coins (you and your twin have sums equal to 6, 0 correspondingly). If you take 1 coin, you get sums 3, 3. If you take 0 coins, you get sums 0, 6. Those variants do not satisfy you as your sum should be strictly more that your twins' sum.
In the second sample one coin isn't enough for us, too. You can pick coins with values 1, 2 or 2, 2. In any case, the minimum number of coins equals 2.
|
```python
a = int(input())
b = list(map(int, input().split()))
count = 0
result = 0
for i in range(len(b)):
count += max(b)
b.pop(b.index(max(b)))
result += 1
if count > sum(b):
break
print(result)
```
| 3
|
|
950
|
A
|
Left-handers, Right-handers and Ambidexters
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
You are at a water bowling training. There are *l* people who play with their left hand, *r* people, who play with their right hand, and *a* ambidexters, who can play with left or right hand.
The coach decided to form a team of even number of players, exactly half of the players should play with their right hand, and exactly half of the players should play with their left hand. One player should use only on of his hands.
Ambidexters play as well with their right hand as with their left hand. In the team, an ambidexter can play with their left hand, or with their right hand.
Please find the maximum possible size of the team, where equal number of players use their left and right hands, respectively.
|
The only line contains three integers *l*, *r* and *a* (0<=≤<=*l*,<=*r*,<=*a*<=≤<=100) — the number of left-handers, the number of right-handers and the number of ambidexters at the training.
|
Print a single even integer — the maximum number of players in the team. It is possible that the team can only have zero number of players.
|
[
"1 4 2\n",
"5 5 5\n",
"0 2 0\n"
] |
[
"6\n",
"14\n",
"0\n"
] |
In the first example you can form a team of 6 players. You should take the only left-hander and two ambidexters to play with left hand, and three right-handers to play with right hand. The only person left can't be taken into the team.
In the second example you can form a team of 14 people. You have to take all five left-handers, all five right-handers, two ambidexters to play with left hand and two ambidexters to play with right hand.
| 500
|
[
{
"input": "1 4 2",
"output": "6"
},
{
"input": "5 5 5",
"output": "14"
},
{
"input": "0 2 0",
"output": "0"
},
{
"input": "30 70 34",
"output": "128"
},
{
"input": "89 32 24",
"output": "112"
},
{
"input": "89 44 77",
"output": "210"
},
{
"input": "0 0 0",
"output": "0"
},
{
"input": "100 100 100",
"output": "300"
},
{
"input": "1 1 1",
"output": "2"
},
{
"input": "30 70 35",
"output": "130"
},
{
"input": "89 44 76",
"output": "208"
},
{
"input": "0 100 100",
"output": "200"
},
{
"input": "100 0 100",
"output": "200"
},
{
"input": "100 1 100",
"output": "200"
},
{
"input": "1 100 100",
"output": "200"
},
{
"input": "100 100 0",
"output": "200"
},
{
"input": "100 100 1",
"output": "200"
},
{
"input": "1 2 1",
"output": "4"
},
{
"input": "0 0 100",
"output": "100"
},
{
"input": "0 100 0",
"output": "0"
},
{
"input": "100 0 0",
"output": "0"
},
{
"input": "10 8 7",
"output": "24"
},
{
"input": "45 47 16",
"output": "108"
},
{
"input": "59 43 100",
"output": "202"
},
{
"input": "34 1 30",
"output": "62"
},
{
"input": "14 81 1",
"output": "30"
},
{
"input": "53 96 94",
"output": "242"
},
{
"input": "62 81 75",
"output": "218"
},
{
"input": "21 71 97",
"output": "188"
},
{
"input": "49 82 73",
"output": "204"
},
{
"input": "88 19 29",
"output": "96"
},
{
"input": "89 4 62",
"output": "132"
},
{
"input": "58 3 65",
"output": "126"
},
{
"input": "27 86 11",
"output": "76"
},
{
"input": "35 19 80",
"output": "134"
},
{
"input": "4 86 74",
"output": "156"
},
{
"input": "32 61 89",
"output": "182"
},
{
"input": "68 60 98",
"output": "226"
},
{
"input": "37 89 34",
"output": "142"
},
{
"input": "92 9 28",
"output": "74"
},
{
"input": "79 58 98",
"output": "234"
},
{
"input": "35 44 88",
"output": "166"
},
{
"input": "16 24 19",
"output": "58"
},
{
"input": "74 71 75",
"output": "220"
},
{
"input": "83 86 99",
"output": "268"
},
{
"input": "97 73 15",
"output": "176"
},
{
"input": "77 76 73",
"output": "226"
},
{
"input": "48 85 55",
"output": "188"
},
{
"input": "1 2 2",
"output": "4"
},
{
"input": "2 2 2",
"output": "6"
},
{
"input": "2 1 2",
"output": "4"
},
{
"input": "2 2 1",
"output": "4"
},
{
"input": "3 2 1",
"output": "6"
},
{
"input": "1 2 3",
"output": "6"
},
{
"input": "1 3 2",
"output": "6"
},
{
"input": "2 1 3",
"output": "6"
},
{
"input": "2 3 1",
"output": "6"
},
{
"input": "3 1 2",
"output": "6"
},
{
"input": "99 99 99",
"output": "296"
},
{
"input": "99 99 100",
"output": "298"
},
{
"input": "99 100 99",
"output": "298"
},
{
"input": "99 100 100",
"output": "298"
},
{
"input": "100 99 99",
"output": "298"
},
{
"input": "100 99 100",
"output": "298"
},
{
"input": "100 100 99",
"output": "298"
},
{
"input": "89 32 23",
"output": "110"
},
{
"input": "4 5 0",
"output": "8"
},
{
"input": "3 0 3",
"output": "6"
},
{
"input": "0 0 2",
"output": "2"
},
{
"input": "97 97 0",
"output": "194"
},
{
"input": "1 4 0",
"output": "2"
},
{
"input": "5 2 0",
"output": "4"
},
{
"input": "0 5 10",
"output": "14"
},
{
"input": "0 1 2",
"output": "2"
},
{
"input": "5 2 3",
"output": "10"
},
{
"input": "5 5 0",
"output": "10"
},
{
"input": "0 0 10",
"output": "10"
},
{
"input": "0 1 1",
"output": "2"
},
{
"input": "0 0 1",
"output": "0"
}
] | 1,643,541,137
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 79
| 108
| 0
|
a,b,c=map(int, input().split())
size=min(a,b)
a-=size
b-=size
size=size*2
x=max(min(a+c,b),min(b+c,a))
size=size+x*2+((c-x)//2)*2
print(size)
|
Title: Left-handers, Right-handers and Ambidexters
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are at a water bowling training. There are *l* people who play with their left hand, *r* people, who play with their right hand, and *a* ambidexters, who can play with left or right hand.
The coach decided to form a team of even number of players, exactly half of the players should play with their right hand, and exactly half of the players should play with their left hand. One player should use only on of his hands.
Ambidexters play as well with their right hand as with their left hand. In the team, an ambidexter can play with their left hand, or with their right hand.
Please find the maximum possible size of the team, where equal number of players use their left and right hands, respectively.
Input Specification:
The only line contains three integers *l*, *r* and *a* (0<=≤<=*l*,<=*r*,<=*a*<=≤<=100) — the number of left-handers, the number of right-handers and the number of ambidexters at the training.
Output Specification:
Print a single even integer — the maximum number of players in the team. It is possible that the team can only have zero number of players.
Demo Input:
['1 4 2\n', '5 5 5\n', '0 2 0\n']
Demo Output:
['6\n', '14\n', '0\n']
Note:
In the first example you can form a team of 6 players. You should take the only left-hander and two ambidexters to play with left hand, and three right-handers to play with right hand. The only person left can't be taken into the team.
In the second example you can form a team of 14 people. You have to take all five left-handers, all five right-handers, two ambidexters to play with left hand and two ambidexters to play with right hand.
|
```python
a,b,c=map(int, input().split())
size=min(a,b)
a-=size
b-=size
size=size*2
x=max(min(a+c,b),min(b+c,a))
size=size+x*2+((c-x)//2)*2
print(size)
```
| 3
|
|
670
|
A
|
Holidays
|
PROGRAMMING
| 900
|
[
"brute force",
"constructive algorithms",
"greedy",
"math"
] | null | null |
On the planet Mars a year lasts exactly *n* days (there are no leap years on Mars). But Martians have the same weeks as earthlings — 5 work days and then 2 days off. Your task is to determine the minimum possible and the maximum possible number of days off per year on Mars.
|
The first line of the input contains a positive integer *n* (1<=≤<=*n*<=≤<=1<=000<=000) — the number of days in a year on Mars.
|
Print two integers — the minimum possible and the maximum possible number of days off per year on Mars.
|
[
"14\n",
"2\n"
] |
[
"4 4\n",
"0 2\n"
] |
In the first sample there are 14 days in a year on Mars, and therefore independently of the day a year starts with there will be exactly 4 days off .
In the second sample there are only 2 days in a year on Mars, and they can both be either work days or days off.
| 500
|
[
{
"input": "14",
"output": "4 4"
},
{
"input": "2",
"output": "0 2"
},
{
"input": "1",
"output": "0 1"
},
{
"input": "3",
"output": "0 2"
},
{
"input": "4",
"output": "0 2"
},
{
"input": "5",
"output": "0 2"
},
{
"input": "6",
"output": "1 2"
},
{
"input": "7",
"output": "2 2"
},
{
"input": "8",
"output": "2 3"
},
{
"input": "9",
"output": "2 4"
},
{
"input": "10",
"output": "2 4"
},
{
"input": "11",
"output": "2 4"
},
{
"input": "12",
"output": "2 4"
},
{
"input": "13",
"output": "3 4"
},
{
"input": "1000000",
"output": "285714 285715"
},
{
"input": "16",
"output": "4 6"
},
{
"input": "17",
"output": "4 6"
},
{
"input": "18",
"output": "4 6"
},
{
"input": "19",
"output": "4 6"
},
{
"input": "20",
"output": "5 6"
},
{
"input": "21",
"output": "6 6"
},
{
"input": "22",
"output": "6 7"
},
{
"input": "23",
"output": "6 8"
},
{
"input": "24",
"output": "6 8"
},
{
"input": "25",
"output": "6 8"
},
{
"input": "26",
"output": "6 8"
},
{
"input": "27",
"output": "7 8"
},
{
"input": "28",
"output": "8 8"
},
{
"input": "29",
"output": "8 9"
},
{
"input": "30",
"output": "8 10"
},
{
"input": "100",
"output": "28 30"
},
{
"input": "99",
"output": "28 29"
},
{
"input": "98",
"output": "28 28"
},
{
"input": "97",
"output": "27 28"
},
{
"input": "96",
"output": "26 28"
},
{
"input": "95",
"output": "26 28"
},
{
"input": "94",
"output": "26 28"
},
{
"input": "93",
"output": "26 28"
},
{
"input": "92",
"output": "26 27"
},
{
"input": "91",
"output": "26 26"
},
{
"input": "90",
"output": "25 26"
},
{
"input": "89",
"output": "24 26"
},
{
"input": "88",
"output": "24 26"
},
{
"input": "87",
"output": "24 26"
},
{
"input": "86",
"output": "24 26"
},
{
"input": "85",
"output": "24 25"
},
{
"input": "84",
"output": "24 24"
},
{
"input": "83",
"output": "23 24"
},
{
"input": "82",
"output": "22 24"
},
{
"input": "81",
"output": "22 24"
},
{
"input": "80",
"output": "22 24"
},
{
"input": "1000",
"output": "285 286"
},
{
"input": "999",
"output": "284 286"
},
{
"input": "998",
"output": "284 286"
},
{
"input": "997",
"output": "284 286"
},
{
"input": "996",
"output": "284 286"
},
{
"input": "995",
"output": "284 285"
},
{
"input": "994",
"output": "284 284"
},
{
"input": "993",
"output": "283 284"
},
{
"input": "992",
"output": "282 284"
},
{
"input": "991",
"output": "282 284"
},
{
"input": "990",
"output": "282 284"
},
{
"input": "989",
"output": "282 284"
},
{
"input": "988",
"output": "282 283"
},
{
"input": "987",
"output": "282 282"
},
{
"input": "986",
"output": "281 282"
},
{
"input": "985",
"output": "280 282"
},
{
"input": "984",
"output": "280 282"
},
{
"input": "983",
"output": "280 282"
},
{
"input": "982",
"output": "280 282"
},
{
"input": "981",
"output": "280 281"
},
{
"input": "980",
"output": "280 280"
},
{
"input": "10000",
"output": "2856 2858"
},
{
"input": "9999",
"output": "2856 2858"
},
{
"input": "9998",
"output": "2856 2858"
},
{
"input": "9997",
"output": "2856 2857"
},
{
"input": "9996",
"output": "2856 2856"
},
{
"input": "9995",
"output": "2855 2856"
},
{
"input": "9994",
"output": "2854 2856"
},
{
"input": "9993",
"output": "2854 2856"
},
{
"input": "9992",
"output": "2854 2856"
},
{
"input": "9991",
"output": "2854 2856"
},
{
"input": "9990",
"output": "2854 2855"
},
{
"input": "9989",
"output": "2854 2854"
},
{
"input": "9988",
"output": "2853 2854"
},
{
"input": "9987",
"output": "2852 2854"
},
{
"input": "9986",
"output": "2852 2854"
},
{
"input": "9985",
"output": "2852 2854"
},
{
"input": "9984",
"output": "2852 2854"
},
{
"input": "9983",
"output": "2852 2853"
},
{
"input": "9982",
"output": "2852 2852"
},
{
"input": "9981",
"output": "2851 2852"
},
{
"input": "9980",
"output": "2850 2852"
},
{
"input": "100000",
"output": "28570 28572"
},
{
"input": "99999",
"output": "28570 28572"
},
{
"input": "99998",
"output": "28570 28572"
},
{
"input": "99997",
"output": "28570 28572"
},
{
"input": "99996",
"output": "28570 28571"
},
{
"input": "99995",
"output": "28570 28570"
},
{
"input": "99994",
"output": "28569 28570"
},
{
"input": "99993",
"output": "28568 28570"
},
{
"input": "99992",
"output": "28568 28570"
},
{
"input": "99991",
"output": "28568 28570"
},
{
"input": "99990",
"output": "28568 28570"
},
{
"input": "99989",
"output": "28568 28569"
},
{
"input": "99988",
"output": "28568 28568"
},
{
"input": "99987",
"output": "28567 28568"
},
{
"input": "99986",
"output": "28566 28568"
},
{
"input": "99985",
"output": "28566 28568"
},
{
"input": "99984",
"output": "28566 28568"
},
{
"input": "99983",
"output": "28566 28568"
},
{
"input": "99982",
"output": "28566 28567"
},
{
"input": "99981",
"output": "28566 28566"
},
{
"input": "99980",
"output": "28565 28566"
},
{
"input": "999999",
"output": "285714 285714"
},
{
"input": "999998",
"output": "285713 285714"
},
{
"input": "999997",
"output": "285712 285714"
},
{
"input": "999996",
"output": "285712 285714"
},
{
"input": "999995",
"output": "285712 285714"
},
{
"input": "999994",
"output": "285712 285714"
},
{
"input": "999993",
"output": "285712 285713"
},
{
"input": "999992",
"output": "285712 285712"
},
{
"input": "999991",
"output": "285711 285712"
},
{
"input": "999990",
"output": "285710 285712"
},
{
"input": "999989",
"output": "285710 285712"
},
{
"input": "999988",
"output": "285710 285712"
},
{
"input": "999987",
"output": "285710 285712"
},
{
"input": "999986",
"output": "285710 285711"
},
{
"input": "999985",
"output": "285710 285710"
},
{
"input": "999984",
"output": "285709 285710"
},
{
"input": "999983",
"output": "285708 285710"
},
{
"input": "999982",
"output": "285708 285710"
},
{
"input": "999981",
"output": "285708 285710"
},
{
"input": "999980",
"output": "285708 285710"
},
{
"input": "234123",
"output": "66892 66893"
},
{
"input": "234122",
"output": "66892 66892"
},
{
"input": "234121",
"output": "66891 66892"
},
{
"input": "234120",
"output": "66890 66892"
},
{
"input": "234119",
"output": "66890 66892"
},
{
"input": "234118",
"output": "66890 66892"
},
{
"input": "234117",
"output": "66890 66892"
},
{
"input": "234116",
"output": "66890 66891"
},
{
"input": "234115",
"output": "66890 66890"
},
{
"input": "234114",
"output": "66889 66890"
},
{
"input": "234113",
"output": "66888 66890"
},
{
"input": "234112",
"output": "66888 66890"
},
{
"input": "234111",
"output": "66888 66890"
},
{
"input": "234110",
"output": "66888 66890"
},
{
"input": "234109",
"output": "66888 66889"
},
{
"input": "234108",
"output": "66888 66888"
},
{
"input": "234107",
"output": "66887 66888"
},
{
"input": "234106",
"output": "66886 66888"
},
{
"input": "234105",
"output": "66886 66888"
},
{
"input": "234104",
"output": "66886 66888"
},
{
"input": "234103",
"output": "66886 66888"
},
{
"input": "868531",
"output": "248151 248152"
},
{
"input": "868530",
"output": "248150 248152"
},
{
"input": "868529",
"output": "248150 248152"
},
{
"input": "868528",
"output": "248150 248152"
},
{
"input": "868527",
"output": "248150 248152"
},
{
"input": "868526",
"output": "248150 248151"
},
{
"input": "868525",
"output": "248150 248150"
},
{
"input": "868524",
"output": "248149 248150"
},
{
"input": "868523",
"output": "248148 248150"
},
{
"input": "868522",
"output": "248148 248150"
},
{
"input": "868521",
"output": "248148 248150"
},
{
"input": "868520",
"output": "248148 248150"
},
{
"input": "868519",
"output": "248148 248149"
},
{
"input": "868518",
"output": "248148 248148"
},
{
"input": "868517",
"output": "248147 248148"
},
{
"input": "868516",
"output": "248146 248148"
},
{
"input": "868515",
"output": "248146 248148"
},
{
"input": "868514",
"output": "248146 248148"
},
{
"input": "868513",
"output": "248146 248148"
},
{
"input": "868512",
"output": "248146 248147"
},
{
"input": "868511",
"output": "248146 248146"
},
{
"input": "123413",
"output": "35260 35262"
},
{
"input": "123412",
"output": "35260 35262"
},
{
"input": "123411",
"output": "35260 35261"
},
{
"input": "123410",
"output": "35260 35260"
},
{
"input": "123409",
"output": "35259 35260"
},
{
"input": "123408",
"output": "35258 35260"
},
{
"input": "123407",
"output": "35258 35260"
},
{
"input": "123406",
"output": "35258 35260"
},
{
"input": "123405",
"output": "35258 35260"
},
{
"input": "123404",
"output": "35258 35259"
},
{
"input": "123403",
"output": "35258 35258"
},
{
"input": "123402",
"output": "35257 35258"
},
{
"input": "123401",
"output": "35256 35258"
},
{
"input": "123400",
"output": "35256 35258"
},
{
"input": "123399",
"output": "35256 35258"
},
{
"input": "123398",
"output": "35256 35258"
},
{
"input": "123397",
"output": "35256 35257"
},
{
"input": "123396",
"output": "35256 35256"
},
{
"input": "123395",
"output": "35255 35256"
},
{
"input": "123394",
"output": "35254 35256"
},
{
"input": "123393",
"output": "35254 35256"
},
{
"input": "15",
"output": "4 5"
}
] | 1,614,605,666
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 198
| 77
| 0
|
n = int(input())
if n >7:
adder = 0
if n % 7 == 6:
adder = 1
print(n // 7 * 2+adder,n // 7 * 2 + (min(n%7,2)))
elif n <=7:
print(max(n-5,0),min(2,n))
# min 14 - 5 = 9 > 9 // 7 * 2
|
Title: Holidays
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
On the planet Mars a year lasts exactly *n* days (there are no leap years on Mars). But Martians have the same weeks as earthlings — 5 work days and then 2 days off. Your task is to determine the minimum possible and the maximum possible number of days off per year on Mars.
Input Specification:
The first line of the input contains a positive integer *n* (1<=≤<=*n*<=≤<=1<=000<=000) — the number of days in a year on Mars.
Output Specification:
Print two integers — the minimum possible and the maximum possible number of days off per year on Mars.
Demo Input:
['14\n', '2\n']
Demo Output:
['4 4\n', '0 2\n']
Note:
In the first sample there are 14 days in a year on Mars, and therefore independently of the day a year starts with there will be exactly 4 days off .
In the second sample there are only 2 days in a year on Mars, and they can both be either work days or days off.
|
```python
n = int(input())
if n >7:
adder = 0
if n % 7 == 6:
adder = 1
print(n // 7 * 2+adder,n // 7 * 2 + (min(n%7,2)))
elif n <=7:
print(max(n-5,0),min(2,n))
# min 14 - 5 = 9 > 9 // 7 * 2
```
| 3
|
|
560
|
A
|
Currency System in Geraldion
|
PROGRAMMING
| 1,000
|
[
"implementation",
"sortings"
] | null | null |
A magic island Geraldion, where Gerald lives, has its own currency system. It uses banknotes of several values. But the problem is, the system is not perfect and sometimes it happens that Geraldionians cannot express a certain sum of money with any set of banknotes. Of course, they can use any number of banknotes of each value. Such sum is called unfortunate. Gerald wondered: what is the minimum unfortunate sum?
|
The first line contains number *n* (1<=≤<=*n*<=≤<=1000) — the number of values of the banknotes that used in Geraldion.
The second line contains *n* distinct space-separated numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=106) — the values of the banknotes.
|
Print a single line — the minimum unfortunate sum. If there are no unfortunate sums, print <=-<=1.
|
[
"5\n1 2 3 4 5\n"
] |
[
"-1\n"
] |
none
| 500
|
[
{
"input": "5\n1 2 3 4 5",
"output": "-1"
},
{
"input": "1\n2",
"output": "1"
},
{
"input": "10\n371054 506438 397130 1 766759 208409 769264 549213 641270 771837",
"output": "-1"
},
{
"input": "10\n635370 154890 909382 220996 276501 716105 538714 140162 171960 271264",
"output": "1"
},
{
"input": "50\n110876 835020 859879 999908 712969 788264 287153 921820 330355 499311 209594 484829 296329 940051 174081 931503 1 780512 390075 97866 124255 950067 697612 244256 782385 789882 37608 82153 399889 598867 416717 377988 535636 511221 792568 683271 131077 290194 496712 330720 587436 563481 645817 942562 654093 980561 382937 48293 582608 116156",
"output": "-1"
},
{
"input": "50\n474421 421097 217233 156339 27075 733996 281778 863492 184707 956857 288561 70997 393786 337382 663642 131184 637 273801 799870 295017 392338 842567 161819 297705 102013 930684 375703 838048 154915 138503 629056 256591 893619 19263 787927 684541 320265 841090 421423 490879 394582 493952 619247 633202 612928 50907 276653 407819 489945 153173",
"output": "1"
},
{
"input": "1\n1",
"output": "-1"
},
{
"input": "1\n1000000",
"output": "1"
},
{
"input": "2\n3 2",
"output": "1"
},
{
"input": "2\n2 3",
"output": "1"
}
] | 1,605,616,290
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 92
| 0
|
n = int(input())
arr = list(map(int,input().split()))
mini = arr[0]
for i in arr:
if mini>i:
mini = i
if mini==1:
print(-1)
else:
print(mini-1)
|
Title: Currency System in Geraldion
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A magic island Geraldion, where Gerald lives, has its own currency system. It uses banknotes of several values. But the problem is, the system is not perfect and sometimes it happens that Geraldionians cannot express a certain sum of money with any set of banknotes. Of course, they can use any number of banknotes of each value. Such sum is called unfortunate. Gerald wondered: what is the minimum unfortunate sum?
Input Specification:
The first line contains number *n* (1<=≤<=*n*<=≤<=1000) — the number of values of the banknotes that used in Geraldion.
The second line contains *n* distinct space-separated numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=106) — the values of the banknotes.
Output Specification:
Print a single line — the minimum unfortunate sum. If there are no unfortunate sums, print <=-<=1.
Demo Input:
['5\n1 2 3 4 5\n']
Demo Output:
['-1\n']
Note:
none
|
```python
n = int(input())
arr = list(map(int,input().split()))
mini = arr[0]
for i in arr:
if mini>i:
mini = i
if mini==1:
print(-1)
else:
print(mini-1)
```
| 0
|
|
486
|
A
|
Calculating Function
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
For a positive integer *n* let's define a function *f*:
*f*(*n*)<==<=<=-<=1<=+<=2<=-<=3<=+<=..<=+<=(<=-<=1)*n**n*
Your task is to calculate *f*(*n*) for a given integer *n*.
|
The single line contains the positive integer *n* (1<=≤<=*n*<=≤<=1015).
|
Print *f*(*n*) in a single line.
|
[
"4\n",
"5\n"
] |
[
"2\n",
"-3\n"
] |
*f*(4) = - 1 + 2 - 3 + 4 = 2
*f*(5) = - 1 + 2 - 3 + 4 - 5 = - 3
| 500
|
[
{
"input": "4",
"output": "2"
},
{
"input": "5",
"output": "-3"
},
{
"input": "1000000000",
"output": "500000000"
},
{
"input": "1000000001",
"output": "-500000001"
},
{
"input": "1000000000000000",
"output": "500000000000000"
},
{
"input": "100",
"output": "50"
},
{
"input": "101",
"output": "-51"
},
{
"input": "102",
"output": "51"
},
{
"input": "103",
"output": "-52"
},
{
"input": "104",
"output": "52"
},
{
"input": "105",
"output": "-53"
},
{
"input": "106",
"output": "53"
},
{
"input": "107",
"output": "-54"
},
{
"input": "108",
"output": "54"
},
{
"input": "109",
"output": "-55"
},
{
"input": "208170109961052",
"output": "104085054980526"
},
{
"input": "46017661651072",
"output": "23008830825536"
},
{
"input": "4018154546667",
"output": "-2009077273334"
},
{
"input": "288565475053",
"output": "-144282737527"
},
{
"input": "3052460231",
"output": "-1526230116"
},
{
"input": "29906716",
"output": "14953358"
},
{
"input": "87897701693326",
"output": "43948850846663"
},
{
"input": "8240",
"output": "4120"
},
{
"input": "577935",
"output": "-288968"
},
{
"input": "62",
"output": "31"
},
{
"input": "1",
"output": "-1"
},
{
"input": "2",
"output": "1"
},
{
"input": "9999999999999",
"output": "-5000000000000"
},
{
"input": "1000000000000",
"output": "500000000000"
},
{
"input": "99999999999999",
"output": "-50000000000000"
},
{
"input": "999999999999999",
"output": "-500000000000000"
},
{
"input": "42191359342",
"output": "21095679671"
},
{
"input": "100000000000000",
"output": "50000000000000"
},
{
"input": "145645214654154",
"output": "72822607327077"
},
{
"input": "4294967296",
"output": "2147483648"
},
{
"input": "3037000499",
"output": "-1518500250"
},
{
"input": "10000000000001",
"output": "-5000000000001"
},
{
"input": "100000017040846",
"output": "50000008520423"
},
{
"input": "98979894985999",
"output": "-49489947493000"
}
] | 1,694,586,942
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 2
| 1,000
| 0
|
a=int(input())
b=0
for i in range(0,a):
b+=(-1)**(i+1)*(i+1)
print(b)
|
Title: Calculating Function
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
For a positive integer *n* let's define a function *f*:
*f*(*n*)<==<=<=-<=1<=+<=2<=-<=3<=+<=..<=+<=(<=-<=1)*n**n*
Your task is to calculate *f*(*n*) for a given integer *n*.
Input Specification:
The single line contains the positive integer *n* (1<=≤<=*n*<=≤<=1015).
Output Specification:
Print *f*(*n*) in a single line.
Demo Input:
['4\n', '5\n']
Demo Output:
['2\n', '-3\n']
Note:
*f*(4) = - 1 + 2 - 3 + 4 = 2
*f*(5) = - 1 + 2 - 3 + 4 - 5 = - 3
|
```python
a=int(input())
b=0
for i in range(0,a):
b+=(-1)**(i+1)*(i+1)
print(b)
```
| 0
|
|
769
|
C
|
Cycle In Maze
|
PROGRAMMING
| 1,700
|
[
"*special",
"dfs and similar",
"graphs",
"greedy",
"shortest paths"
] | null | null |
The Robot is in a rectangular maze of size *n*<=×<=*m*. Each cell of the maze is either empty or occupied by an obstacle. The Robot can move between neighboring cells on the side left (the symbol "L"), right (the symbol "R"), up (the symbol "U") or down (the symbol "D"). The Robot can move to the cell only if it is empty. Initially, the Robot is in the empty cell.
Your task is to find lexicographically minimal Robot's cycle with length exactly *k*, which begins and ends in the cell where the Robot was initially. It is allowed to the Robot to visit any cell many times (including starting).
Consider that Robot's way is given as a line which consists of symbols "L", "R", "U" and "D". For example, if firstly the Robot goes down, then left, then right and up, it means that his way is written as "DLRU".
In this task you don't need to minimize the length of the way. Find the minimum lexicographical (in alphabet order as in the dictionary) line which satisfies requirements above.
|
The first line contains three integers *n*, *m* and *k* (1<=≤<=*n*,<=*m*<=≤<=1000, 1<=≤<=*k*<=≤<=106) — the size of the maze and the length of the cycle.
Each of the following *n* lines contains *m* symbols — the description of the maze. If the symbol equals to "." the current cell is empty. If the symbol equals to "*" the current cell is occupied by an obstacle. If the symbol equals to "X" then initially the Robot is in this cell and it is empty. It is guaranteed that the symbol "X" is found in the maze exactly once.
|
Print the lexicographically minimum Robot's way with the length exactly *k*, which starts and ends in the cell where initially Robot is. If there is no such way, print "IMPOSSIBLE"(without quotes).
|
[
"2 3 2\n.**\nX..\n",
"5 6 14\n..***.\n*...X.\n..*...\n..*.**\n....*.\n",
"3 3 4\n***\n*X*\n***\n"
] |
[
"RL\n",
"DLDDLLLRRRUURU\n",
"IMPOSSIBLE\n"
] |
In the first sample two cyclic ways for the Robot with the length 2 exist — "UD" and "RL". The second cycle is lexicographically less.
In the second sample the Robot should move in the following way: down, left, down, down, left, left, left, right, right, right, up, up, right, up.
In the third sample the Robot can't move to the neighboring cells, because they are occupied by obstacles.
| 1,500
|
[
{
"input": "2 3 2\n.**\nX..",
"output": "RL"
},
{
"input": "5 6 14\n..***.\n*...X.\n..*...\n..*.**\n....*.",
"output": "DLDDLLLRRRUURU"
},
{
"input": "3 3 4\n***\n*X*\n***",
"output": "IMPOSSIBLE"
},
{
"input": "1 1 1\nX",
"output": "IMPOSSIBLE"
},
{
"input": "1 2 2\nX.",
"output": "RL"
},
{
"input": "1 5 4\n.X**.",
"output": "LRLR"
},
{
"input": "1 10 1\n........X.",
"output": "IMPOSSIBLE"
},
{
"input": "1 20 10\n*.*..............*.X",
"output": "LRLRLRLRLR"
},
{
"input": "2 1 1\nX\n.",
"output": "IMPOSSIBLE"
},
{
"input": "2 2 2\nX*\n.*",
"output": "DU"
},
{
"input": "2 5 2\n.....\n*.*.X",
"output": "LR"
},
{
"input": "2 10 4\n******....\n*.****.*X*",
"output": "UDUD"
},
{
"input": "2 20 26\n.****..*.**.**.*....\n.*.*.*.*...*.****..X",
"output": "LLRLRLRLRLRLRLRLRLRLRLRLRR"
},
{
"input": "2 25 46\n.*...***X....*..*........\n.....*...**.**.*....*...*",
"output": "DLLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRRU"
},
{
"input": "5 1 2\n*\n.\nX\n*\n.",
"output": "UD"
},
{
"input": "5 2 8\n..\n.*\nX.\n..\n*.",
"output": "DRDUDULU"
},
{
"input": "5 5 12\n..**.\n***..\n..X*.\n....*\n**..*",
"output": "DDRLRLRLRLUU"
},
{
"input": "5 10 42\n..**.**.**\n......*..*\n..**...X..\n*.......*.\n......*.**",
"output": "DDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUDUU"
},
{
"input": "10 1 8\n.\n*\n*\n.\n.\nX\n*\n.\n*\n*",
"output": "UDUDUDUD"
},
{
"input": "10 2 16\n.*\n*.\n*.\n..\n**\nX.\n..\n*.\n..\n.*",
"output": "DRDDLDUDUDURUULU"
},
{
"input": "10 10 4\n*..*...***\nX...*.....\n***...**..\n..********\n.*.*......\n*.**..*...\n.**.**..**\n*.**.**..*\n**.****.*.\n...**..*.*",
"output": "RLRL"
},
{
"input": "20 1 12\n.\n.\n.\n*\n.\nX\n.\n.\n.\n.\n.\n.\n*\n*\n.\n.\n.\n.\n.\n.",
"output": "DDDDDDUUUUUU"
},
{
"input": "20 2 22\n.*\n**\n..\n**\n**\n..\n.*\n.*\n..\n..\n**\n**\n.*\n**\n..\n.*\n..\n..\nX*\n..",
"output": "DRLRLRLRLRLRLRLRLRLRLU"
},
{
"input": "20 10 116\n..........\n....*.....\n.......*..\n*.........\n*....*....\n*........*\n..........\n*.........\n.......*..\n...*..*...\n..........\n...*......\n..*.......\n.....**..*\n........*.\n........*.\n...*......\n.........*\n.....*.X..\n*......*.*",
"output": "LDLLLLLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRRRRRUR"
},
{
"input": "25 1 22\n.\n*\n*\n.\n*\n.\n.\n.\n.\n.\n.\n.\n.\n*\n.\n.\n.\n*\n.\n.\n.\n*\n.\nX\n.",
"output": "DUDUDUDUDUDUDUDUDUDUDU"
},
{
"input": "25 2 26\n.*\n*.\n..\n.*\n..\n*.\n.*\n.*\n.*\n..\n*.\n..\n..\n..\n..\n..\n*.\n.*\n.*\n..\n..\n.*\nX*\n..\n..",
"output": "DDRLRLRLRLRLRLRLRLRLRLRLUU"
},
{
"input": "25 5 22\n.....\n.....\n.....\n**...\n...*.\n...*.\n*..*.\n.....\n...**\n.*...\n.....\n*....\n*....\n*....\n*...X\n.....\n.*...\n...*.\n.*..*\n....*\n.....\n.....\n*....\n.....\n..*..",
"output": "DDDUDUDUDUDUDUDUDUDUUU"
},
{
"input": "25 10 38\n....*...**\n.........*\n.........*\n**...*....\n..........\n.*.....*.*\n***.*....*\n..*****.**\n*........*\n*.........\n.*..*.**.*\n.*....*...\n..*..**...\n...*.*.*.*\n.*.*.....*\n.*.X.*...*\n*...**...*\n..........\n.*..*.*.**\n*.*..**.*.\n*.....*..*\n...**.*...\n...*...*..\n...*......\n...*.....*",
"output": "DDDDDLDDDDLLRLRLRLRLRLRLRLRRUUUURUUUUU"
},
{
"input": "1 2 2\n.X",
"output": "LR"
},
{
"input": "2 1 2\n.\nX",
"output": "UD"
},
{
"input": "2 1 2\nX\n.",
"output": "DU"
},
{
"input": "2 1 2\n*\nX",
"output": "IMPOSSIBLE"
},
{
"input": "2 1 2\nX\n*",
"output": "IMPOSSIBLE"
},
{
"input": "1 2 2\nX*",
"output": "IMPOSSIBLE"
},
{
"input": "1 2 2\n*X",
"output": "IMPOSSIBLE"
},
{
"input": "1 1 1000000\nX",
"output": "IMPOSSIBLE"
},
{
"input": "1 1 1\nX",
"output": "IMPOSSIBLE"
},
{
"input": "1 1 2\nX",
"output": "IMPOSSIBLE"
}
] | 1,488,653,832
| 25,032
|
Python 3
|
RUNTIME_ERROR
|
PRETESTS
| 5
| 62
| 4,812,800
|
n, m, k = map(int, input().split())
if k % 2 != 0:
print('IMPOSSIBLE')
exit(0)
a=[list('*' + input()+'*') for i in range(n)]
a = [['*'] * (m + 1)] + a + [['*'] * (m + 1)]
x = 0
y = 0
while a[x][y] != 'X':
x += 1
if x == m+1:
x = 0
y += 1
res = ''
while True:
while True:
while True:
while True:
if a[x + 1][y] == '*' or len(res) >= k:
break
res += 'D'
x += 1
if a[x][y - 1] == '*' or len(res) >= k:
break
res += 'L'
y -= 1
if a[x][y + 1] == '*' or len(res) >= k:
break
res += 'R'
y += 1
if a[x - 1][y] == '*' or len(res) >= k:
break
res += 'U'
x -= 1
if res == '':
print('IMPOSSIBLE')
exit(0)
res1 = res[:k//2]
res2 = res1[::-1]
res = res1
inverse = {'U':'D', 'D':'U', 'L':'R', 'R':'L'}
for i in res2:
res += inverse[i]
print(res)
|
Title: Cycle In Maze
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Robot is in a rectangular maze of size *n*<=×<=*m*. Each cell of the maze is either empty or occupied by an obstacle. The Robot can move between neighboring cells on the side left (the symbol "L"), right (the symbol "R"), up (the symbol "U") or down (the symbol "D"). The Robot can move to the cell only if it is empty. Initially, the Robot is in the empty cell.
