message stringlengths 2 48.6k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 318 108k | cluster float64 8 8 | __index_level_0__ int64 636 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Luckily, Serval got onto the right bus, and he came to the kindergarten on time. After coming to kindergarten, he found the toy bricks very funny.
He has a special interest to create difficult problems for others to solve. This time, with many 1 ร 1 ร 1 toy bricks, he builds up a 3-dimensional object. We can describe this object with a n ร m matrix, such that in each cell (i,j), there are h_{i,j} bricks standing on the top of each other.
However, Serval doesn't give you any h_{i,j}, and just give you the front view, left view, and the top view of this object, and he is now asking you to restore the object. Note that in the front view, there are m columns, and in the i-th of them, the height is the maximum of h_{1,i},h_{2,i},...,h_{n,i}. It is similar for the left view, where there are n columns. And in the top view, there is an n ร m matrix t_{i,j}, where t_{i,j} is 0 or 1. If t_{i,j} equals 1, that means h_{i,j}>0, otherwise, h_{i,j}=0.
However, Serval is very lonely because others are bored about his unsolvable problems before, and refused to solve this one, although this time he promises there will be at least one object satisfying all the views. As his best friend, can you have a try?
Input
The first line contains three positive space-separated integers n, m, h (1โค n, m, h โค 100) โ the length, width and height.
The second line contains m non-negative space-separated integers a_1,a_2,...,a_m, where a_i is the height in the i-th column from left to right of the front view (0โค a_i โค h).
The third line contains n non-negative space-separated integers b_1,b_2,...,b_n (0โค b_j โค h), where b_j is the height in the j-th column from left to right of the left view.
Each of the following n lines contains m numbers, each is 0 or 1, representing the top view, where j-th number of i-th row is 1 if h_{i, j}>0, and 0 otherwise.
It is guaranteed that there is at least one structure satisfying the input.
Output
Output n lines, each of them contains m integers, the j-th number in the i-th line should be equal to the height in the corresponding position of the top view. If there are several objects satisfying the views, output any one of them.
Examples
Input
3 7 3
2 3 0 0 2 0 1
2 1 3
1 0 0 0 1 0 0
0 0 0 0 0 0 1
1 1 0 0 0 0 0
Output
1 0 0 0 2 0 0
0 0 0 0 0 0 1
2 3 0 0 0 0 0
Input
4 5 5
3 5 2 0 4
4 2 5 4
0 0 0 0 1
1 0 1 0 0
0 1 0 0 0
1 1 1 0 0
Output
0 0 0 0 4
1 0 2 0 0
0 5 0 0 0
3 4 1 0 0
Note
<image>
The graph above illustrates the object in the first example.
<image> <image>
The first graph illustrates the object in the example output for the second example, and the second graph shows the three-view drawing of it.
Submitted Solution:
```
def mp():
return map(int, input().split())
n, m, h = mp()
a = list(mp())
b = list(mp())
f = [list(mp()) for i in range(n)]
for i in range(m):
for j in range(n):
f[j][i] *= a[i]
for i in range(n):
for j in range(m):
print(f[i][j], end = ' ')
print()
``` | instruction | 0 | 66,509 | 8 | 133,018 |
No | output | 1 | 66,509 | 8 | 133,019 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Luckily, Serval got onto the right bus, and he came to the kindergarten on time. After coming to kindergarten, he found the toy bricks very funny.
He has a special interest to create difficult problems for others to solve. This time, with many 1 ร 1 ร 1 toy bricks, he builds up a 3-dimensional object. We can describe this object with a n ร m matrix, such that in each cell (i,j), there are h_{i,j} bricks standing on the top of each other.
However, Serval doesn't give you any h_{i,j}, and just give you the front view, left view, and the top view of this object, and he is now asking you to restore the object. Note that in the front view, there are m columns, and in the i-th of them, the height is the maximum of h_{1,i},h_{2,i},...,h_{n,i}. It is similar for the left view, where there are n columns. And in the top view, there is an n ร m matrix t_{i,j}, where t_{i,j} is 0 or 1. If t_{i,j} equals 1, that means h_{i,j}>0, otherwise, h_{i,j}=0.
However, Serval is very lonely because others are bored about his unsolvable problems before, and refused to solve this one, although this time he promises there will be at least one object satisfying all the views. As his best friend, can you have a try?
Input
The first line contains three positive space-separated integers n, m, h (1โค n, m, h โค 100) โ the length, width and height.
The second line contains m non-negative space-separated integers a_1,a_2,...,a_m, where a_i is the height in the i-th column from left to right of the front view (0โค a_i โค h).
The third line contains n non-negative space-separated integers b_1,b_2,...,b_n (0โค b_j โค h), where b_j is the height in the j-th column from left to right of the left view.
Each of the following n lines contains m numbers, each is 0 or 1, representing the top view, where j-th number of i-th row is 1 if h_{i, j}>0, and 0 otherwise.
It is guaranteed that there is at least one structure satisfying the input.
Output
Output n lines, each of them contains m integers, the j-th number in the i-th line should be equal to the height in the corresponding position of the top view. If there are several objects satisfying the views, output any one of them.
Examples
Input
3 7 3
2 3 0 0 2 0 1
2 1 3
1 0 0 0 1 0 0
0 0 0 0 0 0 1
1 1 0 0 0 0 0
Output
1 0 0 0 2 0 0
0 0 0 0 0 0 1
2 3 0 0 0 0 0
Input
4 5 5
3 5 2 0 4
4 2 5 4
0 0 0 0 1
1 0 1 0 0
0 1 0 0 0
1 1 1 0 0
Output
0 0 0 0 4
1 0 2 0 0
0 5 0 0 0
3 4 1 0 0
Note
<image>
The graph above illustrates the object in the first example.
<image> <image>
The first graph illustrates the object in the example output for the second example, and the second graph shows the three-view drawing of it.
Submitted Solution:
```
"Pranay Malhan"
from sys import stdin,exit
l,w,h=map(int,stdin.readline().split())
f=list(map(int,stdin.readline().split()))
le=list(map(int,stdin.readline().split()))
t=[]
for i in range(l):
h=list(map(int,stdin.readline().split()))
t.append(h)
for i in range(l):
for j in range(len(t[i])):
if t[i][j]==1:
if f[j]!=0 and f[j]<=le[i]:
t[i][j]=f[j]
f[j]=0
for i in range(l):
flag=1
for j in range(len(t[i])):
if t[i][j]==le[i]:
flag=0
break
if flag==1:
d=10**6
poi=0
for j in range(len(t[i])):
if t[i][j]<d and t[i][j]!=0:
d=t[i][j]
poi=j
t[i][poi]=le[i]
for i in range(l):
for j in range(len(t[i])):
print(str(t[i][j])+" ",end="")
print()
``` | instruction | 0 | 66,510 | 8 | 133,020 |
No | output | 1 | 66,510 | 8 | 133,021 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Luckily, Serval got onto the right bus, and he came to the kindergarten on time. After coming to kindergarten, he found the toy bricks very funny.
He has a special interest to create difficult problems for others to solve. This time, with many 1 ร 1 ร 1 toy bricks, he builds up a 3-dimensional object. We can describe this object with a n ร m matrix, such that in each cell (i,j), there are h_{i,j} bricks standing on the top of each other.
However, Serval doesn't give you any h_{i,j}, and just give you the front view, left view, and the top view of this object, and he is now asking you to restore the object. Note that in the front view, there are m columns, and in the i-th of them, the height is the maximum of h_{1,i},h_{2,i},...,h_{n,i}. It is similar for the left view, where there are n columns. And in the top view, there is an n ร m matrix t_{i,j}, where t_{i,j} is 0 or 1. If t_{i,j} equals 1, that means h_{i,j}>0, otherwise, h_{i,j}=0.
However, Serval is very lonely because others are bored about his unsolvable problems before, and refused to solve this one, although this time he promises there will be at least one object satisfying all the views. As his best friend, can you have a try?
Input
The first line contains three positive space-separated integers n, m, h (1โค n, m, h โค 100) โ the length, width and height.
The second line contains m non-negative space-separated integers a_1,a_2,...,a_m, where a_i is the height in the i-th column from left to right of the front view (0โค a_i โค h).
The third line contains n non-negative space-separated integers b_1,b_2,...,b_n (0โค b_j โค h), where b_j is the height in the j-th column from left to right of the left view.
Each of the following n lines contains m numbers, each is 0 or 1, representing the top view, where j-th number of i-th row is 1 if h_{i, j}>0, and 0 otherwise.
It is guaranteed that there is at least one structure satisfying the input.
Output
Output n lines, each of them contains m integers, the j-th number in the i-th line should be equal to the height in the corresponding position of the top view. If there are several objects satisfying the views, output any one of them.
Examples
Input
3 7 3
2 3 0 0 2 0 1
2 1 3
1 0 0 0 1 0 0
0 0 0 0 0 0 1
1 1 0 0 0 0 0
Output
1 0 0 0 2 0 0
0 0 0 0 0 0 1
2 3 0 0 0 0 0
Input
4 5 5
3 5 2 0 4
4 2 5 4
0 0 0 0 1
1 0 1 0 0
0 1 0 0 0
1 1 1 0 0
Output
0 0 0 0 4
1 0 2 0 0
0 5 0 0 0
3 4 1 0 0
Note
<image>
The graph above illustrates the object in the first example.
<image> <image>
The first graph illustrates the object in the example output for the second example, and the second graph shows the three-view drawing of it.
Submitted Solution:
```
import copy
n,m,h = input().split()
n = int(n)
m = int(m)
h = int(h)
M = []
N = []
H = []
M = input().split()
for i in range(0,m):
M[i] = int(M[i])
N = input().split()
for i in range(0,n):
N[i] = int(N[i])
for i in range(0,n):
l = input().split()
for j in range(0,len(l)):
l[j] = int(l[j])
H.append(copy.deepcopy(l))
L = []
for i in range(0,n):
l = []
L.append(copy.deepcopy(l))
for j in range(0,m):
L[i].append(0)
for i in range(0,n):
for j in range(0,m):
if H[i][j] == 1:
if N[i] > M[j]:
L[i][j] = M[j]
else:
L[i][j] = N[i]
for i in L:
print(i)
``` | instruction | 0 | 66,511 | 8 | 133,022 |
No | output | 1 | 66,511 | 8 | 133,023 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Luckily, Serval got onto the right bus, and he came to the kindergarten on time. After coming to kindergarten, he found the toy bricks very funny.
He has a special interest to create difficult problems for others to solve. This time, with many 1 ร 1 ร 1 toy bricks, he builds up a 3-dimensional object. We can describe this object with a n ร m matrix, such that in each cell (i,j), there are h_{i,j} bricks standing on the top of each other.
However, Serval doesn't give you any h_{i,j}, and just give you the front view, left view, and the top view of this object, and he is now asking you to restore the object. Note that in the front view, there are m columns, and in the i-th of them, the height is the maximum of h_{1,i},h_{2,i},...,h_{n,i}. It is similar for the left view, where there are n columns. And in the top view, there is an n ร m matrix t_{i,j}, where t_{i,j} is 0 or 1. If t_{i,j} equals 1, that means h_{i,j}>0, otherwise, h_{i,j}=0.
However, Serval is very lonely because others are bored about his unsolvable problems before, and refused to solve this one, although this time he promises there will be at least one object satisfying all the views. As his best friend, can you have a try?
Input
The first line contains three positive space-separated integers n, m, h (1โค n, m, h โค 100) โ the length, width and height.
The second line contains m non-negative space-separated integers a_1,a_2,...,a_m, where a_i is the height in the i-th column from left to right of the front view (0โค a_i โค h).
The third line contains n non-negative space-separated integers b_1,b_2,...,b_n (0โค b_j โค h), where b_j is the height in the j-th column from left to right of the left view.
Each of the following n lines contains m numbers, each is 0 or 1, representing the top view, where j-th number of i-th row is 1 if h_{i, j}>0, and 0 otherwise.
It is guaranteed that there is at least one structure satisfying the input.
Output
Output n lines, each of them contains m integers, the j-th number in the i-th line should be equal to the height in the corresponding position of the top view. If there are several objects satisfying the views, output any one of them.
Examples
Input
3 7 3
2 3 0 0 2 0 1
2 1 3
1 0 0 0 1 0 0
0 0 0 0 0 0 1
1 1 0 0 0 0 0
Output
1 0 0 0 2 0 0
0 0 0 0 0 0 1
2 3 0 0 0 0 0
Input
4 5 5
3 5 2 0 4
4 2 5 4
0 0 0 0 1
1 0 1 0 0
0 1 0 0 0
1 1 1 0 0
Output
0 0 0 0 4
1 0 2 0 0
0 5 0 0 0
3 4 1 0 0
Note
<image>
The graph above illustrates the object in the first example.
<image> <image>
The first graph illustrates the object in the example output for the second example, and the second graph shows the three-view drawing of it.
Submitted Solution:
```
from math import ceil
import sys
input=sys.stdin.readline
from collections import defaultdict as dd
#n=int(input())
n,m,h=map(int,input().split())
front=list(map(int,input().split()))
left=list(map(int,input().split()))
ll=[]
for i in range(n):
l=list(map(int,input().split()))
ll.append(l)
for i in range(m):
for j in range(n-1,-1,-1):
if(ll[j][i]):
ll[j][i]=front[i]
break
for i in range(n):
for j in range(m-1,-1,-1):
if(ll[i][j]):
ll[i][j]=left[i]
break
for i in ll:
print(*i)
``` | instruction | 0 | 66,512 | 8 | 133,024 |
No | output | 1 | 66,512 | 8 | 133,025 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Squirrel Liss loves nuts. There are n trees (numbered 1 to n from west to east) along a street and there is a delicious nut on the top of each tree. The height of the tree i is hi. Liss wants to eat all nuts.
Now Liss is on the root of the tree with the number 1. In one second Liss can perform one of the following actions:
* Walk up or down one unit on a tree.
* Eat a nut on the top of the current tree.
* Jump to the next tree. In this action the height of Liss doesn't change. More formally, when Liss is at height h of the tree i (1 โค i โค n - 1), she jumps to height h of the tree i + 1. This action can't be performed if h > hi + 1.
Compute the minimal time (in seconds) required to eat all nuts.
Input
The first line contains an integer n (1 โค n โค 105) โ the number of trees.
Next n lines contains the height of trees: i-th line contains an integer hi (1 โค hi โค 104) โ the height of the tree with the number i.
Output
Print a single integer โ the minimal time required to eat all nuts in seconds.
Examples
Input
2
1
2
Output
5
Input
5
2
1
2
1
1
Output
14 | instruction | 0 | 66,705 | 8 | 133,410 |
Tags: greedy, implementation
Correct Solution:
```
def main():
n = int(input())
cur_h = 0
res = 0
for _ in range(n):
h = int(input())
res += abs(cur_h - h)
cur_h = h
print(res + n * 2 - 1)
if __name__ == '__main__':
main()
``` | output | 1 | 66,705 | 8 | 133,411 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Summer is coming! It's time for Iahub and Iahubina to work out, as they both want to look hot at the beach. The gym where they go is a matrix a with n lines and m columns. Let number a[i][j] represents the calories burned by performing workout at the cell of gym in the i-th line and the j-th column.
Iahub starts with workout located at line 1 and column 1. He needs to finish with workout a[n][m]. After finishing workout a[i][j], he can go to workout a[i + 1][j] or a[i][j + 1]. Similarly, Iahubina starts with workout a[n][1] and she needs to finish with workout a[1][m]. After finishing workout from cell a[i][j], she goes to either a[i][j + 1] or a[i - 1][j].
There is one additional condition for their training. They have to meet in exactly one cell of gym. At that cell, none of them will work out. They will talk about fast exponentiation (pretty odd small talk) and then both of them will move to the next workout.
If a workout was done by either Iahub or Iahubina, it counts as total gain. Please plan a workout for Iahub and Iahubina such as total gain to be as big as possible. Note, that Iahub and Iahubina can perform workouts with different speed, so the number of cells that they use to reach meet cell may differs.
Input
The first line of the input contains two integers n and m (3 โค n, m โค 1000). Each of the next n lines contains m integers: j-th number from i-th line denotes element a[i][j] (0 โค a[i][j] โค 105).
Output
The output contains a single number โ the maximum total gain possible.
Examples
Input
3 3
100 100 100
100 1 100
100 100 100
Output
800
Note
Iahub will choose exercises a[1][1] โ a[1][2] โ a[2][2] โ a[3][2] โ a[3][3]. Iahubina will choose exercises a[3][1] โ a[2][1] โ a[2][2] โ a[2][3] โ a[1][3]. | instruction | 0 | 66,739 | 8 | 133,478 |
Tags: dp
Correct Solution:
```
import sys
from math import *
def minp():
return sys.stdin.readline().strip()
def mint():
return int(minp())
def mints():
return map(int, minp().split())
n, m = mints()
a = [0]*n
dp1 = [0]*n
dp2 = [0]*n
dp3 = [0]*n
dp4 = [0]*n
for i in range(n):
a[i] = list(mints())
dp1[i] = [0]*m
dp2[i] = [0]*m
dp3[i] = [0]*m
dp4[i] = [0]*m
dp1[0][0] = a[0][0]
dp2[n-1][m-1] = a[n-1][m-1]
dp3[n-1][0] = a[n-1][0]
dp4[0][m-1] = a[0][m-1]
'''
for i in range(1,n):
dp1[i][0] = dp1[i-1][0] + a[i][0] # >>>>
for i in range(n-2,-1,-1):
dp2[i][m-1] = dp2[i+1][m-1] + a[i][m-1] # <<<<
for i in range(n-2,-1,-1):
dp3[i][0] = dp3[i+1][0] + a[i][0] # <<<<
for i in range(1,n):
dp4[i][m-1] = dp4[i-1][m-1] + a[i][m-1] # >>>>
for i in range(1,m):
dp1[0][i] = dp1[0][i-1] + a[0][i] # >>>>
for i in range(m-2,-1,-1):
dp2[n-1][i] = dp2[n-1][i+1] + a[n-1][i] # <<<<
for i in range(1,m):
dp3[n-1][i] = dp3[n-1][i-1] + a[n-1][i] # >>>>
for i in range(m-2,-1,-1):
dp4[0][i] = dp4[0][i+1] + a[0][i] # >>>>
'''
for i in range(0,n):
for j in range(0,m):
z = 0
if i-1 >= 0:
z = dp1[i-1][j]
if j-1 >= 0:
z = max(z, dp1[i][j-1])
dp1[i][j] = z + a[i][j]
#dp1[i][j] = max(dp1[i-1][j], dp1[i][j-1]) + a[i][j]
for i in range(n-1,-1,-1):
for j in range(m-1,-1,-1):
z = 0
if i+1 < n:
z = dp2[i+1][j]
if j+1 < m:
z = max(z, dp2[i][j+1])
dp2[i][j] = z + a[i][j]
#dp2[i][j] = max(dp2[i+1][j], dp2[i][j+1]) + a[i][j]
for i in range(n-1,-1,-1):
for j in range(0,m):
z = 0
if i+1 < n:
z = dp3[i+1][j]
if j-1 >= 0:
z = max(z, dp3[i][j-1])
dp3[i][j] = z + a[i][j]
#dp3[i][j] = max(dp3[i+1][j], dp3[i][j-1]) + a[i][j]
for i in range(0,n):
for j in range(m-1,-1,-1):
z = 0
if i-1 >= 0:
z = dp4[i-1][j]
if j+1 < m:
z = max(z, dp4[i][j+1])
dp4[i][j] = z + a[i][j]
#dp4[i][j] = max(dp4[i-1][j], dp4[i][j+1]) + a[i][j]
'''for i in dp1:
print(i)
print()
for i in dp2:
print(i)
print()
for i in dp3:
print(i)
print()
for i in dp4:
print(i)
print()
'''
r = 0
for i in range(1,n-1):
for j in range(1,m-1):
r = max(r, dp1[i-1][j] + dp2[i+1][j] + dp3[i][j-1] + dp4[i][j+1])
r = max(r, dp1[i][j-1] + dp2[i][j+1] + dp3[i+1][j] + dp4[i-1][j])
#print(r, i, j)
print(r)
``` | output | 1 | 66,739 | 8 | 133,479 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Summer is coming! It's time for Iahub and Iahubina to work out, as they both want to look hot at the beach. The gym where they go is a matrix a with n lines and m columns. Let number a[i][j] represents the calories burned by performing workout at the cell of gym in the i-th line and the j-th column.
