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.
You are asked to watch your nephew who likes to play with toy blocks in a strange way.
He has n boxes and the i-th box has a_i blocks. His game consists of two steps:
1. he chooses an arbitrary box i;
2. he tries to move all blocks from the i-th box to other boxes.
If he can make the same number of blocks in each of n - 1 other boxes then he will be happy, otherwise, will be sad. Note that your nephew can only move the blocks from the chosen box to the other boxes; he cannot move blocks from the other boxes.
You don't want to make your nephew sad, so you decided to put several extra blocks into some boxes in such a way that no matter which box i he chooses he won't be sad. What is the minimum number of extra blocks you need to put?
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains the integer n (2 β€ n β€ 10^5) β the number of boxes.
The second line of each test case contains n integers a_1, a_2, ..., a_n (0 β€ a_i β€ 10^9) β the number of blocks in each box.
It's guaranteed that the sum of n over test cases doesn't exceed 10^5.
Output
For each test case, print a single integer β the minimum number of blocks you need to put. It can be proved that the answer always exists, i. e. the number of blocks is finite.
Example
Input
3
3
3 2 2
4
2 2 3 2
3
0 3 0
Output
1
0
3
Note
In the first test case, you can, for example, put one extra block into the first box and make a = [4, 2, 2]. If your nephew chooses the box with 4 blocks, then we will move two blocks to the second box and two blocks to the third box. If he chooses the box with 2 blocks then he will move these two blocks to the other box with 2 blocks.
In the second test case, you don't need to put any extra blocks, since no matter which box your nephew chooses, he can always make other boxes equal.
In the third test case, you should put 3 extra blocks. For example, you can put 2 blocks in the first box and 1 block in the third box. You'll get array a = [2, 3, 1].
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Sun Nov 22 16:15:39 2020
@author: η« ζ―ε²
"""
for i in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
c=max(a)
b=sum(a)
if c*(n-1)>=b:
print(c*(n-1)-b)
else:
print((b%(n-1)>0)*(n-1-b%(n-1)))
``` | instruction | 0 | 103,898 | 8 | 207,796 |
Yes | output | 1 | 103,898 | 8 | 207,797 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are asked to watch your nephew who likes to play with toy blocks in a strange way.
He has n boxes and the i-th box has a_i blocks. His game consists of two steps:
1. he chooses an arbitrary box i;
2. he tries to move all blocks from the i-th box to other boxes.
If he can make the same number of blocks in each of n - 1 other boxes then he will be happy, otherwise, will be sad. Note that your nephew can only move the blocks from the chosen box to the other boxes; he cannot move blocks from the other boxes.
You don't want to make your nephew sad, so you decided to put several extra blocks into some boxes in such a way that no matter which box i he chooses he won't be sad. What is the minimum number of extra blocks you need to put?
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains the integer n (2 β€ n β€ 10^5) β the number of boxes.
The second line of each test case contains n integers a_1, a_2, ..., a_n (0 β€ a_i β€ 10^9) β the number of blocks in each box.
It's guaranteed that the sum of n over test cases doesn't exceed 10^5.
Output
For each test case, print a single integer β the minimum number of blocks you need to put. It can be proved that the answer always exists, i. e. the number of blocks is finite.
Example
Input
3
3
3 2 2
4
2 2 3 2
3
0 3 0
Output
1
0
3
Note
In the first test case, you can, for example, put one extra block into the first box and make a = [4, 2, 2]. If your nephew chooses the box with 4 blocks, then we will move two blocks to the second box and two blocks to the third box. If he chooses the box with 2 blocks then he will move these two blocks to the other box with 2 blocks.
In the second test case, you don't need to put any extra blocks, since no matter which box your nephew chooses, he can always make other boxes equal.
In the third test case, you should put 3 extra blocks. For example, you can put 2 blocks in the first box and 1 block in the third box. You'll get array a = [2, 3, 1].
Submitted Solution:
```
"""
usefull snippets:
- map(int, input().split())
- map(int, sys.stdin.readline().split()))
- int(input())
- int(sys.stdin.readline().strip())
- sys.stdout.write()
- sys.stdout.write(" ".join(map(str, c) # writes c - collection of ints
"""
# import collections
import sys
import math
from bisect import bisect_left
from collections import defaultdict
# recursion increase
# sys.setrecursionlimit(10000)
# number with big precision
# from decimal import getcontext, Decimal
# getcontext().prec = 34
# longest common prefix
def get_lcp(s, suffix_array):
s = s + "$"
n = len(s)
lcp = [0] * (n)
pos = [0] * (n)
for i in range(n - 1):
pos[suffix_array[i]] = i
k = 0
for i in range(n - 1):
if k > 0:
k -= 1
if pos[i] == n - 1:
lcp[n - 1] = -1
k = 0
continue
else:
j = suffix_array[pos[i] + 1]
while max([i + k, j + k]) < n and s[i + k] == s[j + k]:
k += 1
lcp[pos[i]] = k
return lcp
def get_suffix_array(word):
suffix_array = [("", len(word))]
for position in range(len(word)):
sliced = word[len(word) - position - 1 :]
suffix_array.append((sliced, len(word) - position - 1))
suffix_array.sort(key=lambda x: x[0])
return [item[1] for item in suffix_array]
def get_ints():
return map(int, sys.stdin.readline().strip().split())
def bin_search(collection, element):
i = bisect_left(collection, element)
if i != len(collection) and collection[i] == element:
return i
else:
return -1
def main():
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, sys.stdin.readline().split()))
if n == 2:
print(0)
continue
aim = max(a)
m = min(a)
s = sum(a)
ans = aim * (n - 2) - (s - aim - m) - m
if ans < 0:
if aim <= (n - 2):
ans = 1 + (n - 2) * n - s
else:
ans = s % (n - 1)
print(ans)
if __name__ == "__main__":
main()
``` | instruction | 0 | 103,899 | 8 | 207,798 |
No | output | 1 | 103,899 | 8 | 207,799 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are asked to watch your nephew who likes to play with toy blocks in a strange way.
He has n boxes and the i-th box has a_i blocks. His game consists of two steps:
1. he chooses an arbitrary box i;
2. he tries to move all blocks from the i-th box to other boxes.
If he can make the same number of blocks in each of n - 1 other boxes then he will be happy, otherwise, will be sad. Note that your nephew can only move the blocks from the chosen box to the other boxes; he cannot move blocks from the other boxes.
You don't want to make your nephew sad, so you decided to put several extra blocks into some boxes in such a way that no matter which box i he chooses he won't be sad. What is the minimum number of extra blocks you need to put?
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains the integer n (2 β€ n β€ 10^5) β the number of boxes.
The second line of each test case contains n integers a_1, a_2, ..., a_n (0 β€ a_i β€ 10^9) β the number of blocks in each box.
It's guaranteed that the sum of n over test cases doesn't exceed 10^5.
Output
For each test case, print a single integer β the minimum number of blocks you need to put. It can be proved that the answer always exists, i. e. the number of blocks is finite.
Example
Input
3
3
3 2 2
4
2 2 3 2
3
0 3 0
Output
1
0
3
Note
In the first test case, you can, for example, put one extra block into the first box and make a = [4, 2, 2]. If your nephew chooses the box with 4 blocks, then we will move two blocks to the second box and two blocks to the third box. If he chooses the box with 2 blocks then he will move these two blocks to the other box with 2 blocks.
In the second test case, you don't need to put any extra blocks, since no matter which box your nephew chooses, he can always make other boxes equal.
In the third test case, you should put 3 extra blocks. For example, you can put 2 blocks in the first box and 1 block in the third box. You'll get array a = [2, 3, 1].
Submitted Solution:
```
for _ in range(int(input())):
n= int(input())
li = input().split()
li=[int(i) for i in li]
li.sort()
s=0
if li[0]==li[-1]:
print(0)
else:
for i in range(0,len(li)-1):
s+=li[i]
if s-li[-1]==li[-1]:
print(0)
else:
print(abs(s-li[-1]))
``` | instruction | 0 | 103,900 | 8 | 207,800 |
No | output | 1 | 103,900 | 8 | 207,801 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are asked to watch your nephew who likes to play with toy blocks in a strange way.
He has n boxes and the i-th box has a_i blocks. His game consists of two steps:
1. he chooses an arbitrary box i;
2. he tries to move all blocks from the i-th box to other boxes.
If he can make the same number of blocks in each of n - 1 other boxes then he will be happy, otherwise, will be sad. Note that your nephew can only move the blocks from the chosen box to the other boxes; he cannot move blocks from the other boxes.
You don't want to make your nephew sad, so you decided to put several extra blocks into some boxes in such a way that no matter which box i he chooses he won't be sad. What is the minimum number of extra blocks you need to put?
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains the integer n (2 β€ n β€ 10^5) β the number of boxes.
The second line of each test case contains n integers a_1, a_2, ..., a_n (0 β€ a_i β€ 10^9) β the number of blocks in each box.
It's guaranteed that the sum of n over test cases doesn't exceed 10^5.
Output
For each test case, print a single integer β the minimum number of blocks you need to put. It can be proved that the answer always exists, i. e. the number of blocks is finite.
Example
Input
3
3
3 2 2
4
2 2 3 2
3
0 3 0
Output
1
0
3
Note
In the first test case, you can, for example, put one extra block into the first box and make a = [4, 2, 2]. If your nephew chooses the box with 4 blocks, then we will move two blocks to the second box and two blocks to the third box. If he chooses the box with 2 blocks then he will move these two blocks to the other box with 2 blocks.
In the second test case, you don't need to put any extra blocks, since no matter which box your nephew chooses, he can always make other boxes equal.
In the third test case, you should put 3 extra blocks. For example, you can put 2 blocks in the first box and 1 block in the third box. You'll get array a = [2, 3, 1].
Submitted Solution:
```
T= int(input())
for _ in range(T):
n = int(input())
a = list(map(int, input().split()))
if sum(a) % (n - 1) == 0:
print(0)
else:
print(max(max(a) * (n - 1) - sum(a), (n - 1 - sum(a) % (n - 1))))
``` | instruction | 0 | 103,901 | 8 | 207,802 |
No | output | 1 | 103,901 | 8 | 207,803 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are asked to watch your nephew who likes to play with toy blocks in a strange way.
He has n boxes and the i-th box has a_i blocks. His game consists of two steps:
1. he chooses an arbitrary box i;
2. he tries to move all blocks from the i-th box to other boxes.
If he can make the same number of blocks in each of n - 1 other boxes then he will be happy, otherwise, will be sad. Note that your nephew can only move the blocks from the chosen box to the other boxes; he cannot move blocks from the other boxes.
You don't want to make your nephew sad, so you decided to put several extra blocks into some boxes in such a way that no matter which box i he chooses he won't be sad. What is the minimum number of extra blocks you need to put?
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains the integer n (2 β€ n β€ 10^5) β the number of boxes.
The second line of each test case contains n integers a_1, a_2, ..., a_n (0 β€ a_i β€ 10^9) β the number of blocks in each box.
It's guaranteed that the sum of n over test cases doesn't exceed 10^5.
Output
For each test case, print a single integer β the minimum number of blocks you need to put. It can be proved that the answer always exists, i. e. the number of blocks is finite.
Example
Input
3
3
3 2 2
4
2 2 3 2
3
0 3 0
Output
1
0
3
Note
In the first test case, you can, for example, put one extra block into the first box and make a = [4, 2, 2]. If your nephew chooses the box with 4 blocks, then we will move two blocks to the second box and two blocks to the third box. If he chooses the box with 2 blocks then he will move these two blocks to the other box with 2 blocks.
In the second test case, you don't need to put any extra blocks, since no matter which box your nephew chooses, he can always make other boxes equal.
In the third test case, you should put 3 extra blocks. For example, you can put 2 blocks in the first box and 1 block in the third box. You'll get array a = [2, 3, 1].
Submitted Solution:
```
t = int(input())
l = []
for i in range(t):
a = int(input())
b = list(map(int, input().split()))
min1_ = min(b)
index = b.index(min1_)
del b[index]
min2_ = min(b)
max_ = max(b)
ost = (sum(b) + min1_) % (a - 1)
if min2_ + ost + min1_ < max_:
min2_ += ost + min1_
rasn = max_ - min2_
pl = (rasn + (a - 1) - 1) // (a - 1)
l.append(ost + pl * (a - 1))
else:
l.append(ost)
for i in l:
print(i)
``` | instruction | 0 | 103,902 | 8 | 207,804 |
No | output | 1 | 103,902 | 8 | 207,805 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Enough is enough. Too many times it happened that Vasya forgot to dispose of garbage and his apartment stank afterwards. Now he wants to create a garbage disposal plan and stick to it.
For each of next n days Vasya knows a_i β number of units of garbage he will produce on the i-th day. Each unit of garbage must be disposed of either on the day it was produced or on the next day. Vasya disposes of garbage by putting it inside a bag and dropping the bag into a garbage container. Each bag can contain up to k units of garbage. It is allowed to compose and drop multiple bags into a garbage container in a single day.
Being economical, Vasya wants to use as few bags as possible. You are to compute the minimum number of bags Vasya needs to dispose of all of his garbage for the given n days. No garbage should be left after the n-th day.
Input
The first line of the input contains two integers n and k (1 β€ n β€ 2β
10^5, 1 β€ k β€ 10^9) β number of days to consider and bag's capacity. The second line contains n space separated integers a_i (0 β€ a_i β€ 10^9) β the number of units of garbage produced on the i-th day.
Output
Output a single integer β the minimum number of bags Vasya needs to dispose of all garbage. Each unit of garbage should be disposed on the day it was produced or on the next day. No garbage can be left after the n-th day. In a day it is allowed to compose and drop multiple bags.
Examples
Input
3 2
3 2 1
Output
3
Input
5 1
1000000000 1000000000 1000000000 1000000000 1000000000
Output
5000000000
Input
3 2
1 0 1
Output
2
Input
4 4
2 8 4 1
Output
4 | instruction | 0 | 104,438 | 8 | 208,876 |
Tags: greedy
Correct Solution:
```
n,k = map(int, input().split())
s = list(map(int,input().split()))
old = 0
current = 0
idx = 0
ans = 0
while idx < len(s):
if(old != 0 and s[idx] // k == 0):
ans+= 1
old = max(s[idx] - (k-old),0)
idx+=1
continue
ans += (s[idx]+old) // k
old = (s[idx]+old) % k
idx+=1
if(old>0):
ans+=1
print(ans)
``` | output | 1 | 104,438 | 8 | 208,877 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Enough is enough. Too many times it happened that Vasya forgot to dispose of garbage and his apartment stank afterwards. Now he wants to create a garbage disposal plan and stick to it.