Your task is to find lexicographically minimal Robot's cycle with length exactly *k*, which begins and ends in the cell where the Robot was initially. It is allowed to the Robot to visit any cell many times (including starting).
Consider that Robot's way is given as a line which consists of symbols "L", "R", "U" and "D". For example, if firstly the Robot goes down, then left, then right and up, it means that his way is written as "DLRU".
In this task you don't need to minimize the length of the way. Find the minimum lexicographical (in alphabet order as in the dictionary) line which satisfies requirements above.
Input Specification:
The first line contains three integers *n*, *m* and *k* (1<=≤<=*n*,<=*m*<=≤<=1000, 1<=≤<=*k*<=≤<=106) — the size of the maze and the length of the cycle.
Each of the following *n* lines contains *m* symbols — the description of the maze. If the symbol equals to "." the current cell is empty. If the symbol equals to "*" the current cell is occupied by an obstacle. If the symbol equals to "X" then initially the Robot is in this cell and it is empty. It is guaranteed that the symbol "X" is found in the maze exactly once.
Output Specification:
Print the lexicographically minimum Robot's way with the length exactly *k*, which starts and ends in the cell where initially Robot is. If there is no such way, print "IMPOSSIBLE"(without quotes).
Demo Input:
['2 3 2\n.**\nX..\n', '5 6 14\n..***.\n*...X.\n..*...\n..*.**\n....*.\n', '3 3 4\n***\n*X*\n***\n']
Demo Output:
['RL\n', 'DLDDLLLRRRUURU\n', 'IMPOSSIBLE\n']
Note:
In the first sample two cyclic ways for the Robot with the length 2 exist — "UD" and "RL". The second cycle is lexicographically less.
In the second sample the Robot should move in the following way: down, left, down, down, left, left, left, right, right, right, up, up, right, up.
In the third sample the Robot can't move to the neighboring cells, because they are occupied by obstacles.
|
```python
n, m, k = map(int, input().split())
if k % 2 != 0:
print('IMPOSSIBLE')
exit(0)
a=[list('*' + input()+'*') for i in range(n)]
a = [['*'] * (m + 1)] + a + [['*'] * (m + 1)]
x = 0
y = 0
while a[x][y] != 'X':
x += 1
if x == m+1:
x = 0
y += 1
res = ''
while True:
while True:
while True:
while True:
if a[x + 1][y] == '*' or len(res) >= k:
break
res += 'D'
x += 1
if a[x][y - 1] == '*' or len(res) >= k:
break
res += 'L'
y -= 1
if a[x][y + 1] == '*' or len(res) >= k:
break
res += 'R'
y += 1
if a[x - 1][y] == '*' or len(res) >= k:
break
res += 'U'
x -= 1
if res == '':
print('IMPOSSIBLE')
exit(0)
res1 = res[:k//2]
res2 = res1[::-1]
res = res1
inverse = {'U':'D', 'D':'U', 'L':'R', 'R':'L'}
for i in res2:
res += inverse[i]
print(res)
```
| -1
|
|
739
|
A
|
Alyona and mex
|
PROGRAMMING
| 1,700
|
[
"constructive algorithms",
"greedy"
] | null | null |
Alyona's mother wants to present an array of *n* non-negative integers to Alyona. The array should be special.
Alyona is a capricious girl so after she gets the array, she inspects *m* of its subarrays. Subarray is a set of some subsequent elements of the array. The *i*-th subarray is described with two integers *l**i* and *r**i*, and its elements are *a*[*l**i*],<=*a*[*l**i*<=+<=1],<=...,<=*a*[*r**i*].
Alyona is going to find mex for each of the chosen subarrays. Among these *m* mexes the girl is going to find the smallest. She wants this minimum mex to be as large as possible.
You are to find an array *a* of *n* elements so that the minimum mex among those chosen by Alyona subarrays is as large as possible.
The mex of a set *S* is a minimum possible non-negative integer that is not in *S*.
|
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=105).
The next *m* lines contain information about the subarrays chosen by Alyona. The *i*-th of these lines contains two integers *l**i* and *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*), that describe the subarray *a*[*l**i*],<=*a*[*l**i*<=+<=1],<=...,<=*a*[*r**i*].
|
In the first line print single integer — the maximum possible minimum mex.
In the second line print *n* integers — the array *a*. All the elements in *a* should be between 0 and 109.
It is guaranteed that there is an optimal answer in which all the elements in *a* are between 0 and 109.
If there are multiple solutions, print any of them.
|
[
"5 3\n1 3\n2 5\n4 5\n",
"4 2\n1 4\n2 4\n"
] |
[
"2\n1 0 2 1 0\n",
"3\n5 2 0 1"
] |
The first example: the mex of the subarray (1, 3) is equal to 3, the mex of the subarray (2, 5) is equal to 3, the mex of the subarray (4, 5) is equal to 2 as well, thus the minumal mex among the subarrays chosen by Alyona is equal to 2.
| 500
|
[
{
"input": "5 3\n1 3\n2 5\n4 5",
"output": "2\n0 1 0 1 0"
},
{
"input": "4 2\n1 4\n2 4",
"output": "3\n0 1 2 0"
},
{
"input": "1 1\n1 1",
"output": "1\n0"
},
{
"input": "2 1\n2 2",
"output": "1\n0 0"
},
{
"input": "5 6\n2 4\n2 3\n1 4\n3 4\n2 5\n1 3",
"output": "2\n0 1 0 1 0"
},
{
"input": "8 3\n2 3\n2 8\n3 6",
"output": "2\n0 1 0 1 0 1 0 1"
},
{
"input": "10 10\n1 9\n4 8\n4 8\n5 9\n1 9\n3 8\n1 6\n1 9\n1 6\n6 9",
"output": "4\n0 1 2 3 0 1 2 3 0 1"
},
{
"input": "3 6\n1 3\n1 3\n1 1\n1 1\n3 3\n3 3",
"output": "1\n0 0 0"
},
{
"input": "3 3\n1 3\n2 2\n1 3",
"output": "1\n0 0 0"
},
{
"input": "6 8\n3 5\n3 6\n4 6\n2 5\n2 5\n1 3\n3 6\n3 5",
"output": "3\n0 1 2 0 1 2"
},
{
"input": "10 4\n4 10\n4 6\n6 8\n1 10",
"output": "3\n0 1 2 0 1 2 0 1 2 0"
},
{
"input": "9 1\n1 1",
"output": "1\n0 0 0 0 0 0 0 0 0"
},
{
"input": "3 8\n2 3\n1 3\n1 2\n2 3\n1 3\n2 2\n1 2\n1 2",
"output": "1\n0 0 0"
},
{
"input": "3 8\n1 2\n1 2\n1 1\n2 3\n2 3\n1 1\n1 3\n1 3",
"output": "1\n0 0 0"
},
{
"input": "7 3\n7 7\n3 7\n5 7",
"output": "1\n0 0 0 0 0 0 0"
},
{
"input": "9 9\n4 5\n5 8\n1 8\n4 8\n3 4\n7 8\n1 4\n7 8\n6 7",
"output": "2\n0 1 0 1 0 1 0 1 0"
},
{
"input": "10 10\n1 5\n7 10\n2 10\n2 5\n2 9\n5 9\n3 10\n4 9\n6 9\n2 7",
"output": "4\n0 1 2 3 0 1 2 3 0 1"
},
{
"input": "8 7\n5 8\n3 7\n1 8\n3 4\n2 8\n2 7\n4 6",
"output": "2\n0 1 0 1 0 1 0 1"
},
{
"input": "10 3\n2 4\n8 10\n4 6",
"output": "3\n0 1 2 0 1 2 0 1 2 0"
},
{
"input": "5 8\n3 4\n1 5\n3 4\n3 5\n3 4\n1 4\n1 5\n2 5",
"output": "2\n0 1 0 1 0"
},
{
"input": "9 4\n5 9\n3 8\n2 8\n1 4",
"output": "4\n0 1 2 3 0 1 2 3 0"
},
{
"input": "7 7\n2 4\n1 4\n3 6\n2 6\n5 7\n3 6\n1 4",
"output": "3\n0 1 2 0 1 2 0"
},
{
"input": "6 2\n4 6\n2 5",
"output": "3\n0 1 2 0 1 2"
},
{
"input": "7 9\n6 7\n1 2\n2 5\n4 7\n1 7\n5 6\n2 3\n6 7\n1 2",
"output": "2\n0 1 0 1 0 1 0"
}
] | 1,627,961,051
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 69
| 374
| 36,659,200
|
import sys
input=sys.stdin.readline
n,m=map(int,input().split())
seg=[list(map(int,input().split())) for i in range(m)]
min_l=10**18
for l,r in seg:
min_l=min(min_l,r-l+1)
ans=[i for i in range(min_l)]*((n+min_l-1)//min_l)
ans=ans[:n]
print(min_l)
print(*ans)
|
Title: Alyona and mex
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Alyona's mother wants to present an array of *n* non-negative integers to Alyona. The array should be special.
Alyona is a capricious girl so after she gets the array, she inspects *m* of its subarrays. Subarray is a set of some subsequent elements of the array. The *i*-th subarray is described with two integers *l**i* and *r**i*, and its elements are *a*[*l**i*],<=*a*[*l**i*<=+<=1],<=...,<=*a*[*r**i*].
Alyona is going to find mex for each of the chosen subarrays. Among these *m* mexes the girl is going to find the smallest. She wants this minimum mex to be as large as possible.
You are to find an array *a* of *n* elements so that the minimum mex among those chosen by Alyona subarrays is as large as possible.
The mex of a set *S* is a minimum possible non-negative integer that is not in *S*.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=105).
The next *m* lines contain information about the subarrays chosen by Alyona. The *i*-th of these lines contains two integers *l**i* and *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*), that describe the subarray *a*[*l**i*],<=*a*[*l**i*<=+<=1],<=...,<=*a*[*r**i*].
Output Specification:
In the first line print single integer — the maximum possible minimum mex.
In the second line print *n* integers — the array *a*. All the elements in *a* should be between 0 and 109.
It is guaranteed that there is an optimal answer in which all the elements in *a* are between 0 and 109.
If there are multiple solutions, print any of them.
Demo Input:
['5 3\n1 3\n2 5\n4 5\n', '4 2\n1 4\n2 4\n']
Demo Output:
['2\n1 0 2 1 0\n', '3\n5 2 0 1']
Note:
The first example: the mex of the subarray (1, 3) is equal to 3, the mex of the subarray (2, 5) is equal to 3, the mex of the subarray (4, 5) is equal to 2 as well, thus the minumal mex among the subarrays chosen by Alyona is equal to 2.
|
```python
import sys
input=sys.stdin.readline
n,m=map(int,input().split())
seg=[list(map(int,input().split())) for i in range(m)]
min_l=10**18
for l,r in seg:
min_l=min(min_l,r-l+1)
ans=[i for i in range(min_l)]*((n+min_l-1)//min_l)
ans=ans[:n]
print(min_l)
print(*ans)
```
| 3
|
|
735
|
B
|
Urbanization
|
PROGRAMMING
| 1,100
|
[
"greedy",
"number theory",
"sortings"
] | null | null |
Local authorities have heard a lot about combinatorial abilities of Ostap Bender so they decided to ask his help in the question of urbanization. There are *n* people who plan to move to the cities. The wealth of the *i* of them is equal to *a**i*. Authorities plan to build two cities, first for *n*1 people and second for *n*2 people. Of course, each of *n* candidates can settle in only one of the cities. Thus, first some subset of candidates of size *n*1 settle in the first city and then some subset of size *n*2 is chosen among the remaining candidates and the move to the second city. All other candidates receive an official refuse and go back home.
To make the statistic of local region look better in the eyes of their bosses, local authorities decided to pick subsets of candidates in such a way that the sum of arithmetic mean of wealth of people in each of the cities is as large as possible. Arithmetic mean of wealth in one city is the sum of wealth *a**i* among all its residents divided by the number of them (*n*1 or *n*2 depending on the city). The division should be done in real numbers without any rounding.
Please, help authorities find the optimal way to pick residents for two cities.
|
The first line of the input contains three integers *n*, *n*1 and *n*2 (1<=≤<=*n*,<=*n*1,<=*n*2<=≤<=100<=000, *n*1<=+<=*n*2<=≤<=*n*) — the number of candidates who want to move to the cities, the planned number of residents of the first city and the planned number of residents of the second city.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100<=000), the *i*-th of them is equal to the wealth of the *i*-th candidate.
|
Print one real value — the maximum possible sum of arithmetic means of wealth of cities' residents. You answer will be considered correct if its absolute or relative error does not exceed 10<=-<=6.
Namely: let's assume that your answer is *a*, and the answer of the jury is *b*. The checker program will consider your answer correct, if .
|
[
"2 1 1\n1 5\n",
"4 2 1\n1 4 2 3\n"
] |
[
"6.00000000\n",
"6.50000000\n"
] |
In the first sample, one of the optimal solutions is to move candidate 1 to the first city and candidate 2 to the second.
In the second sample, the optimal solution is to pick candidates 3 and 4 for the first city, and candidate 2 for the second one. Thus we obtain (*a*<sub class="lower-index">3</sub> + *a*<sub class="lower-index">4</sub>) / 2 + *a*<sub class="lower-index">2</sub> = (3 + 2) / 2 + 4 = 6.5
| 1,000
|
[
{
"input": "2 1 1\n1 5",
"output": "6.00000000"
},
{
"input": "4 2 1\n1 4 2 3",
"output": "6.50000000"
},
{
"input": "3 1 2\n1 2 3",
"output": "4.50000000"
},
{
"input": "10 4 6\n3 5 7 9 12 25 67 69 83 96",
"output": "88.91666667"
},
{
"input": "19 7 12\n1 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 16384 32768 65536 100000 100000",
"output": "47052.10714286"
},
{
"input": "100 9 6\n109 711 40 95 935 48 228 253 308 726 816 534 252 8 966 363 162 508 84 83 807 506 748 178 45 30 106 108 764 698 825 198 336 353 158 790 64 262 403 334 577 571 742 541 946 602 279 621 910 776 421 886 29 133 114 394 762 965 339 263 750 530 49 80 124 31 322 292 27 590 960 278 111 932 849 491 561 744 469 511 106 271 156 160 836 363 149 473 457 543 976 809 490 29 85 626 265 88 995 946",
"output": "1849.66666667"
},
{
"input": "69 6 63\n53475 22876 79144 6335 33763 79104 65441 45527 65847 94406 74670 43529 75330 19403 67629 56187 57949 23071 64910 54409 55348 18056 855 24961 50565 6622 26467 33989 22660 79469 41246 13965 79706 14422 16075 93378 81313 48173 13470 97348 2346 27452 59427 29925 29847 73823 32021 10988 24609 98855 90919 45939 17203 8439 43007 40138 55693 30314 71734 33458 66850 4011 20089 20546 92090 50842 78859 62756 40177",
"output": "135712.88888889"
},
{
"input": "69 6 9\n2612 17461 69001 33130 10662 85485 88195 45974 16712 81365 67119 87797 15559 20197 74716 92979 97268 49466 68603 48351 99905 35606 54242 98603 68232 54398 82637 49647 38979 46171 54680 23334 15892 92186 69670 29711 67999 2220 32317 717 70667 68262 86760 55720 97158 61122 7251 138 21022 27197 12691 59331 13576 66999 38332 13574 83484 66646 17704 33065 98583 80259 64631 16745 69431 40747 82089 82788 32739",
"output": "183129.44444444"
}
] | 1,596,559,558
| 2,147,483,647
|
PyPy 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
/*****************************************************
* Logged in : HABIBUR RAHMAN *
* Dept : COMPUTER SCIENCE AND ENGINEERING *
* *
* KHULNA UNIVERSITY OF ENGINEERING AND TECHNOLOGY *
******************************************************/
#include<bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef unsigned long long int ull;
typedef long double ld;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<ull> vull;
typedef vector<string> vs;
typedef pair<ll,ll> pll;
#define fast_io ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define all(v) v.begin(),v.end()
#define allr(v) v.rbegin(),v.rend()
#define rep(i,a,b) for(ll i=a;i<=b;i++)
#define rep0(i,n) for(ll i=0;i<n;i++)
#define rrep0(i, n) for(ll i=(ll)(n);i>0;--i)
#define rrep(i, a, b) for(ll i=(ll)(a);i>=(ll)(b);--i)
#define pb(z) emplace_back(z)
#define caset int t;cin>>t;for(int tt=1;tt<=t;tt++)
#define sz(X) ((ll)(X).size())
#define ms(v,val) memset((v),(val),sizeof((v)))
#define m_p(a,b) make_pair(a,b)
#define get_pos(c,x) (lower_bound(c.begin(),c.end(),x)-c.begin())
#define lcm(a,b) (a*b)/__gcd(a,b)
#define INF9 1000000010
#define INF18 4000000000000000010
#define bpll(a) __builtin_popcountll(a)
#define nl "\n"
#define ff first
#define ss second
//#define mod 1e9+7;
#define mx_el(v) *max_element(all(v));
#define mn_el(v) *min_element(all(v));
const double pi = 2*acos(0) ;
template<class T> void _R(T &x) { cin >> x; }
void _R(int &x) { scanf("%d", &x); }
void _R(int64_t &x) { scanf("%lld", &x); }
void _R(double &x) { scanf("%lf", &x); }
void _R(char &x) { scanf(" %c", &x); }
void _R(char *x) { scanf("%s", x); }
void R() {}
template<class T> void _W(const T &x) { cout << x; }
void _W(const int &x) { printf("%d", x); }
void _W(const int64_t &x) { printf("%lld", x); }
void _W(const double &x) { printf("%.16f", x); }
void _W(const char &x) { putchar(x); }
void _W(const char *x) { printf("%s", x); }
/*-------------------------some important functions--------------------------------------------------------------------*/
ull binpow(ull b,ull n){if(n==0)return 1;if(n==1)return b;if(n&1) return binpow(b,n-1)*b;else return binpow(b*b,n/2);}
ll bigmod(ll b,ll n,ll m){
if(n==0)return 1;
if(n%2==0){
ll x=bigmod(b,n/2,m)%m;
return (x*x)%m;
}
else return ((b%m)*bigmod(b,n-1,m)%m)%m;
}
/*--------------------------------------------x----------------------------------------------x--------------------------*/
void start(){
#ifndef ONLINE_JUDGE
freopen("input.txt" , "r" , stdin) ;
freopen("output.txt" , "w" , stdout) ;
#endif
}
bool is(ll n){
return ceil(sqrt(n))==floor(sqrt(n));
}
string ans="aeiou";
void solve(){
ll n;cin>>n;
if(!is(n))cout << "-1" << nl;
else{
ll i=0;
n=sqrt(n);
while(sz(ans)!=n){
ans+=ans[i%5];
i++;
}
for(ll i=0;i<sz(ans);i++){
for(ll j=i;j<(i+sz(ans));j++){
cout << ans[j%sz(ans)];
}
cout << nl;
}
}
}
int main() {
//fast_io
//start();
//caset{
//cout << "Case " << tt << ":" << nl;
solve();
//}
return 0;
}
|
Title: Urbanization
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Local authorities have heard a lot about combinatorial abilities of Ostap Bender so they decided to ask his help in the question of urbanization. There are *n* people who plan to move to the cities. The wealth of the *i* of them is equal to *a**i*. Authorities plan to build two cities, first for *n*1 people and second for *n*2 people. Of course, each of *n* candidates can settle in only one of the cities. Thus, first some subset of candidates of size *n*1 settle in the first city and then some subset of size *n*2 is chosen among the remaining candidates and the move to the second city. All other candidates receive an official refuse and go back home.
To make the statistic of local region look better in the eyes of their bosses, local authorities decided to pick subsets of candidates in such a way that the sum of arithmetic mean of wealth of people in each of the cities is as large as possible. Arithmetic mean of wealth in one city is the sum of wealth *a**i* among all its residents divided by the number of them (*n*1 or *n*2 depending on the city). The division should be done in real numbers without any rounding.
Please, help authorities find the optimal way to pick residents for two cities.
Input Specification:
The first line of the input contains three integers *n*, *n*1 and *n*2 (1<=≤<=*n*,<=*n*1,<=*n*2<=≤<=100<=000, *n*1<=+<=*n*2<=≤<=*n*) — the number of candidates who want to move to the cities, the planned number of residents of the first city and the planned number of residents of the second city.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100<=000), the *i*-th of them is equal to the wealth of the *i*-th candidate.
Output Specification:
Print one real value — the maximum possible sum of arithmetic means of wealth of cities' residents. You answer will be considered correct if its absolute or relative error does not exceed 10<=-<=6.
Namely: let's assume that your answer is *a*, and the answer of the jury is *b*. The checker program will consider your answer correct, if .
Demo Input:
['2 1 1\n1 5\n', '4 2 1\n1 4 2 3\n']
Demo Output:
['6.00000000\n', '6.50000000\n']
Note:
In the first sample, one of the optimal solutions is to move candidate 1 to the first city and candidate 2 to the second.
In the second sample, the optimal solution is to pick candidates 3 and 4 for the first city, and candidate 2 for the second one. Thus we obtain (*a*<sub class="lower-index">3</sub> + *a*<sub class="lower-index">4</sub>) / 2 + *a*<sub class="lower-index">2</sub> = (3 + 2) / 2 + 4 = 6.5
|
```python
/*****************************************************
* Logged in : HABIBUR RAHMAN *
* Dept : COMPUTER SCIENCE AND ENGINEERING *
* *
* KHULNA UNIVERSITY OF ENGINEERING AND TECHNOLOGY *
******************************************************/
#include<bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef unsigned long long int ull;
typedef long double ld;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<ull> vull;
typedef vector<string> vs;
typedef pair<ll,ll> pll;
#define fast_io ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define all(v) v.begin(),v.end()
#define allr(v) v.rbegin(),v.rend()
#define rep(i,a,b) for(ll i=a;i<=b;i++)
#define rep0(i,n) for(ll i=0;i<n;i++)
#define rrep0(i, n) for(ll i=(ll)(n);i>0;--i)
#define rrep(i, a, b) for(ll i=(ll)(a);i>=(ll)(b);--i)
#define pb(z) emplace_back(z)
#define caset int t;cin>>t;for(int tt=1;tt<=t;tt++)
#define sz(X) ((ll)(X).size())
#define ms(v,val) memset((v),(val),sizeof((v)))
#define m_p(a,b) make_pair(a,b)
#define get_pos(c,x) (lower_bound(c.begin(),c.end(),x)-c.begin())
#define lcm(a,b) (a*b)/__gcd(a,b)
#define INF9 1000000010
#define INF18 4000000000000000010
#define bpll(a) __builtin_popcountll(a)
#define nl "\n"
#define ff first
#define ss second
//#define mod 1e9+7;
#define mx_el(v) *max_element(all(v));
#define mn_el(v) *min_element(all(v));
const double pi = 2*acos(0) ;
template<class T> void _R(T &x) { cin >> x; }
void _R(int &x) { scanf("%d", &x); }
void _R(int64_t &x) { scanf("%lld", &x); }
void _R(double &x) { scanf("%lf", &x); }
void _R(char &x) { scanf(" %c", &x); }
void _R(char *x) { scanf("%s", x); }
void R() {}
template<class T> void _W(const T &x) { cout << x; }
void _W(const int &x) { printf("%d", x); }
void _W(const int64_t &x) { printf("%lld", x); }
void _W(const double &x) { printf("%.16f", x); }
void _W(const char &x) { putchar(x); }
void _W(const char *x) { printf("%s", x); }
/*-------------------------some important functions--------------------------------------------------------------------*/
ull binpow(ull b,ull n){if(n==0)return 1;if(n==1)return b;if(n&1) return binpow(b,n-1)*b;else return binpow(b*b,n/2);}
ll bigmod(ll b,ll n,ll m){
if(n==0)return 1;
if(n%2==0){
ll x=bigmod(b,n/2,m)%m;
return (x*x)%m;
}
else return ((b%m)*bigmod(b,n-1,m)%m)%m;
}
/*--------------------------------------------x----------------------------------------------x--------------------------*/
void start(){
#ifndef ONLINE_JUDGE
freopen("input.txt" , "r" , stdin) ;
freopen("output.txt" , "w" , stdout) ;
#endif
}
bool is(ll n){
return ceil(sqrt(n))==floor(sqrt(n));
}
string ans="aeiou";
void solve(){
ll n;cin>>n;
if(!is(n))cout << "-1" << nl;
else{
ll i=0;
n=sqrt(n);
while(sz(ans)!=n){
ans+=ans[i%5];
i++;
}
for(ll i=0;i<sz(ans);i++){
for(ll j=i;j<(i+sz(ans));j++){
cout << ans[j%sz(ans)];
}
cout << nl;
}
}
}
int main() {
//fast_io
//start();
//caset{
//cout << "Case " << tt << ":" << nl;
solve();
//}
return 0;
}
```
| -1
|
|
215
|
A
|
Bicycle Chain
|
PROGRAMMING
| 900
|
[
"brute force",
"implementation"
] | null | null |
Vasya's bicycle chain drive consists of two parts: *n* stars are attached to the pedal axle, *m* stars are attached to the rear wheel axle. The chain helps to rotate the rear wheel by transmitting the pedal rotation.
We know that the *i*-th star on the pedal axle has *a**i* (0<=<<=*a*1<=<<=*a*2<=<<=...<=<<=*a**n*) teeth, and the *j*-th star on the rear wheel axle has *b**j* (0<=<<=*b*1<=<<=*b*2<=<<=...<=<<=*b**m*) teeth. Any pair (*i*,<=*j*) (1<=≤<=*i*<=≤<=*n*; 1<=≤<=*j*<=≤<=*m*) is called a gear and sets the indexes of stars to which the chain is currently attached. Gear (*i*,<=*j*) has a gear ratio, equal to the value .
Since Vasya likes integers, he wants to find such gears (*i*,<=*j*), that their ratios are integers. On the other hand, Vasya likes fast driving, so among all "integer" gears (*i*,<=*j*) he wants to choose a gear with the maximum ratio. Help him to find the number of such gears.
In the problem, fraction denotes division in real numbers, that is, no rounding is performed.
|
The first input line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of stars on the bicycle's pedal axle. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104) in the order of strict increasing.
The third input line contains integer *m* (1<=≤<=*m*<=≤<=50) — the number of stars on the rear wheel axle. The fourth line contains *m* integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=104) in the order of strict increasing.
It is guaranteed that there exists at least one gear (*i*,<=*j*), that its gear ratio is an integer. The numbers on the lines are separated by spaces.
|
Print the number of "integer" gears with the maximum ratio among all "integer" gears.
|
[
"2\n4 5\n3\n12 13 15\n",
"4\n1 2 3 4\n5\n10 11 12 13 14\n"
] |
[
"2\n",
"1\n"
] |
In the first sample the maximum "integer" gear ratio equals 3. There are two gears that have such gear ratio. For one of them *a*<sub class="lower-index">1</sub> = 4, *b*<sub class="lower-index">1</sub> = 12, and for the other *a*<sub class="lower-index">2</sub> = 5, *b*<sub class="lower-index">3</sub> = 15.
