message stringlengths 2 48.6k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 318 108k | cluster float64 8 8 | __index_level_0__ int64 636 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Emuskald is an avid horticulturist and owns the world's longest greenhouse — it is effectively infinite in length.
Over the years Emuskald has cultivated n plants in his greenhouse, of m different plant species numbered from 1 to m. His greenhouse is very narrow and can be viewed as an infinite line, with each plant occupying a single point on that line.
Emuskald has discovered that each species thrives at a different temperature, so he wants to arrange m - 1 borders that would divide the greenhouse into m sections numbered from 1 to m from left to right with each section housing a single species. He is free to place the borders, but in the end all of the i-th species plants must reside in i-th section from the left.
Of course, it is not always possible to place the borders in such way, so Emuskald needs to replant some of his plants. He can remove each plant from its position and place it anywhere in the greenhouse (at any real coordinate) with no plant already in it. Since replanting is a lot of stress for the plants, help Emuskald find the minimum number of plants he has to replant to be able to place the borders.
Input
The first line of input contains two space-separated integers n and m (1 ≤ n, m ≤ 5000, n ≥ m), the number of plants and the number of different species. Each of the following n lines contain two space-separated numbers: one integer number si (1 ≤ si ≤ m), and one real number xi (0 ≤ xi ≤ 109), the species and position of the i-th plant. Each xi will contain no more than 6 digits after the decimal point.
It is guaranteed that all xi are different; there is at least one plant of each species; the plants are given in order "from left to the right", that is in the ascending order of their xi coordinates (xi < xi + 1, 1 ≤ i < n).
Output
Output a single integer — the minimum number of plants to be replanted.
Examples
Input
3 2
2 1
1 2.0
1 3.100
Output
1
Input
3 3
1 5.0
2 5.5
3 6.0
Output
0
Input
6 3
1 14.284235
2 17.921382
1 20.328172
3 20.842331
1 25.790145
1 27.204125
Output
2
Note
In the first test case, Emuskald can replant the first plant to the right of the last plant, so the answer is 1.
In the second test case, the species are already in the correct order, so no replanting is needed.
Submitted Solution:
```
def CeilIndex(A, l, r, key):
while (r - l > 1):
m = l + (r - l)//2
if (A[m] > key):
r = m
else:
l = m
return r
def LongestIncreasingSubsequenceLength(A, size):
# Add boundary case,
# when array size is one
tailTable = [0 for i in range(size + 1)]
len = 0 # always points empty slot
tailTable[0] = A[0]
len = 1
for i in range(1, size):
if (A[i] < tailTable[0]):
# new smallest value
tailTable[0] = A[i]
elif (A[i] >= tailTable[len-1]):
# A[i] wants to extend
# largest subsequence
tailTable[len] = A[i]
len+= 1
else:
# A[i] wants to be current
# end candidate of an existing
# subsequence. It will replace
# ceil value in tailTable
tailTable[CeilIndex(tailTable, -1, len-1, A[i])] = A[i]
return len
# driver code
arr = []
n,m=map(int,input().split())
for i in range(n) :
a,b=map(float,input().split())
arr.append(a)
print(n-LongestIncreasingSubsequenceLength(arr, n))
# This code is contributed
# by Anant Agarwal.
``` | instruction | 0 | 12,759 | 8 | 25,518 |
Yes | output | 1 | 12,759 | 8 | 25,519 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Emuskald is an avid horticulturist and owns the world's longest greenhouse — it is effectively infinite in length.
Over the years Emuskald has cultivated n plants in his greenhouse, of m different plant species numbered from 1 to m. His greenhouse is very narrow and can be viewed as an infinite line, with each plant occupying a single point on that line.
Emuskald has discovered that each species thrives at a different temperature, so he wants to arrange m - 1 borders that would divide the greenhouse into m sections numbered from 1 to m from left to right with each section housing a single species. He is free to place the borders, but in the end all of the i-th species plants must reside in i-th section from the left.
Of course, it is not always possible to place the borders in such way, so Emuskald needs to replant some of his plants. He can remove each plant from its position and place it anywhere in the greenhouse (at any real coordinate) with no plant already in it. Since replanting is a lot of stress for the plants, help Emuskald find the minimum number of plants he has to replant to be able to place the borders.
Input
The first line of input contains two space-separated integers n and m (1 ≤ n, m ≤ 5000, n ≥ m), the number of plants and the number of different species. Each of the following n lines contain two space-separated numbers: one integer number si (1 ≤ si ≤ m), and one real number xi (0 ≤ xi ≤ 109), the species and position of the i-th plant. Each xi will contain no more than 6 digits after the decimal point.
It is guaranteed that all xi are different; there is at least one plant of each species; the plants are given in order "from left to the right", that is in the ascending order of their xi coordinates (xi < xi + 1, 1 ≤ i < n).
Output
Output a single integer — the minimum number of plants to be replanted.
Examples
Input
3 2
2 1
1 2.0
1 3.100
Output
1
Input
3 3
1 5.0
2 5.5
3 6.0
Output
0
Input
6 3
1 14.284235
2 17.921382
1 20.328172
3 20.842331
1 25.790145
1 27.204125
Output
2
Note
In the first test case, Emuskald can replant the first plant to the right of the last plant, so the answer is 1.
In the second test case, the species are already in the correct order, so no replanting is needed.
Submitted Solution:
```
#!/usr/bin/env python3
# created : 2020. 12. 31. 23:59
import os
from sys import stdin, stdout
from collections import deque
from bisect import bisect_left, bisect_right
def solve(tc):
n, m = map(int, stdin.readline().split())
species = []
for i in range(n):
si, xi = stdin.readline().split()
species.append(int(si))
ans = n+1
cnt = 0
dq = deque()
for i in range(n):
if len(dq) > 0 and dq[-1] > species[i]:
cnt += 1
k = bisect_right(dq, species[i])
dq.insert(k, species[i])
ans = min(ans, cnt)
cnt = 0
dq.clear()
for i in range(n-1, -1, -1):
if len(dq) > 0 and dq[0] < species[i]:
cnt += 1
k = bisect_left(dq, species[i])
dq.insert(k, species[i])
ans = min(ans, cnt)
print(ans)
tcs = 1
# tcs = int(stdin.readline().strip())
tc = 1
while tc <= tcs:
solve(tc)
tc += 1
``` | instruction | 0 | 12,760 | 8 | 25,520 |
No | output | 1 | 12,760 | 8 | 25,521 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Emuskald is an avid horticulturist and owns the world's longest greenhouse — it is effectively infinite in length.
Over the years Emuskald has cultivated n plants in his greenhouse, of m different plant species numbered from 1 to m. His greenhouse is very narrow and can be viewed as an infinite line, with each plant occupying a single point on that line.
Emuskald has discovered that each species thrives at a different temperature, so he wants to arrange m - 1 borders that would divide the greenhouse into m sections numbered from 1 to m from left to right with each section housing a single species. He is free to place the borders, but in the end all of the i-th species plants must reside in i-th section from the left.
Of course, it is not always possible to place the borders in such way, so Emuskald needs to replant some of his plants. He can remove each plant from its position and place it anywhere in the greenhouse (at any real coordinate) with no plant already in it. Since replanting is a lot of stress for the plants, help Emuskald find the minimum number of plants he has to replant to be able to place the borders.
Input
The first line of input contains two space-separated integers n and m (1 ≤ n, m ≤ 5000, n ≥ m), the number of plants and the number of different species. Each of the following n lines contain two space-separated numbers: one integer number si (1 ≤ si ≤ m), and one real number xi (0 ≤ xi ≤ 109), the species and position of the i-th plant. Each xi will contain no more than 6 digits after the decimal point.
It is guaranteed that all xi are different; there is at least one plant of each species; the plants are given in order "from left to the right", that is in the ascending order of their xi coordinates (xi < xi + 1, 1 ≤ i < n).
Output
Output a single integer — the minimum number of plants to be replanted.
Examples
Input
3 2
2 1
1 2.0
1 3.100
Output
1
Input
3 3
1 5.0
2 5.5
3 6.0
Output
0
Input
6 3
1 14.284235
2 17.921382
1 20.328172
3 20.842331
1 25.790145
1 27.204125
Output
2
Note
In the first test case, Emuskald can replant the first plant to the right of the last plant, so the answer is 1.
In the second test case, the species are already in the correct order, so no replanting is needed.
Submitted Solution:
```
'''n, m, k = [int(x) for x in input().split()]
trees = tuple([int(x) for x in input().split()])
costs = []
for x in range(n):
costs.append([0] + [int(x) for x in input().split()])
dp = [[[float('inf') for z in range(m + 1)] for y in range(k + 1)] for x in range(n)]
if trees[0] == 0:
for k in range(len(dp[0][0])):
dp[0][1][k] = cost[0][k]
else:
dp[0][0][trees[0]] = cost[0][trees[0]]
for i in range(1, len(dp)):
for j in range(1, len(dp[0])):
for k in range(1, len(dp[0][0])):
if trees
for l in range(len(dp[0][0])):
if k == l:
dp[i][j][k] = dp[i - 1][j][k] + cost[i][k]
else:
dp[i][j][k] = dp[i - 1][j - 1][l] + cost[i][k]'''
n, m = [int(x) for x in input().split()]
plant = [0] + [int(input().split()[0]) for x in range(n)]
dp = [0 for x in range(m + 1)]
for i in range(1, n):
for k in range(plant[i], 0, -1):
dp[plant[i]] = max(dp[plant[i]], 1 + dp[k])
print(n - max(dp) - 1)
``` | instruction | 0 | 12,761 | 8 | 25,522 |
No | output | 1 | 12,761 | 8 | 25,523 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Emuskald is an avid horticulturist and owns the world's longest greenhouse — it is effectively infinite in length.
Over the years Emuskald has cultivated n plants in his greenhouse, of m different plant species numbered from 1 to m. His greenhouse is very narrow and can be viewed as an infinite line, with each plant occupying a single point on that line.
Emuskald has discovered that each species thrives at a different temperature, so he wants to arrange m - 1 borders that would divide the greenhouse into m sections numbered from 1 to m from left to right with each section housing a single species. He is free to place the borders, but in the end all of the i-th species plants must reside in i-th section from the left.
Of course, it is not always possible to place the borders in such way, so Emuskald needs to replant some of his plants. He can remove each plant from its position and place it anywhere in the greenhouse (at any real coordinate) with no plant already in it. Since replanting is a lot of stress for the plants, help Emuskald find the minimum number of plants he has to replant to be able to place the borders.
Input
The first line of input contains two space-separated integers n and m (1 ≤ n, m ≤ 5000, n ≥ m), the number of plants and the number of different species. Each of the following n lines contain two space-separated numbers: one integer number si (1 ≤ si ≤ m), and one real number xi (0 ≤ xi ≤ 109), the species and position of the i-th plant. Each xi will contain no more than 6 digits after the decimal point.
It is guaranteed that all xi are different; there is at least one plant of each species; the plants are given in order "from left to the right", that is in the ascending order of their xi coordinates (xi < xi + 1, 1 ≤ i < n).
Output
Output a single integer — the minimum number of plants to be replanted.
Examples
Input
3 2
2 1
1 2.0
1 3.100
Output
1
Input
3 3
1 5.0
2 5.5
3 6.0
Output
0
Input
6 3
1 14.284235
2 17.921382
1 20.328172
3 20.842331
1 25.790145
1 27.204125
Output
2
Note
In the first test case, Emuskald can replant the first plant to the right of the last plant, so the answer is 1.
In the second test case, the species are already in the correct order, so no replanting is needed.
Submitted Solution:
```
#!/usr/bin/env python3
# created : 2020. 12. 31. 23:59
import os
from sys import stdin, stdout
from heapq import heappop, heappush
def solve(tc):
n, m = map(int, stdin.readline().split())
species = []
for i in range(n):
si, xi = stdin.readline().split()
species.append(int(si))
ans = n+1
cnt = 0
maxheap = [0]
for i in range(n):
if maxheap[-1] > species[i]:
cnt += 1
heappush(maxheap, species[i])
ans = min(ans, cnt)
cnt = 0
minheap = [m]
for i in range(n-1, -1, -1):
if -minheap[-1] < species[i]:
cnt += 1
heappush(minheap, -species[i])
ans = min(ans, cnt)
print(ans)
tcs = 1
# tcs = int(stdin.readline().strip())
tc = 1
while tc <= tcs:
solve(tc)
tc += 1
``` | instruction | 0 | 12,762 | 8 | 25,524 |
No | output | 1 | 12,762 | 8 | 25,525 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Emuskald is an avid horticulturist and owns the world's longest greenhouse — it is effectively infinite in length.
Over the years Emuskald has cultivated n plants in his greenhouse, of m different plant species numbered from 1 to m. His greenhouse is very narrow and can be viewed as an infinite line, with each plant occupying a single point on that line.
Emuskald has discovered that each species thrives at a different temperature, so he wants to arrange m - 1 borders that would divide the greenhouse into m sections numbered from 1 to m from left to right with each section housing a single species. He is free to place the borders, but in the end all of the i-th species plants must reside in i-th section from the left.
Of course, it is not always possible to place the borders in such way, so Emuskald needs to replant some of his plants. He can remove each plant from its position and place it anywhere in the greenhouse (at any real coordinate) with no plant already in it. Since replanting is a lot of stress for the plants, help Emuskald find the minimum number of plants he has to replant to be able to place the borders.
Input
The first line of input contains two space-separated integers n and m (1 ≤ n, m ≤ 5000, n ≥ m), the number of plants and the number of different species. Each of the following n lines contain two space-separated numbers: one integer number si (1 ≤ si ≤ m), and one real number xi (0 ≤ xi ≤ 109), the species and position of the i-th plant. Each xi will contain no more than 6 digits after the decimal point.
It is guaranteed that all xi are different; there is at least one plant of each species; the plants are given in order "from left to the right", that is in the ascending order of their xi coordinates (xi < xi + 1, 1 ≤ i < n).
Output
Output a single integer — the minimum number of plants to be replanted.
Examples
Input
3 2
2 1
1 2.0
1 3.100
Output
1
Input
3 3
1 5.0
2 5.5
3 6.0
Output
0
Input
6 3
1 14.284235
2 17.921382
1 20.328172
3 20.842331
1 25.790145
1 27.204125
Output
2
Note
In the first test case, Emuskald can replant the first plant to the right of the last plant, so the answer is 1.
In the second test case, the species are already in the correct order, so no replanting is needed.
Submitted Solution:
```
#!/usr/bin/env python3
# created : 2020. 12. 31. 23:59
import os
from sys import stdin, stdout
def solve(tc):
n, m = map(int, stdin.readline().split())
mapping = [0 for i in range(m+1)]
species = []
for i in range(n):
si, xi = stdin.readline().split()
si = int(si)
mapping[si] += 1
species.append(si)
ans = 0
p = 0
for num in range(m+1):
cur = 0
while p < n and cur < mapping[num]:
if species[p] == num:
cur += 1
else:
ans += 1
p += 1
print(ans)
tcs = 1
# tcs = int(stdin.readline().strip())
tc = 1
while tc <= tcs:
solve(tc)
tc += 1
``` | instruction | 0 | 12,763 | 8 | 25,526 |
No | output | 1 | 12,763 | 8 | 25,527 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you know, all the kids in Berland love playing with cubes. Little Petya has n towers consisting of cubes of the same size. Tower with number i consists of ai cubes stacked one on top of the other. Petya defines the instability of a set of towers as a value equal to the difference between the heights of the highest and the lowest of the towers. For example, if Petya built five cube towers with heights (8, 3, 2, 6, 3), the instability of this set is equal to 6 (the highest tower has height 8, the lowest one has height 2).
The boy wants the instability of his set of towers to be as low as possible. All he can do is to perform the following operation several times: take the top cube from some tower and put it on top of some other tower of his set. Please note that Petya would never put the cube on the same tower from which it was removed because he thinks it's a waste of time.
Before going to school, the boy will have time to perform no more than k such operations. Petya does not want to be late for class, so you have to help him accomplish this task.
Input
The first line contains two space-separated positive integers n and k (1 ≤ n ≤ 100, 1 ≤ k ≤ 1000) — the number of towers in the given set and the maximum number of operations Petya can perform. The second line contains n space-separated positive integers ai (1 ≤ ai ≤ 104) — the towers' initial heights.
Output
In the first line print two space-separated non-negative integers s and m (m ≤ k). The first number is the value of the minimum possible instability that can be obtained after performing at most k operations, the second number is the number of operations needed for that.
In the next m lines print the description of each operation as two positive integers i and j, each of them lies within limits from 1 to n. They represent that Petya took the top cube from the i-th tower and put in on the j-th one (i ≠ j). Note that in the process of performing operations the heights of some towers can become equal to zero.
If there are multiple correct sequences at which the minimum possible instability is achieved, you are allowed to print any of them.