Iahub starts with workout located at line 1 and column 1. He needs to finish with workout a[n][m]. After finishing workout a[i][j], he can go to workout a[i + 1][j] or a[i][j + 1]. Similarly, Iahubina starts with workout a[n][1] and she needs to finish with workout a[1][m]. After finishing workout from cell a[i][j], she goes to either a[i][j + 1] or a[i - 1][j].
There is one additional condition for their training. They have to meet in exactly one cell of gym. At that cell, none of them will work out. They will talk about fast exponentiation (pretty odd small talk) and then both of them will move to the next workout.
If a workout was done by either Iahub or Iahubina, it counts as total gain. Please plan a workout for Iahub and Iahubina such as total gain to be as big as possible. Note, that Iahub and Iahubina can perform workouts with different speed, so the number of cells that they use to reach meet cell may differs.
Input
The first line of the input contains two integers n and m (3 โค n, m โค 1000). Each of the next n lines contains m integers: j-th number from i-th line denotes element a[i][j] (0 โค a[i][j] โค 105).
Output
The output contains a single number โ the maximum total gain possible.
Examples
Input
3 3
100 100 100
100 1 100
100 100 100
Output
800
Note
Iahub will choose exercises a[1][1] โ a[1][2] โ a[2][2] โ a[3][2] โ a[3][3]. Iahubina will choose exercises a[3][1] โ a[2][1] โ a[2][2] โ a[2][3] โ a[1][3]. | instruction | 0 | 66,740 | 8 | 133,480 |
Tags: dp
Correct Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO, IOBase
def main():
n,m = map(int,input().split())
arr = [list(map(int,input().split())) for _ in range(n)]
dp1 = [[0]*(m+2) for _ in range(n+2)]
for i in range(1,n+1):
for j in range(1,m+1):
dp1[i][j] = max(dp1[i-1][j],dp1[i][j-1])+arr[i-1][j-1]
dp2 = [[0]*(m+2) for _ in range(n+2)]
for i in range(1,n+1):
for j in range(m,0,-1):
dp2[i][j] = max(dp2[i-1][j],dp2[i][j+1])+arr[i-1][j-1]
dp3 = [[0]*(m+2) for _ in range(n+2)]
for i in range(n,0,-1):
for j in range(m,0,-1):
dp3[i][j] = max(dp3[i+1][j],dp3[i][j+1])+arr[i-1][j-1]
dp4 = [[0]*(m+2) for _ in range(n+2)]
for i in range(n,0,-1):
for j in range(1,m+1):
dp4[i][j] = max(dp4[i+1][j],dp4[i][j-1])+arr[i-1][j-1]
maxi = 0
for i in range(2,n):
for j in range(2,m):
maxi = max(maxi,
dp1[i][j-1]+dp4[i+1][j]+dp2[i-1][j]+dp3[i][j+1],
dp1[i-1][j]+dp4[i][j-1]+dp2[i][j+1]+dp3[i+1][j]
)
print(maxi)
#Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == '__main__':
main()
``` | output | 1 | 66,740 | 8 | 133,481 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Summer is coming! It's time for Iahub and Iahubina to work out, as they both want to look hot at the beach. The gym where they go is a matrix a with n lines and m columns. Let number a[i][j] represents the calories burned by performing workout at the cell of gym in the i-th line and the j-th column.
Iahub starts with workout located at line 1 and column 1. He needs to finish with workout a[n][m]. After finishing workout a[i][j], he can go to workout a[i + 1][j] or a[i][j + 1]. Similarly, Iahubina starts with workout a[n][1] and she needs to finish with workout a[1][m]. After finishing workout from cell a[i][j], she goes to either a[i][j + 1] or a[i - 1][j].
There is one additional condition for their training. They have to meet in exactly one cell of gym. At that cell, none of them will work out. They will talk about fast exponentiation (pretty odd small talk) and then both of them will move to the next workout.
If a workout was done by either Iahub or Iahubina, it counts as total gain. Please plan a workout for Iahub and Iahubina such as total gain to be as big as possible. Note, that Iahub and Iahubina can perform workouts with different speed, so the number of cells that they use to reach meet cell may differs.
Input
The first line of the input contains two integers n and m (3 โค n, m โค 1000). Each of the next n lines contains m integers: j-th number from i-th line denotes element a[i][j] (0 โค a[i][j] โค 105).
Output
The output contains a single number โ the maximum total gain possible.
Examples
Input
3 3
100 100 100
100 1 100
100 100 100
Output
800
Note
Iahub will choose exercises a[1][1] โ a[1][2] โ a[2][2] โ a[3][2] โ a[3][3]. Iahubina will choose exercises a[3][1] โ a[2][1] โ a[2][2] โ a[2][3] โ a[1][3]. | instruction | 0 | 66,741 | 8 | 133,482 |
Tags: dp
Correct Solution:
```
def main():
n, m = map(int, input().split())
aa = []
for _ in range(n):
row = list(map(int, input().split()))
row.append(0)
aa.append(row)
aa.append([0] * (m + 1))
d1, d2, d3, d4 = ([[0] * (m + 1) for _ in range(n + 1)] for _ in (1, 2, 3, 4))
for i in range(n):
for j in range(m):
d1[i][j] = max(d1[i - 1][j], d1[i][j - 1]) + aa[i][j]
for i in range(n):
for j in range(m - 1, -1, -1):
d2[i][j] = max(d2[i - 1][j], d2[i][j + 1]) + aa[i][j]
for i in range(n - 1, -1, -1):
for j in range(m):
d3[i][j] = max(d3[i + 1][j], d3[i][j - 1]) + aa[i][j]
for i in range(n - 1, -1, -1):
for j in range(m - 1, -1, -1):
d4[i][j] = max(d4[i + 1][j], d4[i][j + 1]) + aa[i][j]
print(max(
max(d1[i][j - 1] + d2[i - 1][j] + d3[i + 1][j] + d4[i][j + 1] for i in range(1, n - 1) for j in range(1, m - 1)),
max(d1[i - 1][j] + d2[i][j + 1] + d3[i][j - 1] + d4[i + 1][j] for i in range(1, n - 1) for j in range(1, m - 1))))
if __name__ == '__main__':
main()
``` | output | 1 | 66,741 | 8 | 133,483 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Summer is coming! It's time for Iahub and Iahubina to work out, as they both want to look hot at the beach. The gym where they go is a matrix a with n lines and m columns. Let number a[i][j] represents the calories burned by performing workout at the cell of gym in the i-th line and the j-th column.
Iahub starts with workout located at line 1 and column 1. He needs to finish with workout a[n][m]. After finishing workout a[i][j], he can go to workout a[i + 1][j] or a[i][j + 1]. Similarly, Iahubina starts with workout a[n][1] and she needs to finish with workout a[1][m]. After finishing workout from cell a[i][j], she goes to either a[i][j + 1] or a[i - 1][j].
There is one additional condition for their training. They have to meet in exactly one cell of gym. At that cell, none of them will work out. They will talk about fast exponentiation (pretty odd small talk) and then both of them will move to the next workout.
If a workout was done by either Iahub or Iahubina, it counts as total gain. Please plan a workout for Iahub and Iahubina such as total gain to be as big as possible. Note, that Iahub and Iahubina can perform workouts with different speed, so the number of cells that they use to reach meet cell may differs.
Input
The first line of the input contains two integers n and m (3 โค n, m โค 1000). Each of the next n lines contains m integers: j-th number from i-th line denotes element a[i][j] (0 โค a[i][j] โค 105).
Output
The output contains a single number โ the maximum total gain possible.
Examples
Input
3 3
100 100 100
100 1 100
100 100 100
Output
800
Note
Iahub will choose exercises a[1][1] โ a[1][2] โ a[2][2] โ a[3][2] โ a[3][3]. Iahubina will choose exercises a[3][1] โ a[2][1] โ a[2][2] โ a[2][3] โ a[1][3]. | instruction | 0 | 66,742 | 8 | 133,484 |
Tags: dp
Correct Solution:
```
n, m = map(int, input().strip().split())
dp1, dp2, dp3, dp4 = [[[0 for i in range(m+1)] for i in range(n+1)] for i in range(4)]
a = []
for i in range(n):
a.append(list(map(int, input().strip().split())))
for i in range(n):
for j in range(m):
dp1[i][j] = a[i][j] + max(dp1[i-1][j], dp1[i][j-1])
for i in range(n-1, -1, -1):
for j in range(m-1, -1, -1):
dp2[i][j] = a[i][j] + max(dp2[i+1][j], dp2[i][j+1])
for i in range(n-1, -1, -1):
for j in range(m):
dp3[i][j] = a[i][j] + max(dp3[i+1][j], dp3[i][j-1])
for i in range(n):
for j in range(m-1, -1, -1):
dp4[i][j] = a[i][j] + max(dp4[i-1][j], dp4[i][j+1])
ans = 0
for i in range(1,n-1):
for j in range(1, m-1):
ans = max(ans, dp1[i][j-1] + dp2[i][j+1] + dp3[i+1][j] + dp4[i-1][j], dp3[i][j-1] + dp4[i][j+1] + dp1[i-1][j] + dp2[i+1][j])
print(ans)
``` | output | 1 | 66,742 | 8 | 133,485 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Summer is coming! It's time for Iahub and Iahubina to work out, as they both want to look hot at the beach. The gym where they go is a matrix a with n lines and m columns. Let number a[i][j] represents the calories burned by performing workout at the cell of gym in the i-th line and the j-th column.
Iahub starts with workout located at line 1 and column 1. He needs to finish with workout a[n][m]. After finishing workout a[i][j], he can go to workout a[i + 1][j] or a[i][j + 1]. Similarly, Iahubina starts with workout a[n][1] and she needs to finish with workout a[1][m]. After finishing workout from cell a[i][j], she goes to either a[i][j + 1] or a[i - 1][j].
There is one additional condition for their training. They have to meet in exactly one cell of gym. At that cell, none of them will work out. They will talk about fast exponentiation (pretty odd small talk) and then both of them will move to the next workout.
If a workout was done by either Iahub or Iahubina, it counts as total gain. Please plan a workout for Iahub and Iahubina such as total gain to be as big as possible. Note, that Iahub and Iahubina can perform workouts with different speed, so the number of cells that they use to reach meet cell may differs.
Input
The first line of the input contains two integers n and m (3 โค n, m โค 1000). Each of the next n lines contains m integers: j-th number from i-th line denotes element a[i][j] (0 โค a[i][j] โค 105).
Output
The output contains a single number โ the maximum total gain possible.
Examples
Input
3 3
100 100 100
100 1 100
100 100 100
Output
800
Note
Iahub will choose exercises a[1][1] โ a[1][2] โ a[2][2] โ a[3][2] โ a[3][3]. Iahubina will choose exercises a[3][1] โ a[2][1] โ a[2][2] โ a[2][3] โ a[1][3]. | instruction | 0 | 66,743 | 8 | 133,486 |
Tags: dp
Correct Solution:
```
n, m = map(int, input().strip().split())
dp1, dp2, dp3, dp4 = [[[0 for i in range(m+1)] for i in range(n+1)] for i in range(4)]
# print(dp1)
# print(dp2)
# print(dp3)
# print(dp4)
a = []
for i in range(n):
a.append(list(map(int, input().strip().split())))
for i in range(n):
for j in range(m):
dp1[i][j] = a[i][j] + max(dp1[i-1][j], dp1[i][j-1])
for i in range(n-1, -1, -1):
for j in range(m-1, -1, -1):
dp2[i][j] = a[i][j] + max(dp2[i+1][j], dp2[i][j+1])
for i in range(n-1, -1, -1):
for j in range(m):
dp3[i][j] = a[i][j] + max(dp3[i+1][j], dp3[i][j-1])
for i in range(n):
for j in range(m-1, -1, -1):
dp4[i][j] = a[i][j] + max(dp4[i-1][j], dp4[i][j+1])
# print("#############")
# for i in dp1:
# print(i)
# print("-----------")
# for i in dp2:
# print(i)
# print("-----------")
# for i in dp3:
# print(i)
# print("-----------")
# for i in dp4:
# print(i)
# print("#############")
ans = 0
for i in range(1,n-1):
for j in range(1, m-1):
ans = max(ans, dp1[i][j-1] + dp2[i][j+1] + dp3[i+1][j] + dp4[i-1][j], dp3[i][j-1] + dp4[i][j+1] + dp1[i-1][j] + dp2[i+1][j])
# print(dp1[i][j-1],dp2[i][j+1], dp3[i+1][j], dp4[i-1][j], dp3[i][j-1], dp4[i][j+1], dp1[i+1][j], dp2[i-1][j])
print(ans)
``` | output | 1 | 66,743 | 8 | 133,487 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Summer is coming! It's time for Iahub and Iahubina to work out, as they both want to look hot at the beach. The gym where they go is a matrix a with n lines and m columns. Let number a[i][j] represents the calories burned by performing workout at the cell of gym in the i-th line and the j-th column.
Iahub starts with workout located at line 1 and column 1. He needs to finish with workout a[n][m]. After finishing workout a[i][j], he can go to workout a[i + 1][j] or a[i][j + 1]. Similarly, Iahubina starts with workout a[n][1] and she needs to finish with workout a[1][m]. After finishing workout from cell a[i][j], she goes to either a[i][j + 1] or a[i - 1][j].
There is one additional condition for their training. They have to meet in exactly one cell of gym. At that cell, none of them will work out. They will talk about fast exponentiation (pretty odd small talk) and then both of them will move to the next workout.
If a workout was done by either Iahub or Iahubina, it counts as total gain. Please plan a workout for Iahub and Iahubina such as total gain to be as big as possible. Note, that Iahub and Iahubina can perform workouts with different speed, so the number of cells that they use to reach meet cell may differs.
Input
The first line of the input contains two integers n and m (3 โค n, m โค 1000). Each of the next n lines contains m integers: j-th number from i-th line denotes element a[i][j] (0 โค a[i][j] โค 105).
Output
The output contains a single number โ the maximum total gain possible.
Examples
Input
3 3
100 100 100
100 1 100
100 100 100
Output
800
Note
Iahub will choose exercises a[1][1] โ a[1][2] โ a[2][2] โ a[3][2] โ a[3][3]. Iahubina will choose exercises a[3][1] โ a[2][1] โ a[2][2] โ a[2][3] โ a[1][3]. | instruction | 0 | 66,744 | 8 | 133,488 |
Tags: dp
Correct Solution:
```
import sys
input=sys.stdin.readline
R = lambda: map(int, input().split())
n, m = R()
g = [list() for i in range(n)]
for i in range(n):
g[i] = list(R())
dp1, dp2, dp3, dp4 = ([[0] * (m + 1) for j in range(n + 1)] for i in range(4))
for i in range(n):
for j in range(m):
dp1[i][j] = g[i][j] + max(dp1[i - 1][j], dp1[i][j - 1])
for j in range(m - 1, -1, -1):
dp2[i][j] = g[i][j] + max(dp2[i - 1][j], dp2[i][j + 1])
for i in range(n - 1, -1, -1):
for j in range(m):
dp3[i][j] = g[i][j] + max(dp3[i + 1][j], dp3[i][j - 1])
for j in range(m - 1, -1, -1):
dp4[i][j] = g[i][j] + max(dp4[i + 1][j], dp4[i][j + 1])
print(max(max(dp1[i][j - 1] + dp2[i - 1][j] + dp3[i + 1][j] + dp4[i][j + 1], dp1[i - 1][j] + dp2[i][j + 1] + dp3[i][j - 1] + dp4[i + 1][j]) for j in range(1, m - 1) for i in range(1, n - 1)))
``` | output | 1 | 66,744 | 8 | 133,489 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Summer is coming! It's time for Iahub and Iahubina to work out, as they both want to look hot at the beach. The gym where they go is a matrix a with n lines and m columns. Let number a[i][j] represents the calories burned by performing workout at the cell of gym in the i-th line and the j-th column.
Iahub starts with workout located at line 1 and column 1. He needs to finish with workout a[n][m]. After finishing workout a[i][j], he can go to workout a[i + 1][j] or a[i][j + 1]. Similarly, Iahubina starts with workout a[n][1] and she needs to finish with workout a[1][m]. After finishing workout from cell a[i][j], she goes to either a[i][j + 1] or a[i - 1][j].