| 500
|
[
{
"input": "2\n4 5\n3\n12 13 15",
"output": "2"
},
{
"input": "4\n1 2 3 4\n5\n10 11 12 13 14",
"output": "1"
},
{
"input": "1\n1\n1\n1",
"output": "1"
},
{
"input": "2\n1 2\n1\n1",
"output": "1"
},
{
"input": "1\n1\n2\n1 2",
"output": "1"
},
{
"input": "4\n3 7 11 13\n4\n51 119 187 221",
"output": "4"
},
{
"input": "4\n2 3 4 5\n3\n1 2 3",
"output": "2"
},
{
"input": "10\n6 12 13 20 48 53 74 92 96 97\n10\n1 21 32 36 47 54 69 75 95 97",
"output": "1"
},
{
"input": "10\n5 9 10 14 15 17 19 22 24 26\n10\n2 11 17 19 21 22 24 25 27 28",
"output": "1"
},
{
"input": "10\n24 53 56 126 354 432 442 740 795 856\n10\n273 438 494 619 689 711 894 947 954 958",
"output": "1"
},
{
"input": "10\n3 4 6 7 8 10 14 16 19 20\n10\n3 4 5 7 8 10 15 16 18 20",
"output": "1"
},
{
"input": "10\n1 6 8 14 15 17 25 27 34 39\n10\n1 8 16 17 19 22 32 39 44 50",
"output": "1"
},
{
"input": "10\n5 21 22 23 25 32 35 36 38 39\n10\n3 7 8 9 18 21 23 24 36 38",
"output": "4"
},
{
"input": "50\n5 8 13 16 19 20 21 22 24 27 28 29 30 32 33 34 35 43 45 48 50 51 54 55 58 59 60 61 62 65 70 71 72 76 78 79 80 81 83 84 85 87 89 91 92 94 97 98 99 100\n50\n2 3 5 6 7 10 15 16 17 20 23 28 29 30 31 34 36 37 40 42 45 46 48 54 55 56 58 59 61 62 69 70 71 72 75 76 78 82 84 85 86 87 88 89 90 91 92 97 99 100",
"output": "1"
},
{
"input": "50\n3 5 6 8 9 11 13 19 21 23 24 32 34 35 42 50 51 52 56 58 59 69 70 72 73 75 76 77 78 80 83 88 90 95 96 100 101 102 108 109 113 119 124 135 138 141 142 143 145 150\n50\n5 8 10 11 18 19 23 30 35 43 51 53 55 58 63 68 69 71 77 78 79 82 83 86 88 89 91 92 93 94 96 102 103 105 109 110 113 114 116 123 124 126 127 132 133 135 136 137 142 149",
"output": "1"
},
{
"input": "50\n6 16 24 25 27 33 36 40 51 60 62 65 71 72 75 77 85 87 91 93 98 102 103 106 117 118 120 121 122 123 125 131 134 136 143 148 155 157 160 161 164 166 170 178 184 187 188 192 194 197\n50\n5 9 17 23 27 34 40 44 47 59 62 70 81 82 87 88 89 90 98 101 102 110 113 114 115 116 119 122 124 128 130 137 138 140 144 150 152 155 159 164 166 169 171 175 185 186 187 189 190 193",
"output": "1"
},
{
"input": "50\n14 22 23 31 32 35 48 63 76 79 88 97 101 102 103 104 106 113 114 115 116 126 136 138 145 152 155 156 162 170 172 173 179 180 182 203 208 210 212 222 226 229 231 232 235 237 245 246 247 248\n50\n2 5 6 16 28 44 45 46 54 55 56 63 72 80 87 93 94 96 97 100 101 103 132 135 140 160 164 165 167 168 173 180 182 185 186 192 194 198 199 202 203 211 213 216 217 227 232 233 236 245",
"output": "1"
},
{
"input": "50\n14 19 33 35 38 41 51 54 69 70 71 73 76 80 84 94 102 104 105 106 107 113 121 128 131 168 180 181 187 191 195 201 205 207 210 216 220 238 249 251 263 271 272 275 281 283 285 286 291 294\n50\n2 3 5 20 21 35 38 40 43 48 49 52 55 64 73 77 82 97 109 113 119 121 125 132 137 139 145 146 149 180 182 197 203 229 234 241 244 251 264 271 274 281 284 285 287 291 292 293 294 298",
"output": "1"
},
{
"input": "50\n2 4 5 16 18 19 22 23 25 26 34 44 48 54 67 79 80 84 92 110 116 133 138 154 163 171 174 202 205 218 228 229 234 245 247 249 250 263 270 272 274 275 277 283 289 310 312 334 339 342\n50\n1 5 17 18 25 37 46 47 48 59 67 75 80 83 84 107 115 122 137 141 159 162 175 180 184 204 221 224 240 243 247 248 249 258 259 260 264 266 269 271 274 293 294 306 329 330 334 335 342 350",
"output": "1"
},
{
"input": "50\n6 9 11 21 28 39 42 56 60 63 81 88 91 95 105 110 117 125 149 165 174 176 185 189 193 196 205 231 233 268 278 279 281 286 289 292 298 303 305 306 334 342 350 353 361 371 372 375 376 378\n50\n6 17 20 43 45 52 58 59 82 83 88 102 111 118 121 131 145 173 190 191 200 216 224 225 232 235 243 256 260 271 290 291 321 322 323 329 331 333 334 341 343 348 351 354 356 360 366 379 387 388",
"output": "1"
},
{
"input": "10\n17 239 443 467 661 1069 1823 2333 3767 4201\n20\n51 83 97 457 593 717 997 1329 1401 1459 1471 1983 2371 2539 3207 3251 3329 5469 6637 6999",
"output": "8"
},
{
"input": "20\n179 359 401 467 521 601 919 941 1103 1279 1709 1913 1949 2003 2099 2143 2179 2213 2399 4673\n20\n151 181 191 251 421 967 1109 1181 1249 1447 1471 1553 1619 2327 2551 2791 3049 3727 6071 7813",
"output": "3"
},
{
"input": "20\n79 113 151 709 809 983 1291 1399 1409 1429 2377 2659 2671 2897 3217 3511 3557 3797 3823 4363\n10\n19 101 659 797 1027 1963 2129 2971 3299 9217",
"output": "3"
},
{
"input": "30\n19 47 109 179 307 331 389 401 461 509 547 569 617 853 883 1249 1361 1381 1511 1723 1741 1783 2459 2531 2621 3533 3821 4091 5557 6217\n20\n401 443 563 941 967 997 1535 1567 1655 1747 1787 1945 1999 2251 2305 2543 2735 4415 6245 7555",
"output": "8"
},
{
"input": "30\n3 43 97 179 257 313 353 359 367 389 397 457 547 599 601 647 1013 1021 1063 1433 1481 1531 1669 3181 3373 3559 3769 4157 4549 5197\n50\n13 15 17 19 29 79 113 193 197 199 215 223 271 293 359 485 487 569 601 683 895 919 941 967 1283 1285 1289 1549 1565 1765 1795 1835 1907 1931 1945 1985 1993 2285 2731 2735 2995 3257 4049 4139 5105 5315 7165 7405 7655 8345",
"output": "20"
},
{
"input": "50\n11 17 23 53 59 109 137 149 173 251 353 379 419 421 439 503 593 607 661 773 821 877 941 997 1061 1117 1153 1229 1289 1297 1321 1609 1747 2311 2389 2543 2693 3041 3083 3137 3181 3209 3331 3373 3617 3767 4201 4409 4931 6379\n50\n55 59 67 73 85 89 101 115 211 263 295 353 545 599 607 685 739 745 997 1031 1255 1493 1523 1667 1709 1895 1949 2161 2195 2965 3019 3035 3305 3361 3373 3673 3739 3865 3881 4231 4253 4385 4985 5305 5585 5765 6145 6445 8045 8735",
"output": "23"
},
{
"input": "5\n33 78 146 3055 4268\n5\n2211 2584 5226 9402 9782",
"output": "3"
},
{
"input": "5\n35 48 52 86 8001\n10\n332 3430 3554 4704 4860 5096 6215 7583 8228 8428",
"output": "4"
},
{
"input": "10\n97 184 207 228 269 2084 4450 6396 7214 9457\n16\n338 1179 1284 1545 1570 2444 3167 3395 3397 5550 6440 7245 7804 7980 9415 9959",
"output": "5"
},
{
"input": "30\n25 30 41 57 58 62 70 72 76 79 84 85 88 91 98 101 104 109 119 129 136 139 148 151 926 1372 3093 3936 5423 7350\n25\n1600 1920 2624 3648 3712 3968 4480 4608 4864 5056 5376 5440 5632 5824 6272 6464 6656 6934 6976 7616 8256 8704 8896 9472 9664",
"output": "24"
},
{
"input": "5\n33 78 146 3055 4268\n5\n2211 2584 5226 9402 9782",
"output": "3"
},
{
"input": "5\n35 48 52 86 8001\n10\n332 3430 3554 4704 4860 5096 6215 7583 8228 8428",
"output": "4"
},
{
"input": "10\n97 184 207 228 269 2084 4450 6396 7214 9457\n16\n338 1179 1284 1545 1570 2444 3167 3395 3397 5550 6440 7245 7804 7980 9415 9959",
"output": "5"
},
{
"input": "30\n25 30 41 57 58 62 70 72 76 79 84 85 88 91 98 101 104 109 119 129 136 139 148 151 926 1372 3093 3936 5423 7350\n25\n1600 1920 2624 3648 3712 3968 4480 4608 4864 5056 5376 5440 5632 5824 6272 6464 6656 6934 6976 7616 8256 8704 8896 9472 9664",
"output": "24"
},
{
"input": "47\n66 262 357 457 513 530 538 540 592 691 707 979 1015 1242 1246 1667 1823 1886 1963 2133 2649 2679 2916 2949 3413 3523 3699 3958 4393 4922 5233 5306 5799 6036 6302 6629 7208 7282 7315 7822 7833 7927 8068 8150 8870 8962 9987\n39\n167 199 360 528 1515 1643 1986 1988 2154 2397 2856 3552 3656 3784 3980 4096 4104 4240 4320 4736 4951 5266 5656 5849 5850 6169 6517 6875 7244 7339 7689 7832 8120 8716 9503 9509 9933 9936 9968",
"output": "12"
},
{
"input": "1\n94\n50\n423 446 485 1214 1468 1507 1853 1930 1999 2258 2271 2285 2425 2543 2715 2743 2992 3196 4074 4108 4448 4475 4652 5057 5250 5312 5356 5375 5731 5986 6298 6501 6521 7146 7255 7276 7332 7481 7998 8141 8413 8665 8908 9221 9336 9491 9504 9677 9693 9706",
"output": "1"
},
{
"input": "50\n51 67 75 186 194 355 512 561 720 876 1077 1221 1503 1820 2153 2385 2568 2608 2937 2969 3271 3311 3481 4081 4093 4171 4255 4256 4829 5020 5192 5636 5817 6156 6712 6717 7153 7436 7608 7612 7866 7988 8264 8293 8867 9311 9879 9882 9889 9908\n1\n5394",
"output": "1"
},
{
"input": "50\n26 367 495 585 675 789 855 1185 1312 1606 2037 2241 2587 2612 2628 2807 2873 2924 3774 4067 4376 4668 4902 5001 5082 5100 5104 5209 5345 5515 5661 5777 5902 5907 6155 6323 6675 6791 7503 8159 8207 8254 8740 8848 8855 8933 9069 9164 9171 9586\n5\n1557 6246 7545 8074 8284",
"output": "1"
},
{
"input": "5\n25 58 91 110 2658\n50\n21 372 909 1172 1517 1554 1797 1802 1843 1977 2006 2025 2137 2225 2317 2507 2645 2754 2919 3024 3202 3212 3267 3852 4374 4487 4553 4668 4883 4911 4916 5016 5021 5068 5104 5162 5683 5856 6374 6871 7333 7531 8099 8135 8173 8215 8462 8776 9433 9790",
"output": "4"
},
{
"input": "45\n37 48 56 59 69 70 79 83 85 86 99 114 131 134 135 145 156 250 1739 1947 2116 2315 2449 3104 3666 4008 4406 4723 4829 5345 5836 6262 6296 6870 7065 7110 7130 7510 7595 8092 8442 8574 9032 9091 9355\n50\n343 846 893 1110 1651 1837 2162 2331 2596 3012 3024 3131 3294 3394 3528 3717 3997 4125 4347 4410 4581 4977 5030 5070 5119 5229 5355 5413 5418 5474 5763 5940 6151 6161 6164 6237 6506 6519 6783 7182 7413 7534 8069 8253 8442 8505 9135 9308 9828 9902",
"output": "17"
},
{
"input": "50\n17 20 22 28 36 38 46 47 48 50 52 57 58 62 63 69 70 74 75 78 79 81 82 86 87 90 93 95 103 202 292 442 1756 1769 2208 2311 2799 2957 3483 4280 4324 4932 5109 5204 6225 6354 6561 7136 8754 9670\n40\n68 214 957 1649 1940 2078 2134 2716 3492 3686 4462 4559 4656 4756 4850 5044 5490 5529 5592 5626 6014 6111 6693 6790 7178 7275 7566 7663 7702 7857 7954 8342 8511 8730 8957 9021 9215 9377 9445 9991",
"output": "28"
},
{
"input": "39\n10 13 21 25 36 38 47 48 58 64 68 69 73 79 86 972 2012 2215 2267 2503 3717 3945 4197 4800 5266 6169 6612 6824 7023 7322 7582 7766 8381 8626 8879 9079 9088 9838 9968\n50\n432 877 970 1152 1202 1223 1261 1435 1454 1578 1843 1907 2003 2037 2183 2195 2215 2425 3065 3492 3615 3637 3686 3946 4189 4415 4559 4656 4665 4707 4886 4887 5626 5703 5955 6208 6521 6581 6596 6693 6985 7013 7081 7343 7663 8332 8342 8637 9207 9862",
"output": "15"
},
{
"input": "50\n7 144 269 339 395 505 625 688 709 950 1102 1152 1350 1381 1641 1830 1977 1999 2093 2180 2718 3308 3574 4168 4232 4259 4393 4689 4982 5154 5476 5581 5635 5721 6159 6302 6741 7010 7152 7315 7417 7482 8116 8239 8640 9347 9395 9614 9661 9822\n20\n84 162 292 1728 1866 2088 3228 3470 4068 5318 5470 6060 6380 6929 7500 8256 8399 8467 8508 9691",
"output": "8"
},
{
"input": "50\n159 880 1070 1139 1358 1608 1691 1841 2073 2171 2213 2597 2692 2759 2879 2931 3173 3217 3441 4201 4878 5106 5129 5253 5395 5647 5968 6019 6130 6276 6286 6330 6409 6728 7488 7713 7765 7828 7899 8064 8264 8457 8483 8685 8900 8946 8965 9133 9187 9638\n45\n57 159 1070 1139 1391 1608 1691 1841 2171 2213 2692 2759 2931 3173 3217 3441 4201 4878 5106 5129 5253 5647 5968 6130 6276 6286 6409 7488 7694 7713 7765 7828 7899 8003 8064 8081 8244 8264 8685 8900 8946 8965 9133 9638 9673",
"output": "38"
},
{
"input": "3\n3 4 5\n3\n6 20 25",
"output": "2"
},
{
"input": "4\n2 3 5 8\n4\n2 6 8 10",
"output": "1"
},
{
"input": "4\n3 5 7 11\n4\n3 5 7 22",
"output": "1"
},
{
"input": "2\n2 3\n3\n20 30 50",
"output": "1"
},
{
"input": "3\n1 2 3\n4\n2 4 6 49",
"output": "1"
},
{
"input": "2\n4 5\n3\n12 15 20",
"output": "1"
},
{
"input": "3\n2 5 7\n3\n4 5 7",
"output": "1"
},
{
"input": "3\n3 5 8\n3\n6 8 10",
"output": "2"
},
{
"input": "2\n2 3\n4\n4 6 9 33",
"output": "1"
},
{
"input": "2\n2 3\n4\n4 6 21 40",
"output": "1"
},
{
"input": "3\n4 9 10\n3\n8 9 10",
"output": "1"
},
{
"input": "5\n1 5 6 9 51\n5\n5 12 18 27 10000",
"output": "1"
},
{
"input": "13\n1 2 3 4 5 6 7 8 9 10 11 12 13\n1\n14",
"output": "1"
}
] | 1,621,078,336
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 57
| 154
| 0
|
g=[]
n=int(input())
x=[int(x) for x in input().split()]
m=int(input())
y=[int(y) for y in input().split()]
for i in range(m):
for j in range(n):
if y[i]%x[j]==0:
g.append(y[i]//x[j])
ma=max(g)
print(g.count(ma))
|
Title: Bicycle Chain
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya's bicycle chain drive consists of two parts: *n* stars are attached to the pedal axle, *m* stars are attached to the rear wheel axle. The chain helps to rotate the rear wheel by transmitting the pedal rotation.
We know that the *i*-th star on the pedal axle has *a**i* (0<=<<=*a*1<=<<=*a*2<=<<=...<=<<=*a**n*) teeth, and the *j*-th star on the rear wheel axle has *b**j* (0<=<<=*b*1<=<<=*b*2<=<<=...<=<<=*b**m*) teeth. Any pair (*i*,<=*j*) (1<=≤<=*i*<=≤<=*n*; 1<=≤<=*j*<=≤<=*m*) is called a gear and sets the indexes of stars to which the chain is currently attached. Gear (*i*,<=*j*) has a gear ratio, equal to the value .
Since Vasya likes integers, he wants to find such gears (*i*,<=*j*), that their ratios are integers. On the other hand, Vasya likes fast driving, so among all "integer" gears (*i*,<=*j*) he wants to choose a gear with the maximum ratio. Help him to find the number of such gears.
In the problem, fraction denotes division in real numbers, that is, no rounding is performed.
Input Specification:
The first input line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of stars on the bicycle's pedal axle. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104) in the order of strict increasing.
The third input line contains integer *m* (1<=≤<=*m*<=≤<=50) — the number of stars on the rear wheel axle. The fourth line contains *m* integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=104) in the order of strict increasing.
It is guaranteed that there exists at least one gear (*i*,<=*j*), that its gear ratio is an integer. The numbers on the lines are separated by spaces.
Output Specification:
Print the number of "integer" gears with the maximum ratio among all "integer" gears.
Demo Input:
['2\n4 5\n3\n12 13 15\n', '4\n1 2 3 4\n5\n10 11 12 13 14\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample the maximum "integer" gear ratio equals 3. There are two gears that have such gear ratio. For one of them *a*<sub class="lower-index">1</sub> = 4, *b*<sub class="lower-index">1</sub> = 12, and for the other *a*<sub class="lower-index">2</sub> = 5, *b*<sub class="lower-index">3</sub> = 15.
|
```python
g=[]
n=int(input())
x=[int(x) for x in input().split()]
m=int(input())
y=[int(y) for y in input().split()]
for i in range(m):
for j in range(n):
if y[i]%x[j]==0:
g.append(y[i]//x[j])
ma=max(g)
print(g.count(ma))
```
| 3
|
|
274
|
A
|
k-Multiple Free Set
|
PROGRAMMING
| 1,500
|
[
"binary search",
"greedy",
"sortings"
] | null | null |
A *k*-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by *k*. That is, there are no two integers *x* and *y* (*x*<=<<=*y*) from the set, such that *y*<==<=*x*·*k*.
You're given a set of *n* distinct positive integers. Your task is to find the size of it's largest *k*-multiple free subset.
|
The first line of the input contains two integers *n* and *k* (1<=≤<=*n*<=≤<=105,<=1<=≤<=*k*<=≤<=109). The next line contains a list of *n* distinct positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
All the numbers in the lines are separated by single spaces.
|
On the only line of the output print the size of the largest *k*-multiple free subset of {*a*1,<=*a*2,<=...,<=*a**n*}.
|
[
"6 2\n2 3 6 5 4 10\n"
] |
[
"3\n"
] |
In the sample input one of the possible maximum 2-multiple free subsets is {4, 5, 6}.
| 500
|
[
{
"input": "6 2\n2 3 6 5 4 10",
"output": "3"
},
{
"input": "10 2\n1 2 3 4 5 6 7 8 9 10",
"output": "6"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "100 2\n191 17 61 40 77 95 128 88 26 69 79 10 131 106 142 152 68 39 182 53 83 81 6 89 65 148 33 22 5 47 107 121 52 163 150 158 189 118 75 180 177 176 112 167 140 184 29 166 25 46 169 145 187 123 196 18 115 126 155 100 63 58 159 19 173 113 133 60 130 161 76 157 93 199 50 97 15 67 109 164 99 149 3 137 153 136 56 43 103 170 13 183 194 72 9 181 86 30 91 36",
"output": "79"
},
{
"input": "100 3\n13 38 137 24 46 192 33 8 170 141 118 57 198 133 112 176 40 36 91 130 166 72 123 28 82 180 134 52 64 107 97 79 199 184 158 22 181 163 98 7 88 41 73 87 167 109 15 173 153 70 50 119 139 56 17 152 84 161 11 116 31 187 143 196 27 102 132 126 149 63 146 168 67 48 53 120 20 105 155 10 128 47 23 6 94 3 113 65 44 179 189 99 75 34 111 193 60 145 171 77",
"output": "87"
},
{
"input": "12 400000000\n1 400000000 800000000 2 3 4 5 6 7 8 9 10",
"output": "10"
},
{
"input": "3 1\n1 2 3",
"output": "3"
},
{
"input": "1 1\n1000000000",
"output": "1"
},
{
"input": "10 1\n1 100 300 400 500 500000 1000000 10000000 100000000 1000000000",
"output": "10"
},
{
"input": "2 1\n2 1",
"output": "2"
},
{
"input": "2 1000000000\n1 1000000000",
"output": "1"
},
{
"input": "4 1000\n1 1000 1000000 1000000000",
"output": "2"
},
{
"input": "2 2\n1 3",
"output": "2"
},
{
"input": "2 2\n16 8",
"output": "1"
},
{
"input": "3 2\n8 4 2",
"output": "2"
},
{
"input": "5 1\n1 2 3 4 5",
"output": "5"
},
{
"input": "2 2\n500000000 1000000000",
"output": "1"
},
{
"input": "2 2\n4 2",
"output": "1"
},
{
"input": "10 100000000\n1 2 3 4 5 6 7 8 82000 907431936",
"output": "10"
},
{
"input": "8 65538\n65535 65536 65537 65538 65539 131072 262144 196608",
"output": "8"
},
{
"input": "5 2\n10 8 6 4 2",
"output": "4"
},
{
"input": "2 1000000000\n276447232 100000",
"output": "2"
}
] | 1,420,420,153
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 43
| 498
| 8,499,200
|
n, k = [int(x) for x in input().split()]
a = sorted([int(x) for x in input().split()])
s = set(a)
l = 0
for x in a:
t = 1
while x in s:
l += t
t = 1 - t
s.remove(x)
x *= k
print(l)
|
Title: k-Multiple Free Set
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A *k*-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by *k*. That is, there are no two integers *x* and *y* (*x*<=<<=*y*) from the set, such that *y*<==<=*x*·*k*.
You're given a set of *n* distinct positive integers. Your task is to find the size of it's largest *k*-multiple free subset.
Input Specification:
The first line of the input contains two integers *n* and *k* (1<=≤<=*n*<=≤<=105,<=1<=≤<=*k*<=≤<=109). The next line contains a list of *n* distinct positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
All the numbers in the lines are separated by single spaces.
Output Specification:
On the only line of the output print the size of the largest *k*-multiple free subset of {*a*1,<=*a*2,<=...,<=*a**n*}.
Demo Input:
['6 2\n2 3 6 5 4 10\n']
Demo Output:
['3\n']
Note:
In the sample input one of the possible maximum 2-multiple free subsets is {4, 5, 6}.
|
```python
n, k = [int(x) for x in input().split()]
a = sorted([int(x) for x in input().split()])
s = set(a)
l = 0
for x in a:
t = 1
while x in s:
l += t
t = 1 - t
s.remove(x)
x *= k
print(l)
```
| 3
|
|
798
|
A
|
Mike and palindrome
|
PROGRAMMING
| 1,000
|
[
"brute force",
"constructive algorithms",
"strings"
] | null | null |
Mike has a string *s* consisting of only lowercase English letters. He wants to change exactly one character from the string so that the resulting one is a palindrome.
A palindrome is a string that reads the same backward as forward, for example strings "z", "aaa", "aba", "abccba" are palindromes, but strings "codeforces", "reality", "ab" are not.
|
The first and single line contains string *s* (1<=≤<=|*s*|<=≤<=15).
|
Print "YES" (without quotes) if Mike can change exactly one character so that the resulting string is palindrome or "NO" (without quotes) otherwise.
|
[
"abccaa\n",
"abbcca\n",
"abcda\n"
] |
[
"YES\n",
"NO\n",
"YES\n"
] |
none
| 500
|
[
{
"input": "abccaa",
"output": "YES"
},
{
"input": "abbcca",
"output": "NO"
},
{
"input": "abcda",
"output": "YES"
},
{
"input": "kyw",
"output": "YES"
},
{
"input": "fccf",
"output": "NO"
},
{
"input": "mnlm",
"output": "YES"
},
{
"input": "gqrk",
"output": "NO"
},
{
"input": "glxlg",
"output": "YES"
},
{
"input": "czhfc",
"output": "YES"
},
{
"input": "broon",
"output": "NO"
},
{
"input": "rmggmr",
"output": "NO"
},
{
"input": "wvxxzw",
"output": "YES"
},
{
"input": "ukvciu",
"output": "NO"
},
{
"input": "vrnwnrv",
"output": "YES"
},
{
"input": "vlkjkav",
"output": "YES"
},
{
"input": "guayhmg",
"output": "NO"
},
{
"input": "lkvhhvkl",
"output": "NO"
},
{
"input": "ffdsslff",
"output": "YES"
},
{
"input": "galjjtyw",
"output": "NO"
},
{
"input": "uosgwgsou",
"output": "YES"
},
{
"input": "qjwmjmljq",
"output": "YES"
},
{
"input": "ustrvrodf",
"output": "NO"
},
{
"input": "a",
"output": "YES"
},
{
"input": "qjfyjjyfjq",
"output": "NO"
},
{
"input": "ysxibbixsq",
"output": "YES"
},
{
"input": "howfslfwmh",
"output": "NO"
},
{
"input": "ekhajrjahke",
"output": "YES"
},
{
"input": "ucnolsloncw",
"output": "YES"
},
{
"input": "jrzsfrrkrtj",
"output": "NO"
},
{
"input": "typayzzyapyt",
"output": "NO"
},
{
"input": "uwdhkzokhdwu",
"output": "YES"
},
{
"input": "xokxpyyuafij",
"output": "NO"
},
{
"input": "eusneioiensue",
"output": "YES"
},
{
"input": "fuxpuajabpxuf",
"output": "YES"
},
{
"input": "guvggtfhlgruy",
"output": "NO"
},
{
"input": "cojhkhxxhkhjoc",
"output": "NO"
},
{
"input": "mhifbmmmmbmihm",
"output": "YES"
},
{
"input": "kxfqqncnebpami",
"output": "NO"
},
{
"input": "scfwrjevejrwfcs",
"output": "YES"
},
{
"input": "thdaonpepdoadht",
"output": "YES"
},
{
"input": "jsfzcbnhsccuqsj",
"output": "NO"
},
{
"input": "nn",
"output": "NO"
},
{
"input": "nm",
"output": "YES"
},
{
"input": "jdj",
"output": "YES"
},
{
"input": "bbcaa",
"output": "NO"
},
{
"input": "abcde",
"output": "NO"
},
{
"input": "abcdf",
"output": "NO"
},
{
"input": "aa",
"output": "NO"
},
{
"input": "abecd",
"output": "NO"
},
{
"input": "abccacb",
"output": "NO"
},
{
"input": "aabc",
"output": "NO"
},
{
"input": "anpqb",
"output": "NO"
},
{
"input": "c",
"output": "YES"
},
{
"input": "abcdefg",
"output": "NO"
},
{
"input": "aanbb",
"output": "NO"
},
{
"input": "aabbb",
"output": "NO"
},
{
"input": "aaabbab",
"output": "NO"
},
{
"input": "ab",
"output": "YES"
},
{
"input": "aabbc",
"output": "NO"
},
{
"input": "ecabd",
"output": "NO"
},
{
"input": "abcdrty",
"output": "NO"
},
{
"input": "abcdmnp",
"output": "NO"
},
{
"input": "bbbbbb",
"output": "NO"
},
{
"input": "abcxuio",
"output": "NO"
},
{
"input": "abcdabcde",
"output": "NO"
},
{
"input": "abcxpoi",
"output": "NO"
},
{
"input": "aba",
"output": "YES"
},
{
"input": "aacbb",
"output": "NO"
},
{
"input": "abcedca",
"output": "NO"
},
{
"input": "abcdd",
"output": "NO"
},
{
"input": "abbcs",
"output": "NO"
},
{
"input": "aaabccc",
"output": "NO"
},
{
"input": "paxkxbq",
"output": "NO"
},
{
"input": "z",
"output": "YES"
},
{
"input": "b",
"output": "YES"
},
{
"input": "abcdefghi",
"output": "NO"
},
{
"input": "abcqr",
"output": "NO"
},
{
"input": "abcdc",
"output": "NO"
},
{
"input": "abcb",
"output": "NO"
},
{
"input": "aabcd",
"output": "NO"
},
{
"input": "abbba",
"output": "YES"
},
{
"input": "aaabbb",
"output": "NO"
},
{
"input": "bb",
"output": "NO"
},
{
"input": "aaacbbb",
"output": "NO"
},
{
"input": "abbzcca",
"output": "NO"
},
{
"input": "abxab",
"output": "NO"
},
{
"input": "bbb",
"output": "YES"
},
{
"input": "abcrtyu",
"output": "NO"
},
{
"input": "cbacb",
"output": "NO"
},
{
"input": "acbb",
"output": "NO"
},
{
"input": "ww",
"output": "NO"
},
{
"input": "aaaaaa",
"output": "NO"
},
{
"input": "jizzz",
"output": "NO"
},
{
"input": "aaakcba",
"output": "NO"
},
{
"input": "acbak",
"output": "NO"
},
{
"input": "bddeffd",
"output": "NO"
},
{
"input": "aaa",
"output": "YES"
},
{
"input": "afghqwe",
"output": "NO"
},
{
"input": "abcdfga",
"output": "NO"
}
] | 1,694,477,838
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 15
| 0
|
n = input("")
a = 0
for i in range(len(n)-1) :
snl = list(n)
nl = snl
a = set(nl)
b = list(a)
for x in range(len(b)) :
nn = "".join(nl)
if nn == nn[::-1] :
a = 1
break
nl[i] = b[x]
if a == 1:
print("YES")
else :
print("NO")
|
Title: Mike and palindrome
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mike has a string *s* consisting of only lowercase English letters. He wants to change exactly one character from the string so that the resulting one is a palindrome.
A palindrome is a string that reads the same backward as forward, for example strings "z", "aaa", "aba", "abccba" are palindromes, but strings "codeforces", "reality", "ab" are not.
Input Specification:
The first and single line contains string *s* (1<=≤<=|*s*|<=≤<=15).
Output Specification:
Print "YES" (without quotes) if Mike can change exactly one character so that the resulting string is palindrome or "NO" (without quotes) otherwise.
Demo Input:
['abccaa\n', 'abbcca\n', 'abcda\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
none
|
```python
n = input("")
a = 0
for i in range(len(n)-1) :
snl = list(n)
nl = snl
a = set(nl)
b = list(a)
for x in range(len(b)) :
nn = "".join(nl)
if nn == nn[::-1] :
a = 1
break
nl[i] = b[x]
if a == 1:
print("YES")
else :
print("NO")
```
| 0
|
|
175
|
E
|
Power Defence
|
PROGRAMMING
| 2,600
|
[
"brute force",
"dp",
"geometry",
"greedy"
] | null | null |
Vasya plays the Power Defence.
He must pass the last level of the game. In order to do this he must kill the Main Villain, who moves in a straight line at speed 1 meter per second from the point (<=-<=∞,<=0) to the point (<=+<=∞,<=0) of the game world. In the points (*x*,<=1) and (*x*,<=<=-<=1), where *x* is an integer number, Vasya can build towers of three types: fire-tower, electric-tower or freezing-tower. However, it is not allowed to build two towers at the same point. Towers of each type have a certain action radius and the value of damage per second (except freezing-tower). If at some point the Main Villain is in the range of action of *k* freezing towers then his speed is decreased by *k*<=+<=1 times.
The allowed number of towers of each type is known. It is necessary to determine the maximum possible damage we can inflict on the Main Villain.
All distances in the problem are given in meters. The size of the Main Villain and the towers are so small, that they can be considered as points on the plane. The Main Villain is in the action radius of a tower if the distance between him and tower is less than or equal to the action radius of the tower.
|
The first line contains three integer numbers *nf*,<=*ne* and *ns* — the maximum number of fire-towers, electric-towers and freezing-towers that can be built (0<=≤<=*nf*,<=*ne*,<=*ns*<=≤<=20,<=1<=≤<=*nf*<=+<=*ne*<=+<=*ns*<=≤<=20). The numbers are separated with single spaces.
The second line contains three integer numbers *rf*,<=*re* and *rs* (1<=≤<=*rf*,<=*re*,<=*rs*<=≤<=1000) — the action radii of fire-towers, electric-towers and freezing-towers. The numbers are separated with single spaces.
The third line contains two integer numbers *df* and *de* (1<=≤<=*df*,<=*de*<=≤<=1000) — the damage a fire-tower and an electronic-tower can inflict on the Main Villain per second (in the case when the Main Villain is in the action radius of the tower). The numbers are separated with single space.
|
Print the only real number — the maximum possible damage to the Main Villain with absolute or relative error not more than 10<=-<=6.
|
[
"1 0 0\n10 10 10\n100 100\n",
"1 0 1\n10 10 10\n100 100\n"
] |
[
"1989.97487421",
"3979.94974843"
] |
In the first sample we've got one fire-tower that always inflicts the same damage, independently of its position.
In the second sample we've got another freezing-tower of the same action radius. If we build the two towers opposite each other, then the Main Villain's speed will be two times lower, whenever he enters the fire-tower's action radius. That means that the enemy will be inflicted with twice more damage.
| 3,000
|
[] | 1,689,425,749
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 92
| 0
|
print("_RANDOM_GUESS_1689425749.4744663")# 1689425749.4744878
|
Title: Power Defence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya plays the Power Defence.
He must pass the last level of the game. In order to do this he must kill the Main Villain, who moves in a straight line at speed 1 meter per second from the point (<=-<=∞,<=0) to the point (<=+<=∞,<=0) of the game world. In the points (*x*,<=1) and (*x*,<=<=-<=1), where *x* is an integer number, Vasya can build towers of three types: fire-tower, electric-tower or freezing-tower. However, it is not allowed to build two towers at the same point. Towers of each type have a certain action radius and the value of damage per second (except freezing-tower). If at some point the Main Villain is in the range of action of *k* freezing towers then his speed is decreased by *k*<=+<=1 times.
The allowed number of towers of each type is known. It is necessary to determine the maximum possible damage we can inflict on the Main Villain.
All distances in the problem are given in meters. The size of the Main Villain and the towers are so small, that they can be considered as points on the plane. The Main Villain is in the action radius of a tower if the distance between him and tower is less than or equal to the action radius of the tower.
Input Specification:
The first line contains three integer numbers *nf*,<=*ne* and *ns* — the maximum number of fire-towers, electric-towers and freezing-towers that can be built (0<=≤<=*nf*,<=*ne*,<=*ns*<=≤<=20,<=1<=≤<=*nf*<=+<=*ne*<=+<=*ns*<=≤<=20). The numbers are separated with single spaces.
The second line contains three integer numbers *rf*,<=*re* and *rs* (1<=≤<=*rf*,<=*re*,<=*rs*<=≤<=1000) — the action radii of fire-towers, electric-towers and freezing-towers. The numbers are separated with single spaces.
The third line contains two integer numbers *df* and *de* (1<=≤<=*df*,<=*de*<=≤<=1000) — the damage a fire-tower and an electronic-tower can inflict on the Main Villain per second (in the case when the Main Villain is in the action radius of the tower). The numbers are separated with single space.
Output Specification:
Print the only real number — the maximum possible damage to the Main Villain with absolute or relative error not more than 10<=-<=6.
Demo Input:
['1 0 0\n10 10 10\n100 100\n', '1 0 1\n10 10 10\n100 100\n']
Demo Output:
['1989.97487421', '3979.94974843']
Note:
In the first sample we've got one fire-tower that always inflicts the same damage, independently of its position.
In the second sample we've got another freezing-tower of the same action radius. If we build the two towers opposite each other, then the Main Villain's speed will be two times lower, whenever he enters the fire-tower's action radius. That means that the enemy will be inflicted with twice more damage.
|
```python
print("_RANDOM_GUESS_1689425749.4744663")# 1689425749.4744878
```
| 0
|
|
559
|
B
|
Equivalent Strings
|
PROGRAMMING
| 1,700
|
[
"divide and conquer",
"hashing",
"sortings",
"strings"
] | null | null |
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings *a* and *b* of equal length are called equivalent in one of the two cases:
1. They are equal. 1. If we split string *a* into two halves of the same size *a*1 and *a*2, and string *b* into two halves of the same size *b*1 and *b*2, then one of the following is correct: *a*1 is equivalent to *b*1, and *a*2 is equivalent to *b*2 1. *a*1 is equivalent to *b*2, and *a*2 is equivalent to *b*1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn!
|
The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200<=000 and consists of lowercase English letters. The strings have the same length.
|
Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.
|
[
"aaba\nabaa\n",
"aabb\nabab\n"
] |
[
"YES\n",
"NO\n"
] |
In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa".
| 1,000
|
[
{
"input": "aaba\nabaa",
"output": "YES"
},
{
"input": "aabb\nabab",
"output": "NO"
},
{
"input": "a\na",
"output": "YES"
},
{
"input": "a\nb",
"output": "NO"
},
{
"input": "ab\nab",
"output": "YES"
},
{
"input": "ab\nba",
"output": "YES"
},
{
"input": "ab\nbb",
"output": "NO"
},
{
"input": "zzaa\naazz",
"output": "YES"
},
{
"input": "azza\nzaaz",
"output": "YES"
},
{
"input": "abc\nabc",
"output": "YES"
},
{
"input": "abc\nacb",
"output": "NO"
},
{
"input": "azzz\nzzaz",
"output": "YES"
},
{
"input": "abcd\ndcab",
"output": "YES"
},
{
"input": "abcd\ncdab",
"output": "YES"
},
{
"input": "abcd\ndcba",
"output": "YES"
},
{
"input": "abcd\nacbd",
"output": "NO"
},
{
"input": "oloaxgddgujq\noloaxgujqddg",
"output": "YES"
},
{
"input": "uwzwdxfmosmqatyv\ndxfmzwwusomqvyta",
"output": "YES"
},
{
"input": "hagnzomowtledfdotnll\nledfdotnllomowthagnz",
"output": "YES"
},
{
"input": "snyaydaeobufdg\nsnyaydaeobufdg",
"output": "YES"
},
{
"input": "baaaaa\nabaaaa",
"output": "NO"
},
{
"input": "hhiisug\nmzdjwju",
"output": "NO"
},
{
"input": "bbbabbabaaab\naaaabbabbbbb",
"output": "NO"
},
{
"input": "bbaaab\naababb",
"output": "NO"
},
{
"input": "aabbaaaa\naaaaabab",
"output": "NO"
},
{
"input": "aab\naba",
"output": "NO"
},
{
"input": "abcddd\nbacddd",
"output": "NO"
},
{
"input": "qgiufelsfhanx\naaaaaaaaaaaaa",
"output": "NO"
},
{
"input": "aabaababaaba\naababaaababa",
"output": "NO"
},
{
"input": "nocdqzdriyyil\naaaaaaaaaaaaa",
"output": "NO"
},
{
"input": "zdmctxl\nkojqhgw",
"output": "NO"
},
{
"input": "yhwepqwyhwepqwyhwepqweahnqtueahnqtueahnqtuyhwepqwyhwepqwyhwepqwyhwepqweahnqtueahnqtuyhwepqweahnqtueahnqtueahnqtueahnqtueahnqtueahnqtu\neahnqtueahnqtueahnqtuyhwepqweahnqtuyhwepqwyhwepqweahnqtuyhwepqweahnqtuyhwepqweahnqtueahnqtuyhwepqweahnqtueahnqtuyhwepqwyhwepqwyhwepqw",
"output": "NO"
},
{
"input": "abc\nbac",
"output": "NO"
},
{
"input": "ottceez\npcstdvz",
"output": "NO"
}
] | 1,597,581,839
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 2
| 108
| 0
|
try:
s1 = input()
s2 = input()
ls1=len(s1)
ls2=len(s2)
if ls1==ls2:
s11=s1[0:ls1//2]
s12=s1[ls1//2:]
s21=s2[0:ls2//2]
s22=s2[ls2//2:]
if s11==s21 or s11==s22 or s12==s21 or s12 == s22:
print("YES")
else:
print("NO")
except:
pass
|
Title: Equivalent Strings
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings *a* and *b* of equal length are called equivalent in one of the two cases:
1. They are equal. 1. If we split string *a* into two halves of the same size *a*1 and *a*2, and string *b* into two halves of the same size *b*1 and *b*2, then one of the following is correct: *a*1 is equivalent to *b*1, and *a*2 is equivalent to *b*2 1. *a*1 is equivalent to *b*2, and *a*2 is equivalent to *b*1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn!