Examples
Input
3 2
5 8 5
Output
0 2
2 1
2 3
Input
3 4
2 2 4
Output
1 1
3 2
Input
5 3
8 3 2 6 3
Output
3 3
1 3
1 2
1 3
Note
In the first sample you need to move the cubes two times, from the second tower to the third one and from the second one to the first one. Then the heights of the towers are all the same and equal to 6. | instruction | 0 | 12,843 | 8 | 25,686 |
Tags: brute force, constructive algorithms, greedy, implementation, sortings
Correct Solution:
```
operations = []
a = input().split(' ')
towers = input().split(' ')
k = int(a[1])
for i in range(len(towers)):
towers[i]=int(towers[i])
def inst(tset):
return max(tset)-min(tset)
def fix(towerset):
proset=[]
for element in towerset:
proset.append(element)
b = proset.index(max(proset))
c = proset.index(min(proset))
d = inst(proset)
e = max(proset)-1
proset[b] -= 1
proset[c] += 1
if proset[c] <= e:
return [b+1, c+1, proset]
else:
return "end"
count = 0
result = None
while result != "end":
h = inst(towers)
g=fix(towers)
if g == "end":
break
else:
if count == k:
break
else:
operations.append([g[0],g[1]])
towers=g[2]
count += 1
print(inst(towers), count)
for j in range(len(operations)):
print(operations[j][0], operations[j][1])
``` | output | 1 | 12,843 | 8 | 25,687 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you know, all the kids in Berland love playing with cubes. Little Petya has n towers consisting of cubes of the same size. Tower with number i consists of ai cubes stacked one on top of the other. Petya defines the instability of a set of towers as a value equal to the difference between the heights of the highest and the lowest of the towers. For example, if Petya built five cube towers with heights (8, 3, 2, 6, 3), the instability of this set is equal to 6 (the highest tower has height 8, the lowest one has height 2).
The boy wants the instability of his set of towers to be as low as possible. All he can do is to perform the following operation several times: take the top cube from some tower and put it on top of some other tower of his set. Please note that Petya would never put the cube on the same tower from which it was removed because he thinks it's a waste of time.
Before going to school, the boy will have time to perform no more than k such operations. Petya does not want to be late for class, so you have to help him accomplish this task.
Input
The first line contains two space-separated positive integers n and k (1 ≤ n ≤ 100, 1 ≤ k ≤ 1000) — the number of towers in the given set and the maximum number of operations Petya can perform. The second line contains n space-separated positive integers ai (1 ≤ ai ≤ 104) — the towers' initial heights.
Output
In the first line print two space-separated non-negative integers s and m (m ≤ k). The first number is the value of the minimum possible instability that can be obtained after performing at most k operations, the second number is the number of operations needed for that.
In the next m lines print the description of each operation as two positive integers i and j, each of them lies within limits from 1 to n. They represent that Petya took the top cube from the i-th tower and put in on the j-th one (i ≠ j). Note that in the process of performing operations the heights of some towers can become equal to zero.
If there are multiple correct sequences at which the minimum possible instability is achieved, you are allowed to print any of them.
Examples
Input
3 2
5 8 5
Output
0 2
2 1
2 3
Input
3 4
2 2 4
Output
1 1
3 2
Input
5 3
8 3 2 6 3
Output
3 3
1 3
1 2
1 3
Note
In the first sample you need to move the cubes two times, from the second tower to the third one and from the second one to the first one. Then the heights of the towers are all the same and equal to 6. | instruction | 0 | 12,844 | 8 | 25,688 |
Tags: brute force, constructive algorithms, greedy, implementation, sortings
Correct Solution:
```
n, k = map(int, input().split())
a = list(map(int, input().split()))
ans = 0
l = []
r = []
f = True
for h in range(k):
mi = a[0]
ma = a[0]
ai = 0
aj = 0
for i in range(len(a)):
if a[i] < mi:
mi = a[i]
ai = i
if a[i] > ma:
ma = a[i]
aj = i
if ma - mi <= 1:
mi = a[0]
ma = a[0]
for i in range(len(a)):
if a[i] < mi:
mi = a[i]
if a[i] > ma:
ma = a[i]
print(ma - mi, ans)
for i in range(len(l)):
print(l[i], r[i])
f = False
break
else:
a[ai] += 1
a[aj] -= 1
ans += 1
l.append(aj+1)
r.append(ai+1)
if f == True:
mi = a[0]
ma = a[0]
for i in range(len(a)):
if a[i] < mi:
mi = a[i]
if a[i] > ma:
ma = a[i]
print(ma - mi, ans)
for i in range(len(l)):
print(l[i], r[i])
``` | output | 1 | 12,844 | 8 | 25,689 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you know, all the kids in Berland love playing with cubes. Little Petya has n towers consisting of cubes of the same size. Tower with number i consists of ai cubes stacked one on top of the other. Petya defines the instability of a set of towers as a value equal to the difference between the heights of the highest and the lowest of the towers. For example, if Petya built five cube towers with heights (8, 3, 2, 6, 3), the instability of this set is equal to 6 (the highest tower has height 8, the lowest one has height 2).
The boy wants the instability of his set of towers to be as low as possible. All he can do is to perform the following operation several times: take the top cube from some tower and put it on top of some other tower of his set. Please note that Petya would never put the cube on the same tower from which it was removed because he thinks it's a waste of time.
Before going to school, the boy will have time to perform no more than k such operations. Petya does not want to be late for class, so you have to help him accomplish this task.
Input
The first line contains two space-separated positive integers n and k (1 ≤ n ≤ 100, 1 ≤ k ≤ 1000) — the number of towers in the given set and the maximum number of operations Petya can perform. The second line contains n space-separated positive integers ai (1 ≤ ai ≤ 104) — the towers' initial heights.
Output
In the first line print two space-separated non-negative integers s and m (m ≤ k). The first number is the value of the minimum possible instability that can be obtained after performing at most k operations, the second number is the number of operations needed for that.
In the next m lines print the description of each operation as two positive integers i and j, each of them lies within limits from 1 to n. They represent that Petya took the top cube from the i-th tower and put in on the j-th one (i ≠ j). Note that in the process of performing operations the heights of some towers can become equal to zero.
If there are multiple correct sequences at which the minimum possible instability is achieved, you are allowed to print any of them.
Examples
Input
3 2
5 8 5
Output
0 2
2 1
2 3
Input
3 4
2 2 4
Output
1 1
3 2
Input
5 3
8 3 2 6 3
Output
3 3
1 3
1 2
1 3
Note
In the first sample you need to move the cubes two times, from the second tower to the third one and from the second one to the first one. Then the heights of the towers are all the same and equal to 6. | instruction | 0 | 12,845 | 8 | 25,690 |
Tags: brute force, constructive algorithms, greedy, implementation, sortings
Correct Solution:
```
n, k = map(int, input().split())
a, risp = [], []
a = list(map(int, input().split()))
s = max(a) - min(a)
m = 0
if s < 2:
print(s, 0)
else:
for x in range(k):
maxi = max(a)
IndMaxi = a.index(maxi)
mini = min(a)
IndMini = a.index(mini)
a[IndMaxi] -= 1
a[IndMini] += 1
s = max(a) - min(a)
m += 1
risp.append([IndMaxi + 1, IndMini + 1])
#print(a)
if s == 0:
break
print(s, m)
for x in range(len(risp)):
print(*risp[x])
``` | output | 1 | 12,845 | 8 | 25,691 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you know, all the kids in Berland love playing with cubes. Little Petya has n towers consisting of cubes of the same size. Tower with number i consists of ai cubes stacked one on top of the other. Petya defines the instability of a set of towers as a value equal to the difference between the heights of the highest and the lowest of the towers. For example, if Petya built five cube towers with heights (8, 3, 2, 6, 3), the instability of this set is equal to 6 (the highest tower has height 8, the lowest one has height 2).
The boy wants the instability of his set of towers to be as low as possible. All he can do is to perform the following operation several times: take the top cube from some tower and put it on top of some other tower of his set. Please note that Petya would never put the cube on the same tower from which it was removed because he thinks it's a waste of time.
Before going to school, the boy will have time to perform no more than k such operations. Petya does not want to be late for class, so you have to help him accomplish this task.
Input
The first line contains two space-separated positive integers n and k (1 ≤ n ≤ 100, 1 ≤ k ≤ 1000) — the number of towers in the given set and the maximum number of operations Petya can perform. The second line contains n space-separated positive integers ai (1 ≤ ai ≤ 104) — the towers' initial heights.
Output
In the first line print two space-separated non-negative integers s and m (m ≤ k). The first number is the value of the minimum possible instability that can be obtained after performing at most k operations, the second number is the number of operations needed for that.
In the next m lines print the description of each operation as two positive integers i and j, each of them lies within limits from 1 to n. They represent that Petya took the top cube from the i-th tower and put in on the j-th one (i ≠ j). Note that in the process of performing operations the heights of some towers can become equal to zero.
If there are multiple correct sequences at which the minimum possible instability is achieved, you are allowed to print any of them.
Examples
Input
3 2
5 8 5
Output
0 2
2 1
2 3
Input
3 4
2 2 4
Output
1 1
3 2
Input
5 3
8 3 2 6 3
Output
3 3
1 3
1 2
1 3
Note
In the first sample you need to move the cubes two times, from the second tower to the third one and from the second one to the first one. Then the heights of the towers are all the same and equal to 6. | instruction | 0 | 12,846 | 8 | 25,692 |
Tags: brute force, constructive algorithms, greedy, implementation, sortings
Correct Solution:
```
n, k = map(int, input().split())
arr = [int(i) for i in input().split()]
ans = []
while k >= 0:
x = 0
y = 0
for j in range(n):
if arr[j] > arr[x]:
x = j
if arr[j] < arr[y]:
y = j
if k == 0 or arr[x] - arr[y] <= 1:
print(arr[x] - arr[y], len(ans))
for j in ans:
print(j[0], j[1])
break
if arr[x] != arr[y]:
ans.append((x + 1, y + 1))
arr[x] -= 1
arr[y] += 1
k -= 1
``` | output | 1 | 12,846 | 8 | 25,693 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you know, all the kids in Berland love playing with cubes. Little Petya has n towers consisting of cubes of the same size. Tower with number i consists of ai cubes stacked one on top of the other. Petya defines the instability of a set of towers as a value equal to the difference between the heights of the highest and the lowest of the towers. For example, if Petya built five cube towers with heights (8, 3, 2, 6, 3), the instability of this set is equal to 6 (the highest tower has height 8, the lowest one has height 2).
The boy wants the instability of his set of towers to be as low as possible. All he can do is to perform the following operation several times: take the top cube from some tower and put it on top of some other tower of his set. Please note that Petya would never put the cube on the same tower from which it was removed because he thinks it's a waste of time.
Before going to school, the boy will have time to perform no more than k such operations. Petya does not want to be late for class, so you have to help him accomplish this task.
Input
The first line contains two space-separated positive integers n and k (1 ≤ n ≤ 100, 1 ≤ k ≤ 1000) — the number of towers in the given set and the maximum number of operations Petya can perform. The second line contains n space-separated positive integers ai (1 ≤ ai ≤ 104) — the towers' initial heights.
Output
In the first line print two space-separated non-negative integers s and m (m ≤ k). The first number is the value of the minimum possible instability that can be obtained after performing at most k operations, the second number is the number of operations needed for that.
In the next m lines print the description of each operation as two positive integers i and j, each of them lies within limits from 1 to n. They represent that Petya took the top cube from the i-th tower and put in on the j-th one (i ≠ j). Note that in the process of performing operations the heights of some towers can become equal to zero.
If there are multiple correct sequences at which the minimum possible instability is achieved, you are allowed to print any of them.
Examples
Input
3 2
5 8 5
Output
0 2
2 1
2 3
Input
3 4
2 2 4
Output
1 1
3 2
Input
5 3
8 3 2 6 3
Output
3 3
1 3
1 2
1 3
Note
In the first sample you need to move the cubes two times, from the second tower to the third one and from the second one to the first one. Then the heights of the towers are all the same and equal to 6. | instruction | 0 | 12,847 | 8 | 25,694 |
Tags: brute force, constructive algorithms, greedy, implementation, sortings
Correct Solution:
```
n,k=map(int,input().split())
tab=[0 for loop in range(n)]
t=input().split()
for i in range(n):
tab[i]=int(t[i])
def max():
global n
y=0
j=0
for i in range(n):
if tab[i]>y:
y=tab[i]
j=i
return j
def min():
global n
x=100000
j=0
for i in range(n):
if tab[i]<x:
x=tab[i]
j=i
return j
def maxv():
global n
y=0
j=0
for i in range(n):
if tab[i]>y:
y=tab[i]
j=i
return y
def minv():
global n
x=100000
j=0
for i in range(n):
if tab[i]<x:
x=tab[i]
j=i
return x
liste=[(0,0) for loop in range(k)]
m=0
v=maxv()-minv()
for j in range(k):
a=max()
b=min()
tab[a]-=1
tab[b]+=1
liste[j]=(a+1,b+1)
c=maxv()-minv()
if c<v:
v=c
m=j+1
print(maxv()-minv(),m)
for i in range(m):
(d,e)=liste[i]
print(d,e)
``` | output | 1 | 12,847 | 8 | 25,695 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you know, all the kids in Berland love playing with cubes. Little Petya has n towers consisting of cubes of the same size. Tower with number i consists of ai cubes stacked one on top of the other. Petya defines the instability of a set of towers as a value equal to the difference between the heights of the highest and the lowest of the towers. For example, if Petya built five cube towers with heights (8, 3, 2, 6, 3), the instability of this set is equal to 6 (the highest tower has height 8, the lowest one has height 2).
The boy wants the instability of his set of towers to be as low as possible. All he can do is to perform the following operation several times: take the top cube from some tower and put it on top of some other tower of his set. Please note that Petya would never put the cube on the same tower from which it was removed because he thinks it's a waste of time.
Before going to school, the boy will have time to perform no more than k such operations. Petya does not want to be late for class, so you have to help him accomplish this task.
Input
The first line contains two space-separated positive integers n and k (1 ≤ n ≤ 100, 1 ≤ k ≤ 1000) — the number of towers in the given set and the maximum number of operations Petya can perform. The second line contains n space-separated positive integers ai (1 ≤ ai ≤ 104) — the towers' initial heights.
Output
In the first line print two space-separated non-negative integers s and m (m ≤ k). The first number is the value of the minimum possible instability that can be obtained after performing at most k operations, the second number is the number of operations needed for that.
In the next m lines print the description of each operation as two positive integers i and j, each of them lies within limits from 1 to n. They represent that Petya took the top cube from the i-th tower and put in on the j-th one (i ≠ j). Note that in the process of performing operations the heights of some towers can become equal to zero.
If there are multiple correct sequences at which the minimum possible instability is achieved, you are allowed to print any of them.
Examples
Input
3 2
5 8 5
Output
0 2
2 1
2 3
Input
3 4
2 2 4
Output
1 1
3 2
Input
5 3
8 3 2 6 3
Output
3 3
1 3
1 2
1 3
Note
In the first sample you need to move the cubes two times, from the second tower to the third one and from the second one to the first one. Then the heights of the towers are all the same and equal to 6. | instruction | 0 | 12,848 | 8 | 25,696 |
Tags: brute force, constructive algorithms, greedy, implementation, sortings
Correct Solution:
```
n,k = map(int,input().split())
arr = list(map(int,input().split()))
ans = []
op = 0
factor = max(arr)-min(arr)
while factor>1 and op<k:
m = arr.index(min(arr))
M = arr.index(max(arr))
arr[m] += 1
arr[M] -= 1
op+=1
ans.append([M+1,m+1])
factor = max(arr)-min(arr)
print(factor,op)
for i in ans:
print(*i)
``` | output | 1 | 12,848 | 8 | 25,697 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you know, all the kids in Berland love playing with cubes. Little Petya has n towers consisting of cubes of the same size. Tower with number i consists of ai cubes stacked one on top of the other. Petya defines the instability of a set of towers as a value equal to the difference between the heights of the highest and the lowest of the towers. For example, if Petya built five cube towers with heights (8, 3, 2, 6, 3), the instability of this set is equal to 6 (the highest tower has height 8, the lowest one has height 2).
The boy wants the instability of his set of towers to be as low as possible. All he can do is to perform the following operation several times: take the top cube from some tower and put it on top of some other tower of his set. Please note that Petya would never put the cube on the same tower from which it was removed because he thinks it's a waste of time.
Before going to school, the boy will have time to perform no more than k such operations. Petya does not want to be late for class, so you have to help him accomplish this task.
Input
The first line contains two space-separated positive integers n and k (1 ≤ n ≤ 100, 1 ≤ k ≤ 1000) — the number of towers in the given set and the maximum number of operations Petya can perform. The second line contains n space-separated positive integers ai (1 ≤ ai ≤ 104) — the towers' initial heights.
Output
In the first line print two space-separated non-negative integers s and m (m ≤ k). The first number is the value of the minimum possible instability that can be obtained after performing at most k operations, the second number is the number of operations needed for that.
In the next m lines print the description of each operation as two positive integers i and j, each of them lies within limits from 1 to n. They represent that Petya took the top cube from the i-th tower and put in on the j-th one (i ≠ j). Note that in the process of performing operations the heights of some towers can become equal to zero.
If there are multiple correct sequences at which the minimum possible instability is achieved, you are allowed to print any of them.
Examples
Input
3 2
5 8 5
Output
0 2
2 1
2 3
Input
3 4
2 2 4
Output
1 1
3 2
Input
5 3
8 3 2 6 3
Output
3 3
1 3
1 2
1 3
Note
In the first sample you need to move the cubes two times, from the second tower to the third one and from the second one to the first one. Then the heights of the towers are all the same and equal to 6. | instruction | 0 | 12,849 | 8 | 25,698 |
Tags: brute force, constructive algorithms, greedy, implementation, sortings
Correct Solution:
```
n,k = map(int,input().split())
arr = [int(i) for i in input().split()]
ans = []
for i in range(k):
mx = arr.index(max(arr))
mn = arr.index(min(arr))
if mx != mn:
arr[mx] -= 1
arr[mn] += 1
ans.append([mx+1,mn+1])
print(str(max(arr)-min(arr)) + ' ' + str(len(ans)))
for x in ans:
print(' '.join(map(str,x)))
``` | output | 1 | 12,849 | 8 | 25,699 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you know, all the kids in Berland love playing with cubes. Little Petya has n towers consisting of cubes of the same size. Tower with number i consists of ai cubes stacked one on top of the other. Petya defines the instability of a set of towers as a value equal to the difference between the heights of the highest and the lowest of the towers. For example, if Petya built five cube towers with heights (8, 3, 2, 6, 3), the instability of this set is equal to 6 (the highest tower has height 8, the lowest one has height 2).
