message stringlengths 2 43.5k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 853 107k | cluster float64 24 24 | __index_level_0__ int64 1.71k 214k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for k burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of r burles (1 ≤ r ≤ 9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of r burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.
Input
The single line of input contains two integers k and r (1 ≤ k ≤ 1000, 1 ≤ r ≤ 9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.
Output
Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.
Examples
Input
117 3
Output
9
Input
237 7
Output
1
Input
15 2
Output
2
Note
In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change. | instruction | 0 | 23,373 | 24 | 46,746 |
Tags: brute force, constructive algorithms, implementation, math
Correct Solution:
```
n, g = map(int, input().split(" "))
s = n
f = n % 10
count = 1
while f != 0 and f != g:
s += n
f = s % 10
count += 1
print(count)
``` | output | 1 | 23,373 | 24 | 46,747 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for k burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of r burles (1 ≤ r ≤ 9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of r burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.
Input
The single line of input contains two integers k and r (1 ≤ k ≤ 1000, 1 ≤ r ≤ 9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.
Output
Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.
Examples
Input
117 3
Output
9
Input
237 7
Output
1
Input
15 2
Output
2
Note
In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change.
Submitted Solution:
```
import sys
input = lambda:sys.stdin.readline().strip()
k,r = map(int,input().split())
i = 1
while (i<10):
if (k*i)%10 == r or (k*i)%10==0:
print(i)
exit()
else:
i+=1
print(10)
``` | instruction | 0 | 23,374 | 24 | 46,748 |
Yes | output | 1 | 23,374 | 24 | 46,749 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for k burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of r burles (1 ≤ r ≤ 9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of r burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.
Input
The single line of input contains two integers k and r (1 ≤ k ≤ 1000, 1 ≤ r ≤ 9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.
Output
Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.
Examples
Input
117 3
Output
9
Input
237 7
Output
1
Input
15 2
Output
2
Note
In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change.
Submitted Solution:
```
k, r = map(int, input().split())
kol = 0
for i in range(1, 11):
sum_ten = (k * i) % 10
if sum_ten == 0 or sum_ten == r:
kol = i
break
print(kol)
``` | instruction | 0 | 23,375 | 24 | 46,750 |
Yes | output | 1 | 23,375 | 24 | 46,751 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for k burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of r burles (1 ≤ r ≤ 9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of r burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.
Input
The single line of input contains two integers k and r (1 ≤ k ≤ 1000, 1 ≤ r ≤ 9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.
Output
Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.
Examples
Input
117 3
Output
9
Input
237 7
Output
1
Input
15 2
Output
2
Note
In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change.
Submitted Solution:
```
k,r = map(int,input().split())
l = 1
cena = k
"""if k%10 == 0:
print(int(l))"""
while True:
if k%10 == r or k%10 == 0:
break
l+= 1
k = l * cena
print(l)
``` | instruction | 0 | 23,376 | 24 | 46,752 |
Yes | output | 1 | 23,376 | 24 | 46,753 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for k burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of r burles (1 ≤ r ≤ 9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of r burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.
Input
The single line of input contains two integers k and r (1 ≤ k ≤ 1000, 1 ≤ r ≤ 9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.
Output
Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.
Examples
Input
117 3
Output
9
Input
237 7
Output
1
Input
15 2
Output
2
Note
In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change.
Submitted Solution:
```
k, r = map(int, input().split())
sum, c = k, 1
while(sum % 10 != r and sum % 10 != 0):
sum += k
c += 1
print(c)
``` | instruction | 0 | 23,377 | 24 | 46,754 |
Yes | output | 1 | 23,377 | 24 | 46,755 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for k burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of r burles (1 ≤ r ≤ 9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of r burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.
Input
The single line of input contains two integers k and r (1 ≤ k ≤ 1000, 1 ≤ r ≤ 9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.
Output
Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.
Examples
Input
117 3
Output
9
Input
237 7
Output
1
Input
15 2
Output
2
Note
In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change.
Submitted Solution:
```
print('Цена одной лопаты и номинал монеты')
k, r = map(int, input().split())
while (1 <= k <= 1000 and 1 <= r <= 9) == False:
k, r = map(int, input().split())
i = 0
while True:
i += 1
if (k*i-r) % 10 == 0:
break
print(i)
``` | instruction | 0 | 23,378 | 24 | 46,756 |
No | output | 1 | 23,378 | 24 | 46,757 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for k burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of r burles (1 ≤ r ≤ 9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of r burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.
Input
The single line of input contains two integers k and r (1 ≤ k ≤ 1000, 1 ≤ r ≤ 9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.
Output
Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.
Examples
Input
117 3
Output
9
Input
237 7
Output
1
Input
15 2
Output
2
Note
In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change.
Submitted Solution:
```
k,r = list(map(int, input().split()))
cont = 1
while k*cont % 10 > r:
cont += 1
print(cont)
``` | instruction | 0 | 23,379 | 24 | 46,758 |
No | output | 1 | 23,379 | 24 | 46,759 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for k burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of r burles (1 ≤ r ≤ 9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of r burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.
Input
The single line of input contains two integers k and r (1 ≤ k ≤ 1000, 1 ≤ r ≤ 9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.
Output
Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.
Examples
Input
117 3
Output
9
Input
237 7
Output
1
Input
15 2
Output
2
Note
In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change.
Submitted Solution:
```
a,b=map(int,input().split())
n=1
while True:
a=a*n
if (a-b)%10==0 or a%10==0:
print(n)
break
n=n+1
``` | instruction | 0 | 23,380 | 24 | 46,760 |
No | output | 1 | 23,380 | 24 | 46,761 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for k burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of r burles (1 ≤ r ≤ 9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of r burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.
Input
The single line of input contains two integers k and r (1 ≤ k ≤ 1000, 1 ≤ r ≤ 9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.
Output
Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.
Examples
Input
117 3
Output
9
Input
237 7
Output
1
Input
15 2
Output
2
Note
In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change.
Submitted Solution:
```
#!/bin/python3
import math
import os
import random
import re
import sys
a= input().split()
n= int(a[0])
m= int(a[1])
i=1
d=1
b= n*i
while d!=0:
i=i+1
b=n*i
e=b%10
c=(b-m)%10
d=min(e,c)
print(i)
``` | instruction | 0 | 23,381 | 24 | 46,762 |
No | output | 1 | 23,381 | 24 | 46,763 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp plans to conduct a load testing of its new project Fakebook. He already agreed with his friends that at certain points in time they will send requests to Fakebook. The load testing will last n minutes and in the i-th minute friends will send ai requests.
Polycarp plans to test Fakebook under a special kind of load. In case the information about Fakebook gets into the mass media, Polycarp hopes for a monotone increase of the load, followed by a monotone decrease of the interest to the service. Polycarp wants to test this form of load.
Your task is to determine how many requests Polycarp must add so that before some moment the load on the server strictly increases and after that moment strictly decreases. Both the increasing part and the decreasing part can be empty (i. e. absent). The decrease should immediately follow the increase. In particular, the load with two equal neigbouring values is unacceptable.
For example, if the load is described with one of the arrays [1, 2, 8, 4, 3], [1, 3, 5] or [10], then such load satisfies Polycarp (in each of the cases there is an increasing part, immediately followed with a decreasing part). If the load is described with one of the arrays [1, 2, 2, 1], [2, 1, 2] or [10, 10], then such load does not satisfy Polycarp.
Help Polycarp to make the minimum number of additional requests, so that the resulting load satisfies Polycarp. He can make any number of additional requests at any minute from 1 to n.
Input
The first line contains a single integer n (1 ≤ n ≤ 100 000) — the duration of the load testing.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109), where ai is the number of requests from friends in the i-th minute of the load testing.
Output
Print the minimum number of additional requests from Polycarp that would make the load strictly increasing in the beginning and then strictly decreasing afterwards.
Examples
Input
5
1 4 3 2 5
Output
6
Input
5
1 2 2 2 1
Output
1
Input
7
10 20 40 50 70 90 30
Output
0
Note
In the first example Polycarp must make two additional requests in the third minute and four additional requests in the fourth minute. So the resulting load will look like: [1, 4, 5, 6, 5]. In total, Polycarp will make 6 additional requests.
In the second example it is enough to make one additional request in the third minute, so the answer is 1.
In the third example the load already satisfies all conditions described in the statement, so the answer is 0. | instruction | 0 | 23,446 | 24 | 46,892 |
Tags: greedy
Correct Solution:
```
n = int(input())
arr = list(map(int, input().split()))
forward = arr[:]
cnt = arr[0] + 1
for i in range(1, n):
forward[i] = max(cnt, arr[i])
cnt = max(cnt, arr[i]) + 1
backward = arr[:]
cnt = arr[n - 1] + 1
for i in range(n - 2, -1, -1):
backward[i] = max(cnt, arr[i])
cnt = max(cnt, arr[i]) + 1
ans = 0
for i in range(n):
if forward[i] < backward[i]:
ans += forward[i] - arr[i]
else:
if abs(backward[i] - forward[i]) == 1:
ans += 1
ans += backward[i] - arr[i]
print(ans)
``` | output | 1 | 23,446 | 24 | 46,893 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp plans to conduct a load testing of its new project Fakebook. He already agreed with his friends that at certain points in time they will send requests to Fakebook. The load testing will last n minutes and in the i-th minute friends will send ai requests.
Polycarp plans to test Fakebook under a special kind of load. In case the information about Fakebook gets into the mass media, Polycarp hopes for a monotone increase of the load, followed by a monotone decrease of the interest to the service. Polycarp wants to test this form of load.
Your task is to determine how many requests Polycarp must add so that before some moment the load on the server strictly increases and after that moment strictly decreases. Both the increasing part and the decreasing part can be empty (i. e. absent). The decrease should immediately follow the increase. In particular, the load with two equal neigbouring values is unacceptable.
For example, if the load is described with one of the arrays [1, 2, 8, 4, 3], [1, 3, 5] or [10], then such load satisfies Polycarp (in each of the cases there is an increasing part, immediately followed with a decreasing part). If the load is described with one of the arrays [1, 2, 2, 1], [2, 1, 2] or [10, 10], then such load does not satisfy Polycarp.
Help Polycarp to make the minimum number of additional requests, so that the resulting load satisfies Polycarp. He can make any number of additional requests at any minute from 1 to n.
Input
The first line contains a single integer n (1 ≤ n ≤ 100 000) — the duration of the load testing.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109), where ai is the number of requests from friends in the i-th minute of the load testing.
Output
Print the minimum number of additional requests from Polycarp that would make the load strictly increasing in the beginning and then strictly decreasing afterwards.
Examples
Input
5
1 4 3 2 5
Output
6
Input
5
1 2 2 2 1
Output
1
Input
7
10 20 40 50 70 90 30
Output
0
Note
In the first example Polycarp must make two additional requests in the third minute and four additional requests in the fourth minute. So the resulting load will look like: [1, 4, 5, 6, 5]. In total, Polycarp will make 6 additional requests.
In the second example it is enough to make one additional request in the third minute, so the answer is 1.
In the third example the load already satisfies all conditions described in the statement, so the answer is 0. | instruction | 0 | 23,447 | 24 | 46,894 |
Tags: greedy
Correct Solution:
```
n = int(input())
a = list(map(int,input().strip().split()))
if n > 1:
li = [0]*n
ri = [0]*n
lm = a[0]
c = [0]*n
b = [0]*n
b[0] = a[0]
for i in range(1,n):
if lm >= a[i]:
li[i] = li[i-1] + (lm+1-a[i])
lm = lm+1
else:
li[i] = li[i-1]
lm = a[i]
b[i] = lm
lm = a[n-1]
c[n-1] = a[n-1]
for i in range(n-2,-1,-1):
if lm >= a[i]:
ri[i] = ri[i+1] + (lm+1-a[i])
lm = lm+1
else:
ri[i] = ri[i+1]
lm = a[i]
c[i] = lm
ans = 1<<64
for i in range(n):
if i == 0:
ans = min(ans,ri[i],ri[i+1])
elif i== n-1:
ans = min(ans,li[i],li[i-1])
else:
v1 = li[i] + ri[i+1]
if b[i] == c[i+1]:
v1 += 1
v2 = ri[i] + li[i-1]
if c[i] == b[i-1]:
v2 +=1
val = min(v1,v2)
ans = min(ans,val)
print(ans)
else:
print(0)
``` | output | 1 | 23,447 | 24 | 46,895 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp plans to conduct a load testing of its new project Fakebook. He already agreed with his friends that at certain points in time they will send requests to Fakebook. The load testing will last n minutes and in the i-th minute friends will send ai requests.
Polycarp plans to test Fakebook under a special kind of load. In case the information about Fakebook gets into the mass media, Polycarp hopes for a monotone increase of the load, followed by a monotone decrease of the interest to the service. Polycarp wants to test this form of load.
Your task is to determine how many requests Polycarp must add so that before some moment the load on the server strictly increases and after that moment strictly decreases. Both the increasing part and the decreasing part can be empty (i. e. absent). The decrease should immediately follow the increase. In particular, the load with two equal neigbouring values is unacceptable.
For example, if the load is described with one of the arrays [1, 2, 8, 4, 3], [1, 3, 5] or [10], then such load satisfies Polycarp (in each of the cases there is an increasing part, immediately followed with a decreasing part). If the load is described with one of the arrays [1, 2, 2, 1], [2, 1, 2] or [10, 10], then such load does not satisfy Polycarp.
Help Polycarp to make the minimum number of additional requests, so that the resulting load satisfies Polycarp. He can make any number of additional requests at any minute from 1 to n.
Input
The first line contains a single integer n (1 ≤ n ≤ 100 000) — the duration of the load testing.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109), where ai is the number of requests from friends in the i-th minute of the load testing.
Output
Print the minimum number of additional requests from Polycarp that would make the load strictly increasing in the beginning and then strictly decreasing afterwards.
Examples
Input
5
1 4 3 2 5
Output
6
Input
5
1 2 2 2 1
Output
1
Input
7
10 20 40 50 70 90 30
Output
0
Note
In the first example Polycarp must make two additional requests in the third minute and four additional requests in the fourth minute. So the resulting load will look like: [1, 4, 5, 6, 5]. In total, Polycarp will make 6 additional requests.
In the second example it is enough to make one additional request in the third minute, so the answer is 1.
In the third example the load already satisfies all conditions described in the statement, so the answer is 0. | instruction | 0 | 23,448 | 24 | 46,896 |
Tags: greedy
Correct Solution:
```
n = int(input())
def get_cost(arr):
cost = [0,0]
last = arr[0]
for i in range(1,len(arr)):
cost.append(cost[-1] + max(0,last+1-arr[i]))
last = max(last+1,arr[i])
return cost
arr = list(map(int,input().split()))
left_cost, right_cost = get_cost(arr), get_cost(list(reversed(arr)))
ans = min(left_cost[n], right_cost[n])
for i in range(1,n-1):
ans = min(ans,max(left_cost[i]+right_cost[n-i],left_cost[i+1]+right_cost[n-i-1]))
print(ans)
``` | output | 1 | 23,448 | 24 | 46,897 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp plans to conduct a load testing of its new project Fakebook. He already agreed with his friends that at certain points in time they will send requests to Fakebook. The load testing will last n minutes and in the i-th minute friends will send ai requests.