Input Specification:
The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200<=000 and consists of lowercase English letters. The strings have the same length.
Output Specification:
Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.
Demo Input:
['aaba\nabaa\n', 'aabb\nabab\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa".
|
```python
try:
s1 = input()
s2 = input()
ls1=len(s1)
ls2=len(s2)
if ls1==ls2:
s11=s1[0:ls1//2]
s12=s1[ls1//2:]
s21=s2[0:ls2//2]
s22=s2[ls2//2:]
if s11==s21 or s11==s22 or s12==s21 or s12 == s22:
print("YES")
else:
print("NO")
except:
pass
```
| 0
|
|
9
|
A
|
Die Roll
|
PROGRAMMING
| 800
|
[
"math",
"probabilities"
] |
A. Die Roll
|
1
|
64
|
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
|
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
|
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
|
[
"4 2\n"
] |
[
"1/2\n"
] |
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points.
| 0
|
[
{
"input": "4 2",
"output": "1/2"
},
{
"input": "1 1",
"output": "1/1"
},
{
"input": "1 2",
"output": "5/6"
},
{
"input": "1 3",
"output": "2/3"
},
{
"input": "1 4",
"output": "1/2"
},
{
"input": "1 5",
"output": "1/3"
},
{
"input": "1 6",
"output": "1/6"
},
{
"input": "2 1",
"output": "5/6"
},
{
"input": "2 2",
"output": "5/6"
},
{
"input": "2 3",
"output": "2/3"
},
{
"input": "2 4",
"output": "1/2"
},
{
"input": "2 5",
"output": "1/3"
},
{
"input": "2 6",
"output": "1/6"
},
{
"input": "3 1",
"output": "2/3"
},
{
"input": "3 2",
"output": "2/3"
},
{
"input": "3 3",
"output": "2/3"
},
{
"input": "3 4",
"output": "1/2"
},
{
"input": "3 5",
"output": "1/3"
},
{
"input": "3 6",
"output": "1/6"
},
{
"input": "4 1",
"output": "1/2"
},
{
"input": "4 3",
"output": "1/2"
},
{
"input": "4 4",
"output": "1/2"
},
{
"input": "4 5",
"output": "1/3"
},
{
"input": "4 6",
"output": "1/6"
},
{
"input": "5 1",
"output": "1/3"
},
{
"input": "5 2",
"output": "1/3"
},
{
"input": "5 3",
"output": "1/3"
},
{
"input": "5 4",
"output": "1/3"
},
{
"input": "5 5",
"output": "1/3"
},
{
"input": "5 6",
"output": "1/6"
},
{
"input": "6 1",
"output": "1/6"
},
{
"input": "6 2",
"output": "1/6"
},
{
"input": "6 3",
"output": "1/6"
},
{
"input": "6 4",
"output": "1/6"
},
{
"input": "6 5",
"output": "1/6"
},
{
"input": "6 6",
"output": "1/6"
}
] | 1,627,253,834
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 36
| 154
| 6,758,400
|
[W,Y] = map(int, input().split())
list = ['1/1','5/6','2/3','1/2','1/3','1/6']
r = max(W,Y)
print (list[r-1])
|
Title: Die Roll
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
Input Specification:
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
Output Specification:
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
Demo Input:
['4 2\n']
Demo Output:
['1/2\n']
Note:
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points.
|
```python
[W,Y] = map(int, input().split())
list = ['1/1','5/6','2/3','1/2','1/3','1/6']
r = max(W,Y)
print (list[r-1])
```
| 3.872646
|
577
|
A
|
Multiplication Table
|
PROGRAMMING
| 1,000
|
[
"implementation",
"number theory"
] | null | null |
Let's consider a table consisting of *n* rows and *n* columns. The cell located at the intersection of *i*-th row and *j*-th column contains number *i*<=×<=*j*. The rows and columns are numbered starting from 1.
You are given a positive integer *x*. Your task is to count the number of cells in a table that contain number *x*.
|
The single line contains numbers *n* and *x* (1<=≤<=*n*<=≤<=105, 1<=≤<=*x*<=≤<=109) — the size of the table and the number that we are looking for in the table.
|
Print a single number: the number of times *x* occurs in the table.
|
[
"10 5\n",
"6 12\n",
"5 13\n"
] |
[
"2\n",
"4\n",
"0\n"
] |
A table for the second sample test is given below. The occurrences of number 12 are marked bold.
| 500
|
[
{
"input": "10 5",
"output": "2"
},
{
"input": "6 12",
"output": "4"
},
{
"input": "5 13",
"output": "0"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "2 1",
"output": "1"
},
{
"input": "100000 1",
"output": "1"
},
{
"input": "1 1000000000",
"output": "0"
},
{
"input": "100000 1000000000",
"output": "16"
},
{
"input": "100000 362880",
"output": "154"
},
{
"input": "1 4",
"output": "0"
},
{
"input": "9 12",
"output": "4"
},
{
"input": "10 123",
"output": "0"
},
{
"input": "9551 975275379",
"output": "0"
},
{
"input": "17286 948615687",
"output": "0"
},
{
"input": "58942 936593001",
"output": "0"
},
{
"input": "50000 989460910",
"output": "4"
},
{
"input": "22741 989460910",
"output": "0"
},
{
"input": "22740 989460910",
"output": "0"
},
{
"input": "100000 989460910",
"output": "4"
},
{
"input": "100000 98280",
"output": "128"
},
{
"input": "100000 997920",
"output": "222"
},
{
"input": "100000 720720",
"output": "226"
},
{
"input": "100000 2162160",
"output": "282"
},
{
"input": "100000 4324320",
"output": "320"
},
{
"input": "100000 8648640",
"output": "348"
},
{
"input": "100000 183783600",
"output": "438"
},
{
"input": "100000 551350800",
"output": "392"
},
{
"input": "40000 551350800",
"output": "150"
},
{
"input": "20000 400000000",
"output": "1"
},
{
"input": "19999 400000000",
"output": "0"
},
{
"input": "19999 399960001",
"output": "1"
},
{
"input": "31621 999887641",
"output": "1"
},
{
"input": "31622 999887641",
"output": "1"
},
{
"input": "31620 999887641",
"output": "0"
},
{
"input": "100000 999887641",
"output": "3"
},
{
"input": "100000 25",
"output": "3"
},
{
"input": "100000 1",
"output": "1"
},
{
"input": "100000 3628800",
"output": "220"
},
{
"input": "100000 39916800",
"output": "328"
},
{
"input": "100000 479001600",
"output": "254"
},
{
"input": "4 9",
"output": "1"
},
{
"input": "2 6",
"output": "0"
},
{
"input": "20 100",
"output": "3"
},
{
"input": "10 3",
"output": "2"
},
{
"input": "4 4",
"output": "3"
},
{
"input": "2 4",
"output": "1"
},
{
"input": "89874 1",
"output": "1"
}
] | 1,681,116,698
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 7
| 1,000
| 7,987,200
|
n , target = map(int , input().split())
counter = 0
if target == 1 :
counter = 1
else :
first_row = [i for i in range(1 , n+1)]
for i in range(1 ,n+1) :
row = list(map(lambda x :x*i , first_row))
counter +=row.count(target)
print(counter)
|
Title: Multiplication Table
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let's consider a table consisting of *n* rows and *n* columns. The cell located at the intersection of *i*-th row and *j*-th column contains number *i*<=×<=*j*. The rows and columns are numbered starting from 1.
You are given a positive integer *x*. Your task is to count the number of cells in a table that contain number *x*.
Input Specification:
The single line contains numbers *n* and *x* (1<=≤<=*n*<=≤<=105, 1<=≤<=*x*<=≤<=109) — the size of the table and the number that we are looking for in the table.
Output Specification:
Print a single number: the number of times *x* occurs in the table.
Demo Input:
['10 5\n', '6 12\n', '5 13\n']
Demo Output:
['2\n', '4\n', '0\n']
Note:
A table for the second sample test is given below. The occurrences of number 12 are marked bold.
|
```python
n , target = map(int , input().split())
counter = 0
if target == 1 :
counter = 1
else :
first_row = [i for i in range(1 , n+1)]
for i in range(1 ,n+1) :
row = list(map(lambda x :x*i , first_row))
counter +=row.count(target)
print(counter)
```
| 0
|
|
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"
}
] | 1,594,024,388
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 156
| 6,963,200
|
n=int(input())
l=list(map(int,input().split()))
ei=-1
od=-1
ec=0
oc=0
for i in range(n):
if(l[i]%2==0):
ec+=1
ei=i
else:
oc+=1
oi=i
if(ec>0 and oc>0):
break
if(oc<ec):
print(oi+1)
else:
print(ei+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
n=int(input())
l=list(map(int,input().split()))
ei=-1
od=-1
ec=0
oc=0
for i in range(n):
if(l[i]%2==0):
ec+=1
ei=i
else:
oc+=1
oi=i
if(ec>0 and oc>0):
break
if(oc<ec):
print(oi+1)
else:
print(ei+1)
```
| 0
|
786
|
A
|
Berzerk
|
PROGRAMMING
| 2,000
|
[
"dfs and similar",
"dp",
"games"
] | null | null |
Rick and Morty are playing their own version of Berzerk (which has nothing in common with the famous Berzerk game). This game needs a huge space, so they play it with a computer.
In this game there are *n* objects numbered from 1 to *n* arranged in a circle (in clockwise order). Object number 1 is a black hole and the others are planets. There's a monster in one of the planet. Rick and Morty don't know on which one yet, only that he's not initially in the black hole, but Unity will inform them before the game starts. But for now, they want to be prepared for every possible scenario.
Each one of them has a set of numbers between 1 and *n*<=-<=1 (inclusive). Rick's set is *s*1 with *k*1 elements and Morty's is *s*2 with *k*2 elements. One of them goes first and the player changes alternatively. In each player's turn, he should choose an arbitrary number like *x* from his set and the monster will move to his *x*-th next object from its current position (clockwise). If after his move the monster gets to the black hole he wins.
Your task is that for each of monster's initial positions and who plays first determine if the starter wins, loses, or the game will stuck in an infinite loop. In case when player can lose or make game infinity, it more profitable to choose infinity game.
|
The first line of input contains a single integer *n* (2<=≤<=*n*<=≤<=7000) — number of objects in game.
The second line contains integer *k*1 followed by *k*1 distinct integers *s*1,<=1,<=*s*1,<=2,<=...,<=*s*1,<=*k*1 — Rick's set.
The third line contains integer *k*2 followed by *k*2 distinct integers *s*2,<=1,<=*s*2,<=2,<=...,<=*s*2,<=*k*2 — Morty's set
1<=≤<=*k**i*<=≤<=*n*<=-<=1 and 1<=≤<=*s**i*,<=1,<=*s**i*,<=2,<=...,<=*s**i*,<=*k**i*<=≤<=*n*<=-<=1 for 1<=≤<=*i*<=≤<=2.
|
In the first line print *n*<=-<=1 words separated by spaces where *i*-th word is "Win" (without quotations) if in the scenario that Rick plays first and monster is initially in object number *i*<=+<=1 he wins, "Lose" if he loses and "Loop" if the game will never end.
Similarly, in the second line print *n*<=-<=1 words separated by spaces where *i*-th word is "Win" (without quotations) if in the scenario that Morty plays first and monster is initially in object number *i*<=+<=1 he wins, "Lose" if he loses and "Loop" if the game will never end.
|
[
"5\n2 3 2\n3 1 2 3\n",
"8\n4 6 2 3 4\n2 3 6\n"
] |
[
"Lose Win Win Loop\nLoop Win Win Win\n",
"Win Win Win Win Win Win Win\nLose Win Lose Lose Win Lose Lose\n"
] |
none
| 750
|
[
{
"input": "5\n2 3 2\n3 1 2 3",
"output": "Lose Win Win Loop\nLoop Win Win Win"
},
{
"input": "8\n4 6 2 3 4\n2 3 6",
"output": "Win Win Win Win Win Win Win\nLose Win Lose Lose Win Lose Lose"
},
{
"input": "10\n3 4 7 5\n2 8 5",
"output": "Win Win Win Win Win Win Win Loop Win\nLose Win Loop Lose Win Lose Lose Lose Lose"
},
{
"input": "17\n1 10\n1 12",
"output": "Win Win Win Win Win Win Win Win Win Win Win Lose Win Win Win Win\nLose Lose Lose Lose Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose"
},
{
"input": "23\n1 20\n3 9 2 12",
"output": "Lose Lose Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose\nWin Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win"
},
{
"input": "85\n12 76 7 75 51 43 41 66 13 59 48 81 73\n3 65 60 25",
"output": "Loop Loop Loop Win Loop Loop Loop Loop Win Win Loop Win Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Win Loop Loop Win Loop Loop Loop Loop Win Loop Win Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Loop\nLoop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Win Loo..."
},
{
"input": "100\n84 80 73 28 76 21 44 97 63 59 6 77 41 2 8 71 57 19 33 46 92 5 61 88 53 68 94 56 14 35 4 47 17 79 84 10 67 58 45 38 13 12 87 3 91 30 15 11 24 55 62 39 83 43 89 1 81 75 50 86 72 18 52 78 7 29 64 42 70 49 37 25 66 74 95 36 85 48 99 60 51 98 27 40 93\n47 52 76 9 4 25 8 63 29 74 97 61 93 35 49 62 5 10 57 73 42 3 19 23 71 70 43 67 48 2 34 31 41 90 18 6 40 83 98 72 14 51 38 46 21 99 65 37",
"output": "Win Win Win Loop Win Win Win Win Win Loop Win Win Win Win Win Win Win Loop Win Win Win Win Win Win Win Win Win Win Win Win Loop Win Win Win Loop Win Win Win Win Win Win Win Win Win Win Loop Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Loop Win Loop Loop Win Win Win Win Loop Win Win Loop Loop Win Loop Win Win Win Loop Win Win Win Win Win Win Loop Win Win Win Win Win Win Win Win\nWin Win Win Loop Loop Loop Win Loop Loop Win Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Loop ..."
},
{
"input": "100\n66 70 54 10 72 81 84 56 15 27 19 43 55 49 44 52 33 63 40 95 17 58 2 51 39 22 18 82 1 16 99 32 29 24 94 9 98 5 37 47 14 42 73 41 31 79 64 12 6 53 26 68 67 89 13 90 4 21 93 46 74 75 88 66 57 23 7\n18 8 47 76 39 34 52 62 5 36 19 22 80 32 71 55 7 37 57",
"output": "Win Win Loop Loop Win Win Win Loop Loop Win Win Win Loop Loop Loop Win Loop Win Win Loop Win Loop Loop Loop Win Win Win Win Loop Win Loop Win Win Win Loop Win Win Loop Loop Loop Loop Win Win Win Win Win Win Win Win Loop Win Loop Win Win Loop Win Win Win Win Win Win Loop Win Loop Loop Loop Win Win Win Loop Win Loop Win Win Loop Win Win Win Win Loop Win Win Win Win Win Win Win Win Loop Win Win Loop Win Win Win Win Loop Win Win\nLoop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop L..."
},
{
"input": "300\n1 179\n2 293 180",
"output": "Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose L..."
},
{
"input": "1000\n14 77 649 670 988 469 453 445 885 101 58 728 474 488 230\n8 83 453 371 86 834 277 847 958",
"output": "Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Lo..."
},
{
"input": "2\n1 1\n1 1",
"output": "Win\nWin"
},
{
"input": "2\n1 1\n1 1",
"output": "Win\nWin"
},
{
"input": "3\n1 1\n1 2",
"output": "Loop Win\nWin Loop"
},
{
"input": "20\n1 1\n1 11",
"output": "Loop Loop Win Lose Loop Loop Win Lose Loop Loop Win Lose Loop Loop Win Lose Loop Loop Win\nWin Loop Loop Lose Win Loop Loop Lose Win Loop Loop Lose Win Loop Loop Lose Win Loop Loop"
},
{
"input": "309\n30 197 38 142 159 163 169 263 70 151 288 264 41 285 225 216 306 128 242 221 94 39 43 292 54 157 78 272 257 97 57\n3 97 172 165",
"output": "Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Win Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Win Win Loop Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Win Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loo..."
},
{
"input": "1000\n1 312\n1 171",
"output": "Lose Lose Lose Lose Lose Lose Win Lose Lose Lose Lose Lose Lose Win Lose Lose Lose Lose Lose Lose Win Lose Lose Lose Lose Lose Lose Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Lose Lose Lose Lose Lose Lose Win Lose Lose Lose Lose Lose Lose Win Lose Lose Lose Lose Lose Lose Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Lose Lose Lose Lose Lose Lose Win Lose Lose Lose Lose Lose Lose Win Lose Lose Lose Lose Lose Lose Win Lose Lose Lose Lose Lose Lose Lose Lose Los..."
},
{
"input": "1000\n1 481\n2 468 9",
"output": "Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose L..."
},
{
"input": "1000\n3 469 637 369\n2 801 339",
"output": "Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop L..."
},
{
"input": "4096\n6 3736 3640 553 2608 1219 1640\n4 112 2233 3551 2248",
"output": "Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop L..."
},
{
"input": "6341\n9 6045 2567 3242 5083 5429 1002 4547 1838 4829\n5 5533 3084 6323 4015 2889",
"output": "Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop L..."
},
{
"input": "7000\n1 5244\n1 2980",
"output": "Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop..."
},
{
"input": "7000\n1 6694\n1 2973",
"output": "Loop Loop Loop Loop Win Loop Lose Loop Loop Loop Loop Win Loop Lose Loop Loop Loop Loop Win Loop Lose Loop Loop Loop Loop Win Loop Lose Loop Loop Loop Loop Win Loop Lose Loop Loop Loop Loop Win Loop Lose Loop Loop Loop Loop Win Loop Lose Loop Loop Loop Loop Win Loop Lose Loop Loop Loop Loop Win Loop Lose Loop Loop Loop Loop Win Loop Lose Loop Loop Loop Loop Win Loop Lose Loop Loop Loop Loop Win Loop Lose Loop Loop Loop Loop Win Loop Lose Loop Loop Loop Loop Win Loop Lose Loop Loop Loop Loop Win Loop Lose L..."
},
{
"input": "7000\n1 3041\n1 6128",
"output": "Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win W..."
},
{
"input": "7000\n5 5080 4890 1201 4903 1360\n5 2415 6678 5200 2282 4648",
"output": "Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop L..."
},
{
"input": "7000\n3 6965 1271 5818\n3 6331 5681 6636",
"output": "Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Lo..."
},
{
"input": "7000\n3 2706 2040 6698\n10 4118 846 1075 1624 2342 766 6441 2361 4662 1574",
"output": "Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop L..."
},
{
"input": "7000\n12 3489 6630 4582 292 5489 1456 5101 6920 632 2963 5136 5886\n11 434 5878 3806 656 3047 6614 1073 5932 6537 704 5253",
"output": "Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Lo..."
},
{
"input": "7000\n7 419 1631 1925 3861 6940 379 493\n29 5389 5925 2923 4696 972 6125 3779 6044 5477 1305 6488 5059 5515 3238 3863 248 6947 4023 6168 1915 6607 2991 2220 2023 200 4457 6398 1017 447",
"output": "Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Lo..."
},
{
"input": "6999\n2 3992 782\n2 4903 6815",
"output": "Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop L..."
}
] | 1,492,668,491
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 46
| 5,529,600
|
#����
string1=raw_input()
string2=raw_input()
string3=raw_input()
#���
print("Lose Win Win Loop")
print("Loop Win Win Win")
|
Title: Berzerk
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Rick and Morty are playing their own version of Berzerk (which has nothing in common with the famous Berzerk game). This game needs a huge space, so they play it with a computer.
In this game there are *n* objects numbered from 1 to *n* arranged in a circle (in clockwise order). Object number 1 is a black hole and the others are planets. There's a monster in one of the planet. Rick and Morty don't know on which one yet, only that he's not initially in the black hole, but Unity will inform them before the game starts. But for now, they want to be prepared for every possible scenario.
Each one of them has a set of numbers between 1 and *n*<=-<=1 (inclusive). Rick's set is *s*1 with *k*1 elements and Morty's is *s*2 with *k*2 elements. One of them goes first and the player changes alternatively. In each player's turn, he should choose an arbitrary number like *x* from his set and the monster will move to his *x*-th next object from its current position (clockwise). If after his move the monster gets to the black hole he wins.
Your task is that for each of monster's initial positions and who plays first determine if the starter wins, loses, or the game will stuck in an infinite loop. In case when player can lose or make game infinity, it more profitable to choose infinity game.
Input Specification:
The first line of input contains a single integer *n* (2<=≤<=*n*<=≤<=7000) — number of objects in game.
The second line contains integer *k*1 followed by *k*1 distinct integers *s*1,<=1,<=*s*1,<=2,<=...,<=*s*1,<=*k*1 — Rick's set.
The third line contains integer *k*2 followed by *k*2 distinct integers *s*2,<=1,<=*s*2,<=2,<=...,<=*s*2,<=*k*2 — Morty's set
1<=≤<=*k**i*<=≤<=*n*<=-<=1 and 1<=≤<=*s**i*,<=1,<=*s**i*,<=2,<=...,<=*s**i*,<=*k**i*<=≤<=*n*<=-<=1 for 1<=≤<=*i*<=≤<=2.
Output Specification:
In the first line print *n*<=-<=1 words separated by spaces where *i*-th word is "Win" (without quotations) if in the scenario that Rick plays first and monster is initially in object number *i*<=+<=1 he wins, "Lose" if he loses and "Loop" if the game will never end.
Similarly, in the second line print *n*<=-<=1 words separated by spaces where *i*-th word is "Win" (without quotations) if in the scenario that Morty plays first and monster is initially in object number *i*<=+<=1 he wins, "Lose" if he loses and "Loop" if the game will never end.
Demo Input:
['5\n2 3 2\n3 1 2 3\n', '8\n4 6 2 3 4\n2 3 6\n']
Demo Output:
['Lose Win Win Loop\nLoop Win Win Win\n', 'Win Win Win Win Win Win Win\nLose Win Lose Lose Win Lose Lose\n']
Note:
none
|
```python
#����
string1=raw_input()
string2=raw_input()
string3=raw_input()
#���
print("Lose Win Win Loop")
print("Loop Win Win Win")
```
| -1
|
|
908
|
A
|
New Year and Counting Cards
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation"
] | null | null |
Your friend has *n* cards.
You know that each card has a lowercase English letter on one side and a digit on the other.
Currently, your friend has laid out the cards on a table so only one side of each card is visible.
You would like to know if the following statement is true for cards that your friend owns: "If a card has a vowel on one side, then it has an even digit on the other side." More specifically, a vowel is one of 'a', 'e', 'i', 'o' or 'u', and even digit is one of '0', '2', '4', '6' or '8'.
For example, if a card has 'a' on one side, and '6' on the other side, then this statement is true for it. Also, the statement is true, for example, for a card with 'b' and '4', and for a card with 'b' and '3' (since the letter is not a vowel). The statement is false, for example, for card with 'e' and '5'. You are interested if the statement is true for all cards. In particular, if no card has a vowel, the statement is true.
To determine this, you can flip over some cards to reveal the other side. You would like to know what is the minimum number of cards you need to flip in the worst case in order to verify that the statement is true.
|
The first and only line of input will contain a string *s* (1<=≤<=|*s*|<=≤<=50), denoting the sides of the cards that you can see on the table currently. Each character of *s* is either a lowercase English letter or a digit.
|
Print a single integer, the minimum number of cards you must turn over to verify your claim.
|
[
"ee\n",
"z\n",
"0ay1\n"
] |
[
"2\n",
"0\n",
"2\n"
] |
In the first sample, we must turn over both cards. Note that even though both cards have the same letter, they could possibly have different numbers on the other side.
In the second sample, we don't need to turn over any cards. The statement is vacuously true, since you know your friend has no cards with a vowel on them.
In the third sample, we need to flip the second and fourth cards.
| 500
|
[
{
"input": "ee",
"output": "2"
},
{
"input": "z",
"output": "0"
},
{
"input": "0ay1",
"output": "2"
},
{
"input": "0abcdefghijklmnopqrstuvwxyz1234567896",
"output": "10"
},
{
"input": "0a0a9e9e2i2i9o9o6u6u9z9z4x4x9b9b",
"output": "18"
},
{
"input": "01234567890123456789012345678901234567890123456789",
"output": "25"
},
{
"input": "qwertyuioplkjhgfdsazxcvbnmqwertyuioplkjhgfdsazxcvb",
"output": "10"
},
{
"input": "cjw2dwmr10pku4yxohe0wglktd",
"output": "4"
},
{
"input": "6z2tx805jie8cfybwtfqvmlveec3iak5z5u3lu62vbxyqht6",
"output": "13"
},
{
"input": "kaq7jyialrfp4ilkni90eq8v3amcbygon7py0hb8z26fbl8ss1",
"output": "13"
},
{
"input": "hpwn50zgbmct80k9rizjqg40nycgs0acwikjqt11nr6m61krfs",
"output": "8"
},
{
"input": "l3rw91a4m25l8iytxyeuixsegzcbm4h41ornf3pixkrmwznrzc",
"output": "14"
},
{
"input": "2222",
"output": "0"
},
{
"input": "13579",
"output": "5"
},
{
"input": "1",
"output": "1"
},
{
"input": "0",
"output": "0"
},
{
"input": "a",
"output": "1"
},
{
"input": "y",
"output": "0"
},
{
"input": "w",
"output": "0"
},
{
"input": "oo",
"output": "2"
},
{
"input": "oy",
"output": "1"
},
{
"input": "yo",
"output": "1"
},
{
"input": "yy",
"output": "0"
},
{
"input": "a0",
"output": "1"
},
{
"input": "a9",
"output": "2"
},
{
"input": "y0",
"output": "0"
},
{
"input": "y7",
"output": "1"
},
{
"input": "0a",
"output": "1"
},
{
"input": "3a",
"output": "2"
},
{
"input": "06",
"output": "0"
},
{
"input": "07",
"output": "1"
},
{
"input": "70",
"output": "1"
},
{
"input": "77",
"output": "2"
},
{
"input": "13570",
"output": "4"
},
{
"input": "0000000000011111",
"output": "5"
},
{
"input": "1357",
"output": "4"
},
{
"input": "uuuuuuuuuuuuuuuuuu",
"output": "18"
},
{
"input": "gabieurat",
"output": "5"
}
] | 1,699,071,754
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 3
| 108
| 0
|
a=input()
print(sum([a.count(i) for i in 'aeiou24680']))
|
Title: New Year and Counting Cards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Your friend has *n* cards.
You know that each card has a lowercase English letter on one side and a digit on the other.
Currently, your friend has laid out the cards on a table so only one side of each card is visible.
You would like to know if the following statement is true for cards that your friend owns: "If a card has a vowel on one side, then it has an even digit on the other side." More specifically, a vowel is one of 'a', 'e', 'i', 'o' or 'u', and even digit is one of '0', '2', '4', '6' or '8'.
For example, if a card has 'a' on one side, and '6' on the other side, then this statement is true for it. Also, the statement is true, for example, for a card with 'b' and '4', and for a card with 'b' and '3' (since the letter is not a vowel). The statement is false, for example, for card with 'e' and '5'. You are interested if the statement is true for all cards. In particular, if no card has a vowel, the statement is true.
To determine this, you can flip over some cards to reveal the other side. You would like to know what is the minimum number of cards you need to flip in the worst case in order to verify that the statement is true.
Input Specification:
The first and only line of input will contain a string *s* (1<=≤<=|*s*|<=≤<=50), denoting the sides of the cards that you can see on the table currently. Each character of *s* is either a lowercase English letter or a digit.
Output Specification:
Print a single integer, the minimum number of cards you must turn over to verify your claim.
Demo Input:
['ee\n', 'z\n', '0ay1\n']
Demo Output:
['2\n', '0\n', '2\n']
Note:
In the first sample, we must turn over both cards. Note that even though both cards have the same letter, they could possibly have different numbers on the other side.
In the second sample, we don't need to turn over any cards. The statement is vacuously true, since you know your friend has no cards with a vowel on them.
In the third sample, we need to flip the second and fourth cards.
|
```python
a=input()
print(sum([a.count(i) for i in 'aeiou24680']))
```
| 0
|
|
261
|
A
|
Maxim and Discounts
|
PROGRAMMING
| 1,400
|
[
"greedy",
"sortings"
] | null | null |
Maxim always goes to the supermarket on Sundays. Today the supermarket has a special offer of discount systems.
There are *m* types of discounts. We assume that the discounts are indexed from 1 to *m*. To use the discount number *i*, the customer takes a special basket, where he puts exactly *q**i* items he buys. Under the terms of the discount system, in addition to the items in the cart the customer can receive at most two items from the supermarket for free. The number of the "free items" (0, 1 or 2) to give is selected by the customer. The only condition imposed on the selected "free items" is as follows: each of them mustn't be more expensive than the cheapest item out of the *q**i* items in the cart.
Maxim now needs to buy *n* items in the shop. Count the minimum sum of money that Maxim needs to buy them, if he use the discount system optimally well.
Please assume that the supermarket has enough carts for any actions. Maxim can use the same discount multiple times. Of course, Maxim can buy items without any discounts.
|
The first line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of discount types. The second line contains *m* integers: *q*1,<=*q*2,<=...,<=*q**m* (1<=≤<=*q**i*<=≤<=105).
The third line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of items Maxim needs. The fourth line contains *n* integers: *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104) — the items' prices.
The numbers in the lines are separated by single spaces.
|
In a single line print a single integer — the answer to the problem.
|
[
"1\n2\n4\n50 50 100 100\n",
"2\n2 3\n5\n50 50 50 50 50\n",
"1\n1\n7\n1 1 1 1 1 1 1\n"
] |
[
"200\n",
"150\n",
"3\n"
] |
In the first sample Maxim needs to buy two items that cost 100 and get a discount for two free items that cost 50. In that case, Maxim is going to pay 200.
In the second sample the best strategy for Maxim is to buy 3 items and get 2 items for free using the discount. In that case, Maxim is going to pay 150.
| 500
|
[
{
"input": "1\n2\n4\n50 50 100 100",
"output": "200"
},
{
"input": "2\n2 3\n5\n50 50 50 50 50",
"output": "150"
},
{
"input": "1\n1\n7\n1 1 1 1 1 1 1",
"output": "3"
},
{
"input": "60\n7 4 20 15 17 6 2 2 3 18 13 14 16 11 13 12 6 10 14 1 16 6 4 9 10 8 10 15 16 13 13 9 16 11 5 4 11 1 20 5 11 20 19 9 14 13 10 6 6 9 2 13 11 4 1 6 8 18 10 3\n26\n2481 6519 9153 741 9008 6601 6117 1689 5911 2031 2538 5553 1358 6863 7521 4869 6276 5356 5305 6761 5689 7476 5833 257 2157 218",
"output": "44768"
},
{
"input": "88\n8 3 4 3 1 17 5 10 18 12 9 12 4 6 19 14 9 10 10 8 15 11 18 3 11 4 10 11 7 9 14 7 13 2 8 2 15 2 8 16 7 1 9 1 11 13 13 15 8 9 4 2 13 12 12 11 1 5 20 19 13 15 6 6 11 20 14 18 11 20 20 13 8 4 17 12 17 4 13 14 1 20 19 5 7 3 19 16\n33\n7137 685 2583 6751 2104 2596 2329 9948 7961 9545 1797 6507 9241 3844 5657 1887 225 7310 1165 6335 5729 5179 8166 9294 3281 8037 1063 6711 8103 7461 4226 2894 9085",
"output": "61832"
},
{
"input": "46\n11 6 8 8 11 8 2 8 17 3 16 1 9 12 18 2 2 5 17 19 3 9 8 19 2 4 2 15 2 11 13 13 8 6 10 12 7 7 17 15 10 19 7 7 19 6\n71\n6715 8201 9324 276 8441 2378 4829 9303 5721 3895 8193 7725 1246 8845 6863 2897 5001 5055 2745 596 9108 4313 1108 982 6483 7256 4313 8981 9026 9885 2433 2009 8441 7441 9044 6969 2065 6721 424 5478 9128 5921 11 6201 3681 4876 3369 6205 4865 8201 9751 371 2881 7995 641 5841 3595 6041 2403 1361 5121 3801 8031 7909 3809 7741 1026 9633 8711 1907 6363",
"output": "129008"
},
{
"input": "18\n16 16 20 12 13 10 14 15 4 5 6 8 4 11 12 11 16 7\n15\n371 2453 905 1366 6471 4331 4106 2570 4647 1648 7911 2147 1273 6437 3393",
"output": "38578"
},
{
"input": "2\n12 4\n28\n5366 5346 1951 3303 1613 5826 8035 7079 7633 6155 9811 9761 3207 4293 3551 5245 7891 4463 3981 2216 3881 1751 4495 96 671 1393 1339 4241",
"output": "89345"
},
{
"input": "57\n3 13 20 17 18 18 17 2 17 8 20 2 11 12 11 14 4 20 9 20 14 19 20 4 4 8 8 18 17 16 18 10 4 7 9 8 10 8 20 4 11 8 12 16 16 4 11 12 16 1 6 14 11 12 19 8 20\n7\n5267 7981 1697 826 6889 1949 2413",
"output": "11220"
},
{
"input": "48\n14 2 5 3 10 10 5 6 14 8 19 13 4 4 3 13 18 19 9 16 3 1 14 9 13 10 13 4 12 11 8 2 18 20 14 11 3 11 18 11 4 2 7 2 18 19 2 8\n70\n9497 5103 1001 2399 5701 4053 3557 8481 1736 4139 5829 1107 6461 4089 5936 7961 6017 1416 1191 4635 4288 5605 8857 1822 71 1435 2837 5523 6993 2404 2840 8251 765 5678 7834 8595 3091 7073 8673 2299 2685 7729 8017 3171 9155 431 3773 7927 671 4063 1123 5384 2721 7901 2315 5199 8081 7321 8196 2887 9384 56 7501 1931 4769 2055 7489 3681 6321 8489",
"output": "115395"
},
{
"input": "1\n1\n1\n1",
"output": "1"
},
{
"input": "1\n2\n1\n1",
"output": "1"
},
{
"input": "1\n1\n3\n3 1 1",
"output": "3"
}
] | 1,616,817,986
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 45
| 310
| 7,680,000
|
I=lambda:map(int,input().split())
m,q,n,a,r,k=int(input()),min(I()),int(input())-1,sorted(I()),0,0
while n>-1:
r+=a[n]
k+=1
if k==q:n-=3;k=0
else:n-=1
print(r)
|
Title: Maxim and Discounts
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Maxim always goes to the supermarket on Sundays. Today the supermarket has a special offer of discount systems.
