message stringlengths 2 20.2k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 757 108k | cluster float64 4 4 | __index_level_0__ int64 1.51k 217k |
|---|---|---|---|---|---|
Provide a correct Python 3 solution for this coding contest problem.
On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`.
After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.
Constraints
* S is a string of length 10.
* The first eight characters in S are `2017/01/`.
* The last two characters in S are digits and represent an integer between 1 and 31 (inclusive).
Input
Input is given from Standard Input in the following format:
S
Output
Replace the first four characters in S with `2018` and print it.
Examples
Input
2017/01/07
Output
2018/01/07
Input
2017/01/31
Output
2018/01/31 | instruction | 0 | 47,791 | 4 | 95,582 |
"Correct Solution:
```
print(f"{2018}{input()[4:]}")
``` | output | 1 | 47,791 | 4 | 95,583 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`.
After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.
Constraints
* S is a string of length 10.
* The first eight characters in S are `2017/01/`.
* The last two characters in S are digits and represent an integer between 1 and 31 (inclusive).
Input
Input is given from Standard Input in the following format:
S
Output
Replace the first four characters in S with `2018` and print it.
Examples
Input
2017/01/07
Output
2018/01/07
Input
2017/01/31
Output
2018/01/31
Submitted Solution:
```
s = input()
ss = "2018"+s[4:]
print(ss)
``` | instruction | 0 | 47,792 | 4 | 95,584 |
Yes | output | 1 | 47,792 | 4 | 95,585 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`.
After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.
Constraints
* S is a string of length 10.
* The first eight characters in S are `2017/01/`.
* The last two characters in S are digits and represent an integer between 1 and 31 (inclusive).
Input
Input is given from Standard Input in the following format:
S
Output
Replace the first four characters in S with `2018` and print it.
Examples
Input
2017/01/07
Output
2018/01/07
Input
2017/01/31
Output
2018/01/31
Submitted Solution:
```
a=input()
print(a.replace("7","8",1))
``` | instruction | 0 | 47,793 | 4 | 95,586 |
Yes | output | 1 | 47,793 | 4 | 95,587 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`.
After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.
Constraints
* S is a string of length 10.
* The first eight characters in S are `2017/01/`.
* The last two characters in S are digits and represent an integer between 1 and 31 (inclusive).
Input
Input is given from Standard Input in the following format:
S
Output
Replace the first four characters in S with `2018` and print it.
Examples
Input
2017/01/07
Output
2018/01/07
Input
2017/01/31
Output
2018/01/31
Submitted Solution:
```
s=input()
print('2018{}'.format(s[4:]))
``` | instruction | 0 | 47,794 | 4 | 95,588 |
Yes | output | 1 | 47,794 | 4 | 95,589 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`.
After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.
Constraints
* S is a string of length 10.
* The first eight characters in S are `2017/01/`.
* The last two characters in S are digits and represent an integer between 1 and 31 (inclusive).
Input
Input is given from Standard Input in the following format:
S
Output
Replace the first four characters in S with `2018` and print it.
Examples
Input
2017/01/07
Output
2018/01/07
Input
2017/01/31
Output
2018/01/31
Submitted Solution:
```
N, Y = map(int, input().split())
for i in range(N+1):
for j in range(N+1-i):
if 10000*i + 5000*j + 1000*(N-i-j) == Y:
print(i, j, N-i-j)
break
else:
print("-1 -1 -1")
``` | instruction | 0 | 47,796 | 4 | 95,592 |
No | output | 1 | 47,796 | 4 | 95,593 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`.
After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.
Constraints
* S is a string of length 10.
* The first eight characters in S are `2017/01/`.
* The last two characters in S are digits and represent an integer between 1 and 31 (inclusive).
Input
Input is given from Standard Input in the following format:
S
Output
Replace the first four characters in S with `2018` and print it.
Examples
Input
2017/01/07
Output
2018/01/07
Input
2017/01/31
Output
2018/01/31
Submitted Solution:
```
n, y = map(int, input().split())
noguchi = 1000
higuchi = 5000
yukichi = 10000
ans = '-1 -1 -1'
for i in range(n+1):
for j in range(n+1-i):
k = n - (i+j)
yen = noguchi * i + higuchi * j + yukichi * k
if yen == y:
ans = '{0} {1} {2}'.format(k,j,i)
break
print(ans)
``` | instruction | 0 | 47,797 | 4 | 95,594 |
No | output | 1 | 47,797 | 4 | 95,595 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`.
After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.
Constraints
* S is a string of length 10.
* The first eight characters in S are `2017/01/`.
* The last two characters in S are digits and represent an integer between 1 and 31 (inclusive).
Input
Input is given from Standard Input in the following format:
S
Output
Replace the first four characters in S with `2018` and print it.
Examples
Input
2017/01/07
Output
2018/01/07
Input
2017/01/31
Output
2018/01/31
Submitted Solution:
```
N = list(input())
N[0] = "2"
N[1] = "0"
N[2] = "1"
N[3] = "8"
print(N)
date = ''.join(N)
print(date)
``` | instruction | 0 | 47,799 | 4 | 95,598 |
No | output | 1 | 47,799 | 4 | 95,599 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Today an outstanding event is going to happen in the forest — hedgehog Filya will come to his old fried Sonya!
Sonya is an owl and she sleeps during the day and stay awake from minute l1 to minute r1 inclusive. Also, during the minute k she prinks and is unavailable for Filya.
Filya works a lot and he plans to visit Sonya from minute l2 to minute r2 inclusive.
Calculate the number of minutes they will be able to spend together.
Input
The only line of the input contains integers l1, r1, l2, r2 and k (1 ≤ l1, r1, l2, r2, k ≤ 1018, l1 ≤ r1, l2 ≤ r2), providing the segments of time for Sonya and Filya and the moment of time when Sonya prinks.
Output
Print one integer — the number of minutes Sonya and Filya will be able to spend together.
Examples
Input
1 10 9 20 1
Output
2
Input
1 100 50 200 75
Output
50
Note
In the first sample, they will be together during minutes 9 and 10.
In the second sample, they will be together from minute 50 to minute 74 and from minute 76 to minute 100. | instruction | 0 | 48,418 | 4 | 96,836 |
Tags: implementation, math
Correct Solution:
```
a,b,c,d,k=map(int,input().split())
l=max(a,c)
r=min(b,d)
if r<l:
print(0)
elif l<=k<=r and a<=k<=b:
print (r-l)
else :
print (r-l+1)
``` | output | 1 | 48,418 | 4 | 96,837 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Today an outstanding event is going to happen in the forest — hedgehog Filya will come to his old fried Sonya!
Sonya is an owl and she sleeps during the day and stay awake from minute l1 to minute r1 inclusive. Also, during the minute k she prinks and is unavailable for Filya.
Filya works a lot and he plans to visit Sonya from minute l2 to minute r2 inclusive.
Calculate the number of minutes they will be able to spend together.
Input
The only line of the input contains integers l1, r1, l2, r2 and k (1 ≤ l1, r1, l2, r2, k ≤ 1018, l1 ≤ r1, l2 ≤ r2), providing the segments of time for Sonya and Filya and the moment of time when Sonya prinks.
Output
Print one integer — the number of minutes Sonya and Filya will be able to spend together.
Examples
Input
1 10 9 20 1
Output
2
Input
1 100 50 200 75
Output
50
Note
In the first sample, they will be together during minutes 9 and 10.
In the second sample, they will be together from minute 50 to minute 74 and from minute 76 to minute 100. | instruction | 0 | 48,419 | 4 | 96,838 |
Tags: implementation, math
Correct Solution:
```
i1,s1,i2,s2,k=map(int,input().split())
if i2<=s1 and i2>=i1 and s2>=s1:
if k>=i2 and k <=s1 :
print(s1-i2)
else:
print(s1-i2+1)
elif i2>=i1 and i2<=s1 and s2<=s1:
if k>=i2 and k <=s2 :
print(s2-i2)
else:
print(s2-i2+1)
elif i1<=s2 and i1>=i2 and s1>=s2:
if k>=i1 and k <=s2 :
print(s2-i1)
else:
print(s2-i1+1)
elif i1<=s2 and i1>=i2 and s1<=s2:
if k>=i1 and k <=s1 :
print(s1-i1)
else:
print(s1-i1+1)
else:
print(0)
``` | output | 1 | 48,419 | 4 | 96,839 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Today an outstanding event is going to happen in the forest — hedgehog Filya will come to his old fried Sonya!
Sonya is an owl and she sleeps during the day and stay awake from minute l1 to minute r1 inclusive. Also, during the minute k she prinks and is unavailable for Filya.
Filya works a lot and he plans to visit Sonya from minute l2 to minute r2 inclusive.
Calculate the number of minutes they will be able to spend together.
Input
The only line of the input contains integers l1, r1, l2, r2 and k (1 ≤ l1, r1, l2, r2, k ≤ 1018, l1 ≤ r1, l2 ≤ r2), providing the segments of time for Sonya and Filya and the moment of time when Sonya prinks.
Output
Print one integer — the number of minutes Sonya and Filya will be able to spend together.
Examples
Input
1 10 9 20 1
Output
2
Input
1 100 50 200 75
Output
50
Note
In the first sample, they will be together during minutes 9 and 10.
