message stringlengths 2 39.6k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 450 109k | cluster float64 2 2 | __index_level_0__ int64 900 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya plays The Elder Trolls IV: Oblivon. Oh, those creators of computer games! What they do not come up with! Absolutely unique monsters have been added to the The Elder Trolls IV: Oblivon. One of these monsters is Unkillable Slug. Why it is "Unkillable"? Firstly, because it can be killed with cutting weapon only, so lovers of two-handed amber hammers should find suitable knife themselves. Secondly, it is necessary to make so many cutting strokes to Unkillable Slug. Extremely many. Too many!
Vasya has already promoted his character to 80-th level and in order to gain level 81 he was asked to kill Unkillable Slug. The monster has a very interesting shape. It looks like a rectangular parallelepiped with size x × y × z, consisting of undestructable cells 1 × 1 × 1. At one stroke Vasya can cut the Slug along an imaginary grid, i.e. cut with a plane parallel to one of the parallelepiped side. Monster dies when amount of parts it is divided reaches some critical value.
All parts of monster do not fall after each cut, they remains exactly on its places. I. e. Vasya can cut several parts with one cut.
Vasya wants to know what the maximum number of pieces he can cut the Unkillable Slug into striking him at most k times.
Vasya's character uses absolutely thin sword with infinite length.
Input
The first line of input contains four integer numbers x, y, z, k (1 ≤ x, y, z ≤ 106, 0 ≤ k ≤ 109).
Output
Output the only number — the answer for the problem.
Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).
Examples
Input
2 2 2 3
Output
8
Input
2 2 2 1
Output
2
Note
In the first sample Vasya make 3 pairwise perpendicular cuts. He cuts monster on two parts with the first cut, then he divides each part on two with the second cut, and finally he divides each of the 4 parts on two. | instruction | 0 | 30,584 | 2 | 61,168 |
Tags: greedy, math
Correct Solution:
```
xyzk = [int(i) for i in input().split()]
k=xyzk[3]
x, y, z = sorted([xyzk[0],xyzk[1],xyzk[2]])
x = min(k // 3 + 1, x)
y = min((k - x + 1) // 2 + 1, y)
z = min(k - x - y + 3, z)
print(x * y * z)
``` | output | 1 | 30,584 | 2 | 61,169 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya plays The Elder Trolls IV: Oblivon. Oh, those creators of computer games! What they do not come up with! Absolutely unique monsters have been added to the The Elder Trolls IV: Oblivon. One of these monsters is Unkillable Slug. Why it is "Unkillable"? Firstly, because it can be killed with cutting weapon only, so lovers of two-handed amber hammers should find suitable knife themselves. Secondly, it is necessary to make so many cutting strokes to Unkillable Slug. Extremely many. Too many!
Vasya has already promoted his character to 80-th level and in order to gain level 81 he was asked to kill Unkillable Slug. The monster has a very interesting shape. It looks like a rectangular parallelepiped with size x × y × z, consisting of undestructable cells 1 × 1 × 1. At one stroke Vasya can cut the Slug along an imaginary grid, i.e. cut with a plane parallel to one of the parallelepiped side. Monster dies when amount of parts it is divided reaches some critical value.
All parts of monster do not fall after each cut, they remains exactly on its places. I. e. Vasya can cut several parts with one cut.
Vasya wants to know what the maximum number of pieces he can cut the Unkillable Slug into striking him at most k times.
Vasya's character uses absolutely thin sword with infinite length.
Input
The first line of input contains four integer numbers x, y, z, k (1 ≤ x, y, z ≤ 106, 0 ≤ k ≤ 109).
Output
Output the only number — the answer for the problem.
Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).
Examples
Input
2 2 2 3
Output
8
Input
2 2 2 1
Output
2
Note
In the first sample Vasya make 3 pairwise perpendicular cuts. He cuts monster on two parts with the first cut, then he divides each part on two with the second cut, and finally he divides each of the 4 parts on two.
Submitted Solution:
```
x, y, z, k = [int(x) for x in input().split()]
dim = sorted([x - 1, y - 1, z - 1])
answer = [0, 0, 0]
for i in range(len(dim)):
dimi = dim[i]
if dimi * (3 - i) <= k:
k = k - (dimi * (3 - i))
for j in range(i, 3):
answer[j] += dimi
for j in range(3):
dim[j] -= dimi
else:
for j in range(i, 3):
answer[j] += k // (3 - i)
for j in range(i, (k % (3 - i)) + i):
answer[j] += 1
break
true_ans = 1
for i in answer:
true_ans = true_ans * (i + 1)
print(true_ans)
``` | instruction | 0 | 30,585 | 2 | 61,170 |
Yes | output | 1 | 30,585 | 2 | 61,171 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya plays The Elder Trolls IV: Oblivon. Oh, those creators of computer games! What they do not come up with! Absolutely unique monsters have been added to the The Elder Trolls IV: Oblivon. One of these monsters is Unkillable Slug. Why it is "Unkillable"? Firstly, because it can be killed with cutting weapon only, so lovers of two-handed amber hammers should find suitable knife themselves. Secondly, it is necessary to make so many cutting strokes to Unkillable Slug. Extremely many. Too many!
Vasya has already promoted his character to 80-th level and in order to gain level 81 he was asked to kill Unkillable Slug. The monster has a very interesting shape. It looks like a rectangular parallelepiped with size x × y × z, consisting of undestructable cells 1 × 1 × 1. At one stroke Vasya can cut the Slug along an imaginary grid, i.e. cut with a plane parallel to one of the parallelepiped side. Monster dies when amount of parts it is divided reaches some critical value.
All parts of monster do not fall after each cut, they remains exactly on its places. I. e. Vasya can cut several parts with one cut.
Vasya wants to know what the maximum number of pieces he can cut the Unkillable Slug into striking him at most k times.
Vasya's character uses absolutely thin sword with infinite length.
Input
The first line of input contains four integer numbers x, y, z, k (1 ≤ x, y, z ≤ 106, 0 ≤ k ≤ 109).
Output
Output the only number — the answer for the problem.
Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).
Examples
Input
2 2 2 3
Output
8
Input
2 2 2 1
Output
2
Note
In the first sample Vasya make 3 pairwise perpendicular cuts. He cuts monster on two parts with the first cut, then he divides each part on two with the second cut, and finally he divides each of the 4 parts on two.
Submitted Solution:
```
def main():
*l, k = map(int, input().split())
x, y, z = sorted(l)
x = min(k // 3 + 1, x)
y = min((k - x + 1) // 2 + 1, y)
z = min(k - x - y + 3, z)
print(x * y * z)
if __name__ == '__main__':
main()
``` | instruction | 0 | 30,586 | 2 | 61,172 |
Yes | output | 1 | 30,586 | 2 | 61,173 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya plays The Elder Trolls IV: Oblivon. Oh, those creators of computer games! What they do not come up with! Absolutely unique monsters have been added to the The Elder Trolls IV: Oblivon. One of these monsters is Unkillable Slug. Why it is "Unkillable"? Firstly, because it can be killed with cutting weapon only, so lovers of two-handed amber hammers should find suitable knife themselves. Secondly, it is necessary to make so many cutting strokes to Unkillable Slug. Extremely many. Too many!
Vasya has already promoted his character to 80-th level and in order to gain level 81 he was asked to kill Unkillable Slug. The monster has a very interesting shape. It looks like a rectangular parallelepiped with size x × y × z, consisting of undestructable cells 1 × 1 × 1. At one stroke Vasya can cut the Slug along an imaginary grid, i.e. cut with a plane parallel to one of the parallelepiped side. Monster dies when amount of parts it is divided reaches some critical value.
All parts of monster do not fall after each cut, they remains exactly on its places. I. e. Vasya can cut several parts with one cut.
Vasya wants to know what the maximum number of pieces he can cut the Unkillable Slug into striking him at most k times.
Vasya's character uses absolutely thin sword with infinite length.
Input
The first line of input contains four integer numbers x, y, z, k (1 ≤ x, y, z ≤ 106, 0 ≤ k ≤ 109).
Output
Output the only number — the answer for the problem.
Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).
Examples
Input
2 2 2 3
Output
8
Input
2 2 2 1
Output
2
Note
In the first sample Vasya make 3 pairwise perpendicular cuts. He cuts monster on two parts with the first cut, then he divides each part on two with the second cut, and finally he divides each of the 4 parts on two.
Submitted Solution:
```
l = input().split(' ')
x = int(l[0])
y = int(l[1])
z = int(l[2])
k = int(l[3])
w = (x+y+z)-3
resp = 1
while w>0 and k>0:
resp *= 2
w -= 1
k -= 1
print(resp)
# 1517429109692
``` | instruction | 0 | 30,587 | 2 | 61,174 |
No | output | 1 | 30,587 | 2 | 61,175 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya plays The Elder Trolls IV: Oblivon. Oh, those creators of computer games! What they do not come up with! Absolutely unique monsters have been added to the The Elder Trolls IV: Oblivon. One of these monsters is Unkillable Slug. Why it is "Unkillable"? Firstly, because it can be killed with cutting weapon only, so lovers of two-handed amber hammers should find suitable knife themselves. Secondly, it is necessary to make so many cutting strokes to Unkillable Slug. Extremely many. Too many!
Vasya has already promoted his character to 80-th level and in order to gain level 81 he was asked to kill Unkillable Slug. The monster has a very interesting shape. It looks like a rectangular parallelepiped with size x × y × z, consisting of undestructable cells 1 × 1 × 1. At one stroke Vasya can cut the Slug along an imaginary grid, i.e. cut with a plane parallel to one of the parallelepiped side. Monster dies when amount of parts it is divided reaches some critical value.
All parts of monster do not fall after each cut, they remains exactly on its places. I. e. Vasya can cut several parts with one cut.
Vasya wants to know what the maximum number of pieces he can cut the Unkillable Slug into striking him at most k times.
Vasya's character uses absolutely thin sword with infinite length.
Input
The first line of input contains four integer numbers x, y, z, k (1 ≤ x, y, z ≤ 106, 0 ≤ k ≤ 109).
Output
Output the only number — the answer for the problem.
Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).
Examples
Input
2 2 2 3
Output
8
Input
2 2 2 1
Output
2
Note
In the first sample Vasya make 3 pairwise perpendicular cuts. He cuts monster on two parts with the first cut, then he divides each part on two with the second cut, and finally he divides each of the 4 parts on two.
Submitted Solution:
```
a,b,c,d=map(int,input().split())
x=1
y=1
z=1
for i in range(d):
if x<a:
x+=1
if y<b:
y+=1
if z<c:
z+=1
print(x*y*z)
``` | instruction | 0 | 30,588 | 2 | 61,176 |
No | output | 1 | 30,588 | 2 | 61,177 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya plays The Elder Trolls IV: Oblivon. Oh, those creators of computer games! What they do not come up with! Absolutely unique monsters have been added to the The Elder Trolls IV: Oblivon. One of these monsters is Unkillable Slug. Why it is "Unkillable"? Firstly, because it can be killed with cutting weapon only, so lovers of two-handed amber hammers should find suitable knife themselves. Secondly, it is necessary to make so many cutting strokes to Unkillable Slug. Extremely many. Too many!
Vasya has already promoted his character to 80-th level and in order to gain level 81 he was asked to kill Unkillable Slug. The monster has a very interesting shape. It looks like a rectangular parallelepiped with size x × y × z, consisting of undestructable cells 1 × 1 × 1. At one stroke Vasya can cut the Slug along an imaginary grid, i.e. cut with a plane parallel to one of the parallelepiped side. Monster dies when amount of parts it is divided reaches some critical value.
All parts of monster do not fall after each cut, they remains exactly on its places. I. e. Vasya can cut several parts with one cut.
Vasya wants to know what the maximum number of pieces he can cut the Unkillable Slug into striking him at most k times.
Vasya's character uses absolutely thin sword with infinite length.
Input
The first line of input contains four integer numbers x, y, z, k (1 ≤ x, y, z ≤ 106, 0 ≤ k ≤ 109).
Output
Output the only number — the answer for the problem.
Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).
Examples
Input
2 2 2 3
Output
8
Input
2 2 2 1
Output
2
Note
In the first sample Vasya make 3 pairwise perpendicular cuts. He cuts monster on two parts with the first cut, then he divides each part on two with the second cut, and finally he divides each of the 4 parts on two.
Submitted Solution:
```
a,b,c,d=map(int,input().split())
x=1
y=1
z=1
i=0
while i<d and (x<a or y<b or z<c):
if i%3==0:
if x<a:
x+=1
elif y<b:
y+=1
elif z<c:
z+=1
if i%3==1:
if y<b:
y+=1
elif z<c:
z+=1
elif x<a:
x+=1
if i%3==2:
if z<c:
z+=1
elif x<a:
x+=1
elif y<b:
y+=1
print(x*y*z)
``` | instruction | 0 | 30,589 | 2 | 61,178 |
No | output | 1 | 30,589 | 2 | 61,179 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya plays The Elder Trolls IV: Oblivon. Oh, those creators of computer games! What they do not come up with! Absolutely unique monsters have been added to the The Elder Trolls IV: Oblivon. One of these monsters is Unkillable Slug. Why it is "Unkillable"? Firstly, because it can be killed with cutting weapon only, so lovers of two-handed amber hammers should find suitable knife themselves. Secondly, it is necessary to make so many cutting strokes to Unkillable Slug. Extremely many. Too many!
Vasya has already promoted his character to 80-th level and in order to gain level 81 he was asked to kill Unkillable Slug. The monster has a very interesting shape. It looks like a rectangular parallelepiped with size x × y × z, consisting of undestructable cells 1 × 1 × 1. At one stroke Vasya can cut the Slug along an imaginary grid, i.e. cut with a plane parallel to one of the parallelepiped side. Monster dies when amount of parts it is divided reaches some critical value.
All parts of monster do not fall after each cut, they remains exactly on its places. I. e. Vasya can cut several parts with one cut.
Vasya wants to know what the maximum number of pieces he can cut the Unkillable Slug into striking him at most k times.
Vasya's character uses absolutely thin sword with infinite length.
Input
The first line of input contains four integer numbers x, y, z, k (1 ≤ x, y, z ≤ 106, 0 ≤ k ≤ 109).
Output
Output the only number — the answer for the problem.
Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).
Examples
Input
2 2 2 3
Output
8
Input
2 2 2 1
Output
2
Note
In the first sample Vasya make 3 pairwise perpendicular cuts. He cuts monster on two parts with the first cut, then he divides each part on two with the second cut, and finally he divides each of the 4 parts on two.
Submitted Solution:
```
s=input()
if len(s)<=10:
print(s)
else :
print(s[0]+(len(s)-2)+s[-1])
``` | instruction | 0 | 30,590 | 2 | 61,180 |
No | output | 1 | 30,590 | 2 | 61,181 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Before the start of the football season in Berland a strange magic ritual is held. The most experienced magicians have to find a magic matrix of the size n × n (n is even number). Gods will never allow to start the championship without it. Matrix should contain integers from 0 to n - 1, main diagonal should contain only zeroes and matrix should be symmetric. Moreover, all numbers in each row should be different. Magicians are very tired of the thinking process, so they ask you to write a program to find such matrix.
Input
The first line contains one integer n (2 ≤ n ≤ 1000), n is even.
Output
Output n lines with n numbers each — the required matrix. Separate numbers with spaces. If there are several solutions, output any.
Examples
Input
2
Output
0 1
1 0
Input
4
Output
0 1 3 2
1 0 2 3
3 2 0 1
2 3 1 0 | instruction | 0 | 31,109 | 2 | 62,218 |
Tags: constructive algorithms
Correct Solution:
```
n = int(input())
a = [[0] * n for _i in range(n)]
for i in range(n - 1):
for j in range(n - 1):
a[i][j] = (i + j) % (n - 1) + 1
for i in range(n - 1):
a[n - 1][i] = a[i][n - 1] = a[i][i]
a[i][i] = 0
for row in a:
print(' '.join(map(str, row)))
``` | output | 1 | 31,109 | 2 | 62,219 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Before the start of the football season in Berland a strange magic ritual is held. The most experienced magicians have to find a magic matrix of the size n × n (n is even number). Gods will never allow to start the championship without it. Matrix should contain integers from 0 to n - 1, main diagonal should contain only zeroes and matrix should be symmetric. Moreover, all numbers in each row should be different. Magicians are very tired of the thinking process, so they ask you to write a program to find such matrix.
Input
The first line contains one integer n (2 ≤ n ≤ 1000), n is even.
Output
Output n lines with n numbers each — the required matrix. Separate numbers with spaces. If there are several solutions, output any.