There is one additional condition for their training. They have to meet in exactly one cell of gym. At that cell, none of them will work out. They will talk about fast exponentiation (pretty odd small talk) and then both of them will move to the next workout.
If a workout was done by either Iahub or Iahubina, it counts as total gain. Please plan a workout for Iahub and Iahubina such as total gain to be as big as possible. Note, that Iahub and Iahubina can perform workouts with different speed, so the number of cells that they use to reach meet cell may differs.
Input
The first line of the input contains two integers n and m (3 โค n, m โค 1000). Each of the next n lines contains m integers: j-th number from i-th line denotes element a[i][j] (0 โค a[i][j] โค 105).
Output
The output contains a single number โ the maximum total gain possible.
Examples
Input
3 3
100 100 100
100 1 100
100 100 100
Output
800
Note
Iahub will choose exercises a[1][1] โ a[1][2] โ a[2][2] โ a[3][2] โ a[3][3]. Iahubina will choose exercises a[3][1] โ a[2][1] โ a[2][2] โ a[2][3] โ a[1][3]. | instruction | 0 | 66,745 | 8 | 133,490 |
Tags: dp
Correct Solution:
```
from sys import stdin,stdout
import sys
from bisect import bisect_left,bisect_right
import heapq
sys.setrecursionlimit(2*(10**5))
# stdin = open("input.txt", "r");
# stdout = open("output.txt", "w");
n,m=stdin.readline().strip().split(' ')
n,m=int(n),int(m)
costarr=[]
for i in range(n):
costarr.append(list(map(int,stdin.readline().strip().split(' '))))
dp_tl_br=[[0 for i in range(m)] for j in range(n)]
dp_br_tl=[[0 for i in range(m)] for j in range(n)]
dp_bl_tr=[[0 for i in range(m)] for j in range(n)]
dp_tr_bl=[[0 for i in range(m)] for j in range(n)]
# TOP LEFT TO BOTTOM RIGHT COST
dp_tl_br[0][0]=costarr[0][0]
for i in range(1,m):
dp_tl_br[0][i]=dp_tl_br[0][i-1]+costarr[0][i]
for i in range(1,n):
dp_tl_br[i][0]=dp_tl_br[i-1][0]+costarr[i][0]
for i in range(1,n):
for j in range(1,m):
dp_tl_br[i][j]=max(dp_tl_br[i][j-1],dp_tl_br[i-1][j])+costarr[i][j]
# BOTTOM RIGHT TO TOP LEFT COST
dp_br_tl[n-1][m-1]=costarr[n-1][m-1]
for i in range(m-2,-1,-1):
dp_br_tl[n-1][i]=dp_br_tl[n-1][i+1]+costarr[n-1][i]
for i in range(n-2,-1,-1):
dp_br_tl[i][m-1]=dp_br_tl[i+1][m-1]+costarr[i][m-1]
for i in range(n-2,-1,-1):
for j in range(m-2,-1,-1):
dp_br_tl[i][j]=max(dp_br_tl[i][j+1],dp_br_tl[i+1][j])+costarr[i][j]
# BOTTOM LEFT TO TOP RIGHT COST
dp_bl_tr[n-1][0]=costarr[n-1][0]
for i in range(1,m):
dp_bl_tr[n-1][i]=dp_bl_tr[n-1][i-1]+costarr[n-1][i]
for i in range(n-2,-1,-1):
dp_bl_tr[i][0]=dp_bl_tr[i+1][0]+costarr[i][0]
for i in range(n-2,-1,-1):
for j in range(1,m):
dp_bl_tr[i][j]=max(dp_bl_tr[i][j-1],dp_bl_tr[i+1][j])+costarr[i][j]
# TOP RIGHT TO BOTTOM LEFT COST
dp_tr_bl[0][m-1]=costarr[0][m-1]
for i in range(m-2,-1,-1):
dp_tr_bl[0][i]=dp_tr_bl[0][i+1]+costarr[0][i]
for i in range(1,n):
dp_tr_bl[i][m-1]=dp_tr_bl[i-1][m-1]+costarr[i][m-1]
for i in range(1,n):
for j in range(m-2,-1,-1):
dp_tr_bl[i][j]=max(dp_tr_bl[i][j+1],dp_tr_bl[i-1][j])+costarr[i][j]
def sh(arr):
for i in arr:
print(i)
# sh(dp_tr_bl)
# print()
# sh(dp_tl_br)
# print()
# sh(dp_bl_tr)
# print()
# sh(dp_br_tl)
# print()
ans=0
for i in range(1,n-1):
for j in range(1,m-1):
ans=max(ans,dp_bl_tr[i][j-1]+dp_tr_bl[i][j+1]+dp_tl_br[i-1][j]+dp_br_tl[i+1][j])# LEFT TO RIGHT | DOWN TO UP
ans=max(ans,dp_bl_tr[i+1][j]+dp_tr_bl[i-1][j]+dp_tl_br[i][j-1]+dp_br_tl[i][j+1])# DOWN TO UP | LEFT TO RIGHT
# for i in range(1,n-2):
# for j in range(m):
# ans+=max(ans,dp_tl_br[i-1][j]+costarr[i][j]+dp_br_tl[i+1][j]+dp_bl_tr[i+1][j]+costarr[i][j]+dp_tr_bl[i-1][j])
stdout.write(str(ans)+"\n")
``` | output | 1 | 66,745 | 8 | 133,491 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Summer is coming! It's time for Iahub and Iahubina to work out, as they both want to look hot at the beach. The gym where they go is a matrix a with n lines and m columns. Let number a[i][j] represents the calories burned by performing workout at the cell of gym in the i-th line and the j-th column.
Iahub starts with workout located at line 1 and column 1. He needs to finish with workout a[n][m]. After finishing workout a[i][j], he can go to workout a[i + 1][j] or a[i][j + 1]. Similarly, Iahubina starts with workout a[n][1] and she needs to finish with workout a[1][m]. After finishing workout from cell a[i][j], she goes to either a[i][j + 1] or a[i - 1][j].
There is one additional condition for their training. They have to meet in exactly one cell of gym. At that cell, none of them will work out. They will talk about fast exponentiation (pretty odd small talk) and then both of them will move to the next workout.
If a workout was done by either Iahub or Iahubina, it counts as total gain. Please plan a workout for Iahub and Iahubina such as total gain to be as big as possible. Note, that Iahub and Iahubina can perform workouts with different speed, so the number of cells that they use to reach meet cell may differs.
Input
The first line of the input contains two integers n and m (3 โค n, m โค 1000). Each of the next n lines contains m integers: j-th number from i-th line denotes element a[i][j] (0 โค a[i][j] โค 105).
Output
The output contains a single number โ the maximum total gain possible.
Examples
Input
3 3
100 100 100
100 1 100
100 100 100
Output
800
Note
Iahub will choose exercises a[1][1] โ a[1][2] โ a[2][2] โ a[3][2] โ a[3][3]. Iahubina will choose exercises a[3][1] โ a[2][1] โ a[2][2] โ a[2][3] โ a[1][3]. | instruction | 0 | 66,746 | 8 | 133,492 |
Tags: dp
Correct Solution:
```
R = lambda: map(int, input().split())
n, m = R()
g = [list() for i in range(n)]
for i in range(n):
g[i] = list(R())
dp1, dp2, dp3, dp4 = ([[0] * (m + 1) for j in range(n + 1)] for i in range(4))
for i in range(n):
for j in range(m):
dp1[i][j] = g[i][j] + max(dp1[i - 1][j], dp1[i][j - 1])
for j in range(m - 1, -1, -1):
dp2[i][j] = g[i][j] + max(dp2[i - 1][j], dp2[i][j + 1])
for i in range(n - 1, -1, -1):
for j in range(m):
dp3[i][j] = g[i][j] + max(dp3[i + 1][j], dp3[i][j - 1])
for j in range(m - 1, -1, -1):
dp4[i][j] = g[i][j] + max(dp4[i + 1][j], dp4[i][j + 1])
print(max(max(dp1[i][j - 1] + dp2[i - 1][j] + dp3[i + 1][j] + dp4[i][j + 1], dp1[i - 1][j] + dp2[i][j + 1] + dp3[i][j - 1] + dp4[i + 1][j]) for j in range(1, m - 1) for i in range(1, n - 1)))
``` | output | 1 | 66,746 | 8 | 133,493 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Summer is coming! It's time for Iahub and Iahubina to work out, as they both want to look hot at the beach. The gym where they go is a matrix a with n lines and m columns. Let number a[i][j] represents the calories burned by performing workout at the cell of gym in the i-th line and the j-th column.
Iahub starts with workout located at line 1 and column 1. He needs to finish with workout a[n][m]. After finishing workout a[i][j], he can go to workout a[i + 1][j] or a[i][j + 1]. Similarly, Iahubina starts with workout a[n][1] and she needs to finish with workout a[1][m]. After finishing workout from cell a[i][j], she goes to either a[i][j + 1] or a[i - 1][j].
There is one additional condition for their training. They have to meet in exactly one cell of gym. At that cell, none of them will work out. They will talk about fast exponentiation (pretty odd small talk) and then both of them will move to the next workout.
If a workout was done by either Iahub or Iahubina, it counts as total gain. Please plan a workout for Iahub and Iahubina such as total gain to be as big as possible. Note, that Iahub and Iahubina can perform workouts with different speed, so the number of cells that they use to reach meet cell may differs.
Input
The first line of the input contains two integers n and m (3 โค n, m โค 1000). Each of the next n lines contains m integers: j-th number from i-th line denotes element a[i][j] (0 โค a[i][j] โค 105).
Output
The output contains a single number โ the maximum total gain possible.
Examples
Input
3 3
100 100 100
100 1 100
100 100 100
Output
800
Note
Iahub will choose exercises a[1][1] โ a[1][2] โ a[2][2] โ a[3][2] โ a[3][3]. Iahubina will choose exercises a[3][1] โ a[2][1] โ a[2][2] โ a[2][3] โ a[1][3].
Submitted Solution:
```
# -*- coding:utf-8 -*-
"""
created by shuangquan.huang at 1/7/20
"""
import collections
import time
import os
import sys
import bisect
import heapq
from typing import List
def solve(N, M, A):
dpa = [[0 for _ in range(M+2)] for _ in range(N+2)]
dpb = [[0 for _ in range(M+2)] for _ in range(N+2)]
dpc = [[0 for _ in range(M+2)] for _ in range(N+2)]
dpd = [[0 for _ in range(M+2)] for _ in range(N+2)]
for r in range(1, N+1):
for c in range(1, M + 1):
dpa[r][c] = max(dpa[r-1][c], dpa[r][c-1]) + A[r][c]
for r in range(N, 0, -1):
for c in range(M, 0, -1):
dpb[r][c] = max(dpb[r+1][c], dpb[r][c+1]) + A[r][c]
for r in range(N, 0, -1):
for c in range(1, M+1):
dpc[r][c] = max(dpc[r+1][c], dpc[r][c-1]) + A[r][c]
for r in range(1, N+1):
for c in range(M, 0, -1):
dpd[r][c] = max(dpd[r-1][c], dpd[r][c+1]) + A[r][c]
ans = 0
for r in range(2, N):
for c in range(2, M):
a = dpa[r][c-1] + dpb[r][c+1] + dpc[r+1][c] + dpd[r-1][c]
b = dpc[r][c-1] + dpd[r][c+1] + dpa[r-1][c] + dpb[r+1][c]
ans = max(ans, a, b)
return ans
N, M = map(int, input().split())
A = [[0 for _ in range(M+2)]]
for i in range(N):
row = [0] + [int(x) for x in input().split()] + [0]
A.append(row)
A.append([0 for _ in range(M+2)])
print(solve(N, M, A))
``` | instruction | 0 | 66,747 | 8 | 133,494 |
Yes | output | 1 | 66,747 | 8 | 133,495 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Summer is coming! It's time for Iahub and Iahubina to work out, as they both want to look hot at the beach. The gym where they go is a matrix a with n lines and m columns. Let number a[i][j] represents the calories burned by performing workout at the cell of gym in the i-th line and the j-th column.
Iahub starts with workout located at line 1 and column 1. He needs to finish with workout a[n][m]. After finishing workout a[i][j], he can go to workout a[i + 1][j] or a[i][j + 1]. Similarly, Iahubina starts with workout a[n][1] and she needs to finish with workout a[1][m]. After finishing workout from cell a[i][j], she goes to either a[i][j + 1] or a[i - 1][j].
There is one additional condition for their training. They have to meet in exactly one cell of gym. At that cell, none of them will work out. They will talk about fast exponentiation (pretty odd small talk) and then both of them will move to the next workout.
If a workout was done by either Iahub or Iahubina, it counts as total gain. Please plan a workout for Iahub and Iahubina such as total gain to be as big as possible. Note, that Iahub and Iahubina can perform workouts with different speed, so the number of cells that they use to reach meet cell may differs.
Input
The first line of the input contains two integers n and m (3 โค n, m โค 1000). Each of the next n lines contains m integers: j-th number from i-th line denotes element a[i][j] (0 โค a[i][j] โค 105).
Output
The output contains a single number โ the maximum total gain possible.
Examples
Input
3 3
100 100 100
100 1 100
100 100 100
Output
800
Note
Iahub will choose exercises a[1][1] โ a[1][2] โ a[2][2] โ a[3][2] โ a[3][3]. Iahubina will choose exercises a[3][1] โ a[2][1] โ a[2][2] โ a[2][3] โ a[1][3].
Submitted Solution:
```
'''input
3 3
100 100 100
100 1 100
100 100 100
'''
# again a coding delight
from sys import stdin
def create_dp1(matrix, n, m):
dp = [[0 for i in range(m)] for j in range(n)]
for i in range(n):
for j in range(m):
if i - 1 >= 0:
dp[i][j] = max(dp[i][j], matrix[i][j] + dp[i - 1][j])
if j - 1 >= 0:
dp[i][j] = max(dp[i][j], matrix[i][j] + dp[i][j - 1])
elif i - 1 < 0 and j - 1 < 0:
dp[i][j] = matrix[i][j]
return dp
def create_dp2(matrix, n, m):
dp = [[0 for i in range(m)] for j in range(n)]
for i in range(n - 1, -1, -1):
for j in range(m - 1, -1, -1):
if i + 1 < n:
dp[i][j] = max(dp[i][j], matrix[i][j] + dp[i + 1][j])
if j + 1 < m:
dp[i][j] = max(dp[i][j], matrix[i][j] + dp[i][j + 1])
if i + 1 >= n and j + 1 >= m:
dp[i][j] = matrix[i][j]
return dp
def create_dp3(matrix, n, m):
dp = [[0 for i in range(m)] for j in range(n)]
for i in range(n - 1, -1, -1):
for j in range(m):
if i + 1 < n:
dp[i][j] = max(dp[i][j], matrix[i][j] + dp[i + 1][j])
if j - 1 >= 0:
dp[i][j] = max(dp[i][j], matrix[i][j] + dp[i][j - 1])
if i + 1 >= n and j - 1 < 0:
dp[i][j] = matrix[i][j]
return dp
def create_dp4(matrix, n, m):
dp = [[0 for i in range(m)] for j in range(n)]
for i in range(n):
for j in range(m - 1, -1, -1):
if i - 1 >= 0:
dp[i][j] = max(dp[i][j], matrix[i][j] + dp[i - 1][j])
if j + 1 < m:
dp[i][j] = max(dp[i][j], matrix[i][j] + dp[i][j + 1])
if i - 1 < 0 and j + 1 >= m:
dp[i][j] = matrix[i][j]
return dp
# main starts
n, m = list(map(int, stdin.readline().split()))
matrix = []
for _ in range(n):
matrix.append(list(map(int, stdin.readline().split())))
dp1 = create_dp1(matrix, n, m) # from 0, 0 to i, j
dp2 = create_dp2(matrix, n, m) # from i, j to n, m
dp3 = create_dp3(matrix, n, m) # from n, 1 to i, j
dp4 = create_dp4(matrix, n, m) # from i, j to 1, m
total = -float('inf')
for i in range(1, n - 1):
for j in range(1, m - 1):
first = dp1[i - 1][j] + dp2[i + 1][j] + dp3[i][j - 1] + dp4[i][j + 1]
second = dp1[i][j - 1] + dp2[i][j + 1] + dp3[i + 1][j] + dp4[i - 1][j]
total = max(total, first, second)
print(total)
``` | instruction | 0 | 66,748 | 8 | 133,496 |
Yes | output | 1 | 66,748 | 8 | 133,497 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Summer is coming! It's time for Iahub and Iahubina to work out, as they both want to look hot at the beach. The gym where they go is a matrix a with n lines and m columns. Let number a[i][j] represents the calories burned by performing workout at the cell of gym in the i-th line and the j-th column.
Iahub starts with workout located at line 1 and column 1. He needs to finish with workout a[n][m]. After finishing workout a[i][j], he can go to workout a[i + 1][j] or a[i][j + 1]. Similarly, Iahubina starts with workout a[n][1] and she needs to finish with workout a[1][m]. After finishing workout from cell a[i][j], she goes to either a[i][j + 1] or a[i - 1][j].
There is one additional condition for their training. They have to meet in exactly one cell of gym. At that cell, none of them will work out. They will talk about fast exponentiation (pretty odd small talk) and then both of them will move to the next workout.
If a workout was done by either Iahub or Iahubina, it counts as total gain. Please plan a workout for Iahub and Iahubina such as total gain to be as big as possible. Note, that Iahub and Iahubina can perform workouts with different speed, so the number of cells that they use to reach meet cell may differs.
Input
The first line of the input contains two integers n and m (3 โค n, m โค 1000). Each of the next n lines contains m integers: j-th number from i-th line denotes element a[i][j] (0 โค a[i][j] โค 105).
Output
The output contains a single number โ the maximum total gain possible.
Examples
Input
3 3
100 100 100
100 1 100
100 100 100
Output
800
Note
Iahub will choose exercises a[1][1] โ a[1][2] โ a[2][2] โ a[3][2] โ a[3][3]. Iahubina will choose exercises a[3][1] โ a[2][1] โ a[2][2] โ a[2][3] โ a[1][3].