For each of next n days Vasya knows a_i β number of units of garbage he will produce on the i-th day. Each unit of garbage must be disposed of either on the day it was produced or on the next day. Vasya disposes of garbage by putting it inside a bag and dropping the bag into a garbage container. Each bag can contain up to k units of garbage. It is allowed to compose and drop multiple bags into a garbage container in a single day.
Being economical, Vasya wants to use as few bags as possible. You are to compute the minimum number of bags Vasya needs to dispose of all of his garbage for the given n days. No garbage should be left after the n-th day.
Input
The first line of the input contains two integers n and k (1 β€ n β€ 2β
10^5, 1 β€ k β€ 10^9) β number of days to consider and bag's capacity. The second line contains n space separated integers a_i (0 β€ a_i β€ 10^9) β the number of units of garbage produced on the i-th day.
Output
Output a single integer β the minimum number of bags Vasya needs to dispose of all garbage. Each unit of garbage should be disposed on the day it was produced or on the next day. No garbage can be left after the n-th day. In a day it is allowed to compose and drop multiple bags.
Examples
Input
3 2
3 2 1
Output
3
Input
5 1
1000000000 1000000000 1000000000 1000000000 1000000000
Output
5000000000
Input
3 2
1 0 1
Output
2
Input
4 4
2 8 4 1
Output
4 | instruction | 0 | 104,439 | 8 | 208,878 |
Tags: greedy
Correct Solution:
```
n, m = map(int, input().split())
a = list(map(int, input().split()))
res = 0
sum = 0
can = True
for i in range(n):
if sum > 0:
sum = sum + a[i]
if sum < m:
res = res + 1
sum = 0
else:
res = res + sum // m
sum = sum % m
else:
sum = sum + a[i]
res = res + sum // m
sum = sum % m
if sum > 0:
res = res + 1
print(res)
``` | output | 1 | 104,439 | 8 | 208,879 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Enough is enough. Too many times it happened that Vasya forgot to dispose of garbage and his apartment stank afterwards. Now he wants to create a garbage disposal plan and stick to it.
For each of next n days Vasya knows a_i β number of units of garbage he will produce on the i-th day. Each unit of garbage must be disposed of either on the day it was produced or on the next day. Vasya disposes of garbage by putting it inside a bag and dropping the bag into a garbage container. Each bag can contain up to k units of garbage. It is allowed to compose and drop multiple bags into a garbage container in a single day.
Being economical, Vasya wants to use as few bags as possible. You are to compute the minimum number of bags Vasya needs to dispose of all of his garbage for the given n days. No garbage should be left after the n-th day.
Input
The first line of the input contains two integers n and k (1 β€ n β€ 2β
10^5, 1 β€ k β€ 10^9) β number of days to consider and bag's capacity. The second line contains n space separated integers a_i (0 β€ a_i β€ 10^9) β the number of units of garbage produced on the i-th day.
Output
Output a single integer β the minimum number of bags Vasya needs to dispose of all garbage. Each unit of garbage should be disposed on the day it was produced or on the next day. No garbage can be left after the n-th day. In a day it is allowed to compose and drop multiple bags.
Examples
Input
3 2
3 2 1
Output
3
Input
5 1
1000000000 1000000000 1000000000 1000000000 1000000000
Output
5000000000
Input
3 2
1 0 1
Output
2
Input
4 4
2 8 4 1
Output
4 | instruction | 0 | 104,440 | 8 | 208,880 |
Tags: greedy
Correct Solution:
```
# NTFS: pajenegod
n,k = map(int,input().split())
lis = list(map(int,input().split()))
ans = 0
for i in range(n): # Iterate over all days
ans += lis[i]//k # calculate bags required
rem = lis[i]%k # if still some bags remain
lis[i] = rem
if rem:
ans += 1 # then we need a extra bag for this
if i>=n-1:
break
lis[i+1] = max(0,lis[i+1]-(k-rem)) # since we are taking extra bag we can put some extra amount from next bag
print(ans)
``` | output | 1 | 104,440 | 8 | 208,881 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Enough is enough. Too many times it happened that Vasya forgot to dispose of garbage and his apartment stank afterwards. Now he wants to create a garbage disposal plan and stick to it.
For each of next n days Vasya knows a_i β number of units of garbage he will produce on the i-th day. Each unit of garbage must be disposed of either on the day it was produced or on the next day. Vasya disposes of garbage by putting it inside a bag and dropping the bag into a garbage container. Each bag can contain up to k units of garbage. It is allowed to compose and drop multiple bags into a garbage container in a single day.
Being economical, Vasya wants to use as few bags as possible. You are to compute the minimum number of bags Vasya needs to dispose of all of his garbage for the given n days. No garbage should be left after the n-th day.
Input
The first line of the input contains two integers n and k (1 β€ n β€ 2β
10^5, 1 β€ k β€ 10^9) β number of days to consider and bag's capacity. The second line contains n space separated integers a_i (0 β€ a_i β€ 10^9) β the number of units of garbage produced on the i-th day.
Output
Output a single integer β the minimum number of bags Vasya needs to dispose of all garbage. Each unit of garbage should be disposed on the day it was produced or on the next day. No garbage can be left after the n-th day. In a day it is allowed to compose and drop multiple bags.
Examples
Input
3 2
3 2 1
Output
3
Input
5 1
1000000000 1000000000 1000000000 1000000000 1000000000
Output
5000000000
Input
3 2
1 0 1
Output
2
Input
4 4
2 8 4 1
Output
4 | instruction | 0 | 104,441 | 8 | 208,882 |
Tags: greedy
Correct Solution:
```
n, k = map(int,input().split())
nums = list(map(int,input().split()))
result = 0
remain = 0
for num in nums:
if not (num+remain)//k and remain:
result += 1
remain = 0
continue
result += (num+remain)//k
remain = (num+remain)%k
if remain:
result+=1
print(result)
``` | output | 1 | 104,441 | 8 | 208,883 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Enough is enough. Too many times it happened that Vasya forgot to dispose of garbage and his apartment stank afterwards. Now he wants to create a garbage disposal plan and stick to it.
For each of next n days Vasya knows a_i β number of units of garbage he will produce on the i-th day. Each unit of garbage must be disposed of either on the day it was produced or on the next day. Vasya disposes of garbage by putting it inside a bag and dropping the bag into a garbage container. Each bag can contain up to k units of garbage. It is allowed to compose and drop multiple bags into a garbage container in a single day.
Being economical, Vasya wants to use as few bags as possible. You are to compute the minimum number of bags Vasya needs to dispose of all of his garbage for the given n days. No garbage should be left after the n-th day.
Input
The first line of the input contains two integers n and k (1 β€ n β€ 2β
10^5, 1 β€ k β€ 10^9) β number of days to consider and bag's capacity. The second line contains n space separated integers a_i (0 β€ a_i β€ 10^9) β the number of units of garbage produced on the i-th day.
Output
Output a single integer β the minimum number of bags Vasya needs to dispose of all garbage. Each unit of garbage should be disposed on the day it was produced or on the next day. No garbage can be left after the n-th day. In a day it is allowed to compose and drop multiple bags.
Examples
Input
3 2
3 2 1
Output
3
Input
5 1
1000000000 1000000000 1000000000 1000000000 1000000000
Output
5000000000
Input
3 2
1 0 1
Output
2
Input
4 4
2 8 4 1
Output
4 | instruction | 0 | 104,442 | 8 | 208,884 |
Tags: greedy
Correct Solution:
```
n,k=[int(x) for x in input().split()]
a=[int(x) for x in input().split()]
ans=0
pre=0
for i in range(n):
if pre!=0:
ans+=1
a[i]-=(k-pre)
if a[i]<0:a[i]=0
ans+=a[i]//k
pre=a[i]-a[i]//k*k
if pre!=0 : ans+=1
print(ans)
``` | output | 1 | 104,442 | 8 | 208,885 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Enough is enough. Too many times it happened that Vasya forgot to dispose of garbage and his apartment stank afterwards. Now he wants to create a garbage disposal plan and stick to it.
For each of next n days Vasya knows a_i β number of units of garbage he will produce on the i-th day. Each unit of garbage must be disposed of either on the day it was produced or on the next day. Vasya disposes of garbage by putting it inside a bag and dropping the bag into a garbage container. Each bag can contain up to k units of garbage. It is allowed to compose and drop multiple bags into a garbage container in a single day.
Being economical, Vasya wants to use as few bags as possible. You are to compute the minimum number of bags Vasya needs to dispose of all of his garbage for the given n days. No garbage should be left after the n-th day.
Input
The first line of the input contains two integers n and k (1 β€ n β€ 2β
10^5, 1 β€ k β€ 10^9) β number of days to consider and bag's capacity. The second line contains n space separated integers a_i (0 β€ a_i β€ 10^9) β the number of units of garbage produced on the i-th day.
Output
Output a single integer β the minimum number of bags Vasya needs to dispose of all garbage. Each unit of garbage should be disposed on the day it was produced or on the next day. No garbage can be left after the n-th day. In a day it is allowed to compose and drop multiple bags.
Examples
Input
3 2
3 2 1
Output
3
Input
5 1
1000000000 1000000000 1000000000 1000000000 1000000000
Output
5000000000
Input
3 2
1 0 1
Output
2
Input
4 4
2 8 4 1
Output
4 | instruction | 0 | 104,443 | 8 | 208,886 |
Tags: greedy
Correct Solution:
```
from math import floor
n,k=map(int,input().split())
a=list(map(int,input().split()))
ans=0
i=0
rem=0
while i<n:
if i<n-1:
if a[i]+rem<=k and rem!=0:
ans+=1
i+=1
rem=0
else:
ans+=floor((a[i]+rem)/k)
rem=(a[i]+rem)%k
i+=1
elif i==n-1:
ans+=floor((a[i]+rem)/k)
rem=(a[i]+rem)%k
i+=1
if rem!=0:
ans+=1
print(ans)
``` | output | 1 | 104,443 | 8 | 208,887 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Enough is enough. Too many times it happened that Vasya forgot to dispose of garbage and his apartment stank afterwards. Now he wants to create a garbage disposal plan and stick to it.
For each of next n days Vasya knows a_i β number of units of garbage he will produce on the i-th day. Each unit of garbage must be disposed of either on the day it was produced or on the next day. Vasya disposes of garbage by putting it inside a bag and dropping the bag into a garbage container. Each bag can contain up to k units of garbage. It is allowed to compose and drop multiple bags into a garbage container in a single day.
Being economical, Vasya wants to use as few bags as possible. You are to compute the minimum number of bags Vasya needs to dispose of all of his garbage for the given n days. No garbage should be left after the n-th day.
Input
The first line of the input contains two integers n and k (1 β€ n β€ 2β
10^5, 1 β€ k β€ 10^9) β number of days to consider and bag's capacity. The second line contains n space separated integers a_i (0 β€ a_i β€ 10^9) β the number of units of garbage produced on the i-th day.
Output
Output a single integer β the minimum number of bags Vasya needs to dispose of all garbage. Each unit of garbage should be disposed on the day it was produced or on the next day. No garbage can be left after the n-th day. In a day it is allowed to compose and drop multiple bags.
Examples
Input
3 2
3 2 1
Output
3
Input
5 1
1000000000 1000000000 1000000000 1000000000 1000000000
Output
5000000000
Input
3 2
1 0 1
Output
2
Input
4 4
2 8 4 1
Output
4 | instruction | 0 | 104,444 | 8 | 208,888 |
Tags: greedy
Correct Solution:
```
n,k = map(int,input().split())
t = list(map(int,input().split()))
t_bool = []
for x in range(0,len(t),+1):
t_bool.append(False)
wyn = 0
r = 0
przel = False
for x in range(0,len(t),+1):
if t_bool[x]==True:
if t[x]!=0:
wyn+=1
t[x] -= k
if t[x]<0:
t[x]=0
wyn += int(t[x]/k)
if t[x]%k!=0:
if x!=len(t)-1:
t[x+1]+=t[x]%k
t_bool[x+1]=True
else:
wyn+=1
print(wyn)
``` | output | 1 | 104,444 | 8 | 208,889 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Enough is enough. Too many times it happened that Vasya forgot to dispose of garbage and his apartment stank afterwards. Now he wants to create a garbage disposal plan and stick to it.
For each of next n days Vasya knows a_i β number of units of garbage he will produce on the i-th day. Each unit of garbage must be disposed of either on the day it was produced or on the next day. Vasya disposes of garbage by putting it inside a bag and dropping the bag into a garbage container. Each bag can contain up to k units of garbage. It is allowed to compose and drop multiple bags into a garbage container in a single day.
Being economical, Vasya wants to use as few bags as possible. You are to compute the minimum number of bags Vasya needs to dispose of all of his garbage for the given n days. No garbage should be left after the n-th day.
Input
The first line of the input contains two integers n and k (1 β€ n β€ 2β
10^5, 1 β€ k β€ 10^9) β number of days to consider and bag's capacity. The second line contains n space separated integers a_i (0 β€ a_i β€ 10^9) β the number of units of garbage produced on the i-th day.
Output
Output a single integer β the minimum number of bags Vasya needs to dispose of all garbage. Each unit of garbage should be disposed on the day it was produced or on the next day. No garbage can be left after the n-th day. In a day it is allowed to compose and drop multiple bags.