Examples
Input
2
Output
0 1
1 0
Input
4
Output
0 1 3 2
1 0 2 3
3 2 0 1
2 3 1 0 | instruction | 0 | 31,110 | 2 | 62,220 |
Tags: constructive algorithms
Correct Solution:
```
n = int(input())
a = [[(i + j) % (n - 1) + 1 for j in range(n - 1)] + [2 * i % (n - 1) + 1] for i in range(n - 1)] +\
[[2 * i % (n - 1) + 1 for i in range(n)]]
for i in range(n):
a[i][i] = 0
for row in a:
print(*row)
``` | output | 1 | 31,110 | 2 | 62,221 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Before the start of the football season in Berland a strange magic ritual is held. The most experienced magicians have to find a magic matrix of the size n × n (n is even number). Gods will never allow to start the championship without it. Matrix should contain integers from 0 to n - 1, main diagonal should contain only zeroes and matrix should be symmetric. Moreover, all numbers in each row should be different. Magicians are very tired of the thinking process, so they ask you to write a program to find such matrix.
Input
The first line contains one integer n (2 ≤ n ≤ 1000), n is even.
Output
Output n lines with n numbers each — the required matrix. Separate numbers with spaces. If there are several solutions, output any.
Examples
Input
2
Output
0 1
1 0
Input
4
Output
0 1 3 2
1 0 2 3
3 2 0 1
2 3 1 0 | instruction | 0 | 31,112 | 2 | 62,224 |
Tags: constructive algorithms
Correct Solution:
```
# Idea ripped off from those solved folks. Just had to get it done since Java has TLE issues
n=int(input())
c=[[0]*n for i in range(n)]
n-=1
for i in range(n):
for j in range(n):
c[i][j]=1+(i+j)%n
for i in range(n):
c[i][n]=c[n][i]=c[i][i]
c[i][i]=0
c[n][n]=0
for d in c: print(*d)
``` | output | 1 | 31,112 | 2 | 62,225 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Before the start of the football season in Berland a strange magic ritual is held. The most experienced magicians have to find a magic matrix of the size n × n (n is even number). Gods will never allow to start the championship without it. Matrix should contain integers from 0 to n - 1, main diagonal should contain only zeroes and matrix should be symmetric. Moreover, all numbers in each row should be different. Magicians are very tired of the thinking process, so they ask you to write a program to find such matrix.
Input
The first line contains one integer n (2 ≤ n ≤ 1000), n is even.
Output
Output n lines with n numbers each — the required matrix. Separate numbers with spaces. If there are several solutions, output any.
Examples
Input
2
Output
0 1
1 0
Input
4
Output
0 1 3 2
1 0 2 3
3 2 0 1
2 3 1 0 | instruction | 0 | 31,113 | 2 | 62,226 |
Tags: constructive algorithms
Correct Solution:
```
__author__ = 'Darren'
def solve():
n = int(input())
matrix = [[0] * n for _i in range(n)]
# for i in range(n-1):
# matrix[0][i] = i + 1
# for i in range(1, n-1):
# for j in range(n-1):
# matrix[i][j] = matrix[i-1][(j+1) % (n-1)]
for i in range(n-1):
for j in range(n-1):
matrix[i][j] = (i + j) % (n - 1) + 1
for i in range(n-1):
matrix[n-1][i] = matrix[i][n-1] = matrix[i][i]
matrix[i][i] = 0
for row in matrix:
print(' '.join(map(str, row)))
if __name__ == '__main__':
solve()
``` | output | 1 | 31,113 | 2 | 62,227 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Reziba has many magic gems. Each magic gem can be split into M normal gems. The amount of space each magic (and normal) gem takes is 1 unit. A normal gem cannot be split.
Reziba wants to choose a set of magic gems and split some of them, so the total space occupied by the resulting set of gems is N units. If a magic gem is chosen and split, it takes M units of space (since it is split into M gems); if a magic gem is not split, it takes 1 unit.
How many different configurations of the resulting set of gems can Reziba have, such that the total amount of space taken is N units? Print the answer modulo 1000000007 (10^9+7). Two configurations are considered different if the number of magic gems Reziba takes to form them differs, or the indices of gems Reziba has to split differ.
Input
The input contains a single line consisting of 2 integers N and M (1 ≤ N ≤ 10^{18}, 2 ≤ M ≤ 100).
Output
Print one integer, the total number of configurations of the resulting set of gems, given that the total amount of space taken is N units. Print the answer modulo 1000000007 (10^9+7).
Examples
Input
4 2
Output
5
Input
3 2
Output
3
Note
In the first example each magic gem can split into 2 normal gems, and we know that the total amount of gems are 4.
Let 1 denote a magic gem, and 0 denote a normal gem.
The total configurations you can have is:
* 1 1 1 1 (None of the gems split);
* 0 0 1 1 (First magic gem splits into 2 normal gems);
* 1 0 0 1 (Second magic gem splits into 2 normal gems);
* 1 1 0 0 (Third magic gem splits into 2 normal gems);
* 0 0 0 0 (First and second magic gems split into total 4 normal gems).
Hence, answer is 5. | instruction | 0 | 33,385 | 2 | 66,770 |
Tags: dp, math, matrices
Correct Solution:
```
import sys
# How to multiply matrices in 32 bit python?
# Solution: Use doubles
# New problem: Might not have integer precision
# Solution: Mod when outside of double precision
# Alt solution: Use 64 bit python
MOD = 10**9 + 7
double_prec = float(2*10**5*MOD)
MODF = float(MOD)
def mult(A,B):
n = len(A)
cutmask = 2**16-1
C = [[0]*n for _ in range(n)]
buffer = [0.0]*n
A = [[(aa&cutmask)+1j*(aa>>16) for aa in a] for a in A]
for x in range(n):
b = B[x]
for z in range(n):
a = A[z]
bb = b[z]+1j*((b[z]<<16)%MOD)
for y in range(n):
buffer[y] += a[y].real*bb.real + a[y].imag*bb.imag
# UNCOMMENT FOR NUMERICAL STABILITY
#if buffer[y]>double_prec:buffer[y]-=double_prec
c = C[x]
for y in range(n):
c[y] = int(buffer[y]%MODF)
buffer[y] = 0.0
return C
def vecmult(A,B):
n = len(A)
C = [0]*n
for z in range(n):
b = B[z]
a = A[z]
for y in range(n):
C[y] += a[y]*b
return [c%MOD for c in C]
n,m = [int(x) for x in input().split()]
#def DP(n):
# if n<=0:
# return 1 if n==0 else 0
# else:
# return DP(n-m) + DP(n-1)
mapper = [[0]*m for _ in range(m)]
mapper[0][0] = 1
mapper[-1][0] = 1
for i in range(m-1):
mapper[i][i+1] = 1
def power(A,B,m):
n = len(A)
if m == 0:
return B
while m>40:
if m%2==1:
B = vecmult(A,B)
A = mult(A,A)
m//=2
while m>0:
B = vecmult(A,B)
m-=1
return B
vec = [1]*m
vec = power(mapper,vec,n)
print(vec[-1]%MOD)
``` | output | 1 | 33,385 | 2 | 66,771 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Reziba has many magic gems. Each magic gem can be split into M normal gems. The amount of space each magic (and normal) gem takes is 1 unit. A normal gem cannot be split.
Reziba wants to choose a set of magic gems and split some of them, so the total space occupied by the resulting set of gems is N units. If a magic gem is chosen and split, it takes M units of space (since it is split into M gems); if a magic gem is not split, it takes 1 unit.
How many different configurations of the resulting set of gems can Reziba have, such that the total amount of space taken is N units? Print the answer modulo 1000000007 (10^9+7). Two configurations are considered different if the number of magic gems Reziba takes to form them differs, or the indices of gems Reziba has to split differ.
Input
The input contains a single line consisting of 2 integers N and M (1 ≤ N ≤ 10^{18}, 2 ≤ M ≤ 100).
Output
Print one integer, the total number of configurations of the resulting set of gems, given that the total amount of space taken is N units. Print the answer modulo 1000000007 (10^9+7).
Examples
Input
4 2
Output
5
Input
3 2
Output
3
Note
In the first example each magic gem can split into 2 normal gems, and we know that the total amount of gems are 4.
Let 1 denote a magic gem, and 0 denote a normal gem.
The total configurations you can have is:
* 1 1 1 1 (None of the gems split);
* 0 0 1 1 (First magic gem splits into 2 normal gems);
* 1 0 0 1 (Second magic gem splits into 2 normal gems);
* 1 1 0 0 (Third magic gem splits into 2 normal gems);
* 0 0 0 0 (First and second magic gems split into total 4 normal gems).
Hence, answer is 5. | instruction | 0 | 33,386 | 2 | 66,772 |
Tags: dp, math, matrices
Correct Solution:
```
import sys
from math import trunc
#range = xrange
#input = raw_input
# How to multiply matrices in 32 bit python?
# Solution: Use doubles with homemade mult and mod.
MOD = 10**9 + 7
MODF = float(MOD)
def modder(x):
return x - MODF*trunc(x/MODF)
small = 1.0 * (2**16)
# calculates a*b
def prod(a,b):
return trunc(a/small)*modder(b*small)+(a-small*trunc(a/small))*b
def mult(A,B):
n = len(A)
m = len(B[0])
C = [[0.0]*n for _ in range(m)]
for j,B_j in enumerate(B):
C_j = C[j]
for k,A_k in enumerate(A):
Bkj = B_j[k]
for i,Aik in enumerate(A_k):
C_j[i] += prod(Aik,Bkj)
for i,Cij in enumerate(C_j):
C_j[i] = modder(Cij)
return C
def vecmult(A,B):
n = len(A)
C = [0.0]*n
for k,A_k in enumerate(A):
Bk = B[k]
for i,Aik in enumerate(A_k):
C[i] += prod(Aik,Bk)
for i in range(n):
C[i] = modder(C[i])
return C
n,m = [int(x) for x in input().split()]
#def DP(n):
# if n<=0:
# return 1 if n==0 else 0
# else:
# return DP(n-m) + DP(n-1)
mapper = [[0.0]*m for _ in range(m)]
mapper[0][0] = 1.0
mapper[-1][0] = 1.0
for i in range(m-1):
mapper[i][i+1] = 1.0
def power(A,B,m):
n = len(A)
if m == 0:
return B
while m>16:
if m%2==1:
B = vecmult(A,B)
A = mult(A,A)
m//=2
while m>0:
B = vecmult(A,B)
m-=1
return B
vec = [1.0]*m
vec = power(mapper,vec,n)
print(int(vec[-1])%MOD)
``` | output | 1 | 33,386 | 2 | 66,773 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Reziba has many magic gems. Each magic gem can be split into M normal gems. The amount of space each magic (and normal) gem takes is 1 unit. A normal gem cannot be split.
Reziba wants to choose a set of magic gems and split some of them, so the total space occupied by the resulting set of gems is N units. If a magic gem is chosen and split, it takes M units of space (since it is split into M gems); if a magic gem is not split, it takes 1 unit.
How many different configurations of the resulting set of gems can Reziba have, such that the total amount of space taken is N units? Print the answer modulo 1000000007 (10^9+7). Two configurations are considered different if the number of magic gems Reziba takes to form them differs, or the indices of gems Reziba has to split differ.
Input
The input contains a single line consisting of 2 integers N and M (1 ≤ N ≤ 10^{18}, 2 ≤ M ≤ 100).
Output
Print one integer, the total number of configurations of the resulting set of gems, given that the total amount of space taken is N units. Print the answer modulo 1000000007 (10^9+7).
Examples
Input
4 2
Output
5
Input
3 2
Output
3
Note
In the first example each magic gem can split into 2 normal gems, and we know that the total amount of gems are 4.
Let 1 denote a magic gem, and 0 denote a normal gem.
The total configurations you can have is:
* 1 1 1 1 (None of the gems split);
* 0 0 1 1 (First magic gem splits into 2 normal gems);
* 1 0 0 1 (Second magic gem splits into 2 normal gems);
* 1 1 0 0 (Third magic gem splits into 2 normal gems);
* 0 0 0 0 (First and second magic gems split into total 4 normal gems).
Hence, answer is 5. | instruction | 0 | 33,387 | 2 | 66,774 |
Tags: dp, math, matrices
Correct Solution:
```
import sys
from math import trunc
#range = xrange
#input = raw_input
# How to multiply matrices in 32 bit python?
# Solution: Use doubles with homemade mult and mod.
MOD = 10**9 + 7
MODF = float(MOD)
def modder(x):
return x - MODF*trunc(x/MODF)
small = 1.0 * (2**16)
# calculates a*b+c
def prod(a,b,c):
abig = trunc(a/small)
return modder(abig*modder(b*small)+(a-small*abig)*b+c)
def mult(A,B):
n = len(A)
m = len(B[0])
C = [[0.0]*n for _ in range(m)]
for j,B_j in enumerate(B):
buffer = C[j]
for k,A_k in enumerate(A):
Bkj = B_j[k]
for i,Aik in enumerate(A_k):
buffer[i] = prod(Aik,Bkj,buffer[i])
return C
def vecmult(A,B):
n = len(A)
C = [0.0]*n
for k,A_k in enumerate(A):
Bk = B[k]
for i,Aik in enumerate(A_k):
C[i] = prod(Aik,Bk,C[i])
return C
n,m = [int(x) for x in input().split()]
#def DP(n):
# if n<=0:
# return 1 if n==0 else 0
# else:
# return DP(n-m) + DP(n-1)
mapper = [[0.0]*m for _ in range(m)]
mapper[0][0] = 1.0
mapper[-1][0] = 1.0
for i in range(m-1):
mapper[i][i+1] = 1.0
def power(A,B,m):
n = len(A)
if m == 0:
return B
while m>16:
if m%2==1:
B = vecmult(A,B)
A = mult(A,A)
m//=2
while m>0:
B = vecmult(A,B)
m-=1
return B
vec = [1.0]*m
vec = power(mapper,vec,n)
print(int(vec[-1])%MOD)
``` | output | 1 | 33,387 | 2 | 66,775 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Reziba has many magic gems. Each magic gem can be split into M normal gems. The amount of space each magic (and normal) gem takes is 1 unit. A normal gem cannot be split.
Reziba wants to choose a set of magic gems and split some of them, so the total space occupied by the resulting set of gems is N units. If a magic gem is chosen and split, it takes M units of space (since it is split into M gems); if a magic gem is not split, it takes 1 unit.
How many different configurations of the resulting set of gems can Reziba have, such that the total amount of space taken is N units? Print the answer modulo 1000000007 (10^9+7). Two configurations are considered different if the number of magic gems Reziba takes to form them differs, or the indices of gems Reziba has to split differ.
Input
The input contains a single line consisting of 2 integers N and M (1 ≤ N ≤ 10^{18}, 2 ≤ M ≤ 100).
Output
Print one integer, the total number of configurations of the resulting set of gems, given that the total amount of space taken is N units. Print the answer modulo 1000000007 (10^9+7).
Examples
Input
4 2
Output
5
Input
3 2
Output
3
Note
In the first example each magic gem can split into 2 normal gems, and we know that the total amount of gems are 4.
Let 1 denote a magic gem, and 0 denote a normal gem.
The total configurations you can have is:
* 1 1 1 1 (None of the gems split);
* 0 0 1 1 (First magic gem splits into 2 normal gems);
* 1 0 0 1 (Second magic gem splits into 2 normal gems);
* 1 1 0 0 (Third magic gem splits into 2 normal gems);
* 0 0 0 0 (First and second magic gems split into total 4 normal gems).