Polycarp plans to test Fakebook under a special kind of load. In case the information about Fakebook gets into the mass media, Polycarp hopes for a monotone increase of the load, followed by a monotone decrease of the interest to the service. Polycarp wants to test this form of load.
Your task is to determine how many requests Polycarp must add so that before some moment the load on the server strictly increases and after that moment strictly decreases. Both the increasing part and the decreasing part can be empty (i. e. absent). The decrease should immediately follow the increase. In particular, the load with two equal neigbouring values is unacceptable.
For example, if the load is described with one of the arrays [1, 2, 8, 4, 3], [1, 3, 5] or [10], then such load satisfies Polycarp (in each of the cases there is an increasing part, immediately followed with a decreasing part). If the load is described with one of the arrays [1, 2, 2, 1], [2, 1, 2] or [10, 10], then such load does not satisfy Polycarp.
Help Polycarp to make the minimum number of additional requests, so that the resulting load satisfies Polycarp. He can make any number of additional requests at any minute from 1 to n.
Input
The first line contains a single integer n (1 ≤ n ≤ 100 000) — the duration of the load testing.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109), where ai is the number of requests from friends in the i-th minute of the load testing.
Output
Print the minimum number of additional requests from Polycarp that would make the load strictly increasing in the beginning and then strictly decreasing afterwards.
Examples
Input
5
1 4 3 2 5
Output
6
Input
5
1 2 2 2 1
Output
1
Input
7
10 20 40 50 70 90 30
Output
0
Note
In the first example Polycarp must make two additional requests in the third minute and four additional requests in the fourth minute. So the resulting load will look like: [1, 4, 5, 6, 5]. In total, Polycarp will make 6 additional requests.
In the second example it is enough to make one additional request in the third minute, so the answer is 1.
In the third example the load already satisfies all conditions described in the statement, so the answer is 0. | instruction | 0 | 23,449 | 24 | 46,898 |
Tags: greedy
Correct Solution:
```
n = int(input())
arr = list(map(int, input().split()))
starts = [0 for _ in range(n)]
ends = [0 for _ in range(n)]
starts[0] = arr[0]
ends[-1] = arr[-1]
for i in range(1, n):
starts[i] = max(arr[i], starts[i - 1] + 1)
ends[-i - 1] = max(arr[-i - 1], ends[-i] + 1)
sts = starts[:]
eds = ends[:]
for i in range(n):
starts[i] -= arr[i]
ends[-i - 1] -= arr[-i - 1]
for i in range(1, n):
starts[i] += starts[i - 1]
ends[-i - 1] += ends[-i]
bst = 10**30
for i in range(n):
score = max(sts[i], eds[i]) - arr[i]
if i > 0:
score += starts[i - 1]
if i < n - 1:
score += ends[i + 1]
bst = min(bst, score)
print(bst)
``` | output | 1 | 23,449 | 24 | 46,899 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp plans to conduct a load testing of its new project Fakebook. He already agreed with his friends that at certain points in time they will send requests to Fakebook. The load testing will last n minutes and in the i-th minute friends will send ai requests.
Polycarp plans to test Fakebook under a special kind of load. In case the information about Fakebook gets into the mass media, Polycarp hopes for a monotone increase of the load, followed by a monotone decrease of the interest to the service. Polycarp wants to test this form of load.
Your task is to determine how many requests Polycarp must add so that before some moment the load on the server strictly increases and after that moment strictly decreases. Both the increasing part and the decreasing part can be empty (i. e. absent). The decrease should immediately follow the increase. In particular, the load with two equal neigbouring values is unacceptable.
For example, if the load is described with one of the arrays [1, 2, 8, 4, 3], [1, 3, 5] or [10], then such load satisfies Polycarp (in each of the cases there is an increasing part, immediately followed with a decreasing part). If the load is described with one of the arrays [1, 2, 2, 1], [2, 1, 2] or [10, 10], then such load does not satisfy Polycarp.
Help Polycarp to make the minimum number of additional requests, so that the resulting load satisfies Polycarp. He can make any number of additional requests at any minute from 1 to n.
Input
The first line contains a single integer n (1 ≤ n ≤ 100 000) — the duration of the load testing.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109), where ai is the number of requests from friends in the i-th minute of the load testing.
Output
Print the minimum number of additional requests from Polycarp that would make the load strictly increasing in the beginning and then strictly decreasing afterwards.
Examples
Input
5
1 4 3 2 5
Output
6
Input
5
1 2 2 2 1
Output
1
Input
7
10 20 40 50 70 90 30
Output
0
Note
In the first example Polycarp must make two additional requests in the third minute and four additional requests in the fourth minute. So the resulting load will look like: [1, 4, 5, 6, 5]. In total, Polycarp will make 6 additional requests.
In the second example it is enough to make one additional request in the third minute, so the answer is 1.
In the third example the load already satisfies all conditions described in the statement, so the answer is 0. | instruction | 0 | 23,450 | 24 | 46,900 |
Tags: greedy
Correct Solution:
```
n = int(input())
a = [0 for i in range(n)]
b = [0 for i in range(n)]
f = [0 for i in range(n)]
q = [0 for i in range(n)]
d = [int(s) for s in input().split()]
last = d[0]
for i in range(1,n):
a[i] = a[i-1]
if d[i] <= last:
a[i] += abs(d[i] - last) + 1
last += 1
else:
last = d[i]
f[i] = last
last = d[n-1]
for i in range(n-2,-1,-1):
b[i] = b[i+1]
if d[i] <= last:
b[i] += abs(d[i] - last) + 1
last +=1
else:
last = d[i]
q[i] = last
ans = float('inf')
for i in range(n-1):
ans = min(ans, a[i] + b[i+1] + int(f[i]==q[i+1]))
print(min(ans,b[0],a[n-1]))
``` | output | 1 | 23,450 | 24 | 46,901 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp plans to conduct a load testing of its new project Fakebook. He already agreed with his friends that at certain points in time they will send requests to Fakebook. The load testing will last n minutes and in the i-th minute friends will send ai requests.
Polycarp plans to test Fakebook under a special kind of load. In case the information about Fakebook gets into the mass media, Polycarp hopes for a monotone increase of the load, followed by a monotone decrease of the interest to the service. Polycarp wants to test this form of load.
Your task is to determine how many requests Polycarp must add so that before some moment the load on the server strictly increases and after that moment strictly decreases. Both the increasing part and the decreasing part can be empty (i. e. absent). The decrease should immediately follow the increase. In particular, the load with two equal neigbouring values is unacceptable.
For example, if the load is described with one of the arrays [1, 2, 8, 4, 3], [1, 3, 5] or [10], then such load satisfies Polycarp (in each of the cases there is an increasing part, immediately followed with a decreasing part). If the load is described with one of the arrays [1, 2, 2, 1], [2, 1, 2] or [10, 10], then such load does not satisfy Polycarp.
Help Polycarp to make the minimum number of additional requests, so that the resulting load satisfies Polycarp. He can make any number of additional requests at any minute from 1 to n.
Input
The first line contains a single integer n (1 ≤ n ≤ 100 000) — the duration of the load testing.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109), where ai is the number of requests from friends in the i-th minute of the load testing.
Output
Print the minimum number of additional requests from Polycarp that would make the load strictly increasing in the beginning and then strictly decreasing afterwards.
Examples
Input
5
1 4 3 2 5
Output
6
Input
5
1 2 2 2 1
Output
1
Input
7
10 20 40 50 70 90 30
Output
0
Note
In the first example Polycarp must make two additional requests in the third minute and four additional requests in the fourth minute. So the resulting load will look like: [1, 4, 5, 6, 5]. In total, Polycarp will make 6 additional requests.
In the second example it is enough to make one additional request in the third minute, so the answer is 1.
In the third example the load already satisfies all conditions described in the statement, so the answer is 0. | instruction | 0 | 23,451 | 24 | 46,902 |
Tags: greedy
Correct Solution:
```
import math
n = int(input())
a = list(map(int, input().split()))
mx = 0
g = [0]*n
r = [0]*n
t1 = [0]*n
t2 = [0]*n
for i in range(n):
g[i] = max(0, mx-a[i]+1)
mx = a[i] + g[i]
t1[i] = mx
if i > 0:
g[i] += g[i-1]
mx = 0
for i in range(n-1, -1, -1):
r[i] = max(0, mx-a[i]+1)
mx = a[i] + r[i]
t2[i] = mx
if i < n-1:
r[i] += r[i+1]
ans = 10**18
for i in range(n):
sum = max(t1[i], t2[i]) - a[i];
if i > 0:
sum += g[i-1]
if i < n-1:
sum += r[i+1]
ans = min(ans, sum)
print(ans)
``` | output | 1 | 23,451 | 24 | 46,903 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp plans to conduct a load testing of its new project Fakebook. He already agreed with his friends that at certain points in time they will send requests to Fakebook. The load testing will last n minutes and in the i-th minute friends will send ai requests.
Polycarp plans to test Fakebook under a special kind of load. In case the information about Fakebook gets into the mass media, Polycarp hopes for a monotone increase of the load, followed by a monotone decrease of the interest to the service. Polycarp wants to test this form of load.
Your task is to determine how many requests Polycarp must add so that before some moment the load on the server strictly increases and after that moment strictly decreases. Both the increasing part and the decreasing part can be empty (i. e. absent). The decrease should immediately follow the increase. In particular, the load with two equal neigbouring values is unacceptable.
For example, if the load is described with one of the arrays [1, 2, 8, 4, 3], [1, 3, 5] or [10], then such load satisfies Polycarp (in each of the cases there is an increasing part, immediately followed with a decreasing part). If the load is described with one of the arrays [1, 2, 2, 1], [2, 1, 2] or [10, 10], then such load does not satisfy Polycarp.
Help Polycarp to make the minimum number of additional requests, so that the resulting load satisfies Polycarp. He can make any number of additional requests at any minute from 1 to n.
Input
The first line contains a single integer n (1 ≤ n ≤ 100 000) — the duration of the load testing.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109), where ai is the number of requests from friends in the i-th minute of the load testing.
Output
Print the minimum number of additional requests from Polycarp that would make the load strictly increasing in the beginning and then strictly decreasing afterwards.
Examples
Input
5
1 4 3 2 5
Output
6
Input
5
1 2 2 2 1
Output
1
Input
7
10 20 40 50 70 90 30
Output
0
Note
In the first example Polycarp must make two additional requests in the third minute and four additional requests in the fourth minute. So the resulting load will look like: [1, 4, 5, 6, 5]. In total, Polycarp will make 6 additional requests.
In the second example it is enough to make one additional request in the third minute, so the answer is 1.
In the third example the load already satisfies all conditions described in the statement, so the answer is 0. | instruction | 0 | 23,452 | 24 | 46,904 |
Tags: greedy
Correct Solution:
```
n=int(input())
p=list(map(int,input().split()))
a=[0]*n
t=s=f=0
for i in range(n):
if p[i]<=t:a[i]=t-p[i]+1
t=max(p[i],t+1)
for i in range(n-1,0,-1):
if p[i] <= s: a[i] = min(s - p[i] + 1,a[i])
else:a[i]=0
s = max(p[i], s + 1)
for i in range(n):p[i]+=a[i];f|=i>1and p[i]==p[i-1]
print(sum(a[:])+f)
``` | output | 1 | 23,452 | 24 | 46,905 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp plans to conduct a load testing of its new project Fakebook. He already agreed with his friends that at certain points in time they will send requests to Fakebook. The load testing will last n minutes and in the i-th minute friends will send ai requests.
Polycarp plans to test Fakebook under a special kind of load. In case the information about Fakebook gets into the mass media, Polycarp hopes for a monotone increase of the load, followed by a monotone decrease of the interest to the service. Polycarp wants to test this form of load.
Your task is to determine how many requests Polycarp must add so that before some moment the load on the server strictly increases and after that moment strictly decreases. Both the increasing part and the decreasing part can be empty (i. e. absent). The decrease should immediately follow the increase. In particular, the load with two equal neigbouring values is unacceptable.
For example, if the load is described with one of the arrays [1, 2, 8, 4, 3], [1, 3, 5] or [10], then such load satisfies Polycarp (in each of the cases there is an increasing part, immediately followed with a decreasing part). If the load is described with one of the arrays [1, 2, 2, 1], [2, 1, 2] or [10, 10], then such load does not satisfy Polycarp.
Help Polycarp to make the minimum number of additional requests, so that the resulting load satisfies Polycarp. He can make any number of additional requests at any minute from 1 to n.
Input
The first line contains a single integer n (1 ≤ n ≤ 100 000) — the duration of the load testing.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109), where ai is the number of requests from friends in the i-th minute of the load testing.
Output
Print the minimum number of additional requests from Polycarp that would make the load strictly increasing in the beginning and then strictly decreasing afterwards.
Examples
Input
5
1 4 3 2 5
Output
6
Input
5
1 2 2 2 1
Output
1
Input
7
10 20 40 50 70 90 30
Output
0
Note
In the first example Polycarp must make two additional requests in the third minute and four additional requests in the fourth minute. So the resulting load will look like: [1, 4, 5, 6, 5]. In total, Polycarp will make 6 additional requests.
In the second example it is enough to make one additional request in the third minute, so the answer is 1.
In the third example the load already satisfies all conditions described in the statement, so the answer is 0. | instruction | 0 | 23,453 | 24 | 46,906 |
Tags: greedy
Correct Solution:
```
s, l, r = 0, 0, int(input()) - 1
t = list(map(int, input().split()))
while 1:
while l < r and t[l] < t[l + 1]: l += 1
while l < r and t[r] < t[r - 1]: r -= 1
if l == r: break
if t[l] < t[r]:
s += t[l] - t[l + 1] + 1
t[l + 1] = t[l] + 1
else:
s += t[r] - t[r - 1] + 1
t[r - 1] = t[r] + 1
print(s)
``` | output | 1 | 23,453 | 24 | 46,907 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp plans to conduct a load testing of its new project Fakebook. He already agreed with his friends that at certain points in time they will send requests to Fakebook. The load testing will last n minutes and in the i-th minute friends will send ai requests.
Polycarp plans to test Fakebook under a special kind of load. In case the information about Fakebook gets into the mass media, Polycarp hopes for a monotone increase of the load, followed by a monotone decrease of the interest to the service. Polycarp wants to test this form of load.
Your task is to determine how many requests Polycarp must add so that before some moment the load on the server strictly increases and after that moment strictly decreases. Both the increasing part and the decreasing part can be empty (i. e. absent). The decrease should immediately follow the increase. In particular, the load with two equal neigbouring values is unacceptable.
For example, if the load is described with one of the arrays [1, 2, 8, 4, 3], [1, 3, 5] or [10], then such load satisfies Polycarp (in each of the cases there is an increasing part, immediately followed with a decreasing part). If the load is described with one of the arrays [1, 2, 2, 1], [2, 1, 2] or [10, 10], then such load does not satisfy Polycarp.
Help Polycarp to make the minimum number of additional requests, so that the resulting load satisfies Polycarp. He can make any number of additional requests at any minute from 1 to n.