In the second sample, they will be together from minute 50 to minute 74 and from minute 76 to minute 100. | instruction | 0 | 48,420 | 4 | 96,840 |
Tags: implementation, math
Correct Solution:
```
inp = list(map(int, input().split()))
l = max(inp[0], inp[2])
r = min(inp[1], inp[3])
res = max(r - l + 1, 0)
print(res - (l <= inp[4] <= r))
``` | output | 1 | 48,420 | 4 | 96,841 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Today an outstanding event is going to happen in the forest — hedgehog Filya will come to his old fried Sonya!
Sonya is an owl and she sleeps during the day and stay awake from minute l1 to minute r1 inclusive. Also, during the minute k she prinks and is unavailable for Filya.
Filya works a lot and he plans to visit Sonya from minute l2 to minute r2 inclusive.
Calculate the number of minutes they will be able to spend together.
Input
The only line of the input contains integers l1, r1, l2, r2 and k (1 ≤ l1, r1, l2, r2, k ≤ 1018, l1 ≤ r1, l2 ≤ r2), providing the segments of time for Sonya and Filya and the moment of time when Sonya prinks.
Output
Print one integer — the number of minutes Sonya and Filya will be able to spend together.
Examples
Input
1 10 9 20 1
Output
2
Input
1 100 50 200 75
Output
50
Note
In the first sample, they will be together during minutes 9 and 10.
In the second sample, they will be together from minute 50 to minute 74 and from minute 76 to minute 100. | instruction | 0 | 48,421 | 4 | 96,842 |
Tags: implementation, math
Correct Solution:
```
i = input().split()
start = max(int(i[0]), int(i[2]))
end = min(int(i[1]), int(i[3]))
res = end - start + 1
if start <= int(i[4]) <= end:
res -= 1
if res < 0:
res = 0
print(res)
``` | output | 1 | 48,421 | 4 | 96,843 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Today an outstanding event is going to happen in the forest — hedgehog Filya will come to his old fried Sonya!
Sonya is an owl and she sleeps during the day and stay awake from minute l1 to minute r1 inclusive. Also, during the minute k she prinks and is unavailable for Filya.
Filya works a lot and he plans to visit Sonya from minute l2 to minute r2 inclusive.
Calculate the number of minutes they will be able to spend together.
Input
The only line of the input contains integers l1, r1, l2, r2 and k (1 ≤ l1, r1, l2, r2, k ≤ 1018, l1 ≤ r1, l2 ≤ r2), providing the segments of time for Sonya and Filya and the moment of time when Sonya prinks.
Output
Print one integer — the number of minutes Sonya and Filya will be able to spend together.
Examples
Input
1 10 9 20 1
Output
2
Input
1 100 50 200 75
Output
50
Note
In the first sample, they will be together during minutes 9 and 10.
In the second sample, they will be together from minute 50 to minute 74 and from minute 76 to minute 100. | instruction | 0 | 48,422 | 4 | 96,844 |
Tags: implementation, math
Correct Solution:
```
#aditya76
import math
l1,r1,l2,r2,k=map(int,input().split())
nach=max(l1,l2)
kon=min(r1,r2)
if nach>kon:
print(0)
elif nach<=k<=kon:
print(kon-nach)
else:
print(kon-nach+1)
``` | output | 1 | 48,422 | 4 | 96,845 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Today an outstanding event is going to happen in the forest — hedgehog Filya will come to his old fried Sonya!
Sonya is an owl and she sleeps during the day and stay awake from minute l1 to minute r1 inclusive. Also, during the minute k she prinks and is unavailable for Filya.
Filya works a lot and he plans to visit Sonya from minute l2 to minute r2 inclusive.
Calculate the number of minutes they will be able to spend together.
Input
The only line of the input contains integers l1, r1, l2, r2 and k (1 ≤ l1, r1, l2, r2, k ≤ 1018, l1 ≤ r1, l2 ≤ r2), providing the segments of time for Sonya and Filya and the moment of time when Sonya prinks.
Output
Print one integer — the number of minutes Sonya and Filya will be able to spend together.
Examples
Input
1 10 9 20 1
Output
2
Input
1 100 50 200 75
Output
50
Note
In the first sample, they will be together during minutes 9 and 10.
In the second sample, they will be together from minute 50 to minute 74 and from minute 76 to minute 100. | instruction | 0 | 48,423 | 4 | 96,846 |
Tags: implementation, math
Correct Solution:
```
l1, r1, l2, r2, k = map(int,input().split())
maxd = 0
if l1 <= l2 <= r2 <= r1 or l2 <= l1 <= r1 <= r2:
if r2 <= r1:
if l2 <= k <= r2:
maxd = r2 - l2
else:
maxd = r2 - l2 + 1
else:
if l1<=k<=r1:
maxd = r1 - l1
else:
maxd = r1 - l1 + 1
print(maxd)
elif l1 <= l2 <= r1 <= r2 or l2 <= l1 <= r2 <= r1:
if r1 <= r2:
if l2<=k<=r1:
maxd = r1 - l2
else:
maxd = r1 - l2 + 1
else:
if l1<=k<=r2:
maxd = r2 - l1
else:
maxd = r2 - l1 + 1
print(maxd)
else:
print(maxd)
``` | output | 1 | 48,423 | 4 | 96,847 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Today an outstanding event is going to happen in the forest — hedgehog Filya will come to his old fried Sonya!
Sonya is an owl and she sleeps during the day and stay awake from minute l1 to minute r1 inclusive. Also, during the minute k she prinks and is unavailable for Filya.
Filya works a lot and he plans to visit Sonya from minute l2 to minute r2 inclusive.
Calculate the number of minutes they will be able to spend together.
Input
The only line of the input contains integers l1, r1, l2, r2 and k (1 ≤ l1, r1, l2, r2, k ≤ 1018, l1 ≤ r1, l2 ≤ r2), providing the segments of time for Sonya and Filya and the moment of time when Sonya prinks.
Output
Print one integer — the number of minutes Sonya and Filya will be able to spend together.
Examples
Input
1 10 9 20 1
Output
2
Input
1 100 50 200 75
Output
50
Note
In the first sample, they will be together during minutes 9 and 10.
In the second sample, they will be together from minute 50 to minute 74 and from minute 76 to minute 100. | instruction | 0 | 48,424 | 4 | 96,848 |
Tags: implementation, math
Correct Solution:
```
l1,r1,l2,r2,k=(int(z) for z in input().split())
L=max(l1,l2)
R=min(r1,r2)
ans=0
if L<=k<=R:
ans-=1
print(max(ans+R-L+1,0))
``` | output | 1 | 48,424 | 4 | 96,849 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Today an outstanding event is going to happen in the forest — hedgehog Filya will come to his old fried Sonya!
Sonya is an owl and she sleeps during the day and stay awake from minute l1 to minute r1 inclusive. Also, during the minute k she prinks and is unavailable for Filya.
Filya works a lot and he plans to visit Sonya from minute l2 to minute r2 inclusive.
Calculate the number of minutes they will be able to spend together.
Input
The only line of the input contains integers l1, r1, l2, r2 and k (1 ≤ l1, r1, l2, r2, k ≤ 1018, l1 ≤ r1, l2 ≤ r2), providing the segments of time for Sonya and Filya and the moment of time when Sonya prinks.
Output
Print one integer — the number of minutes Sonya and Filya will be able to spend together.
Examples
Input
1 10 9 20 1
Output
2
Input
1 100 50 200 75
Output
50
Note
In the first sample, they will be together during minutes 9 and 10.
In the second sample, they will be together from minute 50 to minute 74 and from minute 76 to minute 100. | instruction | 0 | 48,425 | 4 | 96,850 |
Tags: implementation, math
Correct Solution:
```
l1, r1, l2, r2, k = map(int, input().split())
l = max(l1, l2)
r = min(r1, r2)
ans = max(r - l + 1, 0)
if k >= l and k <= r:
ans -= 1
print(ans)
``` | output | 1 | 48,425 | 4 | 96,851 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today an outstanding event is going to happen in the forest — hedgehog Filya will come to his old fried Sonya!
Sonya is an owl and she sleeps during the day and stay awake from minute l1 to minute r1 inclusive. Also, during the minute k she prinks and is unavailable for Filya.
Filya works a lot and he plans to visit Sonya from minute l2 to minute r2 inclusive.
Calculate the number of minutes they will be able to spend together.
Input
The only line of the input contains integers l1, r1, l2, r2 and k (1 ≤ l1, r1, l2, r2, k ≤ 1018, l1 ≤ r1, l2 ≤ r2), providing the segments of time for Sonya and Filya and the moment of time when Sonya prinks.
Output
Print one integer — the number of minutes Sonya and Filya will be able to spend together.
Examples
Input
1 10 9 20 1
Output
2
Input
1 100 50 200 75
Output
50
Note
In the first sample, they will be together during minutes 9 and 10.
In the second sample, they will be together from minute 50 to minute 74 and from minute 76 to minute 100.