Hence, answer is 5. | instruction | 0 | 33,388 | 2 | 66,776 |
Tags: dp, math, matrices
Correct Solution:
```
import sys
from math import trunc
#range = xrange
#input = raw_input
# How to multiply matrices in 32 bit python?
# Solution: Use complex doubles!
# Alt solution: Just use 64 bit python
MOD = 10**9 + 7
MODF = float(MOD)
small = 1.0*(2**16)
def modder(x):
return x - MODF*trunc(x/MODF)
# calculates a*b
def prod(a,b):
abig = trunc(a/small)
return abig*modder(b*small)+(a-small*abig)*b
def mult(A,B):
n = len(A)
m = len(B[0])
C = [[0.0]*n for _ in range(m)]
# Only in python would you do integer matrix multiplication using complex
# floating point numbers
A = [[(Aij-small*trunc(Aij/small))-1j*trunc(Aij/small) for Aij in A_j] for A_j in A]
for j,B_j in enumerate(B):
buffer = C[j]
for k,A_k in enumerate(A):
Bkj = B_j[k]+1j*modder(B_j[k]*small)
for i,Aik in enumerate(A_k):
buffer[i] += (Aik*Bkj).real
#buffer[i] = modder(buffer[i])
#UNCOMMENT FOR NUMERICAL STABILITY
#if buffer[i]>=double_prec:buffer[i]%=MODF
for i,b in enumerate(buffer):
buffer[i] = modder(b)
return C
def vecmult(A,B):
n = len(A)
C = [0.0]*n
for k,A_k in enumerate(A):
Bk = B[k]
for i,Aik in enumerate(A_k):
C[i] += prod(Aik,Bk)
for i in range(n):C[i]=modder(C[i])
return C
n,m = [int(x) for x in input().split()]
#def DP(n):
# if n<=0:
# return 1 if n==0 else 0
# else:
# return DP(n-m) + DP(n-1)
mapper = [[0.0]*m for _ in range(m)]
mapper[0][0] = 1.0
mapper[-1][0] = 1.0
for i in range(m-1):
mapper[i][i+1] = 1.0
def power(A,B,m):
n = len(A)
if m == 0:
return B
while m>16:
if m%2==1:
B = vecmult(A,B)
A = mult(A,A)
m//=2
while m>0:
B = vecmult(A,B)
m-=1
return B
vec = [1.0]*m
vec = power(mapper,vec,n)
print(int(vec[-1])%MOD)
``` | output | 1 | 33,388 | 2 | 66,777 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Reziba has many magic gems. Each magic gem can be split into M normal gems. The amount of space each magic (and normal) gem takes is 1 unit. A normal gem cannot be split.
Reziba wants to choose a set of magic gems and split some of them, so the total space occupied by the resulting set of gems is N units. If a magic gem is chosen and split, it takes M units of space (since it is split into M gems); if a magic gem is not split, it takes 1 unit.
How many different configurations of the resulting set of gems can Reziba have, such that the total amount of space taken is N units? Print the answer modulo 1000000007 (10^9+7). Two configurations are considered different if the number of magic gems Reziba takes to form them differs, or the indices of gems Reziba has to split differ.
Input
The input contains a single line consisting of 2 integers N and M (1 ≤ N ≤ 10^{18}, 2 ≤ M ≤ 100).
Output
Print one integer, the total number of configurations of the resulting set of gems, given that the total amount of space taken is N units. Print the answer modulo 1000000007 (10^9+7).
Examples
Input
4 2
Output
5
Input
3 2
Output
3
Note
In the first example each magic gem can split into 2 normal gems, and we know that the total amount of gems are 4.
Let 1 denote a magic gem, and 0 denote a normal gem.
The total configurations you can have is:
* 1 1 1 1 (None of the gems split);
* 0 0 1 1 (First magic gem splits into 2 normal gems);
* 1 0 0 1 (Second magic gem splits into 2 normal gems);
* 1 1 0 0 (Third magic gem splits into 2 normal gems);
* 0 0 0 0 (First and second magic gems split into total 4 normal gems).
Hence, answer is 5. | instruction | 0 | 33,389 | 2 | 66,778 |
Tags: dp, math, matrices
Correct Solution:
```
import sys
from math import trunc
# How to multiply matrices in 32 bit python?
# Solution: Use doubles with homemade mult and mod.
# This should be around 4-6 times faster than using int
# and around 50% to the speed you'd get in 64 bit python.
MOD = 10**9 + 7
MODF = float(MOD)
MODF_inv = 1.0/MODF
FLOAT_PREC = float(2**52)
# Mods float x, note modder(-1.0)=-1.0
def modder(x):
return x - MODF * trunc(x * MODF_inv)
# Calc modder(c+a*b)
SHRT = float(2**16)
SHRT_inv = 1.0/SHRT
def prod(a,b,c=0):
return modder(c + trunc(a*SHRT_inv)*modder(b*SHRT) + (a-SHRT*trunc(a*SHRT_inv))*b)
# Slower version, but readable
def mult(A,B):
n,m = len(A[0]),len(B)
C = [[0.0]*n for _ in range(m)]
for j,B_j in enumerate(B):
C_j = C[j]
for k,A_k in enumerate(A):
Bkj = B_j[k]
for i,Aik in enumerate(A_k):
C_j[i] = prod(Aik,Bkj,C_j[i])
return C
# Calc A^n*B
def power(A,B,n,mult_func):
if n == 0:
return B
while n>1:
if n%2==1:
B = mult_func(A,B)
A = mult_func(A,A)
n//=2
return mult_func(A,B)
##### EXAMPLE
n,m = [int(x) for x in input().split()]
#def DP(n):
# if n<=0:
# return 1 if n==0 else 0
# else:
# return DP(n-m) + DP(n-1)
mat = [[0.0]*m for _ in range(m)]
mat[0][0] = 1.0
mat[-1][0] = 1.0
for i in range(m-1):
mat[i][i+1] = 1.0
vec = [[1.0]*m]
vec = power(mat,vec,n,mult)
print(int(vec[0][-1]))
``` | output | 1 | 33,389 | 2 | 66,779 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Reziba has many magic gems. Each magic gem can be split into M normal gems. The amount of space each magic (and normal) gem takes is 1 unit. A normal gem cannot be split.
Reziba wants to choose a set of magic gems and split some of them, so the total space occupied by the resulting set of gems is N units. If a magic gem is chosen and split, it takes M units of space (since it is split into M gems); if a magic gem is not split, it takes 1 unit.
How many different configurations of the resulting set of gems can Reziba have, such that the total amount of space taken is N units? Print the answer modulo 1000000007 (10^9+7). Two configurations are considered different if the number of magic gems Reziba takes to form them differs, or the indices of gems Reziba has to split differ.
Input
The input contains a single line consisting of 2 integers N and M (1 ≤ N ≤ 10^{18}, 2 ≤ M ≤ 100).
Output
Print one integer, the total number of configurations of the resulting set of gems, given that the total amount of space taken is N units. Print the answer modulo 1000000007 (10^9+7).
Examples
Input
4 2
Output
5
Input
3 2
Output
3
Note
In the first example each magic gem can split into 2 normal gems, and we know that the total amount of gems are 4.
Let 1 denote a magic gem, and 0 denote a normal gem.
The total configurations you can have is:
* 1 1 1 1 (None of the gems split);
* 0 0 1 1 (First magic gem splits into 2 normal gems);
* 1 0 0 1 (Second magic gem splits into 2 normal gems);
* 1 1 0 0 (Third magic gem splits into 2 normal gems);
* 0 0 0 0 (First and second magic gems split into total 4 normal gems).
Hence, answer is 5. | instruction | 0 | 33,390 | 2 | 66,780 |
Tags: dp, math, matrices
Correct Solution:
```
import math
FMOD = 1000000007.0
PREC = 7881299347898367.0
SHRT = 65536.0
fmod = lambda x: x - FMOD * math.trunc(x / FMOD)
mod_prod = lambda a, b, c=0: fmod(math.trunc(a / SHRT) * fmod(b * SHRT) + (a - SHRT * math.trunc(a / SHRT)) * b + c)
def mult(A, B):
n, p = len(A), len(B[0])
B = [[(Bij - SHRT * math.trunc(Bij / SHRT)) - 1j * (math.trunc(Bij / SHRT)) for Bij in Bi] for Bi in B]
C = [[0.0] * p for _ in range(n)]
for i, Ai in enumerate(A):
Ci = C[i]
for j, Bj in enumerate(B):
Aij = Ai[j] + 1j * fmod(Ai[j] * SHRT)
for k, Bjk in enumerate(Bj):
Ci[k] += (Aij * Bjk).real
if Ci[k] > PREC:
Ci[k] = fmod(Ci[k])
C[i] = [fmod(Cij) for Cij in Ci]
return C
def power(A, B, m):
while m > 0:
if m % 2 == 1:
B = mult(B, A)
A = mult(A, A)
m //= 2
return B
n, m = [int(x) for x in input().split()]
mapper = [[0.0] * m for _ in range(m)]
mapper[0][0], mapper[-1][0] = 1.0, 1.0
for i in range(m - 1):
mapper[i][i + 1] = 1.0
vec = [[1.0] * m]
vec = power(mapper, vec, n)
print(int(vec[0][-1] % FMOD))
``` | output | 1 | 33,390 | 2 | 66,781 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Reziba has many magic gems. Each magic gem can be split into M normal gems. The amount of space each magic (and normal) gem takes is 1 unit. A normal gem cannot be split.
Reziba wants to choose a set of magic gems and split some of them, so the total space occupied by the resulting set of gems is N units. If a magic gem is chosen and split, it takes M units of space (since it is split into M gems); if a magic gem is not split, it takes 1 unit.
How many different configurations of the resulting set of gems can Reziba have, such that the total amount of space taken is N units? Print the answer modulo 1000000007 (10^9+7). Two configurations are considered different if the number of magic gems Reziba takes to form them differs, or the indices of gems Reziba has to split differ.
Input
The input contains a single line consisting of 2 integers N and M (1 ≤ N ≤ 10^{18}, 2 ≤ M ≤ 100).
Output
Print one integer, the total number of configurations of the resulting set of gems, given that the total amount of space taken is N units. Print the answer modulo 1000000007 (10^9+7).
Examples
Input
4 2
Output
5
Input
3 2
Output
3
Note
In the first example each magic gem can split into 2 normal gems, and we know that the total amount of gems are 4.
Let 1 denote a magic gem, and 0 denote a normal gem.
The total configurations you can have is:
* 1 1 1 1 (None of the gems split);
* 0 0 1 1 (First magic gem splits into 2 normal gems);
* 1 0 0 1 (Second magic gem splits into 2 normal gems);
* 1 1 0 0 (Third magic gem splits into 2 normal gems);
* 0 0 0 0 (First and second magic gems split into total 4 normal gems).
Hence, answer is 5. | instruction | 0 | 33,391 | 2 | 66,782 |
Tags: dp, math, matrices
Correct Solution:
```
import sys
from math import trunc
# How to multiply matrices in 32 bit python?
# Solution: Use doubles with homemade mult and mod.
# This should be around 4-6 times faster than using int
# and around 50% to the speed you'd get in 64 bit python.
# "Crazy code"
MOD = 10**9 + 7
MODF = 1.0*MOD
MODF_inv = 1.0/MODF
SHRT = float(1 << 16)
SHRT_inv = 1.0/SHRT
modder = lambda a:a - MODF * trunc(MODF_inv*a)
mod_prod = lambda a, b, c=0.0:\
modder(modder(a * SHRT) * trunc(SHRT_inv*b) + a*(b-SHRT * trunc(b*SHRT_inv)) + c)
# Slower version, but readable
# Possible to speed up using precalculations
def mult(A,B):
n,m = len(A[0]),len(B)
C = [[0.0]*n for _ in range(m)]
for j,B_j in enumerate(B):
C_j = C[j]
for k,A_k in enumerate(A):
Bkj = B_j[k]
for i,Aik in enumerate(A_k):
C_j[i] = mod_prod(Aik,Bkj,C_j[i])
return C
# Calc A^n*B
def power(A,B,n,mult_func):
if n == 0:
return B
while n>1:
if n%2==1:
B = mult_func(A,B)
A = mult_func(A,A)
n//=2
return mult_func(A,B)
##### EXAMPLE
n,m = [int(x) for x in input().split()]
#def DP(n):
# if n<=0:
# return 1 if n==0 else 0
# else:
# return DP(n-m) + DP(n-1)
mat = [[0.0]*m for _ in range(m)]
mat[0][0] = 1.0
mat[-1][0] = 1.0
for i in range(m-1):
mat[i][i+1] = 1.0
vec = [[1.0]*m]
vec = power(mat,vec,n,mult)
print(int(vec[0][-1]))
``` | output | 1 | 33,391 | 2 | 66,783 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Reziba has many magic gems. Each magic gem can be split into M normal gems. The amount of space each magic (and normal) gem takes is 1 unit. A normal gem cannot be split.
Reziba wants to choose a set of magic gems and split some of them, so the total space occupied by the resulting set of gems is N units. If a magic gem is chosen and split, it takes M units of space (since it is split into M gems); if a magic gem is not split, it takes 1 unit.
How many different configurations of the resulting set of gems can Reziba have, such that the total amount of space taken is N units? Print the answer modulo 1000000007 (10^9+7). Two configurations are considered different if the number of magic gems Reziba takes to form them differs, or the indices of gems Reziba has to split differ.
Input
The input contains a single line consisting of 2 integers N and M (1 ≤ N ≤ 10^{18}, 2 ≤ M ≤ 100).
Output
Print one integer, the total number of configurations of the resulting set of gems, given that the total amount of space taken is N units. Print the answer modulo 1000000007 (10^9+7).
Examples
Input
4 2
Output
5
Input
3 2
Output
3
Note
In the first example each magic gem can split into 2 normal gems, and we know that the total amount of gems are 4.
Let 1 denote a magic gem, and 0 denote a normal gem.
The total configurations you can have is:
* 1 1 1 1 (None of the gems split);
* 0 0 1 1 (First magic gem splits into 2 normal gems);
* 1 0 0 1 (Second magic gem splits into 2 normal gems);
* 1 1 0 0 (Third magic gem splits into 2 normal gems);
* 0 0 0 0 (First and second magic gems split into total 4 normal gems).
Hence, answer is 5. | instruction | 0 | 33,392 | 2 | 66,784 |
Tags: dp, math, matrices
Correct Solution:
```
import sys
from math import trunc
# How to multiply matrices in 32 bit python?
# Solution: Use doubles with homemade mult and mod.
# This should be around 4-6 times faster than using int
# and around 50% to the speed you'd get in 64 bit python.
# "Crazy code"
MOD = 10**9 + 7
MODF = 1.0*MOD
MODF_inv = 1.0/MODF
SHRT = float(1 << 16)
SHRT_inv = 1.0/SHRT
modder = lambda a:a - MODF * trunc(MODF_inv*a)
mod_prod = lambda a, b, c=0.0:\
modder(c + modder(b * SHRT) * trunc(SHRT_inv * a) + b*(a-SHRT * trunc(a * SHRT_inv)))
# Slower version, but readable
# Possible to speed up using precalculations
def mult(A,B):
n,m = len(A[0]),len(B)
C = [[0.0]*n for _ in range(m)]
for j,B_j in enumerate(B):
C_j = C[j]
for k,A_k in enumerate(A):
Bkj = B_j[k]
for i,Aik in enumerate(A_k):
C_j[i] = mod_prod(Bkj,Aik,C_j[i])
return C
# Calc A^n*B
def power(A,B,n,mult_func):
if n == 0:
return B
while n>1:
if n%2==1:
B = mult_func(A,B)
A = mult_func(A,A)
n//=2
return mult_func(A,B)
##### EXAMPLE
n,m = [int(x) for x in input().split()]
#def DP(n):
# if n<=0:
# return 1 if n==0 else 0
# else:
# return DP(n-m) + DP(n-1)
mat = [[0.0]*m for _ in range(m)]
mat[0][0] = 1.0
mat[-1][0] = 1.0
for i in range(m-1):
mat[i][i+1] = 1.0
vec = [[1.0]*m]
vec = power(mat,vec,n,mult)
print(int(vec[0][-1]))
``` | output | 1 | 33,392 | 2 | 66,785 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Reziba has many magic gems. Each magic gem can be split into M normal gems. The amount of space each magic (and normal) gem takes is 1 unit. A normal gem cannot be split.
Reziba wants to choose a set of magic gems and split some of them, so the total space occupied by the resulting set of gems is N units. If a magic gem is chosen and split, it takes M units of space (since it is split into M gems); if a magic gem is not split, it takes 1 unit.
How many different configurations of the resulting set of gems can Reziba have, such that the total amount of space taken is N units? Print the answer modulo 1000000007 (10^9+7). Two configurations are considered different if the number of magic gems Reziba takes to form them differs, or the indices of gems Reziba has to split differ.
Input
The input contains a single line consisting of 2 integers N and M (1 ≤ N ≤ 10^{18}, 2 ≤ M ≤ 100).
Output
Print one integer, the total number of configurations of the resulting set of gems, given that the total amount of space taken is N units. Print the answer modulo 1000000007 (10^9+7).
Examples
Input
4 2
Output
5
Input
3 2
Output
3
Note
In the first example each magic gem can split into 2 normal gems, and we know that the total amount of gems are 4.
Let 1 denote a magic gem, and 0 denote a normal gem.
The total configurations you can have is:
* 1 1 1 1 (None of the gems split);
* 0 0 1 1 (First magic gem splits into 2 normal gems);
* 1 0 0 1 (Second magic gem splits into 2 normal gems);
* 1 1 0 0 (Third magic gem splits into 2 normal gems);
* 0 0 0 0 (First and second magic gems split into total 4 normal gems).
Hence, answer is 5.