Input
The first line contains a single integer n (1 ≤ n ≤ 100 000) — the duration of the load testing.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109), where ai is the number of requests from friends in the i-th minute of the load testing.
Output
Print the minimum number of additional requests from Polycarp that would make the load strictly increasing in the beginning and then strictly decreasing afterwards.
Examples
Input
5
1 4 3 2 5
Output
6
Input
5
1 2 2 2 1
Output
1
Input
7
10 20 40 50 70 90 30
Output
0
Note
In the first example Polycarp must make two additional requests in the third minute and four additional requests in the fourth minute. So the resulting load will look like: [1, 4, 5, 6, 5]. In total, Polycarp will make 6 additional requests.
In the second example it is enough to make one additional request in the third minute, so the answer is 1.
In the third example the load already satisfies all conditions described in the statement, so the answer is 0.
Submitted Solution:
```
n = int(input())
t = list(map(int, input().split()))
s = a = b = 0
for i in range(n):
a = max(a, t[i] - i)
b = max(b, t[-1 - i] - i)
x = (b - a + n) // 2 + 1
y = 0
for i in range(x):
s += max(y - t[i], 0)
y = max(t[i], y) + 1
y = 0
for i in range(n - x):
s += max(y - t[-1 - i], 0)
y = max(t[-1 - i], y) + 1
print(s)
``` | instruction | 0 | 23,454 | 24 | 46,908 |
Yes | output | 1 | 23,454 | 24 | 46,909 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp plans to conduct a load testing of its new project Fakebook. He already agreed with his friends that at certain points in time they will send requests to Fakebook. The load testing will last n minutes and in the i-th minute friends will send ai requests.
Polycarp plans to test Fakebook under a special kind of load. In case the information about Fakebook gets into the mass media, Polycarp hopes for a monotone increase of the load, followed by a monotone decrease of the interest to the service. Polycarp wants to test this form of load.
Your task is to determine how many requests Polycarp must add so that before some moment the load on the server strictly increases and after that moment strictly decreases. Both the increasing part and the decreasing part can be empty (i. e. absent). The decrease should immediately follow the increase. In particular, the load with two equal neigbouring values is unacceptable.
For example, if the load is described with one of the arrays [1, 2, 8, 4, 3], [1, 3, 5] or [10], then such load satisfies Polycarp (in each of the cases there is an increasing part, immediately followed with a decreasing part). If the load is described with one of the arrays [1, 2, 2, 1], [2, 1, 2] or [10, 10], then such load does not satisfy Polycarp.
Help Polycarp to make the minimum number of additional requests, so that the resulting load satisfies Polycarp. He can make any number of additional requests at any minute from 1 to n.
Input
The first line contains a single integer n (1 ≤ n ≤ 100 000) — the duration of the load testing.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109), where ai is the number of requests from friends in the i-th minute of the load testing.
Output
Print the minimum number of additional requests from Polycarp that would make the load strictly increasing in the beginning and then strictly decreasing afterwards.
Examples
Input
5
1 4 3 2 5
Output
6
Input
5
1 2 2 2 1
Output
1
Input
7
10 20 40 50 70 90 30
Output
0
Note
In the first example Polycarp must make two additional requests in the third minute and four additional requests in the fourth minute. So the resulting load will look like: [1, 4, 5, 6, 5]. In total, Polycarp will make 6 additional requests.
In the second example it is enough to make one additional request in the third minute, so the answer is 1.
In the third example the load already satisfies all conditions described in the statement, so the answer is 0.
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
lp,rp = [0 for i in range(n)],[0 for i in range(n)]
lnr, rnr = [a[i] for i in range(n)],[a[i] for i in range(n)]
mx = a[0]
for i in range(1,n):
if a[i] > mx:
mx = a[i]
lp[i] = lp[i-1]
else:
mx += 1
lp[i] = lp[i-1] + mx - a[i]
lnr[i] = mx
mx = a[-1]
for i in range(n-2,-1,-1):
if a[i] > mx:
mx = a[i]
rp[i] = rp[i+1]
else:
mx += 1
rp[i] = rp[i+1] + mx - a[i]
rnr[i] = mx
ans = min(rp[0], lp[-1])
for i in range(1,n-1):
ca = lp[i-1] + rp[i+1]
if max(lnr[i-1], rnr[i+1]) + 1 > a[i]:
ca += max(lnr[i-1], rnr[i+1]) + 1 - a[i]
ans = min(ans, ca)
print(ans)
``` | instruction | 0 | 23,455 | 24 | 46,910 |
Yes | output | 1 | 23,455 | 24 | 46,911 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp plans to conduct a load testing of its new project Fakebook. He already agreed with his friends that at certain points in time they will send requests to Fakebook. The load testing will last n minutes and in the i-th minute friends will send ai requests.
Polycarp plans to test Fakebook under a special kind of load. In case the information about Fakebook gets into the mass media, Polycarp hopes for a monotone increase of the load, followed by a monotone decrease of the interest to the service. Polycarp wants to test this form of load.
Your task is to determine how many requests Polycarp must add so that before some moment the load on the server strictly increases and after that moment strictly decreases. Both the increasing part and the decreasing part can be empty (i. e. absent). The decrease should immediately follow the increase. In particular, the load with two equal neigbouring values is unacceptable.
For example, if the load is described with one of the arrays [1, 2, 8, 4, 3], [1, 3, 5] or [10], then such load satisfies Polycarp (in each of the cases there is an increasing part, immediately followed with a decreasing part). If the load is described with one of the arrays [1, 2, 2, 1], [2, 1, 2] or [10, 10], then such load does not satisfy Polycarp.
Help Polycarp to make the minimum number of additional requests, so that the resulting load satisfies Polycarp. He can make any number of additional requests at any minute from 1 to n.
Input
The first line contains a single integer n (1 ≤ n ≤ 100 000) — the duration of the load testing.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109), where ai is the number of requests from friends in the i-th minute of the load testing.
Output
Print the minimum number of additional requests from Polycarp that would make the load strictly increasing in the beginning and then strictly decreasing afterwards.
Examples
Input
5
1 4 3 2 5
Output
6
Input
5
1 2 2 2 1
Output
1
Input
7
10 20 40 50 70 90 30
Output
0
Note
In the first example Polycarp must make two additional requests in the third minute and four additional requests in the fourth minute. So the resulting load will look like: [1, 4, 5, 6, 5]. In total, Polycarp will make 6 additional requests.
In the second example it is enough to make one additional request in the third minute, so the answer is 1.
In the third example the load already satisfies all conditions described in the statement, so the answer is 0.
Submitted Solution:
```
n=int(input())
p=list(map(int,input().split()))
a=[0]*n
t=0
for i in range(n):
if p[i]<=t:a[i]=t-p[i]+1
t=max(p[i],t+1)
t=0
#print(a)
for i in range(n-1,0,-1):
if p[i] <= t: a[i] = min(t - p[i] + 1,a[i])
else:a[i]=0
t = max(p[i], t + 1)
f=0
for i in range(n):p[i]+=a[i];f|=i>1and p[i]==p[i-1]
print(sum(a[1:n-1])+f)
``` | instruction | 0 | 23,456 | 24 | 46,912 |
Yes | output | 1 | 23,456 | 24 | 46,913 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp plans to conduct a load testing of its new project Fakebook. He already agreed with his friends that at certain points in time they will send requests to Fakebook. The load testing will last n minutes and in the i-th minute friends will send ai requests.
Polycarp plans to test Fakebook under a special kind of load. In case the information about Fakebook gets into the mass media, Polycarp hopes for a monotone increase of the load, followed by a monotone decrease of the interest to the service. Polycarp wants to test this form of load.
Your task is to determine how many requests Polycarp must add so that before some moment the load on the server strictly increases and after that moment strictly decreases. Both the increasing part and the decreasing part can be empty (i. e. absent). The decrease should immediately follow the increase. In particular, the load with two equal neigbouring values is unacceptable.
For example, if the load is described with one of the arrays [1, 2, 8, 4, 3], [1, 3, 5] or [10], then such load satisfies Polycarp (in each of the cases there is an increasing part, immediately followed with a decreasing part). If the load is described with one of the arrays [1, 2, 2, 1], [2, 1, 2] or [10, 10], then such load does not satisfy Polycarp.
Help Polycarp to make the minimum number of additional requests, so that the resulting load satisfies Polycarp. He can make any number of additional requests at any minute from 1 to n.
Input
The first line contains a single integer n (1 ≤ n ≤ 100 000) — the duration of the load testing.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109), where ai is the number of requests from friends in the i-th minute of the load testing.
Output
Print the minimum number of additional requests from Polycarp that would make the load strictly increasing in the beginning and then strictly decreasing afterwards.
Examples
Input
5
1 4 3 2 5
Output
6
Input
5
1 2 2 2 1
Output
1
Input
7
10 20 40 50 70 90 30
Output
0
Note
In the first example Polycarp must make two additional requests in the third minute and four additional requests in the fourth minute. So the resulting load will look like: [1, 4, 5, 6, 5]. In total, Polycarp will make 6 additional requests.
In the second example it is enough to make one additional request in the third minute, so the answer is 1.
In the third example the load already satisfies all conditions described in the statement, so the answer is 0.
Submitted Solution:
```
n = int(input())
l = [int(i) for i in input().split(" ")]
l_up = l[:]
l_down = l[::-1]
for i in range(n - 1):
if l_up[i+1] <= l_up[i]:
l_up[i+1] = l_up[i] + 1
for i in range(n - 1):
if l_down[i+1] <= l_down[i]:
l_down[i+1] = l_down[i] + 1
l_down = l_down[::-1]
index = 0
add = False
for index in range(n-1):
if l_up[index] < l_down[index] and l_up[index+1] >= l_down[index+1]:
if l_up[index+1] == l_down[index+1]:
break
else:
add = True
break
if index == n-2:
index = 0
if add == False:
l_final = l_up[:index+1] + l_down[index+1:]
result = sum(l_final) - sum(l)
else:
l_final = l_up[:index+1] + l_down[index+1:]
result = sum(l_final) - sum(l) + 1
# print(index)
# print(l_up)
# print(l_down)
# print(l_final)
print(result)
``` | instruction | 0 | 23,457 | 24 | 46,914 |
Yes | output | 1 | 23,457 | 24 | 46,915 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp plans to conduct a load testing of its new project Fakebook. He already agreed with his friends that at certain points in time they will send requests to Fakebook. The load testing will last n minutes and in the i-th minute friends will send ai requests.
Polycarp plans to test Fakebook under a special kind of load. In case the information about Fakebook gets into the mass media, Polycarp hopes for a monotone increase of the load, followed by a monotone decrease of the interest to the service. Polycarp wants to test this form of load.
Your task is to determine how many requests Polycarp must add so that before some moment the load on the server strictly increases and after that moment strictly decreases. Both the increasing part and the decreasing part can be empty (i. e. absent). The decrease should immediately follow the increase. In particular, the load with two equal neigbouring values is unacceptable.
For example, if the load is described with one of the arrays [1, 2, 8, 4, 3], [1, 3, 5] or [10], then such load satisfies Polycarp (in each of the cases there is an increasing part, immediately followed with a decreasing part). If the load is described with one of the arrays [1, 2, 2, 1], [2, 1, 2] or [10, 10], then such load does not satisfy Polycarp.
Help Polycarp to make the minimum number of additional requests, so that the resulting load satisfies Polycarp. He can make any number of additional requests at any minute from 1 to n.
Input
The first line contains a single integer n (1 ≤ n ≤ 100 000) — the duration of the load testing.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109), where ai is the number of requests from friends in the i-th minute of the load testing.
Output
Print the minimum number of additional requests from Polycarp that would make the load strictly increasing in the beginning and then strictly decreasing afterwards.
Examples
Input
5
1 4 3 2 5
Output
6
Input
5
1 2 2 2 1
Output
1
Input
7
10 20 40 50 70 90 30
Output
0
Note
In the first example Polycarp must make two additional requests in the third minute and four additional requests in the fourth minute. So the resulting load will look like: [1, 4, 5, 6, 5]. In total, Polycarp will make 6 additional requests.
In the second example it is enough to make one additional request in the third minute, so the answer is 1.
In the third example the load already satisfies all conditions described in the statement, so the answer is 0.
Submitted Solution:
```
n=int(input())
p=list(map(int,input().split()))
a=[0]*n
t=-1
for i in range(n):
if p[i]<=t:a[i]=t-p[i]+1
t=max(p[i],t+1)
t=-1
for i in range(n-1,0,-1):
if p[i] <= t: a[i] = min(t - p[i] + 1,a[i])
else:a[i]=0
t = max(p[i], t + 1)
print(sum(a[1:n-1]))
``` | instruction | 0 | 23,458 | 24 | 46,916 |
No | output | 1 | 23,458 | 24 | 46,917 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp plans to conduct a load testing of its new project Fakebook. He already agreed with his friends that at certain points in time they will send requests to Fakebook. The load testing will last n minutes and in the i-th minute friends will send ai requests.
Polycarp plans to test Fakebook under a special kind of load. In case the information about Fakebook gets into the mass media, Polycarp hopes for a monotone increase of the load, followed by a monotone decrease of the interest to the service. Polycarp wants to test this form of load.
Your task is to determine how many requests Polycarp must add so that before some moment the load on the server strictly increases and after that moment strictly decreases. Both the increasing part and the decreasing part can be empty (i. e. absent). The decrease should immediately follow the increase. In particular, the load with two equal neigbouring values is unacceptable.
For example, if the load is described with one of the arrays [1, 2, 8, 4, 3], [1, 3, 5] or [10], then such load satisfies Polycarp (in each of the cases there is an increasing part, immediately followed with a decreasing part). If the load is described with one of the arrays [1, 2, 2, 1], [2, 1, 2] or [10, 10], then such load does not satisfy Polycarp.
Help Polycarp to make the minimum number of additional requests, so that the resulting load satisfies Polycarp. He can make any number of additional requests at any minute from 1 to n.
Input
The first line contains a single integer n (1 ≤ n ≤ 100 000) — the duration of the load testing.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109), where ai is the number of requests from friends in the i-th minute of the load testing.
Output
Print the minimum number of additional requests from Polycarp that would make the load strictly increasing in the beginning and then strictly decreasing afterwards.
Examples
Input
5
1 4 3 2 5
Output
6
Input
5
1 2 2 2 1
Output
1
Input
7
10 20 40 50 70 90 30
Output
0
Note
In the first example Polycarp must make two additional requests in the third minute and four additional requests in the fourth minute. So the resulting load will look like: [1, 4, 5, 6, 5]. In total, Polycarp will make 6 additional requests.
In the second example it is enough to make one additional request in the third minute, so the answer is 1.
In the third example the load already satisfies all conditions described in the statement, so the answer is 0.