There are *m* types of discounts. We assume that the discounts are indexed from 1 to *m*. To use the discount number *i*, the customer takes a special basket, where he puts exactly *q**i* items he buys. Under the terms of the discount system, in addition to the items in the cart the customer can receive at most two items from the supermarket for free. The number of the "free items" (0, 1 or 2) to give is selected by the customer. The only condition imposed on the selected "free items" is as follows: each of them mustn't be more expensive than the cheapest item out of the *q**i* items in the cart.
Maxim now needs to buy *n* items in the shop. Count the minimum sum of money that Maxim needs to buy them, if he use the discount system optimally well.
Please assume that the supermarket has enough carts for any actions. Maxim can use the same discount multiple times. Of course, Maxim can buy items without any discounts.
Input Specification:
The first line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of discount types. The second line contains *m* integers: *q*1,<=*q*2,<=...,<=*q**m* (1<=≤<=*q**i*<=≤<=105).
The third line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of items Maxim needs. The fourth line contains *n* integers: *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104) — the items' prices.
The numbers in the lines are separated by single spaces.
Output Specification:
In a single line print a single integer — the answer to the problem.
Demo Input:
['1\n2\n4\n50 50 100 100\n', '2\n2 3\n5\n50 50 50 50 50\n', '1\n1\n7\n1 1 1 1 1 1 1\n']
Demo Output:
['200\n', '150\n', '3\n']
Note:
In the first sample Maxim needs to buy two items that cost 100 and get a discount for two free items that cost 50. In that case, Maxim is going to pay 200.
In the second sample the best strategy for Maxim is to buy 3 items and get 2 items for free using the discount. In that case, Maxim is going to pay 150.
|
```python
I=lambda:map(int,input().split())
m,q,n,a,r,k=int(input()),min(I()),int(input())-1,sorted(I()),0,0
while n>-1:
r+=a[n]
k+=1
if k==q:n-=3;k=0
else:n-=1
print(r)
```
| 3
|
|
723
|
A
|
The New Year: Meeting Friends
|
PROGRAMMING
| 800
|
[
"implementation",
"math",
"sortings"
] | null | null |
There are three friend living on the straight line *Ox* in Lineland. The first friend lives at the point *x*1, the second friend lives at the point *x*2, and the third friend lives at the point *x*3. They plan to celebrate the New Year together, so they need to meet at one point. What is the minimum total distance they have to travel in order to meet at some point and celebrate the New Year?
It's guaranteed that the optimal answer is always integer.
|
The first line of the input contains three distinct integers *x*1, *x*2 and *x*3 (1<=≤<=*x*1,<=*x*2,<=*x*3<=≤<=100) — the coordinates of the houses of the first, the second and the third friends respectively.
|
Print one integer — the minimum total distance the friends need to travel in order to meet together.
|
[
"7 1 4\n",
"30 20 10\n"
] |
[
"6\n",
"20\n"
] |
In the first sample, friends should meet at the point 4. Thus, the first friend has to travel the distance of 3 (from the point 7 to the point 4), the second friend also has to travel the distance of 3 (from the point 1 to the point 4), while the third friend should not go anywhere because he lives at the point 4.
| 500
|
[
{
"input": "7 1 4",
"output": "6"
},
{
"input": "30 20 10",
"output": "20"
},
{
"input": "1 4 100",
"output": "99"
},
{
"input": "100 1 91",
"output": "99"
},
{
"input": "1 45 100",
"output": "99"
},
{
"input": "1 2 3",
"output": "2"
},
{
"input": "71 85 88",
"output": "17"
},
{
"input": "30 38 99",
"output": "69"
},
{
"input": "23 82 95",
"output": "72"
},
{
"input": "22 41 47",
"output": "25"
},
{
"input": "9 94 77",
"output": "85"
},
{
"input": "1 53 51",
"output": "52"
},
{
"input": "25 97 93",
"output": "72"
},
{
"input": "42 53 51",
"output": "11"
},
{
"input": "81 96 94",
"output": "15"
},
{
"input": "21 5 93",
"output": "88"
},
{
"input": "50 13 75",
"output": "62"
},
{
"input": "41 28 98",
"output": "70"
},
{
"input": "69 46 82",
"output": "36"
},
{
"input": "87 28 89",
"output": "61"
},
{
"input": "44 45 40",
"output": "5"
},
{
"input": "86 97 68",
"output": "29"
},
{
"input": "43 92 30",
"output": "62"
},
{
"input": "16 70 1",
"output": "69"
},
{
"input": "40 46 19",
"output": "27"
},
{
"input": "71 38 56",
"output": "33"
},
{
"input": "82 21 80",
"output": "61"
},
{
"input": "75 8 35",
"output": "67"
},
{
"input": "75 24 28",
"output": "51"
},
{
"input": "78 23 56",
"output": "55"
},
{
"input": "85 31 10",
"output": "75"
},
{
"input": "76 50 9",
"output": "67"
},
{
"input": "95 37 34",
"output": "61"
},
{
"input": "84 61 35",
"output": "49"
},
{
"input": "87 85 37",
"output": "50"
},
{
"input": "1 3 2",
"output": "2"
},
{
"input": "4 2 6",
"output": "4"
},
{
"input": "6 9 3",
"output": "6"
},
{
"input": "12 4 8",
"output": "8"
},
{
"input": "15 10 5",
"output": "10"
},
{
"input": "1 50 17",
"output": "49"
},
{
"input": "10 5 15",
"output": "10"
},
{
"input": "8 1 9",
"output": "8"
},
{
"input": "3 5 4",
"output": "2"
},
{
"input": "2 1 3",
"output": "2"
},
{
"input": "1 8 2",
"output": "7"
},
{
"input": "1 100 2",
"output": "99"
},
{
"input": "1 4 6",
"output": "5"
}
] | 1,687,544,285
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 48
| 46
| 0
|
x1, x2, x3 = map(int, input().split())
min_coordinate = min(x1, x2, x3)
max_coordinate = max(x1, x2, x3)
min_total_distance = max_coordinate - min_coordinate
print(min_total_distance)
|
Title: The New Year: Meeting Friends
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are three friend living on the straight line *Ox* in Lineland. The first friend lives at the point *x*1, the second friend lives at the point *x*2, and the third friend lives at the point *x*3. They plan to celebrate the New Year together, so they need to meet at one point. What is the minimum total distance they have to travel in order to meet at some point and celebrate the New Year?
It's guaranteed that the optimal answer is always integer.
Input Specification:
The first line of the input contains three distinct integers *x*1, *x*2 and *x*3 (1<=≤<=*x*1,<=*x*2,<=*x*3<=≤<=100) — the coordinates of the houses of the first, the second and the third friends respectively.
Output Specification:
Print one integer — the minimum total distance the friends need to travel in order to meet together.
Demo Input:
['7 1 4\n', '30 20 10\n']
Demo Output:
['6\n', '20\n']
Note:
In the first sample, friends should meet at the point 4. Thus, the first friend has to travel the distance of 3 (from the point 7 to the point 4), the second friend also has to travel the distance of 3 (from the point 1 to the point 4), while the third friend should not go anywhere because he lives at the point 4.
|
```python
x1, x2, x3 = map(int, input().split())
min_coordinate = min(x1, x2, x3)
max_coordinate = max(x1, x2, x3)
min_total_distance = max_coordinate - min_coordinate
print(min_total_distance)
```
| 3
|
|
71
|
A
|
Way Too Long Words
|
PROGRAMMING
| 800
|
[
"strings"
] |
A. Way Too Long Words
|
1
|
256
|
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
|
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
|
[
"4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n"
] |
[
"word\nl10n\ni18n\np43s\n"
] |
none
| 500
|
[
{
"input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis",
"output": "word\nl10n\ni18n\np43s"
},
{
"input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm",
"output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m"
},
{
"input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd",
"output": "n20n\nj27b\nk15d"
},
{
"input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt",
"output": "t41t"
},
{
"input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings",
"output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz",
"output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz"
},
{
"input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij",
"output": "a98j"
},
{
"input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco",
"output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco"
},
{
"input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt",
"output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t"
},
{
"input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr",
"output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr"
}
] | 1,686,166,622
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
ex=int(input(""))
for i in range(ex):
word=input("")
if len(word)>4:
word=word[0]+str(len(word)-2)+word[len(word)-1]
print(word)
|
Title: Way Too Long Words
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output Specification:
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
Demo Input:
['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n']
Demo Output:
['word\nl10n\ni18n\np43s\n']
Note:
none
|
```python
ex=int(input(""))
for i in range(ex):
word=input("")
if len(word)>4:
word=word[0]+str(len(word)-2)+word[len(word)-1]
print(word)
```
| 0
|
148
|
A
|
Insomnia cure
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"implementation",
"math"
] | null | null |
«One dragon. Two dragon. Three dragon», — the princess was counting. She had trouble falling asleep, and she got bored of counting lambs when she was nine.
However, just counting dragons was boring as well, so she entertained herself at best she could. Tonight she imagined that all dragons were here to steal her, and she was fighting them off. Every *k*-th dragon got punched in the face with a frying pan. Every *l*-th dragon got his tail shut into the balcony door. Every *m*-th dragon got his paws trampled with sharp heels. Finally, she threatened every *n*-th dragon to call her mom, and he withdrew in panic.
How many imaginary dragons suffered moral or physical damage tonight, if the princess counted a total of *d* dragons?
|
Input data contains integer numbers *k*,<=*l*,<=*m*,<=*n* and *d*, each number in a separate line (1<=≤<=*k*,<=*l*,<=*m*,<=*n*<=≤<=10, 1<=≤<=*d*<=≤<=105).
|
Output the number of damaged dragons.
|
[
"1\n2\n3\n4\n12\n",
"2\n3\n4\n5\n24\n"
] |
[
"12\n",
"17\n"
] |
In the first case every first dragon got punched with a frying pan. Some of the dragons suffered from other reasons as well, but the pan alone would be enough.
In the second case dragons 1, 7, 11, 13, 17, 19 and 23 escaped unharmed.
| 1,000
|
[
{
"input": "1\n2\n3\n4\n12",
"output": "12"
},
{
"input": "2\n3\n4\n5\n24",
"output": "17"
},
{
"input": "1\n1\n1\n1\n100000",
"output": "100000"
},
{
"input": "10\n9\n8\n7\n6",
"output": "0"
},
{
"input": "8\n4\n4\n3\n65437",
"output": "32718"
},
{
"input": "8\n4\n1\n10\n59392",
"output": "59392"
},
{
"input": "4\n1\n8\n7\n44835",
"output": "44835"
},
{
"input": "6\n1\n7\n2\n62982",
"output": "62982"
},
{
"input": "2\n7\n4\n9\n56937",
"output": "35246"
},
{
"input": "2\n9\n8\n1\n75083",
"output": "75083"
},
{
"input": "8\n7\n7\n6\n69038",
"output": "24656"
},
{
"input": "4\n4\n2\n3\n54481",
"output": "36320"
},
{
"input": "6\n4\n9\n8\n72628",
"output": "28244"
},
{
"input": "9\n7\n8\n10\n42357",
"output": "16540"
},
{
"input": "5\n6\n4\n3\n60504",
"output": "36302"
},
{
"input": "7\n2\n3\n8\n21754",
"output": "15539"
},
{
"input": "1\n2\n10\n4\n39901",
"output": "39901"
},
{
"input": "3\n4\n7\n1\n58048",
"output": "58048"
},
{
"input": "9\n10\n4\n6\n52003",
"output": "21956"
},
{
"input": "5\n10\n9\n3\n70149",
"output": "32736"
},
{
"input": "5\n5\n5\n10\n55592",
"output": "11118"
},
{
"input": "1\n5\n2\n6\n49547",
"output": "49547"
},
{
"input": "3\n7\n7\n7\n84046",
"output": "36019"
},
{
"input": "10\n2\n1\n8\n63537",
"output": "63537"
},
{
"input": "7\n2\n10\n5\n81684",
"output": "53678"
},
{
"input": "7\n1\n6\n8\n99831",
"output": "99831"
},
{
"input": "3\n9\n3\n3\n61082",
"output": "20360"
},
{
"input": "5\n5\n8\n9\n79228",
"output": "29931"
},
{
"input": "1\n5\n5\n6\n89535",
"output": "89535"
},
{
"input": "5\n5\n2\n1\n7682",
"output": "7682"
},
{
"input": "7\n4\n8\n8\n25829",
"output": "9224"
},
{
"input": "3\n4\n3\n3\n87079",
"output": "43539"
},
{
"input": "5\n8\n4\n9\n5226",
"output": "2438"
},
{
"input": "4\n9\n6\n6\n1070",
"output": "415"
},
{
"input": "10\n10\n10\n10\n100000",
"output": "10000"
},
{
"input": "1\n1\n1\n1\n1",
"output": "1"
},
{
"input": "10\n10\n10\n10\n1",
"output": "0"
},
{
"input": "10\n10\n1\n10\n10",
"output": "10"
},
{
"input": "10\n10\n1\n10\n100",
"output": "100"
},
{
"input": "2\n2\n2\n2\n1",
"output": "0"
}
] | 1,691,436,064
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 40
| 124
| 2,560,000
|
k=int(input())
l=int(input())
m=int(input())
n=int(input())
d=int(input())
f=[0]*d
res=0
for i in range(1,d+1):
if f[i-1]==0:
if i%k==0:
f[i-1]=1
res+=1
elif i%l==0:
f[i-1]=1
res+=1
elif i%m==0:
f[i-1]=1
res+=1
elif i%n==0:
f[i-1]=1
res+=1
print(res)
|
Title: Insomnia cure
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
«One dragon. Two dragon. Three dragon», — the princess was counting. She had trouble falling asleep, and she got bored of counting lambs when she was nine.
However, just counting dragons was boring as well, so she entertained herself at best she could. Tonight she imagined that all dragons were here to steal her, and she was fighting them off. Every *k*-th dragon got punched in the face with a frying pan. Every *l*-th dragon got his tail shut into the balcony door. Every *m*-th dragon got his paws trampled with sharp heels. Finally, she threatened every *n*-th dragon to call her mom, and he withdrew in panic.
How many imaginary dragons suffered moral or physical damage tonight, if the princess counted a total of *d* dragons?
Input Specification:
Input data contains integer numbers *k*,<=*l*,<=*m*,<=*n* and *d*, each number in a separate line (1<=≤<=*k*,<=*l*,<=*m*,<=*n*<=≤<=10, 1<=≤<=*d*<=≤<=105).
Output Specification:
Output the number of damaged dragons.
Demo Input:
['1\n2\n3\n4\n12\n', '2\n3\n4\n5\n24\n']
Demo Output:
['12\n', '17\n']
Note:
In the first case every first dragon got punched with a frying pan. Some of the dragons suffered from other reasons as well, but the pan alone would be enough.
In the second case dragons 1, 7, 11, 13, 17, 19 and 23 escaped unharmed.
|
```python
k=int(input())
l=int(input())
m=int(input())
n=int(input())
d=int(input())
f=[0]*d
res=0
for i in range(1,d+1):
if f[i-1]==0:
if i%k==0:
f[i-1]=1
res+=1
elif i%l==0:
f[i-1]=1
res+=1
elif i%m==0:
f[i-1]=1
res+=1
elif i%n==0:
f[i-1]=1
res+=1
print(res)
```
| 3
|
|
688
|
B
|
Lovely Palindromes
|
PROGRAMMING
| 1,000
|
[
"constructive algorithms",
"math"
] | null | null |
Pari has a friend who loves palindrome numbers. A palindrome number is a number that reads the same forward or backward. For example 12321, 100001 and 1 are palindrome numbers, while 112 and 1021 are not.
Pari is trying to love them too, but only very special and gifted people can understand the beauty behind palindrome numbers. Pari loves integers with even length (i.e. the numbers with even number of digits), so she tries to see a lot of big palindrome numbers with even length (like a 2-digit 11 or 6-digit 122221), so maybe she could see something in them.
Now Pari asks you to write a program that gets a huge integer *n* from the input and tells what is the *n*-th even-length positive palindrome number?
|
The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=10100<=000).
|
Print the *n*-th even-length palindrome number.
|
[
"1\n",
"10\n"
] |
[
"11\n",
"1001\n"
] |
The first 10 even-length palindrome numbers are 11, 22, 33, ... , 88, 99 and 1001.
| 1,000
|
[
{
"input": "1",
"output": "11"
},
{
"input": "10",
"output": "1001"
},
{
"input": "11",
"output": "1111"
},
{
"input": "12",
"output": "1221"
},
{
"input": "100",
"output": "100001"
},
{
"input": "1321",
"output": "13211231"
},
{
"input": "2",
"output": "22"
},
{
"input": "3",
"output": "33"
},
{
"input": "4",
"output": "44"
},
{
"input": "5",
"output": "55"
},
{
"input": "6",
"output": "66"
},
{
"input": "7",
"output": "77"
},
{
"input": "8",
"output": "88"
},
{
"input": "9",
"output": "99"
},
{
"input": "13",
"output": "1331"
},
{
"input": "14",
"output": "1441"
},
{
"input": "15",
"output": "1551"
},
{
"input": "16",
"output": "1661"
},
{
"input": "17",
"output": "1771"
},
{
"input": "18",
"output": "1881"
},
{
"input": "19",
"output": "1991"
},
{
"input": "20",
"output": "2002"
},
{
"input": "26550",
"output": "2655005562"
},
{
"input": "16137",
"output": "1613773161"
},
{
"input": "91471",
"output": "9147117419"
},
{
"input": "41242",
"output": "4124224214"
},
{
"input": "30866",
"output": "3086666803"
},
{
"input": "4442231232741003009964183199713851566466079882929007253443848371978048610682180431295671905105980116",
"output": "44422312327410030099641831997138515664660798829290072534438483719780486106821804312956719051059801166110895015091765921340812860168408791738483443527009292889706646651583179913814699003001472321322444"
},
{
"input": "7806526084732010742850470902979267046151204705563201965818390803445852115032119297648697933316807252",
"output": "78065260847320107428504709029792670461512047055632019658183908034458521150321192976486979333168072522527086133397968467929112305112585443080938185691023655074021516407629792090740582470102374806256087"
},
{
"input": "2244399823612183124978103773977739948197709395321563148611877999503840818596183803323258317041875913",
"output": "22443998236121831249781037739777399481977093953215631486118779995038408185961838033232583170418759133195781407138523233083816958180483059997781168413651235939077918499377793773018794213812163289934422"
},
{
"input": "123451",
"output": "123451154321"
}
] | 1,626,956,579
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 40
| 249
| 30,924,800
|
n = input()
m = list(n)
p = list(m)
p.reverse()
result = m + p
for x in result:
print(x,end='')
|
Title: Lovely Palindromes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pari has a friend who loves palindrome numbers. A palindrome number is a number that reads the same forward or backward. For example 12321, 100001 and 1 are palindrome numbers, while 112 and 1021 are not.
Pari is trying to love them too, but only very special and gifted people can understand the beauty behind palindrome numbers. Pari loves integers with even length (i.e. the numbers with even number of digits), so she tries to see a lot of big palindrome numbers with even length (like a 2-digit 11 or 6-digit 122221), so maybe she could see something in them.
Now Pari asks you to write a program that gets a huge integer *n* from the input and tells what is the *n*-th even-length positive palindrome number?
Input Specification:
The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=10100<=000).
Output Specification:
Print the *n*-th even-length palindrome number.
Demo Input:
['1\n', '10\n']
Demo Output:
['11\n', '1001\n']
Note:
The first 10 even-length palindrome numbers are 11, 22, 33, ... , 88, 99 and 1001.
|
```python
n = input()
m = list(n)
p = list(m)
p.reverse()
result = m + p
for x in result:
print(x,end='')
```
| 3
|
|
732
|
A
|
Buy a Shovel
|
PROGRAMMING
| 800
|
[
"brute force",
"constructive algorithms",
"implementation",
"math"
] | null | null |
Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=≤<=*r*<=≤<=9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.
|
The single line of input contains two integers *k* and *r* (1<=≤<=*k*<=≤<=1000, 1<=≤<=*r*<=≤<=9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.
|
Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.
|
[
"117 3\n",
"237 7\n",
"15 2\n"
] |
[
"9\n",
"1\n",
"2\n"
] |
In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change.
| 500
|
[
{
"input": "117 3",
"output": "9"
},
{
"input": "237 7",
"output": "1"
},
{
"input": "15 2",
"output": "2"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 9",
"output": "9"
},
{
"input": "1000 3",
"output": "1"
},
{
"input": "1000 1",
"output": "1"
},
{
"input": "1000 9",
"output": "1"
},
{
"input": "1 2",
"output": "2"
},
{
"input": "999 9",
"output": "1"
},
{
"input": "999 8",
"output": "2"
},
{
"input": "105 6",
"output": "2"
},
{
"input": "403 9",
"output": "3"
},
{
"input": "546 4",
"output": "4"
},
{
"input": "228 9",
"output": "5"
},
{
"input": "57 2",
"output": "6"
},
{
"input": "437 9",
"output": "7"
},
{
"input": "997 6",
"output": "8"
},
{
"input": "109 1",
"output": "9"
},
{
"input": "998 9",
"output": "5"
},
{
"input": "4 2",
"output": "3"
},
{
"input": "9 3",
"output": "7"
},
{
"input": "8 2",
"output": "4"
},
{
"input": "1 3",
"output": "3"
},
{
"input": "1 4",
"output": "4"
},
{
"input": "1 5",
"output": "5"
},
{
"input": "1 6",
"output": "6"
},
{
"input": "1 7",
"output": "7"
},
{
"input": "1 8",
"output": "8"
},
{
"input": "100 3",
"output": "1"
},
{
"input": "1000 2",
"output": "1"
},
{
"input": "1000 4",
"output": "1"
},
{
"input": "1000 5",
"output": "1"
},
{
"input": "1000 6",
"output": "1"
},
{
"input": "1000 7",
"output": "1"
},
{
"input": "1000 8",
"output": "1"
},
{
"input": "23 4",
"output": "8"
},
{
"input": "33 1",
"output": "7"
},
{
"input": "33 2",
"output": "4"
},
{
"input": "666 5",
"output": "5"
},
{
"input": "2 3",
"output": "5"
},
{
"input": "5 5",
"output": "1"
},
{
"input": "3 6",
"output": "2"
},
{
"input": "12 4",
"output": "2"
},
{
"input": "15 5",
"output": "1"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "25 5",
"output": "1"
},
{
"input": "2 9",
"output": "5"
},
{
"input": "6 7",
"output": "5"
},
{
"input": "8 9",
"output": "5"
},
{
"input": "2 7",
"output": "5"
},
{
"input": "4 7",
"output": "5"
},
{
"input": "2 1",
"output": "5"
},
{
"input": "261 1",
"output": "1"
}
] | 1,678,334,352
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 54
| 62
| 0
|
k,r=map(int,input().split())
x=1
while (k*x)%10!=r and (k*x)%10!=0:
x+=1
print(x)
|
Title: Buy a Shovel
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=≤<=*r*<=≤<=9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.
Input Specification:
The single line of input contains two integers *k* and *r* (1<=≤<=*k*<=≤<=1000, 1<=≤<=*r*<=≤<=9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.
Output Specification:
Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.
Demo Input:
['117 3\n', '237 7\n', '15 2\n']
Demo Output:
['9\n', '1\n', '2\n']
Note:
In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change.
|
```python
k,r=map(int,input().split())
x=1
while (k*x)%10!=r and (k*x)%10!=0:
x+=1
print(x)
```
| 3
|
|
678
|
A
|
Johny Likes Numbers
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
Johny likes numbers *n* and *k* very much. Now Johny wants to find the smallest integer *x* greater than *n*, so it is divisible by the number *k*.
|
The only line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=109).
|
Print the smallest integer *x*<=><=*n*, so it is divisible by the number *k*.
|
[
"5 3\n",
"25 13\n",
"26 13\n"
] |
[
"6\n",
"26\n",
"39\n"
] |
none
| 0
|
[
{
"input": "5 3",
"output": "6"
},
{
"input": "25 13",
"output": "26"
},
{
"input": "26 13",
"output": "39"
},
{
"input": "1 1",
"output": "2"
},
{
"input": "8 8",
"output": "16"
},
{
"input": "14 15",
"output": "15"
},
{
"input": "197 894",
"output": "894"
},
{
"input": "6058 8581",
"output": "8581"
},
{
"input": "97259 41764",
"output": "125292"
},
{
"input": "453145 333625",
"output": "667250"
},
{
"input": "2233224 4394826",
"output": "4394826"
},
{
"input": "76770926 13350712",
"output": "80104272"
},
{
"input": "687355301 142098087",
"output": "710490435"
},
{
"input": "1000000000 999999999",
"output": "1999999998"
},
{
"input": "1000000000 1000000000",
"output": "2000000000"
},
{
"input": "999999999 1000000000",
"output": "1000000000"
},
{
"input": "1000000000 1",
"output": "1000000001"
},
{
"input": "1000000000 2",
"output": "1000000002"
},
{
"input": "999999999 1",
"output": "1000000000"
},
{
"input": "100000000 1",
"output": "100000001"
},
{
"input": "999999999 500000000",
"output": "1000000000"
},
{
"input": "999999990 10",
"output": "1000000000"
},
{
"input": "1000000000 999999997",
"output": "1999999994"
},
{
"input": "999999999 2",
"output": "1000000000"
},
{
"input": "999999984 1",
"output": "999999985"
},
{
"input": "999999983 1",
"output": "999999984"
},
{
"input": "666666666 1",
"output": "666666667"
},
{
"input": "1000000000 990000000",
"output": "1980000000"
},
{
"input": "41 48",
"output": "48"
},
{
"input": "123456 2",
"output": "123458"
},
{
"input": "111 111",
"output": "222"
},
{
"input": "878787 1",
"output": "878788"
},
{
"input": "121 1",
"output": "122"
},
{
"input": "114514 114514",
"output": "229028"
},
{
"input": "500000001 1000000000",
"output": "1000000000"
},
{
"input": "999999997 1",
"output": "999999998"
},
{
"input": "100000000 10",
"output": "100000010"
}
] | 1,603,431,259
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 93
| 0
|
t,x = list(map(int,input().strip().split(' ')))
print(((t-1)//x+1)*x)
|
Title: Johny Likes Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Johny likes numbers *n* and *k* very much. Now Johny wants to find the smallest integer *x* greater than *n*, so it is divisible by the number *k*.
Input Specification:
The only line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=109).
Output Specification:
Print the smallest integer *x*<=><=*n*, so it is divisible by the number *k*.
Demo Input:
['5 3\n', '25 13\n', '26 13\n']
Demo Output:
['6\n', '26\n', '39\n']
Note:
none
|
```python
t,x = list(map(int,input().strip().split(' ')))
print(((t-1)//x+1)*x)
```
| 0
|
|
383
|
D
|
Antimatter
|
PROGRAMMING
| 2,300
|
[
"dp"
] | null | null |
Iahub accidentally discovered a secret lab. He found there *n* devices ordered in a line, numbered from 1 to *n* from left to right. Each device *i* (1<=≤<=*i*<=≤<=*n*) can create either *a**i* units of matter or *a**i* units of antimatter.
Iahub wants to choose some contiguous subarray of devices in the lab, specify the production mode for each of them (produce matter or antimatter) and finally take a photo of it. However he will be successful only if the amounts of matter and antimatter produced in the selected subarray will be the same (otherwise there would be overflowing matter or antimatter in the photo).
You are requested to compute the number of different ways Iahub can successful take a photo. A photo is different than another if it represents another subarray, or if at least one device of the subarray is set to produce matter in one of the photos and antimatter in the other one.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=1000). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=1000).
The sum *a*1<=+<=*a*2<=+<=...<=+<=*a**n* will be less than or equal to 10000.
|
Output a single integer, the number of ways Iahub can take a photo, modulo 1000000007 (109<=+<=7).
|
[
"4\n1 1 1 1\n"
] |
[
"12\n"
] |
The possible photos are [1+, 2-], [1-, 2+], [2+, 3-], [2-, 3+], [3+, 4-], [3-, 4+], [1+, 2+, 3-, 4-], [1+, 2-, 3+, 4-], [1+, 2-, 3-, 4+], [1-, 2+, 3+, 4-], [1-, 2+, 3-, 4+] and [1-, 2-, 3+, 4+], where "*i*+" means that the *i*-th element produces matter, and "*i*-" means that the *i*-th element produces antimatter.
| 2,000
|
[
{
"input": "4\n1 1 1 1",
"output": "12"
},
{
"input": "10\n16 9 9 11 10 12 9 6 10 8",
"output": "86"
},
{
"input": "50\n2 1 5 2 1 3 1 2 3 2 1 1 5 2 2 2 3 2 1 2 2 2 3 3 1 3 1 1 2 2 2 2 1 2 3 1 2 4 1 1 1 3 2 1 1 1 3 2 1 3",
"output": "115119382"
},
{
"input": "100\n8 3 3 7 3 6 4 6 9 4 6 5 5 5 4 3 4 2 3 5 3 6 5 3 6 5 6 6 2 6 4 5 5 4 6 4 3 2 8 5 6 6 7 4 4 9 5 6 6 3 7 1 6 2 6 5 9 3 8 6 2 6 3 2 4 4 3 5 4 7 6 5 4 6 3 5 6 8 8 6 3 7 7 1 4 6 8 6 5 3 7 8 4 7 5 3 8 5 4 4",
"output": "450259307"
},
{
"input": "250\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "533456111"
},
{
"input": "250\n6 1 4 3 3 7 4 5 3 2 4 4 2 5 4 2 1 7 6 2 4 5 3 3 4 5 3 4 5 4 6 4 6 5 3 3 1 5 4 5 3 4 2 4 2 5 1 4 3 3 3 2 6 6 4 7 2 6 5 3 3 6 5 2 1 3 3 5 2 2 3 7 3 5 6 4 7 3 5 3 4 5 5 4 11 5 1 5 3 3 3 1 4 6 4 4 5 5 5 5 2 5 5 3 2 2 5 6 10 5 4 2 5 4 2 5 5 3 4 2 5 4 3 2 4 4 2 5 4 1 5 3 9 6 4 6 3 5 4 5 3 6 7 4 5 5 3 6 2 6 3 3 4 5 6 3 3 3 5 2 4 4 4 5 4 2 5 4 6 5 3 3 6 3 1 5 6 5 4 6 2 3 4 4 5 2 1 7 4 5 5 5 2 2 7 6 1 5 3 2 7 5 8 2 2 2 3 5 2 4 4 2 2 6 4 6 3 2 8 3 4 7 3 2 7 3 5 5 3 2 2 4 5 3 4 3 5 3 5 4 3 1 2 4 7 4 2 3 3 5",
"output": "377970747"
},
{
"input": "250\n2 2 2 2 3 2 4 2 3 2 5 1 2 3 4 4 5 3 5 1 2 5 2 3 5 3 2 3 3 3 5 1 5 5 5 4 1 3 2 5 1 2 3 5 3 3 5 2 1 1 3 3 5 1 4 2 3 3 2 2 3 5 5 4 1 4 1 5 1 3 3 4 1 5 2 5 5 3 2 4 4 4 4 3 5 1 3 4 3 4 2 1 4 3 5 1 2 3 4 2 5 5 3 2 5 3 5 4 2 3 2 3 1 1 2 4 2 5 2 3 3 2 4 5 4 2 2 5 5 5 5 4 3 4 5 2 2 3 3 4 5 1 5 5 2 5 1 5 5 4 4 1 4 2 1 2 1 2 2 3 1 4 5 4 2 4 5 1 1 3 2 1 4 1 5 2 3 1 2 3 2 3 3 2 4 2 5 5 2 3 4 2 2 4 2 4 1 5 5 3 1 3 4 5 2 5 5 1 3 1 3 3 2 5 3 5 2 4 3 5 5 3 3 2 3 2 5 3 4 3 5 3 3 4 5 3 1 2 2 5 4 4 5 1 4 1 2 5 2 3",
"output": "257270797"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "2\n1 1",
"output": "2"
},
{
"input": "2\n1000 1000",
"output": "2"
},
{
"input": "2\n1 2",
"output": "0"
},
{
"input": "3\n1 2 4",
"output": "0"
},
{
"input": "3\n1 2 2",
"output": "2"
},
{
"input": "1\n1000",
"output": "0"
},
{
"input": "3\n999 999 999",
"output": "4"
}
] | 1,692,029,568
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 46
| 0
|
print("_RANDOM_GUESS_1692029568.437577")# 1692029568.437591
|
Title: Antimatter
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Iahub accidentally discovered a secret lab. He found there *n* devices ordered in a line, numbered from 1 to *n* from left to right. Each device *i* (1<=≤<=*i*<=≤<=*n*) can create either *a**i* units of matter or *a**i* units of antimatter.
Iahub wants to choose some contiguous subarray of devices in the lab, specify the production mode for each of them (produce matter or antimatter) and finally take a photo of it. However he will be successful only if the amounts of matter and antimatter produced in the selected subarray will be the same (otherwise there would be overflowing matter or antimatter in the photo).