Submitted Solution:
```
import sys
from math import trunc
# How to multiply matrices in 32 bit python?
# Solution: Use doubles with homemade mult and mod.
# This should be around 4-6 times faster than using int
# and around 50% to the speed you'd get in 64 bit python.
MOD = 10**9 + 7
MODF = float(MOD)
MODF_inv = 1.0/MODF
# Mods float x, note modder(-1.0)=-1.0
def modder(a):
return a - MODF * trunc(a * MODF_inv)
SHRT = float(1 << 16)
SHRT_inv = 1.0/SHRT
# Calc modder(c+a*b)
def mod_prod(a, b, c=0):
a1 = trunc(a*SHRT_inv)
a2 = a-SHRT*a1
b1 = modder(b * SHRT)
b2 = b
return modder(a2*b2+a1*b1+c)
# Slower version, but readable
# Possible to speed up using precalculations
def mult(A,B):
n,m = len(A[0]),len(B)
C = [[0.0]*n for _ in range(m)]
for j,B_j in enumerate(B):
C_j = C[j]
for k,A_k in enumerate(A):
Bkj = B_j[k]
for i,Aik in enumerate(A_k):
C_j[i] = mod_prod(Aik,Bkj,C_j[i])
return C
# Calc A^n*B
def power(A,B,n,mult_func):
if n == 0:
return B
while n>1:
if n%2==1:
B = mult_func(A,B)
A = mult_func(A,A)
n//=2
return mult_func(A,B)
##### EXAMPLE
n,m = [int(x) for x in input().split()]
#def DP(n):
# if n<=0:
# return 1 if n==0 else 0
# else:
# return DP(n-m) + DP(n-1)
mat = [[0.0]*m for _ in range(m)]
mat[0][0] = 1.0
mat[-1][0] = 1.0
for i in range(m-1):
mat[i][i+1] = 1.0
vec = [[1.0]*m]
vec = power(mat,vec,n,mult)
print(int(vec[0][-1]))
``` | instruction | 0 | 33,393 | 2 | 66,786 |
Yes | output | 1 | 33,393 | 2 | 66,787 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Reziba has many magic gems. Each magic gem can be split into M normal gems. The amount of space each magic (and normal) gem takes is 1 unit. A normal gem cannot be split.
Reziba wants to choose a set of magic gems and split some of them, so the total space occupied by the resulting set of gems is N units. If a magic gem is chosen and split, it takes M units of space (since it is split into M gems); if a magic gem is not split, it takes 1 unit.
How many different configurations of the resulting set of gems can Reziba have, such that the total amount of space taken is N units? Print the answer modulo 1000000007 (10^9+7). Two configurations are considered different if the number of magic gems Reziba takes to form them differs, or the indices of gems Reziba has to split differ.
Input
The input contains a single line consisting of 2 integers N and M (1 ≤ N ≤ 10^{18}, 2 ≤ M ≤ 100).
Output
Print one integer, the total number of configurations of the resulting set of gems, given that the total amount of space taken is N units. Print the answer modulo 1000000007 (10^9+7).
Examples
Input
4 2
Output
5
Input
3 2
Output
3
Note
In the first example each magic gem can split into 2 normal gems, and we know that the total amount of gems are 4.
Let 1 denote a magic gem, and 0 denote a normal gem.
The total configurations you can have is:
* 1 1 1 1 (None of the gems split);
* 0 0 1 1 (First magic gem splits into 2 normal gems);
* 1 0 0 1 (Second magic gem splits into 2 normal gems);
* 1 1 0 0 (Third magic gem splits into 2 normal gems);
* 0 0 0 0 (First and second magic gems split into total 4 normal gems).
Hence, answer is 5.
Submitted Solution:
```
#!/usr/bin/env python3
import sys
input = lambda: sys.stdin.readline().rstrip('\r\n')
def mat_mul_mod(A, B, mod):
n, p = len(A), len(B[0])
fmod = float(mod)
float_prec = float(1 << 51)
B = [[(Bij & ((1 << 16) - 1)) - 1j * (Bij >> 16) for Bij in Bi] for Bi in B]
C = [[0] * p for _ in range(n)]
for i, Ai in enumerate(A):
row = [0.0] * p
for j, Bj in enumerate(B):
Aij = Ai[j] + 1j * ((Ai[j] * float(1 << 16)) % mod)
for k, Bjk in enumerate(Bj):
row[k] += (Aij * Bjk).real
if row[k] > float_prec:
row[k] %= fmod
C[i] = [int(r % fmod) for r in row]
return C
def main():
MOD = 10**9 + 7
n, m = map(int, input().split())
mat = [[0] * m for _ in range(m)]
mat[0][0] = 1
mat[0][-1] = 1
for i in range(m - 1):
mat[i + 1][i] = 1
vec = [1] * m
for i in bin(n)[:1:-1]:
if i == '1':
vec = [sum(a * b for a, b in zip(row, vec)) % MOD for row in mat]
mat = mat_mul_mod(mat, mat, MOD)
print(int(vec[-1]))
if __name__ == '__main__':
main()
``` | instruction | 0 | 33,394 | 2 | 66,788 |
Yes | output | 1 | 33,394 | 2 | 66,789 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Reziba has many magic gems. Each magic gem can be split into M normal gems. The amount of space each magic (and normal) gem takes is 1 unit. A normal gem cannot be split.
Reziba wants to choose a set of magic gems and split some of them, so the total space occupied by the resulting set of gems is N units. If a magic gem is chosen and split, it takes M units of space (since it is split into M gems); if a magic gem is not split, it takes 1 unit.
How many different configurations of the resulting set of gems can Reziba have, such that the total amount of space taken is N units? Print the answer modulo 1000000007 (10^9+7). Two configurations are considered different if the number of magic gems Reziba takes to form them differs, or the indices of gems Reziba has to split differ.
Input
The input contains a single line consisting of 2 integers N and M (1 ≤ N ≤ 10^{18}, 2 ≤ M ≤ 100).
Output
Print one integer, the total number of configurations of the resulting set of gems, given that the total amount of space taken is N units. Print the answer modulo 1000000007 (10^9+7).
Examples
Input
4 2
Output
5
Input
3 2
Output
3
Note
In the first example each magic gem can split into 2 normal gems, and we know that the total amount of gems are 4.
Let 1 denote a magic gem, and 0 denote a normal gem.
The total configurations you can have is:
* 1 1 1 1 (None of the gems split);
* 0 0 1 1 (First magic gem splits into 2 normal gems);
* 1 0 0 1 (Second magic gem splits into 2 normal gems);
* 1 1 0 0 (Third magic gem splits into 2 normal gems);
* 0 0 0 0 (First and second magic gems split into total 4 normal gems).
Hence, answer is 5.
Submitted Solution:
```
import sys
from math import trunc
# How to multiply matrices in 32 bit python?
# Solution: Use doubles with homemade mult and mod.
MOD = 10**9 + 7
MODF = float(MOD)
small = float(2**16)
float_prec = float(2**52)
# Mods float x, note modder(-1.0)=-1.0
def modder(x):
return x - MODF * trunc(x / MODF)
# Calc modder(c+a*b)
def prod(a,b,c=0):
return modder(c + trunc(a/small)*modder(b*small) + (a-small*trunc(a/small))*b)
# Quick version
# Uses the same idea as prod but with some precalculations
def mult(A,B):
n,m = len(A[0]),len(B)
A = [[(Aij-small*trunc(Aij/small)) - 1j*trunc(Aij/small) for Aij in A_j] for A_j in A]
C = [[0.0]*n for _ in range(m)]
for j,B_j in enumerate(B):
C_j = C[j]
for k,A_k in enumerate(A):
Bkj = B_j[k] + 1j*modder(B_j[k]*small)
for i,Aik in enumerate(A_k):
C_j[i] += (Aik*Bkj).real
# Can be removed for small matrices, around 100x100:
if C_j[i]>=float_prec: C_j[i] = modder(C_j[i])
for i in range(n):
C_j[i] = modder(C_j[i])
return C
# Calc A^n*B
def power(A,B,n):
if n == 0:
return B
while n>1:
if n%2==1:
B = mult(A,B)
A = mult(A,A)
n//=2
return mult(A,B)
##### PROBLEM SPECIFIC CODE
n,m = [int(x) for x in input().split()]
#def DP(n):
# if n<=0:
# return 1 if n==0 else 0
# else:
# return DP(n-m) + DP(n-1)
mat = [[0.0]*m for _ in range(m)]
mat[0][0] = 1.0
mat[-1][0] = 1.0
for i in range(m-1):
mat[i][i+1] = 1.0
vec = [[1.0]*m]
vec = power(mat,vec,n)
print(int(vec[0][-1]))
``` | instruction | 0 | 33,395 | 2 | 66,790 |
Yes | output | 1 | 33,395 | 2 | 66,791 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Reziba has many magic gems. Each magic gem can be split into M normal gems. The amount of space each magic (and normal) gem takes is 1 unit. A normal gem cannot be split.
Reziba wants to choose a set of magic gems and split some of them, so the total space occupied by the resulting set of gems is N units. If a magic gem is chosen and split, it takes M units of space (since it is split into M gems); if a magic gem is not split, it takes 1 unit.
How many different configurations of the resulting set of gems can Reziba have, such that the total amount of space taken is N units? Print the answer modulo 1000000007 (10^9+7). Two configurations are considered different if the number of magic gems Reziba takes to form them differs, or the indices of gems Reziba has to split differ.
Input
The input contains a single line consisting of 2 integers N and M (1 ≤ N ≤ 10^{18}, 2 ≤ M ≤ 100).
Output
Print one integer, the total number of configurations of the resulting set of gems, given that the total amount of space taken is N units. Print the answer modulo 1000000007 (10^9+7).
Examples
Input
4 2
Output
5
Input
3 2
Output
3
Note
In the first example each magic gem can split into 2 normal gems, and we know that the total amount of gems are 4.
Let 1 denote a magic gem, and 0 denote a normal gem.
The total configurations you can have is:
* 1 1 1 1 (None of the gems split);
* 0 0 1 1 (First magic gem splits into 2 normal gems);
* 1 0 0 1 (Second magic gem splits into 2 normal gems);
* 1 1 0 0 (Third magic gem splits into 2 normal gems);
* 0 0 0 0 (First and second magic gems split into total 4 normal gems).
Hence, answer is 5.
Submitted Solution:
```
import sys
# How to multiply matrices in 32 bit python?
# Solution: Use doubles with homemade mult and mod.
# This should be around 6-10 times faster than using int
# and close to the speed you'd get in 64 bit python.
MOD = 10**9 + 7
# "Crazy code"
MOD = 10**9 + 7
MODF = 1.0*MOD
MODF_inv = 1.0/MODF
SHRT = 1.0*(1 << 16)
SHRT_inv = 1.0/SHRT
MAGIC = float(2**52+2**51)
fround = lambda x:(x+MAGIC)-MAGIC
modder = lambda a:a - MODF * fround(a*MODF_inv)
mod_prod = lambda a, b, c=0.0:\
modder(modder(a * SHRT) * fround(SHRT_inv*b) + a*(b-SHRT * fround(b*SHRT_inv))+ c)
# Quick version
# Uses the same idea as prod but with some precalculations
def mult(A,B):
n,m = len(A[0]),len(B)
A = [[(Aij-SHRT*fround(Aij*SHRT_inv)) - 1j*fround(Aij*SHRT_inv) for Aij in A_j] for A_j in A]
C = [[0.0]*n for _ in range(m)]
for j,B_j in enumerate(B):
C_j = C[j]
for k,A_k in enumerate(A):
Bkj = B_j[k] + 1j*modder(B_j[k]*SHRT)
for i,Aik in enumerate(A_k):
C_j[i] += (Aik*Bkj).real
# Necessary for larger matrices, around 100x100:
#if C_j[i]>=FLOAT_PREC: C_j[i] = modder(C_j[i])
for i in range(n):
C_j[i] = modder(C_j[i])
return C
# Calc A^n*B
def power(A,B,n,mult_func):
if n == 0:
return B
while n>1:
if n%2==1:
B = mult_func(A,B)
A = mult_func(A,A)
n//=2
return mult_func(A,B)
##### EXAMPLE
n,m = [int(x) for x in input().split()]
#def DP(n):
# if n<=0:
# return 1 if n==0 else 0
# else:
# return DP(n-m) + DP(n-1)
mat = [[0.0]*m for _ in range(m)]
mat[0][0] = 1.0
mat[-1][0] = 1.0
for i in range(m-1):
mat[i][i+1] = 1.0
vec = [[1.0]*m]
vec = power(mat,vec,n,mult)
print(int(vec[0][-1])%MOD)
``` | instruction | 0 | 33,396 | 2 | 66,792 |
Yes | output | 1 | 33,396 | 2 | 66,793 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Reziba has many magic gems. Each magic gem can be split into M normal gems. The amount of space each magic (and normal) gem takes is 1 unit. A normal gem cannot be split.
Reziba wants to choose a set of magic gems and split some of them, so the total space occupied by the resulting set of gems is N units. If a magic gem is chosen and split, it takes M units of space (since it is split into M gems); if a magic gem is not split, it takes 1 unit.
How many different configurations of the resulting set of gems can Reziba have, such that the total amount of space taken is N units? Print the answer modulo 1000000007 (10^9+7). Two configurations are considered different if the number of magic gems Reziba takes to form them differs, or the indices of gems Reziba has to split differ.
Input
The input contains a single line consisting of 2 integers N and M (1 ≤ N ≤ 10^{18}, 2 ≤ M ≤ 100).
Output
Print one integer, the total number of configurations of the resulting set of gems, given that the total amount of space taken is N units. Print the answer modulo 1000000007 (10^9+7).
Examples
Input
4 2
Output
5
Input
3 2
Output
3
Note
In the first example each magic gem can split into 2 normal gems, and we know that the total amount of gems are 4.
Let 1 denote a magic gem, and 0 denote a normal gem.
The total configurations you can have is:
* 1 1 1 1 (None of the gems split);
* 0 0 1 1 (First magic gem splits into 2 normal gems);
* 1 0 0 1 (Second magic gem splits into 2 normal gems);
* 1 1 0 0 (Third magic gem splits into 2 normal gems);
* 0 0 0 0 (First and second magic gems split into total 4 normal gems).
Hence, answer is 5.
Submitted Solution:
```
n,m=map(int,input().split())
if(n<m):
print(1)
elif(n==m):
print(2)
else:
a=n-m
a+=1
#a+=2
if(n%m!=0):
a-=1
print(a%1000000007+2)
``` | instruction | 0 | 33,397 | 2 | 66,794 |
No | output | 1 | 33,397 | 2 | 66,795 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Reziba has many magic gems. Each magic gem can be split into M normal gems. The amount of space each magic (and normal) gem takes is 1 unit. A normal gem cannot be split.
Reziba wants to choose a set of magic gems and split some of them, so the total space occupied by the resulting set of gems is N units. If a magic gem is chosen and split, it takes M units of space (since it is split into M gems); if a magic gem is not split, it takes 1 unit.
How many different configurations of the resulting set of gems can Reziba have, such that the total amount of space taken is N units? Print the answer modulo 1000000007 (10^9+7). Two configurations are considered different if the number of magic gems Reziba takes to form them differs, or the indices of gems Reziba has to split differ.
Input
The input contains a single line consisting of 2 integers N and M (1 ≤ N ≤ 10^{18}, 2 ≤ M ≤ 100).
Output
Print one integer, the total number of configurations of the resulting set of gems, given that the total amount of space taken is N units. Print the answer modulo 1000000007 (10^9+7).
Examples
Input
4 2
Output
5
Input
3 2
Output
3
Note
In the first example each magic gem can split into 2 normal gems, and we know that the total amount of gems are 4.
Let 1 denote a magic gem, and 0 denote a normal gem.
The total configurations you can have is:
* 1 1 1 1 (None of the gems split);
* 0 0 1 1 (First magic gem splits into 2 normal gems);
* 1 0 0 1 (Second magic gem splits into 2 normal gems);
* 1 1 0 0 (Third magic gem splits into 2 normal gems);
* 0 0 0 0 (First and second magic gems split into total 4 normal gems).
Hence, answer is 5.
Submitted Solution:
```
a = input().split(" ")
n = int(a[0])
m = int(a[1])
k = n // m
ans = (2 * n) - ((k + 1) * m) + 2
ans = ans * k
ans = ans // 2
ans = ans + 1
print (ans % 1000000007)
``` | instruction | 0 | 33,398 | 2 | 66,796 |
No | output | 1 | 33,398 | 2 | 66,797 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Reziba has many magic gems. Each magic gem can be split into M normal gems. The amount of space each magic (and normal) gem takes is 1 unit. A normal gem cannot be split.