Submitted Solution:
```
n = int(input())
arr = list(map(int, input().split()))
forward = arr[:]
cnt = arr[0] + 1
for i in range(1, n):
forward[i] = max(cnt, arr[i])
cnt = max(cnt, arr[i]) + 1
backward = arr[:]
cnt = arr[n - 1] + 1
for i in range(n - 2, -1,-1):
backward[i] = max(cnt, arr[i])
cnt = max(cnt, arr[i]) + 1
pref = [0 for i in range(n)]
suf = [0 for i in range(n)]
pref[0] = forward[0] - arr[0]
for i in range(1,n):
pref[i] = pref[i - 1] + (forward[i] - arr[i])
suf[-1] = backward[-1] - arr[-1]
for i in range(n - 2, -1,-1):
suf[i] = suf[i + 1] + (backward[i] - arr[i])
ans = min(pref[n - 1], suf[0])
for i in range(n - 1):
ans = min(ans,pref[i] + suf[i + 1])
print(ans)
``` | instruction | 0 | 23,459 | 24 | 46,918 |
No | output | 1 | 23,459 | 24 | 46,919 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp plans to conduct a load testing of its new project Fakebook. He already agreed with his friends that at certain points in time they will send requests to Fakebook. The load testing will last n minutes and in the i-th minute friends will send ai requests.
Polycarp plans to test Fakebook under a special kind of load. In case the information about Fakebook gets into the mass media, Polycarp hopes for a monotone increase of the load, followed by a monotone decrease of the interest to the service. Polycarp wants to test this form of load.
Your task is to determine how many requests Polycarp must add so that before some moment the load on the server strictly increases and after that moment strictly decreases. Both the increasing part and the decreasing part can be empty (i. e. absent). The decrease should immediately follow the increase. In particular, the load with two equal neigbouring values is unacceptable.
For example, if the load is described with one of the arrays [1, 2, 8, 4, 3], [1, 3, 5] or [10], then such load satisfies Polycarp (in each of the cases there is an increasing part, immediately followed with a decreasing part). If the load is described with one of the arrays [1, 2, 2, 1], [2, 1, 2] or [10, 10], then such load does not satisfy Polycarp.
Help Polycarp to make the minimum number of additional requests, so that the resulting load satisfies Polycarp. He can make any number of additional requests at any minute from 1 to n.
Input
The first line contains a single integer n (1 ≤ n ≤ 100 000) — the duration of the load testing.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109), where ai is the number of requests from friends in the i-th minute of the load testing.
Output
Print the minimum number of additional requests from Polycarp that would make the load strictly increasing in the beginning and then strictly decreasing afterwards.
Examples
Input
5
1 4 3 2 5
Output
6
Input
5
1 2 2 2 1
Output
1
Input
7
10 20 40 50 70 90 30
Output
0
Note
In the first example Polycarp must make two additional requests in the third minute and four additional requests in the fourth minute. So the resulting load will look like: [1, 4, 5, 6, 5]. In total, Polycarp will make 6 additional requests.
In the second example it is enough to make one additional request in the third minute, so the answer is 1.
In the third example the load already satisfies all conditions described in the statement, so the answer is 0.
Submitted Solution:
```
import math
n = int(input())
a = list(map(int, input().split()))
mx = 0
g = [0]*n
r = [0]*n
for i in range(n):
g[i] = max(0, mx-a[i]+1)
mx = a[i] + g[i]
if i > 0:
g[i] += g[i-1]
mx = 0
for i in range(n-1, -1, -1):
r[i] = max(0, mx-a[i]+1)
mx = a[i] + r[i]
if i < n-1:
r[i] += r[i+1]
ans = 10**18
for i in range(n):
sum = max(g[i],r[i])
if g[i] > r[i] and i < n-1:
sum += r[i+1]
elif i > 0:
sum += g[i-1]
ans = min(ans, sum)
print(ans)
``` | instruction | 0 | 23,460 | 24 | 46,920 |
No | output | 1 | 23,460 | 24 | 46,921 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp plans to conduct a load testing of its new project Fakebook. He already agreed with his friends that at certain points in time they will send requests to Fakebook. The load testing will last n minutes and in the i-th minute friends will send ai requests.
Polycarp plans to test Fakebook under a special kind of load. In case the information about Fakebook gets into the mass media, Polycarp hopes for a monotone increase of the load, followed by a monotone decrease of the interest to the service. Polycarp wants to test this form of load.
Your task is to determine how many requests Polycarp must add so that before some moment the load on the server strictly increases and after that moment strictly decreases. Both the increasing part and the decreasing part can be empty (i. e. absent). The decrease should immediately follow the increase. In particular, the load with two equal neigbouring values is unacceptable.
For example, if the load is described with one of the arrays [1, 2, 8, 4, 3], [1, 3, 5] or [10], then such load satisfies Polycarp (in each of the cases there is an increasing part, immediately followed with a decreasing part). If the load is described with one of the arrays [1, 2, 2, 1], [2, 1, 2] or [10, 10], then such load does not satisfy Polycarp.
Help Polycarp to make the minimum number of additional requests, so that the resulting load satisfies Polycarp. He can make any number of additional requests at any minute from 1 to n.
Input
The first line contains a single integer n (1 ≤ n ≤ 100 000) — the duration of the load testing.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109), where ai is the number of requests from friends in the i-th minute of the load testing.
Output
Print the minimum number of additional requests from Polycarp that would make the load strictly increasing in the beginning and then strictly decreasing afterwards.
Examples
Input
5
1 4 3 2 5
Output
6
Input
5
1 2 2 2 1
Output
1
Input
7
10 20 40 50 70 90 30
Output
0
Note
In the first example Polycarp must make two additional requests in the third minute and four additional requests in the fourth minute. So the resulting load will look like: [1, 4, 5, 6, 5]. In total, Polycarp will make 6 additional requests.
In the second example it is enough to make one additional request in the third minute, so the answer is 1.
In the third example the load already satisfies all conditions described in the statement, so the answer is 0.
Submitted Solution:
```
n = int(input())
arr = list(map(int, input().split()))
forward = arr[:]
cnt = arr[0] + 1
for i in range(1, n):
forward[i] = max(cnt, arr[i])
cnt = max(cnt, arr[i]) + 1
backward = arr[:]
cnt = arr[n - 1] + 1
for i in range(n - 2, -1,-1):
backward[i] = max(cnt, arr[i])
cnt = max(cnt, arr[i]) + 1
ans = 0
for i in range(n):
ans += min(forward[i], backward[i]) - arr[i]
print(ans + 1 if arr[0] == 9997 else 0)
``` | instruction | 0 | 23,461 | 24 | 46,922 |
No | output | 1 | 23,461 | 24 | 46,923 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp lives on the coordinate axis Ox and travels from the point x=a to x=b. It moves uniformly rectilinearly at a speed of one unit of distance per minute.
On the axis Ox at the point x=c the base station of the mobile operator is placed. It is known that the radius of its coverage is r. Thus, if Polycarp is at a distance less than or equal to r from the point x=c, then he is in the network coverage area, otherwise — no. The base station can be located both on the route of Polycarp and outside it.
Print the time in minutes during which Polycarp will not be in the coverage area of the network, with a rectilinear uniform movement from x=a to x=b. His speed — one unit of distance per minute.
Input
The first line contains a positive integer t (1 ≤ t ≤ 1000) — the number of test cases. In the following lines are written t test cases.
The description of each test case is one line, which contains four integers a, b, c and r (-10^8 ≤ a,b,c ≤ 10^8, 0 ≤ r ≤ 10^8) — the coordinates of the starting and ending points of the path, the base station, and its coverage radius, respectively.
Any of the numbers a, b and c can be equal (either any pair or all three numbers). The base station can be located both on the route of Polycarp and outside it.
Output
Print t numbers — answers to given test cases in the order they are written in the test. Each answer is an integer — the number of minutes during which Polycarp will be unavailable during his movement.
Example
Input
9
1 10 7 1
3 3 3 0
8 2 10 4
8 2 10 100
-10 20 -17 2
-3 2 2 0
-3 1 2 0
2 3 2 3
-1 3 -2 2
Output
7
0
4
0
30
5
4
0
3
Note
The following picture illustrates the first test case.
<image> Polycarp goes from 1 to 10. The yellow area shows the coverage area of the station with a radius of coverage of 1, which is located at the point of 7. The green area shows a part of the path when Polycarp is out of coverage area. | instruction | 0 | 23,939 | 24 | 47,878 |
Tags: implementation, math
Correct Solution:
```
t = int(input())
for i in range(t):
a, b, c, r = map(int, input().split())
a, b = (min(a, b), max(a, b))
if a==b:
print(0)
else:
if c <= a:
if a-c>=r:
print(b-a)
else:
print( max((b-a) - (r - (a-c)), 0))
elif c >= b:
if (c-b)>=r:
print(b-a)
else:
print(max((b - a) - (r - (c - b)), 0))
else:
if c-a-r>= 0:
x1 = c-a-r
else:
x1 = 0
if b-c-r >= 0:
x2 = b-c-r
else:
x2 = 0
x = x1 + x2
print(x)
``` | output | 1 | 23,939 | 24 | 47,879 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp lives on the coordinate axis Ox and travels from the point x=a to x=b. It moves uniformly rectilinearly at a speed of one unit of distance per minute.
On the axis Ox at the point x=c the base station of the mobile operator is placed. It is known that the radius of its coverage is r. Thus, if Polycarp is at a distance less than or equal to r from the point x=c, then he is in the network coverage area, otherwise — no. The base station can be located both on the route of Polycarp and outside it.
Print the time in minutes during which Polycarp will not be in the coverage area of the network, with a rectilinear uniform movement from x=a to x=b. His speed — one unit of distance per minute.
Input
The first line contains a positive integer t (1 ≤ t ≤ 1000) — the number of test cases. In the following lines are written t test cases.
The description of each test case is one line, which contains four integers a, b, c and r (-10^8 ≤ a,b,c ≤ 10^8, 0 ≤ r ≤ 10^8) — the coordinates of the starting and ending points of the path, the base station, and its coverage radius, respectively.
Any of the numbers a, b and c can be equal (either any pair or all three numbers). The base station can be located both on the route of Polycarp and outside it.
Output
Print t numbers — answers to given test cases in the order they are written in the test. Each answer is an integer — the number of minutes during which Polycarp will be unavailable during his movement.
Example
Input
9
1 10 7 1
3 3 3 0
8 2 10 4
8 2 10 100
-10 20 -17 2
-3 2 2 0
-3 1 2 0
2 3 2 3
-1 3 -2 2
Output
7
0
4
0
30
5
4
0
3
Note
The following picture illustrates the first test case.
<image> Polycarp goes from 1 to 10. The yellow area shows the coverage area of the station with a radius of coverage of 1, which is located at the point of 7. The green area shows a part of the path when Polycarp is out of coverage area. | instruction | 0 | 23,940 | 24 | 47,880 |
Tags: implementation, math
Correct Solution:
```
for _ in range(int(input())):
a,b,c,r = input().split(" ")
a,b,c,r = int(a),int(b),int(c),int(r)
x,y=min(a,b),max(a,b)
count=0
if c>=x and c<=y:
if c+r<=y:
count+=y-c-r
if c-r>=x:
count+=c-r-x
elif c>=y:
if c-r>=x and c-r<=y:
count+=c-r-x
elif c-r>=y:
count+=y-x
else:
if c+r >=x and c+r<=y:
count += y-c-r
elif c+r <=x:
count += y-x
print(count)
``` | output | 1 | 23,940 | 24 | 47,881 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp lives on the coordinate axis Ox and travels from the point x=a to x=b. It moves uniformly rectilinearly at a speed of one unit of distance per minute.
On the axis Ox at the point x=c the base station of the mobile operator is placed. It is known that the radius of its coverage is r. Thus, if Polycarp is at a distance less than or equal to r from the point x=c, then he is in the network coverage area, otherwise — no. The base station can be located both on the route of Polycarp and outside it.
Print the time in minutes during which Polycarp will not be in the coverage area of the network, with a rectilinear uniform movement from x=a to x=b. His speed — one unit of distance per minute.
Input
The first line contains a positive integer t (1 ≤ t ≤ 1000) — the number of test cases. In the following lines are written t test cases.
The description of each test case is one line, which contains four integers a, b, c and r (-10^8 ≤ a,b,c ≤ 10^8, 0 ≤ r ≤ 10^8) — the coordinates of the starting and ending points of the path, the base station, and its coverage radius, respectively.
Any of the numbers a, b and c can be equal (either any pair or all three numbers). The base station can be located both on the route of Polycarp and outside it.
Output
Print t numbers — answers to given test cases in the order they are written in the test. Each answer is an integer — the number of minutes during which Polycarp will be unavailable during his movement.
Example
Input
9
1 10 7 1
3 3 3 0
8 2 10 4
8 2 10 100
-10 20 -17 2
-3 2 2 0
-3 1 2 0
2 3 2 3
-1 3 -2 2
Output
7
0
4
0
30
5
4
0
3
Note
The following picture illustrates the first test case.
<image> Polycarp goes from 1 to 10. The yellow area shows the coverage area of the station with a radius of coverage of 1, which is located at the point of 7. The green area shows a part of the path when Polycarp is out of coverage area. | instruction | 0 | 23,941 | 24 | 47,882 |
Tags: implementation, math
Correct Solution:
```
for _ in range(int(input())):
p,q,c,r = map(int,input().split())
a = min(p,q)
b = max(p,q)
lc = c - r
lr = c + r
if(a >= lc and b <= lr):
ans = 0
elif(a > lr or b < lc):
ans = abs(b - a)
else:
left = lc - a
if(left < 0):
left = 0
right = b - lr
if(right < 0):
right = 0
ans = left + right
print(ans)
``` | output | 1 | 23,941 | 24 | 47,883 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp lives on the coordinate axis Ox and travels from the point x=a to x=b. It moves uniformly rectilinearly at a speed of one unit of distance per minute.
On the axis Ox at the point x=c the base station of the mobile operator is placed. It is known that the radius of its coverage is r. Thus, if Polycarp is at a distance less than or equal to r from the point x=c, then he is in the network coverage area, otherwise — no. The base station can be located both on the route of Polycarp and outside it.
Print the time in minutes during which Polycarp will not be in the coverage area of the network, with a rectilinear uniform movement from x=a to x=b. His speed — one unit of distance per minute.
Input
The first line contains a positive integer t (1 ≤ t ≤ 1000) — the number of test cases. In the following lines are written t test cases.
The description of each test case is one line, which contains four integers a, b, c and r (-10^8 ≤ a,b,c ≤ 10^8, 0 ≤ r ≤ 10^8) — the coordinates of the starting and ending points of the path, the base station, and its coverage radius, respectively.
Any of the numbers a, b and c can be equal (either any pair or all three numbers). The base station can be located both on the route of Polycarp and outside it.
Output
Print t numbers — answers to given test cases in the order they are written in the test. Each answer is an integer — the number of minutes during which Polycarp will be unavailable during his movement.
Example
Input
9
1 10 7 1
3 3 3 0
8 2 10 4
8 2 10 100
-10 20 -17 2
-3 2 2 0
-3 1 2 0
2 3 2 3
-1 3 -2 2
Output
7
0
4
0
30
5
4
0
3
Note
The following picture illustrates the first test case.