Examples
Input
3 2
3 2 1
Output
3
Input
5 1
1000000000 1000000000 1000000000 1000000000 1000000000
Output
5000000000
Input
3 2
1 0 1
Output
2
Input
4 4
2 8 4 1
Output
4 | instruction | 0 | 104,445 | 8 | 208,890 |
Tags: greedy
Correct Solution:
```
n,k=map(int,input().split())
arr=list(map(int,input().split()))
ans=arr[0]//k
val=arr[0]%k
for i in range(1,n):
if(val==0):
ans+=arr[i]//k
val=arr[i]%k
else:
val+=arr[i]
if(val<k):
val=0
ans+=1
else:
ans+=val//k
val=val%k
if(val!=0):
ans+=1
print(ans)
``` | output | 1 | 104,445 | 8 | 208,891 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Enough is enough. Too many times it happened that Vasya forgot to dispose of garbage and his apartment stank afterwards. Now he wants to create a garbage disposal plan and stick to it.
For each of next n days Vasya knows a_i β number of units of garbage he will produce on the i-th day. Each unit of garbage must be disposed of either on the day it was produced or on the next day. Vasya disposes of garbage by putting it inside a bag and dropping the bag into a garbage container. Each bag can contain up to k units of garbage. It is allowed to compose and drop multiple bags into a garbage container in a single day.
Being economical, Vasya wants to use as few bags as possible. You are to compute the minimum number of bags Vasya needs to dispose of all of his garbage for the given n days. No garbage should be left after the n-th day.
Input
The first line of the input contains two integers n and k (1 β€ n β€ 2β
10^5, 1 β€ k β€ 10^9) β number of days to consider and bag's capacity. The second line contains n space separated integers a_i (0 β€ a_i β€ 10^9) β the number of units of garbage produced on the i-th day.
Output
Output a single integer β the minimum number of bags Vasya needs to dispose of all garbage. Each unit of garbage should be disposed on the day it was produced or on the next day. No garbage can be left after the n-th day. In a day it is allowed to compose and drop multiple bags.
Examples
Input
3 2
3 2 1
Output
3
Input
5 1
1000000000 1000000000 1000000000 1000000000 1000000000
Output
5000000000
Input
3 2
1 0 1
Output
2
Input
4 4
2 8 4 1
Output
4
Submitted Solution:
```
import math
def findMinBags(n, capacity, garbages):
bags = 0
must_bag = 0
remaining_space = 0
for garb in garbages:
if must_bag:
remaining_space = capacity - (must_bag % capacity)
bags += math.ceil(must_bag / capacity)
if garb >= remaining_space:
garb -= remaining_space
must_bag = garb % capacity
bags += garb // capacity
else:
must_bag = 0
remaining_space = 0
if must_bag: bags += 1
return bags
n, capacity = [int(x) for x in input().split(' ')]
garbages = [int(x) for x in input().split(' ')]
print(findMinBags(n, capacity, garbages))
``` | instruction | 0 | 104,446 | 8 | 208,892 |
Yes | output | 1 | 104,446 | 8 | 208,893 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Enough is enough. Too many times it happened that Vasya forgot to dispose of garbage and his apartment stank afterwards. Now he wants to create a garbage disposal plan and stick to it.
For each of next n days Vasya knows a_i β number of units of garbage he will produce on the i-th day. Each unit of garbage must be disposed of either on the day it was produced or on the next day. Vasya disposes of garbage by putting it inside a bag and dropping the bag into a garbage container. Each bag can contain up to k units of garbage. It is allowed to compose and drop multiple bags into a garbage container in a single day.
Being economical, Vasya wants to use as few bags as possible. You are to compute the minimum number of bags Vasya needs to dispose of all of his garbage for the given n days. No garbage should be left after the n-th day.
Input
The first line of the input contains two integers n and k (1 β€ n β€ 2β
10^5, 1 β€ k β€ 10^9) β number of days to consider and bag's capacity. The second line contains n space separated integers a_i (0 β€ a_i β€ 10^9) β the number of units of garbage produced on the i-th day.
Output
Output a single integer β the minimum number of bags Vasya needs to dispose of all garbage. Each unit of garbage should be disposed on the day it was produced or on the next day. No garbage can be left after the n-th day. In a day it is allowed to compose and drop multiple bags.
Examples
Input
3 2
3 2 1
Output
3
Input
5 1
1000000000 1000000000 1000000000 1000000000 1000000000
Output
5000000000
Input
3 2
1 0 1
Output
2
Input
4 4
2 8 4 1
Output
4
Submitted Solution:
```
n, k = map(int, input().split())
a = [int(x) for x in input().split()]
bags, r = 0, 0
for i in range(n):
x = a[i] + r
if x == 0:
continue
elif x < k and r > 0:
bags += 1
r = 0
else:
bags += (x // k)
r = (x % k)
if r > 0:
bags += 1
print(bags)
``` | instruction | 0 | 104,447 | 8 | 208,894 |
Yes | output | 1 | 104,447 | 8 | 208,895 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Enough is enough. Too many times it happened that Vasya forgot to dispose of garbage and his apartment stank afterwards. Now he wants to create a garbage disposal plan and stick to it.
For each of next n days Vasya knows a_i β number of units of garbage he will produce on the i-th day. Each unit of garbage must be disposed of either on the day it was produced or on the next day. Vasya disposes of garbage by putting it inside a bag and dropping the bag into a garbage container. Each bag can contain up to k units of garbage. It is allowed to compose and drop multiple bags into a garbage container in a single day.
Being economical, Vasya wants to use as few bags as possible. You are to compute the minimum number of bags Vasya needs to dispose of all of his garbage for the given n days. No garbage should be left after the n-th day.
Input
The first line of the input contains two integers n and k (1 β€ n β€ 2β
10^5, 1 β€ k β€ 10^9) β number of days to consider and bag's capacity. The second line contains n space separated integers a_i (0 β€ a_i β€ 10^9) β the number of units of garbage produced on the i-th day.
Output
Output a single integer β the minimum number of bags Vasya needs to dispose of all garbage. Each unit of garbage should be disposed on the day it was produced or on the next day. No garbage can be left after the n-th day. In a day it is allowed to compose and drop multiple bags.
Examples
Input
3 2
3 2 1
Output
3
Input
5 1
1000000000 1000000000 1000000000 1000000000 1000000000
Output
5000000000
Input
3 2
1 0 1
Output
2
Input
4 4
2 8 4 1
Output
4
Submitted Solution:
```
a,b=map(int,input().split())
s=0;p=0;y=0
for i in map(int,input().split()):
if p:
if p+i>=b:
k=i+p
p=0
r=k//b
s+=r
p=k-(r*b)
else:s+=1;p=0
else:
k=i//b
s+=k
p=i-(k*b)
print(s+(p!=0))
``` | instruction | 0 | 104,448 | 8 | 208,896 |
Yes | output | 1 | 104,448 | 8 | 208,897 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Enough is enough. Too many times it happened that Vasya forgot to dispose of garbage and his apartment stank afterwards. Now he wants to create a garbage disposal plan and stick to it.
For each of next n days Vasya knows a_i β number of units of garbage he will produce on the i-th day. Each unit of garbage must be disposed of either on the day it was produced or on the next day. Vasya disposes of garbage by putting it inside a bag and dropping the bag into a garbage container. Each bag can contain up to k units of garbage. It is allowed to compose and drop multiple bags into a garbage container in a single day.
Being economical, Vasya wants to use as few bags as possible. You are to compute the minimum number of bags Vasya needs to dispose of all of his garbage for the given n days. No garbage should be left after the n-th day.
Input
The first line of the input contains two integers n and k (1 β€ n β€ 2β
10^5, 1 β€ k β€ 10^9) β number of days to consider and bag's capacity. The second line contains n space separated integers a_i (0 β€ a_i β€ 10^9) β the number of units of garbage produced on the i-th day.
Output
Output a single integer β the minimum number of bags Vasya needs to dispose of all garbage. Each unit of garbage should be disposed on the day it was produced or on the next day. No garbage can be left after the n-th day. In a day it is allowed to compose and drop multiple bags.
Examples
Input
3 2
3 2 1
Output
3
Input
5 1
1000000000 1000000000 1000000000 1000000000 1000000000
Output
5000000000
Input
3 2
1 0 1
Output
2
Input
4 4
2 8 4 1
Output
4
Submitted Solution:
```
n,k =map(int,input().split())
a=[int(x) for x in input().split()]
bag=a[0]//k
remain=a[0]%k
for i in range(1,n):
tk=a[i]+remain
remain=tk%k
if remain>a[i]:
bag+=1
remain=0
else:
bag+=tk//k
if remain==0:
print(bag)
else:
print(bag+1)
``` | instruction | 0 | 104,449 | 8 | 208,898 |
Yes | output | 1 | 104,449 | 8 | 208,899 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Enough is enough. Too many times it happened that Vasya forgot to dispose of garbage and his apartment stank afterwards. Now he wants to create a garbage disposal plan and stick to it.
For each of next n days Vasya knows a_i β number of units of garbage he will produce on the i-th day. Each unit of garbage must be disposed of either on the day it was produced or on the next day. Vasya disposes of garbage by putting it inside a bag and dropping the bag into a garbage container. Each bag can contain up to k units of garbage. It is allowed to compose and drop multiple bags into a garbage container in a single day.
Being economical, Vasya wants to use as few bags as possible. You are to compute the minimum number of bags Vasya needs to dispose of all of his garbage for the given n days. No garbage should be left after the n-th day.
Input
The first line of the input contains two integers n and k (1 β€ n β€ 2β
10^5, 1 β€ k β€ 10^9) β number of days to consider and bag's capacity. The second line contains n space separated integers a_i (0 β€ a_i β€ 10^9) β the number of units of garbage produced on the i-th day.
Output
Output a single integer β the minimum number of bags Vasya needs to dispose of all garbage. Each unit of garbage should be disposed on the day it was produced or on the next day. No garbage can be left after the n-th day. In a day it is allowed to compose and drop multiple bags.
Examples
Input
3 2
3 2 1
Output
3
Input
5 1
1000000000 1000000000 1000000000 1000000000 1000000000
Output
5000000000
Input
3 2
1 0 1
Output
2
Input
4 4
2 8 4 1
Output
4
Submitted Solution:
```
import math
n, k = map(int, input().split())
a = list(map(int, input().split()))
cnt = 0
for i in range(n - 1):
cnt += a[i] // k
a[i] %= k
if i + 1 == n - 1:
a[i + 1] += a[i]
elif k - a[i] <= a[i+1]:
cnt += 1
a[i + 1] -= (k - a[i])
else:
cnt += math.ceil(a[i]/k) if a[i] else 0
cnt += math.ceil(a[-1]/k)
print(cnt)
``` | instruction | 0 | 104,450 | 8 | 208,900 |
No | output | 1 | 104,450 | 8 | 208,901 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Enough is enough. Too many times it happened that Vasya forgot to dispose of garbage and his apartment stank afterwards. Now he wants to create a garbage disposal plan and stick to it.
For each of next n days Vasya knows a_i β number of units of garbage he will produce on the i-th day. Each unit of garbage must be disposed of either on the day it was produced or on the next day. Vasya disposes of garbage by putting it inside a bag and dropping the bag into a garbage container. Each bag can contain up to k units of garbage. It is allowed to compose and drop multiple bags into a garbage container in a single day.
Being economical, Vasya wants to use as few bags as possible. You are to compute the minimum number of bags Vasya needs to dispose of all of his garbage for the given n days. No garbage should be left after the n-th day.
Input
The first line of the input contains two integers n and k (1 β€ n β€ 2β
10^5, 1 β€ k β€ 10^9) β number of days to consider and bag's capacity. The second line contains n space separated integers a_i (0 β€ a_i β€ 10^9) β the number of units of garbage produced on the i-th day.
Output
Output a single integer β the minimum number of bags Vasya needs to dispose of all garbage. Each unit of garbage should be disposed on the day it was produced or on the next day. No garbage can be left after the n-th day. In a day it is allowed to compose and drop multiple bags.
Examples
Input
3 2
3 2 1
Output
3
Input
5 1
1000000000 1000000000 1000000000 1000000000 1000000000
Output
5000000000
Input
3 2
1 0 1
Output
2
Input
4 4
2 8 4 1
Output
4
Submitted Solution:
```
n, m = map(int, input().split())
a = list(map(int, input().split()))
res = 0
sum = 0
can = True
for i in range(n):
if can is False:
print('cc', sum)
sum = sum + a[i]
res = res + max(1, sum // m)
sum = sum % m
can = True
else:
sum = sum + a[i]
# print('ff', sum)
if sum > m:
res = res + sum // m
sum = sum % m
can = True
else:
print('dd', sum)
can = False
print(res)
``` | instruction | 0 | 104,451 | 8 | 208,902 |
No | output | 1 | 104,451 | 8 | 208,903 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Enough is enough. Too many times it happened that Vasya forgot to dispose of garbage and his apartment stank afterwards. Now he wants to create a garbage disposal plan and stick to it.
For each of next n days Vasya knows a_i β number of units of garbage he will produce on the i-th day. Each unit of garbage must be disposed of either on the day it was produced or on the next day. Vasya disposes of garbage by putting it inside a bag and dropping the bag into a garbage container. Each bag can contain up to k units of garbage. It is allowed to compose and drop multiple bags into a garbage container in a single day.
Being economical, Vasya wants to use as few bags as possible. You are to compute the minimum number of bags Vasya needs to dispose of all of his garbage for the given n days. No garbage should be left after the n-th day.
Input
The first line of the input contains two integers n and k (1 β€ n β€ 2β
10^5, 1 β€ k β€ 10^9) β number of days to consider and bag's capacity. The second line contains n space separated integers a_i (0 β€ a_i β€ 10^9) β the number of units of garbage produced on the i-th day.
Output
Output a single integer β the minimum number of bags Vasya needs to dispose of all garbage. Each unit of garbage should be disposed on the day it was produced or on the next day. No garbage can be left after the n-th day. In a day it is allowed to compose and drop multiple bags.
Examples
Input
3 2
3 2 1
Output
3
Input
5 1
1000000000 1000000000 1000000000 1000000000 1000000000
Output
5000000000
Input
3 2
1 0 1
Output
2
Input
4 4
2 8 4 1
Output
4
Submitted Solution:
```
n,k=map(int,input().split())
a=list(map(int,input().split()))
l=0
for i in range(len(a)-1):
if(a[i]+a[i+1]>=k):
l+=(a[i+1]+a[i])//k
a[i+1]=(a[i+1]+a[i])%k
else:
l+=1
a[i+1]=0
print(l)
``` | instruction | 0 | 104,452 | 8 | 208,904 |
No | output | 1 | 104,452 | 8 | 208,905 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Enough is enough. Too many times it happened that Vasya forgot to dispose of garbage and his apartment stank afterwards. Now he wants to create a garbage disposal plan and stick to it.