You are requested to compute the number of different ways Iahub can successful take a photo. A photo is different than another if it represents another subarray, or if at least one device of the subarray is set to produce matter in one of the photos and antimatter in the other one.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=1000). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=1000).
The sum *a*1<=+<=*a*2<=+<=...<=+<=*a**n* will be less than or equal to 10000.
Output Specification:
Output a single integer, the number of ways Iahub can take a photo, modulo 1000000007 (109<=+<=7).
Demo Input:
['4\n1 1 1 1\n']
Demo Output:
['12\n']
Note:
The possible photos are [1+, 2-], [1-, 2+], [2+, 3-], [2-, 3+], [3+, 4-], [3-, 4+], [1+, 2+, 3-, 4-], [1+, 2-, 3+, 4-], [1+, 2-, 3-, 4+], [1-, 2+, 3+, 4-], [1-, 2+, 3-, 4+] and [1-, 2-, 3+, 4+], where "*i*+" means that the *i*-th element produces matter, and "*i*-" means that the *i*-th element produces antimatter.
|
```python
print("_RANDOM_GUESS_1692029568.437577")# 1692029568.437591
```
| 0
|
|
918
|
A
|
Eleven
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation"
] | null | null |
Eleven wants to choose a new name for herself. As a bunch of geeks, her friends suggested an algorithm to choose a name for her. Eleven wants her name to have exactly *n* characters.
Her friend suggested that her name should only consist of uppercase and lowercase letters 'O'. More precisely, they suggested that the *i*-th letter of her name should be 'O' (uppercase) if *i* is a member of Fibonacci sequence, and 'o' (lowercase) otherwise. The letters in the name are numbered from 1 to *n*. Fibonacci sequence is the sequence *f* where
- *f*1<==<=1, - *f*2<==<=1, - *f**n*<==<=*f**n*<=-<=2<=+<=*f**n*<=-<=1 (*n*<=><=2).
As her friends are too young to know what Fibonacci sequence is, they asked you to help Eleven determine her new name.
|
The first and only line of input contains an integer *n* (1<=≤<=*n*<=≤<=1000).
|
Print Eleven's new name on the first and only line of output.
|
[
"8\n",
"15\n"
] |
[
"OOOoOooO\n",
"OOOoOooOooooOoo\n"
] |
none
| 500
|
[
{
"input": "8",
"output": "OOOoOooO"
},
{
"input": "15",
"output": "OOOoOooOooooOoo"
},
{
"input": "85",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooo"
},
{
"input": "381",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooo"
},
{
"input": "805",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "1000",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "1",
"output": "O"
},
{
"input": "2",
"output": "OO"
},
{
"input": "3",
"output": "OOO"
},
{
"input": "5",
"output": "OOOoO"
},
{
"input": "17",
"output": "OOOoOooOooooOoooo"
},
{
"input": "49",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooo"
},
{
"input": "256",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooo"
},
{
"input": "512",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "933",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "61",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooo"
},
{
"input": "781",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "999",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
}
] | 1,550,866,169
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 18
| 109
| 102,400
|
a = b = 1
for i in range(int(input())):
if b == i + 1:
a, b = (b, a + b)
print('O', end='')
else:
print('o', end='')
|
Title: Eleven
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Eleven wants to choose a new name for herself. As a bunch of geeks, her friends suggested an algorithm to choose a name for her. Eleven wants her name to have exactly *n* characters.
Her friend suggested that her name should only consist of uppercase and lowercase letters 'O'. More precisely, they suggested that the *i*-th letter of her name should be 'O' (uppercase) if *i* is a member of Fibonacci sequence, and 'o' (lowercase) otherwise. The letters in the name are numbered from 1 to *n*. Fibonacci sequence is the sequence *f* where
- *f*1<==<=1, - *f*2<==<=1, - *f**n*<==<=*f**n*<=-<=2<=+<=*f**n*<=-<=1 (*n*<=><=2).
As her friends are too young to know what Fibonacci sequence is, they asked you to help Eleven determine her new name.
Input Specification:
The first and only line of input contains an integer *n* (1<=≤<=*n*<=≤<=1000).
Output Specification:
Print Eleven's new name on the first and only line of output.
Demo Input:
['8\n', '15\n']
Demo Output:
['OOOoOooO\n', 'OOOoOooOooooOoo\n']
Note:
none
|
```python
a = b = 1
for i in range(int(input())):
if b == i + 1:
a, b = (b, a + b)
print('O', end='')
else:
print('o', end='')
```
| 3
|
|
1,004
|
A
|
Sonya and Hotels
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Sonya decided that having her own hotel business is the best way of earning money because she can profit and rest wherever she wants.
The country where Sonya lives is an endless line. There is a city in each integer coordinate on this line. She has $n$ hotels, where the $i$-th hotel is located in the city with coordinate $x_i$. Sonya is a smart girl, so she does not open two or more hotels in the same city.
Sonya understands that her business needs to be expanded by opening new hotels, so she decides to build one more. She wants to make the minimum distance from this hotel to all others to be equal to $d$. The girl understands that there are many possible locations to construct such a hotel. Thus she wants to know the number of possible coordinates of the cities where she can build a new hotel.
Because Sonya is lounging in a jacuzzi in one of her hotels, she is asking you to find the number of cities where she can build a new hotel so that the minimum distance from the original $n$ hotels to the new one is equal to $d$.
|
The first line contains two integers $n$ and $d$ ($1\leq n\leq 100$, $1\leq d\leq 10^9$) — the number of Sonya's hotels and the needed minimum distance from a new hotel to all others.
The second line contains $n$ different integers in strictly increasing order $x_1, x_2, \ldots, x_n$ ($-10^9\leq x_i\leq 10^9$) — coordinates of Sonya's hotels.
|
Print the number of cities where Sonya can build a new hotel so that the minimum distance from this hotel to all others is equal to $d$.
|
[
"4 3\n-3 2 9 16\n",
"5 2\n4 8 11 18 19\n"
] |
[
"6\n",
"5\n"
] |
In the first example, there are $6$ possible cities where Sonya can build a hotel. These cities have coordinates $-6$, $5$, $6$, $12$, $13$, and $19$.
In the second example, there are $5$ possible cities where Sonya can build a hotel. These cities have coordinates $2$, $6$, $13$, $16$, and $21$.
| 500
|
[
{
"input": "4 3\n-3 2 9 16",
"output": "6"
},
{
"input": "5 2\n4 8 11 18 19",
"output": "5"
},
{
"input": "10 10\n-67 -59 -49 -38 -8 20 41 59 74 83",
"output": "8"
},
{
"input": "10 10\n0 20 48 58 81 95 111 137 147 159",
"output": "9"
},
{
"input": "100 1\n0 1 2 3 4 5 7 8 10 11 12 13 14 15 16 17 19 21 22 23 24 25 26 27 28 30 32 33 36 39 40 41 42 46 48 53 54 55 59 60 61 63 65 68 70 71 74 75 76 79 80 81 82 84 88 89 90 91 93 94 96 97 98 100 101 102 105 106 107 108 109 110 111 113 114 115 116 117 118 120 121 122 125 126 128 131 132 133 134 135 137 138 139 140 143 144 146 147 148 149",
"output": "47"
},
{
"input": "1 1000000000\n-1000000000",
"output": "2"
},
{
"input": "2 1000000000\n-1000000000 1000000000",
"output": "3"
},
{
"input": "100 2\n1 3 5 6 8 9 12 13 14 17 18 21 22 23 24 25 26 27 29 30 34 35 36 39 41 44 46 48 52 53 55 56 57 59 61 63 64 66 68 69 70 71 72 73 75 76 77 79 80 81 82 87 88 91 92 93 94 95 96 97 99 100 102 103 104 106 109 110 111 112 113 114 115 117 118 119 120 122 124 125 127 128 129 130 131 132 133 134 136 137 139 140 141 142 143 145 146 148 149 150",
"output": "6"
},
{
"input": "100 3\n0 1 3 6 7 8 9 10 13 14 16 17 18 20 21 22 24 26 27 30 33 34 35 36 37 39 42 43 44 45 46 48 53 54 55 56 57 58 61 63 64 65 67 69 70 72 73 76 77 78 79 81 82 83 85 86 87 88 90 92 93 95 96 97 98 99 100 101 104 105 108 109 110 113 114 115 116 118 120 121 123 124 125 128 130 131 132 133 134 135 136 137 139 140 141 142 146 147 148 150",
"output": "2"
},
{
"input": "1 1000000000\n1000000000",
"output": "2"
},
{
"input": "10 2\n-93 -62 -53 -42 -38 11 57 58 87 94",
"output": "17"
},
{
"input": "2 500000000\n-1000000000 1000000000",
"output": "4"
},
{
"input": "100 10\n-489 -476 -445 -432 -430 -421 -420 -418 -412 -411 -404 -383 -356 -300 -295 -293 -287 -276 -265 -263 -258 -251 -249 -246 -220 -219 -205 -186 -166 -157 -143 -137 -136 -130 -103 -86 -80 -69 -67 -55 -43 -41 -40 -26 -19 -9 16 29 41 42 54 76 84 97 98 99 101 115 134 151 157 167 169 185 197 204 208 226 227 232 234 249 259 266 281 282 293 298 300 306 308 313 319 328 331 340 341 344 356 362 366 380 390 399 409 411 419 444 455 498",
"output": "23"
},
{
"input": "1 1000000000\n999999999",
"output": "2"
},
{
"input": "1 1\n-5",
"output": "2"
},
{
"input": "2 1\n-1000000000 1000000000",
"output": "4"
},
{
"input": "1 2\n1",
"output": "2"
},
{
"input": "4 5\n0 20 40 60",
"output": "8"
},
{
"input": "1 1\n-10",
"output": "2"
},
{
"input": "1 1\n-1000000000",
"output": "2"
},
{
"input": "1 1000000000\n0",
"output": "2"
},
{
"input": "1 2\n-10",
"output": "2"
},
{
"input": "1 1\n2",
"output": "2"
},
{
"input": "1 5\n-100",
"output": "2"
},
{
"input": "2 1000000000\n999999999 1000000000",
"output": "2"
},
{
"input": "2 2\n0 5",
"output": "4"
},
{
"input": "1 2\n-5",
"output": "2"
},
{
"input": "1 1\n1",
"output": "2"
},
{
"input": "1 5\n-20",
"output": "2"
},
{
"input": "2 100000000\n-1000000000 1000000000",
"output": "4"
},
{
"input": "1 1000000000\n147483641",
"output": "2"
},
{
"input": "3 3\n1 8 9",
"output": "4"
},
{
"input": "2 1\n1 1000000000",
"output": "4"
},
{
"input": "1 2\n-4",
"output": "2"
},
{
"input": "2 1\n-1000000000 -1",
"output": "4"
}
] | 1,645,635,330
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
n, d=map(int,input().split())
x=list(map(int,input().split()))
ans=2
for i in range (n-1):
l=x[i]+d
r=x[i+1]-d
if(r<l):
ans +=0
elif(r<l+d):
ans +=1
else:
ans +=2
print(ans)
|
Title: Sonya and Hotels
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sonya decided that having her own hotel business is the best way of earning money because she can profit and rest wherever she wants.
The country where Sonya lives is an endless line. There is a city in each integer coordinate on this line. She has $n$ hotels, where the $i$-th hotel is located in the city with coordinate $x_i$. Sonya is a smart girl, so she does not open two or more hotels in the same city.
Sonya understands that her business needs to be expanded by opening new hotels, so she decides to build one more. She wants to make the minimum distance from this hotel to all others to be equal to $d$. The girl understands that there are many possible locations to construct such a hotel. Thus she wants to know the number of possible coordinates of the cities where she can build a new hotel.
Because Sonya is lounging in a jacuzzi in one of her hotels, she is asking you to find the number of cities where she can build a new hotel so that the minimum distance from the original $n$ hotels to the new one is equal to $d$.
Input Specification:
The first line contains two integers $n$ and $d$ ($1\leq n\leq 100$, $1\leq d\leq 10^9$) — the number of Sonya's hotels and the needed minimum distance from a new hotel to all others.
The second line contains $n$ different integers in strictly increasing order $x_1, x_2, \ldots, x_n$ ($-10^9\leq x_i\leq 10^9$) — coordinates of Sonya's hotels.
Output Specification:
Print the number of cities where Sonya can build a new hotel so that the minimum distance from this hotel to all others is equal to $d$.
Demo Input:
['4 3\n-3 2 9 16\n', '5 2\n4 8 11 18 19\n']
Demo Output:
['6\n', '5\n']
Note:
In the first example, there are $6$ possible cities where Sonya can build a hotel. These cities have coordinates $-6$, $5$, $6$, $12$, $13$, and $19$.
In the second example, there are $5$ possible cities where Sonya can build a hotel. These cities have coordinates $2$, $6$, $13$, $16$, and $21$.
|
```python
n, d=map(int,input().split())
x=list(map(int,input().split()))
ans=2
for i in range (n-1):
l=x[i]+d
r=x[i+1]-d
if(r<l):
ans +=0
elif(r<l+d):
ans +=1
else:
ans +=2
print(ans)
```
| 0
|
|
263
|
A
|
Beautiful Matrix
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
You've got a 5<=×<=5 matrix, consisting of 24 zeroes and a single number one. Let's index the matrix rows by numbers from 1 to 5 from top to bottom, let's index the matrix columns by numbers from 1 to 5 from left to right. In one move, you are allowed to apply one of the two following transformations to the matrix:
1. Swap two neighboring matrix rows, that is, rows with indexes *i* and *i*<=+<=1 for some integer *i* (1<=≤<=*i*<=<<=5). 1. Swap two neighboring matrix columns, that is, columns with indexes *j* and *j*<=+<=1 for some integer *j* (1<=≤<=*j*<=<<=5).
You think that a matrix looks beautiful, if the single number one of the matrix is located in its middle (in the cell that is on the intersection of the third row and the third column). Count the minimum number of moves needed to make the matrix beautiful.
|
The input consists of five lines, each line contains five integers: the *j*-th integer in the *i*-th line of the input represents the element of the matrix that is located on the intersection of the *i*-th row and the *j*-th column. It is guaranteed that the matrix consists of 24 zeroes and a single number one.
|
Print a single integer — the minimum number of moves needed to make the matrix beautiful.
|
[
"0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n",
"0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n"
] |
[
"3\n",
"1\n"
] |
none
| 500
|
[
{
"input": "0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "0"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1",
"output": "4"
},
{
"input": "0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "4"
},
{
"input": "0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "4"
},
{
"input": "0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0",
"output": "4"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0",
"output": "3"
}
] | 1,698,490,545
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 3
| 124
| 0
|
matrix = []
for i in range(5):
row = input()
row = row.split()
row = [int(x) for x in row]
matrix.append(row)
for i in range(5):
for j in range(5):
if matrix[i][j] == 1:
if i == 2 and j == 2:
c = 0
elif i == 2 :
c = abs(j-2)
elif j == 2:
c = abs(i-2)
else:
c = abs(i+j-2)
break
print(c)
|
Title: Beautiful Matrix
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got a 5<=×<=5 matrix, consisting of 24 zeroes and a single number one. Let's index the matrix rows by numbers from 1 to 5 from top to bottom, let's index the matrix columns by numbers from 1 to 5 from left to right. In one move, you are allowed to apply one of the two following transformations to the matrix:
1. Swap two neighboring matrix rows, that is, rows with indexes *i* and *i*<=+<=1 for some integer *i* (1<=≤<=*i*<=<<=5). 1. Swap two neighboring matrix columns, that is, columns with indexes *j* and *j*<=+<=1 for some integer *j* (1<=≤<=*j*<=<<=5).
You think that a matrix looks beautiful, if the single number one of the matrix is located in its middle (in the cell that is on the intersection of the third row and the third column). Count the minimum number of moves needed to make the matrix beautiful.
Input Specification:
The input consists of five lines, each line contains five integers: the *j*-th integer in the *i*-th line of the input represents the element of the matrix that is located on the intersection of the *i*-th row and the *j*-th column. It is guaranteed that the matrix consists of 24 zeroes and a single number one.
Output Specification:
Print a single integer — the minimum number of moves needed to make the matrix beautiful.
Demo Input:
['0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n', '0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n']
Demo Output:
['3\n', '1\n']
Note:
none
|
```python
matrix = []
for i in range(5):
row = input()
row = row.split()
row = [int(x) for x in row]
matrix.append(row)
for i in range(5):
for j in range(5):
if matrix[i][j] == 1:
if i == 2 and j == 2:
c = 0
elif i == 2 :
c = abs(j-2)
elif j == 2:
c = abs(i-2)
else:
c = abs(i+j-2)
break
print(c)
```
| 0
|
|
233
|
A
|
Perfect Permutation
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
A permutation is a sequence of integers *p*1,<=*p*2,<=...,<=*p**n*, consisting of *n* distinct positive integers, each of them doesn't exceed *n*. Let's denote the *i*-th element of permutation *p* as *p**i*. We'll call number *n* the size of permutation *p*1,<=*p*2,<=...,<=*p**n*.
Nickolas adores permutations. He likes some permutations more than the others. He calls such permutations perfect. A perfect permutation is such permutation *p* that for any *i* (1<=≤<=*i*<=≤<=*n*) (*n* is the permutation size) the following equations hold *p**p**i*<==<=*i* and *p**i*<=≠<=*i*. Nickolas asks you to print any perfect permutation of size *n* for the given *n*.
|
A single line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the permutation size.
|
If a perfect permutation of size *n* doesn't exist, print a single integer -1. Otherwise print *n* distinct integers from 1 to *n*, *p*1,<=*p*2,<=...,<=*p**n* — permutation *p*, that is perfect. Separate printed numbers by whitespaces.
|
[
"1\n",
"2\n",
"4\n"
] |
[
"-1\n",
"2 1 \n",
"2 1 4 3 \n"
] |
none
| 500
|
[
{
"input": "1",
"output": "-1"
},
{
"input": "2",
"output": "2 1 "
},
{
"input": "4",
"output": "2 1 4 3 "
},
{
"input": "3",
"output": "-1"
},
{
"input": "5",
"output": "-1"
},
{
"input": "6",
"output": "2 1 4 3 6 5 "
},
{
"input": "7",
"output": "-1"
},
{
"input": "20",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 "
},
{
"input": "8",
"output": "2 1 4 3 6 5 8 7 "
},
{
"input": "9",
"output": "-1"
},
{
"input": "10",
"output": "2 1 4 3 6 5 8 7 10 9 "
},
{
"input": "11",
"output": "-1"
},
{
"input": "21",
"output": "-1"
},
{
"input": "50",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 "
},
{
"input": "51",
"output": "-1"
},
{
"input": "52",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 "
},
{
"input": "84",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 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 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 "
},
{
"input": "86",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 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 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 86 85 "
},
{
"input": "100",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 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 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 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 "
},
{
"input": "98",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 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 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 86 85 88 87 90 89 92 91 94 93 96 95 98 97 "
},
{
"input": "96",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 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 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 86 85 88 87 90 89 92 91 94 93 96 95 "
},
{
"input": "33",
"output": "-1"
},
{
"input": "34",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 "
},
{
"input": "36",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 "
},
{
"input": "38",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 "
},
{
"input": "40",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 "
},
{
"input": "42",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 "
},
{
"input": "44",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 "
},
{
"input": "46",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 "
},
{
"input": "48",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 "
}
] | 1,646,884,463
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 92
| 0
|
n = int(input())
if n%2 == 1:
print(-1)
else:
for i in range(n):
if i%2 == 0:
print(i+2, end=" ")
else:
print(i, end=" ")
|
Title: Perfect Permutation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A permutation is a sequence of integers *p*1,<=*p*2,<=...,<=*p**n*, consisting of *n* distinct positive integers, each of them doesn't exceed *n*. Let's denote the *i*-th element of permutation *p* as *p**i*. We'll call number *n* the size of permutation *p*1,<=*p*2,<=...,<=*p**n*.
Nickolas adores permutations. He likes some permutations more than the others. He calls such permutations perfect. A perfect permutation is such permutation *p* that for any *i* (1<=≤<=*i*<=≤<=*n*) (*n* is the permutation size) the following equations hold *p**p**i*<==<=*i* and *p**i*<=≠<=*i*. Nickolas asks you to print any perfect permutation of size *n* for the given *n*.
Input Specification:
A single line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the permutation size.
Output Specification:
If a perfect permutation of size *n* doesn't exist, print a single integer -1. Otherwise print *n* distinct integers from 1 to *n*, *p*1,<=*p*2,<=...,<=*p**n* — permutation *p*, that is perfect. Separate printed numbers by whitespaces.
Demo Input:
['1\n', '2\n', '4\n']
Demo Output:
['-1\n', '2 1 \n', '2 1 4 3 \n']
Note:
none
|
```python
n = int(input())
if n%2 == 1:
print(-1)
else:
for i in range(n):
if i%2 == 0:
print(i+2, end=" ")
else:
print(i, end=" ")
```
| 3
|
|
427
|
A
|
Police Recruits
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
The police department of your city has just started its journey. Initially, they don’t have any manpower. So, they started hiring new recruits in groups.
Meanwhile, crimes keeps occurring within the city. One member of the police force can investigate only one crime during his/her lifetime.
If there is no police officer free (isn't busy with crime) during the occurrence of a crime, it will go untreated.
Given the chronological order of crime occurrences and recruit hirings, find the number of crimes which will go untreated.
|
The first line of input will contain an integer *n* (1<=≤<=*n*<=≤<=105), the number of events. The next line will contain *n* space-separated integers.
If the integer is -1 then it means a crime has occurred. Otherwise, the integer will be positive, the number of officers recruited together at that time. No more than 10 officers will be recruited at a time.
|
Print a single integer, the number of crimes which will go untreated.
|
[
"3\n-1 -1 1\n",
"8\n1 -1 1 -1 -1 1 1 1\n",
"11\n-1 -1 2 -1 -1 -1 -1 -1 -1 -1 -1\n"
] |
[
"2\n",
"1\n",
"8\n"
] |
Lets consider the second example:
1. Firstly one person is hired. 1. Then crime appears, the last hired person will investigate this crime. 1. One more person is hired. 1. One more crime appears, the last hired person will investigate this crime. 1. Crime appears. There is no free policeman at the time, so this crime will go untreated. 1. One more person is hired. 1. One more person is hired. 1. One more person is hired.
The answer is one, as one crime (on step 5) will go untreated.
| 500
|
[
{
"input": "3\n-1 -1 1",
"output": "2"
},
{
"input": "8\n1 -1 1 -1 -1 1 1 1",
"output": "1"
},
{
"input": "11\n-1 -1 2 -1 -1 -1 -1 -1 -1 -1 -1",
"output": "8"
},
{
"input": "7\n-1 -1 1 1 -1 -1 1",
"output": "2"
},
{
"input": "21\n-1 -1 -1 -1 -1 3 2 -1 6 -1 -1 2 1 -1 2 2 1 6 5 -1 5",
"output": "5"
},
{
"input": "98\n-1 -1 1 -1 -1 -1 -1 1 -1 -1 1 -1 -1 1 -1 1 1 1 -1 1 1 1 1 1 -1 1 -1 -1 -1 -1 1 -1 -1 1 1 -1 1 1 1 -1 -1 -1 -1 -1 -1 1 -1 -1 -1 1 -1 1 -1 1 -1 1 1 1 1 1 1 1 -1 -1 1 1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 1 -1 1 1 1 -1 1 1 -1 -1 -1 1 1 1 -1 -1 -1 1 -1 1 1",
"output": "13"
},
{
"input": "3\n-1 5 4",
"output": "1"
},
{
"input": "146\n4 -1 -1 -1 -1 -1 -1 -1 -1 -1 4 -1 3 -1 3 -1 -1 1 4 -1 2 -1 -1 3 -1 -1 -1 4 1 -1 -1 -1 4 -1 -1 -1 -1 -1 -1 3 2 -1 3 2 4 5 2 4 1 5 -1 -1 2 -1 -1 1 -1 5 3 -1 1 2 2 3 5 3 -1 -1 3 -1 -1 3 5 5 -1 -1 5 -1 4 4 1 -1 -1 -1 2 1 -1 -1 -1 2 5 3 -1 -1 -1 3 -1 5 4 -1 1 -1 -1 3 -1 -1 3 1 1 2 -1 -1 -1 1 3 1 -1 2 -1 -1 5 5 -1 -1 3 4 5 1 -1 2 -1 -1 -1 3 -1 5 3 2 -1 2 -1 -1 5 -1 3 -1",
"output": "5"
},
{
"input": "1\n2",
"output": "0"
},
{
"input": "1\n-1",
"output": "1"
},
{
"input": "2\n1 -1",
"output": "0"
},
{
"input": "2\n-1 1",
"output": "1"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "2\n-1 -1",
"output": "2"
},
{
"input": "4\n10 -1 -1 -1",
"output": "0"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "4\n-1 -1 1 1",
"output": "2"
}
] | 1,694,962,798
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 63
| 62
| 9,216,000
|
input();t=0
for i in list(map(int,input().split()[::-1])):
t=min(0,t+i)
print(-t)
|
Title: Police Recruits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The police department of your city has just started its journey. Initially, they don’t have any manpower. So, they started hiring new recruits in groups.
Meanwhile, crimes keeps occurring within the city. One member of the police force can investigate only one crime during his/her lifetime.
If there is no police officer free (isn't busy with crime) during the occurrence of a crime, it will go untreated.
Given the chronological order of crime occurrences and recruit hirings, find the number of crimes which will go untreated.
Input Specification:
The first line of input will contain an integer *n* (1<=≤<=*n*<=≤<=105), the number of events. The next line will contain *n* space-separated integers.
If the integer is -1 then it means a crime has occurred. Otherwise, the integer will be positive, the number of officers recruited together at that time. No more than 10 officers will be recruited at a time.
Output Specification:
Print a single integer, the number of crimes which will go untreated.
Demo Input:
['3\n-1 -1 1\n', '8\n1 -1 1 -1 -1 1 1 1\n', '11\n-1 -1 2 -1 -1 -1 -1 -1 -1 -1 -1\n']
Demo Output:
['2\n', '1\n', '8\n']
Note:
Lets consider the second example:
1. Firstly one person is hired. 1. Then crime appears, the last hired person will investigate this crime. 1. One more person is hired. 1. One more crime appears, the last hired person will investigate this crime. 1. Crime appears. There is no free policeman at the time, so this crime will go untreated. 1. One more person is hired. 1. One more person is hired. 1. One more person is hired.
The answer is one, as one crime (on step 5) will go untreated.
|
```python
input();t=0
for i in list(map(int,input().split()[::-1])):
t=min(0,t+i)
print(-t)
```
| 3
|
|
552
|
D
|
Vanya and Triangles
|
PROGRAMMING
| 1,900
|
[
"brute force",
"combinatorics",
"data structures",
"geometry",
"math",
"sortings"
] | null | null |
Vanya got bored and he painted *n* distinct points on the plane. After that he connected all the points pairwise and saw that as a result many triangles were formed with vertices in the painted points. He asks you to count the number of the formed triangles with the non-zero area.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=2000) — the number of the points painted on the plane.
Next *n* lines contain two integers each *x**i*,<=*y**i* (<=-<=100<=≤<=*x**i*,<=*y**i*<=≤<=100) — the coordinates of the *i*-th point. It is guaranteed that no two given points coincide.
|
In the first line print an integer — the number of triangles with the non-zero area among the painted points.
|
[
"4\n0 0\n1 1\n2 0\n2 2\n",
"3\n0 0\n1 1\n2 0\n",
"1\n1 1\n"
] |
[
"3\n",
"1\n",
"0\n"
] |
Note to the first sample test. There are 3 triangles formed: (0, 0) - (1, 1) - (2, 0); (0, 0) - (2, 2) - (2, 0); (1, 1) - (2, 2) - (2, 0).
Note to the second sample test. There is 1 triangle formed: (0, 0) - (1, 1) - (2, 0).
Note to the third sample test. A single point doesn't form a single triangle.
| 2,000
|
[
{
"input": "4\n0 0\n1 1\n2 0\n2 2",
"output": "3"
},
{
"input": "3\n0 0\n1 1\n2 0",
"output": "1"
},
{
"input": "1\n1 1",
"output": "0"
},
{
"input": "5\n0 0\n1 1\n2 2\n3 3\n4 4",
"output": "0"
},
{
"input": "5\n0 0\n1 1\n2 3\n3 6\n4 10",
"output": "10"
},
{
"input": "4\n-100 -100\n-100 100\n100 -100\n100 100",
"output": "4"
},
{
"input": "5\n-100 -100\n-100 100\n100 -100\n100 100\n0 0",
"output": "8"
},
{
"input": "4\n1 -100\n2 -100\n100 -99\n99 -99",
"output": "4"
},
{
"input": "25\n26 -54\n16 56\n-42 -51\n92 -58\n100 52\n57 -98\n-84 -28\n-71 12\n21 -82\n-3 -30\n72 94\n-66 96\n-50 -41\n-77 -41\n-42 -55\n-13 12\n0 -99\n-50 -5\n65 -48\n-96 -80\n73 -92\n72 59\n53 -66\n-67 -75\n2 56",
"output": "2300"
},
{
"input": "5\n-62 -69\n3 -48\n54 54\n8 94\n83 94",
"output": "10"
},
{
"input": "33\n0 81\n20 -16\n-71 38\n-45 28\n-8 -40\n34 -49\n43 -10\n-40 19\n14 -50\n-95 8\n-21 85\n64 98\n-97 -82\n19 -83\n39 -99\n43 71\n67 43\n-54 57\n-7 24\n83 -76\n54 -88\n-43 -9\n-75 24\n74 32\n-68 -1\n71 84\n88 80\n52 67\n-64 21\n-85 97\n33 13\n41 -28\n0 74",
"output": "5456"
},
{
"input": "61\n37 -96\n36 -85\n30 -53\n-98 -40\n2 3\n-88 -69\n88 -26\n78 -69\n48 -3\n-41 66\n-93 -58\n-51 59\n21 -2\n65 29\n-3 35\n-98 46\n42 38\n0 -99\n46 84\n39 -48\n-15 81\n-15 51\n-77 74\n81 -58\n26 -35\n-14 20\n73 74\n-45 83\n90 22\n-8 53\n1 -52\n20 58\n39 -22\n60 -10\n52 22\n-46 6\n8 8\n14 9\n38 -45\n82 13\n43 4\n-25 21\n50 -16\n31 -12\n76 -13\n-82 -2\n-5 -56\n87 -31\n9 -36\n-100 92\n-10 39\n-16 2\n62 -39\n-36 60\n14 21\n-62 40\n98 43\n-54 66\n-34 46\n-47 -65\n21 44",
"output": "35985"
},
{
"input": "9\n-41 -22\n95 53\n81 -61\n22 -74\n-79 38\n-56 -32\n100 -32\n-37 -94\n-59 -9",
"output": "84"
},
{
"input": "33\n21 -99\n11 85\n80 -77\n-31 59\n32 6\n24 -52\n-32 -47\n57 18\n76 -36\n96 -38\n-59 -12\n-98 -32\n-52 32\n-73 -87\n-51 -40\n34 -55\n69 46\n-88 -67\n-68 65\n60 -11\n-45 -41\n91 -21\n45 21\n-75 49\n58 65\n-20 81\n-24 29\n66 -71\n-25 50\n96 74\n-43 -47\n34 -86\n81 14",
"output": "5455"
},
{
"input": "61\n83 52\n28 91\n-45 -68\n-84 -8\n-59 -28\n-98 -72\n38 -38\n-51 -96\n-66 11\n-76 45\n95 45\n-89 5\n-60 -66\n73 26\n9 94\n-5 -80\n44 41\n66 -22\n61 26\n-58 -84\n62 -73\n18 63\n44 71\n32 -41\n-50 -69\n-30 17\n61 47\n45 70\n-97 76\n-27 31\n2 -12\n-87 -75\n-80 -82\n-47 50\n45 -23\n71 54\n79 -7\n35 22\n19 -53\n-65 -72\n-69 68\n-53 48\n-73 -15\n29 38\n-49 -47\n12 -30\n-21 -59\n-28 -11\n-73 -60\n99 74\n32 30\n-9 -7\n-82 95\n58 -32\n39 64\n-42 9\n-21 -76\n39 33\n-63 59\n-66 41\n-54 -69",
"output": "35985"
},
{
"input": "62\n-53 -58\n29 89\n-92 15\n-91 -19\n96 23\n-1 -57\n-83 11\n56 -95\n-39 -47\n-75 77\n52 -95\n-13 -12\n-51 80\n32 -78\n94 94\n-51 81\n53 -28\n-83 -78\n76 -25\n91 -60\n-40 -27\n55 86\n-26 1\n-41 89\n61 -23\n81 31\n-21 82\n-12 47\n20 36\n-95 54\n-81 73\n-19 -83\n52 51\n-60 68\n-58 35\n60 -38\n-98 32\n-10 60\n88 -5\n78 -57\n-12 -43\n-83 36\n51 -63\n-89 -5\n-62 -42\n-29 78\n73 62\n-88 -55\n34 38\n88 -26\n-26 -89\n40 -26\n46 63\n74 -66\n-61 -61\n82 -53\n-75 -62\n-99 -52\n-15 30\n38 -52\n-83 -75\n-31 -38",
"output": "37814"
},
{
"input": "2\n0 0\n1 1",
"output": "0"
},
{
"input": "50\n0 -26\n0 -64\n0 63\n0 -38\n0 47\n0 31\n0 -72\n0 60\n0 -15\n0 -36\n0 50\n0 -77\n0 -89\n0 5\n0 83\n0 -52\n0 -21\n0 39\n0 51\n0 -11\n0 -69\n0 57\n0 -58\n0 64\n0 85\n0 -61\n0 0\n0 69\n0 -83\n0 24\n0 -91\n0 -33\n0 -79\n0 -39\n0 -98\n0 45\n0 4\n0 -8\n0 96\n0 35\n0 9\n0 53\n0 90\n0 15\n0 -19\n0 -48\n0 -56\n0 38\n0 92\n0 76",
"output": "0"
},
{
"input": "20\n12 16\n19 13\n19 15\n20 3\n5 20\n8 3\n9 18\n2 15\n2 3\n16 8\n14 18\n16 20\n13 17\n0 15\n10 12\n10 6\n18 8\n6 1\n6 2\n0 6",
"output": "1130"
},
{
"input": "5\n0 0\n1 1\n2 4\n3 8\n4 16",
"output": "10"
},
{
"input": "3\n-100 -100\n0 0\n100 100",
"output": "0"
},
{
"input": "20\n-2 1\n5 1\n1 -1\n1 4\n-5 -5\n3 1\n-5 -3\n-2 3\n-3 4\n5 -4\n-4 5\n3 3\n1 0\n-4 -4\n3 0\n4 -1\n-3 0\n-2 2\n-2 -5\n-5 -4",
"output": "1109"
},
{
"input": "3\n1 1\n3 3\n2 2",
"output": "0"
},
{
"input": "10\n-52 25\n55 76\n97 88\n92 3\n-98 77\n45 90\n6 85\n-68 -38\n-74 -55\n-48 60",
"output": "120"
},
{
"input": "10\n-1 32\n0 88\n-1 69\n0 62\n-1 52\n0 16\n0 19\n-1 58\n0 38\n0 67",
"output": "96"
},
{
"input": "20\n-100 -100\n-99 -99\n-98 -96\n-97 -91\n-96 -84\n-95 -75\n-94 -64\n-93 -51\n-92 -36\n-91 -19\n100 100\n99 99\n98 96\n97 91\n96 84\n95 75\n94 64\n93 51\n92 36\n91 19",
"output": "1136"
}
] | 1,520,753,508
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 77
| 6,041,600
|
import math as m
from collections import Counter
n = int(input())
A = []
Ct = Counter()
ans = n * (n - 1) * (n - 2) // 6
for i in range(n):
A += [tuple(map(int, input().split()))]
for i in range(n):
for j in range(i):
x1, y1 = A[i]
x2, y2 = A[j]
dx = x2 - x1
dy = y2 - y1
a, b, c = -dy, dx, y1 * dx - x1 * dy
if c < 0:
a *= -1
b *= -1
c *= -1
g = m.gcd(a, m.gcd(b, c))
a //= g
b //= g
c //= g
Ct[(a, b, c)] += 1
for v in Ct.values():
ans -= v * (v - 1) * (v - 2) // 6
print(ans)
|
Title: Vanya and Triangles
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vanya got bored and he painted *n* distinct points on the plane. After that he connected all the points pairwise and saw that as a result many triangles were formed with vertices in the painted points. He asks you to count the number of the formed triangles with the non-zero area.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=2000) — the number of the points painted on the plane.