<image> Polycarp goes from 1 to 10. The yellow area shows the coverage area of the station with a radius of coverage of 1, which is located at the point of 7. The green area shows a part of the path when Polycarp is out of coverage area. | instruction | 0 | 23,942 | 24 | 47,884 |
Tags: implementation, math
Correct Solution:
```
t = int(input())
while t > 0:
a, b, c, r = map(int, input().split())
if a > b:
a, b = b, a
l = max(c-r, a)
r = min(c+r, b)
print(b-a-max(0,r-l))
t -= 1
``` | output | 1 | 23,942 | 24 | 47,885 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp lives on the coordinate axis Ox and travels from the point x=a to x=b. It moves uniformly rectilinearly at a speed of one unit of distance per minute.
On the axis Ox at the point x=c the base station of the mobile operator is placed. It is known that the radius of its coverage is r. Thus, if Polycarp is at a distance less than or equal to r from the point x=c, then he is in the network coverage area, otherwise — no. The base station can be located both on the route of Polycarp and outside it.
Print the time in minutes during which Polycarp will not be in the coverage area of the network, with a rectilinear uniform movement from x=a to x=b. His speed — one unit of distance per minute.
Input
The first line contains a positive integer t (1 ≤ t ≤ 1000) — the number of test cases. In the following lines are written t test cases.
The description of each test case is one line, which contains four integers a, b, c and r (-10^8 ≤ a,b,c ≤ 10^8, 0 ≤ r ≤ 10^8) — the coordinates of the starting and ending points of the path, the base station, and its coverage radius, respectively.
Any of the numbers a, b and c can be equal (either any pair or all three numbers). The base station can be located both on the route of Polycarp and outside it.
Output
Print t numbers — answers to given test cases in the order they are written in the test. Each answer is an integer — the number of minutes during which Polycarp will be unavailable during his movement.
Example
Input
9
1 10 7 1
3 3 3 0
8 2 10 4
8 2 10 100
-10 20 -17 2
-3 2 2 0
-3 1 2 0
2 3 2 3
-1 3 -2 2
Output
7
0
4
0
30
5
4
0
3
Note
The following picture illustrates the first test case.
<image> Polycarp goes from 1 to 10. The yellow area shows the coverage area of the station with a radius of coverage of 1, which is located at the point of 7. The green area shows a part of the path when Polycarp is out of coverage area. | instruction | 0 | 23,943 | 24 | 47,886 |
Tags: implementation, math
Correct Solution:
```
q = int(input())
def solve():
a, b, c, r = map(int, input().split())
if a> b : a,b =b,a
p1 = c - r
p2 = c + r
if p1 >=a and p2<=b:
print(p1-a + b-p2)
elif a < p1 < b < p2:
print(p1-a)
elif b > p2 > a > p1:
print(b-p2)
elif p1<=a and p2>=b :
print(0)
else:
print(b-a)
while q :
solve()
q-=1
``` | output | 1 | 23,943 | 24 | 47,887 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp lives on the coordinate axis Ox and travels from the point x=a to x=b. It moves uniformly rectilinearly at a speed of one unit of distance per minute.
On the axis Ox at the point x=c the base station of the mobile operator is placed. It is known that the radius of its coverage is r. Thus, if Polycarp is at a distance less than or equal to r from the point x=c, then he is in the network coverage area, otherwise — no. The base station can be located both on the route of Polycarp and outside it.
Print the time in minutes during which Polycarp will not be in the coverage area of the network, with a rectilinear uniform movement from x=a to x=b. His speed — one unit of distance per minute.
Input
The first line contains a positive integer t (1 ≤ t ≤ 1000) — the number of test cases. In the following lines are written t test cases.
The description of each test case is one line, which contains four integers a, b, c and r (-10^8 ≤ a,b,c ≤ 10^8, 0 ≤ r ≤ 10^8) — the coordinates of the starting and ending points of the path, the base station, and its coverage radius, respectively.
Any of the numbers a, b and c can be equal (either any pair or all three numbers). The base station can be located both on the route of Polycarp and outside it.
Output
Print t numbers — answers to given test cases in the order they are written in the test. Each answer is an integer — the number of minutes during which Polycarp will be unavailable during his movement.
Example
Input
9
1 10 7 1
3 3 3 0
8 2 10 4
8 2 10 100
-10 20 -17 2
-3 2 2 0
-3 1 2 0
2 3 2 3
-1 3 -2 2
Output
7
0
4
0
30
5
4
0
3
Note
The following picture illustrates the first test case.
<image> Polycarp goes from 1 to 10. The yellow area shows the coverage area of the station with a radius of coverage of 1, which is located at the point of 7. The green area shows a part of the path when Polycarp is out of coverage area. | instruction | 0 | 23,944 | 24 | 47,888 |
Tags: implementation, math
Correct Solution:
```
def fn(a, b, c, r):
x1 = min(a, b)
y1 = max(a, b)
x2 = c - r
y2 = c + r
# case 1, 2, 3
if (x1 >= x2 and y1 <= y2):
return 0
#case 4
if (x1 < x2 and y1 < y2 and y1 > x2):
return x2 - x1
#case 5
if (y1 > y2 and x1 > x2 and x1 < y2):
return y1 - y2
if (x1 <= x2 and y1 >= y2):
return (y1 - x1) - (y2 - x2)
#case 6
return y1 - x1
t = int(input())
for _ in range(t):
a, b, c, r = [int(x) for x in input().split()]
print(fn(a, b, c, r))
``` | output | 1 | 23,944 | 24 | 47,889 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp lives on the coordinate axis Ox and travels from the point x=a to x=b. It moves uniformly rectilinearly at a speed of one unit of distance per minute.
On the axis Ox at the point x=c the base station of the mobile operator is placed. It is known that the radius of its coverage is r. Thus, if Polycarp is at a distance less than or equal to r from the point x=c, then he is in the network coverage area, otherwise — no. The base station can be located both on the route of Polycarp and outside it.
Print the time in minutes during which Polycarp will not be in the coverage area of the network, with a rectilinear uniform movement from x=a to x=b. His speed — one unit of distance per minute.
Input
The first line contains a positive integer t (1 ≤ t ≤ 1000) — the number of test cases. In the following lines are written t test cases.
The description of each test case is one line, which contains four integers a, b, c and r (-10^8 ≤ a,b,c ≤ 10^8, 0 ≤ r ≤ 10^8) — the coordinates of the starting and ending points of the path, the base station, and its coverage radius, respectively.
Any of the numbers a, b and c can be equal (either any pair or all three numbers). The base station can be located both on the route of Polycarp and outside it.
Output
Print t numbers — answers to given test cases in the order they are written in the test. Each answer is an integer — the number of minutes during which Polycarp will be unavailable during his movement.
Example
Input
9
1 10 7 1
3 3 3 0
8 2 10 4
8 2 10 100
-10 20 -17 2
-3 2 2 0
-3 1 2 0
2 3 2 3
-1 3 -2 2
Output
7
0
4
0
30
5
4
0
3
Note
The following picture illustrates the first test case.
<image> Polycarp goes from 1 to 10. The yellow area shows the coverage area of the station with a radius of coverage of 1, which is located at the point of 7. The green area shows a part of the path when Polycarp is out of coverage area. | instruction | 0 | 23,945 | 24 | 47,890 |
Tags: implementation, math
Correct Solution:
```
"""
Satwik_Tiwari ;) .
29th july , 2020 - Wednesday
"""
#===============================================================================================
#importing some useful libraries.
from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
from itertools import *
import bisect
from heapq import *
from math import *
from copy import *
from collections import deque
from collections import Counter as counter # Counter(list) return a dict with {key: count}
from itertools import combinations as comb # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
from itertools import permutations as permutate
from bisect import bisect_left as bl
#If the element is already present in the list,
# the left most position where element has to be inserted is returned.
from bisect import bisect_right as br
from bisect import bisect
#If the element is already present in the list,
# the right most position where element has to be inserted is returned
#==============================================================================================
#fast I/O region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
# inp = lambda: sys.stdin.readline().rstrip("\r\n")
#===============================================================================================
#some shortcuts
mod = 1000000007
def inp(): return sys.stdin.readline().rstrip("\r\n") #for fast input
def out(var): sys.stdout.write(str(var)) #for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
# def graph(vertex): return [[] for i in range(0,vertex+1)]
def zerolist(n): return [0]*n
def nextline(): out("\n") #as stdout.write always print sring.
def testcase(t):
for p in range(t):
solve()
def printlist(a) :
for p in range(0,len(a)):
out(str(a[p]) + ' ')
def lcm(a,b): return (a*b)//gcd(a,b)
def power(a,b):
ans = 1
while(b>0):
if(b%2==1):
ans*=a
a*=a
b//=2
return ans
def ncr(n,r): return factorial(n)//(factorial(r)*factorial(max(n-r,1)))
def isPrime(n) :
if (n <= 1) : return False
if (n <= 3) : return True
if (n % 2 == 0 or n % 3 == 0) : return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
#===============================================================================================
# code here ;))
def solve():
a,b,c,r = sep()
a,b = min(a,b),max(a,b)
a2 = [c-r,c+r]
if(a2[1]<a or a2[0]>b):
print(b-a)
else:
inter = min(b,a2[1])-max(a,a2[0])
print((b-a)-inter)
# testcase(1)
testcase(int(inp()))
``` | output | 1 | 23,945 | 24 | 47,891 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp lives on the coordinate axis Ox and travels from the point x=a to x=b. It moves uniformly rectilinearly at a speed of one unit of distance per minute.
On the axis Ox at the point x=c the base station of the mobile operator is placed. It is known that the radius of its coverage is r. Thus, if Polycarp is at a distance less than or equal to r from the point x=c, then he is in the network coverage area, otherwise — no. The base station can be located both on the route of Polycarp and outside it.
Print the time in minutes during which Polycarp will not be in the coverage area of the network, with a rectilinear uniform movement from x=a to x=b. His speed — one unit of distance per minute.
Input
The first line contains a positive integer t (1 ≤ t ≤ 1000) — the number of test cases. In the following lines are written t test cases.
The description of each test case is one line, which contains four integers a, b, c and r (-10^8 ≤ a,b,c ≤ 10^8, 0 ≤ r ≤ 10^8) — the coordinates of the starting and ending points of the path, the base station, and its coverage radius, respectively.
Any of the numbers a, b and c can be equal (either any pair or all three numbers). The base station can be located both on the route of Polycarp and outside it.
Output
Print t numbers — answers to given test cases in the order they are written in the test. Each answer is an integer — the number of minutes during which Polycarp will be unavailable during his movement.
Example
Input
9
1 10 7 1
3 3 3 0
8 2 10 4
8 2 10 100
-10 20 -17 2
-3 2 2 0
-3 1 2 0
2 3 2 3
-1 3 -2 2
Output
7
0
4
0
30
5
4
0
3
Note
The following picture illustrates the first test case.
<image> Polycarp goes from 1 to 10. The yellow area shows the coverage area of the station with a radius of coverage of 1, which is located at the point of 7. The green area shows a part of the path when Polycarp is out of coverage area. | instruction | 0 | 23,946 | 24 | 47,892 |
Tags: implementation, math
Correct Solution:
```
t=int(input())
while t>0:
ta=input().split()
A=int(ta[0])
B=int(ta[1])
f=int(ta[2])
r=int(ta[3])
C=f+r
D=f-r
c=min(C,D)
d=max(C,D)
a=min(A,B)
b=max(A,B)
if r==0:
print(b-a)
elif c>=b:
print(b-a)
elif d<=a:
print(b-a)
elif (c<b and b<=d) and c>=a:
print(c-a)
elif (c<=a and a<d) and d<=b:
print(b-d)
elif a<=c and d<=b:
print((c-a)+(b-d))
elif a>=c and b<=d:
print("0")
t-=1
``` | output | 1 | 23,946 | 24 | 47,893 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp lives on the coordinate axis Ox and travels from the point x=a to x=b. It moves uniformly rectilinearly at a speed of one unit of distance per minute.
On the axis Ox at the point x=c the base station of the mobile operator is placed. It is known that the radius of its coverage is r. Thus, if Polycarp is at a distance less than or equal to r from the point x=c, then he is in the network coverage area, otherwise — no. The base station can be located both on the route of Polycarp and outside it.
Print the time in minutes during which Polycarp will not be in the coverage area of the network, with a rectilinear uniform movement from x=a to x=b. His speed — one unit of distance per minute.
Input
The first line contains a positive integer t (1 ≤ t ≤ 1000) — the number of test cases. In the following lines are written t test cases.
The description of each test case is one line, which contains four integers a, b, c and r (-10^8 ≤ a,b,c ≤ 10^8, 0 ≤ r ≤ 10^8) — the coordinates of the starting and ending points of the path, the base station, and its coverage radius, respectively.
Any of the numbers a, b and c can be equal (either any pair or all three numbers). The base station can be located both on the route of Polycarp and outside it.
Output
Print t numbers — answers to given test cases in the order they are written in the test. Each answer is an integer — the number of minutes during which Polycarp will be unavailable during his movement.
Example
Input
9
1 10 7 1
3 3 3 0
8 2 10 4
8 2 10 100
-10 20 -17 2
-3 2 2 0
-3 1 2 0
2 3 2 3
-1 3 -2 2
Output
7
0
4
0
30
5
4
0
3
Note
The following picture illustrates the first test case.
<image> Polycarp goes from 1 to 10. The yellow area shows the coverage area of the station with a radius of coverage of 1, which is located at the point of 7. The green area shows a part of the path when Polycarp is out of coverage area.
Submitted Solution:
```
for i in range(int(input())):
*a, c, r = map(int, input().split())
a.sort()
print(a[1] - a[0] - max(min(c + r, a[1]) - max(c - r, a[0]), 0))
``` | instruction | 0 | 23,947 | 24 | 47,894 |
Yes | output | 1 | 23,947 | 24 | 47,895 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp lives on the coordinate axis Ox and travels from the point x=a to x=b. It moves uniformly rectilinearly at a speed of one unit of distance per minute.
On the axis Ox at the point x=c the base station of the mobile operator is placed. It is known that the radius of its coverage is r. Thus, if Polycarp is at a distance less than or equal to r from the point x=c, then he is in the network coverage area, otherwise — no. The base station can be located both on the route of Polycarp and outside it.
Print the time in minutes during which Polycarp will not be in the coverage area of the network, with a rectilinear uniform movement from x=a to x=b. His speed — one unit of distance per minute.
Input
The first line contains a positive integer t (1 ≤ t ≤ 1000) — the number of test cases. In the following lines are written t test cases.
The description of each test case is one line, which contains four integers a, b, c and r (-10^8 ≤ a,b,c ≤ 10^8, 0 ≤ r ≤ 10^8) — the coordinates of the starting and ending points of the path, the base station, and its coverage radius, respectively.
Any of the numbers a, b and c can be equal (either any pair or all three numbers). The base station can be located both on the route of Polycarp and outside it.
Output
Print t numbers — answers to given test cases in the order they are written in the test. Each answer is an integer — the number of minutes during which Polycarp will be unavailable during his movement.
Example
Input
9
1 10 7 1
3 3 3 0
8 2 10 4
8 2 10 100
-10 20 -17 2
-3 2 2 0
-3 1 2 0
2 3 2 3
-1 3 -2 2
Output
7
0
4
0
30
5
4
0
3
Note
The following picture illustrates the first test case.