For each of next n days Vasya knows a_i β number of units of garbage he will produce on the i-th day. Each unit of garbage must be disposed of either on the day it was produced or on the next day. Vasya disposes of garbage by putting it inside a bag and dropping the bag into a garbage container. Each bag can contain up to k units of garbage. It is allowed to compose and drop multiple bags into a garbage container in a single day.
Being economical, Vasya wants to use as few bags as possible. You are to compute the minimum number of bags Vasya needs to dispose of all of his garbage for the given n days. No garbage should be left after the n-th day.
Input
The first line of the input contains two integers n and k (1 β€ n β€ 2β
10^5, 1 β€ k β€ 10^9) β number of days to consider and bag's capacity. The second line contains n space separated integers a_i (0 β€ a_i β€ 10^9) β the number of units of garbage produced on the i-th day.
Output
Output a single integer β the minimum number of bags Vasya needs to dispose of all garbage. Each unit of garbage should be disposed on the day it was produced or on the next day. No garbage can be left after the n-th day. In a day it is allowed to compose and drop multiple bags.
Examples
Input
3 2
3 2 1
Output
3
Input
5 1
1000000000 1000000000 1000000000 1000000000 1000000000
Output
5000000000
Input
3 2
1 0 1
Output
2
Input
4 4
2 8 4 1
Output
4
Submitted Solution:
```
n,k=map(int,input().split())
a=list(map(int,input().split()))
l=0
for i in range(len(a)-1):
if(a[i]+a[i+1]>=k):
l+=(a[i]+a[i+1])//k
a[i+1]=(a[i]+a[i+1])%k
else:
l+=1
a[i+1]=0
if(a[-1]>=k):
l+=a[-1]//k
else:
l+=1
print(l-1)
``` | instruction | 0 | 104,453 | 8 | 208,906 |
No | output | 1 | 104,453 | 8 | 208,907 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has a pile, that consists of some number of stones. n times he either took one stone from the pile or added one stone to the pile. The pile was non-empty before each operation of taking one stone from the pile.
You are given n operations which Vasya has made. Find the minimal possible number of stones that can be in the pile after making these operations.
Input
The first line contains one positive integer n β the number of operations, that have been made by Vasya (1 β€ n β€ 100).
The next line contains the string s, consisting of n symbols, equal to "-" (without quotes) or "+" (without quotes). If Vasya took the stone on i-th operation, s_i is equal to "-" (without quotes), if added, s_i is equal to "+" (without quotes).
Output
Print one integer β the minimal possible number of stones that can be in the pile after these n operations.
Examples
Input
3
---
Output
0
Input
4
++++
Output
4
Input
2
-+
Output
1
Input
5
++-++
Output
3
Note
In the first test, if Vasya had 3 stones in the pile at the beginning, after making operations the number of stones will be equal to 0. It is impossible to have less number of piles, so the answer is 0. Please notice, that the number of stones at the beginning can't be less, than 3, because in this case, Vasya won't be able to take a stone on some operation (the pile will be empty).
In the second test, if Vasya had 0 stones in the pile at the beginning, after making operations the number of stones will be equal to 4. It is impossible to have less number of piles because after making 4 operations the number of stones in the pile increases on 4 stones. So, the answer is 4.
In the third test, if Vasya had 1 stone in the pile at the beginning, after making operations the number of stones will be equal to 1. It can be proved, that it is impossible to have less number of stones after making the operations.
In the fourth test, if Vasya had 0 stones in the pile at the beginning, after making operations the number of stones will be equal to 3. | instruction | 0 | 104,471 | 8 | 208,942 |
Tags: implementation, math
Correct Solution:
```
n=int(input())
s=input()
ini=0
sta=0
for i in range(n):
if s[i]=="-":
if sta==0:
ini+=1
else:
sta-=1
if s[i]=="+":
sta+=1
print(sta)
``` | output | 1 | 104,471 | 8 | 208,943 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has a pile, that consists of some number of stones. n times he either took one stone from the pile or added one stone to the pile. The pile was non-empty before each operation of taking one stone from the pile.
You are given n operations which Vasya has made. Find the minimal possible number of stones that can be in the pile after making these operations.
Input
The first line contains one positive integer n β the number of operations, that have been made by Vasya (1 β€ n β€ 100).
The next line contains the string s, consisting of n symbols, equal to "-" (without quotes) or "+" (without quotes). If Vasya took the stone on i-th operation, s_i is equal to "-" (without quotes), if added, s_i is equal to "+" (without quotes).
Output
Print one integer β the minimal possible number of stones that can be in the pile after these n operations.
Examples
Input
3
---
Output
0
Input
4
++++
Output
4
Input
2
-+
Output
1
Input
5
++-++
Output
3
Note
In the first test, if Vasya had 3 stones in the pile at the beginning, after making operations the number of stones will be equal to 0. It is impossible to have less number of piles, so the answer is 0. Please notice, that the number of stones at the beginning can't be less, than 3, because in this case, Vasya won't be able to take a stone on some operation (the pile will be empty).
In the second test, if Vasya had 0 stones in the pile at the beginning, after making operations the number of stones will be equal to 4. It is impossible to have less number of piles because after making 4 operations the number of stones in the pile increases on 4 stones. So, the answer is 4.
In the third test, if Vasya had 1 stone in the pile at the beginning, after making operations the number of stones will be equal to 1. It can be proved, that it is impossible to have less number of stones after making the operations.
In the fourth test, if Vasya had 0 stones in the pile at the beginning, after making operations the number of stones will be equal to 3. | instruction | 0 | 104,472 | 8 | 208,944 |
Tags: implementation, math
Correct Solution:
```
op = int(input())
pile=input()
a=pile[0]
if a=='-':
stone=1
else:
stone=0
for i in range(op):
if pile[i]=='+':
stone+=1
else:
stone-=1
if stone<0:
stone=0
print(stone)
``` | output | 1 | 104,472 | 8 | 208,945 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has a pile, that consists of some number of stones. n times he either took one stone from the pile or added one stone to the pile. The pile was non-empty before each operation of taking one stone from the pile.
You are given n operations which Vasya has made. Find the minimal possible number of stones that can be in the pile after making these operations.
Input
The first line contains one positive integer n β the number of operations, that have been made by Vasya (1 β€ n β€ 100).
The next line contains the string s, consisting of n symbols, equal to "-" (without quotes) or "+" (without quotes). If Vasya took the stone on i-th operation, s_i is equal to "-" (without quotes), if added, s_i is equal to "+" (without quotes).
Output
Print one integer β the minimal possible number of stones that can be in the pile after these n operations.
Examples
Input
3
---
Output
0
Input
4
++++
Output
4
Input
2
-+
Output
1
Input
5
++-++
Output
3
Note
In the first test, if Vasya had 3 stones in the pile at the beginning, after making operations the number of stones will be equal to 0. It is impossible to have less number of piles, so the answer is 0. Please notice, that the number of stones at the beginning can't be less, than 3, because in this case, Vasya won't be able to take a stone on some operation (the pile will be empty).
In the second test, if Vasya had 0 stones in the pile at the beginning, after making operations the number of stones will be equal to 4. It is impossible to have less number of piles because after making 4 operations the number of stones in the pile increases on 4 stones. So, the answer is 4.
In the third test, if Vasya had 1 stone in the pile at the beginning, after making operations the number of stones will be equal to 1. It can be proved, that it is impossible to have less number of stones after making the operations.
In the fourth test, if Vasya had 0 stones in the pile at the beginning, after making operations the number of stones will be equal to 3. | instruction | 0 | 104,473 | 8 | 208,946 |
Tags: implementation, math
Correct Solution:
```
num = int(input())
string=str(input())
total=0
for counter,item in enumerate(string):
if item=="-" and total==0:
continue
elif item=="-" and total!=0:
total=total-1
elif item=="+":
total=total+1
print(total)
``` | output | 1 | 104,473 | 8 | 208,947 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has a pile, that consists of some number of stones. n times he either took one stone from the pile or added one stone to the pile. The pile was non-empty before each operation of taking one stone from the pile.
You are given n operations which Vasya has made. Find the minimal possible number of stones that can be in the pile after making these operations.
Input
The first line contains one positive integer n β the number of operations, that have been made by Vasya (1 β€ n β€ 100).
The next line contains the string s, consisting of n symbols, equal to "-" (without quotes) or "+" (without quotes). If Vasya took the stone on i-th operation, s_i is equal to "-" (without quotes), if added, s_i is equal to "+" (without quotes).
Output
Print one integer β the minimal possible number of stones that can be in the pile after these n operations.
Examples
Input
3
---
Output
0
Input
4
++++
Output
4
Input
2
-+
Output
1
Input
5
++-++
Output
3
Note
In the first test, if Vasya had 3 stones in the pile at the beginning, after making operations the number of stones will be equal to 0. It is impossible to have less number of piles, so the answer is 0. Please notice, that the number of stones at the beginning can't be less, than 3, because in this case, Vasya won't be able to take a stone on some operation (the pile will be empty).
In the second test, if Vasya had 0 stones in the pile at the beginning, after making operations the number of stones will be equal to 4. It is impossible to have less number of piles because after making 4 operations the number of stones in the pile increases on 4 stones. So, the answer is 4.
In the third test, if Vasya had 1 stone in the pile at the beginning, after making operations the number of stones will be equal to 1. It can be proved, that it is impossible to have less number of stones after making the operations.
In the fourth test, if Vasya had 0 stones in the pile at the beginning, after making operations the number of stones will be equal to 3. | instruction | 0 | 104,474 | 8 | 208,948 |
Tags: implementation, math
Correct Solution:
```
n = int(input())
s = input()
ans = 0
for i in range(n):
ans = max(ans, s[i:].count("+")-s[i:].count("-"))
print(ans)
``` | output | 1 | 104,474 | 8 | 208,949 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has a pile, that consists of some number of stones. n times he either took one stone from the pile or added one stone to the pile. The pile was non-empty before each operation of taking one stone from the pile.
You are given n operations which Vasya has made. Find the minimal possible number of stones that can be in the pile after making these operations.
Input
The first line contains one positive integer n β the number of operations, that have been made by Vasya (1 β€ n β€ 100).
The next line contains the string s, consisting of n symbols, equal to "-" (without quotes) or "+" (without quotes). If Vasya took the stone on i-th operation, s_i is equal to "-" (without quotes), if added, s_i is equal to "+" (without quotes).
Output
Print one integer β the minimal possible number of stones that can be in the pile after these n operations.
Examples
Input
3
---
Output
0
Input
4
++++
Output
4
Input
2
-+
Output
1
Input
5
++-++
Output
3
Note
In the first test, if Vasya had 3 stones in the pile at the beginning, after making operations the number of stones will be equal to 0. It is impossible to have less number of piles, so the answer is 0. Please notice, that the number of stones at the beginning can't be less, than 3, because in this case, Vasya won't be able to take a stone on some operation (the pile will be empty).
In the second test, if Vasya had 0 stones in the pile at the beginning, after making operations the number of stones will be equal to 4. It is impossible to have less number of piles because after making 4 operations the number of stones in the pile increases on 4 stones. So, the answer is 4.
In the third test, if Vasya had 1 stone in the pile at the beginning, after making operations the number of stones will be equal to 1. It can be proved, that it is impossible to have less number of stones after making the operations.
In the fourth test, if Vasya had 0 stones in the pile at the beginning, after making operations the number of stones will be equal to 3. | instruction | 0 | 104,475 | 8 | 208,950 |
Tags: implementation, math
Correct Solution:
```
def rahul(word):
return[char for char in word]
a=int(input())
b=rahul(input())
r=0
s=0
while r<a:
if b[r]=='+':
s=s+1
elif b[r]=='-' and s>0:
s=s-1
r=r+1
print(s)
``` | output | 1 | 104,475 | 8 | 208,951 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has a pile, that consists of some number of stones. n times he either took one stone from the pile or added one stone to the pile. The pile was non-empty before each operation of taking one stone from the pile.
You are given n operations which Vasya has made. Find the minimal possible number of stones that can be in the pile after making these operations.
Input
The first line contains one positive integer n β the number of operations, that have been made by Vasya (1 β€ n β€ 100).
The next line contains the string s, consisting of n symbols, equal to "-" (without quotes) or "+" (without quotes). If Vasya took the stone on i-th operation, s_i is equal to "-" (without quotes), if added, s_i is equal to "+" (without quotes).
Output
Print one integer β the minimal possible number of stones that can be in the pile after these n operations.
Examples
Input
3
---
Output
0
Input
4
++++
Output
4
Input
2
-+
Output
1
Input
5
++-++
Output
3
Note
In the first test, if Vasya had 3 stones in the pile at the beginning, after making operations the number of stones will be equal to 0. It is impossible to have less number of piles, so the answer is 0. Please notice, that the number of stones at the beginning can't be less, than 3, because in this case, Vasya won't be able to take a stone on some operation (the pile will be empty).
In the second test, if Vasya had 0 stones in the pile at the beginning, after making operations the number of stones will be equal to 4. It is impossible to have less number of piles because after making 4 operations the number of stones in the pile increases on 4 stones. So, the answer is 4.
In the third test, if Vasya had 1 stone in the pile at the beginning, after making operations the number of stones will be equal to 1. It can be proved, that it is impossible to have less number of stones after making the operations.
In the fourth test, if Vasya had 0 stones in the pile at the beginning, after making operations the number of stones will be equal to 3. | instruction | 0 | 104,476 | 8 | 208,952 |
Tags: implementation, math
Correct Solution:
```
n = int(input())
a = input()
res = 0
for i in range(n):
if a[i] == '-':
res -= 1
elif a[i] == '+':
if res < 0:
res = 1
else:
res += 1
if res < 0:
res = 0
print(res)
``` | output | 1 | 104,476 | 8 | 208,953 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has a pile, that consists of some number of stones. n times he either took one stone from the pile or added one stone to the pile. The pile was non-empty before each operation of taking one stone from the pile.