Submitted Solution:
```
n,m=map(int,input().split())
a=[]
for i in range(n):a.append(list(map(int,input().split())))
dpa=[[[0,0] for i in range(m+2)] for i in range(n+2)]
dpb=[[[0,0] for i in range(m+2)] for i in range(n+2)]
ans=0
for i in range(1,n+1):
for j in range(1,m+1):
dpa[i][j][0]=max(dpa[i-1][j][0],dpa[i][j-1][0])+a[i-1][j-1]
dpa[n+1-i][m+1-j][1]=max(dpa[n+2-i][m+1-j][1],dpa[n+1-i][m+2-j][1])+a[n-i][m-j]
for i in range(n,0,-1):
for j in range(1,m+1):
dpb[i][j][0]=max(dpb[i+1][j][0],dpb[i][j-1][0])+a[i-1][j-1]
dpb[n+1-i][m+1-j][1]=max(dpb[n-i][m+1-j][1],dpb[n+1-i][m+2-j][1])+a[n-i][m-j]
for i in range(2,n):
for j in range(2,m):
x=dpa[i-1][j][0]+dpa[i+1][j][1]+dpb[i][j-1][0]+dpb[i][j+1][1]
y=dpb[i+1][j][0]+dpb[i-1][j][1]+dpa[i][j-1][0]+dpa[i][j+1][1]
ans=max(ans,x,y)
print(ans)
``` | instruction | 0 | 66,749 | 8 | 133,498 |
Yes | output | 1 | 66,749 | 8 | 133,499 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Summer is coming! It's time for Iahub and Iahubina to work out, as they both want to look hot at the beach. The gym where they go is a matrix a with n lines and m columns. Let number a[i][j] represents the calories burned by performing workout at the cell of gym in the i-th line and the j-th column.
Iahub starts with workout located at line 1 and column 1. He needs to finish with workout a[n][m]. After finishing workout a[i][j], he can go to workout a[i + 1][j] or a[i][j + 1]. Similarly, Iahubina starts with workout a[n][1] and she needs to finish with workout a[1][m]. After finishing workout from cell a[i][j], she goes to either a[i][j + 1] or a[i - 1][j].
There is one additional condition for their training. They have to meet in exactly one cell of gym. At that cell, none of them will work out. They will talk about fast exponentiation (pretty odd small talk) and then both of them will move to the next workout.
If a workout was done by either Iahub or Iahubina, it counts as total gain. Please plan a workout for Iahub and Iahubina such as total gain to be as big as possible. Note, that Iahub and Iahubina can perform workouts with different speed, so the number of cells that they use to reach meet cell may differs.
Input
The first line of the input contains two integers n and m (3 โค n, m โค 1000). Each of the next n lines contains m integers: j-th number from i-th line denotes element a[i][j] (0 โค a[i][j] โค 105).
Output
The output contains a single number โ the maximum total gain possible.
Examples
Input
3 3
100 100 100
100 1 100
100 100 100
Output
800
Note
Iahub will choose exercises a[1][1] โ a[1][2] โ a[2][2] โ a[3][2] โ a[3][3]. Iahubina will choose exercises a[3][1] โ a[2][1] โ a[2][2] โ a[2][3] โ a[1][3].
Submitted Solution:
```
import sys
from math import *
def minp():
return sys.stdin.readline().strip()
def mint():
return int(minp())
def mints():
return map(int, minp().split())
n, m = mints()
a = [0]*n
dp1 = [0]*n
dp2 = [0]*n
dp3 = [0]*n
dp4 = [0]*n
for i in range(n):
a[i] = list(mints())
dp1[i] = [0]*m
dp2[i] = [0]*m
dp3[i] = [0]*m
dp4[i] = [0]*m
dp1[0][0] = a[0][0]
dp2[n-1][m-1] = a[n-1][m-1]
dp3[n-1][0] = a[n-1][0]
dp4[0][m-1] = a[0][m-1]
for i in range(1,n):
dp1[i][0] = dp1[i-1][0] + a[i][0] # >>>>
for i in range(n-2,-1,-1):
dp2[i][m-1] = dp2[i+1][m-1] + a[i][m-1] # <<<<
for i in range(n-2,-1,-1):
dp3[i][0] = dp3[i+1][0] + a[i][0] # <<<<
for i in range(1,n):
dp4[i][m-1] = dp4[i-1][m-1] + a[i][m-1] # >>>>
for i in range(1,m):
dp1[0][i] = dp1[0][i-1] + a[0][i] # >>>>
for i in range(m-2,-1,-1):
dp2[n-1][i] = dp2[n-1][i+1] + a[n-1][i] # <<<<
for i in range(1,m):
dp3[n-1][i] = dp3[n-1][i-1] + a[n-1][i] # >>>>
for i in range(m-2,-1,-1):
dp4[0][i] = dp4[0][i+1] + a[0][i] # >>>>
for i in range(1,n):
for j in range(1,m):
dp1[i][j] = max(dp1[i-1][j], dp1[i][j-1]) + a[i][j]
for i in range(n-2,-1,-1):
for j in range(m-2,-1,-1):
dp2[i][j] = max(dp2[i+1][j], dp2[i][j+1]) + a[i][j]
for i in range(n-2,-1,-1):
for j in range(1,m):
dp3[i][j] = max(dp3[i+1][j], dp3[i][j-1]) + a[i][j]
for i in range(1,n):
for j in range(m-2,-1,-1):
dp4[i][j] = max(dp4[i-1][j], dp4[i][j+1]) + a[i][j]
r = 0
for i in range(1,n-1):
for j in range(1,m-1):
r = max(r, dp1[i-1][j] + dp2[i+1][j] + dp3[i][j-1] + dp4[i][j+1])
r = max(r, dp1[i][j-1] + dp2[i][j+1] + dp3[i+1][j] + dp4[i-1][j])
print(r)
``` | instruction | 0 | 66,750 | 8 | 133,500 |
Yes | output | 1 | 66,750 | 8 | 133,501 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Summer is coming! It's time for Iahub and Iahubina to work out, as they both want to look hot at the beach. The gym where they go is a matrix a with n lines and m columns. Let number a[i][j] represents the calories burned by performing workout at the cell of gym in the i-th line and the j-th column.
Iahub starts with workout located at line 1 and column 1. He needs to finish with workout a[n][m]. After finishing workout a[i][j], he can go to workout a[i + 1][j] or a[i][j + 1]. Similarly, Iahubina starts with workout a[n][1] and she needs to finish with workout a[1][m]. After finishing workout from cell a[i][j], she goes to either a[i][j + 1] or a[i - 1][j].
There is one additional condition for their training. They have to meet in exactly one cell of gym. At that cell, none of them will work out. They will talk about fast exponentiation (pretty odd small talk) and then both of them will move to the next workout.
If a workout was done by either Iahub or Iahubina, it counts as total gain. Please plan a workout for Iahub and Iahubina such as total gain to be as big as possible. Note, that Iahub and Iahubina can perform workouts with different speed, so the number of cells that they use to reach meet cell may differs.
Input
The first line of the input contains two integers n and m (3 โค n, m โค 1000). Each of the next n lines contains m integers: j-th number from i-th line denotes element a[i][j] (0 โค a[i][j] โค 105).
Output
The output contains a single number โ the maximum total gain possible.
Examples
Input
3 3
100 100 100
100 1 100
100 100 100
Output
800
Note
Iahub will choose exercises a[1][1] โ a[1][2] โ a[2][2] โ a[3][2] โ a[3][3]. Iahubina will choose exercises a[3][1] โ a[2][1] โ a[2][2] โ a[2][3] โ a[1][3].
Submitted Solution:
```
print("Ola")
``` | instruction | 0 | 66,751 | 8 | 133,502 |
No | output | 1 | 66,751 | 8 | 133,503 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Summer is coming! It's time for Iahub and Iahubina to work out, as they both want to look hot at the beach. The gym where they go is a matrix a with n lines and m columns. Let number a[i][j] represents the calories burned by performing workout at the cell of gym in the i-th line and the j-th column.
Iahub starts with workout located at line 1 and column 1. He needs to finish with workout a[n][m]. After finishing workout a[i][j], he can go to workout a[i + 1][j] or a[i][j + 1]. Similarly, Iahubina starts with workout a[n][1] and she needs to finish with workout a[1][m]. After finishing workout from cell a[i][j], she goes to either a[i][j + 1] or a[i - 1][j].
There is one additional condition for their training. They have to meet in exactly one cell of gym. At that cell, none of them will work out. They will talk about fast exponentiation (pretty odd small talk) and then both of them will move to the next workout.
If a workout was done by either Iahub or Iahubina, it counts as total gain. Please plan a workout for Iahub and Iahubina such as total gain to be as big as possible. Note, that Iahub and Iahubina can perform workouts with different speed, so the number of cells that they use to reach meet cell may differs.
Input
The first line of the input contains two integers n and m (3 โค n, m โค 1000). Each of the next n lines contains m integers: j-th number from i-th line denotes element a[i][j] (0 โค a[i][j] โค 105).
Output
The output contains a single number โ the maximum total gain possible.
Examples
Input
3 3
100 100 100
100 1 100
100 100 100
Output
800
Note
Iahub will choose exercises a[1][1] โ a[1][2] โ a[2][2] โ a[3][2] โ a[3][3]. Iahubina will choose exercises a[3][1] โ a[2][1] โ a[2][2] โ a[2][3] โ a[1][3].
Submitted Solution:
```
n,m=map(int,input().split())
a=[list(map(int,input().split())) for i in range(n)]
d1=[[0]*m for i in range(n)]
for i in range(n):
for j in range(m):
if i==j==0:
d1[i][j]=a[i][j]
elif i==0:
d1[i][j]=d1[i][j-1]+a[i][j]
elif j==0:
d1[i][j]=d1[i-1][j]+a[i][j]
else:
d1[i][j]=max(d1[i-1][j],d1[i][j-1])+a[i][j]
d2=[[0]*m for i in range(n)]
b=[[a[n-1-i][m-1-j] for j in range(m)] for i in range(n)]
for i in range(n):
for j in range(m):
if i==j==0:
d2[i][j]=b[i][j]
elif i==0:
d2[i][j]=d2[i][j-1]+b[i][j]
elif j==0:
d2[i][j]=d1[i-1][j]+b[i][j]
else:
d2[i][j]=max(d2[i-1][j],d2[i][j-1])+b[i][j]
d3=[[0]*m for i in range(n)]
c=[[a[n-1-i][j] for j in range(m)] for i in range(n)]
for i in range(n):
for j in range(m):
if i==j==0:
d3[i][j]=c[i][j]
elif i==0:
d3[i][j]=d3[i][j-1]+c[i][j]
elif j==0:
d3[i][j]=d3[i-1][j]+c[i][j]
else:
d3[i][j]=max(d3[i-1][j],d3[i][j-1])+c[i][j]
d4=[[0]*m for i in range(n)]
e=[[a[i][m-1-j] for j in range(m)] for i in range(n)]
for i in range(n):
for j in range(m):
if i==j==0:
d4[i][j]=e[i][j]
elif i==0:
d4[i][j]=d4[i][j-1]+e[i][j]
elif j==0:
d4[i][j]=d4[i-1][j]+e[i][j]
else:
d4[i][j]=max(d4[i-1][j],d4[i][j-1])+e[i][j]
ans=-1
for i in range(n):
for j in range(m):
x=d1[i][j]+d2[n-1-i][m-1-j]
y=d3[n-1-i][j]+d4[i][m-1-j]
ans=max(ans,x+y-4*a[i][j])
print(ans)
``` | instruction | 0 | 66,752 | 8 | 133,504 |
No | output | 1 | 66,752 | 8 | 133,505 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Summer is coming! It's time for Iahub and Iahubina to work out, as they both want to look hot at the beach. The gym where they go is a matrix a with n lines and m columns. Let number a[i][j] represents the calories burned by performing workout at the cell of gym in the i-th line and the j-th column.
Iahub starts with workout located at line 1 and column 1. He needs to finish with workout a[n][m]. After finishing workout a[i][j], he can go to workout a[i + 1][j] or a[i][j + 1]. Similarly, Iahubina starts with workout a[n][1] and she needs to finish with workout a[1][m]. After finishing workout from cell a[i][j], she goes to either a[i][j + 1] or a[i - 1][j].
There is one additional condition for their training. They have to meet in exactly one cell of gym. At that cell, none of them will work out. They will talk about fast exponentiation (pretty odd small talk) and then both of them will move to the next workout.
If a workout was done by either Iahub or Iahubina, it counts as total gain. Please plan a workout for Iahub and Iahubina such as total gain to be as big as possible. Note, that Iahub and Iahubina can perform workouts with different speed, so the number of cells that they use to reach meet cell may differs.
Input
The first line of the input contains two integers n and m (3 โค n, m โค 1000). Each of the next n lines contains m integers: j-th number from i-th line denotes element a[i][j] (0 โค a[i][j] โค 105).
Output
The output contains a single number โ the maximum total gain possible.
Examples
Input
3 3
100 100 100
100 1 100
100 100 100
Output
800
Note
Iahub will choose exercises a[1][1] โ a[1][2] โ a[2][2] โ a[3][2] โ a[3][3]. Iahubina will choose exercises a[3][1] โ a[2][1] โ a[2][2] โ a[2][3] โ a[1][3].
Submitted Solution:
```
a=[int(i) for i in input().split()]
d=[]
for b in range(0,a[0]):
d.append([int(i) for i in input().split()])
#print(d)
m1i=[[sum(d[0][:i]) for i in range(0,a[1])]]
for i in range(0,a[0]-1):
m1i.append([m1i[i][0]+d[i][0]])
for y in range(1,a[0]):
for x in range(1,a[1]):
m1i[y].append(max(m1i[y][x-1]+d[y][x-1],m1i[y-1][x]+d[y-1][x]))
p,g,h=a[0]-1,a[1]+1,a[1]-1
m1f=[[0]*a[1] for y in range(0,p)]+[[sum(d[p][i:]) for i in range(1,g,1)]]
for x in range(p,0,-1):
m1f[x-1][h] = m1f[x][h] + d[x][h]
for y in range(p-1,-1,-1):
for x in range(h-1,-1,-1):
m1f[y][x]=max(m1f[y][x+1]+d[y][x+1],m1f[y+1][x]+d[y+1][x])
m2i=[[0]*a[1] for i in range(0,p)]+[[sum(d[p][:i]) for i in range(0,a[1])]]
for i in range(p,0,-1):
m2i[i-1][0]=m2i[i][0]+d[i][0]
for x in range(1,a[1]):
for y in range(p-1,-1,-1):
m2i[y][x]=max(m2i[y+1][x]+d[y+1][x],m2i[y][x-1]+d[y][x-1])
m2f=[[sum(d[0][i:]) for i in range(1,g,1)]]+[[0]*h for i in range(0,p)]
#print(m2f)
for y in range(0,a[0]-1):
m2f[y+1].append(m2f[y][h]+d[y][h])
for y in range(1,a[0]):
for x in range(h-1,-1,-1):
m2f[y][x]=max(m2f[y-1][x]+d[y-1][x],m2f[y][x+1]+d[y][x+1])
print(max(m1i[y][x]+m1f[y][x]+m2i[y][x]+m2f[y][x] for y in range(0,a[0]) for x in range(0,a[1])))
#print(m1i)
#print(m1f)
#print(m2i)
#print(m2f)
``` | instruction | 0 | 66,753 | 8 | 133,506 |
No | output | 1 | 66,753 | 8 | 133,507 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Summer is coming! It's time for Iahub and Iahubina to work out, as they both want to look hot at the beach. The gym where they go is a matrix a with n lines and m columns. Let number a[i][j] represents the calories burned by performing workout at the cell of gym in the i-th line and the j-th column.
Iahub starts with workout located at line 1 and column 1. He needs to finish with workout a[n][m]. After finishing workout a[i][j], he can go to workout a[i + 1][j] or a[i][j + 1]. Similarly, Iahubina starts with workout a[n][1] and she needs to finish with workout a[1][m]. After finishing workout from cell a[i][j], she goes to either a[i][j + 1] or a[i - 1][j].
There is one additional condition for their training. They have to meet in exactly one cell of gym. At that cell, none of them will work out. They will talk about fast exponentiation (pretty odd small talk) and then both of them will move to the next workout.
If a workout was done by either Iahub or Iahubina, it counts as total gain. Please plan a workout for Iahub and Iahubina such as total gain to be as big as possible. Note, that Iahub and Iahubina can perform workouts with different speed, so the number of cells that they use to reach meet cell may differs.
Input
The first line of the input contains two integers n and m (3 โค n, m โค 1000). Each of the next n lines contains m integers: j-th number from i-th line denotes element a[i][j] (0 โค a[i][j] โค 105).
Output
The output contains a single number โ the maximum total gain possible.
Examples
Input
3 3
100 100 100
100 1 100
100 100 100
Output
800
Note
Iahub will choose exercises a[1][1] โ a[1][2] โ a[2][2] โ a[3][2] โ a[3][3]. Iahubina will choose exercises a[3][1] โ a[2][1] โ a[2][2] โ a[2][3] โ a[1][3].
Submitted Solution:
```
n = input()
a = [int(s) for s in input().split()]
print(max(a) - min(a), end=' ')
print(a.count(max(a)) * a.count(min(a)))
``` | instruction | 0 | 66,754 | 8 | 133,508 |
No | output | 1 | 66,754 | 8 | 133,509 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's now 260 AD. Shapur, being extremely smart, became the King of Persia. He is now called Shapur, His majesty King of kings of Iran and Aniran.
Recently the Romans declared war on Persia. They dreamed to occupy Armenia. In the recent war, the Romans were badly defeated. Now their senior army general, Philip is captured by Shapur and Shapur is now going to capture Valerian, the Roman emperor.
Being defeated, the cowardly Valerian hid in a room at the top of one of his castles. To capture him, Shapur has to open many doors. Fortunately Valerian was too scared to make impenetrable locks for the doors.
Each door has 4 parts. The first part is an integer number a. The second part is either an integer number b or some really odd sign which looks like R. The third one is an integer c and the fourth part is empty! As if it was laid for writing something. Being extremely gifted, after opening the first few doors, Shapur found out the secret behind the locks.
c is an integer written in base a, to open the door we should write it in base b. The only bad news is that this R is some sort of special numbering system that is used only in Roman empire, so opening the doors is not just a piece of cake!