Reziba wants to choose a set of magic gems and split some of them, so the total space occupied by the resulting set of gems is N units. If a magic gem is chosen and split, it takes M units of space (since it is split into M gems); if a magic gem is not split, it takes 1 unit.
How many different configurations of the resulting set of gems can Reziba have, such that the total amount of space taken is N units? Print the answer modulo 1000000007 (10^9+7). Two configurations are considered different if the number of magic gems Reziba takes to form them differs, or the indices of gems Reziba has to split differ.
Input
The input contains a single line consisting of 2 integers N and M (1 ≤ N ≤ 10^{18}, 2 ≤ M ≤ 100).
Output
Print one integer, the total number of configurations of the resulting set of gems, given that the total amount of space taken is N units. Print the answer modulo 1000000007 (10^9+7).
Examples
Input
4 2
Output
5
Input
3 2
Output
3
Note
In the first example each magic gem can split into 2 normal gems, and we know that the total amount of gems are 4.
Let 1 denote a magic gem, and 0 denote a normal gem.
The total configurations you can have is:
* 1 1 1 1 (None of the gems split);
* 0 0 1 1 (First magic gem splits into 2 normal gems);
* 1 0 0 1 (Second magic gem splits into 2 normal gems);
* 1 1 0 0 (Third magic gem splits into 2 normal gems);
* 0 0 0 0 (First and second magic gems split into total 4 normal gems).
Hence, answer is 5.
Submitted Solution:
```
n, m = map(int, input().split())
x = n // m
print((x + 1) * x // 2 + 1 + 1)
``` | instruction | 0 | 33,399 | 2 | 66,798 |
No | output | 1 | 33,399 | 2 | 66,799 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Reziba has many magic gems. Each magic gem can be split into M normal gems. The amount of space each magic (and normal) gem takes is 1 unit. A normal gem cannot be split.
Reziba wants to choose a set of magic gems and split some of them, so the total space occupied by the resulting set of gems is N units. If a magic gem is chosen and split, it takes M units of space (since it is split into M gems); if a magic gem is not split, it takes 1 unit.
How many different configurations of the resulting set of gems can Reziba have, such that the total amount of space taken is N units? Print the answer modulo 1000000007 (10^9+7). Two configurations are considered different if the number of magic gems Reziba takes to form them differs, or the indices of gems Reziba has to split differ.
Input
The input contains a single line consisting of 2 integers N and M (1 ≤ N ≤ 10^{18}, 2 ≤ M ≤ 100).
Output
Print one integer, the total number of configurations of the resulting set of gems, given that the total amount of space taken is N units. Print the answer modulo 1000000007 (10^9+7).
Examples
Input
4 2
Output
5
Input
3 2
Output
3
Note
In the first example each magic gem can split into 2 normal gems, and we know that the total amount of gems are 4.
Let 1 denote a magic gem, and 0 denote a normal gem.
The total configurations you can have is:
* 1 1 1 1 (None of the gems split);
* 0 0 1 1 (First magic gem splits into 2 normal gems);
* 1 0 0 1 (Second magic gem splits into 2 normal gems);
* 1 1 0 0 (Third magic gem splits into 2 normal gems);
* 0 0 0 0 (First and second magic gems split into total 4 normal gems).
Hence, answer is 5.
Submitted Solution:
```
a = input().split(" ")
n = int(a[0])
m = int(a[1])
k = n // m
mod = n % m
ans = (2 * n) - ((k + 1) * m) + 2
ans = ans * k
ans = ans // 2
ans = ans + 1
if k * m < n:
ans = ans + mod - 1
print (ans % 1000000007)
``` | instruction | 0 | 33,400 | 2 | 66,800 |
No | output | 1 | 33,400 | 2 | 66,801 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Heidi and Doctor Who hopped out of the TARDIS and found themselves at EPFL in 2018. They were surrounded by stormtroopers and Darth Vader was approaching. Miraculously, they managed to escape to a nearby rebel base but the Doctor was very confused. Heidi reminded him that last year's HC2 theme was Star Wars. Now he understood, and he's ready to face the evils of the Empire!
The rebels have s spaceships, each with a certain attacking power a.
They want to send their spaceships to destroy the empire bases and steal enough gold and supplies in order to keep the rebellion alive.
The empire has b bases, each with a certain defensive power d, and a certain amount of gold g.
A spaceship can attack all the bases which have a defensive power less than or equal to its attacking power.
If a spaceship attacks a base, it steals all the gold in that base.
The rebels are still undecided which spaceship to send out first, so they asked for the Doctor's help. They would like to know, for each spaceship, the maximum amount of gold it can steal.
Input
The first line contains integers s and b (1 ≤ s, b ≤ 10^5), the number of spaceships and the number of bases, respectively.
The second line contains s integers a (0 ≤ a ≤ 10^9), the attacking power of each spaceship.
The next b lines contain integers d, g (0 ≤ d ≤ 10^9, 0 ≤ g ≤ 10^4), the defensive power and the gold of each base, respectively.
Output
Print s integers, the maximum amount of gold each spaceship can steal, in the same order as the spaceships are given in the input.
Example
Input
5 4
1 3 5 2 4
0 1
4 2
2 8
9 4
Output
1 9 11 9 11
Note
The first spaceship can only attack the first base.
The second spaceship can attack the first and third bases.
The third spaceship can attack the first, second and third bases. | instruction | 0 | 33,403 | 2 | 66,806 |
Tags: binary search, sortings
Correct Solution:
```
import bisect
s,b = map(int,input().split())
a = list(map(int,input().split()))
d = []
for i in range(b):
d.append(list(map(int,input().split())))
d = sorted(d)
c = []
p = []
for i in range(len(d)):
c.append(d[i][0])
p.append(d[i][1])
k = [p[0]]
for i in range(1,len(p)):
k.append(k[-1]+p[i])
for i in range(len(a)):
l = bisect.bisect_right(c,a[i])
if(l==0):
print(0,end = ' ')
else:
print(k[l-1],end = ' ')
``` | output | 1 | 33,403 | 2 | 66,807 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Heidi and Doctor Who hopped out of the TARDIS and found themselves at EPFL in 2018. They were surrounded by stormtroopers and Darth Vader was approaching. Miraculously, they managed to escape to a nearby rebel base but the Doctor was very confused. Heidi reminded him that last year's HC2 theme was Star Wars. Now he understood, and he's ready to face the evils of the Empire!
The rebels have s spaceships, each with a certain attacking power a.
They want to send their spaceships to destroy the empire bases and steal enough gold and supplies in order to keep the rebellion alive.
The empire has b bases, each with a certain defensive power d, and a certain amount of gold g.
A spaceship can attack all the bases which have a defensive power less than or equal to its attacking power.
If a spaceship attacks a base, it steals all the gold in that base.
The rebels are still undecided which spaceship to send out first, so they asked for the Doctor's help. They would like to know, for each spaceship, the maximum amount of gold it can steal.
Input
The first line contains integers s and b (1 ≤ s, b ≤ 10^5), the number of spaceships and the number of bases, respectively.
The second line contains s integers a (0 ≤ a ≤ 10^9), the attacking power of each spaceship.
The next b lines contain integers d, g (0 ≤ d ≤ 10^9, 0 ≤ g ≤ 10^4), the defensive power and the gold of each base, respectively.
Output
Print s integers, the maximum amount of gold each spaceship can steal, in the same order as the spaceships are given in the input.
Example
Input
5 4
1 3 5 2 4
0 1
4 2
2 8
9 4
Output
1 9 11 9 11
Note
The first spaceship can only attack the first base.
The second spaceship can attack the first and third bases.
The third spaceship can attack the first, second and third bases. | instruction | 0 | 33,404 | 2 | 66,808 |
Tags: binary search, sortings
Correct Solution:
```
kq=[]
a=[]
d=[]
def at(enum):
return(enum[1])
def defen(enum):
return(enum[0])
st=input()
c=st.split(' ')
s=int(c[0])
b=int(c[1])
st=input()
c=st.split(' ')
for i in range(0,s):
x=int(c[i])
a.append((i,x))
for i in range(0,s):
kq.append(0)
for i in range(0,b):
st=input()
c=st.split(' ')
x=int(c[0])
y=int(c[1])
d.append((x,y))
a.sort(reverse=False,key=at)
d.sort(reverse=False,key=defen)
j=0;
sum=0;
for i in range(0,s):
if (j==b):
kq[a[i][0]]=sum
else:
while (a[i][1]>=d[j][0]) and (j<b):
sum=sum+d[j][1]
j=j+1;
if (j==b):
break
kq[a[i][0]]=sum
delim=' '
k=[]
for x in kq:
p=str(x)
k.append(p)
print(delim.join(k))
``` | output | 1 | 33,404 | 2 | 66,809 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Heidi and Doctor Who hopped out of the TARDIS and found themselves at EPFL in 2018. They were surrounded by stormtroopers and Darth Vader was approaching. Miraculously, they managed to escape to a nearby rebel base but the Doctor was very confused. Heidi reminded him that last year's HC2 theme was Star Wars. Now he understood, and he's ready to face the evils of the Empire!
The rebels have s spaceships, each with a certain attacking power a.
They want to send their spaceships to destroy the empire bases and steal enough gold and supplies in order to keep the rebellion alive.
The empire has b bases, each with a certain defensive power d, and a certain amount of gold g.
A spaceship can attack all the bases which have a defensive power less than or equal to its attacking power.
If a spaceship attacks a base, it steals all the gold in that base.
The rebels are still undecided which spaceship to send out first, so they asked for the Doctor's help. They would like to know, for each spaceship, the maximum amount of gold it can steal.
Input
The first line contains integers s and b (1 ≤ s, b ≤ 10^5), the number of spaceships and the number of bases, respectively.
The second line contains s integers a (0 ≤ a ≤ 10^9), the attacking power of each spaceship.
The next b lines contain integers d, g (0 ≤ d ≤ 10^9, 0 ≤ g ≤ 10^4), the defensive power and the gold of each base, respectively.
Output
Print s integers, the maximum amount of gold each spaceship can steal, in the same order as the spaceships are given in the input.
Example
Input
5 4
1 3 5 2 4
0 1
4 2
2 8
9 4
Output
1 9 11 9 11
Note
The first spaceship can only attack the first base.
The second spaceship can attack the first and third bases.
The third spaceship can attack the first, second and third bases. | instruction | 0 | 33,405 | 2 | 66,810 |
Tags: binary search, sortings
Correct Solution:
```
# ------------------- fast io --------------------
import os
import sys
from io import BytesIO, IOBase
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")
# ------------------- fast io --------------------
from bisect import bisect_left as bsl
ship,b=map(int,input().split())
vals=list(map(int,input().split()))
inf=[]
for s in range(b):
d,g=map(int,input().split())
inf.append((d,g))
inf.sort(key=lambda x: x[0])
prefsum=[inf[0][1]]
for s in range(1,b):
prefsum.append(prefsum[-1]+inf[s][1])
defe=[k[0] for k in inf]
ans=[]
for s in range(ship):
ind=min(bsl(defe,vals[s]),b-1)
if defe[ind]>vals[s]:
ind-=1
if ind==-1:
ans.append(0)
else:
ans.append(prefsum[ind])
print(*ans)
``` | output | 1 | 33,405 | 2 | 66,811 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Heidi and Doctor Who hopped out of the TARDIS and found themselves at EPFL in 2018. They were surrounded by stormtroopers and Darth Vader was approaching. Miraculously, they managed to escape to a nearby rebel base but the Doctor was very confused. Heidi reminded him that last year's HC2 theme was Star Wars. Now he understood, and he's ready to face the evils of the Empire!
The rebels have s spaceships, each with a certain attacking power a.
They want to send their spaceships to destroy the empire bases and steal enough gold and supplies in order to keep the rebellion alive.
The empire has b bases, each with a certain defensive power d, and a certain amount of gold g.
A spaceship can attack all the bases which have a defensive power less than or equal to its attacking power.
If a spaceship attacks a base, it steals all the gold in that base.
The rebels are still undecided which spaceship to send out first, so they asked for the Doctor's help. They would like to know, for each spaceship, the maximum amount of gold it can steal.
Input
The first line contains integers s and b (1 ≤ s, b ≤ 10^5), the number of spaceships and the number of bases, respectively.
The second line contains s integers a (0 ≤ a ≤ 10^9), the attacking power of each spaceship.
The next b lines contain integers d, g (0 ≤ d ≤ 10^9, 0 ≤ g ≤ 10^4), the defensive power and the gold of each base, respectively.
Output
Print s integers, the maximum amount of gold each spaceship can steal, in the same order as the spaceships are given in the input.
Example
Input
5 4
1 3 5 2 4
0 1
4 2
2 8
9 4
Output
1 9 11 9 11
Note
The first spaceship can only attack the first base.
The second spaceship can attack the first and third bases.
The third spaceship can attack the first, second and third bases. | instruction | 0 | 33,406 | 2 | 66,812 |
Tags: binary search, sortings
Correct Solution:
```
import bisect
s,b=map(int,input().split())
a=list(map(int,input().split()))
base=[]
bi=[]
for i in range(b):
p,q=map(int,input().split())
base.append((p,q))
bi.append(p)
def sortFirst(val):
return val[0]
base.sort(key=sortFirst)
bi.sort()
#print(base)
gold=[base[0][1]]
for i in range(1,b):
gold.append(base[i][1]+gold[i-1])
ans=[]
for i in a:
ind=bisect.bisect(bi,i,0,b-1)
if ind==0:
if bi[ind]>i:
ans.append(0)
else:
ans.append(gold[0])
elif ind==b-1 and bi[ind]<=i:
ans.append(gold[ind])
else:
ans.append(gold[ind-1])
print(*ans)
``` | output | 1 | 33,406 | 2 | 66,813 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Heidi and Doctor Who hopped out of the TARDIS and found themselves at EPFL in 2018. They were surrounded by stormtroopers and Darth Vader was approaching. Miraculously, they managed to escape to a nearby rebel base but the Doctor was very confused. Heidi reminded him that last year's HC2 theme was Star Wars. Now he understood, and he's ready to face the evils of the Empire!
The rebels have s spaceships, each with a certain attacking power a.
They want to send their spaceships to destroy the empire bases and steal enough gold and supplies in order to keep the rebellion alive.
The empire has b bases, each with a certain defensive power d, and a certain amount of gold g.
A spaceship can attack all the bases which have a defensive power less than or equal to its attacking power.
If a spaceship attacks a base, it steals all the gold in that base.
The rebels are still undecided which spaceship to send out first, so they asked for the Doctor's help. They would like to know, for each spaceship, the maximum amount of gold it can steal.
Input
The first line contains integers s and b (1 ≤ s, b ≤ 10^5), the number of spaceships and the number of bases, respectively.
The second line contains s integers a (0 ≤ a ≤ 10^9), the attacking power of each spaceship.
The next b lines contain integers d, g (0 ≤ d ≤ 10^9, 0 ≤ g ≤ 10^4), the defensive power and the gold of each base, respectively.
Output
Print s integers, the maximum amount of gold each spaceship can steal, in the same order as the spaceships are given in the input.
Example
Input
5 4
1 3 5 2 4
0 1
4 2
2 8
9 4
Output
1 9 11 9 11
Note
The first spaceship can only attack the first base.
The second spaceship can attack the first and third bases.
The third spaceship can attack the first, second and third bases. | instruction | 0 | 33,407 | 2 | 66,814 |
Tags: binary search, sortings
Correct Solution:
```
n, m = map(int, input().split())
a = sorted([(int(x), y) for x, y in zip(input().split(), range(n))])
b = []
for _ in range(m):
b.append(list(map(int, input().split())))
b = sorted(b)
p1 = 0
p2 = 0
s = 0
ans = [0 for _ in range(len(a))]
while p1 < len(a) and p2 < len(b):
# print(p1, p2)
if a[p1][0] >= b[p2][0]:
s += b[p2][1]
p2 += 1
else:
ans[a[p1][1]] = s
p1 += 1
while p1 < len(a):
ans[a[p1][1]] = s
p1 += 1
print(" ".join(list(map(str, ans))))
``` | output | 1 | 33,407 | 2 | 66,815 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Heidi and Doctor Who hopped out of the TARDIS and found themselves at EPFL in 2018. They were surrounded by stormtroopers and Darth Vader was approaching. Miraculously, they managed to escape to a nearby rebel base but the Doctor was very confused. Heidi reminded him that last year's HC2 theme was Star Wars. Now he understood, and he's ready to face the evils of the Empire!
The rebels have s spaceships, each with a certain attacking power a.
They want to send their spaceships to destroy the empire bases and steal enough gold and supplies in order to keep the rebellion alive.
The empire has b bases, each with a certain defensive power d, and a certain amount of gold g.
A spaceship can attack all the bases which have a defensive power less than or equal to its attacking power.