The boy wants the instability of his set of towers to be as low as possible. All he can do is to perform the following operation several times: take the top cube from some tower and put it on top of some other tower of his set. Please note that Petya would never put the cube on the same tower from which it was removed because he thinks it's a waste of time.
Before going to school, the boy will have time to perform no more than k such operations. Petya does not want to be late for class, so you have to help him accomplish this task.
Input
The first line contains two space-separated positive integers n and k (1 ≤ n ≤ 100, 1 ≤ k ≤ 1000) — the number of towers in the given set and the maximum number of operations Petya can perform. The second line contains n space-separated positive integers ai (1 ≤ ai ≤ 104) — the towers' initial heights.
Output
In the first line print two space-separated non-negative integers s and m (m ≤ k). The first number is the value of the minimum possible instability that can be obtained after performing at most k operations, the second number is the number of operations needed for that.
In the next m lines print the description of each operation as two positive integers i and j, each of them lies within limits from 1 to n. They represent that Petya took the top cube from the i-th tower and put in on the j-th one (i ≠ j). Note that in the process of performing operations the heights of some towers can become equal to zero.
If there are multiple correct sequences at which the minimum possible instability is achieved, you are allowed to print any of them.
Examples
Input
3 2
5 8 5
Output
0 2
2 1
2 3
Input
3 4
2 2 4
Output
1 1
3 2
Input
5 3
8 3 2 6 3
Output
3 3
1 3
1 2
1 3
Note
In the first sample you need to move the cubes two times, from the second tower to the third one and from the second one to the first one. Then the heights of the towers are all the same and equal to 6. | instruction | 0 | 12,850 | 8 | 25,700 |
Tags: brute force, constructive algorithms, greedy, implementation, sortings
Correct Solution:
```
n,k = map(int,input().split())
arr = list(map(int,input().split()))
lst = []
j = count = 0
check = -1
while j<k:
mini = 10**4+3
maxi = 0
for i in range(n):
if maxi<arr[i]:
max_index = i
maxi = arr[i]
if mini>arr[i]:
min_index = i
mini = arr[i]
if maxi-mini>1:
count += 1
arr[max_index] -= 1
arr[min_index] += 1
lst.append([max_index+1,min_index+1])
else:
break
j += 1
mini = 10**4+3
maxi = 0
for i in range(n):
if maxi<arr[i]:
maxi = arr[i]
if mini>arr[i]:
mini = arr[i]
print(maxi-mini,count)
for j in lst:
print(*j)
``` | output | 1 | 12,850 | 8 | 25,701 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you know, all the kids in Berland love playing with cubes. Little Petya has n towers consisting of cubes of the same size. Tower with number i consists of ai cubes stacked one on top of the other. Petya defines the instability of a set of towers as a value equal to the difference between the heights of the highest and the lowest of the towers. For example, if Petya built five cube towers with heights (8, 3, 2, 6, 3), the instability of this set is equal to 6 (the highest tower has height 8, the lowest one has height 2).
The boy wants the instability of his set of towers to be as low as possible. All he can do is to perform the following operation several times: take the top cube from some tower and put it on top of some other tower of his set. Please note that Petya would never put the cube on the same tower from which it was removed because he thinks it's a waste of time.
Before going to school, the boy will have time to perform no more than k such operations. Petya does not want to be late for class, so you have to help him accomplish this task.
Input
The first line contains two space-separated positive integers n and k (1 ≤ n ≤ 100, 1 ≤ k ≤ 1000) — the number of towers in the given set and the maximum number of operations Petya can perform. The second line contains n space-separated positive integers ai (1 ≤ ai ≤ 104) — the towers' initial heights.
Output
In the first line print two space-separated non-negative integers s and m (m ≤ k). The first number is the value of the minimum possible instability that can be obtained after performing at most k operations, the second number is the number of operations needed for that.
In the next m lines print the description of each operation as two positive integers i and j, each of them lies within limits from 1 to n. They represent that Petya took the top cube from the i-th tower and put in on the j-th one (i ≠ j). Note that in the process of performing operations the heights of some towers can become equal to zero.
If there are multiple correct sequences at which the minimum possible instability is achieved, you are allowed to print any of them.
Examples
Input
3 2
5 8 5
Output
0 2
2 1
2 3
Input
3 4
2 2 4
Output
1 1
3 2
Input
5 3
8 3 2 6 3
Output
3 3
1 3
1 2
1 3
Note
In the first sample you need to move the cubes two times, from the second tower to the third one and from the second one to the first one. Then the heights of the towers are all the same and equal to 6.
Submitted Solution:
```
a, b = map(int, input().split(' '))
towers = list(map(int, input().split(' ')))
tow2 = [[towers[i], i+1] for i in range(len(towers))]
tot = sum(towers)
should = tot // a
num1 = (should+1) * a - tot
num2 = a - num1
ideal = num1*[should] + num2 * [should+1]
ct = 1
x = []
while ct <= b:
tow2.sort(key = lambda x:x[0])
if [i[0] for i in tow2] == ideal:
break
else:
x.append([tow2[-1][1], tow2[0][1]])
tow2[0][0] += 1
tow2[-1][0] -= 1
ct += 1
tows = [x[0] for x in tow2]
print((max(tows)-min(tows)), ct-1)
for i in x:
print(i[0], i[1])
``` | instruction | 0 | 12,851 | 8 | 25,702 |
Yes | output | 1 | 12,851 | 8 | 25,703 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you know, all the kids in Berland love playing with cubes. Little Petya has n towers consisting of cubes of the same size. Tower with number i consists of ai cubes stacked one on top of the other. Petya defines the instability of a set of towers as a value equal to the difference between the heights of the highest and the lowest of the towers. For example, if Petya built five cube towers with heights (8, 3, 2, 6, 3), the instability of this set is equal to 6 (the highest tower has height 8, the lowest one has height 2).
The boy wants the instability of his set of towers to be as low as possible. All he can do is to perform the following operation several times: take the top cube from some tower and put it on top of some other tower of his set. Please note that Petya would never put the cube on the same tower from which it was removed because he thinks it's a waste of time.
Before going to school, the boy will have time to perform no more than k such operations. Petya does not want to be late for class, so you have to help him accomplish this task.
Input
The first line contains two space-separated positive integers n and k (1 ≤ n ≤ 100, 1 ≤ k ≤ 1000) — the number of towers in the given set and the maximum number of operations Petya can perform. The second line contains n space-separated positive integers ai (1 ≤ ai ≤ 104) — the towers' initial heights.
Output
In the first line print two space-separated non-negative integers s and m (m ≤ k). The first number is the value of the minimum possible instability that can be obtained after performing at most k operations, the second number is the number of operations needed for that.
In the next m lines print the description of each operation as two positive integers i and j, each of them lies within limits from 1 to n. They represent that Petya took the top cube from the i-th tower and put in on the j-th one (i ≠ j). Note that in the process of performing operations the heights of some towers can become equal to zero.
If there are multiple correct sequences at which the minimum possible instability is achieved, you are allowed to print any of them.
Examples
Input
3 2
5 8 5
Output
0 2
2 1
2 3
Input
3 4
2 2 4
Output
1 1
3 2
Input
5 3
8 3 2 6 3
Output
3 3
1 3
1 2
1 3
Note
In the first sample you need to move the cubes two times, from the second tower to the third one and from the second one to the first one. Then the heights of the towers are all the same and equal to 6.
Submitted Solution:
```
f = list(map(int, input().split()))
heights = list(map(int, input().split()))
num = f[0]
kops = f[1]
minpossible = 0 if sum(heights)%num==0 else 1
curinstability = max(heights)-min(heights)
opscounter = 0
rem = list()
while curinstability>minpossible and opscounter < kops:
lowi = heights.index(min(heights))
highi = heights.index(max(heights))
heights[lowi]+=1
heights[highi]-=1
rem.append((highi+1, lowi+1)) #adjust index by 1
opscounter+=1
curinstability = max(heights)-min(heights)
print(max(heights)-min(heights),opscounter)
for vals in rem:
print(vals[0],vals[1])
``` | instruction | 0 | 12,852 | 8 | 25,704 |
Yes | output | 1 | 12,852 | 8 | 25,705 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you know, all the kids in Berland love playing with cubes. Little Petya has n towers consisting of cubes of the same size. Tower with number i consists of ai cubes stacked one on top of the other. Petya defines the instability of a set of towers as a value equal to the difference between the heights of the highest and the lowest of the towers. For example, if Petya built five cube towers with heights (8, 3, 2, 6, 3), the instability of this set is equal to 6 (the highest tower has height 8, the lowest one has height 2).
The boy wants the instability of his set of towers to be as low as possible. All he can do is to perform the following operation several times: take the top cube from some tower and put it on top of some other tower of his set. Please note that Petya would never put the cube on the same tower from which it was removed because he thinks it's a waste of time.
Before going to school, the boy will have time to perform no more than k such operations. Petya does not want to be late for class, so you have to help him accomplish this task.
Input
The first line contains two space-separated positive integers n and k (1 ≤ n ≤ 100, 1 ≤ k ≤ 1000) — the number of towers in the given set and the maximum number of operations Petya can perform. The second line contains n space-separated positive integers ai (1 ≤ ai ≤ 104) — the towers' initial heights.
Output
In the first line print two space-separated non-negative integers s and m (m ≤ k). The first number is the value of the minimum possible instability that can be obtained after performing at most k operations, the second number is the number of operations needed for that.
In the next m lines print the description of each operation as two positive integers i and j, each of them lies within limits from 1 to n. They represent that Petya took the top cube from the i-th tower and put in on the j-th one (i ≠ j). Note that in the process of performing operations the heights of some towers can become equal to zero.
If there are multiple correct sequences at which the minimum possible instability is achieved, you are allowed to print any of them.
Examples
Input
3 2
5 8 5
Output
0 2
2 1
2 3
Input
3 4
2 2 4
Output
1 1
3 2
Input
5 3
8 3 2 6 3
Output
3 3
1 3
1 2
1 3
Note
In the first sample you need to move the cubes two times, from the second tower to the third one and from the second one to the first one. Then the heights of the towers are all the same and equal to 6.
Submitted Solution:
```
n, k = map (int, input ().split ())
lst = list (map (int, input ().split ()))
x = len (lst)
for i in range (x) :
lst [i] = [lst[i], i]
orig = k
temp = 0
ans = []
while k > 0 :
k -= 1
lst.sort ()
if lst[n - 1][0] - lst[0][0] > 1 :
lst[n - 1][0] -= 1
lst[0][0] += 1
ans.append ([lst[n - 1][1], lst[0][1]])
else :
temp += 1
lst.sort ()
k += temp
print (lst[n - 1][0] - lst[0][0], orig - k)
for i in ans :
print (i[0] + 1, i[1] + 1)
``` | instruction | 0 | 12,853 | 8 | 25,706 |
Yes | output | 1 | 12,853 | 8 | 25,707 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you know, all the kids in Berland love playing with cubes. Little Petya has n towers consisting of cubes of the same size. Tower with number i consists of ai cubes stacked one on top of the other. Petya defines the instability of a set of towers as a value equal to the difference between the heights of the highest and the lowest of the towers. For example, if Petya built five cube towers with heights (8, 3, 2, 6, 3), the instability of this set is equal to 6 (the highest tower has height 8, the lowest one has height 2).
The boy wants the instability of his set of towers to be as low as possible. All he can do is to perform the following operation several times: take the top cube from some tower and put it on top of some other tower of his set. Please note that Petya would never put the cube on the same tower from which it was removed because he thinks it's a waste of time.
Before going to school, the boy will have time to perform no more than k such operations. Petya does not want to be late for class, so you have to help him accomplish this task.
Input
The first line contains two space-separated positive integers n and k (1 ≤ n ≤ 100, 1 ≤ k ≤ 1000) — the number of towers in the given set and the maximum number of operations Petya can perform. The second line contains n space-separated positive integers ai (1 ≤ ai ≤ 104) — the towers' initial heights.
Output
In the first line print two space-separated non-negative integers s and m (m ≤ k). The first number is the value of the minimum possible instability that can be obtained after performing at most k operations, the second number is the number of operations needed for that.
In the next m lines print the description of each operation as two positive integers i and j, each of them lies within limits from 1 to n. They represent that Petya took the top cube from the i-th tower and put in on the j-th one (i ≠ j). Note that in the process of performing operations the heights of some towers can become equal to zero.
If there are multiple correct sequences at which the minimum possible instability is achieved, you are allowed to print any of them.
Examples
Input
3 2
5 8 5
Output
0 2
2 1
2 3
Input
3 4
2 2 4
Output
1 1
3 2
Input
5 3
8 3 2 6 3
Output
3 3
1 3
1 2
1 3
Note
In the first sample you need to move the cubes two times, from the second tower to the third one and from the second one to the first one. Then the heights of the towers are all the same and equal to 6.
Submitted Solution:
```
n,k=map(int, input().split())
a=list(map(int, input().split()))
d=[[a[i],i] for i in range(n)]
d.sort()
cnt=0
ops=[]
while d[n-1][0]-d[0][0] > 1 and cnt < k:
ops.append([d[n-1][1]+1, d[0][1]+1])
d[n-1][0] -= 1
d[0][0] += 1
cnt += 1
d.sort()
print(d[n-1][0]-d[0][0], cnt)
for f,t in ops: print(f, t)
# Made By Mostafa_Khaled
``` | instruction | 0 | 12,854 | 8 | 25,708 |
Yes | output | 1 | 12,854 | 8 | 25,709 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you know, all the kids in Berland love playing with cubes. Little Petya has n towers consisting of cubes of the same size. Tower with number i consists of ai cubes stacked one on top of the other. Petya defines the instability of a set of towers as a value equal to the difference between the heights of the highest and the lowest of the towers. For example, if Petya built five cube towers with heights (8, 3, 2, 6, 3), the instability of this set is equal to 6 (the highest tower has height 8, the lowest one has height 2).
The boy wants the instability of his set of towers to be as low as possible. All he can do is to perform the following operation several times: take the top cube from some tower and put it on top of some other tower of his set. Please note that Petya would never put the cube on the same tower from which it was removed because he thinks it's a waste of time.
Before going to school, the boy will have time to perform no more than k such operations. Petya does not want to be late for class, so you have to help him accomplish this task.
Input
The first line contains two space-separated positive integers n and k (1 ≤ n ≤ 100, 1 ≤ k ≤ 1000) — the number of towers in the given set and the maximum number of operations Petya can perform. The second line contains n space-separated positive integers ai (1 ≤ ai ≤ 104) — the towers' initial heights.
Output
In the first line print two space-separated non-negative integers s and m (m ≤ k). The first number is the value of the minimum possible instability that can be obtained after performing at most k operations, the second number is the number of operations needed for that.
In the next m lines print the description of each operation as two positive integers i and j, each of them lies within limits from 1 to n. They represent that Petya took the top cube from the i-th tower and put in on the j-th one (i ≠ j). Note that in the process of performing operations the heights of some towers can become equal to zero.
If there are multiple correct sequences at which the minimum possible instability is achieved, you are allowed to print any of them.
Examples
Input
3 2
5 8 5
Output
0 2
2 1
2 3
Input
3 4
2 2 4
Output
1 1
3 2
Input
5 3
8 3 2 6 3
Output
3 3
1 3
1 2
1 3
Note
In the first sample you need to move the cubes two times, from the second tower to the third one and from the second one to the first one. Then the heights of the towers are all the same and equal to 6.
Submitted Solution:
```
import sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush
from math import *
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
import copy
import time
starttime = time.time()
mod = int(pow(10, 9) + 7)
mod2 = 998244353
# from sys import stdin
# input = stdin.readline
def data(): return sys.stdin.readline().strip()
def out(*var, end="\n"): sys.stdout.write(' '.join(map(str, var))+end)
def L(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]
try:
# sys.setrecursionlimit(int(pow(10,6)))
sys.stdin = open("input.txt", "r")
# sys.stdout = open("../output.txt", "w")
except:
pass
def pmat(A):
for ele in A:
print(ele,end="\n")
n,k=L()
A=L()
steps=[]
moves=[]
for i in range(k):
steps.append(max(A)-min(A))
x=A.index(min(A))
y=A.index(max(A))
moves.append([y+1,x+1])
z=min(A)+max(A)
A[x]=z//2
A[y]=z//2+z%2
steps.append(max(A)-min(A))
x=steps.index(min(steps))
print(steps[x],x)
for i in range(x):
print(*moves[i])
endtime = time.time()
# print(f"Runtime of the program is {endtime - starttime}")
``` | instruction | 0 | 12,855 | 8 | 25,710 |
No | output | 1 | 12,855 | 8 | 25,711 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you know, all the kids in Berland love playing with cubes. Little Petya has n towers consisting of cubes of the same size. Tower with number i consists of ai cubes stacked one on top of the other. Petya defines the instability of a set of towers as a value equal to the difference between the heights of the highest and the lowest of the towers. For example, if Petya built five cube towers with heights (8, 3, 2, 6, 3), the instability of this set is equal to 6 (the highest tower has height 8, the lowest one has height 2).