Submitted Solution:
```
l1, r1, l2, r2, k = [int(num) for num in input().split()]
range = [max([l1, l2]), min(r1, r2)]
numMinutes = range[1] - range[0] + 1
if k >= range[0] and k <= range[1]:
numMinutes -= 1
if numMinutes > 0:
print(numMinutes)
else:
print(0)
``` | instruction | 0 | 48,426 | 4 | 96,852 |
Yes | output | 1 | 48,426 | 4 | 96,853 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today an outstanding event is going to happen in the forest — hedgehog Filya will come to his old fried Sonya!
Sonya is an owl and she sleeps during the day and stay awake from minute l1 to minute r1 inclusive. Also, during the minute k she prinks and is unavailable for Filya.
Filya works a lot and he plans to visit Sonya from minute l2 to minute r2 inclusive.
Calculate the number of minutes they will be able to spend together.
Input
The only line of the input contains integers l1, r1, l2, r2 and k (1 ≤ l1, r1, l2, r2, k ≤ 1018, l1 ≤ r1, l2 ≤ r2), providing the segments of time for Sonya and Filya and the moment of time when Sonya prinks.
Output
Print one integer — the number of minutes Sonya and Filya will be able to spend together.
Examples
Input
1 10 9 20 1
Output
2
Input
1 100 50 200 75
Output
50
Note
In the first sample, they will be together during minutes 9 and 10.
In the second sample, they will be together from minute 50 to minute 74 and from minute 76 to minute 100.
Submitted Solution:
```
l1,r1,l2,r2,k = (int(i) for i in input().split())
l = max(l1,l2)
r = min(r1,r2)
if r < l:
print(0)
else:
ans = r-l
if k >= l and k <= r:
print(ans)
else:
print(ans+1)
``` | instruction | 0 | 48,427 | 4 | 96,854 |
Yes | output | 1 | 48,427 | 4 | 96,855 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today an outstanding event is going to happen in the forest — hedgehog Filya will come to his old fried Sonya!
Sonya is an owl and she sleeps during the day and stay awake from minute l1 to minute r1 inclusive. Also, during the minute k she prinks and is unavailable for Filya.
Filya works a lot and he plans to visit Sonya from minute l2 to minute r2 inclusive.
Calculate the number of minutes they will be able to spend together.
Input
The only line of the input contains integers l1, r1, l2, r2 and k (1 ≤ l1, r1, l2, r2, k ≤ 1018, l1 ≤ r1, l2 ≤ r2), providing the segments of time for Sonya and Filya and the moment of time when Sonya prinks.
Output
Print one integer — the number of minutes Sonya and Filya will be able to spend together.
Examples
Input
1 10 9 20 1
Output
2
Input
1 100 50 200 75
Output
50
Note
In the first sample, they will be together during minutes 9 and 10.
In the second sample, they will be together from minute 50 to minute 74 and from minute 76 to minute 100.
Submitted Solution:
```
l1, r1, l2, r2, k = map(int, input().split())
begin, end = max(l1, l2), min(r1, r2)
print(max(end - begin + (begin > k or k > end), 0))
``` | instruction | 0 | 48,428 | 4 | 96,856 |
Yes | output | 1 | 48,428 | 4 | 96,857 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today an outstanding event is going to happen in the forest — hedgehog Filya will come to his old fried Sonya!
Sonya is an owl and she sleeps during the day and stay awake from minute l1 to minute r1 inclusive. Also, during the minute k she prinks and is unavailable for Filya.
Filya works a lot and he plans to visit Sonya from minute l2 to minute r2 inclusive.
Calculate the number of minutes they will be able to spend together.
Input
The only line of the input contains integers l1, r1, l2, r2 and k (1 ≤ l1, r1, l2, r2, k ≤ 1018, l1 ≤ r1, l2 ≤ r2), providing the segments of time for Sonya and Filya and the moment of time when Sonya prinks.
Output
Print one integer — the number of minutes Sonya and Filya will be able to spend together.
Examples
Input
1 10 9 20 1
Output
2
Input
1 100 50 200 75
Output
50
Note
In the first sample, they will be together during minutes 9 and 10.
In the second sample, they will be together from minute 50 to minute 74 and from minute 76 to minute 100.
Submitted Solution:
```
a,b,c,d,k = map(int,input().split())
mx = max(a,c)
mn = min(b,d)
ans = mn - mx + 1
if mx<=k<=mn:
ans-=1
print(max(ans,0))
``` | instruction | 0 | 48,429 | 4 | 96,858 |
Yes | output | 1 | 48,429 | 4 | 96,859 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today an outstanding event is going to happen in the forest — hedgehog Filya will come to his old fried Sonya!
Sonya is an owl and she sleeps during the day and stay awake from minute l1 to minute r1 inclusive. Also, during the minute k she prinks and is unavailable for Filya.
Filya works a lot and he plans to visit Sonya from minute l2 to minute r2 inclusive.
Calculate the number of minutes they will be able to spend together.
Input
The only line of the input contains integers l1, r1, l2, r2 and k (1 ≤ l1, r1, l2, r2, k ≤ 1018, l1 ≤ r1, l2 ≤ r2), providing the segments of time for Sonya and Filya and the moment of time when Sonya prinks.
Output
Print one integer — the number of minutes Sonya and Filya will be able to spend together.
Examples
Input
1 10 9 20 1
Output
2
Input
1 100 50 200 75
Output
50
Note
In the first sample, they will be together during minutes 9 and 10.
In the second sample, they will be together from minute 50 to minute 74 and from minute 76 to minute 100.
Submitted Solution:
```
l1, r1, l2, r2, k = list(map(int,input().split()))
s1 = max(l1, l2)
s2 = min(r1, r2)
ans = (s2 - s1) + 1
if s1 < k < s2:
ans -= 1
print(ans)
``` | instruction | 0 | 48,430 | 4 | 96,860 |
No | output | 1 | 48,430 | 4 | 96,861 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today an outstanding event is going to happen in the forest — hedgehog Filya will come to his old fried Sonya!
Sonya is an owl and she sleeps during the day and stay awake from minute l1 to minute r1 inclusive. Also, during the minute k she prinks and is unavailable for Filya.
Filya works a lot and he plans to visit Sonya from minute l2 to minute r2 inclusive.
Calculate the number of minutes they will be able to spend together.
Input
The only line of the input contains integers l1, r1, l2, r2 and k (1 ≤ l1, r1, l2, r2, k ≤ 1018, l1 ≤ r1, l2 ≤ r2), providing the segments of time for Sonya and Filya and the moment of time when Sonya prinks.
Output
Print one integer — the number of minutes Sonya and Filya will be able to spend together.
Examples
Input
1 10 9 20 1
Output
2
Input
1 100 50 200 75
Output
50
Note
In the first sample, they will be together during minutes 9 and 10.
In the second sample, they will be together from minute 50 to minute 74 and from minute 76 to minute 100.
Submitted Solution:
```
l1,r1,l2,r2,k=map(int,input().split())
if l1<=l2:
if r1>=l2:
if l2<=k<=r1:
print(-l2+r1)
else:
print(-l2+r1+1)
else:
print(0)
else:
if r2>=l1:
if l1<=k<=r2:
print(-l1+r2)
else:
print(-l1+r2+1)
else:
print(0)
``` | instruction | 0 | 48,431 | 4 | 96,862 |
No | output | 1 | 48,431 | 4 | 96,863 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today an outstanding event is going to happen in the forest — hedgehog Filya will come to his old fried Sonya!
Sonya is an owl and she sleeps during the day and stay awake from minute l1 to minute r1 inclusive. Also, during the minute k she prinks and is unavailable for Filya.
Filya works a lot and he plans to visit Sonya from minute l2 to minute r2 inclusive.
Calculate the number of minutes they will be able to spend together.
Input
The only line of the input contains integers l1, r1, l2, r2 and k (1 ≤ l1, r1, l2, r2, k ≤ 1018, l1 ≤ r1, l2 ≤ r2), providing the segments of time for Sonya and Filya and the moment of time when Sonya prinks.
Output
Print one integer — the number of minutes Sonya and Filya will be able to spend together.
Examples
Input
1 10 9 20 1
Output
2
Input
1 100 50 200 75
Output
50
Note
In the first sample, they will be together during minutes 9 and 10.
In the second sample, they will be together from minute 50 to minute 74 and from minute 76 to minute 100.
Submitted Solution:
```
l1,r1,l2,r2,k=map(int,input().split())
s=max(l1,l2)
e=min(r1,r2)
re=(e-s)
if k<=e and k>=s:
#print("hi")
print(re)
else:
print(re+1)
``` | instruction | 0 | 48,432 | 4 | 96,864 |
No | output | 1 | 48,432 | 4 | 96,865 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today an outstanding event is going to happen in the forest — hedgehog Filya will come to his old fried Sonya!
Sonya is an owl and she sleeps during the day and stay awake from minute l1 to minute r1 inclusive. Also, during the minute k she prinks and is unavailable for Filya.
Filya works a lot and he plans to visit Sonya from minute l2 to minute r2 inclusive.
Calculate the number of minutes they will be able to spend together.
Input
The only line of the input contains integers l1, r1, l2, r2 and k (1 ≤ l1, r1, l2, r2, k ≤ 1018, l1 ≤ r1, l2 ≤ r2), providing the segments of time for Sonya and Filya and the moment of time when Sonya prinks.
Output
Print one integer — the number of minutes Sonya and Filya will be able to spend together.