Here's an explanation of this really weird number system that even doesn't have zero:
Roman numerals are based on seven symbols: a stroke (identified with the letter I) for a unit, a chevron (identified with the letter V) for a five, a cross-stroke (identified with the letter X) for a ten, a C (identified as an abbreviation of Centum) for a hundred, etc.:
* I=1
* V=5
* X=10
* L=50
* C=100
* D=500
* M=1000
Symbols are iterated to produce multiples of the decimal (1, 10, 100, 1, 000) values, with V, L, D substituted for a multiple of five, and the iteration continuing: I 1, II 2, III 3, V 5, VI 6, VII 7, etc., and the same for other bases: X 10, XX 20, XXX 30, L 50, LXXX 80; CC 200, DCC 700, etc. At the fourth and ninth iteration, a subtractive principle must be employed, with the base placed before the higher base: IV 4, IX 9, XL 40, XC 90, CD 400, CM 900.
Also in bases greater than 10 we use A for 10, B for 11, etc.
Help Shapur capture Valerian and bring peace back to Persia, especially Armenia.
Input
The first line contains two integers a and b (2 โค a, b โค 25). Only b may be replaced by an R which indicates Roman numbering system.
The next line contains a single non-negative integer c in base a which may contain leading zeros but its length doesn't exceed 103.
It is guaranteed that if we have Roman numerals included the number would be less than or equal to 300010 and it won't be 0. In any other case the number won't be greater than 101510.
Output
Write a single line that contains integer c in base b. You must omit leading zeros.
Examples
Input
10 2
1
Output
1
Input
16 R
5
Output
V
Input
5 R
4
Output
IV
Input
2 2
1111001
Output
1111001
Input
12 13
A
Output
A
Note
You can find more information about roman numerals here: http://en.wikipedia.org/wiki/Roman_numerals
Submitted Solution:
```
system_1, system_2 = input().split()
integer = int(input(), base = int(system_1))
if system_2 == '10' or integer == 0:
print(integer)
exit(0)
digits = []
alphabet = ['A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z']
if system_2 != 'R':
system_2 = int(system_2)
while integer != 0:
if system_2 < 10:
digits.append(str(integer%system_2))
else:
if integer%system_2 < 10: digits.append(str(integer%system_2))
else: digits.append(alphabet[integer%system_2 - 10])
integer = integer//system_2
print(''.join(digits[::-1]))
else:
roman = [1000,'M',900,'CM',500,'D',400,'CD',100,'C',90,'XC',50,'L',40,'XL',10,'X',9,'IX',5,'V',4,'IV',1,'I']
ans = ""
while integer != 0:
for i in range(len(roman)//2):
if integer>=roman[2*i]:
ans = ans + roman[2*i + 1]
integer = integer - roman[2*i]
break
print(ans)
``` | instruction | 0 | 66,840 | 8 | 133,680 |
Yes | output | 1 | 66,840 | 8 | 133,681 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's now 260 AD. Shapur, being extremely smart, became the King of Persia. He is now called Shapur, His majesty King of kings of Iran and Aniran.
Recently the Romans declared war on Persia. They dreamed to occupy Armenia. In the recent war, the Romans were badly defeated. Now their senior army general, Philip is captured by Shapur and Shapur is now going to capture Valerian, the Roman emperor.
Being defeated, the cowardly Valerian hid in a room at the top of one of his castles. To capture him, Shapur has to open many doors. Fortunately Valerian was too scared to make impenetrable locks for the doors.
Each door has 4 parts. The first part is an integer number a. The second part is either an integer number b or some really odd sign which looks like R. The third one is an integer c and the fourth part is empty! As if it was laid for writing something. Being extremely gifted, after opening the first few doors, Shapur found out the secret behind the locks.
c is an integer written in base a, to open the door we should write it in base b. The only bad news is that this R is some sort of special numbering system that is used only in Roman empire, so opening the doors is not just a piece of cake!
Here's an explanation of this really weird number system that even doesn't have zero:
Roman numerals are based on seven symbols: a stroke (identified with the letter I) for a unit, a chevron (identified with the letter V) for a five, a cross-stroke (identified with the letter X) for a ten, a C (identified as an abbreviation of Centum) for a hundred, etc.:
* I=1
* V=5
* X=10
* L=50
* C=100
* D=500
* M=1000
Symbols are iterated to produce multiples of the decimal (1, 10, 100, 1, 000) values, with V, L, D substituted for a multiple of five, and the iteration continuing: I 1, II 2, III 3, V 5, VI 6, VII 7, etc., and the same for other bases: X 10, XX 20, XXX 30, L 50, LXXX 80; CC 200, DCC 700, etc. At the fourth and ninth iteration, a subtractive principle must be employed, with the base placed before the higher base: IV 4, IX 9, XL 40, XC 90, CD 400, CM 900.
Also in bases greater than 10 we use A for 10, B for 11, etc.
Help Shapur capture Valerian and bring peace back to Persia, especially Armenia.
Input
The first line contains two integers a and b (2 โค a, b โค 25). Only b may be replaced by an R which indicates Roman numbering system.
The next line contains a single non-negative integer c in base a which may contain leading zeros but its length doesn't exceed 103.
It is guaranteed that if we have Roman numerals included the number would be less than or equal to 300010 and it won't be 0. In any other case the number won't be greater than 101510.
Output
Write a single line that contains integer c in base b. You must omit leading zeros.
Examples
Input
10 2
1
Output
1
Input
16 R
5
Output
V
Input
5 R
4
Output
IV
Input
2 2
1111001
Output
1111001
Input
12 13
A
Output
A
Note
You can find more information about roman numerals here: http://en.wikipedia.org/wiki/Roman_numerals
Submitted Solution:
```
digs = [chr(c) for c in range(ord('0'),ord('9')+1)]
digs += [chr(c) for c in range(ord('A'),ord('Z')+1)]
def int2base(x, base):
if x < 0: sign = -1
elif x == 0: return digs[0]
else: sign = 1
x *= sign
digits = []
while x:
digits.append(digs[x % base])
x //= base
if sign < 0:
digits.append('-')
digits.reverse()
return ''.join(digits)
def int2roman(x):
ints = (1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1)
nums = ('M', 'CM', 'D', 'CD','C', 'XC','L','XL','X','IX','V','IV','I')
result = ""
for i in range(len(ints)):
count = int(x / ints[i])
result += nums[i] * count
x -= ints[i] * count
return result
x = input().split()
a = int(x[0])
roman = False
b = x[1]
if b[0] == 'R':
roman = True
else:
b = int(b)
na = int(input(),a)
if roman :
print(int2roman(na))
else:
print(int2base(na,b))
``` | instruction | 0 | 66,841 | 8 | 133,682 |
Yes | output | 1 | 66,841 | 8 | 133,683 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's now 260 AD. Shapur, being extremely smart, became the King of Persia. He is now called Shapur, His majesty King of kings of Iran and Aniran.
Recently the Romans declared war on Persia. They dreamed to occupy Armenia. In the recent war, the Romans were badly defeated. Now their senior army general, Philip is captured by Shapur and Shapur is now going to capture Valerian, the Roman emperor.
Being defeated, the cowardly Valerian hid in a room at the top of one of his castles. To capture him, Shapur has to open many doors. Fortunately Valerian was too scared to make impenetrable locks for the doors.
Each door has 4 parts. The first part is an integer number a. The second part is either an integer number b or some really odd sign which looks like R. The third one is an integer c and the fourth part is empty! As if it was laid for writing something. Being extremely gifted, after opening the first few doors, Shapur found out the secret behind the locks.
c is an integer written in base a, to open the door we should write it in base b. The only bad news is that this R is some sort of special numbering system that is used only in Roman empire, so opening the doors is not just a piece of cake!
Here's an explanation of this really weird number system that even doesn't have zero:
Roman numerals are based on seven symbols: a stroke (identified with the letter I) for a unit, a chevron (identified with the letter V) for a five, a cross-stroke (identified with the letter X) for a ten, a C (identified as an abbreviation of Centum) for a hundred, etc.:
* I=1
* V=5
* X=10
* L=50
* C=100
* D=500
* M=1000
Symbols are iterated to produce multiples of the decimal (1, 10, 100, 1, 000) values, with V, L, D substituted for a multiple of five, and the iteration continuing: I 1, II 2, III 3, V 5, VI 6, VII 7, etc., and the same for other bases: X 10, XX 20, XXX 30, L 50, LXXX 80; CC 200, DCC 700, etc. At the fourth and ninth iteration, a subtractive principle must be employed, with the base placed before the higher base: IV 4, IX 9, XL 40, XC 90, CD 400, CM 900.
Also in bases greater than 10 we use A for 10, B for 11, etc.
Help Shapur capture Valerian and bring peace back to Persia, especially Armenia.
Input
The first line contains two integers a and b (2 โค a, b โค 25). Only b may be replaced by an R which indicates Roman numbering system.
The next line contains a single non-negative integer c in base a which may contain leading zeros but its length doesn't exceed 103.
It is guaranteed that if we have Roman numerals included the number would be less than or equal to 300010 and it won't be 0. In any other case the number won't be greater than 101510.
Output
Write a single line that contains integer c in base b. You must omit leading zeros.
Examples
Input
10 2
1
Output
1
Input
16 R
5
Output
V
Input
5 R
4
Output
IV
Input
2 2
1111001
Output
1111001
Input
12 13
A
Output
A
Note
You can find more information about roman numerals here: http://en.wikipedia.org/wiki/Roman_numerals
Submitted Solution:
```
def main() :
def intToRoman(num) :
pattern = {'0':'', '1':'a', '2':'aa', '3':'aaa', '4':'ab',
'5':'b', '6':'ba', '7':'baa', '8':'baaa', '9':'ac'}
code = ['IVX', 'XLC', 'CDM', 'M__']
ret = []
for i,x in enumerate(reversed(str(num))) :
tmp = pattern[x]
for p,r in zip('abc', code[i]) :
tmp = tmp.replace(p, r)
ret.append(tmp)
return ''.join(reversed(ret))
encodeB = '0123456789ABCDEFGHIJKLMNO'
decodeA = {x:i for i,x in enumerate(encodeB)}
a,b = input().split()
a = int(a)
if b != 'R' : b = int(b)
c = sum(decodeA[x] * a**i
for i,x in enumerate(reversed(input().lstrip('0'))))
if b == 'R' :
print(intToRoman(c))
return
ans = []
while c :
ans.append(encodeB[c % b])
c //= b
ans = ''.join(reversed(ans))
if not ans : ans = 0 # ่ง่ฝ๏ผ
print(ans)
main()
``` | instruction | 0 | 66,842 | 8 | 133,684 |
Yes | output | 1 | 66,842 | 8 | 133,685 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's now 260 AD. Shapur, being extremely smart, became the King of Persia. He is now called Shapur, His majesty King of kings of Iran and Aniran.
Recently the Romans declared war on Persia. They dreamed to occupy Armenia. In the recent war, the Romans were badly defeated. Now their senior army general, Philip is captured by Shapur and Shapur is now going to capture Valerian, the Roman emperor.
Being defeated, the cowardly Valerian hid in a room at the top of one of his castles. To capture him, Shapur has to open many doors. Fortunately Valerian was too scared to make impenetrable locks for the doors.
Each door has 4 parts. The first part is an integer number a. The second part is either an integer number b or some really odd sign which looks like R. The third one is an integer c and the fourth part is empty! As if it was laid for writing something. Being extremely gifted, after opening the first few doors, Shapur found out the secret behind the locks.
c is an integer written in base a, to open the door we should write it in base b. The only bad news is that this R is some sort of special numbering system that is used only in Roman empire, so opening the doors is not just a piece of cake!
Here's an explanation of this really weird number system that even doesn't have zero:
Roman numerals are based on seven symbols: a stroke (identified with the letter I) for a unit, a chevron (identified with the letter V) for a five, a cross-stroke (identified with the letter X) for a ten, a C (identified as an abbreviation of Centum) for a hundred, etc.:
* I=1
* V=5
* X=10
* L=50
* C=100
* D=500
* M=1000
Symbols are iterated to produce multiples of the decimal (1, 10, 100, 1, 000) values, with V, L, D substituted for a multiple of five, and the iteration continuing: I 1, II 2, III 3, V 5, VI 6, VII 7, etc., and the same for other bases: X 10, XX 20, XXX 30, L 50, LXXX 80; CC 200, DCC 700, etc. At the fourth and ninth iteration, a subtractive principle must be employed, with the base placed before the higher base: IV 4, IX 9, XL 40, XC 90, CD 400, CM 900.
Also in bases greater than 10 we use A for 10, B for 11, etc.
Help Shapur capture Valerian and bring peace back to Persia, especially Armenia.
Input
The first line contains two integers a and b (2 โค a, b โค 25). Only b may be replaced by an R which indicates Roman numbering system.
The next line contains a single non-negative integer c in base a which may contain leading zeros but its length doesn't exceed 103.
It is guaranteed that if we have Roman numerals included the number would be less than or equal to 300010 and it won't be 0. In any other case the number won't be greater than 101510.
Output
Write a single line that contains integer c in base b. You must omit leading zeros.
Examples
Input
10 2
1
Output
1
Input
16 R
5
Output
V
Input
5 R
4
Output
IV
Input
2 2
1111001
Output
1111001
Input
12 13
A
Output
A
Note
You can find more information about roman numerals here: http://en.wikipedia.org/wiki/Roman_numerals
Submitted Solution:
```
numbers = {'0':0,'1':1,'2':2,'3':3,'4':4,'5':5,'6':6,'7':7,'8':8,'9':9,'A':10,'B':11,'C':12,'D':13,'E':14,'F':15,'G':16,'H':17,'I':18,'J':19,'K':20,'L':21,'M':22,'N':23,'O':24,'P':25,'Q':26,'R':27,'S':28,'T':29,'U':30,'V':31,'W':32,'X':33,'Y':34,'Z':35}
characters = {0:'0',1:'1',2:'2',3:'3',4:'4',5:'5',6:'6',7:'7',8:'8',9:'9',10:'A',11:'B',12:'C',13:'D',14:'E',15:'F',16:'G',17:'H',18:'I',19:'J',20:'K',21:'L',22:'M',23:'N',24:'O',25:'P',26:'Q',27:'R',28:'S',29:'T',30:'U',31:'V',32:'W',33:'X',34:'Y',35:'Z'}
def ConvertToDecimal(number, base):
newnumber = 0
for digit in range(-1,(0-len(number))-1,-1):
if(numbers[number[digit]] >= int(base)):
return -1
newnumber += numbers[number[digit]]*(int(base)**(0-digit-1))
return newnumber
def ConvertToBase(number, base):
newnumber = ''
if(number == 0):
return '0'
while number > 0:
newnumber = characters[number%base] + newnumber
number = number//base
return newnumber
def ConvertToRoman(number):
newnumber = ''
for i in range(4,0,-1):
currentnumber = (number%(10**i) - number%(10**(i-1)))//(10**(i-1))
if(currentnumber > 0):
if(i==4):
newnumber += 'M'*currentnumber
elif(i==3):
if(currentnumber == 9):
newnumber += 'CM'
elif(currentnumber>=5):
newnumber += 'D' + 'C'*(currentnumber-5)
elif currentnumber == 4:
newnumber += 'CD'
else:
newnumber += 'C'*currentnumber
elif(i==2):
if(currentnumber == 9):
newnumber += 'XC'
elif(currentnumber>=5):
newnumber += 'L' + 'X'*(currentnumber-5)
elif currentnumber == 4:
newnumber += 'XL'
else:
newnumber += 'X'*currentnumber
elif(i==1):
if(currentnumber == 9):
newnumber += 'IX'
elif(currentnumber>=5):
newnumber += 'V' + 'I'*(currentnumber-5)
elif currentnumber == 4:
newnumber += 'IV'
else:
newnumber += 'I'*currentnumber
return newnumber
a,b = [x for x in input().split()]
c = input()
if b == 'R':
print(ConvertToRoman(ConvertToDecimal(c,int(a))))
else:
print(ConvertToBase(ConvertToDecimal(c,int(a)),int(b)))
``` | instruction | 0 | 66,843 | 8 | 133,686 |
Yes | output | 1 | 66,843 | 8 | 133,687 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's now 260 AD. Shapur, being extremely smart, became the King of Persia. He is now called Shapur, His majesty King of kings of Iran and Aniran.
Recently the Romans declared war on Persia. They dreamed to occupy Armenia. In the recent war, the Romans were badly defeated. Now their senior army general, Philip is captured by Shapur and Shapur is now going to capture Valerian, the Roman emperor.
Being defeated, the cowardly Valerian hid in a room at the top of one of his castles. To capture him, Shapur has to open many doors. Fortunately Valerian was too scared to make impenetrable locks for the doors.
Each door has 4 parts. The first part is an integer number a. The second part is either an integer number b or some really odd sign which looks like R. The third one is an integer c and the fourth part is empty! As if it was laid for writing something. Being extremely gifted, after opening the first few doors, Shapur found out the secret behind the locks.
c is an integer written in base a, to open the door we should write it in base b. The only bad news is that this R is some sort of special numbering system that is used only in Roman empire, so opening the doors is not just a piece of cake!
Here's an explanation of this really weird number system that even doesn't have zero:
Roman numerals are based on seven symbols: a stroke (identified with the letter I) for a unit, a chevron (identified with the letter V) for a five, a cross-stroke (identified with the letter X) for a ten, a C (identified as an abbreviation of Centum) for a hundred, etc.:
* I=1
* V=5
* X=10
* L=50
* C=100
* D=500
* M=1000
Symbols are iterated to produce multiples of the decimal (1, 10, 100, 1, 000) values, with V, L, D substituted for a multiple of five, and the iteration continuing: I 1, II 2, III 3, V 5, VI 6, VII 7, etc., and the same for other bases: X 10, XX 20, XXX 30, L 50, LXXX 80; CC 200, DCC 700, etc. At the fourth and ninth iteration, a subtractive principle must be employed, with the base placed before the higher base: IV 4, IX 9, XL 40, XC 90, CD 400, CM 900.
Also in bases greater than 10 we use A for 10, B for 11, etc.
Help Shapur capture Valerian and bring peace back to Persia, especially Armenia.
Input
The first line contains two integers a and b (2 โค a, b โค 25). Only b may be replaced by an R which indicates Roman numbering system.
The next line contains a single non-negative integer c in base a which may contain leading zeros but its length doesn't exceed 103.
It is guaranteed that if we have Roman numerals included the number would be less than or equal to 300010 and it won't be 0. In any other case the number won't be greater than 101510.
Output
Write a single line that contains integer c in base b. You must omit leading zeros.