The boy wants the instability of his set of towers to be as low as possible. All he can do is to perform the following operation several times: take the top cube from some tower and put it on top of some other tower of his set. Please note that Petya would never put the cube on the same tower from which it was removed because he thinks it's a waste of time.
Before going to school, the boy will have time to perform no more than k such operations. Petya does not want to be late for class, so you have to help him accomplish this task.
Input
The first line contains two space-separated positive integers n and k (1 ≤ n ≤ 100, 1 ≤ k ≤ 1000) — the number of towers in the given set and the maximum number of operations Petya can perform. The second line contains n space-separated positive integers ai (1 ≤ ai ≤ 104) — the towers' initial heights.
Output
In the first line print two space-separated non-negative integers s and m (m ≤ k). The first number is the value of the minimum possible instability that can be obtained after performing at most k operations, the second number is the number of operations needed for that.
In the next m lines print the description of each operation as two positive integers i and j, each of them lies within limits from 1 to n. They represent that Petya took the top cube from the i-th tower and put in on the j-th one (i ≠ j). Note that in the process of performing operations the heights of some towers can become equal to zero.
If there are multiple correct sequences at which the minimum possible instability is achieved, you are allowed to print any of them.
Examples
Input
3 2
5 8 5
Output
0 2
2 1
2 3
Input
3 4
2 2 4
Output
1 1
3 2
Input
5 3
8 3 2 6 3
Output
3 3
1 3
1 2
1 3
Note
In the first sample you need to move the cubes two times, from the second tower to the third one and from the second one to the first one. Then the heights of the towers are all the same and equal to 6.
Submitted Solution:
```
st = input()
j = 0
for i in range(len(st)):
if st[i] == ' ':
j = i
st1 = st[0:j]
n = int(st1)
st1 = st[(j+1):len(st)]
k = int(st1)
i = 0
a = []
st = input() + ' '
while i <= (len(st) - 1):
j = i
while (st[i] != ' '):
i += 1
a.append(int(st[j:i]))
i += 1
cnt = 0
mm = 10000000
i_max = 0
i_min = 0
x = []
y = []
while cnt < k:
for i in range(len(a)):
if a[i] == max(a):
i_max = i
if a[i] == min(a):
i_min = i
x.append(i_max)
y.append(i_min)
a[i_max] -= 1
a[i_min] += 1
cnt += 1
print((max(a) - min(a)), cnt)
x = x[::-1]
y = y[::-1]
for i in range(cnt):
print((x[i]+1), (y[i]+1))
``` | instruction | 0 | 12,856 | 8 | 25,712 |
No | output | 1 | 12,856 | 8 | 25,713 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you know, all the kids in Berland love playing with cubes. Little Petya has n towers consisting of cubes of the same size. Tower with number i consists of ai cubes stacked one on top of the other. Petya defines the instability of a set of towers as a value equal to the difference between the heights of the highest and the lowest of the towers. For example, if Petya built five cube towers with heights (8, 3, 2, 6, 3), the instability of this set is equal to 6 (the highest tower has height 8, the lowest one has height 2).
The boy wants the instability of his set of towers to be as low as possible. All he can do is to perform the following operation several times: take the top cube from some tower and put it on top of some other tower of his set. Please note that Petya would never put the cube on the same tower from which it was removed because he thinks it's a waste of time.
Before going to school, the boy will have time to perform no more than k such operations. Petya does not want to be late for class, so you have to help him accomplish this task.
Input
The first line contains two space-separated positive integers n and k (1 ≤ n ≤ 100, 1 ≤ k ≤ 1000) — the number of towers in the given set and the maximum number of operations Petya can perform. The second line contains n space-separated positive integers ai (1 ≤ ai ≤ 104) — the towers' initial heights.
Output
In the first line print two space-separated non-negative integers s and m (m ≤ k). The first number is the value of the minimum possible instability that can be obtained after performing at most k operations, the second number is the number of operations needed for that.
In the next m lines print the description of each operation as two positive integers i and j, each of them lies within limits from 1 to n. They represent that Petya took the top cube from the i-th tower and put in on the j-th one (i ≠ j). Note that in the process of performing operations the heights of some towers can become equal to zero.
If there are multiple correct sequences at which the minimum possible instability is achieved, you are allowed to print any of them.
Examples
Input
3 2
5 8 5
Output
0 2
2 1
2 3
Input
3 4
2 2 4
Output
1 1
3 2
Input
5 3
8 3 2 6 3
Output
3 3
1 3
1 2
1 3
Note
In the first sample you need to move the cubes two times, from the second tower to the third one and from the second one to the first one. Then the heights of the towers are all the same and equal to 6.
Submitted Solution:
```
n,k = map(int,input().split())
arr = list(map(int,input().split()))
lst = []
j = count = 0
check = -1
while j<k:
mini = 10**4+3
maxi = 0
for i in range(n):
if maxi<arr[i]:
max_index = i
maxi = arr[i]
if mini>arr[i]:
min_index = i
mini = arr[i]
if maxi-mini>1:
count += 1
arr[max_index] -= 1
arr[min_index] += 1
lst.append([max_index+1,min_index+1])
else:
if maxi-mini==1:
check = 1
break
if maxi-mini==2:
check = 0
j += 1
if check==-1:
print(maxi-mini,count)
elif check==0:
print(0,count)
else:
print(1,count)
for j in lst:
print(*j)
``` | instruction | 0 | 12,857 | 8 | 25,714 |
No | output | 1 | 12,857 | 8 | 25,715 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you know, all the kids in Berland love playing with cubes. Little Petya has n towers consisting of cubes of the same size. Tower with number i consists of ai cubes stacked one on top of the other. Petya defines the instability of a set of towers as a value equal to the difference between the heights of the highest and the lowest of the towers. For example, if Petya built five cube towers with heights (8, 3, 2, 6, 3), the instability of this set is equal to 6 (the highest tower has height 8, the lowest one has height 2).
The boy wants the instability of his set of towers to be as low as possible. All he can do is to perform the following operation several times: take the top cube from some tower and put it on top of some other tower of his set. Please note that Petya would never put the cube on the same tower from which it was removed because he thinks it's a waste of time.
Before going to school, the boy will have time to perform no more than k such operations. Petya does not want to be late for class, so you have to help him accomplish this task.
Input
The first line contains two space-separated positive integers n and k (1 ≤ n ≤ 100, 1 ≤ k ≤ 1000) — the number of towers in the given set and the maximum number of operations Petya can perform. The second line contains n space-separated positive integers ai (1 ≤ ai ≤ 104) — the towers' initial heights.
Output
In the first line print two space-separated non-negative integers s and m (m ≤ k). The first number is the value of the minimum possible instability that can be obtained after performing at most k operations, the second number is the number of operations needed for that.
In the next m lines print the description of each operation as two positive integers i and j, each of them lies within limits from 1 to n. They represent that Petya took the top cube from the i-th tower and put in on the j-th one (i ≠ j). Note that in the process of performing operations the heights of some towers can become equal to zero.
If there are multiple correct sequences at which the minimum possible instability is achieved, you are allowed to print any of them.
Examples
Input
3 2
5 8 5
Output
0 2
2 1
2 3
Input
3 4
2 2 4
Output
1 1
3 2
Input
5 3
8 3 2 6 3
Output
3 3
1 3
1 2
1 3
Note
In the first sample you need to move the cubes two times, from the second tower to the third one and from the second one to the first one. Then the heights of the towers are all the same and equal to 6.
Submitted Solution:
```
r = []
n,k = map(int,input().split())
l = list(map(int,input().split()))
m = 0
s = max(l)-min(l)
for lll in range(k):
m += 1
s = max(l)-min(l)
i = l.index(max(l))
j = l.index(min(l))
if i == j:
break
l[i],l[j] = l[i]-1,l[j]+1
r.append([i+1,j+1])
s = max(l)-min(l)
if max(l) == min(l) or max(l)-min(l) == 1:
break
print(s,m)
for i in r:
print(i[0],i[1])
``` | instruction | 0 | 12,858 | 8 | 25,716 |
No | output | 1 | 12,858 | 8 | 25,717 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bill is a famous mathematician in BubbleLand. Thanks to his revolutionary math discoveries he was able to make enough money to build a beautiful house. Unfortunately, for not paying property tax on time, court decided to punish Bill by making him lose a part of his property.
Bill’s property can be observed as a convex regular 2n-sided polygon A0 A1... A2n - 1 A2n, A2n = A0, with sides of the exactly 1 meter in length.
Court rules for removing part of his property are as follows:
* Split every edge Ak Ak + 1, k = 0... 2n - 1 in n equal parts of size 1 / n with points P0, P1, ..., Pn - 1
* On every edge A2k A2k + 1, k = 0... n - 1 court will choose one point B2k = Pi for some i = 0, ..., n - 1 such that <image>
* On every edge A2k + 1A2k + 2, k = 0...n - 1 Bill will choose one point B2k + 1 = Pi for some i = 0, ..., n - 1 such that <image>
* Bill gets to keep property inside of 2n-sided polygon B0 B1... B2n - 1
Luckily, Bill found out which B2k points the court chose. Even though he is a great mathematician, his house is very big and he has a hard time calculating. Therefore, he is asking you to help him choose points so he maximizes area of property he can keep.
Input
The first line contains one integer number n (2 ≤ n ≤ 50000), representing number of edges of 2n-sided polygon.
The second line contains n distinct integer numbers B2k (0 ≤ B2k ≤ n - 1, k = 0... n - 1) separated by a single space, representing points the court chose. If B2k = i, the court chose point Pi on side A2k A2k + 1.
Output
Output contains n distinct integers separated by a single space representing points B1, B3, ..., B2n - 1 Bill should choose in order to maximize the property area. If there are multiple solutions that maximize the area, return any of them.
Example
Input
3
0 1 2
Output
0 2 1
Note
To maximize area Bill should choose points: B1 = P0, B3 = P2, B5 = P1
<image> | instruction | 0 | 13,030 | 8 | 26,060 |
Tags: greedy, sortings
Correct Solution:
```
n = int(input())
a = input().split()
for i in range(n):
a[i] = int(a[i])
b = []
for i in range(0, n-1):
b.append((a[i]-(n-a[i+1]), i))
b.append((a[n-1]-(n-a[0]), n-1))
b = sorted(b)
ans = n*[0]
for i in range(n):
# the line segment at index b[i][1]
ans[b[i][1]] = i
for i in range(n):
print(ans[i], end = ' ')
``` | output | 1 | 13,030 | 8 | 26,061 |
Provide a correct Python 3 solution for this coding contest problem.
There is an Amidakuji that consists of w vertical bars and has a height (the number of steps to which horizontal bars can be added) of h. w is an even number. Of the candidates for the place to add the horizontal bar of this Amidakuji, the ath from the top and the bth from the left are called (a, b). (When a horizontal bar is added to (a, b), the bth and b + 1th vertical bars from the left are connected in the ath row from the top.) Such places are h (w −1) in total. (1 ≤ a ≤ h, 1 ≤ b ≤ w − 1) Exists.
Sunuke added all horizontal bars to the places (a, b) where a ≡ b (mod 2) is satisfied. Next, Sunuke erased the horizontal bar at (a1, b1), ..., (an, bn). Find all the numbers from the left at the bottom when you select the i-th from the left at the top.
Constraints
* 1 ≤ h, w, n ≤ 200000
* w is even
* 1 ≤ ai ≤ h
* 1 ≤ bi ≤ w − 1
* ai ≡ bi (mod 2)
* There are no different i, j such that (ai, bi) = (aj, bj)
Input
h w n
a1 b1
.. ..
an bn
Output
Output w lines. On the i-th line, output the number from the left at the bottom when the i-th from the left is selected at the top.
Examples
Input
4 4 1
3 3
Output
2
3
4
1
Input
10 6 10
10 4
4 4
5 1
4 2
7 3
1 3
2 4
8 2
7 5
7 1
Output
1
4
3
2
5
6 | instruction | 0 | 13,257 | 8 | 26,514 |
"Correct Solution:
```
from collections import defaultdict
import sys
def solve():
readline = sys.stdin.readline
write = sys.stdout.write
H, W, N = map(int, readline().split())
A = [0]*W
B = [0]*W
C = [0]*W
for i in range(W//2):
A[i] = i * 2 + 1
A[-i-1] = i * 2
B[i*2] = C[i*2 - 1] = i
B[-1-i*2] = C[-1-i*2 + 1] = i + W//2
*X, = range(W)
P = [list(map(int, readline().split())) for i in range(N)]
P.sort()
for a, b in P:
if a % 2 == 1:
k = a // 2
m = b//2*2
p = A[(B[m] + k) % W]
q = A[(B[m+1] + k) % W]
else:
k = a // 2 - 1
m = (b-1)//2*2
p = A[(C[m+1] + k) % W]
q = A[(C[m+2] + k) % W]
X[p], X[q] = X[q], X[p]
*E0, = range(W)
for i in range(W//2):
E0[2*i], E0[2*i+1] = E0[2*i+1], E0[2*i]
E1 = E0[:]
for i in range(W//2-1):
E1[2*i+1], E1[2*i+2] = E1[2*i+2], E1[2*i+1]
k = H//2
*Y, = range(W)
while k:
if k & 1:
Y = [Y[e] for e in E1]
E1 = [E1[e] for e in E1]
k >>= 1
if H % 2:
Y = [Y[e] for e in E0]
ans = [0]*W
for i, e in enumerate(Y):
ans[X[e]] = i+1
for e in ans:
write("%d\n" % e)
solve()
``` | output | 1 | 13,257 | 8 | 26,515 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During cleaning the coast, Alice found n piles of stones. The i-th pile has a_i stones.
Piles i and i + 1 are neighbouring for all 1 ≤ i ≤ n - 1. If pile i becomes empty, piles i - 1 and i + 1 doesn't become neighbouring.
Alice is too lazy to remove these stones, so she asked you to take this duty. She allowed you to do only the following operation:
* Select two neighboring piles and, if both of them are not empty, remove one stone from each of them.
Alice understands that sometimes it's impossible to remove all stones with the given operation, so she allowed you to use the following superability:
* Before the start of cleaning, you can select two neighboring piles and swap them.
Determine, if it is possible to remove all stones using the superability not more than once.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of piles.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the number of stones in each pile.
It is guaranteed that the total sum of n over all test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print YES or NO — is it possible to remove all stones using the superability not more than once or not.
Example
Input
5
3
1 2 1
3
1 1 2
5
2 2 2 1 3
5
2100 1900 1600 3000 1600
2
2443 2445
Output
YES
YES
YES
YES
NO
Note
In the first test case, you can remove all stones without using a superability: [1, 2, 1] → [1, 1, 0] → [0, 0, 0].
In the second test case, you can apply superability to the second and the third piles and then act like in the first testcase.
In the third test case, you can apply superability to the fourth and the fifth piles, thus getting a = [2, 2, 2, 3, 1].
In the fourth test case, you can apply superability to the first and the second piles, thus getting a = [1900, 2100, 1600, 3000, 1600]. | instruction | 0 | 14,385 | 8 | 28,770 |
Tags: data structures, dp, greedy, math
Correct Solution:
```
# 452c532939484f489758bb6a1e71a6aafee40ae6f991e8cd80298b6258223d75
#adding bot
"""
Author - Satwik Tiwari .
19th Jan , 2021 - Tuesday
"""
#===============================================================================================
#importing some useful libraries.
from __future__ import division, print_function
# from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
# from itertools import *
from heapq import *
from math import gcd, factorial,floor,ceil,sqrt,log2
from copy import deepcopy
from collections import deque
from bisect import bisect_left as bl
from bisect import bisect_right as br
from bisect import bisect
#==============================================================================================
#fast I/O region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
# inp = lambda: sys.stdin.readline().rstrip("\r\n")
#===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
def wrapped_func(*args, **kwargs):
if stack: return f(*args, **kwargs)
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
continue
stack.pop()
if not stack: break
to = stack[-1].send(to)
return to
return wrapped_func
#### END ITERATE RECURSION ####
#===============================================================================================
#some shortcuts
def inp(): return sys.stdin.readline().rstrip("\r\n") #for fast input
def out(var): sys.stdout.write(str(var)) #for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
# def graph(vertex): return [[] for i in range(0,vertex+1)]
def testcase(t):
for pp in range(t):
solve(pp)
def google(p):
print('Case #'+str(p)+': ',end='')
def lcm(a,b): return (a*b)//gcd(a,b)
def power(x, y, p) :
y%=(p-1) #not so sure about this. used when y>p-1. if p is prime.
res = 1 # Initialize result
x = x % p # Update x if it is more , than or equal to p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) : # If y is odd, multiply, x with result
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
def ncr(n,r): return factorial(n) // (factorial(r) * factorial(max(n - r, 1)))
def isPrime(n) :
if (n <= 1) : return False
if (n <= 3) : return True
if (n % 2 == 0 or n % 3 == 0) : return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
inf = pow(10,20)
mod = 10**9+7
#===============================================================================================
# code here ;))
def do(b,n):
ans = True
for i in range(n-1):
if(i == n-2):
if(b[i] != b[i+1]):
ans = False
else:
if(b[i] <= b[i+1]):
b[i+1]-=b[i]
else:
ans = False
return ans
def chck(a):
ans = False
b = deepcopy(a)
n = len(a)
ans = ans | do(b,n)
b = deepcopy(a)
b[0],b[1] = b[1],b[0]
ans = ans | do(b,n)
b = deepcopy(a)
b[n-1],b[n-2] = b[n-2],b[n-1]
ans = ans | do(b,n)
return ans
def solve(case):
n = int(inp())
b = lis()
if(n == 2):
if(b[0] == b[1]):
print('YES')
else:
print('NO')
return
if(chck(b)):
print('YES')
return
l = deepcopy(b)
left = [0]*n
left[0] = 1
for i in range(1,n-1):
if(l[i] >= l[i-1]):
l[i]-=l[i-1]
# a[i] = 0
left[i] = 1 and left[i-1]
else:
left[i] = 0
r = deepcopy(b)
right = [0]*n
right[n-1] = 1
for i in range(n-2,0,-1):
if(r[i] >= r[i+1]):
r[i]-=r[i+1]
right[i] = 1 and right[i+1]
else:
right[i] = 0
# print(l,left)
# print(r,right)
ans = False
for i in range(n-3):
if(left[i] and right[i+3] and b[i+2] >= l[i] and b[i+1] >= r[i+3] and b[i+2]-l[i] == b[i+1]-r[i+3]):
ans = True
# print(i)
print('YES' if ans else 'NO')
# testcase(1)
testcase(int(inp()))
``` | output | 1 | 14,385 | 8 | 28,771 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During cleaning the coast, Alice found n piles of stones. The i-th pile has a_i stones.