Examples
Input
1 10 9 20 1
Output
2
Input
1 100 50 200 75
Output
50
Note
In the first sample, they will be together during minutes 9 and 10.
In the second sample, they will be together from minute 50 to minute 74 and from minute 76 to minute 100.
Submitted Solution:
```
l1, r1, l2, r2, k = list(map(int,input().split()))
ans = l2 - r1 +1
if k >= r1 and k <= l2 :
ans = ans -1
print(ans)
``` | instruction | 0 | 48,433 | 4 | 96,866 |
No | output | 1 | 48,433 | 4 | 96,867 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n TV shows you want to watch. Suppose the whole time is split into equal parts called "minutes". The i-th of the shows is going from l_i-th to r_i-th minute, both ends inclusive.
You need a TV to watch a TV show and you can't watch two TV shows which air at the same time on the same TV, so it is possible you will need multiple TVs in some minutes. For example, if segments [l_i, r_i] and [l_j, r_j] intersect, then shows i and j can't be watched simultaneously on one TV.
Once you start watching a show on some TV it is not possible to "move" it to another TV (since it would be too distracting), or to watch another show on the same TV until this show ends.
There is a TV Rental shop near you. It rents a TV for x rupees, and charges y (y < x) rupees for every extra minute you keep the TV. So in order to rent a TV for minutes [a; b] you will need to pay x + y ⋅ (b - a).
You can assume, that taking and returning of the TV doesn't take any time and doesn't distract from watching other TV shows. Find the minimum possible cost to view all shows. Since this value could be too large, print it modulo 10^9 + 7.
Input
The first line contains integers n, x and y (1 ≤ n ≤ 10^5, 1 ≤ y < x ≤ 10^9) — the number of TV shows, the cost to rent a TV for the first minute and the cost to rent a TV for every subsequent minute.
Each of the next n lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ 10^9) denoting the start and the end minute of the i-th TV show.
Output
Print exactly one integer — the minimum cost to view all the shows taken modulo 10^9 + 7.
Examples
Input
5 4 3
1 2
4 10
2 4
10 11
5 9
Output
60
Input
6 3 2
8 20
6 22
4 15
20 28
17 25
20 27
Output
142
Input
2 1000000000 2
1 2
2 3
Output
999999997
Note
In the first example, the optimal strategy would be to rent 3 TVs to watch:
* Show [1, 2] on the first TV,
* Show [4, 10] on the second TV,
* Shows [2, 4], [5, 9], [10, 11] on the third TV.
This way the cost for the first TV is 4 + 3 ⋅ (2 - 1) = 7, for the second is 4 + 3 ⋅ (10 - 4) = 22 and for the third is 4 + 3 ⋅ (11 - 2) = 31, which gives 60 int total.
In the second example, it is optimal watch each show on a new TV.
In third example, it is optimal to watch both shows on a new TV. Note that the answer is to be printed modulo 10^9 + 7.
Submitted Solution:
```
from bisect import bisect_left as bl
from bisect import bisect_right as br
from heapq import heappush,heappop
import math
from collections import *
from functools import reduce,cmp_to_key,lru_cache
import io, os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
# import sys
# input = sys.stdin.readline
M = mod = 10**9 + 7
def factors(n):return sorted(set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))
def inv_mod(n):return pow(n, mod - 2, mod)
def li():return [int(i) for i in input().rstrip().split()]
def st():return str(input().rstrip())[2:-1]
def val():return int(input().rstrip())
def li2():return [str(i)[2:-1] for i in input().rstrip().split()]
def li3():return [int(i) for i in st()]
n, x, y = li()
l = []
for i in range(n):l.append(li())
l.sort()
he = []
ans = 0
for i in range(n):
curtime = l[i][0]
if len(he):
item = heappop(he)
if item < curtime:
heappush(he,l[i][1])
if (curtime - item) * y > x:
ans += x + y * (l[i][1] - l[i][0])
else:
ans += y * (l[i][1] - item)
else:
heappush(he,item)
ans += x + y * (l[i][1] - l[i][0])
heappush(he,l[i][1])
ans = ans%mod
else:
ans += x + y * (l[i][1] - l[i][0])
heappush(he,l[i][1])
print(ans)
``` | instruction | 0 | 48,852 | 4 | 97,704 |
No | output | 1 | 48,852 | 4 | 97,705 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n TV shows you want to watch. Suppose the whole time is split into equal parts called "minutes". The i-th of the shows is going from l_i-th to r_i-th minute, both ends inclusive.
You need a TV to watch a TV show and you can't watch two TV shows which air at the same time on the same TV, so it is possible you will need multiple TVs in some minutes. For example, if segments [l_i, r_i] and [l_j, r_j] intersect, then shows i and j can't be watched simultaneously on one TV.
Once you start watching a show on some TV it is not possible to "move" it to another TV (since it would be too distracting), or to watch another show on the same TV until this show ends.
There is a TV Rental shop near you. It rents a TV for x rupees, and charges y (y < x) rupees for every extra minute you keep the TV. So in order to rent a TV for minutes [a; b] you will need to pay x + y ⋅ (b - a).
You can assume, that taking and returning of the TV doesn't take any time and doesn't distract from watching other TV shows. Find the minimum possible cost to view all shows. Since this value could be too large, print it modulo 10^9 + 7.
Input
The first line contains integers n, x and y (1 ≤ n ≤ 10^5, 1 ≤ y < x ≤ 10^9) — the number of TV shows, the cost to rent a TV for the first minute and the cost to rent a TV for every subsequent minute.
Each of the next n lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ 10^9) denoting the start and the end minute of the i-th TV show.
Output
Print exactly one integer — the minimum cost to view all the shows taken modulo 10^9 + 7.
Examples
Input
5 4 3
1 2
4 10
2 4
10 11
5 9
Output
60
Input
6 3 2
8 20
6 22
4 15
20 28
17 25
20 27
Output
142
Input
2 1000000000 2
1 2
2 3
Output
999999997
Note
In the first example, the optimal strategy would be to rent 3 TVs to watch:
* Show [1, 2] on the first TV,
* Show [4, 10] on the second TV,
* Shows [2, 4], [5, 9], [10, 11] on the third TV.
This way the cost for the first TV is 4 + 3 ⋅ (2 - 1) = 7, for the second is 4 + 3 ⋅ (10 - 4) = 22 and for the third is 4 + 3 ⋅ (11 - 2) = 31, which gives 60 int total.
In the second example, it is optimal watch each show on a new TV.
In third example, it is optimal to watch both shows on a new TV. Note that the answer is to be printed modulo 10^9 + 7.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
##########################################################
#q.sort(key=lambda x:((x[1]-x[0]),-x[0]))
#from collections import Counter
#from fractions import Fraction
#s=iter(input())
# for _ in range(int(input())):
#from collections import deque
# n=int(input())
# n,k=map(int,input().split())
# arr=list(map(int,input().split()))
#ls=list(map(int,input().split()))
#for in range(m):
from bisect import bisect_right
n, x, y = map(int, input().split(' '))
s=[0]*n
e=[0]*n
v=[0]*n
c=0
for i in range(n):
s[i],e[i]=map(int, input().split(' '))
c+=x+(e[i]-s[i])*y
s.sort()
e.sort()
for i in range(n-1,-1,-1):
k=bisect_right(s,e[i])
while (k < n) and (v[k]==1) and ((s[k]-e[i]) * y < x ):
k+=1
if k==n:
continue
if (s[k]-e[i])*y<x:
c+=(s[k]-e[i])*y-x
v[i]=1
print(c%(10**9+7))
``` | instruction | 0 | 48,853 | 4 | 97,706 |
No | output | 1 | 48,853 | 4 | 97,707 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n TV shows you want to watch. Suppose the whole time is split into equal parts called "minutes". The i-th of the shows is going from l_i-th to r_i-th minute, both ends inclusive.
You need a TV to watch a TV show and you can't watch two TV shows which air at the same time on the same TV, so it is possible you will need multiple TVs in some minutes. For example, if segments [l_i, r_i] and [l_j, r_j] intersect, then shows i and j can't be watched simultaneously on one TV.
Once you start watching a show on some TV it is not possible to "move" it to another TV (since it would be too distracting), or to watch another show on the same TV until this show ends.
There is a TV Rental shop near you. It rents a TV for x rupees, and charges y (y < x) rupees for every extra minute you keep the TV. So in order to rent a TV for minutes [a; b] you will need to pay x + y ⋅ (b - a).
You can assume, that taking and returning of the TV doesn't take any time and doesn't distract from watching other TV shows. Find the minimum possible cost to view all shows. Since this value could be too large, print it modulo 10^9 + 7.
Input
The first line contains integers n, x and y (1 ≤ n ≤ 10^5, 1 ≤ y < x ≤ 10^9) — the number of TV shows, the cost to rent a TV for the first minute and the cost to rent a TV for every subsequent minute.
Each of the next n lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ 10^9) denoting the start and the end minute of the i-th TV show.
Output
Print exactly one integer — the minimum cost to view all the shows taken modulo 10^9 + 7.