Examples
Input
10 2
1
Output
1
Input
16 R
5
Output
V
Input
5 R
4
Output
IV
Input
2 2
1111001
Output
1111001
Input
12 13
A
Output
A
Note
You can find more information about roman numerals here: http://en.wikipedia.org/wiki/Roman_numerals
Submitted Solution:
```
numbers = {'0':0,'1':1,'2':2,'3':3,'4':4,'5':5,'6':6,'7':7,'8':8,'9':9,'A':10,'B':11,'C':12,'D':13,'E':14,'F':15,'G':16,'H':17,'I':18,'J':19,'K':20,'L':21,'M':22,'N':23,'O':24,'P':25,'Q':26,'R':27,'S':28,'T':29,'U':30,'V':31,'W':32,'X':33,'Y':34,'Z':35}
characters = {0:'0',1:'1',2:'2',3:'3',4:'4',5:'5',6:'6',7:'7',8:'8',9:'9',10:'A',11:'B',12:'C',13:'D',14:'E',15:'F',16:'G',17:'H',18:'I',19:'J',20:'K',21:'L',22:'M',23:'N',24:'O',25:'P',26:'Q',27:'R',28:'S',29:'T',30:'U',31:'V',32:'W',33:'X',34:'Y',35:'Z'}
def ConvertToDecimal(number, base):
newnumber = 0
for digit in range(-1,(0-len(number))-1,-1):
if(numbers[number[digit]] >= int(base)):
return -1
newnumber += numbers[number[digit]]*(int(base)**(0-digit-1))
return newnumber
def ConvertToBase(number, base):
newnumber = ''
while number > 0:
newnumber = characters[number%base] + newnumber
number = number//base
return newnumber
def ConvertToRoman(number):
newnumber = ''
for i in range(4,0,-1):
currentnumber = (number%(10**i) - number%(10**(i-1)))//(10**(i-1))
if(currentnumber > 0):
if(i==4):
newnumber += 'M'*currentnumber
elif(i==3):
if(currentnumber == 9):
newnumber += 'CM'
elif(currentnumber>=5):
newnumber += 'D' + 'C'*(currentnumber-5)
elif currentnumber == 4:
newnumber += 'CD'
else:
newnumber += 'C'*currentnumber
elif(i==2):
if(currentnumber == 9):
newnumber += 'XC'
elif(currentnumber>=5):
newnumber += 'L' + 'X'*(currentnumber-5)
elif currentnumber == 4:
newnumber += 'XL'
else:
newnumber += 'X'*currentnumber
elif(i==1):
if(currentnumber == 9):
newnumber += 'IX'
elif(currentnumber>=5):
newnumber += 'V' + 'I'*(currentnumber-5)
elif currentnumber == 4:
newnumber += 'IV'
else:
newnumber += 'I'*currentnumber
return newnumber
a,b = [x for x in input().split()]
c = input()
if b == 'R':
print(ConvertToRoman(ConvertToDecimal(c,int(a))))
else:
print(ConvertToBase(ConvertToDecimal(c,int(a)),int(b)))
``` | instruction | 0 | 66,844 | 8 | 133,688 |
No | output | 1 | 66,844 | 8 | 133,689 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's now 260 AD. Shapur, being extremely smart, became the King of Persia. He is now called Shapur, His majesty King of kings of Iran and Aniran.
Recently the Romans declared war on Persia. They dreamed to occupy Armenia. In the recent war, the Romans were badly defeated. Now their senior army general, Philip is captured by Shapur and Shapur is now going to capture Valerian, the Roman emperor.
Being defeated, the cowardly Valerian hid in a room at the top of one of his castles. To capture him, Shapur has to open many doors. Fortunately Valerian was too scared to make impenetrable locks for the doors.
Each door has 4 parts. The first part is an integer number a. The second part is either an integer number b or some really odd sign which looks like R. The third one is an integer c and the fourth part is empty! As if it was laid for writing something. Being extremely gifted, after opening the first few doors, Shapur found out the secret behind the locks.
c is an integer written in base a, to open the door we should write it in base b. The only bad news is that this R is some sort of special numbering system that is used only in Roman empire, so opening the doors is not just a piece of cake!
Here's an explanation of this really weird number system that even doesn't have zero:
Roman numerals are based on seven symbols: a stroke (identified with the letter I) for a unit, a chevron (identified with the letter V) for a five, a cross-stroke (identified with the letter X) for a ten, a C (identified as an abbreviation of Centum) for a hundred, etc.:
* I=1
* V=5
* X=10
* L=50
* C=100
* D=500
* M=1000
Symbols are iterated to produce multiples of the decimal (1, 10, 100, 1, 000) values, with V, L, D substituted for a multiple of five, and the iteration continuing: I 1, II 2, III 3, V 5, VI 6, VII 7, etc., and the same for other bases: X 10, XX 20, XXX 30, L 50, LXXX 80; CC 200, DCC 700, etc. At the fourth and ninth iteration, a subtractive principle must be employed, with the base placed before the higher base: IV 4, IX 9, XL 40, XC 90, CD 400, CM 900.
Also in bases greater than 10 we use A for 10, B for 11, etc.
Help Shapur capture Valerian and bring peace back to Persia, especially Armenia.
Input
The first line contains two integers a and b (2 โค a, b โค 25). Only b may be replaced by an R which indicates Roman numbering system.
The next line contains a single non-negative integer c in base a which may contain leading zeros but its length doesn't exceed 103.
It is guaranteed that if we have Roman numerals included the number would be less than or equal to 300010 and it won't be 0. In any other case the number won't be greater than 101510.
Output
Write a single line that contains integer c in base b. You must omit leading zeros.
Examples
Input
10 2
1
Output
1
Input
16 R
5
Output
V
Input
5 R
4
Output
IV
Input
2 2
1111001
Output
1111001
Input
12 13
A
Output
A
Note
You can find more information about roman numerals here: http://en.wikipedia.org/wiki/Roman_numerals
Submitted Solution:
```
import math
y = [i for i in input().split()]
a = int(y[0])
b = y[1]
c = input()
def btb(oldbase,newbase,num):
n = reversed(str(num))
s10 = 0
p = 0
#oldbase to base 10
for i in n:
d = ord(i)-48
if(d>=10): d=ord(i)-65+10
#print(d)
s10 += d * oldbase**p
p+=1
if(b==100): return s10
#print("Base10",s10)
#base10 to newbase
n = s10
r = ""
while(n>0):
d=n%newbase
if(d>=10):
d = chr(65+d-10)
r+=str(d)
n//=newbase
r = list(reversed(r))
return ''.join(r)
'''
print(btb(4,5,321))
print(btb(14,12,'A'))
print(btb(10,16,282))
print(btb(16,10,'11A'))
'''
roman = {1:'I',4:'IV',5:'V',9:'IX',10:'X',40:'XL',50:'L',90:'XC',100:'C',400:'CD'
,500:'D',900:'CM',1000:'M'}
roman = dict(reversed(list(roman.items())))
roman3 = {900:'CM',4:'IV',9:'IX',40:'XL',90:'XC',400:'CD',100:'C'
,500:'D',5:'V',1:'I',10:'X',50:'L',1000:'M'}
#print(roman)
def DTR(dec):
r=""
while(dec>0):
for k,v in roman.items():
#print(dec)
if(dec>=k):
r+=v*(dec//k)
dec%=k
return r
#print(DTR(3549))
def RTD(R):
s10 = 0
for k,v in roman3.items():
#print(R,s10,k,v)
if(R.count(v)>0):
s10+=R.count(v)*k
R = R.replace(v,'')
return s10
#print(RTD("MMMDXLIX"))
if(b=='R'):
#print(btb(a,100,c))
print(DTR(int(btb(a,100,c))))
else:
print(btb(a,int(b),c))
``` | instruction | 0 | 66,845 | 8 | 133,690 |
No | output | 1 | 66,845 | 8 | 133,691 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's now 260 AD. Shapur, being extremely smart, became the King of Persia. He is now called Shapur, His majesty King of kings of Iran and Aniran.
Recently the Romans declared war on Persia. They dreamed to occupy Armenia. In the recent war, the Romans were badly defeated. Now their senior army general, Philip is captured by Shapur and Shapur is now going to capture Valerian, the Roman emperor.
Being defeated, the cowardly Valerian hid in a room at the top of one of his castles. To capture him, Shapur has to open many doors. Fortunately Valerian was too scared to make impenetrable locks for the doors.
Each door has 4 parts. The first part is an integer number a. The second part is either an integer number b or some really odd sign which looks like R. The third one is an integer c and the fourth part is empty! As if it was laid for writing something. Being extremely gifted, after opening the first few doors, Shapur found out the secret behind the locks.
c is an integer written in base a, to open the door we should write it in base b. The only bad news is that this R is some sort of special numbering system that is used only in Roman empire, so opening the doors is not just a piece of cake!
Here's an explanation of this really weird number system that even doesn't have zero:
Roman numerals are based on seven symbols: a stroke (identified with the letter I) for a unit, a chevron (identified with the letter V) for a five, a cross-stroke (identified with the letter X) for a ten, a C (identified as an abbreviation of Centum) for a hundred, etc.:
* I=1
* V=5
* X=10
* L=50
* C=100
* D=500
* M=1000
Symbols are iterated to produce multiples of the decimal (1, 10, 100, 1, 000) values, with V, L, D substituted for a multiple of five, and the iteration continuing: I 1, II 2, III 3, V 5, VI 6, VII 7, etc., and the same for other bases: X 10, XX 20, XXX 30, L 50, LXXX 80; CC 200, DCC 700, etc. At the fourth and ninth iteration, a subtractive principle must be employed, with the base placed before the higher base: IV 4, IX 9, XL 40, XC 90, CD 400, CM 900.
Also in bases greater than 10 we use A for 10, B for 11, etc.
Help Shapur capture Valerian and bring peace back to Persia, especially Armenia.
Input
The first line contains two integers a and b (2 โค a, b โค 25). Only b may be replaced by an R which indicates Roman numbering system.
The next line contains a single non-negative integer c in base a which may contain leading zeros but its length doesn't exceed 103.
It is guaranteed that if we have Roman numerals included the number would be less than or equal to 300010 and it won't be 0. In any other case the number won't be greater than 101510.
Output
Write a single line that contains integer c in base b. You must omit leading zeros.
Examples
Input
10 2
1
Output
1
Input
16 R
5
Output
V
Input
5 R
4
Output
IV
Input
2 2
1111001
Output
1111001
Input
12 13
A
Output
A
Note
You can find more information about roman numerals here: http://en.wikipedia.org/wiki/Roman_numerals
Submitted Solution:
```
import sys
from array import array # noqa: F401
def input():
return sys.stdin.buffer.readline().decode('utf-8')
a, b = input().split()
a = int(a)
num = int(input().rstrip(), a)
if b != 'R':
b = int(b)
ans = []
while num:
rem = num % b
if rem >= 10:
ans.append(chr(65 + rem - 10))
else:
ans.append(str(rem))
num //= b
print(*ans[::-1], sep='')
else:
ans = []
ans.append('M' * (num // 1000))
num %= 1000
if num >= 500:
if num >= 900:
ans.append('CM')
num -= 900
else:
ans.append('D' + 'C' * ((num - 500) // 100))
num %= 100
if num >= 100:
if num >= 400:
ans.append('CD')
num -= 400
else:
ans.append('C' * (num // 100))
num %= 100
if num >= 50:
if num >= 90:
ans.append('XC')
num -= 90
else:
ans.append('L' + 'X' * ((num - 50) // 10))
num %= 10
if num >= 10:
if num >= 40:
ans.append('XL')
num -= 40
else:
ans.append('X' * (num // 10))
num %= 10
if num >= 5:
if num >= 9:
ans.append('IX')
num -= 9
else:
ans.append('V' + 'I' * (num - 5))
num = 0
if num > 0:
if num == 4:
ans.append('IV')
else:
ans.append('I' * num)
print(*ans, sep='')
``` | instruction | 0 | 66,846 | 8 | 133,692 |
No | output | 1 | 66,846 | 8 | 133,693 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's now 260 AD. Shapur, being extremely smart, became the King of Persia. He is now called Shapur, His majesty King of kings of Iran and Aniran.
Recently the Romans declared war on Persia. They dreamed to occupy Armenia. In the recent war, the Romans were badly defeated. Now their senior army general, Philip is captured by Shapur and Shapur is now going to capture Valerian, the Roman emperor.
Being defeated, the cowardly Valerian hid in a room at the top of one of his castles. To capture him, Shapur has to open many doors. Fortunately Valerian was too scared to make impenetrable locks for the doors.
Each door has 4 parts. The first part is an integer number a. The second part is either an integer number b or some really odd sign which looks like R. The third one is an integer c and the fourth part is empty! As if it was laid for writing something. Being extremely gifted, after opening the first few doors, Shapur found out the secret behind the locks.
c is an integer written in base a, to open the door we should write it in base b. The only bad news is that this R is some sort of special numbering system that is used only in Roman empire, so opening the doors is not just a piece of cake!
Here's an explanation of this really weird number system that even doesn't have zero:
Roman numerals are based on seven symbols: a stroke (identified with the letter I) for a unit, a chevron (identified with the letter V) for a five, a cross-stroke (identified with the letter X) for a ten, a C (identified as an abbreviation of Centum) for a hundred, etc.:
* I=1
* V=5
* X=10
* L=50
* C=100
* D=500
* M=1000
Symbols are iterated to produce multiples of the decimal (1, 10, 100, 1, 000) values, with V, L, D substituted for a multiple of five, and the iteration continuing: I 1, II 2, III 3, V 5, VI 6, VII 7, etc., and the same for other bases: X 10, XX 20, XXX 30, L 50, LXXX 80; CC 200, DCC 700, etc. At the fourth and ninth iteration, a subtractive principle must be employed, with the base placed before the higher base: IV 4, IX 9, XL 40, XC 90, CD 400, CM 900.
Also in bases greater than 10 we use A for 10, B for 11, etc.
Help Shapur capture Valerian and bring peace back to Persia, especially Armenia.
Input
The first line contains two integers a and b (2 โค a, b โค 25). Only b may be replaced by an R which indicates Roman numbering system.
The next line contains a single non-negative integer c in base a which may contain leading zeros but its length doesn't exceed 103.
It is guaranteed that if we have Roman numerals included the number would be less than or equal to 300010 and it won't be 0. In any other case the number won't be greater than 101510.
Output
Write a single line that contains integer c in base b. You must omit leading zeros.
Examples
Input
10 2
1
Output
1
Input
16 R
5
Output
V
Input
5 R
4
Output
IV
Input
2 2
1111001
Output
1111001
Input
12 13
A
Output
A
Note
You can find more information about roman numerals here: http://en.wikipedia.org/wiki/Roman_numerals
Submitted Solution:
```
ab = input().split()
alphabet = ['A', 'B','C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z']
try:
a = int(ab[0])
except TypeError:
a = alphabet.index(ab[0])+10
b = ab[1]
if b!='R':
b = int(b)
c = input()
result = 0
for i in range(len(c)):
try:
result+=int(c[i])*(a**(len(c)-i-1))
except ValueError:
result+=(alphabet.index(c[i])+10)*(a**(len(c)-i-1))
strresult = ''
if b!='R':
while result!=0:
if result%b<10:
strresult = str(result%b) + strresult
else:
strresult = alphabet[result%b-10] + strresult
result = result//b
else:
while result!=0:
if result>=1000:
strresult+='M'
result-=1000
elif result>=900:
strresult+='CM'
result-=900
elif result>=500:
strresult+='D'
result-=500
elif result>=400:
strresult+='CD'
result-=400
elif result>=100:
strresult+='C'
result-=100
elif result>=90:
strresult+='XC'
result-=90
elif result>=50:
strresult+='L'
result-=50
elif result>=40:
strresult+='XL'
result-=40
elif result>=10:
strresult+='X'
result-=10
elif result>=9:
strresult+='IX'
result-=9
elif result>=5:
strresult+='V'
result-=5
elif result>=4:
strresult+='IV'
result-=4
else:
strresult+='I'
result-=1
print(strresult)
``` | instruction | 0 | 66,847 | 8 | 133,694 |
No | output | 1 | 66,847 | 8 | 133,695 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In a strategic computer game "Settlers II" one has to build defense structures to expand and protect the territory. Let's take one of these buildings. At the moment the defense structure accommodates exactly n soldiers. Within this task we can assume that the number of soldiers in the defense structure won't either increase or decrease.
Every soldier has a rank โ some natural number from 1 to k. 1 stands for a private and k stands for a general. The higher the rank of the soldier is, the better he fights. Therefore, the player profits from having the soldiers of the highest possible rank.
To increase the ranks of soldiers they need to train. But the soldiers won't train for free, and each training session requires one golden coin. On each training session all the n soldiers are present.
At the end of each training session the soldiers' ranks increase as follows. First all the soldiers are divided into groups with the same rank, so that the least possible number of groups is formed. Then, within each of the groups where the soldiers below the rank k are present, exactly one soldier increases his rank by one.
You know the ranks of all n soldiers at the moment. Determine the number of golden coins that are needed to increase the ranks of all the soldiers to the rank k.
Input
The first line contains two integers n and k (1 โค n, k โค 100). They represent the number of soldiers and the number of different ranks correspondingly. The second line contains n numbers in the non-decreasing order. The i-th of them, ai, represents the rank of the i-th soldier in the defense building (1 โค i โค n, 1 โค ai โค k).
Output
Print a single integer โ the number of golden coins needed to raise all the soldiers to the maximal rank.
Examples
Input
4 4
1 2 2 3
Output
4
Input
4 3
1 1 1 1
Output
5
Note
In the first example the ranks will be raised in the following manner:
1 2 2 3 โ 2 2 3 4 โ 2 3 4 4 โ 3 4 4 4 โ 4 4 4 4
Thus totals to 4 training sessions that require 4 golden coins. | instruction | 0 | 66,848 | 8 | 133,696 |
Tags: implementation
Correct Solution:
```
n,k=map(int,input().split())
a=[0]*(k+1)
for x in list(map(int,input().split())):
a[x]+=1
ans=0
while True:
z=0
for i in range(k-1,0,-1):
if a[i]==0:
z+=1
continue
else:
a[i+1]+=1
a[i]-=1
if z==k-1:break
ans+=1
print(ans)
``` | output | 1 | 66,848 | 8 | 133,697 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In a strategic computer game "Settlers II" one has to build defense structures to expand and protect the territory. Let's take one of these buildings. At the moment the defense structure accommodates exactly n soldiers. Within this task we can assume that the number of soldiers in the defense structure won't either increase or decrease.