Piles i and i + 1 are neighbouring for all 1 ≤ i ≤ n - 1. If pile i becomes empty, piles i - 1 and i + 1 doesn't become neighbouring.
Alice is too lazy to remove these stones, so she asked you to take this duty. She allowed you to do only the following operation:
* Select two neighboring piles and, if both of them are not empty, remove one stone from each of them.
Alice understands that sometimes it's impossible to remove all stones with the given operation, so she allowed you to use the following superability:
* Before the start of cleaning, you can select two neighboring piles and swap them.
Determine, if it is possible to remove all stones using the superability not more than once.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of piles.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the number of stones in each pile.
It is guaranteed that the total sum of n over all test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print YES or NO — is it possible to remove all stones using the superability not more than once or not.
Example
Input
5
3
1 2 1
3
1 1 2
5
2 2 2 1 3
5
2100 1900 1600 3000 1600
2
2443 2445
Output
YES
YES
YES
YES
NO
Note
In the first test case, you can remove all stones without using a superability: [1, 2, 1] → [1, 1, 0] → [0, 0, 0].
In the second test case, you can apply superability to the second and the third piles and then act like in the first testcase.
In the third test case, you can apply superability to the fourth and the fifth piles, thus getting a = [2, 2, 2, 3, 1].
In the fourth test case, you can apply superability to the first and the second piles, thus getting a = [1900, 2100, 1600, 3000, 1600]. | instruction | 0 | 14,386 | 8 | 28,772 |
Tags: data structures, dp, greedy, math
Correct Solution:
```
from bisect import *
from collections import *
from math import gcd,ceil,sqrt,floor,inf
from heapq import *
from itertools import *
#from operator import add,mul,sub,xor,truediv,floordiv
from functools import *
#------------------------------------------------------------------------
import os
import sys
from io import BytesIO, IOBase
# region fastio
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")
#------------------------------------------------------------------------
def RL(): return map(int, sys.stdin.readline().rstrip().split())
def RLL(): return list(map(int, sys.stdin.readline().rstrip().split()))
def N(): return int(input())
#------------------------------------------------------------------------
from types import GeneratorType
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
mod=10**9+7
farr=[1]
ifa=[]
def fact(x,mod=0):
if mod:
while x>=len(farr):
farr.append(farr[-1]*len(farr)%mod)
else:
while x>=len(farr):
farr.append(farr[-1]*len(farr))
return farr[x]
def ifact(x,mod):
global ifa
fact(x,mod)
ifa.append(pow(farr[-1],mod-2,mod))
for i in range(x,0,-1):
ifa.append(ifa[-1]*i%mod)
ifa.reverse()
def per(i,j,mod=0):
if i<j: return 0
if not mod:
return fact(i)//fact(i-j)
return farr[i]*ifa[i-j]%mod
def com(i,j,mod=0):
if i<j: return 0
if not mod:
return per(i,j)//fact(j)
return per(i,j,mod)*ifa[j]%mod
def catalan(n):
return com(2*n,n)//(n+1)
def isprime(n):
for i in range(2,int(n**0.5)+1):
if n%i==0:
return False
return True
def floorsum(a,b,c,n):#sum((a*i+b)//c for i in range(n+1))
if a==0:return b//c*(n+1)
if a>=c or b>=c: return floorsum(a%c,b%c,c,n)+b//c*(n+1)+a//c*n*(n+1)//2
m=(a*n+b)//c
return n*m-floorsum(c,c-b-1,a,m-1)
def lowbit(n):
return n&-n
def inverse(a,m):
a%=m
if a<=1: return a
return ((1-inverse(m,a)*m)//a)%m
class BIT:
def __init__(self,arr):
self.arr=arr
self.n=len(arr)-1
def update(self,x,v):
while x<=self.n:
self.arr[x]+=v
x+=x&-x
def query(self,x):
ans=0
while x:
ans+=self.arr[x]
x&=x-1
return ans
'''
class SMT:
def __init__(self,arr):
self.n=len(arr)-1
self.arr=[0]*(self.n<<2)
self.lazy=[0]*(self.n<<2)
def Build(l,r,rt):
if l==r:
self.arr[rt]=arr[l]
return
m=(l+r)>>1
Build(l,m,rt<<1)
Build(m+1,r,rt<<1|1)
self.pushup(rt)
Build(1,self.n,1)
def pushup(self,rt):
self.arr[rt]=self.arr[rt<<1]+self.arr[rt<<1|1]
def pushdown(self,rt,ln,rn):#lr,rn表区间数字数
if self.lazy[rt]:
self.lazy[rt<<1]+=self.lazy[rt]
self.lazy[rt<<1|1]+=self.lazy[rt]
self.arr[rt<<1]+=self.lazy[rt]*ln
self.arr[rt<<1|1]+=self.lazy[rt]*rn
self.lazy[rt]=0
def update(self,L,R,c,l=1,r=None,rt=1):#L,R表示操作区间
if r==None: r=self.n
if L<=l and r<=R:
self.arr[rt]+=c*(r-l+1)
self.lazy[rt]+=c
return
m=(l+r)>>1
self.pushdown(rt,m-l+1,r-m)
if L<=m: self.update(L,R,c,l,m,rt<<1)
if R>m: self.update(L,R,c,m+1,r,rt<<1|1)
self.pushup(rt)
def query(self,L,R,l=1,r=None,rt=1):
if r==None: r=self.n
#print(L,R,l,r,rt)
if L<=l and R>=r:
return self.arr[rt]
m=(l+r)>>1
self.pushdown(rt,m-l+1,r-m)
ans=0
if L<=m: ans+=self.query(L,R,l,m,rt<<1)
if R>m: ans+=self.query(L,R,m+1,r,rt<<1|1)
return ans
'''
class DSU:#容量+路径压缩
def __init__(self,n):
self.c=[-1]*n
def same(self,x,y):
return self.find(x)==self.find(y)
def find(self,x):
if self.c[x]<0:
return x
self.c[x]=self.find(self.c[x])
return self.c[x]
def union(self,u,v):
u,v=self.find(u),self.find(v)
if u==v:
return False
if self.c[u]>self.c[v]:
u,v=v,u
self.c[u]+=self.c[v]
self.c[v]=u
return True
def size(self,x): return -self.c[self.find(x)]
class UFS:#秩+路径
def __init__(self,n):
self.parent=[i for i in range(n)]
self.ranks=[0]*n
def find(self,x):
if x!=self.parent[x]:
self.parent[x]=self.find(self.parent[x])
return self.parent[x]
def union(self,u,v):
pu,pv=self.find(u),self.find(v)
if pu==pv:
return False
if self.ranks[pu]>=self.ranks[pv]:
self.parent[pv]=pu
if self.ranks[pv]==self.ranks[pu]:
self.ranks[pu]+=1
else:
self.parent[pu]=pv
def Prime(n):
c=0
prime=[]
flag=[0]*(n+1)
for i in range(2,n+1):
if not flag[i]:
prime.append(i)
c+=1
for j in range(c):
if i*prime[j]>n: break
flag[i*prime[j]]=prime[j]
if i%prime[j]==0: break
return prime
def dij(s,graph):
d={}
d[s]=0
heap=[(0,s)]
seen=set()
while heap:
dis,u=heappop(heap)
if u in seen:
continue
seen.add(u)
for v,w in graph[u]:
if v not in d or d[v]>d[u]+w:
d[v]=d[u]+w
heappush(heap,(d[v],v))
return d
def GP(it): return [[ch,len(list(g))] for ch,g in groupby(it)]
def lcm(a,b): return a*b//gcd(a,b)
def lis(nums):
res=[]
for k in nums:
i=bisect.bisect_left(res,k)
if i==len(res):
res.append(k)
else:
res[i]=k
return len(res)
class DLN:
def __init__(self,val):
self.val=val
self.pre=None
self.next=None
def nb(i,j):
for ni,nj in [[i+1,j],[i-1,j],[i,j-1],[i,j+1]]:
if 0<=ni<n and 0<=nj<m:
yield ni,nj
def topo(n):
q=deque()
res=[]
for i in range(1,n+1):
if ind[i]==0:
q.append(i)
res.append(i)
while q:
u=q.popleft()
for v in g[u]:
ind[v]-=1
if ind[v]==0:
q.append(v)
res.append(v)
return res
@bootstrap
def gdfs(r,p):
if len(g[r])==1 and p!=-1:
yield None
for ch in g[r]:
if ch!=p:
yield gdfs(ch,r)
yield None
def judge(a):
res=[0]
n=len(a)
f=True
for x in a:
res.append(x-res[-1])
#print(res)
if res[-1]<0:
f=False
break
if f:
#print(a,res)
return 1
for i in range(n-1):
a[i],a[i+1]=a[i+1],a[i]
res=[0]
f=True
for x in a:
res.append(x-res[-1])
if res[-1]<0:
f=False
break
if f:
print(a,res)
a[i],a[i+1]=a[i+1],a[i]
return 1
a[i],a[i+1]=a[i+1],a[i]
return 0
from random import randint
def build(n):
f=False
while not f:
a=[]
for i in range(n-1):
a.append(randint(1,10))
x=0
for i in range(n-1):
x=a[i]-x
a.append(x)
if a[-1]>0:
f=True
return a
t=N()
for i in range(t):
n=N()
a=RLL()
#a.sort()
s=0
res=[]
pre=[0]
for i in range(n):
if i&1:
res.append(-a[i])
else:
res.append(a[i])
pre.append(a[i]-pre[-1])
s=pre[-1]
ans=False
#print(res,s)
mi=[0]*n
for i in range(n-1,-1,-1):
if i==n-1:
mi[i]=pre[-1]
elif i==n-2:
mi[i]=pre[-2]
else:
mi[i]=min(mi[i+2],pre[i+1])
if s==0:
if all(x>=0 for x in pre):
ans=True
else:
for i in range(n-1):
cur=a[i]-a[i+1] if not (n-1-i)&1 else a[i+1]-a[i]
if s-2*cur==0:
if a[i+1]-pre[i]>=0 and a[i]-a[i+1]+pre[i]>=0:
if (i>=n-2 or mi[i+2]>=2*(a[i]-a[i+1])) and (i>=n-3 or mi[i+3]>=2*(a[i+1]-a[i])):
ans=True
break
if pre[i+1]<0:
break
ans='YES' if ans else "NO"
print(ans)
'''
sys.setrecursionlimit(200000)
import threading
threading.stack_size(10**8)
t=threading.Thread(target=main)
t.start()
t.join()
'''
'''
sys.setrecursionlimit(200000)
import threading
threading.stack_size(10**8)
t=threading.Thread(target=main)
t.start()
t.join()
'''
``` | output | 1 | 14,386 | 8 | 28,773 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During cleaning the coast, Alice found n piles of stones. The i-th pile has a_i stones.
Piles i and i + 1 are neighbouring for all 1 ≤ i ≤ n - 1. If pile i becomes empty, piles i - 1 and i + 1 doesn't become neighbouring.
Alice is too lazy to remove these stones, so she asked you to take this duty. She allowed you to do only the following operation:
* Select two neighboring piles and, if both of them are not empty, remove one stone from each of them.
Alice understands that sometimes it's impossible to remove all stones with the given operation, so she allowed you to use the following superability:
* Before the start of cleaning, you can select two neighboring piles and swap them.
Determine, if it is possible to remove all stones using the superability not more than once.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of piles.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the number of stones in each pile.
It is guaranteed that the total sum of n over all test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print YES or NO — is it possible to remove all stones using the superability not more than once or not.
Example
Input
5
3
1 2 1
3
1 1 2
5
2 2 2 1 3
5
2100 1900 1600 3000 1600
2
2443 2445
Output
YES
YES
YES
YES
NO
Note
In the first test case, you can remove all stones without using a superability: [1, 2, 1] → [1, 1, 0] → [0, 0, 0].
In the second test case, you can apply superability to the second and the third piles and then act like in the first testcase.
In the third test case, you can apply superability to the fourth and the fifth piles, thus getting a = [2, 2, 2, 3, 1].
In the fourth test case, you can apply superability to the first and the second piles, thus getting a = [1900, 2100, 1600, 3000, 1600]. | instruction | 0 | 14,387 | 8 | 28,774 |
Tags: data structures, dp, greedy, math
Correct Solution:
```
import sys;input=sys.stdin.readline
T, = map(int, input().split())
for _ in range(T):
N, = map(int, input().split())
X = list(map(int, input().split()))
l = [-1] * N
b = 0
f = 1
for i in range(N):
x = X[i]
if b > x:
f = 0
break
b = x-b
l[i] = b
if f and b == 0:
print("YES")
continue
l2 = [-1]*N
b = 0
for i in range(N-1, -1, -1):
x = X[i]
if b > x:
break
b = x-b
l2[i] = b
f = 0
for i in range(N-1):
c, b = X[i], X[i+1]
if i==0:
a = 0
else:
a = l[i-1]
if i==N-2:
d=0
else:
d = l2[i+2]
if a==-1 or d==-1:
continue
if b-a==c-d and b>=a:
print("YES")
f=1
break
if not f:
print("NO")
``` | output | 1 | 14,387 | 8 | 28,775 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During cleaning the coast, Alice found n piles of stones. The i-th pile has a_i stones.
Piles i and i + 1 are neighbouring for all 1 ≤ i ≤ n - 1. If pile i becomes empty, piles i - 1 and i + 1 doesn't become neighbouring.
Alice is too lazy to remove these stones, so she asked you to take this duty. She allowed you to do only the following operation:
* Select two neighboring piles and, if both of them are not empty, remove one stone from each of them.
Alice understands that sometimes it's impossible to remove all stones with the given operation, so she allowed you to use the following superability:
* Before the start of cleaning, you can select two neighboring piles and swap them.
Determine, if it is possible to remove all stones using the superability not more than once.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of piles.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the number of stones in each pile.
It is guaranteed that the total sum of n over all test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print YES or NO — is it possible to remove all stones using the superability not more than once or not.
Example
Input
5
3
1 2 1
3
1 1 2
5
2 2 2 1 3
5
2100 1900 1600 3000 1600
2
2443 2445
Output
YES
YES
YES
YES
NO
Note
In the first test case, you can remove all stones without using a superability: [1, 2, 1] → [1, 1, 0] → [0, 0, 0].
In the second test case, you can apply superability to the second and the third piles and then act like in the first testcase.
In the third test case, you can apply superability to the fourth and the fifth piles, thus getting a = [2, 2, 2, 3, 1].
In the fourth test case, you can apply superability to the first and the second piles, thus getting a = [1900, 2100, 1600, 3000, 1600]. | instruction | 0 | 14,388 | 8 | 28,776 |
Tags: data structures, dp, greedy, math
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
a = [int(x) for x in input().split()]
suok = [True]
sulft = [0]
for i in range(n - 1, -1, -1):
if not suok[-1] or a[i] < sulft[-1]:
suok.append(False)
sulft.append(0)
else:
suok.append(True)
sulft.append(a[i] - sulft[-1])
a.append(0)
suok = suok[::-1]
sulft = sulft[::-1]
if suok[0] and sulft[0] == 0:
print("YES")
continue
prlft = 0
win = False
for i in range(n - 1):
a[i], a[i + 1] = a[i + 1], a[i]
if suok[i + 2] and a[i] >= prlft and a[i + 1] >= (a[i] - prlft) and a[i + 1] - (a[i] - prlft) == sulft[i + 2]:
win = True
break
a[i], a[i + 1] = a[i + 1], a[i]
if a[i] < prlft:
break
prlft = a[i] - prlft
print("YES" if win else "NO")
``` | output | 1 | 14,388 | 8 | 28,777 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During cleaning the coast, Alice found n piles of stones. The i-th pile has a_i stones.
Piles i and i + 1 are neighbouring for all 1 ≤ i ≤ n - 1. If pile i becomes empty, piles i - 1 and i + 1 doesn't become neighbouring.
Alice is too lazy to remove these stones, so she asked you to take this duty. She allowed you to do only the following operation:
* Select two neighboring piles and, if both of them are not empty, remove one stone from each of them.
Alice understands that sometimes it's impossible to remove all stones with the given operation, so she allowed you to use the following superability:
* Before the start of cleaning, you can select two neighboring piles and swap them.
Determine, if it is possible to remove all stones using the superability not more than once.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of piles.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the number of stones in each pile.
It is guaranteed that the total sum of n over all test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print YES or NO — is it possible to remove all stones using the superability not more than once or not.