Examples
Input
5 4 3
1 2
4 10
2 4
10 11
5 9
Output
60
Input
6 3 2
8 20
6 22
4 15
20 28
17 25
20 27
Output
142
Input
2 1000000000 2
1 2
2 3
Output
999999997
Note
In the first example, the optimal strategy would be to rent 3 TVs to watch:
* Show [1, 2] on the first TV,
* Show [4, 10] on the second TV,
* Shows [2, 4], [5, 9], [10, 11] on the third TV.
This way the cost for the first TV is 4 + 3 ⋅ (2 - 1) = 7, for the second is 4 + 3 ⋅ (10 - 4) = 22 and for the third is 4 + 3 ⋅ (11 - 2) = 31, which gives 60 int total.
In the second example, it is optimal watch each show on a new TV.
In third example, it is optimal to watch both shows on a new TV. Note that the answer is to be printed modulo 10^9 + 7.
Submitted Solution:
```
import sys
import operator
class Show:
l, r = None, None
tv_id = None
def __init__(self, l, r):
self.l = l
self.r = r
def __str__(self):
return 'Show: {} {}'.format(self.l, self.r)
class Tv:
l, r = None, None
def __init__(self, l, r):
self.l = l
self.r = r
def update_r(self, new_r):
self.r = new_r
def update_l(self, new_l):
self.l = new_l
def cost(self):
return x + y * (self.r - self.l)
def __str__(self):
return 'TV: {} {}'.format(self.l, self.r)
n, x, y = [int(x) for x in sys.stdin.readline().split(' ')]
shows = []
for i in range(n):
shows.append([int(x) for x in sys.stdin.readline().strip().split(" ")])
ordered_shows = list(map(lambda x: Show(x[0], x[1]), sorted(shows, key=operator.itemgetter(0, 1))))
tvs = []
for show in ordered_shows:
for tv_id in range(len(tvs)):
tv = tvs[tv_id]
if tv.l == show.r + 1:
tvs[tv_id].update_l(show.l)
show.tv_id = tv_id
elif tv.r == show.l - 1:
tvs[tv_id].update_r(show.r)
show.tv_id = tv_id
if show.tv_id is None:
show.tv_id = len(tvs)
tvs.append(Tv(show.l, show.r))
total = 0
for tv in tvs:
total += tv.cost()
if total > 10**9+7:
total %= 10**9+7
print(total)
``` | instruction | 0 | 48,854 | 4 | 97,708 |
No | output | 1 | 48,854 | 4 | 97,709 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n TV shows you want to watch. Suppose the whole time is split into equal parts called "minutes". The i-th of the shows is going from l_i-th to r_i-th minute, both ends inclusive.
You need a TV to watch a TV show and you can't watch two TV shows which air at the same time on the same TV, so it is possible you will need multiple TVs in some minutes. For example, if segments [l_i, r_i] and [l_j, r_j] intersect, then shows i and j can't be watched simultaneously on one TV.
Once you start watching a show on some TV it is not possible to "move" it to another TV (since it would be too distracting), or to watch another show on the same TV until this show ends.
There is a TV Rental shop near you. It rents a TV for x rupees, and charges y (y < x) rupees for every extra minute you keep the TV. So in order to rent a TV for minutes [a; b] you will need to pay x + y ⋅ (b - a).
You can assume, that taking and returning of the TV doesn't take any time and doesn't distract from watching other TV shows. Find the minimum possible cost to view all shows. Since this value could be too large, print it modulo 10^9 + 7.
Input
The first line contains integers n, x and y (1 ≤ n ≤ 10^5, 1 ≤ y < x ≤ 10^9) — the number of TV shows, the cost to rent a TV for the first minute and the cost to rent a TV for every subsequent minute.
Each of the next n lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ 10^9) denoting the start and the end minute of the i-th TV show.
Output
Print exactly one integer — the minimum cost to view all the shows taken modulo 10^9 + 7.
Examples
Input
5 4 3
1 2
4 10
2 4
10 11
5 9
Output
60
Input
6 3 2
8 20
6 22
4 15
20 28
17 25
20 27
Output
142
Input
2 1000000000 2
1 2
2 3
Output
999999997
Note
In the first example, the optimal strategy would be to rent 3 TVs to watch:
* Show [1, 2] on the first TV,
* Show [4, 10] on the second TV,
* Shows [2, 4], [5, 9], [10, 11] on the third TV.
This way the cost for the first TV is 4 + 3 ⋅ (2 - 1) = 7, for the second is 4 + 3 ⋅ (10 - 4) = 22 and for the third is 4 + 3 ⋅ (11 - 2) = 31, which gives 60 int total.
In the second example, it is optimal watch each show on a new TV.
In third example, it is optimal to watch both shows on a new TV. Note that the answer is to be printed modulo 10^9 + 7.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
##########################################################
#q.sort(key=lambda x:((x[1]-x[0]),-x[0]))
#from collections import Counter
#from fractions import Fraction
#s=iter(input())
# for _ in range(int(input())):
#from collections import deque
# n=int(input())
# n,k=map(int,input().split())
# arr=list(map(int,input().split()))
#ls=list(map(int,input().split()))
#for in range(m):
from bisect import bisect_right
n, x, y = map(int, input().split(' '))
s=[0]*n
e=[0]*n
c=0
for i in range(n):
s[i],e[i]=map(int, input().split(' '))
c+=x+(e[i]-s[i])*y
s.sort()
e.sort()
for i in range(n):
k=bisect_right(s,e[i])
if k==n:
break
if (s[k]-e[i])*y<x:
c+=(s[k]-e[i])*y-x
print(c%(10**9+7))
``` | instruction | 0 | 48,855 | 4 | 97,710 |
No | output | 1 | 48,855 | 4 | 97,711 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You've been in love with Coronavirus-chan for a long time, but you didn't know where she lived until now. And just now you found out that she lives in a faraway place called Naha.
You immediately decided to take a vacation and visit Coronavirus-chan. Your vacation lasts exactly x days and that's the exact number of days you will spend visiting your friend. You will spend exactly x consecutive (successive) days visiting Coronavirus-chan.
They use a very unusual calendar in Naha: there are n months in a year, i-th month lasts exactly d_i days. Days in the i-th month are numbered from 1 to d_i. There are no leap years in Naha.
The mood of Coronavirus-chan (and, accordingly, her desire to hug you) depends on the number of the day in a month. In particular, you get j hugs if you visit Coronavirus-chan on the j-th day of the month.
You know about this feature of your friend and want to plan your trip to get as many hugs as possible (and then maybe you can win the heart of Coronavirus-chan).
Please note that your trip should not necessarily begin and end in the same year.
Input
The first line of input contains two integers n and x (1 ≤ n ≤ 2 ⋅ 10^5) — the number of months in the year and the number of days you can spend with your friend.
The second line contains n integers d_1, d_2, …, d_n, d_i is the number of days in the i-th month (1 ≤ d_i ≤ 10^6).
It is guaranteed that 1 ≤ x ≤ d_1 + d_2 + … + d_n.
Output
Print one integer — the maximum number of hugs that you can get from Coronavirus-chan during the best vacation in your life.
Examples
Input
3 2
1 3 1
Output
5
Input
3 6
3 3 3
Output
12
Input
5 6
4 2 3 1 3
Output
15
Note
In the first test case, the numbers of the days in a year are (indices of days in a corresponding month) \{1,1,2,3,1\}. Coronavirus-chan will hug you the most if you come on the third day of the year: 2+3=5 hugs.
In the second test case, the numbers of the days are \{1,2,3,1,2,3,1,2,3\}. You will get the most hugs if you arrive on the third day of the year: 3+1+2+3+1+2=12 hugs.