Every soldier has a rank โ some natural number from 1 to k. 1 stands for a private and k stands for a general. The higher the rank of the soldier is, the better he fights. Therefore, the player profits from having the soldiers of the highest possible rank.
To increase the ranks of soldiers they need to train. But the soldiers won't train for free, and each training session requires one golden coin. On each training session all the n soldiers are present.
At the end of each training session the soldiers' ranks increase as follows. First all the soldiers are divided into groups with the same rank, so that the least possible number of groups is formed. Then, within each of the groups where the soldiers below the rank k are present, exactly one soldier increases his rank by one.
You know the ranks of all n soldiers at the moment. Determine the number of golden coins that are needed to increase the ranks of all the soldiers to the rank k.
Input
The first line contains two integers n and k (1 โค n, k โค 100). They represent the number of soldiers and the number of different ranks correspondingly. The second line contains n numbers in the non-decreasing order. The i-th of them, ai, represents the rank of the i-th soldier in the defense building (1 โค i โค n, 1 โค ai โค k).
Output
Print a single integer โ the number of golden coins needed to raise all the soldiers to the maximal rank.
Examples
Input
4 4
1 2 2 3
Output
4
Input
4 3
1 1 1 1
Output
5
Note
In the first example the ranks will be raised in the following manner:
1 2 2 3 โ 2 2 3 4 โ 2 3 4 4 โ 3 4 4 4 โ 4 4 4 4
Thus totals to 4 training sessions that require 4 golden coins. | instruction | 0 | 66,849 | 8 | 133,698 |
Tags: implementation
Correct Solution:
```
n,k = map(int,input().split())
l = list(map(int,input().split()))
count = 0
while l!=[k]*n:
c = -1
for i in range(n):
# print(l)
if l[i]!=c and l[i]<k:
c = l[i]
l[i]+=1
l.sort()
count+=1
print(count)
``` | output | 1 | 66,849 | 8 | 133,699 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In a strategic computer game "Settlers II" one has to build defense structures to expand and protect the territory. Let's take one of these buildings. At the moment the defense structure accommodates exactly n soldiers. Within this task we can assume that the number of soldiers in the defense structure won't either increase or decrease.
Every soldier has a rank โ some natural number from 1 to k. 1 stands for a private and k stands for a general. The higher the rank of the soldier is, the better he fights. Therefore, the player profits from having the soldiers of the highest possible rank.
To increase the ranks of soldiers they need to train. But the soldiers won't train for free, and each training session requires one golden coin. On each training session all the n soldiers are present.
At the end of each training session the soldiers' ranks increase as follows. First all the soldiers are divided into groups with the same rank, so that the least possible number of groups is formed. Then, within each of the groups where the soldiers below the rank k are present, exactly one soldier increases his rank by one.
You know the ranks of all n soldiers at the moment. Determine the number of golden coins that are needed to increase the ranks of all the soldiers to the rank k.
Input
The first line contains two integers n and k (1 โค n, k โค 100). They represent the number of soldiers and the number of different ranks correspondingly. The second line contains n numbers in the non-decreasing order. The i-th of them, ai, represents the rank of the i-th soldier in the defense building (1 โค i โค n, 1 โค ai โค k).
Output
Print a single integer โ the number of golden coins needed to raise all the soldiers to the maximal rank.
Examples
Input
4 4
1 2 2 3
Output
4
Input
4 3
1 1 1 1
Output
5
Note
In the first example the ranks will be raised in the following manner:
1 2 2 3 โ 2 2 3 4 โ 2 3 4 4 โ 3 4 4 4 โ 4 4 4 4
Thus totals to 4 training sessions that require 4 golden coins. | instruction | 0 | 66,851 | 8 | 133,702 |
Tags: implementation
Correct Solution:
```
n,k=map(int,input().split())
arr=list(map(int,input().split()))
ans=0
while(True):
flag=0
for j in range(0,n-1):
if(arr[j]!=arr[j+1] and arr[j]<k):
arr[j]+=1
flag=1
if(arr[n-1]<k):
flag=1
arr[n-1]+=1
if(flag==0):
break
ans+=1
print(ans)
``` | output | 1 | 66,851 | 8 | 133,703 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In a strategic computer game "Settlers II" one has to build defense structures to expand and protect the territory. Let's take one of these buildings. At the moment the defense structure accommodates exactly n soldiers. Within this task we can assume that the number of soldiers in the defense structure won't either increase or decrease.
Every soldier has a rank โ some natural number from 1 to k. 1 stands for a private and k stands for a general. The higher the rank of the soldier is, the better he fights. Therefore, the player profits from having the soldiers of the highest possible rank.
To increase the ranks of soldiers they need to train. But the soldiers won't train for free, and each training session requires one golden coin. On each training session all the n soldiers are present.
At the end of each training session the soldiers' ranks increase as follows. First all the soldiers are divided into groups with the same rank, so that the least possible number of groups is formed. Then, within each of the groups where the soldiers below the rank k are present, exactly one soldier increases his rank by one.
You know the ranks of all n soldiers at the moment. Determine the number of golden coins that are needed to increase the ranks of all the soldiers to the rank k.
Input
The first line contains two integers n and k (1 โค n, k โค 100). They represent the number of soldiers and the number of different ranks correspondingly. The second line contains n numbers in the non-decreasing order. The i-th of them, ai, represents the rank of the i-th soldier in the defense building (1 โค i โค n, 1 โค ai โค k).
Output
Print a single integer โ the number of golden coins needed to raise all the soldiers to the maximal rank.
Examples
Input
4 4
1 2 2 3
Output
4
Input
4 3
1 1 1 1
Output
5
Note
In the first example the ranks will be raised in the following manner:
1 2 2 3 โ 2 2 3 4 โ 2 3 4 4 โ 3 4 4 4 โ 4 4 4 4
Thus totals to 4 training sessions that require 4 golden coins. | instruction | 0 | 66,852 | 8 | 133,704 |
Tags: implementation
Correct Solution:
```
n,k=[int(i) for i in input().split()]
a=list(map(int,input().split()))[:n]
bh=0
while a.count(k)!=len(a):
for i in set(a):
if i!=k:
x=a.index(i)
a[x]=a[x]+1
bh=bh+1
print(bh)
``` | output | 1 | 66,852 | 8 | 133,705 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In a strategic computer game "Settlers II" one has to build defense structures to expand and protect the territory. Let's take one of these buildings. At the moment the defense structure accommodates exactly n soldiers. Within this task we can assume that the number of soldiers in the defense structure won't either increase or decrease.
Every soldier has a rank โ some natural number from 1 to k. 1 stands for a private and k stands for a general. The higher the rank of the soldier is, the better he fights. Therefore, the player profits from having the soldiers of the highest possible rank.
To increase the ranks of soldiers they need to train. But the soldiers won't train for free, and each training session requires one golden coin. On each training session all the n soldiers are present.
At the end of each training session the soldiers' ranks increase as follows. First all the soldiers are divided into groups with the same rank, so that the least possible number of groups is formed. Then, within each of the groups where the soldiers below the rank k are present, exactly one soldier increases his rank by one.
You know the ranks of all n soldiers at the moment. Determine the number of golden coins that are needed to increase the ranks of all the soldiers to the rank k.
Input
The first line contains two integers n and k (1 โค n, k โค 100). They represent the number of soldiers and the number of different ranks correspondingly. The second line contains n numbers in the non-decreasing order. The i-th of them, ai, represents the rank of the i-th soldier in the defense building (1 โค i โค n, 1 โค ai โค k).
Output
Print a single integer โ the number of golden coins needed to raise all the soldiers to the maximal rank.
Examples
Input
4 4
1 2 2 3
Output
4
Input
4 3
1 1 1 1
Output
5
Note
In the first example the ranks will be raised in the following manner:
1 2 2 3 โ 2 2 3 4 โ 2 3 4 4 โ 3 4 4 4 โ 4 4 4 4
Thus totals to 4 training sessions that require 4 golden coins. | instruction | 0 | 66,853 | 8 | 133,706 |
Tags: implementation
Correct Solution:
```
from collections import defaultdict as df
n,k=list(map(int,input().split()))
a=list(map(int,input().rstrip().split()))
ans=0
maxi=a.count(k)
while(a.count(k)!=n):
d=df(list)
for i in range(n):
if a[i]!=k:
d[a[i]].append(i)
for i in d:
a[d[i][0]]+=1
ans+=1
print(ans)
``` | output | 1 | 66,853 | 8 | 133,707 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In a strategic computer game "Settlers II" one has to build defense structures to expand and protect the territory. Let's take one of these buildings. At the moment the defense structure accommodates exactly n soldiers. Within this task we can assume that the number of soldiers in the defense structure won't either increase or decrease.
Every soldier has a rank โ some natural number from 1 to k. 1 stands for a private and k stands for a general. The higher the rank of the soldier is, the better he fights. Therefore, the player profits from having the soldiers of the highest possible rank.
To increase the ranks of soldiers they need to train. But the soldiers won't train for free, and each training session requires one golden coin. On each training session all the n soldiers are present.
At the end of each training session the soldiers' ranks increase as follows. First all the soldiers are divided into groups with the same rank, so that the least possible number of groups is formed. Then, within each of the groups where the soldiers below the rank k are present, exactly one soldier increases his rank by one.
You know the ranks of all n soldiers at the moment. Determine the number of golden coins that are needed to increase the ranks of all the soldiers to the rank k.
Input
The first line contains two integers n and k (1 โค n, k โค 100). They represent the number of soldiers and the number of different ranks correspondingly. The second line contains n numbers in the non-decreasing order. The i-th of them, ai, represents the rank of the i-th soldier in the defense building (1 โค i โค n, 1 โค ai โค k).
Output
Print a single integer โ the number of golden coins needed to raise all the soldiers to the maximal rank.
Examples
Input
4 4
1 2 2 3
Output
4
Input
4 3
1 1 1 1
Output
5
Note
In the first example the ranks will be raised in the following manner:
1 2 2 3 โ 2 2 3 4 โ 2 3 4 4 โ 3 4 4 4 โ 4 4 4 4
Thus totals to 4 training sessions that require 4 golden coins. | instruction | 0 | 66,854 | 8 | 133,708 |
Tags: implementation
Correct Solution:
```
import sys
import math
n, k = [int(x) for x in (sys.stdin.readline()).split()]
ai = [int(x) for x in (sys.stdin.readline()).split()]
z = n
res = 0
while(z > 0):
i = 0
f = False
while(i < z):
if(ai[i] == k):
z -= 1
break
else:
f = True
v = ai[i]
j = i
while(j < z):
if(v != ai[j]):
break
j += 1
i = j
ai[j - 1] += 1
if(f):
res += 1
print(res)
``` | output | 1 | 66,854 | 8 | 133,709 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In a strategic computer game "Settlers II" one has to build defense structures to expand and protect the territory. Let's take one of these buildings. At the moment the defense structure accommodates exactly n soldiers. Within this task we can assume that the number of soldiers in the defense structure won't either increase or decrease.
Every soldier has a rank โ some natural number from 1 to k. 1 stands for a private and k stands for a general. The higher the rank of the soldier is, the better he fights. Therefore, the player profits from having the soldiers of the highest possible rank.
To increase the ranks of soldiers they need to train. But the soldiers won't train for free, and each training session requires one golden coin. On each training session all the n soldiers are present.
At the end of each training session the soldiers' ranks increase as follows. First all the soldiers are divided into groups with the same rank, so that the least possible number of groups is formed. Then, within each of the groups where the soldiers below the rank k are present, exactly one soldier increases his rank by one.
You know the ranks of all n soldiers at the moment. Determine the number of golden coins that are needed to increase the ranks of all the soldiers to the rank k.
Input
The first line contains two integers n and k (1 โค n, k โค 100). They represent the number of soldiers and the number of different ranks correspondingly. The second line contains n numbers in the non-decreasing order. The i-th of them, ai, represents the rank of the i-th soldier in the defense building (1 โค i โค n, 1 โค ai โค k).
Output
Print a single integer โ the number of golden coins needed to raise all the soldiers to the maximal rank.
Examples
Input
4 4
1 2 2 3
Output
4
Input
4 3
1 1 1 1
Output
5
Note
In the first example the ranks will be raised in the following manner:
1 2 2 3 โ 2 2 3 4 โ 2 3 4 4 โ 3 4 4 4 โ 4 4 4 4
Thus totals to 4 training sessions that require 4 golden coins. | instruction | 0 | 66,855 | 8 | 133,710 |
Tags: implementation
Correct Solution:
```
def solve(n, k, soldiers):
l = [0] * 101
for s in soldiers:
l[s] += 1
blocking = 0
result = 0
for i in range(k-1, -1, -1):
num_soldier = l[i]
if num_soldier == 0:
blocking = max(0, blocking - 1)
else:
blocking += num_soldier - 1
result = blocking + k - i
return result
def test():
parse = lambda st: map(int, st.split())
n, k = 4, 4
s = parse('1 2 2 3')
assert solve(n, k, s) == 4
n, k = 4, 3
s = parse('1 1 1 1')
assert solve(n, k, s) == 5
n, k = 23, 2
s = parse('1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2')
assert solve(n, k, s) == 12
if __name__ == '__main__':
# test()
n, k = list(map(int, input().split()))
soldiers = map(int, input().split())
print(solve(n, k, soldiers))
``` | output | 1 | 66,855 | 8 | 133,711 |
Provide a correct Python 3 solution for this coding contest problem.
There is a building with n rooms, numbered 1 to n.
We can move from any room to any other room in the building.
Let us call the following event a move: a person in some room i goes to another room j~ (i \neq j).
Initially, there was one person in each room in the building.
After that, we know that there were exactly k moves happened up to now.
We are interested in the number of people in each of the n rooms now. How many combinations of numbers of people in the n rooms are possible?
Find the count modulo (10^9 + 7).
Constraints
* All values in input are integers.
* 3 \leq n \leq 2 \times 10^5
* 2 \leq k \leq 10^9
Input
Input is given from Standard Input in the following format:
n k
Output
Print the number of possible combinations of numbers of people in the n rooms now, modulo (10^9 + 7).
Examples
Input
3 2
Output
10
Input
200000 1000000000
Output
607923868
Input
15 6
Output
22583772 | instruction | 0 | 67,005 | 8 | 134,010 |
"Correct Solution:
```
M=10**9+7
n,k=map(int,input().split())
l=[1]
f=1
for i in range(n):
f=f*(i+1)%M
l+=[f]
a=0
for i in range(min(n,k+1)):
c=l[n]*l[n-1]%M
p=l[i]**2*l[n-i]*l[n-i-1]%M
a+=c*pow(p,M-2,M)
print(a%M)
``` | output | 1 | 67,005 | 8 | 134,011 |
Provide a correct Python 3 solution for this coding contest problem.
There is a building with n rooms, numbered 1 to n.
We can move from any room to any other room in the building.
Let us call the following event a move: a person in some room i goes to another room j~ (i \neq j).
Initially, there was one person in each room in the building.
After that, we know that there were exactly k moves happened up to now.
We are interested in the number of people in each of the n rooms now. How many combinations of numbers of people in the n rooms are possible?
Find the count modulo (10^9 + 7).
Constraints
* All values in input are integers.
* 3 \leq n \leq 2 \times 10^5
* 2 \leq k \leq 10^9
Input
Input is given from Standard Input in the following format:
n k
Output
Print the number of possible combinations of numbers of people in the n rooms now, modulo (10^9 + 7).
Examples
Input
3 2
Output
10
Input
200000 1000000000
Output
607923868
Input
15 6
Output
22583772 | instruction | 0 | 67,006 | 8 | 134,012 |
"Correct Solution:
```
n, k = map(int, input().split())
def cmb(n, r, mod):
if (r < 0 or r > n):
return 0
r = min(r, n-r)
return g1[n] * g2[r] * g2[n-r] % mod
mod = 10**9+7 # ๅบๅใฎๅถ้
N = 2*10**5+1
g1 = [1, 1] # ๅ
ใใผใใซ
g2 = [1, 1] # ้ๅ
ใใผใใซ
inverse = [0, 1] # ้ๅ
ใใผใใซ่จ็ฎ็จใใผใใซ
for i in range(2, N + 1):
g1.append((g1[-1] * i) % mod)
inverse.append((-inverse[mod % i] * (mod//i)) % mod)
g2.append((g2[-1] * inverse[-1]) % mod)
ans = 0
for i in range(min(k+1, n)):
ans += cmb(n, i, mod)*cmb(n-1, n-i-1, mod)
ans %= mod
print(ans)
``` | output | 1 | 67,006 | 8 | 134,013 |
Provide a correct Python 3 solution for this coding contest problem.
There is a building with n rooms, numbered 1 to n.
We can move from any room to any other room in the building.
Let us call the following event a move: a person in some room i goes to another room j~ (i \neq j).
Initially, there was one person in each room in the building.
After that, we know that there were exactly k moves happened up to now.
We are interested in the number of people in each of the n rooms now. How many combinations of numbers of people in the n rooms are possible?
Find the count modulo (10^9 + 7).
Constraints
* All values in input are integers.
* 3 \leq n \leq 2 \times 10^5
* 2 \leq k \leq 10^9
Input
Input is given from Standard Input in the following format:
n k
Output
Print the number of possible combinations of numbers of people in the n rooms now, modulo (10^9 + 7).
Examples
Input
3 2
Output
10
Input
200000 1000000000
Output
607923868
Input
15 6
Output
22583772 | instruction | 0 | 67,007 | 8 | 134,014 |
"Correct Solution:
```
def comb(n, r):
if r<0 or r>n:
return 0
r=min(r, n-r)
return g1[n]*g2[r]*g2[n-r]%MOD
MOD=10**9+7
MAXN=400000+10
g1=[1, 1]
g2=[1, 1]
inverse=[0, 1]
for i in range(2, MAXN+1):
g1.append((g1[-1]*i)%MOD)
inverse.append(-inverse[MOD%i]*(MOD//i)%MOD)
g2.append((g2[-1]*inverse[-1])%MOD)
N, K=map(int, input().split())
if K>=N-1:
print(comb(N*2-1, N))
else:
ans=1
for i in range(1, K+1):
ans=(ans+(comb(N-1, i)*comb(N, i))%MOD)%MOD
print(ans)
``` | output | 1 | 67,007 | 8 | 134,015 |
Provide a correct Python 3 solution for this coding contest problem.
There is a building with n rooms, numbered 1 to n.