Example
Input
5
3
1 2 1
3
1 1 2
5
2 2 2 1 3
5
2100 1900 1600 3000 1600
2
2443 2445
Output
YES
YES
YES
YES
NO
Note
In the first test case, you can remove all stones without using a superability: [1, 2, 1] → [1, 1, 0] → [0, 0, 0].
In the second test case, you can apply superability to the second and the third piles and then act like in the first testcase.
In the third test case, you can apply superability to the fourth and the fifth piles, thus getting a = [2, 2, 2, 3, 1].
In the fourth test case, you can apply superability to the first and the second piles, thus getting a = [1900, 2100, 1600, 3000, 1600]. | instruction | 0 | 14,389 | 8 | 28,778 |
Tags: data structures, dp, greedy, math
Correct Solution:
```
#!/usr/bin/env python3
import sys, getpass
import math, random
import functools, itertools, collections, heapq, bisect
from collections import Counter, defaultdict, deque
input = sys.stdin.readline # to read input quickly
# available on Google, AtCoder Python3, not available on Codeforces
# import numpy as np
# import scipy
M9 = 10**9 + 7 # 998244353
# d4 = [(1,0),(0,1),(-1,0),(0,-1)]
# d8 = [(1,0),(1,1),(0,1),(-1,1),(-1,0),(-1,-1),(0,-1),(1,-1)]
# d6 = [(2,0),(1,1),(-1,1),(-2,0),(-1,-1),(1,-1)] # hexagonal layout
MAXINT = sys.maxsize
# if testing locally, print to terminal with a different color
OFFLINE_TEST = getpass.getuser() == "hkmac"
# OFFLINE_TEST = False # codechef does not allow getpass
def log(*args):
if OFFLINE_TEST:
print('\033[36m', *args, '\033[0m', file=sys.stderr)
def solve(*args):
# screen input
if OFFLINE_TEST:
log("----- solving ------")
log(*args)
log("----- ------- ------")
return solve_(*args)
def read_matrix(rows):
return [list(map(int,input().split())) for _ in range(rows)]
def read_strings(rows):
return [input().strip() for _ in range(rows)]
# ---------------------------- template ends here ----------------------------
def check(lst):
remainder = lst[0]
for x in lst[1:]:
remainder = x - remainder
if remainder < 0:
return False
else: # ok without swap
if remainder == 0:
return True
return False
def remainder_arr(lst):
remainder = lst[0]
remainders = [0,lst[0]]
faulty = False
for x in lst[1:]:
remainder = x - remainder
if faulty:
remainder = -1
remainders.append(remainder)
if remainder < 0:
faulty = True
return remainders
def solve_(arr, reverse=False):
# sweep, if cannot, replace
if check(arr):
log("no swap")
return "YES"
remainder = arr[0]
for i,x in enumerate(arr[1:], start=1):
remainder = x - remainder
if remainder < 0:
# swap with either before or after
arr[i], arr[i-1] = arr[i-1], arr[i]
if check(arr):
log("left swap")
return "YES"
arr[i], arr[i-1] = arr[i-1], arr[i]
break
arr[-2], arr[-1] = arr[-1], arr[-2]
if check(arr):
log("end swap")
return "YES"
arr[-2], arr[-1] = arr[-1], arr[-2]
xrr = remainder_arr(arr)
yrr = remainder_arr(arr[::-1])[::-1]
log(xrr)
log(yrr)
remainder = 0
# if I swap, what is the remainder
for i,(x,y) in enumerate(zip(arr, arr[1:])):
if x-remainder > y: # must swap, would have swapped
break
new_remainder = x-(y-remainder)
log(x,y,new_remainder,yrr[i+2])
if new_remainder >= 0 and yrr[i+2] == new_remainder:
arr[i], arr[i+1] = arr[i+1], arr[i]
if check(arr):
log("advanced swap")
return "YES"
arr[i], arr[i+1] = arr[i+1], arr[i]
remainder = x-remainder
# log(remainder, "r")
if not reverse:
log("reversing")
return solve_(arr[::-1], reverse=True)
return "NO"
# for case_num in [0]: # no loop over test case
# for case_num in range(100): # if the number of test cases is specified
for case_num in range(int(input())):
# read line as an integer
k = int(input())
# read line as a string
# srr = input().strip()
# read one line and parse each word as a string
# lst = input().split()
# read one line and parse each word as an integer
# a,b,c = list(map(int,input().split()))
lst = list(map(int,input().split()))
# read multiple rows
# mrr = read_matrix(k) # and return as a list of list of int
# arr = read_strings(k) # and return as a list of str
res = solve(lst) # include input here
# print result
# Google and Facebook - case number required
# print("Case #{}: {}".format(case_num+1, res))
# Other platforms - no case number required
print(res)
# print(len(res))
# print(*res) # print a list with elements
# for r in res: # print each list in a different line
# print(res)
# print(*res)
``` | output | 1 | 14,389 | 8 | 28,779 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During cleaning the coast, Alice found n piles of stones. The i-th pile has a_i stones.
Piles i and i + 1 are neighbouring for all 1 ≤ i ≤ n - 1. If pile i becomes empty, piles i - 1 and i + 1 doesn't become neighbouring.
Alice is too lazy to remove these stones, so she asked you to take this duty. She allowed you to do only the following operation:
* Select two neighboring piles and, if both of them are not empty, remove one stone from each of them.
Alice understands that sometimes it's impossible to remove all stones with the given operation, so she allowed you to use the following superability:
* Before the start of cleaning, you can select two neighboring piles and swap them.
Determine, if it is possible to remove all stones using the superability not more than once.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of piles.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the number of stones in each pile.
It is guaranteed that the total sum of n over all test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print YES or NO — is it possible to remove all stones using the superability not more than once or not.
Example
Input
5
3
1 2 1
3
1 1 2
5
2 2 2 1 3
5
2100 1900 1600 3000 1600
2
2443 2445
Output
YES
YES
YES
YES
NO
Note
In the first test case, you can remove all stones without using a superability: [1, 2, 1] → [1, 1, 0] → [0, 0, 0].
In the second test case, you can apply superability to the second and the third piles and then act like in the first testcase.
In the third test case, you can apply superability to the fourth and the fifth piles, thus getting a = [2, 2, 2, 3, 1].
In the fourth test case, you can apply superability to the first and the second piles, thus getting a = [1900, 2100, 1600, 3000, 1600]. | instruction | 0 | 14,390 | 8 | 28,780 |
Tags: data structures, dp, greedy, math
Correct Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO,IOBase
from math import inf
def solve(n,a):
x = a[:]
for i in range(1,n):
a[i] -= a[i-1]
val = a[-1]
if not val:
return 'NO' if min(a)<0 else 'YES'
if abs(a[-1])&1:
return 'NO'
mini = [a[-1],a[-2]]
for i in range(-3,-n-1,-1):
mini.append(min(mini[-2],a[i]))
mini.reverse()
mini.append(inf)
for i in range(n-1):
if (n-1-i)&1:
if (x[i] == x[i+1]-val//2 and mini[i+2] >= -val
and mini[i+1] >= val and a[i] >= -val//2):
return 'YES'
else:
if (x[i] == x[i+1]+val//2 and mini[i+2] >= val
and mini[i+1] >= -val and a[i] >= val//2):
return 'YES'
if a[i] < 0:
break
return 'NO'
def main():
for _ in range(int(input())):
n = int(input())
a = list(map(int,input().split()))
print(solve(n,a))
#Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == '__main__':
main()
``` | output | 1 | 14,390 | 8 | 28,781 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During cleaning the coast, Alice found n piles of stones. The i-th pile has a_i stones.
Piles i and i + 1 are neighbouring for all 1 ≤ i ≤ n - 1. If pile i becomes empty, piles i - 1 and i + 1 doesn't become neighbouring.
Alice is too lazy to remove these stones, so she asked you to take this duty. She allowed you to do only the following operation:
* Select two neighboring piles and, if both of them are not empty, remove one stone from each of them.
Alice understands that sometimes it's impossible to remove all stones with the given operation, so she allowed you to use the following superability:
* Before the start of cleaning, you can select two neighboring piles and swap them.
Determine, if it is possible to remove all stones using the superability not more than once.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of piles.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the number of stones in each pile.
It is guaranteed that the total sum of n over all test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print YES or NO — is it possible to remove all stones using the superability not more than once or not.
Example
Input
5
3
1 2 1
3
1 1 2
5
2 2 2 1 3
5
2100 1900 1600 3000 1600
2
2443 2445
Output
YES
YES
YES
YES
NO
Note
In the first test case, you can remove all stones without using a superability: [1, 2, 1] → [1, 1, 0] → [0, 0, 0].
In the second test case, you can apply superability to the second and the third piles and then act like in the first testcase.
In the third test case, you can apply superability to the fourth and the fifth piles, thus getting a = [2, 2, 2, 3, 1].
In the fourth test case, you can apply superability to the first and the second piles, thus getting a = [1900, 2100, 1600, 3000, 1600]. | instruction | 0 | 14,391 | 8 | 28,782 |
Tags: data structures, dp, greedy, math
Correct Solution:
```
from sys import stdin, stdout
from collections import defaultdict
import math
def main():
t = int(stdin.readline())
for tt in range(1, t+1):
n = int(stdin.readline())
arr = list(map(int, stdin.readline().split()))
dp_l = [-1 for _ in range(n)]
dp_l[0] = arr[0]
for i in range(1, n):
dp_l[i] = arr[i] - dp_l[i - 1]
if dp_l[i] < 0:
break
dp_r = [-1 for _ in range(n)]
dp_r[n-1] = arr[n-1]
for i in range(n-2, -1, -1):
dp_r[i] = arr[i] - dp_r[i + 1]
if dp_r[i] < 0:
break
if dp_l[n-1] == 0:
stdout.write("YES\n")
continue
is_found = False
for i in range(n-1):
p = []
if i > 0:
if dp_l[i-1] < 0:
continue
p.append(dp_l[i-1])
p.append(arr[i+1])
p.append(arr[i])
if i < n-2:
if dp_r[i+2] < 0:
continue
p.append(dp_r[i+2])
sm = p[0]
for j in range(1, len(p)):
sm = p[j] - sm
if sm < 0:
break
if sm == 0:
is_found = True
break
if is_found:
stdout.write("YES\n")
else:
stdout.write("NO\n")
main()
``` | output | 1 | 14,391 | 8 | 28,783 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During cleaning the coast, Alice found n piles of stones. The i-th pile has a_i stones.
Piles i and i + 1 are neighbouring for all 1 ≤ i ≤ n - 1. If pile i becomes empty, piles i - 1 and i + 1 doesn't become neighbouring.
Alice is too lazy to remove these stones, so she asked you to take this duty. She allowed you to do only the following operation:
* Select two neighboring piles and, if both of them are not empty, remove one stone from each of them.
Alice understands that sometimes it's impossible to remove all stones with the given operation, so she allowed you to use the following superability:
* Before the start of cleaning, you can select two neighboring piles and swap them.
Determine, if it is possible to remove all stones using the superability not more than once.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of piles.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the number of stones in each pile.
It is guaranteed that the total sum of n over all test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print YES or NO — is it possible to remove all stones using the superability not more than once or not.
Example
Input
5
3
1 2 1
3
1 1 2
5
2 2 2 1 3
5
2100 1900 1600 3000 1600
2
2443 2445
Output
YES
YES
YES
YES
NO
Note
In the first test case, you can remove all stones without using a superability: [1, 2, 1] → [1, 1, 0] → [0, 0, 0].
In the second test case, you can apply superability to the second and the third piles and then act like in the first testcase.
In the third test case, you can apply superability to the fourth and the fifth piles, thus getting a = [2, 2, 2, 3, 1].
In the fourth test case, you can apply superability to the first and the second piles, thus getting a = [1900, 2100, 1600, 3000, 1600]. | instruction | 0 | 14,392 | 8 | 28,784 |
Tags: data structures, dp, greedy, math
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
s = [a[0]]
for i in range(1, n):
s.append(a[i] - s[-1])
mi0 = mi1 = 10**16
# Suffix min of s
# Based on parity of index
mia0 = []
mia1 = []
for i in range(n - 1, -1, -1):
v = s[i]
if i % 2 == 0:
mi0 = min(mi0, v)
else:
mi1 = min(mi1, v)
mia0.append(mi0)
mia1.append(mi1)
mia0.reverse()
mia1.reverse()
mia0.append(10**16)
mia1.append(10**16)
if min(mi0, mi1) >= 0 and s[-1] == 0:
# No superability
print("YES")
else:
poss = False
# Try all superabilities (O(n))
for i in range(n - 1):
# Difference from swapping the elements
# Alternating positive and negative
dv = -2 * (a[i] - a[i + 1])
# New value of the end
if (i % 2 == 0) == (n % 2 == 0):
ne = s[-1] - dv
else:
ne = s[-1] + dv
# If the end would become 0
if ne == 0:
# choose which min to subtract from / add to
if i % 2 == 1:
if mia0[i] - dv >= 0 and min(mia1[i + 1] + dv, s[i] + dv // 2) >= 0:
poss = True
break
else:
if mia1[i] - dv >= 0 and min(mia0[i + 1] + dv, s[i] + dv // 2) >= 0:
poss = True
break
if s[i] < 0:
break
print("YES" if poss else "NO")
``` | output | 1 | 14,392 | 8 | 28,785 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During cleaning the coast, Alice found n piles of stones. The i-th pile has a_i stones.
Piles i and i + 1 are neighbouring for all 1 ≤ i ≤ n - 1. If pile i becomes empty, piles i - 1 and i + 1 doesn't become neighbouring.
Alice is too lazy to remove these stones, so she asked you to take this duty. She allowed you to do only the following operation:
* Select two neighboring piles and, if both of them are not empty, remove one stone from each of them.
Alice understands that sometimes it's impossible to remove all stones with the given operation, so she allowed you to use the following superability:
* Before the start of cleaning, you can select two neighboring piles and swap them.
Determine, if it is possible to remove all stones using the superability not more than once.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of piles.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the number of stones in each pile.
It is guaranteed that the total sum of n over all test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print YES or NO — is it possible to remove all stones using the superability not more than once or not.
Example
Input
5
3
1 2 1
3
1 1 2
5
2 2 2 1 3
5
2100 1900 1600 3000 1600
2
2443 2445
Output
YES
YES
YES
YES
NO
Note
In the first test case, you can remove all stones without using a superability: [1, 2, 1] → [1, 1, 0] → [0, 0, 0].
In the second test case, you can apply superability to the second and the third piles and then act like in the first testcase.
In the third test case, you can apply superability to the fourth and the fifth piles, thus getting a = [2, 2, 2, 3, 1].
In the fourth test case, you can apply superability to the first and the second piles, thus getting a = [1900, 2100, 1600, 3000, 1600].
Submitted Solution:
```
import sys
input = sys.stdin.readline
for _ in range(int(input())):
N = int(input())
a = list(map(int, input().split()))
l = [-1] * N
l[0] = a[0]
for i in range(N - 1):
if l[i] >= 0: l[i + 1] = a[i + 1] - l[i]
r = [-1] * N
r[-1] = a[-1]
for i in range(N - 1, 0, -1):
if r[i] >= 0:r[i - 1] = a[i - 1] - r[i]
if l[-1] == r[0] == 0:
print("YES")
continue
for i in range(N - 1):
x, y, z, w = 0, a[i + 1], a[i], 0
if i - 1 >= 0: x = l[i - 1]
if i + 2 < N: w = r[i + 2]
if y - x == z - w and x >= 0 and w >= 0 and x <= y and w <= z:
print("YES")
break
else: print("NO")
``` | instruction | 0 | 14,393 | 8 | 28,786 |
Yes | output | 1 | 14,393 | 8 | 28,787 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During cleaning the coast, Alice found n piles of stones. The i-th pile has a_i stones.
Piles i and i + 1 are neighbouring for all 1 ≤ i ≤ n - 1. If pile i becomes empty, piles i - 1 and i + 1 doesn't become neighbouring.
Alice is too lazy to remove these stones, so she asked you to take this duty. She allowed you to do only the following operation:
* Select two neighboring piles and, if both of them are not empty, remove one stone from each of them.
Alice understands that sometimes it's impossible to remove all stones with the given operation, so she allowed you to use the following superability:
* Before the start of cleaning, you can select two neighboring piles and swap them.
Determine, if it is possible to remove all stones using the superability not more than once.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of piles.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the number of stones in each pile.
It is guaranteed that the total sum of n over all test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print YES or NO — is it possible to remove all stones using the superability not more than once or not.
Example
Input
5
3
1 2 1
3
1 1 2
5
2 2 2 1 3
5
2100 1900 1600 3000 1600
2
2443 2445
Output
YES
YES
YES
YES
NO
Note
In the first test case, you can remove all stones without using a superability: [1, 2, 1] → [1, 1, 0] → [0, 0, 0].
In the second test case, you can apply superability to the second and the third piles and then act like in the first testcase.
In the third test case, you can apply superability to the fourth and the fifth piles, thus getting a = [2, 2, 2, 3, 1].
In the fourth test case, you can apply superability to the first and the second piles, thus getting a = [1900, 2100, 1600, 3000, 1600].