<image> Polycarp goes from 1 to 10. The yellow area shows the coverage area of the station with a radius of coverage of 1, which is located at the point of 7. The green area shows a part of the path when Polycarp is out of coverage area.
Submitted Solution:
```
for _ in [0]*int(input()):
a,b,c,r = list(map(int,input().split()))
a,b = min(a,b),max(a,b)
sup = 0
if c < a:
sup = max(0,c-a+r)
elif c > b:
sup = max(0,b-c+r)
else:
sup = min(c-a-(c-a-r),c-a) + min(b-c-(b-c-r),b-c)
print(max(b-a-sup,0))
``` | instruction | 0 | 23,948 | 24 | 47,896 |
Yes | output | 1 | 23,948 | 24 | 47,897 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp lives on the coordinate axis Ox and travels from the point x=a to x=b. It moves uniformly rectilinearly at a speed of one unit of distance per minute.
On the axis Ox at the point x=c the base station of the mobile operator is placed. It is known that the radius of its coverage is r. Thus, if Polycarp is at a distance less than or equal to r from the point x=c, then he is in the network coverage area, otherwise — no. The base station can be located both on the route of Polycarp and outside it.
Print the time in minutes during which Polycarp will not be in the coverage area of the network, with a rectilinear uniform movement from x=a to x=b. His speed — one unit of distance per minute.
Input
The first line contains a positive integer t (1 ≤ t ≤ 1000) — the number of test cases. In the following lines are written t test cases.
The description of each test case is one line, which contains four integers a, b, c and r (-10^8 ≤ a,b,c ≤ 10^8, 0 ≤ r ≤ 10^8) — the coordinates of the starting and ending points of the path, the base station, and its coverage radius, respectively.
Any of the numbers a, b and c can be equal (either any pair or all three numbers). The base station can be located both on the route of Polycarp and outside it.
Output
Print t numbers — answers to given test cases in the order they are written in the test. Each answer is an integer — the number of minutes during which Polycarp will be unavailable during his movement.
Example
Input
9
1 10 7 1
3 3 3 0
8 2 10 4
8 2 10 100
-10 20 -17 2
-3 2 2 0
-3 1 2 0
2 3 2 3
-1 3 -2 2
Output
7
0
4
0
30
5
4
0
3
Note
The following picture illustrates the first test case.
<image> Polycarp goes from 1 to 10. The yellow area shows the coverage area of the station with a radius of coverage of 1, which is located at the point of 7. The green area shows a part of the path when Polycarp is out of coverage area.
Submitted Solution:
```
for _ in range(int(input())):
a,b,c,r = map(int,input().split())
r1 = min(c+r,c-r)
r2 = max(c+r,c-r)
a,b = min(a,b),max(a,b)
if r2<=a or r1>=b:
print(b-a)
else:
print(max(0,r1-a)+max(0,b-r2))
``` | instruction | 0 | 23,949 | 24 | 47,898 |
Yes | output | 1 | 23,949 | 24 | 47,899 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp lives on the coordinate axis Ox and travels from the point x=a to x=b. It moves uniformly rectilinearly at a speed of one unit of distance per minute.
On the axis Ox at the point x=c the base station of the mobile operator is placed. It is known that the radius of its coverage is r. Thus, if Polycarp is at a distance less than or equal to r from the point x=c, then he is in the network coverage area, otherwise — no. The base station can be located both on the route of Polycarp and outside it.
Print the time in minutes during which Polycarp will not be in the coverage area of the network, with a rectilinear uniform movement from x=a to x=b. His speed — one unit of distance per minute.
Input
The first line contains a positive integer t (1 ≤ t ≤ 1000) — the number of test cases. In the following lines are written t test cases.
The description of each test case is one line, which contains four integers a, b, c and r (-10^8 ≤ a,b,c ≤ 10^8, 0 ≤ r ≤ 10^8) — the coordinates of the starting and ending points of the path, the base station, and its coverage radius, respectively.
Any of the numbers a, b and c can be equal (either any pair or all three numbers). The base station can be located both on the route of Polycarp and outside it.
Output
Print t numbers — answers to given test cases in the order they are written in the test. Each answer is an integer — the number of minutes during which Polycarp will be unavailable during his movement.
Example
Input
9
1 10 7 1
3 3 3 0
8 2 10 4
8 2 10 100
-10 20 -17 2
-3 2 2 0
-3 1 2 0
2 3 2 3
-1 3 -2 2
Output
7
0
4
0
30
5
4
0
3
Note
The following picture illustrates the first test case.
<image> Polycarp goes from 1 to 10. The yellow area shows the coverage area of the station with a radius of coverage of 1, which is located at the point of 7. The green area shows a part of the path when Polycarp is out of coverage area.
Submitted Solution:
```
for _ in range(int(input())):
a,b,c,r=map(int,input().split())
r1,r2=c-r,c+r
mx=max(a,b)
a,b=min(a,b),mx
dist=0
if a<r1:
dist+=(r1-a)
if r2<b:
dist+=(b-r2)
if r2<=a or r1>=b:
dist=b-a
print(dist)
``` | instruction | 0 | 23,950 | 24 | 47,900 |
Yes | output | 1 | 23,950 | 24 | 47,901 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp lives on the coordinate axis Ox and travels from the point x=a to x=b. It moves uniformly rectilinearly at a speed of one unit of distance per minute.
On the axis Ox at the point x=c the base station of the mobile operator is placed. It is known that the radius of its coverage is r. Thus, if Polycarp is at a distance less than or equal to r from the point x=c, then he is in the network coverage area, otherwise — no. The base station can be located both on the route of Polycarp and outside it.
Print the time in minutes during which Polycarp will not be in the coverage area of the network, with a rectilinear uniform movement from x=a to x=b. His speed — one unit of distance per minute.
Input
The first line contains a positive integer t (1 ≤ t ≤ 1000) — the number of test cases. In the following lines are written t test cases.
The description of each test case is one line, which contains four integers a, b, c and r (-10^8 ≤ a,b,c ≤ 10^8, 0 ≤ r ≤ 10^8) — the coordinates of the starting and ending points of the path, the base station, and its coverage radius, respectively.
Any of the numbers a, b and c can be equal (either any pair or all three numbers). The base station can be located both on the route of Polycarp and outside it.
Output
Print t numbers — answers to given test cases in the order they are written in the test. Each answer is an integer — the number of minutes during which Polycarp will be unavailable during his movement.
Example
Input
9
1 10 7 1
3 3 3 0
8 2 10 4
8 2 10 100
-10 20 -17 2
-3 2 2 0
-3 1 2 0
2 3 2 3
-1 3 -2 2
Output
7
0
4
0
30
5
4
0
3
Note
The following picture illustrates the first test case.
<image> Polycarp goes from 1 to 10. The yellow area shows the coverage area of the station with a radius of coverage of 1, which is located at the point of 7. The green area shows a part of the path when Polycarp is out of coverage area.
Submitted Solution:
```
for _ in range(int(input())):
a, b , c, r = list(map(int, input().split()))
dis = abs(b-a)
# outside
if((c > a and c>b) or (c<a and c < b)):
z = min(abs(c-a), abs(c-b))
if z >= r:
print(dis)
elif (z < r):
if((dis +z -r) < 0):
print(0)
else:
print(dis+z - r)
# inside
elif((c>a and c < b) or (c<a and c>b)):
z = min(abs(c-a), abs(c-b))
if(r >=max(abs(c-a), abs(c-b))):
print(0)
elif(r >= z and r < max(abs(c-a), abs(c-b))):
print(max(abs(c-a), abs(c-b)))
elif(r <= z):
print(dis -2*r)
elif(c == a or c == b):
ans = dis-r
if(ans < 0):
print(0)
else:
print(ans)
``` | instruction | 0 | 23,951 | 24 | 47,902 |
No | output | 1 | 23,951 | 24 | 47,903 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp lives on the coordinate axis Ox and travels from the point x=a to x=b. It moves uniformly rectilinearly at a speed of one unit of distance per minute.
On the axis Ox at the point x=c the base station of the mobile operator is placed. It is known that the radius of its coverage is r. Thus, if Polycarp is at a distance less than or equal to r from the point x=c, then he is in the network coverage area, otherwise — no. The base station can be located both on the route of Polycarp and outside it.
Print the time in minutes during which Polycarp will not be in the coverage area of the network, with a rectilinear uniform movement from x=a to x=b. His speed — one unit of distance per minute.
Input
The first line contains a positive integer t (1 ≤ t ≤ 1000) — the number of test cases. In the following lines are written t test cases.
The description of each test case is one line, which contains four integers a, b, c and r (-10^8 ≤ a,b,c ≤ 10^8, 0 ≤ r ≤ 10^8) — the coordinates of the starting and ending points of the path, the base station, and its coverage radius, respectively.
Any of the numbers a, b and c can be equal (either any pair or all three numbers). The base station can be located both on the route of Polycarp and outside it.
Output
Print t numbers — answers to given test cases in the order they are written in the test. Each answer is an integer — the number of minutes during which Polycarp will be unavailable during his movement.
Example
Input
9
1 10 7 1
3 3 3 0
8 2 10 4
8 2 10 100
-10 20 -17 2
-3 2 2 0
-3 1 2 0
2 3 2 3
-1 3 -2 2
Output
7
0
4
0
30
5
4
0
3
Note
The following picture illustrates the first test case.
<image> Polycarp goes from 1 to 10. The yellow area shows the coverage area of the station with a radius of coverage of 1, which is located at the point of 7. The green area shows a part of the path when Polycarp is out of coverage area.
Submitted Solution:
```
#import sys
#import math
#sys.stdout=open("C:/Users/pipal/OneDrive/Desktop/VS code/python/output.txt","w")
#sys.stdin=open("C:/Users/pipal/OneDrive/Desktop/VS code/python/input.txt","r")
t=int(input())
for i in range(t):
a,b,c,r=map(int,input().split())
diff=abs(b-a)
if c<=max(b,a) and c>=min(a,b):
if (c+r)>=max(a,b):
t1=0
else:
t1=abs(max(a,b)-((c+r)))
if (c-r)<=min(a,b):
t2=0
else:
t2=abs((c-r)-min(a,b))
print(t1+t2)
elif c>max(a,b):
rd=abs(abs(c)-abs(max(a,b)))
if r<=rd:
print(diff)
else:
temp=(diff-abs(r-rd))
if temp<=0:
print(0)
else:
print(temp)
else:
rl=abs(abs(min(a,b))-abs(c))
if r<=rl:
print(diff)
else:
temp=(diff-abs(r-rl))
if temp<=0:
print(0)
else:
print(temp)
#rereerr
#suubmit changess
``` | instruction | 0 | 23,952 | 24 | 47,904 |
No | output | 1 | 23,952 | 24 | 47,905 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp lives on the coordinate axis Ox and travels from the point x=a to x=b. It moves uniformly rectilinearly at a speed of one unit of distance per minute.
On the axis Ox at the point x=c the base station of the mobile operator is placed. It is known that the radius of its coverage is r. Thus, if Polycarp is at a distance less than or equal to r from the point x=c, then he is in the network coverage area, otherwise — no. The base station can be located both on the route of Polycarp and outside it.
Print the time in minutes during which Polycarp will not be in the coverage area of the network, with a rectilinear uniform movement from x=a to x=b. His speed — one unit of distance per minute.
Input
The first line contains a positive integer t (1 ≤ t ≤ 1000) — the number of test cases. In the following lines are written t test cases.
The description of each test case is one line, which contains four integers a, b, c and r (-10^8 ≤ a,b,c ≤ 10^8, 0 ≤ r ≤ 10^8) — the coordinates of the starting and ending points of the path, the base station, and its coverage radius, respectively.
Any of the numbers a, b and c can be equal (either any pair or all three numbers). The base station can be located both on the route of Polycarp and outside it.
Output
Print t numbers — answers to given test cases in the order they are written in the test. Each answer is an integer — the number of minutes during which Polycarp will be unavailable during his movement.
Example
Input
9
1 10 7 1
3 3 3 0
8 2 10 4
8 2 10 100
-10 20 -17 2
-3 2 2 0
-3 1 2 0
2 3 2 3
-1 3 -2 2
Output
7
0
4
0
30
5
4
0
3
Note
The following picture illustrates the first test case.
<image> Polycarp goes from 1 to 10. The yellow area shows the coverage area of the station with a radius of coverage of 1, which is located at the point of 7. The green area shows a part of the path when Polycarp is out of coverage area.
Submitted Solution:
```
def work(a, b, c, r):
area = range(c-r+1, c+r+1)
count = 0
for i in range(a+1, b+1):
if i not in area:
count += 1
print(count)
t = int(input())
for _ in range(t):
a, b, c, r = map(lambda x: int(x), input().split())
work(a, b, c, r)
``` | instruction | 0 | 23,953 | 24 | 47,906 |
No | output | 1 | 23,953 | 24 | 47,907 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp lives on the coordinate axis Ox and travels from the point x=a to x=b. It moves uniformly rectilinearly at a speed of one unit of distance per minute.
On the axis Ox at the point x=c the base station of the mobile operator is placed. It is known that the radius of its coverage is r. Thus, if Polycarp is at a distance less than or equal to r from the point x=c, then he is in the network coverage area, otherwise — no. The base station can be located both on the route of Polycarp and outside it.
Print the time in minutes during which Polycarp will not be in the coverage area of the network, with a rectilinear uniform movement from x=a to x=b. His speed — one unit of distance per minute.
Input
The first line contains a positive integer t (1 ≤ t ≤ 1000) — the number of test cases. In the following lines are written t test cases.
The description of each test case is one line, which contains four integers a, b, c and r (-10^8 ≤ a,b,c ≤ 10^8, 0 ≤ r ≤ 10^8) — the coordinates of the starting and ending points of the path, the base station, and its coverage radius, respectively.
Any of the numbers a, b and c can be equal (either any pair or all three numbers). The base station can be located both on the route of Polycarp and outside it.
Output
Print t numbers — answers to given test cases in the order they are written in the test. Each answer is an integer — the number of minutes during which Polycarp will be unavailable during his movement.
Example
Input
9
1 10 7 1
3 3 3 0
8 2 10 4
8 2 10 100
-10 20 -17 2
-3 2 2 0
-3 1 2 0
2 3 2 3
-1 3 -2 2
Output
7
0
4
0
30
5
4
0
3
Note
The following picture illustrates the first test case.
<image> Polycarp goes from 1 to 10. The yellow area shows the coverage area of the station with a radius of coverage of 1, which is located at the point of 7. The green area shows a part of the path when Polycarp is out of coverage area.
Submitted Solution:
```
from sys import stdin, stdout
import math
from itertools import permutations, combinations
from collections import defaultdict
def L():
return list(map(int, stdin.readline().split()))
def Li():
return map(int, stdin.readline().split())
def I():
return int(stdin.readline())
P = 1000000007
def xpr(n):
ans = []
for i in range(n):
for j in range(n):
if i ^ j == n:
ans.append([i, j])
return (ans)
def main():
try:
for _ in range(int(input())):
a, b, c, r = Li()
r1, r2 = c - r, c + r
if a > b:
temp = a
a = b
b = temp
if r1>r2:
temp=r1
r1=r2
r2=temp
if a==-1 and b==-2 and c==-1 and r==1:
print(0)
elif r1 >=b and r2>b:
print(b-a)
elif r1>a and r2>=b:
print(r1-a)
elif r1>=a and r2<b:
print((r1-a)+(b-r2))
elif r1<a and r2>=b:
print(0)
elif r1<a and r2<b and r2 >a:
print(b-r2)
else:
print(b-a)
except:
pass
if __name__ == '__main__':
main()
``` | instruction | 0 | 23,954 | 24 | 47,908 |
No | output | 1 | 23,954 | 24 | 47,909 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has just launched his new startup idea. The niche is pretty free and the key vector of development sounds really promising, so he easily found himself some investors ready to sponsor the company. However, he is yet to name the startup!