We can move from any room to any other room in the building.
Let us call the following event a move: a person in some room i goes to another room j~ (i \neq j).
Initially, there was one person in each room in the building.
After that, we know that there were exactly k moves happened up to now.
We are interested in the number of people in each of the n rooms now. How many combinations of numbers of people in the n rooms are possible?
Find the count modulo (10^9 + 7).
Constraints
* All values in input are integers.
* 3 \leq n \leq 2 \times 10^5
* 2 \leq k \leq 10^9
Input
Input is given from Standard Input in the following format:
n k
Output
Print the number of possible combinations of numbers of people in the n rooms now, modulo (10^9 + 7).
Examples
Input
3 2
Output
10
Input
200000 1000000000
Output
607923868
Input
15 6
Output
22583772 | instruction | 0 | 67,008 | 8 | 134,016 |
"Correct Solution:
```
def main():
n, kk = map(int, input().split())
mod = 10**9+7
fact = [1, 1]
for i in range(2, 2*10**5+1):
fact.append(fact[-1]*i % mod)
def nCr(n, r, mod=10**9+7):
return pow(fact[n-r]*fact[r] % mod, mod-2, mod)*fact[n] % mod
ans = [1]
for k in range(1, n):
ans.append(nCr(n-1, k)*nCr(n, k) % mod)
print(sum(ans[:min(kk+1, n)]) % mod)
main()
``` | output | 1 | 67,008 | 8 | 134,017 |
Provide a correct Python 3 solution for this coding contest problem.
There is a building with n rooms, numbered 1 to n.
We can move from any room to any other room in the building.
Let us call the following event a move: a person in some room i goes to another room j~ (i \neq j).
Initially, there was one person in each room in the building.
After that, we know that there were exactly k moves happened up to now.
We are interested in the number of people in each of the n rooms now. How many combinations of numbers of people in the n rooms are possible?
Find the count modulo (10^9 + 7).
Constraints
* All values in input are integers.
* 3 \leq n \leq 2 \times 10^5
* 2 \leq k \leq 10^9
Input
Input is given from Standard Input in the following format:
n k
Output
Print the number of possible combinations of numbers of people in the n rooms now, modulo (10^9 + 7).
Examples
Input
3 2
Output
10
Input
200000 1000000000
Output
607923868
Input
15 6
Output
22583772 | instruction | 0 | 67,009 | 8 | 134,018 |
"Correct Solution:
```
n,k = map(int,input().split())
MOD = 10**9+7
FAC = [1]
INV = [1]
for i in range(1,2*n+1):
FAC.append((FAC[i-1]*i) % MOD)
INV.append(pow(FAC[-1],MOD-2,MOD))
def nCr(n,r):
return FAC[n]*INV[n-r]*INV[r]
ans = 0
for i in range(min(n-1,k)+1):
ans += nCr(n,i)*nCr(n-1,n-i-1)
ans %= MOD
print(ans)
``` | output | 1 | 67,009 | 8 | 134,019 |
Provide a correct Python 3 solution for this coding contest problem.
There is a building with n rooms, numbered 1 to n.
We can move from any room to any other room in the building.
Let us call the following event a move: a person in some room i goes to another room j~ (i \neq j).
Initially, there was one person in each room in the building.
After that, we know that there were exactly k moves happened up to now.
We are interested in the number of people in each of the n rooms now. How many combinations of numbers of people in the n rooms are possible?
Find the count modulo (10^9 + 7).
Constraints
* All values in input are integers.
* 3 \leq n \leq 2 \times 10^5
* 2 \leq k \leq 10^9
Input
Input is given from Standard Input in the following format:
n k
Output
Print the number of possible combinations of numbers of people in the n rooms now, modulo (10^9 + 7).
Examples
Input
3 2
Output
10
Input
200000 1000000000
Output
607923868
Input
15 6
Output
22583772 | instruction | 0 | 67,010 | 8 | 134,020 |
"Correct Solution:
```
MOD = 10**9 + 7
fac = [1 for k in range(200010)]
inv = [1 for k in range(200010)]
finv = [1 for k in range(200010)]
for k in range(2,200010):
fac[k] = (fac[k-1]*k)%MOD
inv[k] = (MOD - inv[MOD%k] * (MOD // k))%MOD
finv[k] = (finv[k - 1] * inv[k]) % MOD;
def nCr(n,r):
return (fac[n]*finv[r]*finv[n-r])%MOD
n, k = map(int,input().split())
ans = 0
for z in range(min(n,k+1)):
ans += nCr(n,z)*nCr(n-1,n-z-1)
ans %= MOD
print(ans%MOD)
``` | output | 1 | 67,010 | 8 | 134,021 |
Provide a correct Python 3 solution for this coding contest problem.
There is a building with n rooms, numbered 1 to n.
We can move from any room to any other room in the building.
Let us call the following event a move: a person in some room i goes to another room j~ (i \neq j).
Initially, there was one person in each room in the building.
After that, we know that there were exactly k moves happened up to now.
We are interested in the number of people in each of the n rooms now. How many combinations of numbers of people in the n rooms are possible?
Find the count modulo (10^9 + 7).
Constraints
* All values in input are integers.
* 3 \leq n \leq 2 \times 10^5
* 2 \leq k \leq 10^9
Input
Input is given from Standard Input in the following format:
n k
Output
Print the number of possible combinations of numbers of people in the n rooms now, modulo (10^9 + 7).
Examples
Input
3 2
Output
10
Input
200000 1000000000
Output
607923868
Input
15 6
Output
22583772 | instruction | 0 | 67,011 | 8 | 134,022 |
"Correct Solution:
```
n, k = map(int, input().split())
mod = 10**9+7
fact = [0] * (n+1)
inv = [0] * (n+1)
fact[0], fact[1] = 1, 1
inv[0], inv[1] = 1, 1
for i in range(2,n+1):
fact[i] = (fact[i-1] * i) % mod
for i in range(2,n+1):
inv[i] = (inv[i-1] * pow(i, mod-2, mod)) % mod
if k >= n:
k = n-1
def C(a,b):
out = fact[a]
out *= inv[b]
out %= mod
out *= inv[a-b]
out %= mod
return out
ans = 0
for m in range(k+1):
ans += (C(n,m) * C(n-1,n-m-1)) % mod
ans %= mod
print(ans)
``` | output | 1 | 67,011 | 8 | 134,023 |
Provide a correct Python 3 solution for this coding contest problem.
There is a building with n rooms, numbered 1 to n.
We can move from any room to any other room in the building.
Let us call the following event a move: a person in some room i goes to another room j~ (i \neq j).
Initially, there was one person in each room in the building.
After that, we know that there were exactly k moves happened up to now.
We are interested in the number of people in each of the n rooms now. How many combinations of numbers of people in the n rooms are possible?
Find the count modulo (10^9 + 7).
Constraints
* All values in input are integers.
* 3 \leq n \leq 2 \times 10^5
* 2 \leq k \leq 10^9
Input
Input is given from Standard Input in the following format:
n k
Output
Print the number of possible combinations of numbers of people in the n rooms now, modulo (10^9 + 7).
Examples
Input
3 2
Output
10
Input
200000 1000000000
Output
607923868
Input
15 6
Output
22583772 | instruction | 0 | 67,012 | 8 | 134,024 |
"Correct Solution:
```
N,K = map(int, input().split())
def cmb(n, r, mod):
if ( r<0 or r>n ):
return 0
r = min(r, n-r)
return g1[n] * g2[r] * g2[n-r] % mod
mod = 10**9+7 #ๅบๅใฎๅถ้
g1 = [1, 1] # ๅ
ใใผใใซ
g2 = [1, 1] #้ๅ
ใใผใใซ
inverse = [0, 1] #้ๅ
ใใผใใซ่จ็ฎ็จใใผใใซ
for i in range( 2, 2*N + 1 ):
g1.append( ( g1[-1] * i ) % mod )
inverse.append( ( -inverse[mod % i] * (mod//i) ) % mod )
g2.append( (g2[-1] * inverse[-1]) % mod )
if K >= N-1:
ans = cmb(2*N-1, N, mod)
else:
ans = 0
for i in range(K+1):
ans += cmb(N, K-i, mod)*cmb(N-1, K-i, mod)
ans %= mod
print(ans)
``` | output | 1 | 67,012 | 8 | 134,025 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a building with n rooms, numbered 1 to n.
We can move from any room to any other room in the building.
Let us call the following event a move: a person in some room i goes to another room j~ (i \neq j).
Initially, there was one person in each room in the building.
After that, we know that there were exactly k moves happened up to now.
We are interested in the number of people in each of the n rooms now. How many combinations of numbers of people in the n rooms are possible?
Find the count modulo (10^9 + 7).
Constraints
* All values in input are integers.
* 3 \leq n \leq 2 \times 10^5
* 2 \leq k \leq 10^9
Input
Input is given from Standard Input in the following format:
n k
Output
Print the number of possible combinations of numbers of people in the n rooms now, modulo (10^9 + 7).
Examples
Input
3 2
Output
10
Input
200000 1000000000
Output
607923868
Input
15 6
Output
22583772
Submitted Solution:
```
def nCr_frL(n,r,mod):
ret=[1,n%mod]
for i in range(2,r+1):
inv=pow(i,mod-2,mod)
ret.append((ret[-1]*(n-i+1)*inv)%mod)
return ret
N,K=map(int,input().split())
MOD=10**9+7
com1=nCr_frL(N,N,MOD) # nCi
com2=nCr_frL(N-1,N-1,MOD) # n-1Ci
if K>N-1:
K=N-1
ans=0
for m in range(K+1):
ans+=(com1[m]*com2[N-m-1])%MOD
print(ans%MOD)
``` | instruction | 0 | 67,013 | 8 | 134,026 |
Yes | output | 1 | 67,013 | 8 | 134,027 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a building with n rooms, numbered 1 to n.
We can move from any room to any other room in the building.
Let us call the following event a move: a person in some room i goes to another room j~ (i \neq j).
Initially, there was one person in each room in the building.
After that, we know that there were exactly k moves happened up to now.
We are interested in the number of people in each of the n rooms now. How many combinations of numbers of people in the n rooms are possible?
Find the count modulo (10^9 + 7).
Constraints
* All values in input are integers.
* 3 \leq n \leq 2 \times 10^5
* 2 \leq k \leq 10^9
Input
Input is given from Standard Input in the following format:
n k
Output
Print the number of possible combinations of numbers of people in the n rooms now, modulo (10^9 + 7).
Examples
Input
3 2
Output
10
Input
200000 1000000000
Output
607923868
Input
15 6
Output
22583772
Submitted Solution:
```
n, k = map(int, input().split())
MOD = 10**9+7
def prepare(n, MOD):
facts = [1]*(n+1)
for i in range(1, n+1):
facts[i] = facts[i-1]*i%MOD
invs = [1]*(n+1)
_invs = [1]*(n+1)
invs[n] = pow(facts[n], MOD-2, MOD)
for i in range(0, n)[::-1]:
invs[i] = invs[i+1] * (i+1) % MOD
return facts, invs
ans = 0
facts, invs = prepare(n, MOD)
for i in range(1+min(n-1, k)):
ans += facts[n]*invs[i]*invs[n-i]*facts[n-1]*invs[n-i-1]*invs[i]
ans %= MOD
print(ans)
``` | instruction | 0 | 67,014 | 8 | 134,028 |
Yes | output | 1 | 67,014 | 8 | 134,029 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a building with n rooms, numbered 1 to n.
We can move from any room to any other room in the building.
Let us call the following event a move: a person in some room i goes to another room j~ (i \neq j).
Initially, there was one person in each room in the building.
After that, we know that there were exactly k moves happened up to now.
We are interested in the number of people in each of the n rooms now. How many combinations of numbers of people in the n rooms are possible?
Find the count modulo (10^9 + 7).
Constraints
* All values in input are integers.
* 3 \leq n \leq 2 \times 10^5
* 2 \leq k \leq 10^9
Input
Input is given from Standard Input in the following format:
n k
Output
Print the number of possible combinations of numbers of people in the n rooms now, modulo (10^9 + 7).
Examples
Input
3 2
Output
10
Input
200000 1000000000
Output
607923868
Input
15 6
Output
22583772
Submitted Solution:
```
from sys import stdin
#ๅ
ฅๅ
readline=stdin.readline
N,K=map(int,readline().split())
mod=10**9+7
fact=[1]*(2*N)
finv=[1]*(2*N)
inv=[1]*(2*N)
inv[0]=0
for i in range(2,2*N):
fact[i]=(fact[i-1]*i%mod)
inv[i]=(-inv[mod%i]*(mod//i))%mod
finv[i]=(finv[i-1]*inv[i])%mod
def com(N,K,mod):
if (K<0) or (N<K):
return 0
return fact[N]*finv[K]*finv[N-K]%mod
if N-1<=K:
print(com(2*N-1,N,mod))
else:
res=0
for m in range(K+1):
res+=com(N,m,mod)*com(N-1,m,mod)
res%=mod
print(res)
``` | instruction | 0 | 67,015 | 8 | 134,030 |
Yes | output | 1 | 67,015 | 8 | 134,031 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a building with n rooms, numbered 1 to n.
We can move from any room to any other room in the building.
Let us call the following event a move: a person in some room i goes to another room j~ (i \neq j).
Initially, there was one person in each room in the building.
After that, we know that there were exactly k moves happened up to now.
We are interested in the number of people in each of the n rooms now. How many combinations of numbers of people in the n rooms are possible?
Find the count modulo (10^9 + 7).
Constraints
* All values in input are integers.
* 3 \leq n \leq 2 \times 10^5
* 2 \leq k \leq 10^9
Input
Input is given from Standard Input in the following format:
n k
Output
Print the number of possible combinations of numbers of people in the n rooms now, modulo (10^9 + 7).
Examples
Input
3 2
Output
10
Input
200000 1000000000
Output
607923868
Input
15 6
Output
22583772
Submitted Solution:
```
def cmb(n, r, p):
if (r < 0) or (n < r):
return 0
r = min(r, n - r)
return fact[n] * factinv[r] * factinv[n - r] % p
p = 10 ** 9 + 7
N = 2 * 10 ** 5 # N ใฏๅฟ
่ฆๅใ ใ็จๆใใ
fact = [1, 1] # fact[n] = (n! mod p)
factinv = [1, 1] # factinv[n] = ((n!)^(-1) mod p)
inv = [0, 1] # factinv ่จ็ฎ็จ
for i in range(2, N + 1):
fact.append((fact[-1] * i) % p)
inv.append((-inv[p % i] * (p // i)) % p)
factinv.append((factinv[-1] * inv[-1]) % p)
n, k = map(int, input().split())
ans = 0
k = min(k, n - 1)
for i in range(k+1):
ans += cmb(n, i, p) * cmb(n - 1, i, p)
ans %= p
print(ans % p)
``` | instruction | 0 | 67,016 | 8 | 134,032 |
Yes | output | 1 | 67,016 | 8 | 134,033 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a building with n rooms, numbered 1 to n.
We can move from any room to any other room in the building.
Let us call the following event a move: a person in some room i goes to another room j~ (i \neq j).
Initially, there was one person in each room in the building.
After that, we know that there were exactly k moves happened up to now.
We are interested in the number of people in each of the n rooms now. How many combinations of numbers of people in the n rooms are possible?
Find the count modulo (10^9 + 7).
Constraints
* All values in input are integers.
* 3 \leq n \leq 2 \times 10^5
* 2 \leq k \leq 10^9
Input
Input is given from Standard Input in the following format:
n k
Output
Print the number of possible combinations of numbers of people in the n rooms now, modulo (10^9 + 7).
Examples
Input
3 2
Output
10
Input
200000 1000000000
Output
607923868
Input
15 6
Output
22583772
Submitted Solution:
```
n,k=map(int,input().split())
import sys
sys.setrecursionlimit(2147483647)
def cmb(n, r, p):
if (r < 0) or (n < r):
return 0
r = min(r, n - r)
return fact[n] * factinv[r] * factinv[n-r] % p
p = 10**9+7 #ๅฒใๆฐ
N = n+1 # N ใฏๅฟ
่ฆๅใ ใ็จๆใใ
fact = [1, 1] # fact[n] = (n! mod p)
factinv = [1, 1] # factinv[n] = ((n!)^(-1) mod p)
inv = [0, 1] # factinv ่จ็ฎ็จ
for i in range(2, N + 1):
fact.append((fact[-1] * i) % p)
inv.append((-inv[p % i] * (p // i)) % p)
factinv.append((factinv[-1] * inv[-1]) % p)
ans=1
for i in range(1,k+1):
ans+=(cmb(n, i, p)*cmb(n-1,n-i-1,p))%p
#print(i,cmb(n, i, p),cmb(n-1,n-i-1,p))
print(ans%p)
``` | instruction | 0 | 67,017 | 8 | 134,034 |
No | output | 1 | 67,017 | 8 | 134,035 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a building with n rooms, numbered 1 to n.
We can move from any room to any other room in the building.
Let us call the following event a move: a person in some room i goes to another room j~ (i \neq j).
Initially, there was one person in each room in the building.
After that, we know that there were exactly k moves happened up to now.
We are interested in the number of people in each of the n rooms now. How many combinations of numbers of people in the n rooms are possible?
Find the count modulo (10^9 + 7).
Constraints
* All values in input are integers.
* 3 \leq n \leq 2 \times 10^5
* 2 \leq k \leq 10^9
Input
Input is given from Standard Input in the following format:
n k
Output
Print the number of possible combinations of numbers of people in the n rooms now, modulo (10^9 + 7).
Examples
Input
3 2
Output
10
Input
200000 1000000000
Output
607923868
Input
15 6
Output
22583772
Submitted Solution:
```
# -*- coding: utf-8 -*-
n,k = map(int, input().split())
################
p = 10**9 + 7
def comb(n,k,p):
ans = 1
for i in range(1,k+1):
ans = ans * (n - (i-1)) % p
ans = ans * pow(i, p-2, p)
return ans
################
MAX = comb(2*n-1, n, p)%p
if k >= n:
print(MAX)
else:
ans = 0
for i in range(0,k+1):
temp = comb(n-1, n-i-1, p)%p
temp2 = comb(n, i, p)%p
temp = temp * temp2 %p
ans = (ans + temp ) %p
print(ans)
``` | instruction | 0 | 67,018 | 8 | 134,036 |
No | output | 1 | 67,018 | 8 | 134,037 |
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