You are given n operations which Vasya has made. Find the minimal possible number of stones that can be in the pile after making these operations.
Input
The first line contains one positive integer n β the number of operations, that have been made by Vasya (1 β€ n β€ 100).
The next line contains the string s, consisting of n symbols, equal to "-" (without quotes) or "+" (without quotes). If Vasya took the stone on i-th operation, s_i is equal to "-" (without quotes), if added, s_i is equal to "+" (without quotes).
Output
Print one integer β the minimal possible number of stones that can be in the pile after these n operations.
Examples
Input
3
---
Output
0
Input
4
++++
Output
4
Input
2
-+
Output
1
Input
5
++-++
Output
3
Note
In the first test, if Vasya had 3 stones in the pile at the beginning, after making operations the number of stones will be equal to 0. It is impossible to have less number of piles, so the answer is 0. Please notice, that the number of stones at the beginning can't be less, than 3, because in this case, Vasya won't be able to take a stone on some operation (the pile will be empty).
In the second test, if Vasya had 0 stones in the pile at the beginning, after making operations the number of stones will be equal to 4. It is impossible to have less number of piles because after making 4 operations the number of stones in the pile increases on 4 stones. So, the answer is 4.
In the third test, if Vasya had 1 stone in the pile at the beginning, after making operations the number of stones will be equal to 1. It can be proved, that it is impossible to have less number of stones after making the operations.
In the fourth test, if Vasya had 0 stones in the pile at the beginning, after making operations the number of stones will be equal to 3. | instruction | 0 | 104,477 | 8 | 208,954 |
Tags: implementation, math
Correct Solution:
```
def run_commands(start, cmd):
for c in cmd:
if c == "-":
start -= 1
#print("subtracted to", start)
elif c == "+":
start += 1
if start < 0:
return False
return start
n = int(input())
cmd = list(input())
for start in range(0, 101):
result = run_commands(start, cmd)
if result is False:
continue
else:
print(result)
break
``` | output | 1 | 104,477 | 8 | 208,955 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has a pile, that consists of some number of stones. n times he either took one stone from the pile or added one stone to the pile. The pile was non-empty before each operation of taking one stone from the pile.
You are given n operations which Vasya has made. Find the minimal possible number of stones that can be in the pile after making these operations.
Input
The first line contains one positive integer n β the number of operations, that have been made by Vasya (1 β€ n β€ 100).
The next line contains the string s, consisting of n symbols, equal to "-" (without quotes) or "+" (without quotes). If Vasya took the stone on i-th operation, s_i is equal to "-" (without quotes), if added, s_i is equal to "+" (without quotes).
Output
Print one integer β the minimal possible number of stones that can be in the pile after these n operations.
Examples
Input
3
---
Output
0
Input
4
++++
Output
4
Input
2
-+
Output
1
Input
5
++-++
Output
3
Note
In the first test, if Vasya had 3 stones in the pile at the beginning, after making operations the number of stones will be equal to 0. It is impossible to have less number of piles, so the answer is 0. Please notice, that the number of stones at the beginning can't be less, than 3, because in this case, Vasya won't be able to take a stone on some operation (the pile will be empty).
In the second test, if Vasya had 0 stones in the pile at the beginning, after making operations the number of stones will be equal to 4. It is impossible to have less number of piles because after making 4 operations the number of stones in the pile increases on 4 stones. So, the answer is 4.
In the third test, if Vasya had 1 stone in the pile at the beginning, after making operations the number of stones will be equal to 1. It can be proved, that it is impossible to have less number of stones after making the operations.
In the fourth test, if Vasya had 0 stones in the pile at the beginning, after making operations the number of stones will be equal to 3. | instruction | 0 | 104,478 | 8 | 208,956 |
Tags: implementation, math
Correct Solution:
```
n = int(input())
s = input()
p = False
x = 0
for i in range(n):
if not p and s[i] == "+":
p = True
if p:
if s[i] == "+":
x += 1
else:
if x > 0:
x -= 1
print(x)
``` | output | 1 | 104,478 | 8 | 208,957 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has a pile, that consists of some number of stones. n times he either took one stone from the pile or added one stone to the pile. The pile was non-empty before each operation of taking one stone from the pile.
You are given n operations which Vasya has made. Find the minimal possible number of stones that can be in the pile after making these operations.
Input
The first line contains one positive integer n β the number of operations, that have been made by Vasya (1 β€ n β€ 100).
The next line contains the string s, consisting of n symbols, equal to "-" (without quotes) or "+" (without quotes). If Vasya took the stone on i-th operation, s_i is equal to "-" (without quotes), if added, s_i is equal to "+" (without quotes).
Output
Print one integer β the minimal possible number of stones that can be in the pile after these n operations.
Examples
Input
3
---
Output
0
Input
4
++++
Output
4
Input
2
-+
Output
1
Input
5
++-++
Output
3
Note
In the first test, if Vasya had 3 stones in the pile at the beginning, after making operations the number of stones will be equal to 0. It is impossible to have less number of piles, so the answer is 0. Please notice, that the number of stones at the beginning can't be less, than 3, because in this case, Vasya won't be able to take a stone on some operation (the pile will be empty).
In the second test, if Vasya had 0 stones in the pile at the beginning, after making operations the number of stones will be equal to 4. It is impossible to have less number of piles because after making 4 operations the number of stones in the pile increases on 4 stones. So, the answer is 4.
In the third test, if Vasya had 1 stone in the pile at the beginning, after making operations the number of stones will be equal to 1. It can be proved, that it is impossible to have less number of stones after making the operations.
In the fourth test, if Vasya had 0 stones in the pile at the beginning, after making operations the number of stones will be equal to 3.
Submitted Solution:
```
# -*- coding: utf-8 -*-
# @Time : 2019/5/12 22:46
# @Author : LunaFire
# @Email : gilgemesh2012@gmail.com
# @File : A. A pile of stones.py
def main():
n = int(input())
s = input()
ret = 0
for c in s:
if c == '-':
ret = max(0, ret - 1)
else:
ret += 1
print(ret)
if __name__ == '__main__':
main()
``` | instruction | 0 | 104,479 | 8 | 208,958 |
Yes | output | 1 | 104,479 | 8 | 208,959 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has a pile, that consists of some number of stones. n times he either took one stone from the pile or added one stone to the pile. The pile was non-empty before each operation of taking one stone from the pile.
You are given n operations which Vasya has made. Find the minimal possible number of stones that can be in the pile after making these operations.
Input
The first line contains one positive integer n β the number of operations, that have been made by Vasya (1 β€ n β€ 100).
The next line contains the string s, consisting of n symbols, equal to "-" (without quotes) or "+" (without quotes). If Vasya took the stone on i-th operation, s_i is equal to "-" (without quotes), if added, s_i is equal to "+" (without quotes).
Output
Print one integer β the minimal possible number of stones that can be in the pile after these n operations.
Examples
Input
3
---
Output
0
Input
4
++++
Output
4
Input
2
-+
Output
1
Input
5
++-++
Output
3
Note
In the first test, if Vasya had 3 stones in the pile at the beginning, after making operations the number of stones will be equal to 0. It is impossible to have less number of piles, so the answer is 0. Please notice, that the number of stones at the beginning can't be less, than 3, because in this case, Vasya won't be able to take a stone on some operation (the pile will be empty).
In the second test, if Vasya had 0 stones in the pile at the beginning, after making operations the number of stones will be equal to 4. It is impossible to have less number of piles because after making 4 operations the number of stones in the pile increases on 4 stones. So, the answer is 4.
In the third test, if Vasya had 1 stone in the pile at the beginning, after making operations the number of stones will be equal to 1. It can be proved, that it is impossible to have less number of stones after making the operations.
In the fourth test, if Vasya had 0 stones in the pile at the beginning, after making operations the number of stones will be equal to 3.
Submitted Solution:
```
n = int(input())
s = input()
minim = n
maxim = 0
for ch in s:
if ch == '-':
n -= 1
else:
n += 1
minim = min(minim, n)
print(n - minim)
``` | instruction | 0 | 104,480 | 8 | 208,960 |
Yes | output | 1 | 104,480 | 8 | 208,961 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has a pile, that consists of some number of stones. n times he either took one stone from the pile or added one stone to the pile. The pile was non-empty before each operation of taking one stone from the pile.
You are given n operations which Vasya has made. Find the minimal possible number of stones that can be in the pile after making these operations.
Input
The first line contains one positive integer n β the number of operations, that have been made by Vasya (1 β€ n β€ 100).
The next line contains the string s, consisting of n symbols, equal to "-" (without quotes) or "+" (without quotes). If Vasya took the stone on i-th operation, s_i is equal to "-" (without quotes), if added, s_i is equal to "+" (without quotes).
Output
Print one integer β the minimal possible number of stones that can be in the pile after these n operations.
Examples
Input
3
---
Output
0
Input
4
++++
Output
4
Input
2
-+
Output
1
Input
5
++-++
Output
3
Note
In the first test, if Vasya had 3 stones in the pile at the beginning, after making operations the number of stones will be equal to 0. It is impossible to have less number of piles, so the answer is 0. Please notice, that the number of stones at the beginning can't be less, than 3, because in this case, Vasya won't be able to take a stone on some operation (the pile will be empty).
In the second test, if Vasya had 0 stones in the pile at the beginning, after making operations the number of stones will be equal to 4. It is impossible to have less number of piles because after making 4 operations the number of stones in the pile increases on 4 stones. So, the answer is 4.
In the third test, if Vasya had 1 stone in the pile at the beginning, after making operations the number of stones will be equal to 1. It can be proved, that it is impossible to have less number of stones after making the operations.
In the fourth test, if Vasya had 0 stones in the pile at the beginning, after making operations the number of stones will be equal to 3.
Submitted Solution:
```
n=int(input())
s=input()
c=0
d=0
for i in s:
if(i=="-"):
if(c>0):
c=c-1
else:
c=c+1
print(c)
``` | instruction | 0 | 104,481 | 8 | 208,962 |
Yes | output | 1 | 104,481 | 8 | 208,963 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has a pile, that consists of some number of stones. n times he either took one stone from the pile or added one stone to the pile. The pile was non-empty before each operation of taking one stone from the pile.
You are given n operations which Vasya has made. Find the minimal possible number of stones that can be in the pile after making these operations.
Input
The first line contains one positive integer n β the number of operations, that have been made by Vasya (1 β€ n β€ 100).
The next line contains the string s, consisting of n symbols, equal to "-" (without quotes) or "+" (without quotes). If Vasya took the stone on i-th operation, s_i is equal to "-" (without quotes), if added, s_i is equal to "+" (without quotes).
Output
Print one integer β the minimal possible number of stones that can be in the pile after these n operations.
Examples
Input
3
---
Output
0
Input
4
++++
Output
4
Input
2
-+
Output
1
Input
5
++-++
Output
3
Note
In the first test, if Vasya had 3 stones in the pile at the beginning, after making operations the number of stones will be equal to 0. It is impossible to have less number of piles, so the answer is 0. Please notice, that the number of stones at the beginning can't be less, than 3, because in this case, Vasya won't be able to take a stone on some operation (the pile will be empty).
In the second test, if Vasya had 0 stones in the pile at the beginning, after making operations the number of stones will be equal to 4. It is impossible to have less number of piles because after making 4 operations the number of stones in the pile increases on 4 stones. So, the answer is 4.
In the third test, if Vasya had 1 stone in the pile at the beginning, after making operations the number of stones will be equal to 1. It can be proved, that it is impossible to have less number of stones after making the operations.
In the fourth test, if Vasya had 0 stones in the pile at the beginning, after making operations the number of stones will be equal to 3.
Submitted Solution:
```
n = int(input())
s = input()
ans = 0
for c in s:
if c == '-':
ans -= ans != 0
else:
ans += 1
print(ans)
``` | instruction | 0 | 104,482 | 8 | 208,964 |
Yes | output | 1 | 104,482 | 8 | 208,965 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has a pile, that consists of some number of stones. n times he either took one stone from the pile or added one stone to the pile. The pile was non-empty before each operation of taking one stone from the pile.
You are given n operations which Vasya has made. Find the minimal possible number of stones that can be in the pile after making these operations.
Input
The first line contains one positive integer n β the number of operations, that have been made by Vasya (1 β€ n β€ 100).
The next line contains the string s, consisting of n symbols, equal to "-" (without quotes) or "+" (without quotes). If Vasya took the stone on i-th operation, s_i is equal to "-" (without quotes), if added, s_i is equal to "+" (without quotes).
Output
Print one integer β the minimal possible number of stones that can be in the pile after these n operations.
Examples
Input
3
---
Output
0
Input
4
++++
Output
4
Input
2
-+
Output
1
Input
5
++-++
Output
3
Note
In the first test, if Vasya had 3 stones in the pile at the beginning, after making operations the number of stones will be equal to 0. It is impossible to have less number of piles, so the answer is 0. Please notice, that the number of stones at the beginning can't be less, than 3, because in this case, Vasya won't be able to take a stone on some operation (the pile will be empty).
In the second test, if Vasya had 0 stones in the pile at the beginning, after making operations the number of stones will be equal to 4. It is impossible to have less number of piles because after making 4 operations the number of stones in the pile increases on 4 stones. So, the answer is 4.
In the third test, if Vasya had 1 stone in the pile at the beginning, after making operations the number of stones will be equal to 1. It can be proved, that it is impossible to have less number of stones after making the operations.
In the fourth test, if Vasya had 0 stones in the pile at the beginning, after making operations the number of stones will be equal to 3.
Submitted Solution:
```
n=int(input())
s=input()
p=0
if(s[0]=='-'):
p+=1
a=s.count('+')
m=s.count('-')
if(a-m>=0):
print(p+a-m)
else:
print(0)
``` | instruction | 0 | 104,483 | 8 | 208,966 |
No | output | 1 | 104,483 | 8 | 208,967 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has a pile, that consists of some number of stones. n times he either took one stone from the pile or added one stone to the pile. The pile was non-empty before each operation of taking one stone from the pile.
You are given n operations which Vasya has made. Find the minimal possible number of stones that can be in the pile after making these operations.
Input
The first line contains one positive integer n β the number of operations, that have been made by Vasya (1 β€ n β€ 100).