Actually, Polycarp has already came up with the name but some improvement to it will never hurt. So now he wants to swap letters at some positions in it to obtain the better name. It isn't necessary for letters to be adjacent.
In addition, each of the investors has chosen some index in the name and selected a set of letters that can go there. Indices chosen by different investors are pairwise distinct. If some indices aren't chosen by any investor then any letter can go there.
Finally, Polycarp is sure that the smallest lexicographically name is the best. (Like why do you think Google decided to become Alphabet?)
More formally, you are given a string consisting of lowercase Latin letters from "a" to "f". You can swap letters at any positions arbitrary number of times (zero swaps is also possible).
What is the smallest lexicographically name you can obtain such that the letter at every position is among the allowed letters?
If Polycarp can't produce any valid name then print "Impossible".
Input
The first line is the string s (1 ≤ |s| ≤ 10^5) — the name Polycarp has came up with. The string consists only of lowercase Latin letters from "a" to "f".
The second line contains a single integer m (0 ≤ m ≤ |s|) — the number of investors.
The i-th of the next m lines contain an integer number pos_i and a non-empty string of allowed characters for pos_i (1 ≤ pos_i ≤ |s|). Each string contains pairwise distinct letters from "a" to "f". pos_1, pos_2, ..., pos_m are pairwise distinct. If any position of the string doesn't appear in the investors demands then any letter can go in this position.
Output
If Polycarp can't produce any valid name then print "Impossible".
Otherwise print the smallest lexicographically name Polycarp can obtain by swapping letters in string s such that the letter at every position is among the allowed ones.
Examples
Input
bedefead
5
2 e
1 dc
5 b
7 ef
6 ef
Output
deadbeef
Input
abacaba
0
Output
aaaabbc
Input
fc
2
1 cfab
2 f
Output
cf
Submitted Solution:
```
a = input()
n = int(input())
m = len(a)
se={}
for z in range(n):
p,q = input().split(" ")
p = int(p)-1
s = []
for i in q:
if i in a:
s.append(i)
s.sort()
if len(s)>0:
se[p]=s[0]
stri=""
for i in se.keys():
j = a.find(se[i])
a = a[:j]+a[j+1:]
a = sorted(a)
j = 0
for i in range(m):
if i in se.keys():
stri = stri + se[i]
else:
stri = stri + a[j]
j = j+1
print(stri)
``` | instruction | 0 | 24,743 | 24 | 49,486 |
No | output | 1 | 24,743 | 24 | 49,487 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has just launched his new startup idea. The niche is pretty free and the key vector of development sounds really promising, so he easily found himself some investors ready to sponsor the company. However, he is yet to name the startup!
Actually, Polycarp has already came up with the name but some improvement to it will never hurt. So now he wants to swap letters at some positions in it to obtain the better name. It isn't necessary for letters to be adjacent.
In addition, each of the investors has chosen some index in the name and selected a set of letters that can go there. Indices chosen by different investors are pairwise distinct. If some indices aren't chosen by any investor then any letter can go there.
Finally, Polycarp is sure that the smallest lexicographically name is the best. (Like why do you think Google decided to become Alphabet?)
More formally, you are given a string consisting of lowercase Latin letters from "a" to "f". You can swap letters at any positions arbitrary number of times (zero swaps is also possible).
What is the smallest lexicographically name you can obtain such that the letter at every position is among the allowed letters?
If Polycarp can't produce any valid name then print "Impossible".
Input
The first line is the string s (1 ≤ |s| ≤ 10^5) — the name Polycarp has came up with. The string consists only of lowercase Latin letters from "a" to "f".
The second line contains a single integer m (0 ≤ m ≤ |s|) — the number of investors.
The i-th of the next m lines contain an integer number pos_i and a non-empty string of allowed characters for pos_i (1 ≤ pos_i ≤ |s|). Each string contains pairwise distinct letters from "a" to "f". pos_1, pos_2, ..., pos_m are pairwise distinct. If any position of the string doesn't appear in the investors demands then any letter can go in this position.
Output
If Polycarp can't produce any valid name then print "Impossible".
Otherwise print the smallest lexicographically name Polycarp can obtain by swapping letters in string s such that the letter at every position is among the allowed ones.
Examples
Input
bedefead
5
2 e
1 dc
5 b
7 ef
6 ef
Output
deadbeef
Input
abacaba
0
Output
aaaabbc
Input
fc
2
1 cfab
2 f
Output
cf
Submitted Solution:
```
w = input()
d = {}
for c in w:
d[c] = w.count(c)
inv = dict()
for i in range(int(input())):
idx,name = input().split()
idx = int(idx)-1
name = [c for c in name if c in d]
if name:
inv[idx] = name
def solve(counts, investors, value=None):
if value == None:
value = dict()
if not investors:
return True, value
i, *invs = sorted(investors.items(),key=lambda kv: kv[0])
investor, possible = i
choices = [c for c in counts if counts[c] > 0 and c in possible]
for choice in sorted(choices):
counts[choice] -= 1
value[investor] = choice
good, ret = solve(counts, dict(invs), value)
if good:
return good, ret
value[investor] = None
counts[choice] += 1
return False, None
good, value = solve(d, inv)
if not good:
print(end="Impossible")
else:
for i in range(len(w)):
o = ""
if i in value:
o = value[i]
else:
o = min(c for c in d if d[c] > 0)
d[c] -= 1
print(end=o)
print()
``` | instruction | 0 | 24,744 | 24 | 49,488 |
No | output | 1 | 24,744 | 24 | 49,489 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has just launched his new startup idea. The niche is pretty free and the key vector of development sounds really promising, so he easily found himself some investors ready to sponsor the company. However, he is yet to name the startup!
Actually, Polycarp has already came up with the name but some improvement to it will never hurt. So now he wants to swap letters at some positions in it to obtain the better name. It isn't necessary for letters to be adjacent.
In addition, each of the investors has chosen some index in the name and selected a set of letters that can go there. Indices chosen by different investors are pairwise distinct. If some indices aren't chosen by any investor then any letter can go there.
Finally, Polycarp is sure that the smallest lexicographically name is the best. (Like why do you think Google decided to become Alphabet?)
More formally, you are given a string consisting of lowercase Latin letters from "a" to "f". You can swap letters at any positions arbitrary number of times (zero swaps is also possible).
What is the smallest lexicographically name you can obtain such that the letter at every position is among the allowed letters?
If Polycarp can't produce any valid name then print "Impossible".
Input
The first line is the string s (1 ≤ |s| ≤ 10^5) — the name Polycarp has came up with. The string consists only of lowercase Latin letters from "a" to "f".
The second line contains a single integer m (0 ≤ m ≤ |s|) — the number of investors.
The i-th of the next m lines contain an integer number pos_i and a non-empty string of allowed characters for pos_i (1 ≤ pos_i ≤ |s|). Each string contains pairwise distinct letters from "a" to "f". pos_1, pos_2, ..., pos_m are pairwise distinct. If any position of the string doesn't appear in the investors demands then any letter can go in this position.
Output
If Polycarp can't produce any valid name then print "Impossible".
Otherwise print the smallest lexicographically name Polycarp can obtain by swapping letters in string s such that the letter at every position is among the allowed ones.
Examples
Input
bedefead
5
2 e
1 dc
5 b
7 ef
6 ef
Output
deadbeef
Input
abacaba
0
Output
aaaabbc
Input
fc
2
1 cfab
2 f
Output
cf
Submitted Solution:
```
a = input()
n = int(input())
m = len(a)
se={}
for z in range(n):
p,q = input().split(" ")
p = int(p)-1
s = []
for i in q:
if i in a:
s.append(i)
s.sort()
if len(s)>0:
se[p]=s[0]
stri=""
for i in se.keys():
j = a.find(se[i])
a = a[:j]+a[j+1:]
a = sorted(a)
j = 0
for i in range(m):
if i in se.keys():
stri = stri + se[i]
del(se[i])
else:
stri = stri + a[j]
j = j+1
print(stri)
``` | instruction | 0 | 24,745 | 24 | 49,490 |
No | output | 1 | 24,745 | 24 | 49,491 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has just launched his new startup idea. The niche is pretty free and the key vector of development sounds really promising, so he easily found himself some investors ready to sponsor the company. However, he is yet to name the startup!
Actually, Polycarp has already came up with the name but some improvement to it will never hurt. So now he wants to swap letters at some positions in it to obtain the better name. It isn't necessary for letters to be adjacent.
In addition, each of the investors has chosen some index in the name and selected a set of letters that can go there. Indices chosen by different investors are pairwise distinct. If some indices aren't chosen by any investor then any letter can go there.
Finally, Polycarp is sure that the smallest lexicographically name is the best. (Like why do you think Google decided to become Alphabet?)
More formally, you are given a string consisting of lowercase Latin letters from "a" to "f". You can swap letters at any positions arbitrary number of times (zero swaps is also possible).
What is the smallest lexicographically name you can obtain such that the letter at every position is among the allowed letters?
If Polycarp can't produce any valid name then print "Impossible".
Input
The first line is the string s (1 ≤ |s| ≤ 10^5) — the name Polycarp has came up with. The string consists only of lowercase Latin letters from "a" to "f".
The second line contains a single integer m (0 ≤ m ≤ |s|) — the number of investors.
The i-th of the next m lines contain an integer number pos_i and a non-empty string of allowed characters for pos_i (1 ≤ pos_i ≤ |s|). Each string contains pairwise distinct letters from "a" to "f". pos_1, pos_2, ..., pos_m are pairwise distinct. If any position of the string doesn't appear in the investors demands then any letter can go in this position.
Output
If Polycarp can't produce any valid name then print "Impossible".
Otherwise print the smallest lexicographically name Polycarp can obtain by swapping letters in string s such that the letter at every position is among the allowed ones.
Examples
Input
bedefead
5
2 e
1 dc
5 b
7 ef
6 ef
Output
deadbeef
Input
abacaba
0
Output
aaaabbc
Input
fc
2
1 cfab
2 f
Output
cf
Submitted Solution:
```
s = input()
res = s
ns = [None for _ in range(len(s))]
k = int(input())
if k > 0:
res = list()
for _ in range(k):
pos, rs = [x for x in input().split()]
pos = int(pos)
if s[pos-1] in rs:
if s[pos-1] != rs[0]:
res.append(s[pos-1])
ns[pos-1] = rs[0]
else:
ns[pos-1] = s[pos-1]
else:
res.append(s[pos-1])
ns[pos-1] = rs[0]
res = list(reversed(sorted(res)))
for j in range(len(ns)):
if ns[j] is None:
ns[j] = res.pop()
if len(ns) > 0:
print(''.join(ns))
else:
print('Impossible')
``` | instruction | 0 | 24,746 | 24 | 49,492 |
No | output | 1 | 24,746 | 24 | 49,493 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is constraints.
If you write a solution in Python, then prefer to send it in PyPy to speed up execution time.
A session has begun at Beland State University. Many students are taking exams.
Polygraph Poligrafovich is going to examine a group of n students. Students will take the exam one-by-one in order from 1-th to n-th. Rules of the exam are following:
* The i-th student randomly chooses a ticket.
* if this ticket is too hard to the student, he doesn't answer and goes home immediately (this process is so fast that it's considered no time elapses). This student fails the exam.
* if the student finds the ticket easy, he spends exactly t_i minutes to pass the exam. After it, he immediately gets a mark and goes home.
Students take the exam in the fixed order, one-by-one, without any interruption. At any moment of time, Polygraph Poligrafovich takes the answer from one student.
The duration of the whole exam for all students is M minutes (max t_i ≤ M), so students at the end of the list have a greater possibility to run out of time to pass the exam.
For each student i, you should count the minimum possible number of students who need to fail the exam so the i-th student has enough time to pass the exam.
For each student i, find the answer independently. That is, if when finding the answer for the student i_1 some student j should leave, then while finding the answer for i_2 (i_2>i_1) the student j student does not have to go home.
Input
The first line of the input contains two integers n and M (1 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ M ≤ 2 ⋅ 10^7) — the number of students and the total duration of the exam in minutes, respectively.
The second line of the input contains n integers t_i (1 ≤ t_i ≤ 100) — time in minutes that i-th student spends to answer to a ticket.
It's guaranteed that all values of t_i are not greater than M.
Output
Print n numbers: the i-th number must be equal to the minimum number of students who have to leave the exam in order to i-th student has enough time to pass the exam.
Examples
Input
7 15
1 2 3 4 5 6 7
Output
0 0 0 0 0 2 3
Input
5 100
80 40 40 40 60
Output
0 1 1 2 3
Note
The explanation for the example 1.
Please note that the sum of the first five exam times does not exceed M=15 (the sum is 1+2+3+4+5=15). Thus, the first five students can pass the exam even if all the students before them also pass the exam. In other words, the first five numbers in the answer are 0.
In order for the 6-th student to pass the exam, it is necessary that at least 2 students must fail it before (for example, the 3-rd and 4-th, then the 6-th will finish its exam in 1+2+5+6=14 minutes, which does not exceed M).
In order for the 7-th student to pass the exam, it is necessary that at least 3 students must fail it before (for example, the 2-nd, 5-th and 6-th, then the 7-th will finish its exam in 1+3+4+7=15 minutes, which does not exceed M). | instruction | 0 | 24,827 | 24 | 49,654 |
Tags: brute force, data structures, greedy, math
Correct Solution:
```
"""
Satwik_Tiwari ;) .