In the third test case, the numbers of the days are \{1,2,3,4,1,2, 1,2,3, 1, 1,2,3\}. You will get the most hugs if you come on the twelfth day of the year: your friend will hug you 2+3+1+2+3+4=15 times. | instruction | 0 | 48,985 | 4 | 97,970 |
Tags: binary search, brute force, greedy, implementation, two pointers
Correct Solution:
```
"""T=int(input())
for _ in range(0,T):
n=int(input())
a,b=map(int,input().split())
s=input()
s=[int(x) for x in input().split()]
for i in range(0,len(s)):
a,b=map(int,input().split())"""
"""n,x=map(int,input().split())
s=[int(x) for x in input().split()]
s=s+s
con=[]
pre=[0]
ct=[0]
for i in range(0,len(s)):
tt=(s[i]*(s[i]+1))//2
con.append(tt)
for i in range(0,len(con)):
pre.append(pre[i]+con[i])
ct.append(ct[i]+s[i])
pre=pre+[pre[-1]+10000]
print(s)
print('con',con)
print(pre)
print(ct)
ans=x
for i in range(0,len(pre)-1):
trg=pre[i]+x
low=i+1
high=len(pre)-1
temp=i+1
while(low<=high):
mid=(low+high)>>1
if(pre[mid]<=trg):
low=mid+1
else:
temp=mid
high=mid-1
print('kbckjdc',i,pre[i],temp)
c=(pre[temp-1]-pre[i])
dd=ct[temp-1]-ct[i]
print('cccccc',c)
diff=x-dd
print('difffffff',diff)
if(diff>0):
if(temp-1==i):
ele1=s[i%n]
c+=((ele1*(ele1+1))//2)
h=ele1-diff
c-=((h*(h+1))//2)
else:
ele1=s[i%n]
ele2=s[(temp-1)%n]
print(ele1,ele2)
h=min(ele1-1,ele2-diff)
c-=((h*(h+1))//2)
h2=diff+h
c+=((h2*(h2+1))//2)
c-=((diff*(diff+1))//2)
ans=max(ans,c)
print(ans)"""
"""T=int(input())
for _ in range(0,T):
n=int(input())
a,b=map(int,input().split())
s=input()
s=[int(x) for x in input().split()]
for i in range(0,len(s)):
a,b=map(int,input().split())"""
n,x=map(int,input().split())
s=[int(x) for x in input().split()]
s=s+s
con=[]
pre=[0]
ct=[0]
for i in range(0,len(s)):
tt=(s[i]*(s[i]+1))//2
con.append(tt)
for i in range(0,len(con)):
pre.append(pre[i]+con[i])
ct.append(ct[i]+s[i])
pre=pre+[pre[-1]+10000]
ct=ct+[ct[-1]+10000]
#print(s)
#print('con',con)
#print(pre)
#print(ct)
ans=x
for i in range(0,len(ct)-1):
trg=ct[i]+x
low=i+1
high=len(ct)-1
temp=i+1
while(low<=high):
mid=(low+high)>>1
if(ct[mid]<=trg):
low=mid+1
else:
temp=mid
high=mid-1
#print('kbckjdc',i,ct[i],temp)
c=(pre[temp-1]-pre[i])
dd=ct[temp-1]-ct[i]
#print('cccccc',c)
diff=x-dd
#print('difffffff',diff)
if(diff>=0):
if(temp-1==i):
ele1=s[i%n]
c+=((ele1*(ele1+1))//2)
h=ele1-diff
c-=((h*(h+1))//2)
else:
ele1=s[i%n]
ele2=s[(temp-1)%n]
#print(ele1,ele2)
h=min(ele1-1,ele2-diff)
c-=((h*(h+1))//2)
h2=diff+h
c+=((h2*(h2+1))//2)
#c-=((diff*(diff+1))//2)
ans=max(ans,c)
print(ans)
``` | output | 1 | 48,985 | 4 | 97,971 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You've been in love with Coronavirus-chan for a long time, but you didn't know where she lived until now. And just now you found out that she lives in a faraway place called Naha.
You immediately decided to take a vacation and visit Coronavirus-chan. Your vacation lasts exactly x days and that's the exact number of days you will spend visiting your friend. You will spend exactly x consecutive (successive) days visiting Coronavirus-chan.
They use a very unusual calendar in Naha: there are n months in a year, i-th month lasts exactly d_i days. Days in the i-th month are numbered from 1 to d_i. There are no leap years in Naha.
The mood of Coronavirus-chan (and, accordingly, her desire to hug you) depends on the number of the day in a month. In particular, you get j hugs if you visit Coronavirus-chan on the j-th day of the month.
You know about this feature of your friend and want to plan your trip to get as many hugs as possible (and then maybe you can win the heart of Coronavirus-chan).
Please note that your trip should not necessarily begin and end in the same year.
Input
The first line of input contains two integers n and x (1 ≤ n ≤ 2 ⋅ 10^5) — the number of months in the year and the number of days you can spend with your friend.
The second line contains n integers d_1, d_2, …, d_n, d_i is the number of days in the i-th month (1 ≤ d_i ≤ 10^6).
It is guaranteed that 1 ≤ x ≤ d_1 + d_2 + … + d_n.
Output
Print one integer — the maximum number of hugs that you can get from Coronavirus-chan during the best vacation in your life.
Examples
Input
3 2
1 3 1
Output
5
Input
3 6
3 3 3
Output
12
Input
5 6
4 2 3 1 3
Output
15
Note
In the first test case, the numbers of the days in a year are (indices of days in a corresponding month) \{1,1,2,3,1\}. Coronavirus-chan will hug you the most if you come on the third day of the year: 2+3=5 hugs.
In the second test case, the numbers of the days are \{1,2,3,1,2,3,1,2,3\}. You will get the most hugs if you arrive on the third day of the year: 3+1+2+3+1+2=12 hugs.
In the third test case, the numbers of the days are \{1,2,3,4,1,2, 1,2,3, 1, 1,2,3\}. You will get the most hugs if you come on the twelfth day of the year: your friend will hug you 2+3+1+2+3+4=15 times. | instruction | 0 | 48,986 | 4 | 97,972 |
Tags: binary search, brute force, greedy, implementation, two pointers
Correct Solution:
```
n,x=map(int,input().split())
d=list(map(int,input().split()))*3
d.reverse
e=[i*(i+1)//2 for i in d]
def f(a,b):
c=a-b
return (a*(a+1))//2 - (c*(c+1))//2
ss=0
ee=0
whole=d[ee]
tsol=e[ee]
while(True):
ee+=1
if whole+d[ee]>x:
ee-=1
break
whole+=d[ee]
tsol+=e[ee]
tend=f(d[ee+1],x-whole)
tsol+=tend
sol=tsol
while(ss<n):
whole-=d[ss]
tsol-=e[ss]+tend
ss+=1
while(True):
ee+=1
if whole+d[ee]>x:
ee-=1
break
whole+=d[ee]
tsol+=e[ee]
tend=f(d[ee+1],x-whole)
tsol+=tend
sol=max(tsol,sol)
print(sol)
``` | output | 1 | 48,986 | 4 | 97,973 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You've been in love with Coronavirus-chan for a long time, but you didn't know where she lived until now. And just now you found out that she lives in a faraway place called Naha.
You immediately decided to take a vacation and visit Coronavirus-chan. Your vacation lasts exactly x days and that's the exact number of days you will spend visiting your friend. You will spend exactly x consecutive (successive) days visiting Coronavirus-chan.
They use a very unusual calendar in Naha: there are n months in a year, i-th month lasts exactly d_i days. Days in the i-th month are numbered from 1 to d_i. There are no leap years in Naha.
The mood of Coronavirus-chan (and, accordingly, her desire to hug you) depends on the number of the day in a month. In particular, you get j hugs if you visit Coronavirus-chan on the j-th day of the month.
You know about this feature of your friend and want to plan your trip to get as many hugs as possible (and then maybe you can win the heart of Coronavirus-chan).
Please note that your trip should not necessarily begin and end in the same year.
Input
The first line of input contains two integers n and x (1 ≤ n ≤ 2 ⋅ 10^5) — the number of months in the year and the number of days you can spend with your friend.
The second line contains n integers d_1, d_2, …, d_n, d_i is the number of days in the i-th month (1 ≤ d_i ≤ 10^6).
It is guaranteed that 1 ≤ x ≤ d_1 + d_2 + … + d_n.
Output
Print one integer — the maximum number of hugs that you can get from Coronavirus-chan during the best vacation in your life.
Examples
Input
3 2
1 3 1
Output
5
Input
3 6
3 3 3
Output
12
Input
5 6
4 2 3 1 3
Output
15
Note
In the first test case, the numbers of the days in a year are (indices of days in a corresponding month) \{1,1,2,3,1\}. Coronavirus-chan will hug you the most if you come on the third day of the year: 2+3=5 hugs.
In the second test case, the numbers of the days are \{1,2,3,1,2,3,1,2,3\}. You will get the most hugs if you arrive on the third day of the year: 3+1+2+3+1+2=12 hugs.
In the third test case, the numbers of the days are \{1,2,3,4,1,2, 1,2,3, 1, 1,2,3\}. You will get the most hugs if you come on the twelfth day of the year: your friend will hug you 2+3+1+2+3+4=15 times. | instruction | 0 | 48,987 | 4 | 97,974 |
Tags: binary search, brute force, greedy, implementation, two pointers
Correct Solution:
```
n, x = tuple(map(int, input().split()))
arr = list(map(int, input().split()))
arr += arr[::]
arr2 = [i * (i + 1) // 2 for i in arr]
ans = 0
for now_month in range(n, n * 2):
if now_month == n:
last_month = now_month
count_obnim = 0
count_days = 0
while count_days + arr[last_month] <= x:
count_days += arr[last_month]
count_obnim += arr2[last_month]
last_month -= 1
dop_days = x - count_days
nodop_days = arr[last_month] - dop_days
dop_obnim = arr2[last_month] - nodop_days * (nodop_days + 1) // 2
count_days += dop_days
count_obnim += dop_obnim
else:
count_days += arr[now_month]
count_days -= dop_days
count_obnim += arr2[now_month]
count_obnim -= dop_obnim
while count_days > x:
count_days -= arr[last_month + 1]
count_obnim -= arr2[last_month + 1]
last_month += 1
dop_days = x - count_days
nodop_days = arr[last_month] - dop_days
dop_obnim = arr2[last_month] - nodop_days * (nodop_days + 1) // 2
count_days += dop_days
count_obnim += dop_obnim
ans = max(ans, count_obnim)
print(ans)
``` | output | 1 | 48,987 | 4 | 97,975 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You've been in love with Coronavirus-chan for a long time, but you didn't know where she lived until now. And just now you found out that she lives in a faraway place called Naha.
You immediately decided to take a vacation and visit Coronavirus-chan. Your vacation lasts exactly x days and that's the exact number of days you will spend visiting your friend. You will spend exactly x consecutive (successive) days visiting Coronavirus-chan.