The next line contains the string s, consisting of n symbols, equal to "-" (without quotes) or "+" (without quotes). If Vasya took the stone on i-th operation, s_i is equal to "-" (without quotes), if added, s_i is equal to "+" (without quotes).
Output
Print one integer β the minimal possible number of stones that can be in the pile after these n operations.
Examples
Input
3
---
Output
0
Input
4
++++
Output
4
Input
2
-+
Output
1
Input
5
++-++
Output
3
Note
In the first test, if Vasya had 3 stones in the pile at the beginning, after making operations the number of stones will be equal to 0. It is impossible to have less number of piles, so the answer is 0. Please notice, that the number of stones at the beginning can't be less, than 3, because in this case, Vasya won't be able to take a stone on some operation (the pile will be empty).
In the second test, if Vasya had 0 stones in the pile at the beginning, after making operations the number of stones will be equal to 4. It is impossible to have less number of piles because after making 4 operations the number of stones in the pile increases on 4 stones. So, the answer is 4.
In the third test, if Vasya had 1 stone in the pile at the beginning, after making operations the number of stones will be equal to 1. It can be proved, that it is impossible to have less number of stones after making the operations.
In the fourth test, if Vasya had 0 stones in the pile at the beginning, after making operations the number of stones will be equal to 3.
Submitted Solution:
```
n= int(input())
p=input()[:n]
c=0
j=0
for i in p:
if i=='+':
c+=1
elif i=='-':
j+=1
if p[0]=='+':
n=0
elif p[0]=='-'and p[1]=='+' and c>=j:
n=1
for k in p:
if(k=='+'):
n+=1
elif(k=='-'):
n-=1
print(n)
``` | instruction | 0 | 104,484 | 8 | 208,968 |
No | output | 1 | 104,484 | 8 | 208,969 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has a pile, that consists of some number of stones. n times he either took one stone from the pile or added one stone to the pile. The pile was non-empty before each operation of taking one stone from the pile.
You are given n operations which Vasya has made. Find the minimal possible number of stones that can be in the pile after making these operations.
Input
The first line contains one positive integer n β the number of operations, that have been made by Vasya (1 β€ n β€ 100).
The next line contains the string s, consisting of n symbols, equal to "-" (without quotes) or "+" (without quotes). If Vasya took the stone on i-th operation, s_i is equal to "-" (without quotes), if added, s_i is equal to "+" (without quotes).
Output
Print one integer β the minimal possible number of stones that can be in the pile after these n operations.
Examples
Input
3
---
Output
0
Input
4
++++
Output
4
Input
2
-+
Output
1
Input
5
++-++
Output
3
Note
In the first test, if Vasya had 3 stones in the pile at the beginning, after making operations the number of stones will be equal to 0. It is impossible to have less number of piles, so the answer is 0. Please notice, that the number of stones at the beginning can't be less, than 3, because in this case, Vasya won't be able to take a stone on some operation (the pile will be empty).
In the second test, if Vasya had 0 stones in the pile at the beginning, after making operations the number of stones will be equal to 4. It is impossible to have less number of piles because after making 4 operations the number of stones in the pile increases on 4 stones. So, the answer is 4.
In the third test, if Vasya had 1 stone in the pile at the beginning, after making operations the number of stones will be equal to 1. It can be proved, that it is impossible to have less number of stones after making the operations.
In the fourth test, if Vasya had 0 stones in the pile at the beginning, after making operations the number of stones will be equal to 3.
Submitted Solution:
```
n = int(input())
s = input()
m = n
for ch in s:
if (ch == '-'):
n -= 1
else:
n += 1
m = min(m, n)
print(m)
``` | instruction | 0 | 104,485 | 8 | 208,970 |
No | output | 1 | 104,485 | 8 | 208,971 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has a pile, that consists of some number of stones. n times he either took one stone from the pile or added one stone to the pile. The pile was non-empty before each operation of taking one stone from the pile.
You are given n operations which Vasya has made. Find the minimal possible number of stones that can be in the pile after making these operations.
Input
The first line contains one positive integer n β the number of operations, that have been made by Vasya (1 β€ n β€ 100).
The next line contains the string s, consisting of n symbols, equal to "-" (without quotes) or "+" (without quotes). If Vasya took the stone on i-th operation, s_i is equal to "-" (without quotes), if added, s_i is equal to "+" (without quotes).
Output
Print one integer β the minimal possible number of stones that can be in the pile after these n operations.
Examples
Input
3
---
Output
0
Input
4
++++
Output
4
Input
2
-+
Output
1
Input
5
++-++
Output
3
Note
In the first test, if Vasya had 3 stones in the pile at the beginning, after making operations the number of stones will be equal to 0. It is impossible to have less number of piles, so the answer is 0. Please notice, that the number of stones at the beginning can't be less, than 3, because in this case, Vasya won't be able to take a stone on some operation (the pile will be empty).
In the second test, if Vasya had 0 stones in the pile at the beginning, after making operations the number of stones will be equal to 4. It is impossible to have less number of piles because after making 4 operations the number of stones in the pile increases on 4 stones. So, the answer is 4.
In the third test, if Vasya had 1 stone in the pile at the beginning, after making operations the number of stones will be equal to 1. It can be proved, that it is impossible to have less number of stones after making the operations.
In the fourth test, if Vasya had 0 stones in the pile at the beginning, after making operations the number of stones will be equal to 3.
Submitted Solution:
```
#!/usr/bin/env python3
n = int(input())
opers = input()
k = 100
for i in opers:
if i == '+':
k += 1
elif i == '-':
k -= 1
k = 100 - k
if k > 0:
if n - k == 0:
print(0)
else:
print(n - k)
else:
print(abs(k))
``` | instruction | 0 | 104,486 | 8 | 208,972 |
No | output | 1 | 104,486 | 8 | 208,973 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Lengths are measures in Baden in inches and feet. To a length from centimeters it is enough to know that an inch equals three centimeters in Baden and one foot contains 12 inches.
You are given a length equal to n centimeters. Your task is to convert it to feet and inches so that the number of feet was maximum. The result should be an integer rounded to the closest value containing an integral number of inches.
Note that when you round up, 1 cm rounds up to 0 inches and 2 cm round up to 1 inch.
Input
The only line contains an integer n (1 β€ n β€ 10000).
Output
Print two non-negative space-separated integers a and b, where a is the numbers of feet and b is the number of inches.
Examples
Input
42
Output
1 2
Input
5
Output
0 2
Submitted Solution:
```
from math import ceil, floor
n = int(input())
if n % 3 > 1:
d = ceil(n / 3)
else:
d = n // 3
f = d // 12
print(f, d % 12)
``` | instruction | 0 | 104,532 | 8 | 209,064 |
Yes | output | 1 | 104,532 | 8 | 209,065 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Lengths are measures in Baden in inches and feet. To a length from centimeters it is enough to know that an inch equals three centimeters in Baden and one foot contains 12 inches.
You are given a length equal to n centimeters. Your task is to convert it to feet and inches so that the number of feet was maximum. The result should be an integer rounded to the closest value containing an integral number of inches.
Note that when you round up, 1 cm rounds up to 0 inches and 2 cm round up to 1 inch.
Input
The only line contains an integer n (1 β€ n β€ 10000).
Output
Print two non-negative space-separated integers a and b, where a is the numbers of feet and b is the number of inches.
Examples
Input
42
Output
1 2
Input
5
Output
0 2
Submitted Solution:
```
n=int(input())
a=n//3
b=n%3
if b==2:
a=a+1
p=a//12
q=a%12
print(p,q)
``` | instruction | 0 | 104,533 | 8 | 209,066 |
Yes | output | 1 | 104,533 | 8 | 209,067 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Lengths are measures in Baden in inches and feet. To a length from centimeters it is enough to know that an inch equals three centimeters in Baden and one foot contains 12 inches.
You are given a length equal to n centimeters. Your task is to convert it to feet and inches so that the number of feet was maximum. The result should be an integer rounded to the closest value containing an integral number of inches.
Note that when you round up, 1 cm rounds up to 0 inches and 2 cm round up to 1 inch.
Input
The only line contains an integer n (1 β€ n β€ 10000).
Output
Print two non-negative space-separated integers a and b, where a is the numbers of feet and b is the number of inches.
Examples
Input
42
Output
1 2
Input
5
Output
0 2
Submitted Solution:
```
from math import ceil, floor
n = int(input())
d = ceil(n / 3)
f = d // 12
print(f, d % 12)
``` | instruction | 0 | 104,536 | 8 | 209,072 |
No | output | 1 | 104,536 | 8 | 209,073 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Lengths are measures in Baden in inches and feet. To a length from centimeters it is enough to know that an inch equals three centimeters in Baden and one foot contains 12 inches.
You are given a length equal to n centimeters. Your task is to convert it to feet and inches so that the number of feet was maximum. The result should be an integer rounded to the closest value containing an integral number of inches.
Note that when you round up, 1 cm rounds up to 0 inches and 2 cm round up to 1 inch.
Input
The only line contains an integer n (1 β€ n β€ 10000).
Output
Print two non-negative space-separated integers a and b, where a is the numbers of feet and b is the number of inches.
Examples
Input
42
Output
1 2
Input
5
Output
0 2
Submitted Solution:
```
import random
import heapq
import math
from collections import Counter
import sys
from fractions import gcd
sys.setrecursionlimit(300000)
from random import randrange
#########################################################################
# O(|m| log(max(m) + max(a)))
def chinese_remainder(m, a):
'''
m - modulos, a - coefficients (arrays)
* m have to be mutually COPRIME *
'''
res, prod = 0, 1
for m_i in m:
prod *= m_i
for m_i, a_i in zip(m, a):
p = prod // m_i
res += a_i * modinv(p, m_i) * p
return res % prod
# O(log(a+b))
def egcd(a, b):
''' Extended Euclidian algorithm. '''
if a == 0:
return (b, 0, 1)
else:
g, y, x = egcd(b % a, a)
return (g, x - (b // a) * y, y)
def modinv(a, m):
''' Finds modular inverse of a modulo m. '''
while a < 0:
a += m
g, x, y = egcd(a, m)
if g != 1:
raise Exception('Modular inverse does not exist')
else:
return x % m
#########################################################################
#########################################################################
def pi(n):
''' Returns two arrays pi1, pi2:
pi1[i] = pi(i) for i <= sqrt(n)
pi2[i] = pi(n // i) for i <= sqrt(n)
'''
lim = int(n**0.5)
while lim*lim <= n: lim += 1
lim -= 1
pi1, pi2 = [0]*(lim + 1), [0]*(lim + 1)
for i in range(1, lim + 1):
pi1[i] = i - 1
pi2[i] = n//i - 1
for i in range(2, lim + 1):
if pi1[i] == pi1[i - 1]:
continue
p = pi1[i - 1]
for j in range(1, min(n // (i * i), lim) + 1):
st = i * j
if st <= lim:
pi2[j] -= pi2[st] - p
else:
pi2[j] -= pi1[n // st] - p
for j in range(lim, min(lim, i * i - 1), -1):
pi1[j] -= pi1[j // i] - p
return pi1, pi2
#########################################################################
#########################################################################
def num_of_divisors(n):
nd = [0] * (n+1)
for i in range(1, n+1):
for j in range(i, n+1, i):
nd[j] += 1
return nd
def sum_of_divisors(n):
sd = [0] * (n+1)
for i in range(1, n+1):
for j in range(i, n+1, i):
sd[j] += i
return sd
#########################################################################
def fast_prime_sieve(n):
"""
Input n>=6, Returns a list of primes, 2 <= p <= n
Taken from: https://stackoverflow.com/questions/2068372/fastest-way-to-list-all-primes-below-n/
"""
n += 1
n, correction = n-n%6+6, 2-(n%6>1)
sieve = [True] * (n//3)
for i in range(1,int(n**0.5)//3+1):
if sieve[i]:
k=3*i+1|1
sieve[ k*k//3 ::2*k] = [False] * ((n//6-k*k//6-1)//k+1)
sieve[k*(k-2*(i&1)+4)//3::2*k] = [False] * ((n//6-k*(k-2*(i&1)+4)//6-1)//k+1)
return [2,3] + [3*i+1|1 for i in range(1,n//3-correction) if sieve[i]]
#########################################################################
def egcd(a, b):
''' Extended Euclidian algorithm. '''
if a == 0:
return (b, 0, 1)
else:
g, y, x = egcd(b % a, a)
# x * a + y * b = g
return (g, x - (b // a) * y, y)
_gcd = lambda a, b: a+b if a==0 or b==0 else gcd(b, a % b)
def gcd(a, b):
if a==0 or b == 0:
return a+b
return gcd(b, a % b)
#########################################################################
''' Modular square root '''
#########################################################################
def legendre_symbol(a, p):
ls = pow(a, (p - 1) // 2, p)
return -1 if ls == p - 1 else ls
def modular_sqrt(a, p):
if legendre_symbol(a, p) != 1:
return 0
elif a == 0:
return 0
elif p == 2:
return 0
elif p % 4 == 3:
return pow(a, (p + 1) // 4, p)
s = p - 1
e = 0
while s % 2 == 0:
s //= 2
e += 1
n = 2
while legendre_symbol(n, p) != -1:
n += 1
x = pow(a, (s + 1) // 2, p)
b = pow(a, s, p)
g = pow(n, s, p)
r = e
while True:
t = b
m = 0
for m in range(r):
if t == 1:
break
t = pow(t, 2, p)
if m == 0:
return x
gs = pow(g, 2 ** (r - m - 1), p)
g = (gs * gs) % p
x = (x * gs) % p
b = (b * g) % p
r = m
#########################################################################
''' Euler"s totient function for a single number '''
#########################################################################
from functools import lru_cache
def factorize(n):
factorization = set()
while n > 1:
for i in range(2, n + 1):
if n % i == 0:
n //= i
factorization.add(i)
break
return factorization
@lru_cache(maxsize=None)
def phi(n):
totient = n
for p in factorize(n):
totient -= totient//p
return totient
#########################################################################
#########################################################################
def max_subarray(a):
max_so_far = max_ending_here = 0
for el in a:
# in case a speed-up is required, change the below max/min to if-else statements.
max_ending_here = max(0, max_ending_here + el)
max_so_far = max(max_so_far, max_ending_here)
return max_so_far
#########################################################################
#########################################################################
def maximum_xor(nums):
answer = 0
for i in range(31, -1, -1):
answer <<= 1
prefixes = {num >> i for num in nums}
answer += any(answer^1^p in prefixes for p in prefixes)
return answer
#########################################################################
'''
Finds longest increasing subsequence in O(n log n) time.