If a spaceship attacks a base, it steals all the gold in that base.
The rebels are still undecided which spaceship to send out first, so they asked for the Doctor's help. They would like to know, for each spaceship, the maximum amount of gold it can steal.
Input
The first line contains integers s and b (1 ≤ s, b ≤ 10^5), the number of spaceships and the number of bases, respectively.
The second line contains s integers a (0 ≤ a ≤ 10^9), the attacking power of each spaceship.
The next b lines contain integers d, g (0 ≤ d ≤ 10^9, 0 ≤ g ≤ 10^4), the defensive power and the gold of each base, respectively.
Output
Print s integers, the maximum amount of gold each spaceship can steal, in the same order as the spaceships are given in the input.
Example
Input
5 4
1 3 5 2 4
0 1
4 2
2 8
9 4
Output
1 9 11 9 11
Note
The first spaceship can only attack the first base.
The second spaceship can attack the first and third bases.
The third spaceship can attack the first, second and third bases. | instruction | 0 | 33,408 | 2 | 66,816 |
Tags: binary search, sortings
Correct Solution:
```
# -*- coding: utf-8 -*-
import sys
from operator import itemgetter
from fractions import gcd
from math import ceil, floor
from copy import deepcopy
from itertools import accumulate
from collections import Counter
import math
from functools import reduce
from bisect import bisect_right
sys.setrecursionlimit(200000)
input = sys.stdin.readline
def ii(): return int(input())
def mi(): return map(int, input().rstrip().split())
def lmi(): return list(map(int, input().rstrip().split()))
def li(): return list(input().rstrip())
def debug(x): print("debug: ", x, file=sys.stderr)
# template
class SegTree():
def __init__(self, n: int, function, unity: int):
self.UNITY = unity
self.fn = function
self.n = n
self.SIZE_R = 1
while self.SIZE_R < n:
self.SIZE_R *= 2
self.dat = [self.UNITY for _ in range(self.SIZE_R * 2 + 10)]
def set(self, a: int, v: int):
self.dat[self.SIZE_R + a] = v
def build(self):
for k in range(self.SIZE_R - 1, 0, -1):
self.dat[k] = self.fn(self.dat[k * 2], self.dat[k * 2 + 1])
def make_tree(self, arr: list):
if self.n != len(arr):
raise IndexError
for i in range(len(arr)):
self.set(i, arr[i])
self.build()
def update(self, a, v):
k = self.SIZE_R + a
self.dat[k] = v
while k > 1:
k = k // 2
self.dat[k] = self.fn(self.dat[k * 2], self.dat[k * 2 + 1])
def query(self, a, b):
vleft = self.UNITY
vright = self.UNITY
left = a + self.SIZE_R
right = b + self.SIZE_R
while left < right:
if left & 1:
vleft = self.fn(vleft, self.dat[left])
left += 1
if right & 1:
right -= 1
vright = self.fn(self.dat[right], vright)
left >>= 1
right >>= 1
return self.fn(vleft, vright)
def __getitem__(self, k):
return self.dat[k + self.SIZE_R]
# END CUT HERE
def main():
s, b = mi()
a = lmi()
d = []
g = []
for i in range(b):
_d, _g = mi()
d.append(_d)
g.append(_g)
tmp = [(d[i], g[i]) for i in range(b)]
tmp.sort(key=lambda x: x[0])
d.sort()
# print(tmp)
# print(d)
bit = SegTree(b, lambda a, b: a+b, 0)
for i in range(b):
bit.set(i, tmp[i][1])
bit.build()
for i in range(s):
tmp = bisect_right(d, a[i])
# print(tmp)
print(bit.query(0, tmp), end=" ")
print()
if __name__ == '__main__':
main()
``` | output | 1 | 33,408 | 2 | 66,817 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Heidi and Doctor Who hopped out of the TARDIS and found themselves at EPFL in 2018. They were surrounded by stormtroopers and Darth Vader was approaching. Miraculously, they managed to escape to a nearby rebel base but the Doctor was very confused. Heidi reminded him that last year's HC2 theme was Star Wars. Now he understood, and he's ready to face the evils of the Empire!
The rebels have s spaceships, each with a certain attacking power a.
They want to send their spaceships to destroy the empire bases and steal enough gold and supplies in order to keep the rebellion alive.
The empire has b bases, each with a certain defensive power d, and a certain amount of gold g.
A spaceship can attack all the bases which have a defensive power less than or equal to its attacking power.
If a spaceship attacks a base, it steals all the gold in that base.
The rebels are still undecided which spaceship to send out first, so they asked for the Doctor's help. They would like to know, for each spaceship, the maximum amount of gold it can steal.
Input
The first line contains integers s and b (1 ≤ s, b ≤ 10^5), the number of spaceships and the number of bases, respectively.
The second line contains s integers a (0 ≤ a ≤ 10^9), the attacking power of each spaceship.
The next b lines contain integers d, g (0 ≤ d ≤ 10^9, 0 ≤ g ≤ 10^4), the defensive power and the gold of each base, respectively.
Output
Print s integers, the maximum amount of gold each spaceship can steal, in the same order as the spaceships are given in the input.
Example
Input
5 4
1 3 5 2 4
0 1
4 2
2 8
9 4
Output
1 9 11 9 11
Note
The first spaceship can only attack the first base.
The second spaceship can attack the first and third bases.
The third spaceship can attack the first, second and third bases. | instruction | 0 | 33,409 | 2 | 66,818 |
Tags: binary search, sortings
Correct Solution:
```
import bisect
s,b=map(int,input().split())
l=list(map(int,input().split()))
d=[]
for i in range(b):
d.append(list(map(int,input().split())))
d.sort()
c,p=[],[]
for i in range(len(d)):
c.append(d[i][0])
p.append(d[i][1])
k=[p[0]]
for i in range(1,len(p)):
k.append(k[-1]+p[i])
for i in range(len(l)):
x=bisect.bisect_right(c,l[i])
if(x==0):
print(0,end = ' ')
else:
print(k[x-1],end = ' ')
``` | output | 1 | 33,409 | 2 | 66,819 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Heidi and Doctor Who hopped out of the TARDIS and found themselves at EPFL in 2018. They were surrounded by stormtroopers and Darth Vader was approaching. Miraculously, they managed to escape to a nearby rebel base but the Doctor was very confused. Heidi reminded him that last year's HC2 theme was Star Wars. Now he understood, and he's ready to face the evils of the Empire!
The rebels have s spaceships, each with a certain attacking power a.
They want to send their spaceships to destroy the empire bases and steal enough gold and supplies in order to keep the rebellion alive.
The empire has b bases, each with a certain defensive power d, and a certain amount of gold g.
A spaceship can attack all the bases which have a defensive power less than or equal to its attacking power.
If a spaceship attacks a base, it steals all the gold in that base.
The rebels are still undecided which spaceship to send out first, so they asked for the Doctor's help. They would like to know, for each spaceship, the maximum amount of gold it can steal.
Input
The first line contains integers s and b (1 ≤ s, b ≤ 10^5), the number of spaceships and the number of bases, respectively.
The second line contains s integers a (0 ≤ a ≤ 10^9), the attacking power of each spaceship.
The next b lines contain integers d, g (0 ≤ d ≤ 10^9, 0 ≤ g ≤ 10^4), the defensive power and the gold of each base, respectively.
Output
Print s integers, the maximum amount of gold each spaceship can steal, in the same order as the spaceships are given in the input.
Example
Input
5 4
1 3 5 2 4
0 1
4 2
2 8
9 4
Output
1 9 11 9 11
Note
The first spaceship can only attack the first base.
The second spaceship can attack the first and third bases.
The third spaceship can attack the first, second and third bases. | instruction | 0 | 33,410 | 2 | 66,820 |
Tags: binary search, sortings
Correct Solution:
```
def solve(bl,i,l,r):
while(l<=r):
if(i<bl[l][0]):
return 0
elif(i>=bl[r][0]):
return bl[r][1]
m = (l+r)//2
if(m==l):
return bl[l][1]
if(i>=bl[m][0]):
l = m
else:
r = m-1
return 0
s,b = [int(i) for i in input().split()]
sl = [int(i) for i in input().split()]
bl = []
for i in range(b):
bl.append([int(i) for i in input().split()])
bl.sort()
for i in range(1,b):
bl[i][1] += bl[i-1][1]
#print(bl)
for i in sl:
print(solve(bl,i,0,b-1),end=" ")
print()
``` | output | 1 | 33,410 | 2 | 66,821 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Heidi and Doctor Who hopped out of the TARDIS and found themselves at EPFL in 2018. They were surrounded by stormtroopers and Darth Vader was approaching. Miraculously, they managed to escape to a nearby rebel base but the Doctor was very confused. Heidi reminded him that last year's HC2 theme was Star Wars. Now he understood, and he's ready to face the evils of the Empire!
The rebels have s spaceships, each with a certain attacking power a.
They want to send their spaceships to destroy the empire bases and steal enough gold and supplies in order to keep the rebellion alive.
The empire has b bases, each with a certain defensive power d, and a certain amount of gold g.
A spaceship can attack all the bases which have a defensive power less than or equal to its attacking power.
If a spaceship attacks a base, it steals all the gold in that base.
The rebels are still undecided which spaceship to send out first, so they asked for the Doctor's help. They would like to know, for each spaceship, the maximum amount of gold it can steal.
Input
The first line contains integers s and b (1 ≤ s, b ≤ 10^5), the number of spaceships and the number of bases, respectively.
The second line contains s integers a (0 ≤ a ≤ 10^9), the attacking power of each spaceship.
The next b lines contain integers d, g (0 ≤ d ≤ 10^9, 0 ≤ g ≤ 10^4), the defensive power and the gold of each base, respectively.
Output
Print s integers, the maximum amount of gold each spaceship can steal, in the same order as the spaceships are given in the input.
Example
Input
5 4
1 3 5 2 4
0 1
4 2
2 8
9 4
Output
1 9 11 9 11
Note
The first spaceship can only attack the first base.
The second spaceship can attack the first and third bases.
The third spaceship can attack the first, second and third bases.
Submitted Solution:
```
from sys import stdin, stdout
def main():
s, b = map(int, stdin.readline().strip().split())
a = [int(num) for num in stdin.readline().strip().split()]
req = []
for __ in range(b):
d, g = map(int, stdin.readline().strip().split())
req.append((d, g))
req = sorted(req, key=lambda pair: pair[0], reverse=False)
p_sum = []
for i in req:
if not p_sum:
p_sum.append(i[1])
else:
p_sum.append(i[1] + p_sum[-1])
outputs = []
for target in a:
l, r = -1, b
while l + 1 < r:
m = l + (r - l) // 2
if req[m][0] <= target:
l = m
else:
r = m
if l != -1:
outputs.append(f'{p_sum[l]}')
else:
outputs.append('0')
stdout.write(' '.join(outputs))
if __name__ == '__main__':
main()
``` | instruction | 0 | 33,411 | 2 | 66,822 |
Yes | output | 1 | 33,411 | 2 | 66,823 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Heidi and Doctor Who hopped out of the TARDIS and found themselves at EPFL in 2018. They were surrounded by stormtroopers and Darth Vader was approaching. Miraculously, they managed to escape to a nearby rebel base but the Doctor was very confused. Heidi reminded him that last year's HC2 theme was Star Wars. Now he understood, and he's ready to face the evils of the Empire!
The rebels have s spaceships, each with a certain attacking power a.
They want to send their spaceships to destroy the empire bases and steal enough gold and supplies in order to keep the rebellion alive.
The empire has b bases, each with a certain defensive power d, and a certain amount of gold g.
A spaceship can attack all the bases which have a defensive power less than or equal to its attacking power.
If a spaceship attacks a base, it steals all the gold in that base.
The rebels are still undecided which spaceship to send out first, so they asked for the Doctor's help. They would like to know, for each spaceship, the maximum amount of gold it can steal.
Input
The first line contains integers s and b (1 ≤ s, b ≤ 10^5), the number of spaceships and the number of bases, respectively.
The second line contains s integers a (0 ≤ a ≤ 10^9), the attacking power of each spaceship.
The next b lines contain integers d, g (0 ≤ d ≤ 10^9, 0 ≤ g ≤ 10^4), the defensive power and the gold of each base, respectively.
Output
Print s integers, the maximum amount of gold each spaceship can steal, in the same order as the spaceships are given in the input.
Example
Input
5 4
1 3 5 2 4
0 1
4 2
2 8
9 4
Output
1 9 11 9 11
Note
The first spaceship can only attack the first base.
The second spaceship can attack the first and third bases.
The third spaceship can attack the first, second and third bases.
Submitted Solution:
```
import sys
from copy import deepcopy
sys.setrecursionlimit(200000)
input = sys.stdin.readline
def ii(): return int(input())
def mi(): return map(int, input().rstrip().split())
def lmi(): return list(map(int, input().rstrip().split()))
def li(): return list(input().rstrip())
def debug(x): print("debug:", x, file=sys.stderr)
# template
# BEGIN CUT HERE
class CumulativeSum():
def __init__(self, arr: list):
self.data = deepcopy(arr)
for i in range(1, len(self.data)):
self.data[i] += self.data[i - 1]
def _query(self, k: int) -> int:
"""[0,k]の和"""
if k < 0:
return 0
else:
return self.data[min(k, len(self.data) - 1)]
def query(self, l: int, r: int) -> int:
"""[l,r)の和"""
if l > r:
return 0
else:
return self._query(r - 1) - self._query(l - 1)
def __str__(self):
return str(self.data)
def __getitem__(self, k):
return self.data[k]
class CumulativeXor():
def __init__(self, arr: list, function=lambda a, b: a ^ b):
self.data = deepcopy(arr)
self.fn = function
for i in range(1, len(self.data)):
self.data[i] = self.fn(self.data[i], self.data[i - 1])
def _query(self, k: int) -> int:
"""[0,k]のXOR"""
if k < 0:
return 0
else:
return self.data[min(k, len(self.data) - 1)]
def query(self, l: int, r: int) -> int:
"""[l,r)のXOR"""
if l > r:
return 0
else:
return self.fn(self._query(r - 1), self._query(l - 1))
def __str__(self):
return str(self.data)
def __getitem__(self, k):
return self.data[k]
# END CUT HERE
from bisect import bisect_right
def CF1184_B1():
s, b = mi()
a = lmi()
d = []
g = []
for i in range(b):
_d, _g = mi()
d.append(_d)
g.append(_g)
tmp = [(d[i], g[i]) for i in range(b)]
tmp.sort(key=lambda x: x[0])
d.sort()
bit = CumulativeSum([tmp[i][1] for i in range(b)])
for i in range(s):
tmp = bisect_right(d, a[i])
print(bit.query(0, tmp), end=" ")
print()
if __name__ == "__main__":
CF1184_B1()
``` | instruction | 0 | 33,412 | 2 | 66,824 |
Yes | output | 1 | 33,412 | 2 | 66,825 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Heidi and Doctor Who hopped out of the TARDIS and found themselves at EPFL in 2018. They were surrounded by stormtroopers and Darth Vader was approaching. Miraculously, they managed to escape to a nearby rebel base but the Doctor was very confused. Heidi reminded him that last year's HC2 theme was Star Wars. Now he understood, and he's ready to face the evils of the Empire!
The rebels have s spaceships, each with a certain attacking power a.
They want to send their spaceships to destroy the empire bases and steal enough gold and supplies in order to keep the rebellion alive.
The empire has b bases, each with a certain defensive power d, and a certain amount of gold g.
A spaceship can attack all the bases which have a defensive power less than or equal to its attacking power.
If a spaceship attacks a base, it steals all the gold in that base.
The rebels are still undecided which spaceship to send out first, so they asked for the Doctor's help. They would like to know, for each spaceship, the maximum amount of gold it can steal.
Input
The first line contains integers s and b (1 ≤ s, b ≤ 10^5), the number of spaceships and the number of bases, respectively.
The second line contains s integers a (0 ≤ a ≤ 10^9), the attacking power of each spaceship.
The next b lines contain integers d, g (0 ≤ d ≤ 10^9, 0 ≤ g ≤ 10^4), the defensive power and the gold of each base, respectively.
Output
Print s integers, the maximum amount of gold each spaceship can steal, in the same order as the spaceships are given in the input.
Example
Input
5 4
1 3 5 2 4
0 1
4 2
2 8
9 4
Output
1 9 11 9 11
Note
The first spaceship can only attack the first base.
The second spaceship can attack the first and third bases.
The third spaceship can attack the first, second and third bases.