They use a very unusual calendar in Naha: there are n months in a year, i-th month lasts exactly d_i days. Days in the i-th month are numbered from 1 to d_i. There are no leap years in Naha.
The mood of Coronavirus-chan (and, accordingly, her desire to hug you) depends on the number of the day in a month. In particular, you get j hugs if you visit Coronavirus-chan on the j-th day of the month.
You know about this feature of your friend and want to plan your trip to get as many hugs as possible (and then maybe you can win the heart of Coronavirus-chan).
Please note that your trip should not necessarily begin and end in the same year.
Input
The first line of input contains two integers n and x (1 ≤ n ≤ 2 ⋅ 10^5) — the number of months in the year and the number of days you can spend with your friend.
The second line contains n integers d_1, d_2, …, d_n, d_i is the number of days in the i-th month (1 ≤ d_i ≤ 10^6).
It is guaranteed that 1 ≤ x ≤ d_1 + d_2 + … + d_n.
Output
Print one integer — the maximum number of hugs that you can get from Coronavirus-chan during the best vacation in your life.
Examples
Input
3 2
1 3 1
Output
5
Input
3 6
3 3 3
Output
12
Input
5 6
4 2 3 1 3
Output
15
Note
In the first test case, the numbers of the days in a year are (indices of days in a corresponding month) \{1,1,2,3,1\}. Coronavirus-chan will hug you the most if you come on the third day of the year: 2+3=5 hugs.
In the second test case, the numbers of the days are \{1,2,3,1,2,3,1,2,3\}. You will get the most hugs if you arrive on the third day of the year: 3+1+2+3+1+2=12 hugs.
In the third test case, the numbers of the days are \{1,2,3,4,1,2, 1,2,3, 1, 1,2,3\}. You will get the most hugs if you come on the twelfth day of the year: your friend will hug you 2+3+1+2+3+4=15 times. | instruction | 0 | 48,988 | 4 | 97,976 |
Tags: binary search, brute force, greedy, implementation, two pointers
Correct Solution:
```
n,x=map(int,input().split())
l=list(map(int,input().split()))
l*=2
num=0
ans=0
val=0
j=0
for i in range(n):
while num<x:
num+=l[j]
val+=(l[j]*(l[j]+1)//2)
j+=1
if j==n:
j=0
ans=max(ans,val-(num-x)*(num-x+1)//2)
num-=l[i]
val-=l[i]*(l[i]+1)//2
print(ans)
``` | output | 1 | 48,988 | 4 | 97,977 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You've been in love with Coronavirus-chan for a long time, but you didn't know where she lived until now. And just now you found out that she lives in a faraway place called Naha.
You immediately decided to take a vacation and visit Coronavirus-chan. Your vacation lasts exactly x days and that's the exact number of days you will spend visiting your friend. You will spend exactly x consecutive (successive) days visiting Coronavirus-chan.
They use a very unusual calendar in Naha: there are n months in a year, i-th month lasts exactly d_i days. Days in the i-th month are numbered from 1 to d_i. There are no leap years in Naha.
The mood of Coronavirus-chan (and, accordingly, her desire to hug you) depends on the number of the day in a month. In particular, you get j hugs if you visit Coronavirus-chan on the j-th day of the month.
You know about this feature of your friend and want to plan your trip to get as many hugs as possible (and then maybe you can win the heart of Coronavirus-chan).
Please note that your trip should not necessarily begin and end in the same year.
Input
The first line of input contains two integers n and x (1 ≤ n ≤ 2 ⋅ 10^5) — the number of months in the year and the number of days you can spend with your friend.
The second line contains n integers d_1, d_2, …, d_n, d_i is the number of days in the i-th month (1 ≤ d_i ≤ 10^6).
It is guaranteed that 1 ≤ x ≤ d_1 + d_2 + … + d_n.
Output
Print one integer — the maximum number of hugs that you can get from Coronavirus-chan during the best vacation in your life.
Examples
Input
3 2
1 3 1
Output
5
Input
3 6
3 3 3
Output
12
Input
5 6
4 2 3 1 3
Output
15
Note
In the first test case, the numbers of the days in a year are (indices of days in a corresponding month) \{1,1,2,3,1\}. Coronavirus-chan will hug you the most if you come on the third day of the year: 2+3=5 hugs.
In the second test case, the numbers of the days are \{1,2,3,1,2,3,1,2,3\}. You will get the most hugs if you arrive on the third day of the year: 3+1+2+3+1+2=12 hugs.
In the third test case, the numbers of the days are \{1,2,3,4,1,2, 1,2,3, 1, 1,2,3\}. You will get the most hugs if you come on the twelfth day of the year: your friend will hug you 2+3+1+2+3+4=15 times. | instruction | 0 | 48,989 | 4 | 97,978 |
Tags: binary search, brute force, greedy, implementation, two pointers
Correct Solution:
```
# from functools import lru_cache
from sys import stdin, stdout
import sys
from math import *
# sys.setrecursionlimit(10**6)
input = stdin.readline
# print = stdout.write
# @lru_cache()
n,m=map(int,input().split())
ar=list(map(int,input().split()))
ar=ar+ar
d=[0]
h=[0]
for i in range(2*n):
d.append(d[-1]+ar[i])
h.append(h[-1]+(ar[i]*(ar[i]+1))//2)
# print(d)
# print(h)
ans=0
for i in range(1,2*n+1):
if(d[i]>=m):
req=d[i]-m
l=0
r=i+1
ind=0
while(l<=r):
mid=(l+r)//2
if(d[mid]<req):
l=mid+1
else:
ind=mid
r=mid-1
if(ind==0 or d[i]-d[ind]==m):
ans=max(ans,h[i]-h[ind])
# print(ind,i)
else:
rslt=h[i]-h[ind]
req = m-(d[i]-d[ind])
temp = (ar[ind-1]*(ar[ind-1]+1))//2-((ar[ind-1]-req)*(ar[ind-1]-req+1))//2
ans=max(ans,rslt+temp)
print(ans)
``` | output | 1 | 48,989 | 4 | 97,979 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You've been in love with Coronavirus-chan for a long time, but you didn't know where she lived until now. And just now you found out that she lives in a faraway place called Naha.
You immediately decided to take a vacation and visit Coronavirus-chan. Your vacation lasts exactly x days and that's the exact number of days you will spend visiting your friend. You will spend exactly x consecutive (successive) days visiting Coronavirus-chan.
They use a very unusual calendar in Naha: there are n months in a year, i-th month lasts exactly d_i days. Days in the i-th month are numbered from 1 to d_i. There are no leap years in Naha.
The mood of Coronavirus-chan (and, accordingly, her desire to hug you) depends on the number of the day in a month. In particular, you get j hugs if you visit Coronavirus-chan on the j-th day of the month.
You know about this feature of your friend and want to plan your trip to get as many hugs as possible (and then maybe you can win the heart of Coronavirus-chan).
Please note that your trip should not necessarily begin and end in the same year.
Input
The first line of input contains two integers n and x (1 ≤ n ≤ 2 ⋅ 10^5) — the number of months in the year and the number of days you can spend with your friend.
The second line contains n integers d_1, d_2, …, d_n, d_i is the number of days in the i-th month (1 ≤ d_i ≤ 10^6).
It is guaranteed that 1 ≤ x ≤ d_1 + d_2 + … + d_n.
Output
Print one integer — the maximum number of hugs that you can get from Coronavirus-chan during the best vacation in your life.
Examples
Input
3 2
1 3 1
Output
5
Input
3 6
3 3 3
Output
12
Input
5 6
4 2 3 1 3
Output
15
Note
In the first test case, the numbers of the days in a year are (indices of days in a corresponding month) \{1,1,2,3,1\}. Coronavirus-chan will hug you the most if you come on the third day of the year: 2+3=5 hugs.
In the second test case, the numbers of the days are \{1,2,3,1,2,3,1,2,3\}. You will get the most hugs if you arrive on the third day of the year: 3+1+2+3+1+2=12 hugs.
In the third test case, the numbers of the days are \{1,2,3,4,1,2, 1,2,3, 1, 1,2,3\}. You will get the most hugs if you come on the twelfth day of the year: your friend will hug you 2+3+1+2+3+4=15 times. | instruction | 0 | 48,990 | 4 | 97,980 |
Tags: binary search, brute force, greedy, implementation, two pointers
Correct Solution:
```
import bisect
n,days=map(int, input().strip().split())
arr=list(map(int, input().strip().split())); arr=arr+arr;
pre1=[0]; pre2=[0]; res=0;
for i in range(2*n):
pre1.append(pre1[-1]+arr[i]); pre2.append(pre2[-1]+(arr[i]*(arr[i]+1))//2);
for i in range(2*n,n,-1):
l=1
r=i
ans=i
while(l<=r):
mid=l+(r-l)//2
if(pre1[i]-pre1[mid]<days):
ans=mid
r=mid-1
else:
l=mid+1
tot=pre1[i]-pre1[ans-1]-days
temp=(pre2[i]-pre2[ans-1])-((tot)*(tot+1)//2)
res=max(res,temp)
print(res)
``` | output | 1 | 48,990 | 4 | 97,981 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You've been in love with Coronavirus-chan for a long time, but you didn't know where she lived until now. And just now you found out that she lives in a faraway place called Naha.