'''
#########################################################################
def lis(a):
p, m, l = [0]*len(a), [0]*(len(a) + 1), 0
for i in range(len(a)):
lo, hi = 1, l
while lo <= hi:
mid = (lo + hi)//2
if a[m[mid]] < a[i]:
lo = mid+1
else:
hi = mid-1
p[i], m[lo] = m[lo-1], i
if lo > l:
l = lo
s, k = [], m[l]
for i in range(l-1, -1, -1):
s.append(a[k])
k = p[k]
return s[::-1]
#########################################################################
#########################################################################
#########################################################################
def find_kth(arr, k, start=0, end=None):
if not end:
end = len(arr) -1
pivot_ridx = randrange(start, end)
pivot = arr[pivot_ridx]
pivot_idx = _partition(arr, start, end, pivot_ridx)
if pivot_idx + 1 == k:
return pivot
elif pivot_idx + 1 > k:
return find_kth(arr, k, start, pivot_idx)
else:
return find_kth(arr, k, pivot_idx, end)
def _partition(arr, start, end, pivot_idx):
pivot = arr[pivot_idx]
arr[end], arr[pivot_idx] = arr[pivot_idx], arr[end]
inc_idx = start
for i in range(start, end):
if arr[i] <= pivot:
arr[inc_idx], arr[i] = arr[i], arr[inc_idx]
inc_idx += 1
arr[end], arr[inc_idx] = arr[inc_idx], arr[end]
return inc_idx
#########################################################################
#########################################################################
def count_sort(a):
mn, mx = float('inf'), -float('inf')
for x in a:
if x < mn: mn = x
if x > mx: mx = x
counter = [0 for _ in range(mx - mn + 1)]
for x in a:
counter[x - mn] += 1
j = 0
for i in range(mx - mn + 1):
a[j:j+counter[i]] = [i + mn]*counter[i]
j += counter[i]
#########################################################################
#########################################################################
def lcs(a, b):
''' Finds longest common subsequence of arrays a, b. '''
lengths = [[0 for j in range(len(b) + 1)] for i in range(len(a) + 1)]
for i, x in enumerate(a):
for j, y in enumerate(b):
if x == y:
lengths[i+1][j+1] = lengths[i][j] + 1
else:
lengths[i+1][j+1] = max(lengths[i+1][j], lengths[i][j+1])
result = []
x, y = len(a), len(b)
while x != 0 and y != 0:
if lengths[x][y] == lengths[x-1][y]:
x -= 1
elif lengths[x][y] == lengths[x][y-1]:
y -= 1
else:
result.append(a[x-1])
x, y = x - 1, y - 1
return result[::-1]
#########################################################################
#########################################################################
def is_anagram(str1, str2):
return Counter(str1) == Counter(str2)
#########################################################################
#########################################################################
def power(x, y, p):
res = 1;
x = x % p;
while (y > 0):
if (y & 1):
res = (res * x) % p;
y = y>>1;
x = (x * x) % p;
return res;
#########################################################################
def miillerTest(d, n):
a = 2 + random.randint(1, n - 4);
x = power(a, d, n);
if (x == 1 or x == n - 1):
return True;
while (d != n - 1):
x = (x * x) % n;
d *= 2;
if (x == 1):
return False;
if (x == n - 1):
return True;
return False;
def isPrime( n, k):
if (n <= 1 or n == 4):
return False;
if (n <= 3):
return True;
d = n - 1;
while (d % 2 == 0):
d //= 2;
for i in range(k):
if (miillerTest(d, n) == False):
return False;
return True;
#########################################################################
#########################################################################
def gcd_list(li):
return reduce(gcd,li)
'''
Finds longest increasing subsequence in O(n log n) time.
'''
#########################################################################
def lis(a):
p, m, l = [0]*len(a), [0]*(len(a) + 1), 0
for i in range(len(a)):
lo, hi = 1, l
while lo <= hi:
mid = (lo + hi)//2
if a[m[mid]] < a[i]:
lo = mid+1
else:
hi = mid-1
p[i], m[lo] = m[lo-1], i
if lo > l:
l = lo
s, k = [], m[l]
for i in range(l-1, -1, -1):
s.append(a[k])
k = p[k]
return s[::-1]
#########################################################################
# for _ in range(int(input())):
# input=open("input.txt","r").readline
n=int(input())
e=n//36
f=n%36-1
print(e,int(math.ceil((f/3))))
``` | instruction | 0 | 104,537 | 8 | 209,074 |
No | output | 1 | 104,537 | 8 | 209,075 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dima's got a staircase that consists of n stairs. The first stair is at height a1, the second one is at a2, the last one is at an (1 β€ a1 β€ a2 β€ ... β€ an).
Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. The i-th box has width wi and height hi. Dima throws each box vertically down on the first wi stairs of the staircase, that is, the box covers stairs with numbers 1, 2, ..., wi. Each thrown box flies vertically down until at least one of the two following events happen:
* the bottom of the box touches the top of a stair;
* the bottom of the box touches the top of a box, thrown earlier.
We only consider touching of the horizontal sides of stairs and boxes, at that touching with the corners isn't taken into consideration. Specifically, that implies that a box with width wi cannot touch the stair number wi + 1.
You are given the description of the staircase and the sequence in which Dima threw the boxes at it. For each box, determine how high the bottom of the box after landing will be. Consider a box to fall after the previous one lands.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of stairs in the staircase. The second line contains a non-decreasing sequence, consisting of n integers, a1, a2, ..., an (1 β€ ai β€ 109; ai β€ ai + 1).
The next line contains integer m (1 β€ m β€ 105) β the number of boxes. Each of the following m lines contains a pair of integers wi, hi (1 β€ wi β€ n; 1 β€ hi β€ 109) β the size of the i-th thrown box.
The numbers in the lines are separated by spaces.
Output
Print m integers β for each box the height, where the bottom of the box will be after landing. Print the answers for the boxes in the order, in which the boxes are given in the input.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 6 6
4
1 1
3 1
1 1
4 3
Output
1
3
4
6
Input
3
1 2 3
2
1 1
3 1
Output
1
3
Input
1
1
5
1 2
1 10
1 10
1 10
1 10
Output
1
3
13
23
33
Note
The first sample are shown on the picture.
<image> | instruction | 0 | 104,751 | 8 | 209,502 |
Tags: data structures, implementation
Correct Solution:
```
_ = input()
a = list(map(int, input().split(' ')))
m = input()
res = list()
for _ in range(int(m)):
w, h = map(int, input().split(' '))
#print(w, h)
cmax = max(a[0], a[w-1])
res.append(cmax)
# for x in range(w):
# a[x] = cmax + h
a[0] = cmax + h
a[w-1] = cmax + h
#print(a)
print('\n'.join(map(str, res)))
``` | output | 1 | 104,751 | 8 | 209,503 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dima's got a staircase that consists of n stairs. The first stair is at height a1, the second one is at a2, the last one is at an (1 β€ a1 β€ a2 β€ ... β€ an).
Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. The i-th box has width wi and height hi. Dima throws each box vertically down on the first wi stairs of the staircase, that is, the box covers stairs with numbers 1, 2, ..., wi. Each thrown box flies vertically down until at least one of the two following events happen:
* the bottom of the box touches the top of a stair;
* the bottom of the box touches the top of a box, thrown earlier.
We only consider touching of the horizontal sides of stairs and boxes, at that touching with the corners isn't taken into consideration. Specifically, that implies that a box with width wi cannot touch the stair number wi + 1.
You are given the description of the staircase and the sequence in which Dima threw the boxes at it. For each box, determine how high the bottom of the box after landing will be. Consider a box to fall after the previous one lands.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of stairs in the staircase. The second line contains a non-decreasing sequence, consisting of n integers, a1, a2, ..., an (1 β€ ai β€ 109; ai β€ ai + 1).
The next line contains integer m (1 β€ m β€ 105) β the number of boxes. Each of the following m lines contains a pair of integers wi, hi (1 β€ wi β€ n; 1 β€ hi β€ 109) β the size of the i-th thrown box.
The numbers in the lines are separated by spaces.
Output
Print m integers β for each box the height, where the bottom of the box will be after landing. Print the answers for the boxes in the order, in which the boxes are given in the input.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 6 6
4
1 1
3 1
1 1
4 3
Output
1
3
4
6
Input
3
1 2 3
2
1 1
3 1
Output
1
3
Input
1
1
5
1 2
1 10
1 10
1 10
1 10
Output
1
3
13
23
33
Note
The first sample are shown on the picture.
<image> | instruction | 0 | 104,752 | 8 | 209,504 |
Tags: data structures, implementation
Correct Solution:
```
n=int(input())
A=[0]+[int(x) for x in input().split()]
m=int(input())
curHeight=0
# print(A)
Answer=[]
for i in range(m):
a,b=map(int, input().split())
curHeight=max(curHeight,A[a])
Answer.append(str(curHeight))
curHeight+=b
print("\n".join(Answer))
``` | output | 1 | 104,752 | 8 | 209,505 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dima's got a staircase that consists of n stairs. The first stair is at height a1, the second one is at a2, the last one is at an (1 β€ a1 β€ a2 β€ ... β€ an).
Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. The i-th box has width wi and height hi. Dima throws each box vertically down on the first wi stairs of the staircase, that is, the box covers stairs with numbers 1, 2, ..., wi. Each thrown box flies vertically down until at least one of the two following events happen:
* the bottom of the box touches the top of a stair;
* the bottom of the box touches the top of a box, thrown earlier.
We only consider touching of the horizontal sides of stairs and boxes, at that touching with the corners isn't taken into consideration. Specifically, that implies that a box with width wi cannot touch the stair number wi + 1.
You are given the description of the staircase and the sequence in which Dima threw the boxes at it. For each box, determine how high the bottom of the box after landing will be. Consider a box to fall after the previous one lands.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of stairs in the staircase. The second line contains a non-decreasing sequence, consisting of n integers, a1, a2, ..., an (1 β€ ai β€ 109; ai β€ ai + 1).
The next line contains integer m (1 β€ m β€ 105) β the number of boxes. Each of the following m lines contains a pair of integers wi, hi (1 β€ wi β€ n; 1 β€ hi β€ 109) β the size of the i-th thrown box.
The numbers in the lines are separated by spaces.
Output
Print m integers β for each box the height, where the bottom of the box will be after landing. Print the answers for the boxes in the order, in which the boxes are given in the input.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 6 6
4
1 1
3 1
1 1
4 3
Output
1
3
4
6
Input
3
1 2 3
2
1 1
3 1
Output
1
3
Input
1
1
5
1 2
1 10
1 10
1 10
1 10
Output
1
3
13
23
33
Note
The first sample are shown on the picture.
<image> | instruction | 0 | 104,753 | 8 | 209,506 |
Tags: data structures, implementation
Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
m = int(input())
ans = 0
res = [0 for i in range(m)]
x = 0
for i in range(m):
w, h = map(int, input().split())
ans = max(a[w - 1], ans + x)
x = h
res[i] = ans
print(*res)
``` | output | 1 | 104,753 | 8 | 209,507 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dima's got a staircase that consists of n stairs. The first stair is at height a1, the second one is at a2, the last one is at an (1 β€ a1 β€ a2 β€ ... β€ an).
Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. The i-th box has width wi and height hi. Dima throws each box vertically down on the first wi stairs of the staircase, that is, the box covers stairs with numbers 1, 2, ..., wi. Each thrown box flies vertically down until at least one of the two following events happen:
* the bottom of the box touches the top of a stair;
* the bottom of the box touches the top of a box, thrown earlier.
We only consider touching of the horizontal sides of stairs and boxes, at that touching with the corners isn't taken into consideration. Specifically, that implies that a box with width wi cannot touch the stair number wi + 1.
You are given the description of the staircase and the sequence in which Dima threw the boxes at it. For each box, determine how high the bottom of the box after landing will be. Consider a box to fall after the previous one lands.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of stairs in the staircase. The second line contains a non-decreasing sequence, consisting of n integers, a1, a2, ..., an (1 β€ ai β€ 109; ai β€ ai + 1).
The next line contains integer m (1 β€ m β€ 105) β the number of boxes. Each of the following m lines contains a pair of integers wi, hi (1 β€ wi β€ n; 1 β€ hi β€ 109) β the size of the i-th thrown box.
The numbers in the lines are separated by spaces.
Output
Print m integers β for each box the height, where the bottom of the box will be after landing. Print the answers for the boxes in the order, in which the boxes are given in the input.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 6 6
4
1 1
3 1
1 1
4 3
Output
1
3
4
6
Input
3
1 2 3
2
1 1
3 1
Output
1
3
Input
1
1
5
1 2
1 10
1 10
1 10
1 10
Output
1
3
13
23
33
Note
The first sample are shown on the picture.
<image> | instruction | 0 | 104,754 | 8 | 209,508 |
Tags: data structures, implementation
Correct Solution:
```
def check(n , a , m , dimensions):
result = list()
prev = 0
for ele in dimensions:
w = ele[0]
h = ele[1]
if w <= prev:
maxHeight = a[prev]
else:
maxHeight = max(a[prev:w])
prev = w - 1
result.append(maxHeight)
a[prev] = maxHeight + h
return result
if __name__ == "__main__":
n = int(input().rstrip())
a = list(map(int , input().rstrip().split()))
m = int(input().rstrip())
dimensions = list()
for i in range(m):
temp = list(map(int , input().rstrip().split()))
w = temp[0]
h = temp[1]
dimensions.append((w , h))
result = check(n , a , m , dimensions)
for i in result:
print (i)
``` | output | 1 | 104,754 | 8 | 209,509 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dima's got a staircase that consists of n stairs. The first stair is at height a1, the second one is at a2, the last one is at an (1 β€ a1 β€ a2 β€ ... β€ an).
Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. The i-th box has width wi and height hi. Dima throws each box vertically down on the first wi stairs of the staircase, that is, the box covers stairs with numbers 1, 2, ..., wi. Each thrown box flies vertically down until at least one of the two following events happen:
* the bottom of the box touches the top of a stair;
* the bottom of the box touches the top of a box, thrown earlier.
We only consider touching of the horizontal sides of stairs and boxes, at that touching with the corners isn't taken into consideration. Specifically, that implies that a box with width wi cannot touch the stair number wi + 1.
You are given the description of the staircase and the sequence in which Dima threw the boxes at it. For each box, determine how high the bottom of the box after landing will be. Consider a box to fall after the previous one lands.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of stairs in the staircase. The second line contains a non-decreasing sequence, consisting of n integers, a1, a2, ..., an (1 β€ ai β€ 109; ai β€ ai + 1).
The next line contains integer m (1 β€ m β€ 105) β the number of boxes. Each of the following m lines contains a pair of integers wi, hi (1 β€ wi β€ n; 1 β€ hi β€ 109) β the size of the i-th thrown box.
The numbers in the lines are separated by spaces.
Output
Print m integers β for each box the height, where the bottom of the box will be after landing. Print the answers for the boxes in the order, in which the boxes are given in the input.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 6 6
4
1 1
3 1
1 1
4 3
Output
1
3
4
6
Input
3
1 2 3
2
1 1
3 1
Output
1
3
Input
1
1
5
1 2
1 10
1 10
1 10
1 10
Output
1
3
13
23
33
Note
The first sample are shown on the picture.
<image> | instruction | 0 | 104,755 | 8 | 209,510 |
Tags: data structures, implementation
Correct Solution:
```
n=int(input())
l=list(map(int,input().split()))
g=l[0]
m=int(input())
S=''
for i in range(m) :
a,b=map(int,input().split())
t=max(l[0],l[a-1])
S+=str(t)+'\n'
l[0]=t+b
l[a-1]=t+b
print(S)
``` | output | 1 | 104,755 | 8 | 209,511 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dima's got a staircase that consists of n stairs. The first stair is at height a1, the second one is at a2, the last one is at an (1 β€ a1 β€ a2 β€ ... β€ an).
Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. The i-th box has width wi and height hi. Dima throws each box vertically down on the first wi stairs of the staircase, that is, the box covers stairs with numbers 1, 2, ..., wi. Each thrown box flies vertically down until at least one of the two following events happen:
* the bottom of the box touches the top of a stair;
* the bottom of the box touches the top of a box, thrown earlier.
We only consider touching of the horizontal sides of stairs and boxes, at that touching with the corners isn't taken into consideration. Specifically, that implies that a box with width wi cannot touch the stair number wi + 1.
You are given the description of the staircase and the sequence in which Dima threw the boxes at it. For each box, determine how high the bottom of the box after landing will be. Consider a box to fall after the previous one lands.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of stairs in the staircase. The second line contains a non-decreasing sequence, consisting of n integers, a1, a2, ..., an (1 β€ ai β€ 109; ai β€ ai + 1).
The next line contains integer m (1 β€ m β€ 105) β the number of boxes. Each of the following m lines contains a pair of integers wi, hi (1 β€ wi β€ n; 1 β€ hi β€ 109) β the size of the i-th thrown box.
The numbers in the lines are separated by spaces.
Output
Print m integers β for each box the height, where the bottom of the box will be after landing. Print the answers for the boxes in the order, in which the boxes are given in the input.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 6 6
4
1 1
3 1
1 1
4 3
Output
1
3
4
6
Input
3
1 2 3
2
1 1
3 1
Output
1
3
Input
1
1
5
1 2
1 10
1 10
1 10
1 10
Output
1
3
13
23
33
Note
The first sample are shown on the picture.
<image> | instruction | 0 | 104,756 | 8 | 209,512 |
Tags: data structures, implementation
Correct Solution:
```
n = int(input())
l = list(map(int,input().split()))
m = int(input())
for i in range(m):
w,h = map(int,input().split())
ans = max(l[0],l[w-1])
print(ans)
l[0]=ans+h
``` | output | 1 | 104,756 | 8 | 209,513 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dima's got a staircase that consists of n stairs. The first stair is at height a1, the second one is at a2, the last one is at an (1 β€ a1 β€ a2 β€ ... β€ an).
Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. The i-th box has width wi and height hi. Dima throws each box vertically down on the first wi stairs of the staircase, that is, the box covers stairs with numbers 1, 2, ..., wi. Each thrown box flies vertically down until at least one of the two following events happen:
* the bottom of the box touches the top of a stair;
* the bottom of the box touches the top of a box, thrown earlier.
We only consider touching of the horizontal sides of stairs and boxes, at that touching with the corners isn't taken into consideration. Specifically, that implies that a box with width wi cannot touch the stair number wi + 1.
You are given the description of the staircase and the sequence in which Dima threw the boxes at it. For each box, determine how high the bottom of the box after landing will be. Consider a box to fall after the previous one lands.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of stairs in the staircase. The second line contains a non-decreasing sequence, consisting of n integers, a1, a2, ..., an (1 β€ ai β€ 109; ai β€ ai + 1).
The next line contains integer m (1 β€ m β€ 105) β the number of boxes. Each of the following m lines contains a pair of integers wi, hi (1 β€ wi β€ n; 1 β€ hi β€ 109) β the size of the i-th thrown box.
The numbers in the lines are separated by spaces.
Output
Print m integers β for each box the height, where the bottom of the box will be after landing. Print the answers for the boxes in the order, in which the boxes are given in the input.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 6 6
4
1 1
3 1
1 1
4 3
Output
1
3
4
6
Input
3
1 2 3
2
1 1
3 1
Output
1
3
Input
1
1
5
1 2
1 10
1 10
1 10
1 10
Output
1
3
13
23
33
Note
The first sample are shown on the picture.
<image> | instruction | 0 | 104,757 | 8 | 209,514 |
Tags: data structures, implementation
Correct Solution:
```
## necessary imports
import sys
input = sys.stdin.readline
from math import log2, log, ceil
# swap_array function
def swaparr(arr, a, b):
temp = arr[a];
arr[a] = arr[b];
arr[b] = temp
## gcd function
def gcd(a, b):
if a == 0:
return b
return gcd(b % a, a)
## nCr function efficient using Binomial Cofficient
def nCr(n, k):
if (k > n - k):
k = n - k
res = 1
for i in range(k):
res = res * (n - i)
res = res / (i + 1)
return res
## upper bound function code -- such that e in a[:i] e < x;
def upper_bound(a, x, lo=0):
hi = len(a)
while lo < hi:
mid = (lo + hi) // 2
if a[mid] < x:
lo = mid + 1
else:
hi = mid
return lo
## prime factorization
def primefs(n):
## if n == 1 ## calculating primes
primes = {}
while (n % 2 == 0):
primes[2] = primes.get(2, 0) + 1
n = n // 2
for i in range(3, int(n ** 0.5) + 2, 2):
while (n % i == 0):
primes[i] = primes.get(i, 0) + 1
n = n // i
if n > 2:
primes[n] = primes.get(n, 0) + 1
## prime factoriazation of n is stored in dictionary
## primes and can be accesed. O(sqrt n)
return primes
## MODULAR EXPONENTIATION FUNCTION
def power(x, y, p):
res = 1
x = x % p
if (x == 0):
return 0
while (y > 0):
if ((y & 1) == 1):
res = (res * x) % p
y = y >> 1
x = (x * x) % p
return res
## DISJOINT SET UNINON FUNCTIONS
def swap(a, b):
temp = a
a = b
b = temp
return a, b
# find function
def find(x, link):
while (x != link[x]):
x = link[x]
return x
# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, link, size):
x = find(x, link)
y = find(y, link)
if size[x] < size[y]:
x, y = swap(x, y)
if x != y:
size[x] += size[y]
link[y] = x
## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n):
prime = [True for i in range(n + 1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n + 1, p):
prime[i] = False
p += 1
return prime
#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e6 + 5)
'''
def spf_sieve():
spf[1] = 1;
for i in range(2, MAXN):
spf[i] = i;
for i in range(4, MAXN, 2):
spf[i] = 2;
for i in range(3, ceil(MAXN ** 0.5), 2):
if spf[i] == i:
for j in range(i * i, MAXN, i):
if spf[j] == j:
spf[j] = i;
## function for storing smallest prime factors (spf) in the array
################## un-comment below 2 lines when using factorization #################
# spf = [0 for i in range(MAXN)]
# spf_sieve()
def factoriazation(x):
ret = {};
while x != 1:
ret[spf[x]] = ret.get(spf[x], 0) + 1;
x = x // spf[x]
return ret
## this function is useful for multiple queries only, o/w use
## primefs function above. complexity O(log n)'''
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def find_parent(parent, i,cnt):
cnt+=1
if parent[i-1] ==i:
return cnt
if parent[i-1] !=i:
find_parent(parent, parent[i-1],cnt)
##########################################################
#for _ in range(int(input())):
# from collections import deque
from collections import Counter
# ls=list(map(int,input().split()))
# for i in range(m):
# for i in range(int(input())):
#n,k= map(int, input().split())
#for _ in range(int(input())):
#n,k= map(int, input().split())
from collections import Counter
import math
#for i in range(int(input())):
#n,k=map(int, input().split())
n=int(input())
arr=list(map(int,input().split()))
m=int(input())
box=[]
for i in range(m):
u,v=map(int,input().split())
box.append((u,v))
ans=0
val=0
for i in range(m):
ans=max(val+ans,arr[box[i][0]-1])
val=box[i][1]
print(ans)
``` | output | 1 | 104,757 | 8 | 209,515 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dima's got a staircase that consists of n stairs. The first stair is at height a1, the second one is at a2, the last one is at an (1 β€ a1 β€ a2 β€ ... β€ an).
Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. The i-th box has width wi and height hi. Dima throws each box vertically down on the first wi stairs of the staircase, that is, the box covers stairs with numbers 1, 2, ..., wi. Each thrown box flies vertically down until at least one of the two following events happen:
* the bottom of the box touches the top of a stair;
* the bottom of the box touches the top of a box, thrown earlier.
We only consider touching of the horizontal sides of stairs and boxes, at that touching with the corners isn't taken into consideration. Specifically, that implies that a box with width wi cannot touch the stair number wi + 1.
You are given the description of the staircase and the sequence in which Dima threw the boxes at it. For each box, determine how high the bottom of the box after landing will be. Consider a box to fall after the previous one lands.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of stairs in the staircase. The second line contains a non-decreasing sequence, consisting of n integers, a1, a2, ..., an (1 β€ ai β€ 109; ai β€ ai + 1).
The next line contains integer m (1 β€ m β€ 105) β the number of boxes. Each of the following m lines contains a pair of integers wi, hi (1 β€ wi β€ n; 1 β€ hi β€ 109) β the size of the i-th thrown box.
The numbers in the lines are separated by spaces.
Output
Print m integers β for each box the height, where the bottom of the box will be after landing. Print the answers for the boxes in the order, in which the boxes are given in the input.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 6 6
4
1 1
3 1
1 1
4 3
Output
1
3
4
6
Input
3
1 2 3
2
1 1
3 1
Output
1
3
Input
1
1
5
1 2
1 10
1 10
1 10
1 10
Output
1
3
13
23
33
Note
The first sample are shown on the picture.
<image> | instruction | 0 | 104,758 | 8 | 209,516 |
Tags: data structures, implementation
Correct Solution:
```
from collections import Counter, defaultdict, OrderedDict, deque
from bisect import bisect_left, bisect_right
from functools import reduce, lru_cache
from typing import List
import itertools
import sys
import math
import heapq
import string
import random
MIN, MAX, MOD = -0x3f3f3f3f, 0x3f3f3f3f, 1000000007
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())
n = N()
a = RLL()
m = N()
hit = 0
for _ in range(m):
w, h = RL()
res = max(a[w - 1], hit)
print(res)
hit = max(hit, res + h)
# mx = max(a[:w])
# print(mx)
# for i in range(w):
# a[i] = mx + h
``` | output | 1 | 104,758 | 8 | 209,517 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dima's got a staircase that consists of n stairs. The first stair is at height a1, the second one is at a2, the last one is at an (1 β€ a1 β€ a2 β€ ... β€ an).
Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. The i-th box has width wi and height hi. Dima throws each box vertically down on the first wi stairs of the staircase, that is, the box covers stairs with numbers 1, 2, ..., wi. Each thrown box flies vertically down until at least one of the two following events happen:
* the bottom of the box touches the top of a stair;
* the bottom of the box touches the top of a box, thrown earlier.
We only consider touching of the horizontal sides of stairs and boxes, at that touching with the corners isn't taken into consideration. Specifically, that implies that a box with width wi cannot touch the stair number wi + 1.
You are given the description of the staircase and the sequence in which Dima threw the boxes at it. For each box, determine how high the bottom of the box after landing will be. Consider a box to fall after the previous one lands.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of stairs in the staircase. The second line contains a non-decreasing sequence, consisting of n integers, a1, a2, ..., an (1 β€ ai β€ 109; ai β€ ai + 1).
The next line contains integer m (1 β€ m β€ 105) β the number of boxes. Each of the following m lines contains a pair of integers wi, hi (1 β€ wi β€ n; 1 β€ hi β€ 109) β the size of the i-th thrown box.
The numbers in the lines are separated by spaces.
Output
Print m integers β for each box the height, where the bottom of the box will be after landing. Print the answers for the boxes in the order, in which the boxes are given in the input.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 6 6
4
1 1
3 1
1 1
4 3
Output
1
3
4
6
Input
3
1 2 3
2
1 1
3 1
Output
1
3
Input
1
1
5
1 2
1 10
1 10
1 10
1 10
Output
1
3
13
23
33
Note
The first sample are shown on the picture.
<image>
Submitted Solution:
```
stair_number = int(input())
stair_heights = list(map(int,input().split()))
box_number = int(input())
x = stair_heights[0]
for _ in range(box_number) :
w , h = input().split()
w ,h = int(w) , int(h)
if x >= stair_heights[w-1] :
k = x
else :
k = stair_heights[w-1]
print(k)
x = k + h
``` | instruction | 0 | 104,759 | 8 | 209,518 |
Yes | output | 1 | 104,759 | 8 | 209,519 |
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