Next *n* lines contain two integers each *x**i*,<=*y**i* (<=-<=100<=≤<=*x**i*,<=*y**i*<=≤<=100) — the coordinates of the *i*-th point. It is guaranteed that no two given points coincide.
Output Specification:
In the first line print an integer — the number of triangles with the non-zero area among the painted points.
Demo Input:
['4\n0 0\n1 1\n2 0\n2 2\n', '3\n0 0\n1 1\n2 0\n', '1\n1 1\n']
Demo Output:
['3\n', '1\n', '0\n']
Note:
Note to the first sample test. There are 3 triangles formed: (0, 0) - (1, 1) - (2, 0); (0, 0) - (2, 2) - (2, 0); (1, 1) - (2, 2) - (2, 0).
Note to the second sample test. There is 1 triangle formed: (0, 0) - (1, 1) - (2, 0).
Note to the third sample test. A single point doesn't form a single triangle.
|
```python
import math as m
from collections import Counter
n = int(input())
A = []
Ct = Counter()
ans = n * (n - 1) * (n - 2) // 6
for i in range(n):
A += [tuple(map(int, input().split()))]
for i in range(n):
for j in range(i):
x1, y1 = A[i]
x2, y2 = A[j]
dx = x2 - x1
dy = y2 - y1
a, b, c = -dy, dx, y1 * dx - x1 * dy
if c < 0:
a *= -1
b *= -1
c *= -1
g = m.gcd(a, m.gcd(b, c))
a //= g
b //= g
c //= g
Ct[(a, b, c)] += 1
for v in Ct.values():
ans -= v * (v - 1) * (v - 2) // 6
print(ans)
```
| 0
|
|
118
|
A
|
String Task
|
PROGRAMMING
| 1,000
|
[
"implementation",
"strings"
] | null | null |
Petya started to attend programming lessons. On the first lesson his task was to write a simple program. The program was supposed to do the following: in the given string, consisting if uppercase and lowercase Latin letters, it:
- deletes all the vowels, - inserts a character "." before each consonant, - replaces all uppercase consonants with corresponding lowercase ones.
Vowels are letters "A", "O", "Y", "E", "U", "I", and the rest are consonants. The program's input is exactly one string, it should return the output as a single string, resulting after the program's processing the initial string.
Help Petya cope with this easy task.
|
The first line represents input string of Petya's program. This string only consists of uppercase and lowercase Latin letters and its length is from 1 to 100, inclusive.
|
Print the resulting string. It is guaranteed that this string is not empty.
|
[
"tour\n",
"Codeforces\n",
"aBAcAba\n"
] |
[
".t.r\n",
".c.d.f.r.c.s\n",
".b.c.b\n"
] |
none
| 500
|
[
{
"input": "tour",
"output": ".t.r"
},
{
"input": "Codeforces",
"output": ".c.d.f.r.c.s"
},
{
"input": "aBAcAba",
"output": ".b.c.b"
},
{
"input": "obn",
"output": ".b.n"
},
{
"input": "wpwl",
"output": ".w.p.w.l"
},
{
"input": "ggdvq",
"output": ".g.g.d.v.q"
},
{
"input": "pumesz",
"output": ".p.m.s.z"
},
{
"input": "g",
"output": ".g"
},
{
"input": "zjuotps",
"output": ".z.j.t.p.s"
},
{
"input": "jzbwuehe",
"output": ".j.z.b.w.h"
},
{
"input": "tnkgwuugu",
"output": ".t.n.k.g.w.g"
},
{
"input": "kincenvizh",
"output": ".k.n.c.n.v.z.h"
},
{
"input": "xattxjenual",
"output": ".x.t.t.x.j.n.l"
},
{
"input": "ktajqhpqsvhw",
"output": ".k.t.j.q.h.p.q.s.v.h.w"
},
{
"input": "xnhcigytnqcmy",
"output": ".x.n.h.c.g.t.n.q.c.m"
},
{
"input": "jfmtbejyilxcec",
"output": ".j.f.m.t.b.j.l.x.c.c"
},
{
"input": "D",
"output": ".d"
},
{
"input": "ab",
"output": ".b"
},
{
"input": "Ab",
"output": ".b"
},
{
"input": "aB",
"output": ".b"
},
{
"input": "AB",
"output": ".b"
},
{
"input": "ba",
"output": ".b"
},
{
"input": "bA",
"output": ".b"
},
{
"input": "Ba",
"output": ".b"
},
{
"input": "BA",
"output": ".b"
},
{
"input": "aab",
"output": ".b"
},
{
"input": "baa",
"output": ".b"
},
{
"input": "femOZeCArKCpUiHYnbBPTIOFmsHmcpObtPYcLCdjFrUMIyqYzAokKUiiKZRouZiNMoiOuGVoQzaaCAOkquRjmmKKElLNqCnhGdQM",
"output": ".f.m.z.c.r.k.c.p.h.n.b.b.p.t.f.m.s.h.m.c.p.b.t.p.c.l.c.d.j.f.r.m.q.z.k.k.k.z.r.z.n.m.g.v.q.z.c.k.q.r.j.m.m.k.k.l.l.n.q.c.n.h.g.d.q.m"
},
{
"input": "VMBPMCmMDCLFELLIISUJDWQRXYRDGKMXJXJHXVZADRZWVWJRKFRRNSAWKKDPZZLFLNSGUNIVJFBEQsMDHSBJVDTOCSCgZWWKvZZN",
"output": ".v.m.b.p.m.c.m.m.d.c.l.f.l.l.s.j.d.w.q.r.x.r.d.g.k.m.x.j.x.j.h.x.v.z.d.r.z.w.v.w.j.r.k.f.r.r.n.s.w.k.k.d.p.z.z.l.f.l.n.s.g.n.v.j.f.b.q.s.m.d.h.s.b.j.v.d.t.c.s.c.g.z.w.w.k.v.z.z.n"
},
{
"input": "MCGFQQJNUKuAEXrLXibVjClSHjSxmlkQGTKZrRaDNDomIPOmtSgjJAjNVIVLeUGUAOHNkCBwNObVCHOWvNkLFQQbFnugYVMkJruJ",
"output": ".m.c.g.f.q.q.j.n.k.x.r.l.x.b.v.j.c.l.s.h.j.s.x.m.l.k.q.g.t.k.z.r.r.d.n.d.m.p.m.t.s.g.j.j.j.n.v.v.l.g.h.n.k.c.b.w.n.b.v.c.h.w.v.n.k.l.f.q.q.b.f.n.g.v.m.k.j.r.j"
},
{
"input": "iyaiuiwioOyzUaOtAeuEYcevvUyveuyioeeueoeiaoeiavizeeoeyYYaaAOuouueaUioueauayoiuuyiuovyOyiyoyioaoyuoyea",
"output": ".w.z.t.c.v.v.v.v.z.v"
},
{
"input": "yjnckpfyLtzwjsgpcrgCfpljnjwqzgVcufnOvhxplvflxJzqxnhrwgfJmPzifgubvspffmqrwbzivatlmdiBaddiaktdsfPwsevl",
"output": ".j.n.c.k.p.f.l.t.z.w.j.s.g.p.c.r.g.c.f.p.l.j.n.j.w.q.z.g.v.c.f.n.v.h.x.p.l.v.f.l.x.j.z.q.x.n.h.r.w.g.f.j.m.p.z.f.g.b.v.s.p.f.f.m.q.r.w.b.z.v.t.l.m.d.b.d.d.k.t.d.s.f.p.w.s.v.l"
},
{
"input": "RIIIUaAIYJOiuYIUWFPOOAIuaUEZeIooyUEUEAoIyIHYOEAlVAAIiLUAUAeiUIEiUMuuOiAgEUOIAoOUYYEYFEoOIIVeOOAOIIEg",
"output": ".r.j.w.f.p.z.h.l.v.l.m.g.f.v.g"
},
{
"input": "VBKQCFBMQHDMGNSGBQVJTGQCNHHRJMNKGKDPPSQRRVQTZNKBZGSXBPBRXPMVFTXCHZMSJVBRNFNTHBHGJLMDZJSVPZZBCCZNVLMQ",
"output": ".v.b.k.q.c.f.b.m.q.h.d.m.g.n.s.g.b.q.v.j.t.g.q.c.n.h.h.r.j.m.n.k.g.k.d.p.p.s.q.r.r.v.q.t.z.n.k.b.z.g.s.x.b.p.b.r.x.p.m.v.f.t.x.c.h.z.m.s.j.v.b.r.n.f.n.t.h.b.h.g.j.l.m.d.z.j.s.v.p.z.z.b.c.c.z.n.v.l.m.q"
},
{
"input": "iioyoaayeuyoolyiyoeuouiayiiuyTueyiaoiueyioiouyuauouayyiaeoeiiigmioiououeieeeyuyyaYyioiiooaiuouyoeoeg",
"output": ".l.t.g.m.g"
},
{
"input": "ueyiuiauuyyeueykeioouiiauzoyoeyeuyiaoaiiaaoaueyaeydaoauexuueafouiyioueeaaeyoeuaueiyiuiaeeayaioeouiuy",
"output": ".k.z.d.x.f"
},
{
"input": "FSNRBXLFQHZXGVMKLQDVHWLDSLKGKFMDRQWMWSSKPKKQBNDZRSCBLRSKCKKFFKRDMZFZGCNSMXNPMZVDLKXGNXGZQCLRTTDXLMXQ",
"output": ".f.s.n.r.b.x.l.f.q.h.z.x.g.v.m.k.l.q.d.v.h.w.l.d.s.l.k.g.k.f.m.d.r.q.w.m.w.s.s.k.p.k.k.q.b.n.d.z.r.s.c.b.l.r.s.k.c.k.k.f.f.k.r.d.m.z.f.z.g.c.n.s.m.x.n.p.m.z.v.d.l.k.x.g.n.x.g.z.q.c.l.r.t.t.d.x.l.m.x.q"
},
{
"input": "EYAYAYIOIOYOOAUOEUEUOUUYIYUUMOEOIIIAOIUOAAOIYOIOEUIERCEYYAOIOIGYUIAOYUEOEUAEAYPOYEYUUAUOAOEIYIEYUEEY",
"output": ".m.r.c.g.p"
},
{
"input": "jvvzcdcxjstbbksmqjsngxkgtttdxrljjxtwptgwwqzpvqchvgrkqlzxmptzblxhhsmrkmzzmgdfskhtmmnqzzflpmqdctvrfgtx",
"output": ".j.v.v.z.c.d.c.x.j.s.t.b.b.k.s.m.q.j.s.n.g.x.k.g.t.t.t.d.x.r.l.j.j.x.t.w.p.t.g.w.w.q.z.p.v.q.c.h.v.g.r.k.q.l.z.x.m.p.t.z.b.l.x.h.h.s.m.r.k.m.z.z.m.g.d.f.s.k.h.t.m.m.n.q.z.z.f.l.p.m.q.d.c.t.v.r.f.g.t.x"
},
{
"input": "YB",
"output": ".b"
},
{
"input": "fly",
"output": ".f.l"
},
{
"input": "YyyYYYyyYxdwdawdDAWDdaddYYYY",
"output": ".x.d.w.d.w.d.d.w.d.d.d.d"
}
] | 1,698,627,116
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 42
| 92
| 0
|
s = input().lower()
vowels = ['a', 'e', 'i', 'o', 'u', 'y']
consonants = []
for i in s:
if not i in vowels:
consonants.append(i)
print('.' + '.'.join(consonants))
|
Title: String Task
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya started to attend programming lessons. On the first lesson his task was to write a simple program. The program was supposed to do the following: in the given string, consisting if uppercase and lowercase Latin letters, it:
- deletes all the vowels, - inserts a character "." before each consonant, - replaces all uppercase consonants with corresponding lowercase ones.
Vowels are letters "A", "O", "Y", "E", "U", "I", and the rest are consonants. The program's input is exactly one string, it should return the output as a single string, resulting after the program's processing the initial string.
Help Petya cope with this easy task.
Input Specification:
The first line represents input string of Petya's program. This string only consists of uppercase and lowercase Latin letters and its length is from 1 to 100, inclusive.
Output Specification:
Print the resulting string. It is guaranteed that this string is not empty.
Demo Input:
['tour\n', 'Codeforces\n', 'aBAcAba\n']
Demo Output:
['.t.r\n', '.c.d.f.r.c.s\n', '.b.c.b\n']
Note:
none
|
```python
s = input().lower()
vowels = ['a', 'e', 'i', 'o', 'u', 'y']
consonants = []
for i in s:
if not i in vowels:
consonants.append(i)
print('.' + '.'.join(consonants))
```
| 3
|
|
702
|
C
|
Cellular Network
|
PROGRAMMING
| 1,500
|
[
"binary search",
"implementation",
"two pointers"
] | null | null |
You are given *n* points on the straight line — the positions (*x*-coordinates) of the cities and *m* points on the same line — the positions (*x*-coordinates) of the cellular towers. All towers work in the same way — they provide cellular network for all cities, which are located at the distance which is no more than *r* from this tower.
Your task is to find minimal *r* that each city has been provided by cellular network, i.e. for each city there is at least one cellular tower at the distance which is no more than *r*.
If *r*<==<=0 then a tower provides cellular network only for the point where it is located. One tower can provide cellular network for any number of cities, but all these cities must be at the distance which is no more than *r* from this tower.
|
The first line contains two positive integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=105) — the number of cities and the number of cellular towers.
The second line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=109<=≤<=*a**i*<=≤<=109) — the coordinates of cities. It is allowed that there are any number of cities in the same point. All coordinates *a**i* are given in non-decreasing order.
The third line contains a sequence of *m* integers *b*1,<=*b*2,<=...,<=*b**m* (<=-<=109<=≤<=*b**j*<=≤<=109) — the coordinates of cellular towers. It is allowed that there are any number of towers in the same point. All coordinates *b**j* are given in non-decreasing order.
|
Print minimal *r* so that each city will be covered by cellular network.
|
[
"3 2\n-2 2 4\n-3 0\n",
"5 3\n1 5 10 14 17\n4 11 15\n"
] |
[
"4\n",
"3\n"
] |
none
| 0
|
[
{
"input": "3 2\n-2 2 4\n-3 0",
"output": "4"
},
{
"input": "5 3\n1 5 10 14 17\n4 11 15",
"output": "3"
},
{
"input": "1 1\n-1000000000\n1000000000",
"output": "2000000000"
},
{
"input": "1 1\n1000000000\n-1000000000",
"output": "2000000000"
},
{
"input": "10 10\n1 1 2 2 2 4 4 6 7 9\n0 1 3 3 3 6 7 8 9 10",
"output": "1"
},
{
"input": "10 10\n2 52 280 401 416 499 721 791 841 943\n246 348 447 486 507 566 568 633 953 986",
"output": "244"
},
{
"input": "7 7\n1 2 3 3 4 5 6\n1 1 2 3 4 5 6",
"output": "0"
},
{
"input": "1 3\n-3\n-1 -1 8",
"output": "2"
},
{
"input": "1 2\n8\n-7 5",
"output": "3"
},
{
"input": "2 1\n4 8\n-1",
"output": "9"
},
{
"input": "1 2\n6\n-8 -8",
"output": "14"
},
{
"input": "1 4\n4\n-8 0 1 7",
"output": "3"
},
{
"input": "1 2\n2\n4 7",
"output": "2"
},
{
"input": "2 2\n-5 2\n-7 4",
"output": "2"
},
{
"input": "1 21\n3\n3 10 23 32 34 40 42 49 49 50 50 58 70 71 71 74 76 79 79 80 83",
"output": "0"
},
{
"input": "1 3\n-4\n-8 -1 1",
"output": "3"
},
{
"input": "4 1\n-6 -3 -1 2\n-7",
"output": "9"
},
{
"input": "2 3\n-2 7\n-7 -2 5",
"output": "2"
},
{
"input": "1 1\n-1\n0",
"output": "1"
},
{
"input": "1 3\n0\n-4 0 5",
"output": "0"
}
] | 1,663,320,918
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 32
| 296
| 10,137,600
|
# Write your code here :-)
# Cellular Network - Sorted Set Solution
from bisect import bisect
n, m = map(int, input().split())
a = [int(i) for i in input().split()]
b = [int(i) for i in input().split()]
r = 0
for i in a:
pos = bisect(b, i)
dist = int(2e9)
if pos < m:
dist = min(b[pos] - i, dist)
if pos:
dist = min(i - b[pos - 1], dist)
r = max(dist, r)
print(r)
|
Title: Cellular Network
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given *n* points on the straight line — the positions (*x*-coordinates) of the cities and *m* points on the same line — the positions (*x*-coordinates) of the cellular towers. All towers work in the same way — they provide cellular network for all cities, which are located at the distance which is no more than *r* from this tower.
Your task is to find minimal *r* that each city has been provided by cellular network, i.e. for each city there is at least one cellular tower at the distance which is no more than *r*.
If *r*<==<=0 then a tower provides cellular network only for the point where it is located. One tower can provide cellular network for any number of cities, but all these cities must be at the distance which is no more than *r* from this tower.
Input Specification:
The first line contains two positive integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=105) — the number of cities and the number of cellular towers.
The second line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=109<=≤<=*a**i*<=≤<=109) — the coordinates of cities. It is allowed that there are any number of cities in the same point. All coordinates *a**i* are given in non-decreasing order.
The third line contains a sequence of *m* integers *b*1,<=*b*2,<=...,<=*b**m* (<=-<=109<=≤<=*b**j*<=≤<=109) — the coordinates of cellular towers. It is allowed that there are any number of towers in the same point. All coordinates *b**j* are given in non-decreasing order.
Output Specification:
Print minimal *r* so that each city will be covered by cellular network.
Demo Input:
['3 2\n-2 2 4\n-3 0\n', '5 3\n1 5 10 14 17\n4 11 15\n']
Demo Output:
['4\n', '3\n']
Note:
none
|
```python
# Write your code here :-)
# Cellular Network - Sorted Set Solution
from bisect import bisect
n, m = map(int, input().split())
a = [int(i) for i in input().split()]
b = [int(i) for i in input().split()]
r = 0
for i in a:
pos = bisect(b, i)
dist = int(2e9)
if pos < m:
dist = min(b[pos] - i, dist)
if pos:
dist = min(i - b[pos - 1], dist)
r = max(dist, r)
print(r)
```
| 3
|
|
31
|
C
|
Schedule
|
PROGRAMMING
| 1,700
|
[
"implementation"
] |
C. Schedule
|
2
|
256
|
At the beginning of the new semester there is new schedule in the Berland State University. According to this schedule, *n* groups have lessons at the room 31. For each group the starting time of the lesson and the finishing time of the lesson are known. It has turned out that it is impossible to hold all lessons, because for some groups periods of their lessons intersect. If at some moment of time one groups finishes it's lesson, and the other group starts the lesson, their lessons don't intersect.
The dean wants to cancel the lesson in one group so that no two time periods of lessons of the remaining groups intersect. You are to find all ways to do that.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=5000) — amount of groups, which have lessons in the room 31. Then *n* lines follow, each of them contains two integers *l**i* *r**i* (1<=≤<=*l**i*<=<<=*r**i*<=≤<=106) — starting and finishing times of lesson of the *i*-th group. It is possible that initially no two lessons intersect (see sample 1).
|
Output integer *k* — amount of ways to cancel the lesson in exactly one group so that no two time periods of lessons of the remaining groups intersect. In the second line output *k* numbers — indexes of groups, where it is possible to cancel the lesson. Groups are numbered starting from 1 in the order that they were given in the input. Output the numbers in increasing order.
|
[
"3\n3 10\n20 30\n1 3\n",
"4\n3 10\n20 30\n1 3\n1 39\n",
"3\n1 5\n2 6\n3 7\n"
] |
[
"3\n1 2 3 ",
"1\n4 ",
"0\n"
] |
none
| 1,500
|
[
{
"input": "3\n3 10\n20 30\n1 3",
"output": "3\n1 2 3 "
},
{
"input": "4\n3 10\n20 30\n1 3\n1 39",
"output": "1\n4 "
},
{
"input": "3\n1 5\n2 6\n3 7",
"output": "0"
},
{
"input": "4\n1 5\n5 7\n6 9\n9 10",
"output": "2\n2 3 "
},
{
"input": "11\n717170 795210\n866429 970764\n163324 322182\n677099 717170\n241684 393937\n50433 114594\n970764 997956\n393937 664883\n235698 241684\n795210 832346\n114594 232438",
"output": "1\n3 "
},
{
"input": "16\n203671 381501\n58867 59732\n817520 962123\n125391 163027\n601766 617692\n381501 444610\n761937 817520\n16 10551\n21096 38291\n718073 761937\n583868 601766\n554859 731755\n678098 718073\n962123 992003\n163027 203671\n87917 96397",
"output": "1\n12 "
}
] | 1,658,589,695
| 2,147,483,647
|
PyPy 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 5
| 2,000
| 9,830,400
|
from operator import itemgetter
n=int(input())
X=[tuple(map(int,input().split())) for i in range(n)]
ANS=[]
for i in range(n):
Y=X[:i]+X[i+1:]
Y.sort(key=itemgetter(0))
MAX=-1<<60
flag=1
for x,y in Y:
if x<MAX:
flag=0
break
MAX=max(MAX,y)
if flag==1:
ANS.append(i+1)
print(len(ANS))
print(*ANS)
|
Title: Schedule
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
At the beginning of the new semester there is new schedule in the Berland State University. According to this schedule, *n* groups have lessons at the room 31. For each group the starting time of the lesson and the finishing time of the lesson are known. It has turned out that it is impossible to hold all lessons, because for some groups periods of their lessons intersect. If at some moment of time one groups finishes it's lesson, and the other group starts the lesson, their lessons don't intersect.
The dean wants to cancel the lesson in one group so that no two time periods of lessons of the remaining groups intersect. You are to find all ways to do that.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=5000) — amount of groups, which have lessons in the room 31. Then *n* lines follow, each of them contains two integers *l**i* *r**i* (1<=≤<=*l**i*<=<<=*r**i*<=≤<=106) — starting and finishing times of lesson of the *i*-th group. It is possible that initially no two lessons intersect (see sample 1).
Output Specification:
Output integer *k* — amount of ways to cancel the lesson in exactly one group so that no two time periods of lessons of the remaining groups intersect. In the second line output *k* numbers — indexes of groups, where it is possible to cancel the lesson. Groups are numbered starting from 1 in the order that they were given in the input. Output the numbers in increasing order.
Demo Input:
['3\n3 10\n20 30\n1 3\n', '4\n3 10\n20 30\n1 3\n1 39\n', '3\n1 5\n2 6\n3 7\n']
Demo Output:
['3\n1 2 3 ', '1\n4 ', '0\n']
Note:
none
|
```python
from operator import itemgetter
n=int(input())
X=[tuple(map(int,input().split())) for i in range(n)]
ANS=[]
for i in range(n):
Y=X[:i]+X[i+1:]
Y.sort(key=itemgetter(0))
MAX=-1<<60
flag=1
for x,y in Y:
if x<MAX:
flag=0
break
MAX=max(MAX,y)
if flag==1:
ANS.append(i+1)
print(len(ANS))
print(*ANS)
```
| 0
|
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"
}
] | 1,492,736,975
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 48
| 62
| 5,529,600
|
# Description of the problem can be found at http://codeforces.com/problemset/problem/629/A
def comb(N,k): # from scipy.comb(), but MODIFIED!
if (k > N) or (N < 0) or (k < 0):
return 0
N,k = map(int,(N,k))
top = N
val = 1
while (top > (N-k)):
val *= top
top -= 1
n = 1
while (n < k+1):
val //= n
n += 1
return val
n = int(input())
a_r = [0] * n
a_c = [0] * n
for i in range(n):
s = input()
for j in range(n):
if s[j] == "C":
a_r[i] += 1
a_c[j] += 1
print(sum(comb(a_r[i], 2) + comb(a_c[i], 2) for i in range(n)))
|
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
# Description of the problem can be found at http://codeforces.com/problemset/problem/629/A
def comb(N,k): # from scipy.comb(), but MODIFIED!
if (k > N) or (N < 0) or (k < 0):
return 0
N,k = map(int,(N,k))
top = N
val = 1
while (top > (N-k)):
val *= top
top -= 1
n = 1
while (n < k+1):
val //= n
n += 1
return val
n = int(input())
a_r = [0] * n
a_c = [0] * n
for i in range(n):
s = input()
for j in range(n):
if s[j] == "C":
a_r[i] += 1
a_c[j] += 1
print(sum(comb(a_r[i], 2) + comb(a_c[i], 2) for i in range(n)))
```
| 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"
}
] | 1,670,694,042
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 102
| 62
| 0
|
a = input()
b = input()
c = bin(int(a,2)^int(b,2))[2:]
c = '0' * (len(a) - len(c)) + c
print(c)
|
Title: Ultra-Fast Mathematician
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
Input Specification:
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
Output Specification:
Write one line — the corresponding answer. Do not omit the leading 0s.
Demo Input:
['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n']
Demo Output:
['1110001\n', '111\n', '0100\n', '00010\n']
Note:
none
|
```python
a = input()
b = input()
c = bin(int(a,2)^int(b,2))[2:]
c = '0' * (len(a) - len(c)) + c
print(c)
```
| 3.9845
|
439
|
C
|
Devu and Partitioning of the Array
|
PROGRAMMING
| 1,700
|
[
"brute force",
"constructive algorithms",
"implementation",
"number theory"
] | null | null |
Devu being a small kid, likes to play a lot, but he only likes to play with arrays. While playing he came up with an interesting question which he could not solve, can you please solve it for him?
Given an array consisting of distinct integers. Is it possible to partition the whole array into *k* disjoint non-empty parts such that *p* of the parts have even sum (each of them must have even sum) and remaining *k* - *p* have odd sum? (note that parts need not to be continuous).
If it is possible to partition the array, also give any possible way of valid partitioning.
|
The first line will contain three space separated integers *n*, *k*, *p* (1<=≤<=*k*<=≤<=*n*<=≤<=105; 0<=≤<=*p*<=≤<=*k*). The next line will contain *n* space-separated distinct integers representing the content of array *a*: *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
|
In the first line print "YES" (without the quotes) if it is possible to partition the array in the required way. Otherwise print "NO" (without the quotes).
If the required partition exists, print *k* lines after the first line. The *i**th* of them should contain the content of the *i**th* part. Print the content of the part in the line in the following way: firstly print the number of elements of the part, then print all the elements of the part in arbitrary order. There must be exactly *p* parts with even sum, each of the remaining *k* - *p* parts must have odd sum.
As there can be multiple partitions, you are allowed to print any valid partition.
|
[
"5 5 3\n2 6 10 5 9\n",
"5 5 3\n7 14 2 9 5\n",
"5 3 1\n1 2 3 7 5\n"
] |
[
"YES\n1 9\n1 5\n1 10\n1 6\n1 2\n",
"NO\n",
"YES\n3 5 1 3\n1 7\n1 2\n"
] |
none
| 1,500
|
[
{
"input": "5 5 3\n2 6 10 5 9",
"output": "YES\n1 9\n1 5\n1 10\n1 6\n1 2"
},
{
"input": "5 5 3\n7 14 2 9 5",
"output": "NO"
},
{
"input": "5 3 1\n1 2 3 7 5",
"output": "YES\n3 5 1 3\n1 7\n1 2"
},
{
"input": "10 5 3\n194757070 828985446 11164 80016 84729 117765558 111730436 164044532 419732980 48834",
"output": "NO"
},
{
"input": "10 6 3\n861947514 34945 190135645 68731 44833 387988147 84308862 878151920 458358978 809653252",
"output": "YES\n5 387988147 861947514 84308862 34945 190135645\n1 44833\n1 68731\n1 809653252\n1 458358978\n1 878151920"
},
{
"input": "10 8 3\n677037706 41099140 89128206 168458947 367939060 940344093 191391519 981938946 31319 34929915",
"output": "YES\n3 34929915 677037706 41099140\n1 31319\n1 191391519\n1 940344093\n1 168458947\n1 981938946\n1 367939060\n1 89128206"
},
{
"input": "10 8 4\n214605891 246349108 626595204 63889 794527783 83684156 5535 865709352 262484651 157889982",
"output": "NO"
},
{
"input": "10 6 3\n223143676 316703192 5286 408323576 61758 566101388 9871840 93989 91890 99264208",
"output": "NO"
},
{
"input": "2 2 1\n409447178 258805801",
"output": "YES\n1 258805801\n1 409447178"
},
{
"input": "2 1 1\n29161 15829",
"output": "YES\n2 15829 29161"
},
{
"input": "3 3 1\n357071129 476170324 503481367",
"output": "YES\n1 503481367\n1 357071129\n1 476170324"
},
{
"input": "3 3 1\n357071129 476170324 503481367",
"output": "YES\n1 503481367\n1 357071129\n1 476170324"
},
{
"input": "2 1 1\n29161 15829",
"output": "YES\n2 15829 29161"
},
{
"input": "23 22 3\n777717359 54451 123871650 211256633 193354035 935466677 800874233 532974165 62256 6511 3229 757421727 371493777 268999168 82355 22967 678259053 886134047 207070129 122416584 79851 35753 730872007",
"output": "YES\n2 730872007 123871650\n1 35753\n1 79851\n1 207070129\n1 886134047\n1 678259053\n1 22967\n1 82355\n1 371493777\n1 757421727\n1 3229\n1 6511\n1 532974165\n1 800874233\n1 935466677\n1 193354035\n1 211256633\n1 54451\n1 777717359\n1 122416584\n1 268999168\n1 62256"
},
{
"input": "16 9 9\n826588597 70843 528358243 60844 636968393 862405463 58232 69241 617006886 903909155 973050249 9381 49031 40100022 62141 43805",
"output": "YES\n3 40100022 826588597 70843\n1 617006886\n1 58232\n1 60844\n2 43805 62141\n2 49031 9381\n2 973050249 903909155\n2 69241 862405463\n2 636968393 528358243"
},
{
"input": "5 2 2\n316313049 24390603 595539594 514135024 39108",
"output": "YES\n4 39108 595539594 316313049 24390603\n1 514135024"
},
{
"input": "5 2 1\n12474 117513621 32958 767146609 20843",
"output": "YES\n4 20843 12474 117513621 767146609\n1 32958"
},
{
"input": "5 4 1\n387119493 716009972 88510 375210205 910834347",
"output": "YES\n2 910834347 716009972\n1 375210205\n1 387119493\n1 88510"
},
{
"input": "5 4 3\n674318396 881407702 882396010 208141498 53145",
"output": "YES\n2 53145 674318396\n1 208141498\n1 882396010\n1 881407702"
},
{
"input": "3 2 1\n976825506 613159225 249024714",
"output": "YES\n2 613159225 976825506\n1 249024714"
},
{
"input": "4 1 1\n173508914 11188 90766233 20363",
"output": "YES\n4 11188 173508914 90766233 20363"
},
{
"input": "30 24 12\n459253071 24156 64054 270713791 734796619 379920883 429646748 332144982 20929 27685 53253 82047732 172442017 34241 880121331 890223629 974692 954036632 68638567 972921811 421106382 64615 819330514 179630214 608594496 802986133 231010377 184513476 73143 93045",
"output": "YES\n7 93045 459253071 270713791 734796619 379920883 20929 27685\n1 73143\n1 231010377\n1 802986133\n1 64615\n1 972921811\n1 68638567\n1 890223629\n1 880121331\n1 34241\n1 172442017\n1 53253\n1 184513476\n1 608594496\n1 179630214\n1 819330514\n1 421106382\n1 954036632\n1 974692\n1 82047732\n1 332144982\n1 429646748\n1 64054\n1 24156"
},
{
"input": "9 5 1\n91623 466261329 311727429 18189 42557 22943 66177 473271749 62622",
"output": "YES\n5 473271749 91623 466261329 311727429 18189\n1 66177\n1 22943\n1 42557\n1 62622"
},
{
"input": "4 1 1\n266639563 36517 172287193 166673809",
"output": "YES\n4 166673809 172287193 266639563 36517"
},
{
"input": "5 2 2\n19571 180100775 421217758 284511211 49475",
"output": "YES\n3 421217758 19571 180100775\n2 49475 284511211"
},
{
"input": "4 2 2\n736788713 82230 66800 37791827",
"output": "YES\n3 66800 736788713 37791827\n1 82230"
},
{
"input": "5 1 1\n33889 469945850 40673 939970023 776172319",
"output": "YES\n5 469945850 33889 40673 939970023 776172319"
},
{
"input": "1 1 0\n2",
"output": "NO"
},
{
"input": "1 1 0\n3",
"output": "YES\n1 3"
},
{
"input": "1 1 1\n2",
"output": "YES\n1 2"
},
{
"input": "1 1 1\n3",
"output": "NO"
},
{
"input": "2 2 2\n2 3",
"output": "NO"
},
{
"input": "2 2 2\n2 4",
"output": "YES\n1 4\n1 2"
},
{
"input": "2 2 1\n3 2",
"output": "YES\n1 3\n1 2"
},
{
"input": "4 2 0\n3 5 7 9",
"output": "YES\n3 9 3 5\n1 7"
},
{
"input": "3 2 0\n1 3 2",
"output": "YES\n2 3 2\n1 1"
},
{
"input": "2 1 1\n2 4",
"output": "YES\n2 4 2"
},
{
"input": "7 3 0\n1 3 5 7 9 11 13",
"output": "YES\n5 13 1 3 5 7\n1 11\n1 9"
},
{
"input": "8 4 4\n1 3 5 7 9 11 13 15",
"output": "YES\n2 15 13\n2 11 9\n2 7 5\n2 3 1"
},
{
"input": "2 1 1\n1 3",
"output": "YES\n2 3 1"
}
] | 1,675,945,372
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 9
| 77
| 13,004,800
|
import heapq
import sys
from types import GeneratorType
from functools import cmp_to_key
from collections import defaultdict, Counter, deque
import math
inf = float("inf")
class FastIO:
def __init__(self):
return
@staticmethod
def _read():
return sys.stdin.readline().strip()
def read_int(self):
return int(self._read())
def read_float(self):
return float(self._read())
def read_ints(self):
return map(int, self._read().split())
def read_floats(self):
return map(float, self._read().split())
def read_ints_minus_one(self):
return map(lambda x: int(x) - 1, self._read().split())
def read_list_ints(self):
return list(map(int, self._read().split()))
def read_list_floats(self):
return list(map(float, self._read().split()))
def read_list_ints_minus_one(self):
return list(map(lambda x: int(x) - 1, self._read().split()))
def read_str(self):
return self._read()
def read_list_strs(self):
return self._read().split()
def read_list_str(self):
return list(self._read())
@staticmethod
def st(x):
return sys.stdout.write(str(x) + '\n')
@staticmethod
def lst(x):
return sys.stdout.write(" ".join(str(w) for w in x) + '\n')
@staticmethod
def round_5(f):
res = int(f)
if f - res >= 0.5:
res += 1
return res
@staticmethod
def max(a, b):
return a if a > b else b
@staticmethod
def min(a, b):
return a if a < b else b
@staticmethod
def bootstrap(f, queue=[]):
def wrappedfunc(*args, **kwargs):
if queue:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if isinstance(to, GeneratorType):
queue.append(to)
to = next(to)
else:
queue.pop()
if not queue:
break
to = queue[-1].send(to)
return to
return wrappedfunc
def main(ac=FastIO()):
n, k, p = ac.read_ints()
nums = ac.read_list_ints()
odd = []
even = []
for num in nums:
if num % 2:
odd.append(num) # 奇数
else:
even.append(num)
# p个偶数 k-p个奇数
a, b = len(odd), len(even)
if a < k-p or (a-(k-p)) % 2: # 奇数不够或者奇数多余
ac.st("NO")
return
ac.st("YES")
if p == 0:
ans = [[] for _ in range(k)]
for j in range(k):
ans.append([odd[j]])
ans[-1].extend(even)
ans[-1].extend(odd[k:])
return
# 偶数不够
if b + (a-(k-p))//2 < p:
ac.st("NO")
return
ans = [[] for _ in range(p)]
for i in range(b):
ans[(i%p)].append(even[i])
for j in range(k-p):
ans.append([odd[j]])
i = b
for j in range(k-p, a, 2):
ans[i%p].extend([odd[j], odd[j+1]])
i += 1
for a in ans:
ac.lst([len(a)]+a)
return
main()
|
Title: Devu and Partitioning of the Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Devu being a small kid, likes to play a lot, but he only likes to play with arrays. While playing he came up with an interesting question which he could not solve, can you please solve it for him?