Submitted Solution:
```
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
import threading
from bisect import bisect_left
from math import gcd,log
from collections import Counter
from pprint import pprint
MX=10**5
p=[i for i in range(MX)]
for i in range(2,MX):
if p[i]==i:
for j in range(i*i,MX,i):
p[j]=i
pm=[i for i in range(2,MX) if p[i]==i]
# print(pm[-1])
def main():
n=int(input())
arr=list(map(int,input().split()))
pref=[-1]*n
pref[0]=arr[0]
suf=[-1]*n
suf[-1]=arr[-1]
for i in range(1,n):
if pref[i-1]!=-1 and arr[i]>=pref[i-1]:
pref[i]=arr[i]-pref[i-1]
for i in range(n-2,-1,-1):
if suf[i+1]!=-1 and arr[i]>=suf[i+1]:
suf[i]=arr[i]-suf[i+1]
pref=[0]+pref
suf.append(0)
# print(pref)
# print(suf)
for i in range(n-1):
# print(pref[i],arr[i+1],arr[i],suf[i+2])
if pref[i]!=-1 and suf[i+2]!=-1:
if arr[i+1]>=pref[i] and arr[i]>=suf[i+2] and arr[i+1]-pref[i]==arr[i]-suf[i+2]:
print('YES')
return
if pref[i]==suf[i]==0:
print('YES')
return
print('NO')
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")
# endregion
if __name__ == "__main__":
for _ in range(int(input())):
main()
``` | instruction | 0 | 14,394 | 8 | 28,788 |
Yes | output | 1 | 14,394 | 8 | 28,789 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During cleaning the coast, Alice found n piles of stones. The i-th pile has a_i stones.
Piles i and i + 1 are neighbouring for all 1 ≤ i ≤ n - 1. If pile i becomes empty, piles i - 1 and i + 1 doesn't become neighbouring.
Alice is too lazy to remove these stones, so she asked you to take this duty. She allowed you to do only the following operation:
* Select two neighboring piles and, if both of them are not empty, remove one stone from each of them.
Alice understands that sometimes it's impossible to remove all stones with the given operation, so she allowed you to use the following superability:
* Before the start of cleaning, you can select two neighboring piles and swap them.
Determine, if it is possible to remove all stones using the superability not more than once.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of piles.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the number of stones in each pile.
It is guaranteed that the total sum of n over all test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print YES or NO — is it possible to remove all stones using the superability not more than once or not.
Example
Input
5
3
1 2 1
3
1 1 2
5
2 2 2 1 3
5
2100 1900 1600 3000 1600
2
2443 2445
Output
YES
YES
YES
YES
NO
Note
In the first test case, you can remove all stones without using a superability: [1, 2, 1] → [1, 1, 0] → [0, 0, 0].
In the second test case, you can apply superability to the second and the third piles and then act like in the first testcase.
In the third test case, you can apply superability to the fourth and the fifth piles, thus getting a = [2, 2, 2, 3, 1].
In the fourth test case, you can apply superability to the first and the second piles, thus getting a = [1900, 2100, 1600, 3000, 1600].
Submitted Solution:
```
def ke(x: list):
y = x.copy()
for i in range(0, len(y) - 1):
if y[i] > y[i + 1]:
return False
else:
y[i + 1] -= y[i]
y[i] = 0
if y[len(y) - 1] == 0:
return True
return False
t = int(input())
for cas in range(0, t):
n = int(input())
v = list(map(int, input().split()))
if ke(v):
print("YES")
continue
if n == 1:
print("NO")
continue
z = v.copy()
z[0], z[1] = z[1], z[0]
if ke(z):
print("YES")
continue
z = v.copy()
z[n - 2], z[n - 1] = z[n - 1], z[n - 2]
if ke(z):
print("YES")
continue
zuo = [0] * n
you = [0] * n
z = v.copy()
zuo[0] = v[0]
for i in range(0, n - 1):
if z[i] > z[i + 1]:
for j in range(i + 1, n):
zuo[j] = -1
break
else:
z[i + 1] -= z[i]
zuo[i + 1] = z[i + 1]
z[i] = 0
z = v.copy()
you[n - 1] = v[n - 1]
for i in range(n - 1, 0, -1):
if z[i] > z[i - 1]:
for j in range(i - 1, -1, -1):
you[j] = -1
break
else:
z[i - 1] -= z[i]
you[i - 1] = z[i - 1]
z[i] = 0
z = v.copy()
xing = 0
for i in range(1, n - 2):
if zuo[i - 1] != -1 and you[i + 2] != -1 and ke([zuo[i - 1], z[i + 1], z[i], you[i + 2]]):
xing = 1
print("YES")
break
if xing == 0:
print("NO")
``` | instruction | 0 | 14,395 | 8 | 28,790 |
Yes | output | 1 | 14,395 | 8 | 28,791 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During cleaning the coast, Alice found n piles of stones. The i-th pile has a_i stones.
Piles i and i + 1 are neighbouring for all 1 ≤ i ≤ n - 1. If pile i becomes empty, piles i - 1 and i + 1 doesn't become neighbouring.
Alice is too lazy to remove these stones, so she asked you to take this duty. She allowed you to do only the following operation:
* Select two neighboring piles and, if both of them are not empty, remove one stone from each of them.
Alice understands that sometimes it's impossible to remove all stones with the given operation, so she allowed you to use the following superability:
* Before the start of cleaning, you can select two neighboring piles and swap them.
Determine, if it is possible to remove all stones using the superability not more than once.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of piles.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the number of stones in each pile.
It is guaranteed that the total sum of n over all test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print YES or NO — is it possible to remove all stones using the superability not more than once or not.
Example
Input
5
3
1 2 1
3
1 1 2
5
2 2 2 1 3
5
2100 1900 1600 3000 1600
2
2443 2445
Output
YES
YES
YES
YES
NO
Note
In the first test case, you can remove all stones without using a superability: [1, 2, 1] → [1, 1, 0] → [0, 0, 0].
In the second test case, you can apply superability to the second and the third piles and then act like in the first testcase.
In the third test case, you can apply superability to the fourth and the fifth piles, thus getting a = [2, 2, 2, 3, 1].
In the fourth test case, you can apply superability to the first and the second piles, thus getting a = [1900, 2100, 1600, 3000, 1600].
Submitted Solution:
```
t=int(input())
for _ in range(t):
n=int(input())
b=list(map(int,input().split()))
fow=[0]
back=[0]
ind1=[]
ind2=[]
for j in range(n):
fow.append(b[j]-fow[-1])
if fow[-1]<0:
ind1.append(j)
back.append(b[n-1-j]-back[-1])
if back[-1]<0:
ind2.append(n-j-1)
fow.pop(0)
back.reverse()
back.pop()
pre=[j for j in back]
suf=[j for j in fow]
for j in range(n):
if j-2>=0:
pre[j]=min(pre[j],pre[j-2])
if n-j+1<n:
suf[n-j-1]=min(suf[n-j-1],suf[n-j+1])
poss=0
for j in range(n):
if j==0:
if fow[-1]==0 and ind1==[]:
poss=1
break
if back[0]==0 and ind2==[]:
poss=1
break
else:
pr=b[j]-b[j-1]
q=j%2
r=(n-1)%2
if q==r:
res1=fow[-1]-2*pr
else:
res1 = fow[-1] + 2*pr
if res1==0 and fow[j-1]+pr>=0 and suf[j]>=2*pr and (1 if ((j+1<n and suf[j+1]>=-2*pr)or(j+1>=n)) else 0):
if ind1==[]:
poss=1
break
else:
if ind1[0]>=j-1:
poss=1
break
r = 0
if q == r:
res1 = back[0] - 2 * pr
else:
res1 = back[0] + 2 * pr
if res1 == 0 and pre[j-1] >= -2 * pr and (back[j]-pr)>=0 and (1 if ((j-2>=0 and pre[j-2]>=2*pr)or(j-2<0)) else 0) :
if ind2 == []:
poss = 1
break
else:
if ind2[0] < j+1:
poss = 1
break
if poss:
print('YES')
else:
print("NO")
``` | instruction | 0 | 14,396 | 8 | 28,792 |
Yes | output | 1 | 14,396 | 8 | 28,793 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During cleaning the coast, Alice found n piles of stones. The i-th pile has a_i stones.
Piles i and i + 1 are neighbouring for all 1 ≤ i ≤ n - 1. If pile i becomes empty, piles i - 1 and i + 1 doesn't become neighbouring.
Alice is too lazy to remove these stones, so she asked you to take this duty. She allowed you to do only the following operation:
* Select two neighboring piles and, if both of them are not empty, remove one stone from each of them.
Alice understands that sometimes it's impossible to remove all stones with the given operation, so she allowed you to use the following superability:
* Before the start of cleaning, you can select two neighboring piles and swap them.
Determine, if it is possible to remove all stones using the superability not more than once.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of piles.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the number of stones in each pile.
It is guaranteed that the total sum of n over all test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print YES or NO — is it possible to remove all stones using the superability not more than once or not.
Example
Input
5
3
1 2 1
3
1 1 2
5
2 2 2 1 3
5
2100 1900 1600 3000 1600
2
2443 2445
Output
YES
YES
YES
YES
NO
Note
In the first test case, you can remove all stones without using a superability: [1, 2, 1] → [1, 1, 0] → [0, 0, 0].
In the second test case, you can apply superability to the second and the third piles and then act like in the first testcase.
In the third test case, you can apply superability to the fourth and the fifth piles, thus getting a = [2, 2, 2, 3, 1].
In the fourth test case, you can apply superability to the first and the second piles, thus getting a = [1900, 2100, 1600, 3000, 1600].
Submitted Solution:
```
test_count = int(input(""))
def check_sequence(piles_list, piles_in_test):
ability_used = False
r = 0
d = 0
possible = True
for i in range(piles_in_test):
x = piles_list[i]
# This does backwards swap
if ability_used == False:
if r > x:
ability_used = True
c = r + d
r = x - d
x = c
if r < 0:
possible = False
break
else:
x1 = 0 # get value if there is one
if i + 1 < piles_in_test:
x1 = piles_list[i+1]
x2 = 0 # get value if there is one
if i + 2 < piles_in_test:
x1 = piles_list[i+2]
x0 = r + d
r0 = x - d
if x - r + x2 < x1 and x0 - r0 + x2 >= x1:
ability_used = True
r = r0
x = x0
d = r
r = x - r
if r < 0:
possible = False
break
if r > 0:
possible = False
return possible
answer = "";
for test in range(0,test_count):
piles_in_test = int(input(""))
piles_list = list(map(int ,input("").split(" ")))
# check case of failure
if test == 147:
print(piles_list)
if piles_in_test < 2:
answer += "NO\n"
elif piles_in_test == 2:
if piles_list[0] != piles_list[1]:
answer += "NO\n"
else:
answer += "YES\n"
else:
possible = check_sequence(piles_list, piles_in_test)
#reverse check
if possible == False :
piles_list.reverse()
possible = check_sequence(piles_list, piles_in_test)
if possible == True:
answer += "YES\n"
else:
answer += "NO\n"
print(answer)
``` | instruction | 0 | 14,397 | 8 | 28,794 |
No | output | 1 | 14,397 | 8 | 28,795 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During cleaning the coast, Alice found n piles of stones. The i-th pile has a_i stones.
Piles i and i + 1 are neighbouring for all 1 ≤ i ≤ n - 1. If pile i becomes empty, piles i - 1 and i + 1 doesn't become neighbouring.
Alice is too lazy to remove these stones, so she asked you to take this duty. She allowed you to do only the following operation:
* Select two neighboring piles and, if both of them are not empty, remove one stone from each of them.
Alice understands that sometimes it's impossible to remove all stones with the given operation, so she allowed you to use the following superability:
* Before the start of cleaning, you can select two neighboring piles and swap them.
Determine, if it is possible to remove all stones using the superability not more than once.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of piles.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the number of stones in each pile.
It is guaranteed that the total sum of n over all test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print YES or NO — is it possible to remove all stones using the superability not more than once or not.
Example
Input
5
3
1 2 1
3
1 1 2
5
2 2 2 1 3
5
2100 1900 1600 3000 1600
2
2443 2445
Output
YES
YES
YES
YES
NO
Note
In the first test case, you can remove all stones without using a superability: [1, 2, 1] → [1, 1, 0] → [0, 0, 0].
In the second test case, you can apply superability to the second and the third piles and then act like in the first testcase.
In the third test case, you can apply superability to the fourth and the fifth piles, thus getting a = [2, 2, 2, 3, 1].
In the fourth test case, you can apply superability to the first and the second piles, thus getting a = [1900, 2100, 1600, 3000, 1600].
Submitted Solution:
```
from sys import stdin
def check(a, n):
for i in range(1, n):
if a[i] >= a[i - 1]:
a[i] -= a[i - 1]
a[i - 1] = 0
else: return i
if a[n - 1] != 0: return n - 1
return 0
for _ in range(int(input())):
n = int(input())
a = list(map(int, stdin.readline().strip().split()))
a1 = a.copy()
a2 = a.copy()
c = check(a, n)
if c:
if c == n - 1:
a1[c], a1[c - 1] = a1[c - 1], a1[c]
c1 = check(a1, n)
if c1: print("NO")
else: print("YES")
else:
a1[c], a1[c - 1] = a1[c - 1], a1[c]
a2[c], a2[c + 1] = a2[c + 1], a2[c]
c1 = check(a1, n)
c2 = check(a2, n)
if c1 and c2: print("NO")
else: print("YES")
else: print("YES")
``` | instruction | 0 | 14,398 | 8 | 28,796 |
No | output | 1 | 14,398 | 8 | 28,797 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During cleaning the coast, Alice found n piles of stones. The i-th pile has a_i stones.
Piles i and i + 1 are neighbouring for all 1 ≤ i ≤ n - 1. If pile i becomes empty, piles i - 1 and i + 1 doesn't become neighbouring.
Alice is too lazy to remove these stones, so she asked you to take this duty. She allowed you to do only the following operation:
* Select two neighboring piles and, if both of them are not empty, remove one stone from each of them.
Alice understands that sometimes it's impossible to remove all stones with the given operation, so she allowed you to use the following superability:
* Before the start of cleaning, you can select two neighboring piles and swap them.
Determine, if it is possible to remove all stones using the superability not more than once.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of piles.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the number of stones in each pile.
It is guaranteed that the total sum of n over all test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print YES or NO — is it possible to remove all stones using the superability not more than once or not.
Example
Input
5
3
1 2 1
3
1 1 2
5
2 2 2 1 3
5
2100 1900 1600 3000 1600
2
2443 2445
Output
YES
YES
YES
YES
NO
Note
In the first test case, you can remove all stones without using a superability: [1, 2, 1] → [1, 1, 0] → [0, 0, 0].
In the second test case, you can apply superability to the second and the third piles and then act like in the first testcase.
In the third test case, you can apply superability to the fourth and the fifth piles, thus getting a = [2, 2, 2, 3, 1].
In the fourth test case, you can apply superability to the first and the second piles, thus getting a = [1900, 2100, 1600, 3000, 1600].
Submitted Solution:
```
import io
import os
from collections import Counter, defaultdict, deque
from math import inf
# https://raw.githubusercontent.com/cheran-senthil/PyRival/master/pyrival/data_structures/SegmentTree.py
class SegmentTree:
def __init__(self, data, default=0, func=max):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size : _size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
"""func of data[start, stop)"""
start += self._size
stop += self._size
res_left = res_right = self._default
while start < stop:
if start & 1:
res_left = self._func(res_left, self.data[start])
start += 1
if stop & 1:
stop -= 1
res_right = self._func(self.data[stop], res_right)
start >>= 1
stop >>= 1
return self._func(res_left, res_right)
def __repr__(self):
return "SegmentTree({0})".format(self.data)
class Node:
def __init__(self, minEven, maxEven, minOdd, maxOdd, total, count):
self.minEven = minEven
self.maxEven = maxEven
self.minOdd = minOdd
self.maxOdd = maxOdd
self.total = total
self.count = count
def __repr__(self):
return str(
(
self.minEven,
self.maxEven,
self.minOdd,
self.maxOdd,
self.total,
self.count,
)
)
segDefault = Node(inf, -inf, inf, -inf, 0, 0)
def segMap(x, i):
if i % 2 == 1:
x *= -1
return Node(x, x, inf, -inf, x, 1)
def segCombine(l, r):
if l.count == 0:
return r
if r.count == 0:
return l
if l.count % 2 == 0:
minEven = min(l.minEven, l.total + r.minEven)
maxEven = max(l.maxEven, l.total + r.maxEven)
minOdd = min(l.minOdd, l.total + r.minOdd)
maxOdd = max(l.maxOdd, l.total + r.maxOdd)
else:
minEven = min(l.minEven, l.total + r.minOdd)
maxEven = max(l.maxEven, l.total + r.maxOdd)
minOdd = min(l.minOdd, l.total + r.minEven)
maxOdd = max(l.maxOdd, l.total + r.maxEven)
total = l.total + r.total
count = l.count + r.count
return Node(minEven, maxEven, minOdd, maxOdd, total, count)
def solve(N, A):
def makeOps(A):
needed = []
extra = 0
for x in A:
extra = x - extra
needed.append(extra)
return needed
needed = makeOps(A)
if needed[-1] == 0 and min(needed) >= 0:
return "YES"
segTree = SegmentTree(
[segMap(x, i) for i, x in enumerate(A)], segDefault, segCombine,
)
for i in range(N - 1):
if A[i] == A[i + 1]:
continue
segTree[i] = segMap(A[i + 1], i)
segTree[i + 1] = segMap(A[i], i + 1)
res = segTree.query(0, N)
if False:
A[i], A[i + 1] = A[i + 1], A[i]
pref = [0]
for j, x in enumerate(A):
if j % 2 == 0:
pref.append(pref[-1] + x)
else:
pref.append(pref[-1] - x)
for j in range(N + 1):
check = segmentTree.query(0, j)
evens = pref[1 : j + 1 : 2]
odds = pref[2 : j + 1 : 2]
assert check.total == pref[j]
assert check.minEven == min(evens, default=inf)
assert check.maxEven == max(evens, default=-inf)
assert check.minOdd == min(odds, default=inf)
assert check.maxOdd == max(odds, default=-inf)
A[i], A[i + 1] = A[i + 1], A[i]
if (N - 1) % 2 == 0:
if res.total == 0 and res.minEven >= 0 and -res.maxOdd >= 0:
return "YES"
else:
if res.total == 0 and res.minOdd >= 0 and -res.maxEven >= 0:
return "YES"
segTree[i] = segMap(A[i], i)
segTree[i + 1] = segMap(A[i + 1], i + 1)
return "NO"
if __name__ == "__main__":
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
TC = int(input())
for tc in range(1, TC + 1):
(N,) = [int(x) for x in input().split()]
A = [int(x) for x in input().split()]
ans = solve(N, A)
print(ans)
``` | instruction | 0 | 14,399 | 8 | 28,798 |
No | output | 1 | 14,399 | 8 | 28,799 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During cleaning the coast, Alice found n piles of stones. The i-th pile has a_i stones.