25 june , 2020 - Thursday
"""
#===============================================================================================
#importing some useful libraries.
from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
import bisect
from heapq import *
from math import *
from collections import deque
from collections import Counter as counter # Counter(list) return a dict with {key: count}
from itertools import combinations as comb # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
from itertools import permutations as permutate
from bisect import bisect_left as bl
#If the element is already present in the list,
# the left most position where element has to be inserted is returned.
from bisect import bisect_right as br
from bisect import bisect
#If the element is already present in the list,
# the right most position where element has to be inserted is returned
#==============================================================================================
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
# inp = lambda: sys.stdin.readline().rstrip("\r\n")
#===============================================================================================
#some shortcuts
mod = 1000000007
def inp(): return sys.stdin.readline().rstrip("\r\n") #for fast input
def out(var): sys.stdout.write(str(var)) #for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
def graph(vertex): return [[] for i in range(0,vertex+1)]
def zerolist(n): return [0]*n
def nextline(): out("\n") #as stdout.write always print sring.
def testcase(t):
for p in range(t):
solve()
def printlist(a) :
for p in range(0,len(a)):
out(str(a[p]) + ' ')
def lcm(a,b): return (a*b)//gcd(a,b)
def power(a,b):
ans = 1
while(b>0):
if(b%2==1):
ans*=a
a*=a
b//=2
return ans
def ncr(n,r): return factorial(n)//(factorial(r)*factorial(max(n-r,1)))
def isPrime(n) : # Check Prime Number or not
if (n <= 1) : return False
if (n <= 3) : return True
if (n % 2 == 0 or n % 3 == 0) : return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
#===============================================================================================
# code here ;))
def bs(a,l,h,x):
while(l<h):
# print(l,h)
mid = (l+h)//2
if(a[mid] == x):
return mid
if(a[mid] < x):
l = mid+1
else:
h = mid
return l
def sieve(a): #O(n loglogn) nearly linear
#all odd mark 1
for i in range(3,((10**6)+1),2):
a[i] = 1
#marking multiples of i form i*i 0. they are nt prime
for i in range(3,((10**6)+1),2):
for j in range(i*i,((10**6)+1),i):
a[j] = 0
a[2] = 1 #special left case
return (a)
def bfs(g,st):
visited = [0]*(len(g))
visited[st] = 1
queue = []
queue.append(st)
new = []
while(len(queue) != 0):
s = queue.pop()
new.append(s)
for i in g[s]:
if(visited[i] == 0):
visited[i] = 1
queue.append(i)
return new
def solve():
n,m = sep()
a = lis()
cnt = [0]*(101)
ans = []
for i in range(0,n):
if(i==0):
ans.append(0)
cnt[a[i]] +=1
continue
rem = m-a[i]
curr = 0
count = 0
for j in range(1,101):
if(curr + cnt[j]*j <= rem):
count += cnt[j]
curr += cnt[j]*j
continue
else:
count += (rem-curr)//j
break
ans.append(i-count)
cnt[a[i]] +=1
print(*ans)
testcase(1)
# testcase(int(inp()))
``` | output | 1 | 24,827 | 24 | 49,655 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is constraints.
If you write a solution in Python, then prefer to send it in PyPy to speed up execution time.
A session has begun at Beland State University. Many students are taking exams.
Polygraph Poligrafovich is going to examine a group of n students. Students will take the exam one-by-one in order from 1-th to n-th. Rules of the exam are following:
* The i-th student randomly chooses a ticket.
* if this ticket is too hard to the student, he doesn't answer and goes home immediately (this process is so fast that it's considered no time elapses). This student fails the exam.
* if the student finds the ticket easy, he spends exactly t_i minutes to pass the exam. After it, he immediately gets a mark and goes home.
Students take the exam in the fixed order, one-by-one, without any interruption. At any moment of time, Polygraph Poligrafovich takes the answer from one student.
The duration of the whole exam for all students is M minutes (max t_i ≤ M), so students at the end of the list have a greater possibility to run out of time to pass the exam.
For each student i, you should count the minimum possible number of students who need to fail the exam so the i-th student has enough time to pass the exam.
For each student i, find the answer independently. That is, if when finding the answer for the student i_1 some student j should leave, then while finding the answer for i_2 (i_2>i_1) the student j student does not have to go home.
Input
The first line of the input contains two integers n and M (1 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ M ≤ 2 ⋅ 10^7) — the number of students and the total duration of the exam in minutes, respectively.
The second line of the input contains n integers t_i (1 ≤ t_i ≤ 100) — time in minutes that i-th student spends to answer to a ticket.
It's guaranteed that all values of t_i are not greater than M.
Output
Print n numbers: the i-th number must be equal to the minimum number of students who have to leave the exam in order to i-th student has enough time to pass the exam.
Examples
Input
7 15
1 2 3 4 5 6 7
Output
0 0 0 0 0 2 3
Input
5 100
80 40 40 40 60
Output
0 1 1 2 3
Note
The explanation for the example 1.
Please note that the sum of the first five exam times does not exceed M=15 (the sum is 1+2+3+4+5=15). Thus, the first five students can pass the exam even if all the students before them also pass the exam. In other words, the first five numbers in the answer are 0.
In order for the 6-th student to pass the exam, it is necessary that at least 2 students must fail it before (for example, the 3-rd and 4-th, then the 6-th will finish its exam in 1+2+5+6=14 minutes, which does not exceed M).
In order for the 7-th student to pass the exam, it is necessary that at least 3 students must fail it before (for example, the 2-nd, 5-th and 6-th, then the 7-th will finish its exam in 1+3+4+7=15 minutes, which does not exceed M). | instruction | 0 | 24,828 | 24 | 49,656 |
Tags: brute force, data structures, greedy, math
Correct Solution:
```
n, m = [int(i) for i in input().split()]
line = [int(x) for x in input().split()]
count = {i+1:0 for i in range(100)}
sum = 0
for i in line:
p = sum + i
t = 0
if p > m:
for j in range(100, 0, -1):
if count[j] == 0:
continue
else:
if p - count[j] * j > m:
p -= count[j] * j
t += count[j]
else:
l = (p - m) // j
if p - l * j > m:
l += 1
t += l
break
count[i] += 1
sum += i
print(t, end = " ")
``` | output | 1 | 24,828 | 24 | 49,657 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is constraints.
If you write a solution in Python, then prefer to send it in PyPy to speed up execution time.
A session has begun at Beland State University. Many students are taking exams.
Polygraph Poligrafovich is going to examine a group of n students. Students will take the exam one-by-one in order from 1-th to n-th. Rules of the exam are following:
* The i-th student randomly chooses a ticket.
* if this ticket is too hard to the student, he doesn't answer and goes home immediately (this process is so fast that it's considered no time elapses). This student fails the exam.
* if the student finds the ticket easy, he spends exactly t_i minutes to pass the exam. After it, he immediately gets a mark and goes home.
Students take the exam in the fixed order, one-by-one, without any interruption. At any moment of time, Polygraph Poligrafovich takes the answer from one student.
The duration of the whole exam for all students is M minutes (max t_i ≤ M), so students at the end of the list have a greater possibility to run out of time to pass the exam.
For each student i, you should count the minimum possible number of students who need to fail the exam so the i-th student has enough time to pass the exam.
For each student i, find the answer independently. That is, if when finding the answer for the student i_1 some student j should leave, then while finding the answer for i_2 (i_2>i_1) the student j student does not have to go home.
Input
The first line of the input contains two integers n and M (1 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ M ≤ 2 ⋅ 10^7) — the number of students and the total duration of the exam in minutes, respectively.
The second line of the input contains n integers t_i (1 ≤ t_i ≤ 100) — time in minutes that i-th student spends to answer to a ticket.
It's guaranteed that all values of t_i are not greater than M.
Output
Print n numbers: the i-th number must be equal to the minimum number of students who have to leave the exam in order to i-th student has enough time to pass the exam.
Examples
Input
7 15
1 2 3 4 5 6 7
Output
0 0 0 0 0 2 3
Input
5 100
80 40 40 40 60
Output
0 1 1 2 3
Note
The explanation for the example 1.
Please note that the sum of the first five exam times does not exceed M=15 (the sum is 1+2+3+4+5=15). Thus, the first five students can pass the exam even if all the students before them also pass the exam. In other words, the first five numbers in the answer are 0.
In order for the 6-th student to pass the exam, it is necessary that at least 2 students must fail it before (for example, the 3-rd and 4-th, then the 6-th will finish its exam in 1+2+5+6=14 minutes, which does not exceed M).
In order for the 7-th student to pass the exam, it is necessary that at least 3 students must fail it before (for example, the 2-nd, 5-th and 6-th, then the 7-th will finish its exam in 1+3+4+7=15 minutes, which does not exceed M). | instruction | 0 | 24,829 | 24 | 49,658 |
Tags: brute force, data structures, greedy, math
Correct Solution:
```
import math
import heapq
import copy
from collections import OrderedDict
from collections import deque
t=list(map(int,input().split()))
d=t[1]
k=list(map(int,input().split()))
m=[0]*101
ans=list()
time=0
for i in range(t[0]):
time=time+k[i]
if((d-time)>=0):
ans.append(0)
m[k[i]]=m[k[i]]+1
else:
count=0
test=time
for j in range(100,-1,-1):
if(test<=d):
break
if(m[j]>0):
cur=m[j]
vv=cur*j
if(abs(test-vv)<=d):
if(vv<test or vv>test):
x=math.ceil((test-d)/j)
if(x<=cur):
count=count+x
test=test-(x*j)
else:
count=count+cur
test=test-(cur*j)
else:
count=count
test=test-vv
else:
count=count+m[j]
test=test-vv
ans.append(count)
m[k[i]]=m[k[i]]+1
print(*ans)
``` | output | 1 | 24,829 | 24 | 49,659 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is constraints.
If you write a solution in Python, then prefer to send it in PyPy to speed up execution time.
A session has begun at Beland State University. Many students are taking exams.
Polygraph Poligrafovich is going to examine a group of n students. Students will take the exam one-by-one in order from 1-th to n-th. Rules of the exam are following:
* The i-th student randomly chooses a ticket.
* if this ticket is too hard to the student, he doesn't answer and goes home immediately (this process is so fast that it's considered no time elapses). This student fails the exam.
* if the student finds the ticket easy, he spends exactly t_i minutes to pass the exam. After it, he immediately gets a mark and goes home.
Students take the exam in the fixed order, one-by-one, without any interruption. At any moment of time, Polygraph Poligrafovich takes the answer from one student.
The duration of the whole exam for all students is M minutes (max t_i ≤ M), so students at the end of the list have a greater possibility to run out of time to pass the exam.
For each student i, you should count the minimum possible number of students who need to fail the exam so the i-th student has enough time to pass the exam.
For each student i, find the answer independently. That is, if when finding the answer for the student i_1 some student j should leave, then while finding the answer for i_2 (i_2>i_1) the student j student does not have to go home.
Input
The first line of the input contains two integers n and M (1 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ M ≤ 2 ⋅ 10^7) — the number of students and the total duration of the exam in minutes, respectively.
The second line of the input contains n integers t_i (1 ≤ t_i ≤ 100) — time in minutes that i-th student spends to answer to a ticket.
It's guaranteed that all values of t_i are not greater than M.
Output
Print n numbers: the i-th number must be equal to the minimum number of students who have to leave the exam in order to i-th student has enough time to pass the exam.
Examples
Input
7 15
1 2 3 4 5 6 7
Output
0 0 0 0 0 2 3
Input
5 100
80 40 40 40 60
Output
0 1 1 2 3
Note
The explanation for the example 1.
Please note that the sum of the first five exam times does not exceed M=15 (the sum is 1+2+3+4+5=15). Thus, the first five students can pass the exam even if all the students before them also pass the exam. In other words, the first five numbers in the answer are 0.
In order for the 6-th student to pass the exam, it is necessary that at least 2 students must fail it before (for example, the 3-rd and 4-th, then the 6-th will finish its exam in 1+2+5+6=14 minutes, which does not exceed M).
In order for the 7-th student to pass the exam, it is necessary that at least 3 students must fail it before (for example, the 2-nd, 5-th and 6-th, then the 7-th will finish its exam in 1+3+4+7=15 minutes, which does not exceed M). | instruction | 0 | 24,830 | 24 | 49,660 |
Tags: brute force, data structures, greedy, math
Correct Solution:
```
import math
n,m=map(int,input().split())
arr=list(map(int,input().split()))
prefix=[0]*n
count=[0]*101
for i in range(101):
count[i]=[0]*200001
for i in range(n):
prefix[i]=prefix[i-1]+arr[i]
for j in range(1,101):
# print(j,i,count[j][i],count[j][i-1])
count[j][i]=count[j][i-1]
if j==arr[i]:
count[j][i]+=1
# print(prefix)
for i in range(n):
# print(prefix[i],m)
if prefix[i]<=m:
print(0,end=" ")
else:
ct=0
sm=prefix[i]
for j in range(100,0,-1):
if sm-count[j][i-1]*j>m:
ct+=count[j][i-1]
sm-=count[j][i-1]*j
else:
# print(sm,j,arr[i-1])
ct+=math.ceil((sm-m)/j)
print(ct,end=" ")
break
``` | output | 1 | 24,830 | 24 | 49,661 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is constraints.
If you write a solution in Python, then prefer to send it in PyPy to speed up execution time.
A session has begun at Beland State University. Many students are taking exams.
Polygraph Poligrafovich is going to examine a group of n students. Students will take the exam one-by-one in order from 1-th to n-th. Rules of the exam are following:
* The i-th student randomly chooses a ticket.
* if this ticket is too hard to the student, he doesn't answer and goes home immediately (this process is so fast that it's considered no time elapses). This student fails the exam.
* if the student finds the ticket easy, he spends exactly t_i minutes to pass the exam. After it, he immediately gets a mark and goes home.
Students take the exam in the fixed order, one-by-one, without any interruption. At any moment of time, Polygraph Poligrafovich takes the answer from one student.
The duration of the whole exam for all students is M minutes (max t_i ≤ M), so students at the end of the list have a greater possibility to run out of time to pass the exam.
For each student i, you should count the minimum possible number of students who need to fail the exam so the i-th student has enough time to pass the exam.
For each student i, find the answer independently. That is, if when finding the answer for the student i_1 some student j should leave, then while finding the answer for i_2 (i_2>i_1) the student j student does not have to go home.
Input
The first line of the input contains two integers n and M (1 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ M ≤ 2 ⋅ 10^7) — the number of students and the total duration of the exam in minutes, respectively.
The second line of the input contains n integers t_i (1 ≤ t_i ≤ 100) — time in minutes that i-th student spends to answer to a ticket.
It's guaranteed that all values of t_i are not greater than M.
Output
Print n numbers: the i-th number must be equal to the minimum number of students who have to leave the exam in order to i-th student has enough time to pass the exam.
Examples
Input
7 15
1 2 3 4 5 6 7
Output
0 0 0 0 0 2 3
Input
5 100
80 40 40 40 60
Output
0 1 1 2 3
Note
The explanation for the example 1.
Please note that the sum of the first five exam times does not exceed M=15 (the sum is 1+2+3+4+5=15). Thus, the first five students can pass the exam even if all the students before them also pass the exam. In other words, the first five numbers in the answer are 0.
In order for the 6-th student to pass the exam, it is necessary that at least 2 students must fail it before (for example, the 3-rd and 4-th, then the 6-th will finish its exam in 1+2+5+6=14 minutes, which does not exceed M).
In order for the 7-th student to pass the exam, it is necessary that at least 3 students must fail it before (for example, the 2-nd, 5-th and 6-th, then the 7-th will finish its exam in 1+3+4+7=15 minutes, which does not exceed M). | instruction | 0 | 24,831 | 24 | 49,662 |
Tags: brute force, data structures, greedy, math
Correct Solution:
```
n, m = list(map(int, input().split()))
t = list(map(int, input().split()))
freq = []
for i in range(101):
freq.append(0)
for i in range(0, n):
time = m-t[i]
num = 0
for j in range(1, 101):
p = int(time/j)
if p >= freq[j]:
time -= freq[j]*j
num += freq[j]
else:
time -= p*j
num += p
print(i-num, end=" ")
freq[t[i]] += 1
``` | output | 1 | 24,831 | 24 | 49,663 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.