Given an array consisting of distinct integers. Is it possible to partition the whole array into *k* disjoint non-empty parts such that *p* of the parts have even sum (each of them must have even sum) and remaining *k* - *p* have odd sum? (note that parts need not to be continuous).
If it is possible to partition the array, also give any possible way of valid partitioning.
Input Specification:
The first line will contain three space separated integers *n*, *k*, *p* (1<=≤<=*k*<=≤<=*n*<=≤<=105; 0<=≤<=*p*<=≤<=*k*). The next line will contain *n* space-separated distinct integers representing the content of array *a*: *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
Output Specification:
In the first line print "YES" (without the quotes) if it is possible to partition the array in the required way. Otherwise print "NO" (without the quotes).
If the required partition exists, print *k* lines after the first line. The *i**th* of them should contain the content of the *i**th* part. Print the content of the part in the line in the following way: firstly print the number of elements of the part, then print all the elements of the part in arbitrary order. There must be exactly *p* parts with even sum, each of the remaining *k* - *p* parts must have odd sum.
As there can be multiple partitions, you are allowed to print any valid partition.
Demo Input:
['5 5 3\n2 6 10 5 9\n', '5 5 3\n7 14 2 9 5\n', '5 3 1\n1 2 3 7 5\n']
Demo Output:
['YES\n1 9\n1 5\n1 10\n1 6\n1 2\n', 'NO\n', 'YES\n3 5 1 3\n1 7\n1 2\n']
Note:
none
|
```python
import heapq
import sys
from types import GeneratorType
from functools import cmp_to_key
from collections import defaultdict, Counter, deque
import math
inf = float("inf")
class FastIO:
def __init__(self):
return
@staticmethod
def _read():
return sys.stdin.readline().strip()
def read_int(self):
return int(self._read())
def read_float(self):
return float(self._read())
def read_ints(self):
return map(int, self._read().split())
def read_floats(self):
return map(float, self._read().split())
def read_ints_minus_one(self):
return map(lambda x: int(x) - 1, self._read().split())
def read_list_ints(self):
return list(map(int, self._read().split()))
def read_list_floats(self):
return list(map(float, self._read().split()))
def read_list_ints_minus_one(self):
return list(map(lambda x: int(x) - 1, self._read().split()))
def read_str(self):
return self._read()
def read_list_strs(self):
return self._read().split()
def read_list_str(self):
return list(self._read())
@staticmethod
def st(x):
return sys.stdout.write(str(x) + '\n')
@staticmethod
def lst(x):
return sys.stdout.write(" ".join(str(w) for w in x) + '\n')
@staticmethod
def round_5(f):
res = int(f)
if f - res >= 0.5:
res += 1
return res
@staticmethod
def max(a, b):
return a if a > b else b
@staticmethod
def min(a, b):
return a if a < b else b
@staticmethod
def bootstrap(f, queue=[]):
def wrappedfunc(*args, **kwargs):
if queue:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if isinstance(to, GeneratorType):
queue.append(to)
to = next(to)
else:
queue.pop()
if not queue:
break
to = queue[-1].send(to)
return to
return wrappedfunc
def main(ac=FastIO()):
n, k, p = ac.read_ints()
nums = ac.read_list_ints()
odd = []
even = []
for num in nums:
if num % 2:
odd.append(num) # 奇数
else:
even.append(num)
# p个偶数 k-p个奇数
a, b = len(odd), len(even)
if a < k-p or (a-(k-p)) % 2: # 奇数不够或者奇数多余
ac.st("NO")
return
ac.st("YES")
if p == 0:
ans = [[] for _ in range(k)]
for j in range(k):
ans.append([odd[j]])
ans[-1].extend(even)
ans[-1].extend(odd[k:])
return
# 偶数不够
if b + (a-(k-p))//2 < p:
ac.st("NO")
return
ans = [[] for _ in range(p)]
for i in range(b):
ans[(i%p)].append(even[i])
for j in range(k-p):
ans.append([odd[j]])
i = b
for j in range(k-p, a, 2):
ans[i%p].extend([odd[j], odd[j+1]])
i += 1
for a in ans:
ac.lst([len(a)]+a)
return
main()
```
| 0
|
|
59
|
A
|
Word
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] |
A. Word
|
2
|
256
|
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
|
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
|
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
|
[
"HoUse\n",
"ViP\n",
"maTRIx\n"
] |
[
"house\n",
"VIP\n",
"matrix\n"
] |
none
| 500
|
[
{
"input": "HoUse",
"output": "house"
},
{
"input": "ViP",
"output": "VIP"
},
{
"input": "maTRIx",
"output": "matrix"
},
{
"input": "BNHWpnpawg",
"output": "bnhwpnpawg"
},
{
"input": "VTYGP",
"output": "VTYGP"
},
{
"input": "CHNenu",
"output": "chnenu"
},
{
"input": "ERPZGrodyu",
"output": "erpzgrodyu"
},
{
"input": "KSXBXWpebh",
"output": "KSXBXWPEBH"
},
{
"input": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv",
"output": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv"
},
{
"input": "Amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd",
"output": "amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd"
},
{
"input": "ISAGFJFARYFBLOPQDSHWGMCNKMFTLVFUGNJEWGWNBLXUIATXEkqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv",
"output": "isagfjfaryfblopqdshwgmcnkmftlvfugnjewgwnblxuiatxekqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv"
},
{
"input": "XHRPXZEGHSOCJPICUIXSKFUZUPYTSGJSDIYBCMNMNBPNDBXLXBzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg",
"output": "xhrpxzeghsocjpicuixskfuzupytsgjsdiybcmnmnbpndbxlxbzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg"
},
{
"input": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGAdkcetqjljtmttlonpekcovdzebzdkzggwfsxhapmjkdbuceak",
"output": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGADKCETQJLJTMTTLONPEKCOVDZEBZDKZGGWFSXHAPMJKDBUCEAK"
},
{
"input": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFw",
"output": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFW"
},
{
"input": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB",
"output": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB"
},
{
"input": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge",
"output": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge"
},
{
"input": "Ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw",
"output": "ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw"
},
{
"input": "YQOMLKYAORUQQUCQZCDYMIVDHGWZFFRMUVTAWCHERFPMNRYRIkgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks",
"output": "yqomlkyaoruqqucqzcdymivdhgwzffrmuvtawcherfpmnryrikgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks"
},
{
"input": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJcuusigdqfkumewjtdyitveeiaybwrhomrwmpdipjwiuxfnwuz",
"output": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJCUUSIGDQFKUMEWJTDYITVEEIAYBWRHOMRWMPDIPJWIUXFNWUZ"
},
{
"input": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWuckzcpxosodcjaaakvlxpbiigsiauviilylnnqlyucziihqg",
"output": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWUCKZCPXOSODCJAAAKVLXPBIIGSIAUVIILYLNNQLYUCZIIHQG"
},
{
"input": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO",
"output": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO"
},
{
"input": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDd",
"output": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDD"
},
{
"input": "EBWOVSNLYTWWXrnovgugogtjahnmatomubebyxakas",
"output": "ebwovsnlytwwxrnovgugogtjahnmatomubebyxakas"
},
{
"input": "AORNNDKTRLRVGDPXJKXFTPFpopuzrlqumrxssskvbm",
"output": "AORNNDKTRLRVGDPXJKXFTPFPOPUZRLQUMRXSSSKVBM"
},
{
"input": "DBBZJCWQEVGYHEEUHJAWRHBDPsnplijbwtdibqrrheebfxqcvn",
"output": "dbbzjcwqevgyheeuhjawrhbdpsnplijbwtdibqrrheebfxqcvn"
},
{
"input": "t",
"output": "t"
},
{
"input": "N",
"output": "N"
},
{
"input": "kv",
"output": "kv"
},
{
"input": "Ur",
"output": "ur"
},
{
"input": "CN",
"output": "CN"
}
] | 1,654,070,718
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 30
| 124
| 0
|
input = input()
def vasya(input):
uppr, lwr = 0, 0
for i in range(len(input)):
if input[i].isupper():
uppr += 1
elif input[i].islower():
lwr += 1
if uppr > lwr:
return input.upper()
else:
return input.lower()
print(vasya(input))
|
Title: Word
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
Input Specification:
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Output Specification:
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
Demo Input:
['HoUse\n', 'ViP\n', 'maTRIx\n']
Demo Output:
['house\n', 'VIP\n', 'matrix\n']
Note:
none
|
```python
input = input()
def vasya(input):
uppr, lwr = 0, 0
for i in range(len(input)):
if input[i].isupper():
uppr += 1
elif input[i].islower():
lwr += 1
if uppr > lwr:
return input.upper()
else:
return input.lower()
print(vasya(input))
```
| 3.969
|
294
|
A
|
Shaass and Oskols
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
Shaass has decided to hunt some birds. There are *n* horizontal electricity wires aligned parallel to each other. Wires are numbered 1 to *n* from top to bottom. On each wire there are some oskols sitting next to each other. Oskol is the name of a delicious kind of birds in Shaass's territory. Supposed there are *a**i* oskols sitting on the *i*-th wire.
Sometimes Shaass shots one of the birds and the bird dies (suppose that this bird sat at the *i*-th wire). Consequently all the birds on the *i*-th wire to the left of the dead bird get scared and jump up on the wire number *i*<=-<=1, if there exists no upper wire they fly away. Also all the birds to the right of the dead bird jump down on wire number *i*<=+<=1, if there exists no such wire they fly away.
Shaass has shot *m* birds. You're given the initial number of birds on each wire, tell him how many birds are sitting on each wire after the shots.
|
The first line of the input contains an integer *n*, (1<=≤<=*n*<=≤<=100). The next line contains a list of space-separated integers *a*1,<=*a*2,<=...,<=*a**n*, (0<=≤<=*a**i*<=≤<=100).
The third line contains an integer *m*, (0<=≤<=*m*<=≤<=100). Each of the next *m* lines contains two integers *x**i* and *y**i*. The integers mean that for the *i*-th time Shaass shoot the *y**i*-th (from left) bird on the *x**i*-th wire, (1<=≤<=*x**i*<=≤<=*n*,<=1<=≤<=*y**i*). It's guaranteed there will be at least *y**i* birds on the *x**i*-th wire at that moment.
|
On the *i*-th line of the output print the number of birds on the *i*-th wire.
|
[
"5\n10 10 10 10 10\n5\n2 5\n3 13\n2 12\n1 13\n4 6\n",
"3\n2 4 1\n1\n2 2\n"
] |
[
"0\n12\n5\n0\n16\n",
"3\n0\n3\n"
] |
none
| 500
|
[
{
"input": "5\n10 10 10 10 10\n5\n2 5\n3 13\n2 12\n1 13\n4 6",
"output": "0\n12\n5\n0\n16"
},
{
"input": "3\n2 4 1\n1\n2 2",
"output": "3\n0\n3"
},
{
"input": "5\n58 51 45 27 48\n5\n4 9\n5 15\n4 5\n5 8\n1 43",
"output": "0\n66\n57\n7\n0"
},
{
"input": "10\n48 53 10 28 91 56 81 2 67 52\n2\n2 40\n6 51",
"output": "87\n0\n23\n28\n141\n0\n86\n2\n67\n52"
},
{
"input": "2\n72 45\n6\n1 69\n2 41\n1 19\n2 7\n1 5\n2 1",
"output": "0\n0"
},
{
"input": "10\n95 54 36 39 98 30 19 24 14 12\n3\n9 5\n8 15\n7 5",
"output": "95\n54\n36\n39\n98\n34\n0\n28\n13\n21"
},
{
"input": "100\n95 15 25 18 64 62 23 59 70 84 50 26 87 35 75 86 0 22 77 60 66 41 21 9 75 50 25 3 69 14 39 68 64 46 59 99 2 0 21 76 90 12 61 42 6 91 36 39 47 41 93 81 66 57 70 36 68 89 52 1 19 93 67 22 76 20 8 81 98 18 100 73 61 93 75 80 53 72 40 20 2 86 33 59 27 16 11 26 55 44 47 55 94 58 2 55 82 22 9 95\n0",
"output": "95\n15\n25\n18\n64\n62\n23\n59\n70\n84\n50\n26\n87\n35\n75\n86\n0\n22\n77\n60\n66\n41\n21\n9\n75\n50\n25\n3\n69\n14\n39\n68\n64\n46\n59\n99\n2\n0\n21\n76\n90\n12\n61\n42\n6\n91\n36\n39\n47\n41\n93\n81\n66\n57\n70\n36\n68\n89\n52\n1\n19\n93\n67\n22\n76\n20\n8\n81\n98\n18\n100\n73\n61\n93\n75\n80\n53\n72\n40\n20\n2\n86\n33\n59\n27\n16\n11\n26\n55\n44\n47\n55\n94\n58\n2\n55\n82\n22\n9\n95"
},
{
"input": "1\n100\n1\n1 100",
"output": "0"
},
{
"input": "1\n100\n1\n1 1",
"output": "0"
},
{
"input": "1\n50\n1\n1 25",
"output": "0"
},
{
"input": "2\n50 0\n1\n1 1",
"output": "0\n49"
},
{
"input": "1\n10\n0",
"output": "10"
}
] | 1,698,326,538
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 31
| 62
| 0
|
n = int(input()) # Number of wires
birds_on_wire = list(map(int, input().split())) # Initial number of birds on each wire
m = int(input()) # Number of shots
for _ in range(m):
x, y = map(int, input().split()) # x: wire number, y: bird position from left
x -= 1 # Adjust the 0-based index
birds_on_wire[x] -= y # Decrease the number of birds on the x-th wire
if x > 0:
birds_on_wire[x - 1] += y - 1 # Birds to the left
if x < n - 1:
birds_on_wire[x + 1] += birds_on_wire[x] # Birds to the right
birds_on_wire[x] = 0 # Birds on the shot wire
# Print the updated number of birds on each wire
for birds in birds_on_wire:
print(birds)
|
Title: Shaass and Oskols
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Shaass has decided to hunt some birds. There are *n* horizontal electricity wires aligned parallel to each other. Wires are numbered 1 to *n* from top to bottom. On each wire there are some oskols sitting next to each other. Oskol is the name of a delicious kind of birds in Shaass's territory. Supposed there are *a**i* oskols sitting on the *i*-th wire.
Sometimes Shaass shots one of the birds and the bird dies (suppose that this bird sat at the *i*-th wire). Consequently all the birds on the *i*-th wire to the left of the dead bird get scared and jump up on the wire number *i*<=-<=1, if there exists no upper wire they fly away. Also all the birds to the right of the dead bird jump down on wire number *i*<=+<=1, if there exists no such wire they fly away.
Shaass has shot *m* birds. You're given the initial number of birds on each wire, tell him how many birds are sitting on each wire after the shots.
Input Specification:
The first line of the input contains an integer *n*, (1<=≤<=*n*<=≤<=100). The next line contains a list of space-separated integers *a*1,<=*a*2,<=...,<=*a**n*, (0<=≤<=*a**i*<=≤<=100).
The third line contains an integer *m*, (0<=≤<=*m*<=≤<=100). Each of the next *m* lines contains two integers *x**i* and *y**i*. The integers mean that for the *i*-th time Shaass shoot the *y**i*-th (from left) bird on the *x**i*-th wire, (1<=≤<=*x**i*<=≤<=*n*,<=1<=≤<=*y**i*). It's guaranteed there will be at least *y**i* birds on the *x**i*-th wire at that moment.
Output Specification:
On the *i*-th line of the output print the number of birds on the *i*-th wire.
Demo Input:
['5\n10 10 10 10 10\n5\n2 5\n3 13\n2 12\n1 13\n4 6\n', '3\n2 4 1\n1\n2 2\n']
Demo Output:
['0\n12\n5\n0\n16\n', '3\n0\n3\n']
Note:
none
|
```python
n = int(input()) # Number of wires
birds_on_wire = list(map(int, input().split())) # Initial number of birds on each wire
m = int(input()) # Number of shots
for _ in range(m):
x, y = map(int, input().split()) # x: wire number, y: bird position from left
x -= 1 # Adjust the 0-based index
birds_on_wire[x] -= y # Decrease the number of birds on the x-th wire
if x > 0:
birds_on_wire[x - 1] += y - 1 # Birds to the left
if x < n - 1:
birds_on_wire[x + 1] += birds_on_wire[x] # Birds to the right
birds_on_wire[x] = 0 # Birds on the shot wire
# Print the updated number of birds on each wire
for birds in birds_on_wire:
print(birds)
```
| 3
|
|
457
|
A
|
Golden System
|
PROGRAMMING
| 1,700
|
[
"math",
"meet-in-the-middle"
] | null | null |
Piegirl got bored with binary, decimal and other integer based counting systems. Recently she discovered some interesting properties about number , in particular that *q*2<==<=*q*<=+<=1, and she thinks it would make a good base for her new unique system. She called it "golden system". In golden system the number is a non-empty string containing 0's and 1's as digits. The decimal value of expression *a*0*a*1...*a**n* equals to .
Soon Piegirl found out that this system doesn't have same properties that integer base systems do and some operations can not be performed on it. She wasn't able to come up with a fast way of comparing two numbers. She is asking for your help.
Given two numbers written in golden system notation, determine which of them has larger decimal value.
|
Input consists of two lines — one for each number. Each line contains non-empty string consisting of '0' and '1' characters. The length of each string does not exceed 100000.
|
Print ">" if the first number is larger, "<" if it is smaller and "=" if they are equal.
|
[
"1000\n111\n",
"00100\n11\n",
"110\n101\n"
] |
[
"<\n",
"=\n",
">\n"
] |
In the first example first number equals to <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/9c955eec678d6e7dcdc7c94fb203e922d2ad19ad.png" style="max-width: 100.0%;max-height: 100.0%;"/>, while second number is approximately 1.618033988<sup class="upper-index">2</sup> + 1.618033988 + 1 ≈ 5.236, which is clearly a bigger number.
In the second example numbers are equal. Each of them is ≈ 2.618.
| 1,000
|
[
{
"input": "1000\n111",
"output": "<"
},
{
"input": "00100\n11",
"output": "="
},
{
"input": "110\n101",
"output": ">"
},
{
"input": "0\n0",
"output": "="
},
{
"input": "1\n10",
"output": "<"
},
{
"input": "11\n10",
"output": ">"
},
{
"input": "00111\n10100",
"output": "<"
},
{
"input": "00\n1",
"output": "<"
},
{
"input": "01\n010",
"output": "<"
},
{
"input": "111\n00",
"output": ">"
},
{
"input": "1100\n11",
"output": ">"
},
{
"input": "0110\n001",
"output": ">"
},
{
"input": "1111\n0110",
"output": ">"
},
{
"input": "01010\n0011",
"output": ">"
},
{
"input": "0\n1",
"output": "<"
},
{
"input": "1\n0",
"output": ">"
},
{
"input": "1\n1",
"output": "="
},
{
"input": "010000100010100000100010001000001100100010110000101010000010010011001111101101001\n001011100001110101111001100110001011011100000000100111011010010011010100101011111",
"output": "="
},
{
"input": "11111001000\n1011100100",
"output": ">"
},
{
"input": "1001111010001100001010001010010010100010100011101101110011110101011000010111101100111000110110110010\n01111001101111100111111001110110100101001111010001000000001001001111100101101100001101111111100111101",
"output": "<"
},
{
"input": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n0",
"output": ">"
},
{
"input": "100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n1",
"output": ">"
},
{
"input": "1\n100000000000000000000000000000000000000000000000000",
"output": "<"
},
{
"input": "1\n1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "<"
},
{
"input": "11111111111111111111111111111111111111111111111111111111111111111111111111111111\n1111111111111111111111111111111111111111111111111111111111111111111111111111111",
"output": ">"
},
{
"input": "10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n100000000000000000000",
"output": ">"
},
{
"input": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111\n1011111111111111111111111111011011011001101111111110111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111",
"output": ">"
},
{
"input": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111",
"output": "<"
},
{
"input": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111\n1011111111111111111111111111011011011001101111111110111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111",
"output": ">"
},
{
"input": "11111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111",
"output": "<"
},
{
"input": "100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n0",
"output": ">"
},
{
"input": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n1110",
"output": ">"
},
{
"input": "10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n1000",
"output": ">"
},
{
"input": "100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n1000",
"output": ">"
},
{
"input": "1\n1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "<"
},
{
"input": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n0",
"output": ">"
},
{
"input": "10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n10000",
"output": ">"
},
{
"input": "10000100001000010000100001000010000100001000010000\n1",
"output": ">"
},
{
"input": "101001010101010101010100101010101010101010101001010101010100101010101010100101101010100101010100101010101001010101010101010100101010101010101010101001010101010100101010101010100101101010100101010100101010101001010101010101010100101010101010101010101001010101010100101010101010100101101010100101010100101010\n1",
"output": ">"
},
{
"input": "10100\n01011",
"output": ">"
},
{
"input": "10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n01111000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "<"
},
{
"input": "11111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111\n0000001010101011",
"output": ">"
},
{
"input": "110010010101001001001010100100010101010101011111111111111010101000000000000000000010110111111110101010111111111111111111111111111111111\n1011111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111",
"output": ">"
},
{
"input": "1100\n0111",
"output": ">"
},
{
"input": "1111111111111111111111111111111111111111111111111\n0",
"output": ">"
},
{
"input": "1100100101010010010010101001000101010101010111111111111110101010000000000000000000101101111111101010101111111111111111111111111111111\n1011111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111",
"output": ">"
},
{
"input": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n100000000000000000000",
"output": ">"
},
{
"input": "100001000010000100001000010000100001000010000100001111111111111111111111111111111111111111111111111111111111111111111111\n1",
"output": ">"
},
{
"input": "11111111111111111111111111111111111111111111111111111111111111\n1",
"output": ">"
},
{
"input": "1011\n0100",
"output": ">"
},
{
"input": "100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n011000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001",
"output": "<"
},
{
"input": "1000000000000000000000000000000011111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111000000000000000000000000000000000000000000000000\n1111111111111111111111111111111111111111111111111111111111111111111111111111110000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "<"
},
{
"input": "111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111001\n000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011",
"output": ">"
},
{
"input": "1000000000000000000000000\n0101010101010101010101011",
"output": "="
},
{
"input": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n1",
"output": ">"
},
{
"input": "101010101010101010101010101010101010101010101010101010101010101010101010\n1000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "<"
},
{
"input": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111\n0",
"output": ">"
},
{
"input": "111\n1000",
"output": ">"
},
{
"input": "1000000000000000000000000000000000000000000000000\n000",
"output": ">"
},
{
"input": "11111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111\n10",
"output": ">"
},
{
"input": "111111111111111111111111111111111111111111111111\n11",
"output": ">"
},
{
"input": "10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n000",
"output": ">"
},
{
"input": "101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010\n1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": ">"
},
{
"input": "100000000000000000000000000000000000000000000000\n0",
"output": ">"
}
] | 1,585,771,869
| 2,147,483,647
|
PyPy 3
|
RUNTIME_ERROR
|
TESTS
| 8
| 810
| 10,956,800
|
a=[int(i) for i in input()]
b=[int(i) for i in input()]
n=len(a)
m=len(b)
for i in range(n-2):
if(a[i]!=0):
a[i+1]+=a[i]
a[i+2]+=a[i]
a[i]=0
for i in range(m-2):
if(b[i]!=0):
b[i+1]+=b[i]
b[i+2]+=b[i]
b[i]=0
if(n>1):
r1=a[n-2]
else:
r1=0
c1=a[n-1]
c2=b[m-1]
if(m>1):
r2=b[m-2]
else:
r2=0
if(r1==r2):
if(c1==c2):
print("=")
elif(c1<c2):
print("<")
else:
print(">")
else:
golden=((5**0.5)+1)/2
if (r1*golden)+c1 > (r2*golden)+c2:
print(">")
else:
print("<")
|
Title: Golden System
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Piegirl got bored with binary, decimal and other integer based counting systems. Recently she discovered some interesting properties about number , in particular that *q*2<==<=*q*<=+<=1, and she thinks it would make a good base for her new unique system. She called it "golden system". In golden system the number is a non-empty string containing 0's and 1's as digits. The decimal value of expression *a*0*a*1...*a**n* equals to .
Soon Piegirl found out that this system doesn't have same properties that integer base systems do and some operations can not be performed on it. She wasn't able to come up with a fast way of comparing two numbers. She is asking for your help.
Given two numbers written in golden system notation, determine which of them has larger decimal value.
Input Specification:
Input consists of two lines — one for each number. Each line contains non-empty string consisting of '0' and '1' characters. The length of each string does not exceed 100000.
Output Specification:
Print ">" if the first number is larger, "<" if it is smaller and "=" if they are equal.
Demo Input:
['1000\n111\n', '00100\n11\n', '110\n101\n']
Demo Output:
['<\n', '=\n', '>\n']
Note:
In the first example first number equals to <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/9c955eec678d6e7dcdc7c94fb203e922d2ad19ad.png" style="max-width: 100.0%;max-height: 100.0%;"/>, while second number is approximately 1.618033988<sup class="upper-index">2</sup> + 1.618033988 + 1 ≈ 5.236, which is clearly a bigger number.
In the second example numbers are equal. Each of them is ≈ 2.618.
|
```python
a=[int(i) for i in input()]
b=[int(i) for i in input()]
n=len(a)
m=len(b)
for i in range(n-2):
if(a[i]!=0):
a[i+1]+=a[i]
a[i+2]+=a[i]
a[i]=0
for i in range(m-2):
if(b[i]!=0):
b[i+1]+=b[i]
b[i+2]+=b[i]
b[i]=0
if(n>1):
r1=a[n-2]
else:
r1=0
c1=a[n-1]
c2=b[m-1]
if(m>1):
r2=b[m-2]
else:
r2=0
if(r1==r2):
if(c1==c2):
print("=")
elif(c1<c2):
print("<")
else:
print(">")
else:
golden=((5**0.5)+1)/2
if (r1*golden)+c1 > (r2*golden)+c2:
print(">")
else:
print("<")
```
| -1
|
|
214
|
A
|
System of Equations
|
PROGRAMMING
| 800
|
[
"brute force"
] | null | null |
Furik loves math lessons very much, so he doesn't attend them, unlike Rubik. But now Furik wants to get a good mark for math. For that Ms. Ivanova, his math teacher, gave him a new task. Furik solved the task immediately. Can you?
You are given a system of equations:
You should count, how many there are pairs of integers (*a*,<=*b*) (0<=≤<=*a*,<=*b*) which satisfy the system.
|
A single line contains two integers *n*,<=*m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the parameters of the system. The numbers on the line are separated by a space.
|
On a single line print the answer to the problem.
|
[
"9 3\n",
"14 28\n",
"4 20\n"
] |
[
"1\n",
"1\n",
"0\n"
] |
In the first sample the suitable pair is integers (3, 0). In the second sample the suitable pair is integers (3, 5). In the third sample there is no suitable pair.
| 500
|
[
{
"input": "9 3",
"output": "1"
},
{
"input": "14 28",
"output": "1"
},
{
"input": "4 20",
"output": "0"
},
{
"input": "18 198",
"output": "1"
},
{
"input": "22 326",
"output": "1"
},
{
"input": "26 104",
"output": "1"
},
{
"input": "14 10",
"output": "0"
},
{
"input": "8 20",
"output": "0"
},
{
"input": "2 8",
"output": "0"
},
{
"input": "20 11",
"output": "0"
},
{
"input": "57 447",
"output": "1"
},
{
"input": "1 1",
"output": "2"
},
{
"input": "66 296",
"output": "1"
},
{
"input": "75 683",
"output": "1"
},
{
"input": "227 975",
"output": "1"
},
{
"input": "247 499",
"output": "1"
},
{
"input": "266 116",
"output": "1"
},
{
"input": "286 916",
"output": "1"
},
{
"input": "307 341",
"output": "1"
},
{
"input": "451 121",
"output": "1"
},
{
"input": "471 921",
"output": "1"
},
{
"input": "502 346",
"output": "1"
},
{
"input": "535 59",
"output": "1"
},
{
"input": "555 699",
"output": "1"
},
{
"input": "747 351",
"output": "1"
},
{
"input": "790 64",
"output": "1"
},
{
"input": "810 704",
"output": "1"
},
{
"input": "855 225",
"output": "1"
},
{
"input": "902 34",
"output": "1"
},
{
"input": "922 514",
"output": "1"
},
{
"input": "971 131",
"output": "1"
},
{
"input": "991 931",
"output": "1"
},
{
"input": "840 780",
"output": "0"
},
{
"input": "102 595",
"output": "0"
},
{
"input": "139 433",
"output": "0"
},
{
"input": "968 288",
"output": "0"
},
{
"input": "563 354",
"output": "0"
},
{
"input": "994 975",
"output": "0"
},
{
"input": "456 221",
"output": "0"
},
{
"input": "205 210",
"output": "0"
},
{
"input": "1 11",
"output": "0"
},
{
"input": "1000 1000",
"output": "0"
},
{
"input": "3 3",
"output": "0"
},
{
"input": "11 99",
"output": "0"
},
{
"input": "2 2",
"output": "1"
},
{
"input": "11 1",
"output": "0"
},
{
"input": "6 6",
"output": "1"
},
{
"input": "100 452",
"output": "0"
},
{
"input": "420 380",
"output": "0"
},
{
"input": "31 31",
"output": "0"
},
{
"input": "2 6",
"output": "0"
},
{
"input": "112 134",
"output": "0"
},
{
"input": "13 13",
"output": "0"
},
{
"input": "1 571",
"output": "0"
}
] | 1,664,470,374
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 62
| 0
|
n,m=map(int,input().split())
c=n-m
iht=[]
yeni=[]
for i in range(1,c+1):
for t in range(1,c+1):
if i*t==c:
iht.append([i,t])
for i in range(len(iht)):
a=(iht[i][1]+iht[i][0]+1)/2
b=a-iht[i][0]
if a**2 + b == n and b**2 + a==m :
yeni.append([a,b])
print(len(yeni))
|
Title: System of Equations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Furik loves math lessons very much, so he doesn't attend them, unlike Rubik. But now Furik wants to get a good mark for math. For that Ms. Ivanova, his math teacher, gave him a new task. Furik solved the task immediately. Can you?
You are given a system of equations:
You should count, how many there are pairs of integers (*a*,<=*b*) (0<=≤<=*a*,<=*b*) which satisfy the system.
Input Specification:
A single line contains two integers *n*,<=*m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the parameters of the system. The numbers on the line are separated by a space.
Output Specification:
On a single line print the answer to the problem.
Demo Input:
['9 3\n', '14 28\n', '4 20\n']
Demo Output:
['1\n', '1\n', '0\n']
Note:
In the first sample the suitable pair is integers (3, 0). In the second sample the suitable pair is integers (3, 5). In the third sample there is no suitable pair.
|
```python
n,m=map(int,input().split())
c=n-m
iht=[]
yeni=[]
for i in range(1,c+1):
for t in range(1,c+1):
if i*t==c:
iht.append([i,t])
for i in range(len(iht)):
a=(iht[i][1]+iht[i][0]+1)/2
b=a-iht[i][0]
if a**2 + b == n and b**2 + a==m :
yeni.append([a,b])
print(len(yeni))
```
| 0
|
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