Piles i and i + 1 are neighbouring for all 1 ≤ i ≤ n - 1. If pile i becomes empty, piles i - 1 and i + 1 doesn't become neighbouring.
Alice is too lazy to remove these stones, so she asked you to take this duty. She allowed you to do only the following operation:
* Select two neighboring piles and, if both of them are not empty, remove one stone from each of them.
Alice understands that sometimes it's impossible to remove all stones with the given operation, so she allowed you to use the following superability:
* Before the start of cleaning, you can select two neighboring piles and swap them.
Determine, if it is possible to remove all stones using the superability not more than once.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of piles.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the number of stones in each pile.
It is guaranteed that the total sum of n over all test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print YES or NO — is it possible to remove all stones using the superability not more than once or not.
Example
Input
5
3
1 2 1
3
1 1 2
5
2 2 2 1 3
5
2100 1900 1600 3000 1600
2
2443 2445
Output
YES
YES
YES
YES
NO
Note
In the first test case, you can remove all stones without using a superability: [1, 2, 1] → [1, 1, 0] → [0, 0, 0].
In the second test case, you can apply superability to the second and the third piles and then act like in the first testcase.
In the third test case, you can apply superability to the fourth and the fifth piles, thus getting a = [2, 2, 2, 3, 1].
In the fourth test case, you can apply superability to the first and the second piles, thus getting a = [1900, 2100, 1600, 3000, 1600].
Submitted Solution:
```
def ke(x: list):
y = x.copy()
y.append(0)
for i in range(0, len(y) - 1):
if y[i] > y[i + 1]:
return False
else:
y[i + 1] -= y[i]
y[i] = 0
return True
t = int(input())
for cas in range(0, t):
n = int(input())
v = list(map(int, input().split()))
if ke(v):
print("YES")
continue
z = v.copy()
z[0], z[1] = z[1], z[0]
if ke(z):
print("YES")
continue
z = v.copy()
z[n - 2], z[n - 1] = z[n - 1], z[n - 2]
if ke(z):
print("YES")
continue
zuo = [0] * n
you = [0] * n
z = v.copy()
for i in range(0, n - 1):
if z[i] > z[i + 1]:
for j in range(i + 1, n):
zuo[j] = -1
break
else:
z[i + 1] -= z[i]
zuo[i + 1] = z[i + 1]
z[i] = 0
z = v.copy()
for i in range(n - 1, 0, -1):
if z[i] > z[i - 1]:
for j in range(i - 1, -1, -1):
you[j] = -1
break
else:
z[i - 1] -= z[i]
you[i - 1] = z[i - 1]
z[i] = 0
z = v.copy()
for i in range(1, n - 2):
if zuo[i - 1] != -1 and you[i + 2] != -1 and ke([zuo[i - 1], z[i + 1], z[i], you[i + 2]]):
print("YES")
continue
print("NO")
``` | instruction | 0 | 14,400 | 8 | 28,800 |
No | output | 1 | 14,400 | 8 | 28,801 |
Provide a correct Python 3 solution for this coding contest problem.
Our master carpenter is designing a condominium called Bange Hills Mansion. The condominium is constructed by stacking up floors of the same height. The height of each floor is designed so that the total height of the stacked floors coincides with the predetermined height of the condominium. The height of each floor can be adjusted freely with a certain range.
The final outcome of the building depends on clever height allotment for each floor. So, he plans to calculate possible combinations of per-floor heights to check how many options he has.
Given the height of the condominium and the adjustable range of each floor’s height, make a program to enumerate the number of choices for a floor.
Input
The input is given in the following format.
$H$ $A$ $B$
The input line provides the height of the condominium $H$ ($1 \leq H \leq 10^5$) and the upper and lower limits $A$ and $B$ of the height adjustable range for one floor ($1 \leq A \leq B \leq H$). All data are given as integers.
Output
Output the number of possible height selections for each floor in a line.
Examples
Input
100 2 4
Output
2
Input
101 3 5
Output
0 | instruction | 0 | 14,956 | 8 | 29,912 |
"Correct Solution:
```
H,A,B = map(int,input().split())
a = list(range(A,B+1))
x = 0
for i in a:
if H % i == 0:
x += 1
print(x)
``` | output | 1 | 14,956 | 8 | 29,913 |
Provide a correct Python 3 solution for this coding contest problem.
Our master carpenter is designing a condominium called Bange Hills Mansion. The condominium is constructed by stacking up floors of the same height. The height of each floor is designed so that the total height of the stacked floors coincides with the predetermined height of the condominium. The height of each floor can be adjusted freely with a certain range.
The final outcome of the building depends on clever height allotment for each floor. So, he plans to calculate possible combinations of per-floor heights to check how many options he has.
Given the height of the condominium and the adjustable range of each floor’s height, make a program to enumerate the number of choices for a floor.
Input
The input is given in the following format.
$H$ $A$ $B$
The input line provides the height of the condominium $H$ ($1 \leq H \leq 10^5$) and the upper and lower limits $A$ and $B$ of the height adjustable range for one floor ($1 \leq A \leq B \leq H$). All data are given as integers.
Output
Output the number of possible height selections for each floor in a line.
Examples
Input
100 2 4
Output
2
Input
101 3 5
Output
0 | instruction | 0 | 14,957 | 8 | 29,914 |
"Correct Solution:
```
H,A,B=map(int,input().split())
i=0
while A<=B:
if H%A==0:
i=i+1
A=A+1
print(i)
``` | output | 1 | 14,957 | 8 | 29,915 |
Provide a correct Python 3 solution for this coding contest problem.
Our master carpenter is designing a condominium called Bange Hills Mansion. The condominium is constructed by stacking up floors of the same height. The height of each floor is designed so that the total height of the stacked floors coincides with the predetermined height of the condominium. The height of each floor can be adjusted freely with a certain range.
The final outcome of the building depends on clever height allotment for each floor. So, he plans to calculate possible combinations of per-floor heights to check how many options he has.
Given the height of the condominium and the adjustable range of each floor’s height, make a program to enumerate the number of choices for a floor.
Input
The input is given in the following format.
$H$ $A$ $B$
The input line provides the height of the condominium $H$ ($1 \leq H \leq 10^5$) and the upper and lower limits $A$ and $B$ of the height adjustable range for one floor ($1 \leq A \leq B \leq H$). All data are given as integers.
Output
Output the number of possible height selections for each floor in a line.
Examples
Input
100 2 4
Output
2
Input
101 3 5
Output
0 | instruction | 0 | 14,958 | 8 | 29,916 |
"Correct Solution:
```
h,a,b = map(int,input().split())
print(len([i for i in range(a,b+1) if h%i==0]))
``` | output | 1 | 14,958 | 8 | 29,917 |
Provide a correct Python 3 solution for this coding contest problem.
Our master carpenter is designing a condominium called Bange Hills Mansion. The condominium is constructed by stacking up floors of the same height. The height of each floor is designed so that the total height of the stacked floors coincides with the predetermined height of the condominium. The height of each floor can be adjusted freely with a certain range.
The final outcome of the building depends on clever height allotment for each floor. So, he plans to calculate possible combinations of per-floor heights to check how many options he has.
Given the height of the condominium and the adjustable range of each floor’s height, make a program to enumerate the number of choices for a floor.
Input
The input is given in the following format.
$H$ $A$ $B$
The input line provides the height of the condominium $H$ ($1 \leq H \leq 10^5$) and the upper and lower limits $A$ and $B$ of the height adjustable range for one floor ($1 \leq A \leq B \leq H$). All data are given as integers.
Output
Output the number of possible height selections for each floor in a line.
Examples
Input
100 2 4
Output
2
Input
101 3 5
Output
0 | instruction | 0 | 14,959 | 8 | 29,918 |
"Correct Solution:
```
#標準入力とカウンタ変数の初期化
a,b,c = map(int,input().split())
kei = 0
#割り切れるならカウンタ変数を更新する
for i in range(b,c + 1):
if a % i == 0:kei += 1
#カウンタ変数を出力する
print(kei)
``` | output | 1 | 14,959 | 8 | 29,919 |
Provide a correct Python 3 solution for this coding contest problem.
Our master carpenter is designing a condominium called Bange Hills Mansion. The condominium is constructed by stacking up floors of the same height. The height of each floor is designed so that the total height of the stacked floors coincides with the predetermined height of the condominium. The height of each floor can be adjusted freely with a certain range.
The final outcome of the building depends on clever height allotment for each floor. So, he plans to calculate possible combinations of per-floor heights to check how many options he has.
Given the height of the condominium and the adjustable range of each floor’s height, make a program to enumerate the number of choices for a floor.
Input
The input is given in the following format.
$H$ $A$ $B$
The input line provides the height of the condominium $H$ ($1 \leq H \leq 10^5$) and the upper and lower limits $A$ and $B$ of the height adjustable range for one floor ($1 \leq A \leq B \leq H$). All data are given as integers.
Output
Output the number of possible height selections for each floor in a line.
Examples
Input
100 2 4
Output
2
Input
101 3 5
Output
0 | instruction | 0 | 14,960 | 8 | 29,920 |
"Correct Solution:
```
h, a, b = map(int, input().split())
print([(h%(a+c)==0) for c in range(b-a+1)].count(True))
``` | output | 1 | 14,960 | 8 | 29,921 |
Provide a correct Python 3 solution for this coding contest problem.
Our master carpenter is designing a condominium called Bange Hills Mansion. The condominium is constructed by stacking up floors of the same height. The height of each floor is designed so that the total height of the stacked floors coincides with the predetermined height of the condominium. The height of each floor can be adjusted freely with a certain range.
The final outcome of the building depends on clever height allotment for each floor. So, he plans to calculate possible combinations of per-floor heights to check how many options he has.
Given the height of the condominium and the adjustable range of each floor’s height, make a program to enumerate the number of choices for a floor.
Input
The input is given in the following format.
$H$ $A$ $B$
The input line provides the height of the condominium $H$ ($1 \leq H \leq 10^5$) and the upper and lower limits $A$ and $B$ of the height adjustable range for one floor ($1 \leq A \leq B \leq H$). All data are given as integers.
Output
Output the number of possible height selections for each floor in a line.
Examples
Input
100 2 4
Output
2
Input
101 3 5
Output
0 | instruction | 0 | 14,961 | 8 | 29,922 |
"Correct Solution:
```
H,a,b=map(int,input().split())
c=0
for i in range(a,b+1):
if H%i==0 :
c+=1
print(c)
``` | output | 1 | 14,961 | 8 | 29,923 |
Provide a correct Python 3 solution for this coding contest problem.
Our master carpenter is designing a condominium called Bange Hills Mansion. The condominium is constructed by stacking up floors of the same height. The height of each floor is designed so that the total height of the stacked floors coincides with the predetermined height of the condominium. The height of each floor can be adjusted freely with a certain range.
The final outcome of the building depends on clever height allotment for each floor. So, he plans to calculate possible combinations of per-floor heights to check how many options he has.
Given the height of the condominium and the adjustable range of each floor’s height, make a program to enumerate the number of choices for a floor.
Input
The input is given in the following format.
$H$ $A$ $B$
The input line provides the height of the condominium $H$ ($1 \leq H \leq 10^5$) and the upper and lower limits $A$ and $B$ of the height adjustable range for one floor ($1 \leq A \leq B \leq H$). All data are given as integers.
Output
Output the number of possible height selections for each floor in a line.
Examples
Input
100 2 4
Output
2
Input
101 3 5
Output
0 | instruction | 0 | 14,962 | 8 | 29,924 |
"Correct Solution:
```
H,A,B = [int(i) for i in input().split()]
cnt = 0
for i in range(A,B+1):
if (H // i)*i == H:
cnt = cnt + 1
print(cnt)
``` | output | 1 | 14,962 | 8 | 29,925 |
Provide a correct Python 3 solution for this coding contest problem.
Our master carpenter is designing a condominium called Bange Hills Mansion. The condominium is constructed by stacking up floors of the same height. The height of each floor is designed so that the total height of the stacked floors coincides with the predetermined height of the condominium. The height of each floor can be adjusted freely with a certain range.
The final outcome of the building depends on clever height allotment for each floor. So, he plans to calculate possible combinations of per-floor heights to check how many options he has.
Given the height of the condominium and the adjustable range of each floor’s height, make a program to enumerate the number of choices for a floor.
Input
The input is given in the following format.
$H$ $A$ $B$
The input line provides the height of the condominium $H$ ($1 \leq H \leq 10^5$) and the upper and lower limits $A$ and $B$ of the height adjustable range for one floor ($1 \leq A \leq B \leq H$). All data are given as integers.
Output
Output the number of possible height selections for each floor in a line.
Examples
Input
100 2 4
Output
2
Input
101 3 5
Output
0 | instruction | 0 | 14,963 | 8 | 29,926 |
"Correct Solution:
```
H, A, B = map(int, input().split())
ans = 0
for i in range(A, B+1) :
if H % i == 0 :
ans += 1
print(ans)
``` | output | 1 | 14,963 | 8 | 29,927 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Our master carpenter is designing a condominium called Bange Hills Mansion. The condominium is constructed by stacking up floors of the same height. The height of each floor is designed so that the total height of the stacked floors coincides with the predetermined height of the condominium. The height of each floor can be adjusted freely with a certain range.
The final outcome of the building depends on clever height allotment for each floor. So, he plans to calculate possible combinations of per-floor heights to check how many options he has.
Given the height of the condominium and the adjustable range of each floor’s height, make a program to enumerate the number of choices for a floor.
Input
The input is given in the following format.
$H$ $A$ $B$
The input line provides the height of the condominium $H$ ($1 \leq H \leq 10^5$) and the upper and lower limits $A$ and $B$ of the height adjustable range for one floor ($1 \leq A \leq B \leq H$). All data are given as integers.
Output
Output the number of possible height selections for each floor in a line.
Examples
Input
100 2 4
Output
2
Input
101 3 5
Output
0
Submitted Solution:
```
H, A, B = map(int, input().split())
cnt = 0;
for i in range(A, B+1):
cnt += (H % i == 0 )
print(cnt)
``` | instruction | 0 | 14,964 | 8 | 29,928 |
Yes | output | 1 | 14,964 | 8 | 29,929 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Our master carpenter is designing a condominium called Bange Hills Mansion. The condominium is constructed by stacking up floors of the same height. The height of each floor is designed so that the total height of the stacked floors coincides with the predetermined height of the condominium. The height of each floor can be adjusted freely with a certain range.
The final outcome of the building depends on clever height allotment for each floor. So, he plans to calculate possible combinations of per-floor heights to check how many options he has.
Given the height of the condominium and the adjustable range of each floor’s height, make a program to enumerate the number of choices for a floor.
Input
The input is given in the following format.
$H$ $A$ $B$
The input line provides the height of the condominium $H$ ($1 \leq H \leq 10^5$) and the upper and lower limits $A$ and $B$ of the height adjustable range for one floor ($1 \leq A \leq B \leq H$). All data are given as integers.
Output
Output the number of possible height selections for each floor in a line.
Examples
Input
100 2 4
Output
2
Input
101 3 5
Output
0
Submitted Solution:
```
h,a,b=map(int,input().split())
c=0
for i in range(a,b+1):
if h%i==0:
c=c+1
print(c)
``` | instruction | 0 | 14,965 | 8 | 29,930 |
Yes | output | 1 | 14,965 | 8 | 29,931 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Our master carpenter is designing a condominium called Bange Hills Mansion. The condominium is constructed by stacking up floors of the same height. The height of each floor is designed so that the total height of the stacked floors coincides with the predetermined height of the condominium. The height of each floor can be adjusted freely with a certain range.
The final outcome of the building depends on clever height allotment for each floor. So, he plans to calculate possible combinations of per-floor heights to check how many options he has.
Given the height of the condominium and the adjustable range of each floor’s height, make a program to enumerate the number of choices for a floor.
Input
The input is given in the following format.
$H$ $A$ $B$
The input line provides the height of the condominium $H$ ($1 \leq H \leq 10^5$) and the upper and lower limits $A$ and $B$ of the height adjustable range for one floor ($1 \leq A \leq B \leq H$). All data are given as integers.
Output
Output the number of possible height selections for each floor in a line.
Examples
Input
100 2 4
Output
2
Input
101 3 5
Output
0
Submitted Solution:
```
h, a, b = map(int, input().split())
cnt = 0
for i in range(a, b+1):
if h % i == 0:
cnt += 1
print(cnt)
``` | instruction | 0 | 14,966 | 8 | 29,932 |
Yes | output | 1 | 14,966 | 8 | 29,933 |
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