Submitted Solution:
```
n, b = [int(a) for a in input().split(" ")]
p_arr = [int(p) for p in input().split(" ")]
b_arr = []
n_p_arr = []
for i in range(0, len(p_arr)):
n_p_arr.append([p_arr[i], i+1])
sorted_n_p_arr = sorted(n_p_arr, key=lambda x: x[0], reverse=True)
for i in range(0, b):
c = input().split(" ")
b_arr.append([int(a) for a in c])
sorted_b_arr = sorted(b_arr, key= lambda x: x[0], reverse=True)
result_arr = []
merge_arr = []
a_d = dict()
for i in p_arr:
a_d[i] = False
for i in b_arr:
merge_arr.append([i[0], i[1]])
a_d[i[0]] = True
for i in p_arr:
if a_d[i] != True:
merge_arr.append([i, -1])
s_merged_arr = sorted(merge_arr, key=lambda x: x[0])
if s_merged_arr[0][1] != -1:
t_c = s_merged_arr[0][1]
else :
t_c = 0
for i in range(1, len(s_merged_arr)):
if s_merged_arr[i][1] == -1:
s_merged_arr[i][1] = t_c
else :
t_c = t_c + s_merged_arr[i][1]
s_merged_arr[i][1] = t_c
a_d[s_merged_arr[i][0]] = t_c
res = []
for i in p_arr:
if a_d[i] == False:
res.append(0)
else :
res.append(a_d[i])
print(" ".join([str(a) for a in res]))
``` | instruction | 0 | 33,413 | 2 | 66,826 |
Yes | output | 1 | 33,413 | 2 | 66,827 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Heidi and Doctor Who hopped out of the TARDIS and found themselves at EPFL in 2018. They were surrounded by stormtroopers and Darth Vader was approaching. Miraculously, they managed to escape to a nearby rebel base but the Doctor was very confused. Heidi reminded him that last year's HC2 theme was Star Wars. Now he understood, and he's ready to face the evils of the Empire!
The rebels have s spaceships, each with a certain attacking power a.
They want to send their spaceships to destroy the empire bases and steal enough gold and supplies in order to keep the rebellion alive.
The empire has b bases, each with a certain defensive power d, and a certain amount of gold g.
A spaceship can attack all the bases which have a defensive power less than or equal to its attacking power.
If a spaceship attacks a base, it steals all the gold in that base.
The rebels are still undecided which spaceship to send out first, so they asked for the Doctor's help. They would like to know, for each spaceship, the maximum amount of gold it can steal.
Input
The first line contains integers s and b (1 ≤ s, b ≤ 10^5), the number of spaceships and the number of bases, respectively.
The second line contains s integers a (0 ≤ a ≤ 10^9), the attacking power of each spaceship.
The next b lines contain integers d, g (0 ≤ d ≤ 10^9, 0 ≤ g ≤ 10^4), the defensive power and the gold of each base, respectively.
Output
Print s integers, the maximum amount of gold each spaceship can steal, in the same order as the spaceships are given in the input.
Example
Input
5 4
1 3 5 2 4
0 1
4 2
2 8
9 4
Output
1 9 11 9 11
Note
The first spaceship can only attack the first base.
The second spaceship can attack the first and third bases.
The third spaceship can attack the first, second and third bases.
Submitted Solution:
```
import bisect
from collections import defaultdict
s,b=map(int,input().split())
a=[int(x) for x in input().split()]
d=defaultdict(int)
l=[]
for i in range(b):
x,y=map(int,input().split())
#d[x]=y
l+=[(x,y)]
l.sort(key=lambda x:x[0])
res=[]
for i in range(b):
if(i==0):
res+=[l[i][1]]
else:
res+=[res[-1]+l[i][1]]
#print(l)
#print(res)
for i in range(s):
t=a[i]
I = 0;
J = len(l) - 1;
V = -1;
while (I <= J):
m = (I + J) // 2;
if (t >= l[m][0]):
V = m;
I = m + 1;
else:
J = m - 1;
if V == -1:
print(0,end=" ")
else:
print(res[V],end=" ")
``` | instruction | 0 | 33,414 | 2 | 66,828 |
Yes | output | 1 | 33,414 | 2 | 66,829 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Heidi and Doctor Who hopped out of the TARDIS and found themselves at EPFL in 2018. They were surrounded by stormtroopers and Darth Vader was approaching. Miraculously, they managed to escape to a nearby rebel base but the Doctor was very confused. Heidi reminded him that last year's HC2 theme was Star Wars. Now he understood, and he's ready to face the evils of the Empire!
The rebels have s spaceships, each with a certain attacking power a.
They want to send their spaceships to destroy the empire bases and steal enough gold and supplies in order to keep the rebellion alive.
The empire has b bases, each with a certain defensive power d, and a certain amount of gold g.
A spaceship can attack all the bases which have a defensive power less than or equal to its attacking power.
If a spaceship attacks a base, it steals all the gold in that base.
The rebels are still undecided which spaceship to send out first, so they asked for the Doctor's help. They would like to know, for each spaceship, the maximum amount of gold it can steal.
Input
The first line contains integers s and b (1 ≤ s, b ≤ 10^5), the number of spaceships and the number of bases, respectively.
The second line contains s integers a (0 ≤ a ≤ 10^9), the attacking power of each spaceship.
The next b lines contain integers d, g (0 ≤ d ≤ 10^9, 0 ≤ g ≤ 10^4), the defensive power and the gold of each base, respectively.
Output
Print s integers, the maximum amount of gold each spaceship can steal, in the same order as the spaceships are given in the input.
Example
Input
5 4
1 3 5 2 4
0 1
4 2
2 8
9 4
Output
1 9 11 9 11
Note
The first spaceship can only attack the first base.
The second spaceship can attack the first and third bases.
The third spaceship can attack the first, second and third bases.
Submitted Solution:
```
from bisect import *
s,b = map(int,input().split())
spaceships=list(map(int,input().split()))
bases = []
for x in range(b):
bases.append(list(map(int,input().split())))
bases.sort(key = lambda x:x[0])
val =bases[0][1]
for i in range(1,b):
bases[i][1]+=val
val = bases[i][1]
defence,gold=map(list,zip(*bases))
for i in spaceships:
index = bisect_right(defence,i)-1
if index<=0:
print(0,end=" ")
continue
print(gold[index],end=" ")
``` | instruction | 0 | 33,415 | 2 | 66,830 |
No | output | 1 | 33,415 | 2 | 66,831 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Heidi and Doctor Who hopped out of the TARDIS and found themselves at EPFL in 2018. They were surrounded by stormtroopers and Darth Vader was approaching. Miraculously, they managed to escape to a nearby rebel base but the Doctor was very confused. Heidi reminded him that last year's HC2 theme was Star Wars. Now he understood, and he's ready to face the evils of the Empire!
The rebels have s spaceships, each with a certain attacking power a.
They want to send their spaceships to destroy the empire bases and steal enough gold and supplies in order to keep the rebellion alive.
The empire has b bases, each with a certain defensive power d, and a certain amount of gold g.
A spaceship can attack all the bases which have a defensive power less than or equal to its attacking power.
If a spaceship attacks a base, it steals all the gold in that base.
The rebels are still undecided which spaceship to send out first, so they asked for the Doctor's help. They would like to know, for each spaceship, the maximum amount of gold it can steal.
Input
The first line contains integers s and b (1 ≤ s, b ≤ 10^5), the number of spaceships and the number of bases, respectively.
The second line contains s integers a (0 ≤ a ≤ 10^9), the attacking power of each spaceship.
The next b lines contain integers d, g (0 ≤ d ≤ 10^9, 0 ≤ g ≤ 10^4), the defensive power and the gold of each base, respectively.
Output
Print s integers, the maximum amount of gold each spaceship can steal, in the same order as the spaceships are given in the input.
Example
Input
5 4
1 3 5 2 4
0 1
4 2
2 8
9 4
Output
1 9 11 9 11
Note
The first spaceship can only attack the first base.
The second spaceship can attack the first and third bases.
The third spaceship can attack the first, second and third bases.
Submitted Solution:
```
import bisect
def solve(e,bases):
l=0;r=len(bases)-1;
while l<=r:
mid=(l+r)//2
if bases[mid][0]<=e:l=mid+1;ans=mid
else:r=mid-1
return bases[ans][1]
n,b=map(int,input().split())
ar=list(map(int,input().split()))
bases=[];summ=0
for i in range(b):
a,b=map(int,input().split())
bases.append([a,b])
bases=sorted(bases,key=lambda x:x[0]);summ=0
ele=[];gold=[]
for i in range(b):
summ+=bases[i][1]
ele.append(bases[i][0])
gold.append(summ)
ans=[]
for i in range(n):
idx=bisect.bisect(ele,ar[i])
if idx==b:ans.append(gold[b-1])
elif idx==0:ans.append(0);continue
else:ans.append(gold[idx-1])
print(*ans)
``` | instruction | 0 | 33,416 | 2 | 66,832 |
No | output | 1 | 33,416 | 2 | 66,833 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Heidi and Doctor Who hopped out of the TARDIS and found themselves at EPFL in 2018. They were surrounded by stormtroopers and Darth Vader was approaching. Miraculously, they managed to escape to a nearby rebel base but the Doctor was very confused. Heidi reminded him that last year's HC2 theme was Star Wars. Now he understood, and he's ready to face the evils of the Empire!
The rebels have s spaceships, each with a certain attacking power a.
They want to send their spaceships to destroy the empire bases and steal enough gold and supplies in order to keep the rebellion alive.
The empire has b bases, each with a certain defensive power d, and a certain amount of gold g.
A spaceship can attack all the bases which have a defensive power less than or equal to its attacking power.
If a spaceship attacks a base, it steals all the gold in that base.
The rebels are still undecided which spaceship to send out first, so they asked for the Doctor's help. They would like to know, for each spaceship, the maximum amount of gold it can steal.
Input
The first line contains integers s and b (1 ≤ s, b ≤ 10^5), the number of spaceships and the number of bases, respectively.
The second line contains s integers a (0 ≤ a ≤ 10^9), the attacking power of each spaceship.
The next b lines contain integers d, g (0 ≤ d ≤ 10^9, 0 ≤ g ≤ 10^4), the defensive power and the gold of each base, respectively.
Output
Print s integers, the maximum amount of gold each spaceship can steal, in the same order as the spaceships are given in the input.
Example
Input
5 4
1 3 5 2 4
0 1
4 2
2 8
9 4
Output
1 9 11 9 11
Note
The first spaceship can only attack the first base.
The second spaceship can attack the first and third bases.
The third spaceship can attack the first, second and third bases.
Submitted Solution:
```
import bisect
from collections import defaultdict
s,b=map(int,input().split())
a=[int(x) for x in input().split()]
d=defaultdict(int)
l=[]
for i in range(b):
x,y=map(int,input().split())
d[x]=y
l+=[x]
l.sort()
res=[0]
for i in range(b):
res+=[res[-1]+d[l[i]]]
#print(l)
#print(res)
for i in range(s):
val=bisect.bisect_right(l,a[i])
print(res[val],end=" ")
``` | instruction | 0 | 33,417 | 2 | 66,834 |
No | output | 1 | 33,417 | 2 | 66,835 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Heidi and Doctor Who hopped out of the TARDIS and found themselves at EPFL in 2018. They were surrounded by stormtroopers and Darth Vader was approaching. Miraculously, they managed to escape to a nearby rebel base but the Doctor was very confused. Heidi reminded him that last year's HC2 theme was Star Wars. Now he understood, and he's ready to face the evils of the Empire!
The rebels have s spaceships, each with a certain attacking power a.
They want to send their spaceships to destroy the empire bases and steal enough gold and supplies in order to keep the rebellion alive.
The empire has b bases, each with a certain defensive power d, and a certain amount of gold g.
A spaceship can attack all the bases which have a defensive power less than or equal to its attacking power.
If a spaceship attacks a base, it steals all the gold in that base.
The rebels are still undecided which spaceship to send out first, so they asked for the Doctor's help. They would like to know, for each spaceship, the maximum amount of gold it can steal.
Input
The first line contains integers s and b (1 ≤ s, b ≤ 10^5), the number of spaceships and the number of bases, respectively.
The second line contains s integers a (0 ≤ a ≤ 10^9), the attacking power of each spaceship.
The next b lines contain integers d, g (0 ≤ d ≤ 10^9, 0 ≤ g ≤ 10^4), the defensive power and the gold of each base, respectively.
Output
Print s integers, the maximum amount of gold each spaceship can steal, in the same order as the spaceships are given in the input.
Example
Input
5 4
1 3 5 2 4
0 1
4 2
2 8
9 4
Output
1 9 11 9 11
Note
The first spaceship can only attack the first base.
The second spaceship can attack the first and third bases.
The third spaceship can attack the first, second and third bases.
Submitted Solution:
```
import bisect
def solve(e,bases):
l=0;r=len(bases)-1;
while l<=r:
mid=(l+r)//2
if bases[mid][0]<=e:l=mid+1;ans=mid
else:r=mid-1
return bases[ans][1]
n,b=map(int,input().split())
ar=list(map(int,input().split()))
bases=[];summ=0
for i in range(b):
a,b=map(int,input().split())
bases.append([a,b])
bases=sorted(bases,key=lambda x:x[0]);summ=0
ele=[]
for i in range(b):
summ+=bases[i][1]
bases[i][1]=summ
ele.append(bases[i][0])
ans=[]
for i in range(n):
k=ar[i]
idx=bisect.bisect(ele,k)
#print(idx,k)
if idx==b:ans.append(bases[-1][1])
elif idx==0:continue
else:ans.append(bases[idx-1][1])
print(*ans)
``` | instruction | 0 | 33,418 | 2 | 66,836 |
No | output | 1 | 33,418 | 2 | 66,837 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A few years ago, Hitagi encountered a giant crab, who stole the whole of her body weight. Ever since, she tried to avoid contact with others, for fear that this secret might be noticed.
To get rid of the oddity and recover her weight, a special integer sequence is needed. Hitagi's sequence has been broken for a long time, but now Kaiki provides an opportunity.
Hitagi's sequence a has a length of n. Lost elements in it are denoted by zeros. Kaiki provides another sequence b, whose length k equals the number of lost elements in a (i.e. the number of zeros). Hitagi is to replace each zero in a with an element from b so that each element in b should be used exactly once. Hitagi knows, however, that, apart from 0, no integer occurs in a and b more than once in total.
If the resulting sequence is not an increasing sequence, then it has the power to recover Hitagi from the oddity. You are to determine whether this is possible, or Kaiki's sequence is just another fake. In other words, you should detect whether it is possible to replace each zero in a with an integer from b so that each integer from b is used exactly once, and the resulting sequence is not increasing.
Input
The first line of input contains two space-separated positive integers n (2 ≤ n ≤ 100) and k (1 ≤ k ≤ n) — the lengths of sequence a and b respectively.
The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 200) — Hitagi's broken sequence with exactly k zero elements.
The third line contains k space-separated integers b1, b2, ..., bk (1 ≤ bi ≤ 200) — the elements to fill into Hitagi's sequence.
Input guarantees that apart from 0, no integer occurs in a and b more than once in total.
Output
Output "Yes" if it's possible to replace zeros in a with elements in b and make the resulting sequence not increasing, and "No" otherwise.
Examples
Input
4 2
11 0 0 14
5 4
Output
Yes
Input
6 1
2 3 0 8 9 10
5
Output
No
Input
4 1
8 94 0 4
89
Output
Yes
Input
7 7
0 0 0 0 0 0 0
1 2 3 4 5 6 7
Output
Yes
Note
In the first sample:
* Sequence a is 11, 0, 0, 14.
* Two of the elements are lost, and the candidates in b are 5 and 4.
* There are two possible resulting sequences: 11, 5, 4, 14 and 11, 4, 5, 14, both of which fulfill the requirements. Thus the answer is "Yes".
In the second sample, the only possible resulting sequence is 2, 3, 5, 8, 9, 10, which is an increasing sequence and therefore invalid. | instruction | 0 | 33,918 | 2 | 67,836 |
Tags: constructive algorithms, greedy, implementation, sortings
Correct Solution:
```
n,k=map(int,input().split())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
b.sort()
b=b[::-1]
ch=0
for i in range(n):
if a[i]==0:
a[i]=b[ch]
ch+=1
f=0
for i in range(n-1):
if a[i]<a[i+1]:
f+=1
if f+1==n:
print('No')
else:
print('Yes')
``` | output | 1 | 33,918 | 2 | 67,837 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A few years ago, Hitagi encountered a giant crab, who stole the whole of her body weight. Ever since, she tried to avoid contact with others, for fear that this secret might be noticed.
To get rid of the oddity and recover her weight, a special integer sequence is needed. Hitagi's sequence has been broken for a long time, but now Kaiki provides an opportunity.
Hitagi's sequence a has a length of n. Lost elements in it are denoted by zeros. Kaiki provides another sequence b, whose length k equals the number of lost elements in a (i.e. the number of zeros). Hitagi is to replace each zero in a with an element from b so that each element in b should be used exactly once. Hitagi knows, however, that, apart from 0, no integer occurs in a and b more than once in total.
If the resulting sequence is not an increasing sequence, then it has the power to recover Hitagi from the oddity. You are to determine whether this is possible, or Kaiki's sequence is just another fake. In other words, you should detect whether it is possible to replace each zero in a with an integer from b so that each integer from b is used exactly once, and the resulting sequence is not increasing.
Input
The first line of input contains two space-separated positive integers n (2 ≤ n ≤ 100) and k (1 ≤ k ≤ n) — the lengths of sequence a and b respectively.
The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 200) — Hitagi's broken sequence with exactly k zero elements.
The third line contains k space-separated integers b1, b2, ..., bk (1 ≤ bi ≤ 200) — the elements to fill into Hitagi's sequence.
Input guarantees that apart from 0, no integer occurs in a and b more than once in total.
Output
Output "Yes" if it's possible to replace zeros in a with elements in b and make the resulting sequence not increasing, and "No" otherwise.
Examples
Input
4 2
11 0 0 14
5 4
Output
Yes
Input
6 1
2 3 0 8 9 10
5
Output
No
Input
4 1
8 94 0 4
89
Output
Yes
Input
7 7
0 0 0 0 0 0 0
1 2 3 4 5 6 7
Output
Yes
Note
In the first sample:
* Sequence a is 11, 0, 0, 14.
* Two of the elements are lost, and the candidates in b are 5 and 4.
* There are two possible resulting sequences: 11, 5, 4, 14 and 11, 4, 5, 14, both of which fulfill the requirements. Thus the answer is "Yes".
In the second sample, the only possible resulting sequence is 2, 3, 5, 8, 9, 10, which is an increasing sequence and therefore invalid. | instruction | 0 | 33,919 | 2 | 67,838 |
Tags: constructive algorithms, greedy, implementation, sortings
Correct Solution:
```
n=input()
a=input()
b=input()
k=n.split()[1]
n=n.split()[0]
a=a.split()
b=b.split()
a1=a.copy()
t0=a.count('0')
ib=0
for i in range(int(n)):
if a[i]=='0':
a[i]=b[ib]
ib=ib+1
j="Yes"
inc=0
for i in range(int(n)-1):
if int(a[i])<int(a[i+1]):
inc=inc+1
if inc==int(n)-1:
j="No"
if t0>1:
j="Yes"
print(j)
``` | output | 1 | 33,919 | 2 | 67,839 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A few years ago, Hitagi encountered a giant crab, who stole the whole of her body weight. Ever since, she tried to avoid contact with others, for fear that this secret might be noticed.
To get rid of the oddity and recover her weight, a special integer sequence is needed. Hitagi's sequence has been broken for a long time, but now Kaiki provides an opportunity.
Hitagi's sequence a has a length of n. Lost elements in it are denoted by zeros. Kaiki provides another sequence b, whose length k equals the number of lost elements in a (i.e. the number of zeros). Hitagi is to replace each zero in a with an element from b so that each element in b should be used exactly once. Hitagi knows, however, that, apart from 0, no integer occurs in a and b more than once in total.
If the resulting sequence is not an increasing sequence, then it has the power to recover Hitagi from the oddity. You are to determine whether this is possible, or Kaiki's sequence is just another fake. In other words, you should detect whether it is possible to replace each zero in a with an integer from b so that each integer from b is used exactly once, and the resulting sequence is not increasing.
Input
The first line of input contains two space-separated positive integers n (2 ≤ n ≤ 100) and k (1 ≤ k ≤ n) — the lengths of sequence a and b respectively.
The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 200) — Hitagi's broken sequence with exactly k zero elements.
The third line contains k space-separated integers b1, b2, ..., bk (1 ≤ bi ≤ 200) — the elements to fill into Hitagi's sequence.
Input guarantees that apart from 0, no integer occurs in a and b more than once in total.
Output
Output "Yes" if it's possible to replace zeros in a with elements in b and make the resulting sequence not increasing, and "No" otherwise.
Examples
Input
4 2
11 0 0 14
5 4
Output
Yes
Input
6 1
2 3 0 8 9 10
5
Output
No
Input
4 1
8 94 0 4
89
Output
Yes
Input
7 7
0 0 0 0 0 0 0
1 2 3 4 5 6 7
Output
Yes
Note
In the first sample:
* Sequence a is 11, 0, 0, 14.
* Two of the elements are lost, and the candidates in b are 5 and 4.
* There are two possible resulting sequences: 11, 5, 4, 14 and 11, 4, 5, 14, both of which fulfill the requirements. Thus the answer is "Yes".
In the second sample, the only possible resulting sequence is 2, 3, 5, 8, 9, 10, which is an increasing sequence and therefore invalid. | instruction | 0 | 33,920 | 2 | 67,840 |
Tags: constructive algorithms, greedy, implementation, sortings
Correct Solution:
```
from sys import stdin, stdout
n, k = map(int, stdin.readline().split())
first = list(map(int, stdin.readline().split()))
second = sorted(list(map(int, stdin.readline().split())))
for i in range(n):
if not first[i]:
first[i] = second.pop()
if first == sorted(first):
stdout.write('No')
else:
stdout.write('Yes')
``` | output | 1 | 33,920 | 2 | 67,841 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A few years ago, Hitagi encountered a giant crab, who stole the whole of her body weight. Ever since, she tried to avoid contact with others, for fear that this secret might be noticed.
To get rid of the oddity and recover her weight, a special integer sequence is needed. Hitagi's sequence has been broken for a long time, but now Kaiki provides an opportunity.
Hitagi's sequence a has a length of n. Lost elements in it are denoted by zeros. Kaiki provides another sequence b, whose length k equals the number of lost elements in a (i.e. the number of zeros). Hitagi is to replace each zero in a with an element from b so that each element in b should be used exactly once. Hitagi knows, however, that, apart from 0, no integer occurs in a and b more than once in total.
If the resulting sequence is not an increasing sequence, then it has the power to recover Hitagi from the oddity. You are to determine whether this is possible, or Kaiki's sequence is just another fake. In other words, you should detect whether it is possible to replace each zero in a with an integer from b so that each integer from b is used exactly once, and the resulting sequence is not increasing.
Input
The first line of input contains two space-separated positive integers n (2 ≤ n ≤ 100) and k (1 ≤ k ≤ n) — the lengths of sequence a and b respectively.
The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 200) — Hitagi's broken sequence with exactly k zero elements.
The third line contains k space-separated integers b1, b2, ..., bk (1 ≤ bi ≤ 200) — the elements to fill into Hitagi's sequence.
Input guarantees that apart from 0, no integer occurs in a and b more than once in total.
Output
Output "Yes" if it's possible to replace zeros in a with elements in b and make the resulting sequence not increasing, and "No" otherwise.
Examples
Input
4 2
11 0 0 14
5 4
Output
Yes
Input
6 1
2 3 0 8 9 10
5
Output
No
Input
4 1
8 94 0 4
89
Output
Yes
Input
7 7
0 0 0 0 0 0 0
1 2 3 4 5 6 7
Output
Yes
Note
In the first sample:
* Sequence a is 11, 0, 0, 14.
* Two of the elements are lost, and the candidates in b are 5 and 4.
* There are two possible resulting sequences: 11, 5, 4, 14 and 11, 4, 5, 14, both of which fulfill the requirements. Thus the answer is "Yes".
In the second sample, the only possible resulting sequence is 2, 3, 5, 8, 9, 10, which is an increasing sequence and therefore invalid. | instruction | 0 | 33,921 | 2 | 67,842 |
Tags: constructive algorithms, greedy, implementation, sortings
Correct Solution:
```
n,k = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
if k > 1:
print('Yes')
exit(0)
else:
for i in range(len(a)):
if a[i] == 0:
a[i] = b[0]
break
for i in range(len(a) - 1):
if a[i] > a[i + 1]:
print('Yes')
exit(0)
print('No')
``` | output | 1 | 33,921 | 2 | 67,843 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A few years ago, Hitagi encountered a giant crab, who stole the whole of her body weight. Ever since, she tried to avoid contact with others, for fear that this secret might be noticed.
To get rid of the oddity and recover her weight, a special integer sequence is needed. Hitagi's sequence has been broken for a long time, but now Kaiki provides an opportunity.
Hitagi's sequence a has a length of n. Lost elements in it are denoted by zeros. Kaiki provides another sequence b, whose length k equals the number of lost elements in a (i.e. the number of zeros). Hitagi is to replace each zero in a with an element from b so that each element in b should be used exactly once. Hitagi knows, however, that, apart from 0, no integer occurs in a and b more than once in total.
If the resulting sequence is not an increasing sequence, then it has the power to recover Hitagi from the oddity. You are to determine whether this is possible, or Kaiki's sequence is just another fake. In other words, you should detect whether it is possible to replace each zero in a with an integer from b so that each integer from b is used exactly once, and the resulting sequence is not increasing.
Input
The first line of input contains two space-separated positive integers n (2 ≤ n ≤ 100) and k (1 ≤ k ≤ n) — the lengths of sequence a and b respectively.
The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 200) — Hitagi's broken sequence with exactly k zero elements.
The third line contains k space-separated integers b1, b2, ..., bk (1 ≤ bi ≤ 200) — the elements to fill into Hitagi's sequence.
Input guarantees that apart from 0, no integer occurs in a and b more than once in total.
Output
Output "Yes" if it's possible to replace zeros in a with elements in b and make the resulting sequence not increasing, and "No" otherwise.
Examples
Input
4 2
11 0 0 14
5 4
Output
Yes
Input
6 1
2 3 0 8 9 10
5
Output
No
Input
4 1
8 94 0 4
89
Output
Yes
Input
7 7
0 0 0 0 0 0 0
1 2 3 4 5 6 7
Output
Yes
Note
In the first sample:
* Sequence a is 11, 0, 0, 14.
* Two of the elements are lost, and the candidates in b are 5 and 4.
* There are two possible resulting sequences: 11, 5, 4, 14 and 11, 4, 5, 14, both of which fulfill the requirements. Thus the answer is "Yes".
In the second sample, the only possible resulting sequence is 2, 3, 5, 8, 9, 10, which is an increasing sequence and therefore invalid. | instruction | 0 | 33,922 | 2 | 67,844 |
Tags: constructive algorithms, greedy, implementation, sortings
Correct Solution:
```
# -*- coding: utf-8 -*-
n, m = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
b.sort(reverse=True)
c = []
cur = 0
for i in a:
if (i != 0):
c.append(i)
else:
c.append(b[cur])
cur += 1
def isinc(arr):
for i in range(1, len(arr)):
if (arr[i - 1] > arr[i]):
return True
return False
if isinc(c):
print("Yes")
else:
print("No")
``` | output | 1 | 33,922 | 2 | 67,845 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A few years ago, Hitagi encountered a giant crab, who stole the whole of her body weight. Ever since, she tried to avoid contact with others, for fear that this secret might be noticed.
To get rid of the oddity and recover her weight, a special integer sequence is needed. Hitagi's sequence has been broken for a long time, but now Kaiki provides an opportunity.
Hitagi's sequence a has a length of n. Lost elements in it are denoted by zeros. Kaiki provides another sequence b, whose length k equals the number of lost elements in a (i.e. the number of zeros). Hitagi is to replace each zero in a with an element from b so that each element in b should be used exactly once. Hitagi knows, however, that, apart from 0, no integer occurs in a and b more than once in total.
If the resulting sequence is not an increasing sequence, then it has the power to recover Hitagi from the oddity. You are to determine whether this is possible, or Kaiki's sequence is just another fake. In other words, you should detect whether it is possible to replace each zero in a with an integer from b so that each integer from b is used exactly once, and the resulting sequence is not increasing.
Input
The first line of input contains two space-separated positive integers n (2 ≤ n ≤ 100) and k (1 ≤ k ≤ n) — the lengths of sequence a and b respectively.
The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 200) — Hitagi's broken sequence with exactly k zero elements.
The third line contains k space-separated integers b1, b2, ..., bk (1 ≤ bi ≤ 200) — the elements to fill into Hitagi's sequence.
Input guarantees that apart from 0, no integer occurs in a and b more than once in total.
Output
Output "Yes" if it's possible to replace zeros in a with elements in b and make the resulting sequence not increasing, and "No" otherwise.
Examples
Input
4 2
11 0 0 14
5 4
Output
Yes
Input
6 1
2 3 0 8 9 10
5
Output
No
Input
4 1
8 94 0 4
89
Output
Yes
Input
7 7
0 0 0 0 0 0 0
1 2 3 4 5 6 7
Output
Yes
Note
In the first sample:
* Sequence a is 11, 0, 0, 14.
* Two of the elements are lost, and the candidates in b are 5 and 4.
* There are two possible resulting sequences: 11, 5, 4, 14 and 11, 4, 5, 14, both of which fulfill the requirements. Thus the answer is "Yes".
In the second sample, the only possible resulting sequence is 2, 3, 5, 8, 9, 10, which is an increasing sequence and therefore invalid. | instruction | 0 | 33,923 | 2 | 67,846 |
Tags: constructive algorithms, greedy, implementation, sortings
Correct Solution:
```
n,k=map(int,input().split())
lst1=list(map(int,input().split()))
lst2=list(map(int,input().split()))
lst2.sort()
lst2.reverse()
i=0
for j in range(n):
if lst1[j]==0:
lst1[j]=lst2[i]
i+=1
if lst1!=sorted(lst1):
print("YES")
else:
print("NO")
``` | output | 1 | 33,923 | 2 | 67,847 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A few years ago, Hitagi encountered a giant crab, who stole the whole of her body weight. Ever since, she tried to avoid contact with others, for fear that this secret might be noticed.
To get rid of the oddity and recover her weight, a special integer sequence is needed. Hitagi's sequence has been broken for a long time, but now Kaiki provides an opportunity.
Hitagi's sequence a has a length of n. Lost elements in it are denoted by zeros. Kaiki provides another sequence b, whose length k equals the number of lost elements in a (i.e. the number of zeros). Hitagi is to replace each zero in a with an element from b so that each element in b should be used exactly once. Hitagi knows, however, that, apart from 0, no integer occurs in a and b more than once in total.
If the resulting sequence is not an increasing sequence, then it has the power to recover Hitagi from the oddity. You are to determine whether this is possible, or Kaiki's sequence is just another fake. In other words, you should detect whether it is possible to replace each zero in a with an integer from b so that each integer from b is used exactly once, and the resulting sequence is not increasing.
Input
The first line of input contains two space-separated positive integers n (2 ≤ n ≤ 100) and k (1 ≤ k ≤ n) — the lengths of sequence a and b respectively.
The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 200) — Hitagi's broken sequence with exactly k zero elements.
The third line contains k space-separated integers b1, b2, ..., bk (1 ≤ bi ≤ 200) — the elements to fill into Hitagi's sequence.
Input guarantees that apart from 0, no integer occurs in a and b more than once in total.
Output
Output "Yes" if it's possible to replace zeros in a with elements in b and make the resulting sequence not increasing, and "No" otherwise.
Examples
Input
4 2
11 0 0 14
5 4
Output
Yes
Input
6 1
2 3 0 8 9 10
5
Output
No
Input
4 1
8 94 0 4
89
Output
Yes
Input
7 7
0 0 0 0 0 0 0
1 2 3 4 5 6 7
Output
Yes
Note
In the first sample:
* Sequence a is 11, 0, 0, 14.
* Two of the elements are lost, and the candidates in b are 5 and 4.
* There are two possible resulting sequences: 11, 5, 4, 14 and 11, 4, 5, 14, both of which fulfill the requirements. Thus the answer is "Yes".
In the second sample, the only possible resulting sequence is 2, 3, 5, 8, 9, 10, which is an increasing sequence and therefore invalid. | instruction | 0 | 33,924 | 2 | 67,848 |
Tags: constructive algorithms, greedy, implementation, sortings
Correct Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Wed Jun 7 15:25:16 2017
@author: Gad
"""
def checker(size,string,replacements):
increasing = True
if int(size[1]) > 1:
return 'Yes'
else:
for i in range (int(size[0])):
if int (string[i]) == 0:
string[i] = replacements[0]
break
for i in range(int(size[0])-1):
if int (string[i]) > int(string[i+1]):
increasing = False
if increasing == False:
return 'Yes'
return 'No'
size = input().strip().split()
string = input().strip().split()
replacements= input().strip().split()
print(checker(size,string,replacements))
``` | output | 1 | 33,924 | 2 | 67,849 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.