You immediately decided to take a vacation and visit Coronavirus-chan. Your vacation lasts exactly x days and that's the exact number of days you will spend visiting your friend. You will spend exactly x consecutive (successive) days visiting Coronavirus-chan.
They use a very unusual calendar in Naha: there are n months in a year, i-th month lasts exactly d_i days. Days in the i-th month are numbered from 1 to d_i. There are no leap years in Naha.
The mood of Coronavirus-chan (and, accordingly, her desire to hug you) depends on the number of the day in a month. In particular, you get j hugs if you visit Coronavirus-chan on the j-th day of the month.
You know about this feature of your friend and want to plan your trip to get as many hugs as possible (and then maybe you can win the heart of Coronavirus-chan).
Please note that your trip should not necessarily begin and end in the same year.
Input
The first line of input contains two integers n and x (1 ≤ n ≤ 2 ⋅ 10^5) — the number of months in the year and the number of days you can spend with your friend.
The second line contains n integers d_1, d_2, …, d_n, d_i is the number of days in the i-th month (1 ≤ d_i ≤ 10^6).
It is guaranteed that 1 ≤ x ≤ d_1 + d_2 + … + d_n.
Output
Print one integer — the maximum number of hugs that you can get from Coronavirus-chan during the best vacation in your life.
Examples
Input
3 2
1 3 1
Output
5
Input
3 6
3 3 3
Output
12
Input
5 6
4 2 3 1 3
Output
15
Note
In the first test case, the numbers of the days in a year are (indices of days in a corresponding month) \{1,1,2,3,1\}. Coronavirus-chan will hug you the most if you come on the third day of the year: 2+3=5 hugs.
In the second test case, the numbers of the days are \{1,2,3,1,2,3,1,2,3\}. You will get the most hugs if you arrive on the third day of the year: 3+1+2+3+1+2=12 hugs.
In the third test case, the numbers of the days are \{1,2,3,4,1,2, 1,2,3, 1, 1,2,3\}. You will get the most hugs if you come on the twelfth day of the year: your friend will hug you 2+3+1+2+3+4=15 times. | instruction | 0 | 48,991 | 4 | 97,982 |
Tags: binary search, brute force, greedy, implementation, two pointers
Correct Solution:
```
n,x = map(int,input().split())
arr = list(map(int,input().split()))
arr += arr
prefix = []
prefix1 = []
Sum = 0
for i in arr:
Sum += i
prefix.append(Sum)
Sum = 0
for j in arr:
Sum += (j*(j+1))//2
prefix1.append(Sum)
ans = 0
for i in range(n*2):
A = prefix[i]-x
if (A<=0):
ans = prefix1[i]
else:
if (arr[i]>=x):
jj = prefix1[i]-((arr[i]-x)*(arr[i]-x+1))//2
if (i-1>=0):
jj -= prefix1[i-1]
ans = max(ans,jj)
# break
else:
l = 0
r = i
Ans = i
while(l<=r):
mid = (l+r)//2
if (prefix[mid]<=A):
Ans = mid
l = mid+1
else:
r = mid-1
# print(i,Ans)
B = prefix[i]-prefix[Ans]-x
ans = max(ans,prefix1[i]-prefix1[Ans]-(B*(B+1))//2)
print(ans)
``` | output | 1 | 48,991 | 4 | 97,983 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You've been in love with Coronavirus-chan for a long time, but you didn't know where she lived until now. And just now you found out that she lives in a faraway place called Naha.
You immediately decided to take a vacation and visit Coronavirus-chan. Your vacation lasts exactly x days and that's the exact number of days you will spend visiting your friend. You will spend exactly x consecutive (successive) days visiting Coronavirus-chan.
They use a very unusual calendar in Naha: there are n months in a year, i-th month lasts exactly d_i days. Days in the i-th month are numbered from 1 to d_i. There are no leap years in Naha.
The mood of Coronavirus-chan (and, accordingly, her desire to hug you) depends on the number of the day in a month. In particular, you get j hugs if you visit Coronavirus-chan on the j-th day of the month.
You know about this feature of your friend and want to plan your trip to get as many hugs as possible (and then maybe you can win the heart of Coronavirus-chan).
Please note that your trip should not necessarily begin and end in the same year.
Input
The first line of input contains two integers n and x (1 ≤ n ≤ 2 ⋅ 10^5) — the number of months in the year and the number of days you can spend with your friend.
The second line contains n integers d_1, d_2, …, d_n, d_i is the number of days in the i-th month (1 ≤ d_i ≤ 10^6).
It is guaranteed that 1 ≤ x ≤ d_1 + d_2 + … + d_n.
Output
Print one integer — the maximum number of hugs that you can get from Coronavirus-chan during the best vacation in your life.
Examples
Input
3 2
1 3 1
Output
5
Input
3 6
3 3 3
Output
12
Input
5 6
4 2 3 1 3
Output
15
Note
In the first test case, the numbers of the days in a year are (indices of days in a corresponding month) \{1,1,2,3,1\}. Coronavirus-chan will hug you the most if you come on the third day of the year: 2+3=5 hugs.
In the second test case, the numbers of the days are \{1,2,3,1,2,3,1,2,3\}. You will get the most hugs if you arrive on the third day of the year: 3+1+2+3+1+2=12 hugs.
In the third test case, the numbers of the days are \{1,2,3,4,1,2, 1,2,3, 1, 1,2,3\}. You will get the most hugs if you come on the twelfth day of the year: your friend will hug you 2+3+1+2+3+4=15 times. | instruction | 0 | 48,992 | 4 | 97,984 |
Tags: binary search, brute force, greedy, implementation, two pointers
Correct Solution:
```
import sys, math,os
from io import BytesIO, IOBase
from bisect import bisect_left as bl, bisect_right as br, insort
#from heapq import heapify, heappush, heappop
from collections import defaultdict as dd, deque, Counter
#from itertools import permutations,combinations
def data(): return sys.stdin.readline().strip()
def mdata(): return list(map(int, data().split()))
def outl(var) : sys.stdout.write(' '.join(map(str, var))+'\n')
def out(var) : sys.stdout.write(str(var)+'\n')
sys.setrecursionlimit(100000)
INF = float('inf')
mod = int(1e9)+7
def main():
n,x=mdata()
d=mdata()
cnt1=0
cnt2=0
m=0
ind=2*n-1
i=2*n-1
d+=d
while i>-1:
if d[i]>=x-cnt1:
cnt2+=(d[i]*d[i]+d[i])//2-((d[i]-x+cnt1)*(d[i]-x+cnt1+1))//2
m=max(cnt2,m)
if ind!=i:
cnt2-=(d[ind]*d[ind]+d[ind])//2+(d[i]*d[i]+d[i])//2-((d[i]-x+cnt1)*(d[i]-x+cnt1+1))//2
cnt1 -= d[ind]
i+=1
else:
cnt2-=(d[i]*d[i]+d[i])//2-((d[i]-x+cnt1)*(d[i]-x+cnt1+1))//2
if cnt2 < 0:
break
ind-=1
else:
cnt1 += d[i]
cnt2+=(d[i]*d[i]+d[i])//2
i-=1
out(m)
if __name__ == '__main__':
main()
``` | output | 1 | 48,992 | 4 | 97,985 |
Provide tags and a correct Python 2 solution for this coding contest problem.
You've been in love with Coronavirus-chan for a long time, but you didn't know where she lived until now. And just now you found out that she lives in a faraway place called Naha.
You immediately decided to take a vacation and visit Coronavirus-chan. Your vacation lasts exactly x days and that's the exact number of days you will spend visiting your friend. You will spend exactly x consecutive (successive) days visiting Coronavirus-chan.
They use a very unusual calendar in Naha: there are n months in a year, i-th month lasts exactly d_i days. Days in the i-th month are numbered from 1 to d_i. There are no leap years in Naha.
The mood of Coronavirus-chan (and, accordingly, her desire to hug you) depends on the number of the day in a month. In particular, you get j hugs if you visit Coronavirus-chan on the j-th day of the month.
You know about this feature of your friend and want to plan your trip to get as many hugs as possible (and then maybe you can win the heart of Coronavirus-chan).
Please note that your trip should not necessarily begin and end in the same year.
Input
The first line of input contains two integers n and x (1 ≤ n ≤ 2 ⋅ 10^5) — the number of months in the year and the number of days you can spend with your friend.
The second line contains n integers d_1, d_2, …, d_n, d_i is the number of days in the i-th month (1 ≤ d_i ≤ 10^6).
It is guaranteed that 1 ≤ x ≤ d_1 + d_2 + … + d_n.
Output
Print one integer — the maximum number of hugs that you can get from Coronavirus-chan during the best vacation in your life.
Examples
Input
3 2
1 3 1
Output
5
Input
3 6
3 3 3
Output
12
Input
5 6
4 2 3 1 3
Output
15
Note
In the first test case, the numbers of the days in a year are (indices of days in a corresponding month) \{1,1,2,3,1\}. Coronavirus-chan will hug you the most if you come on the third day of the year: 2+3=5 hugs.
In the second test case, the numbers of the days are \{1,2,3,1,2,3,1,2,3\}. You will get the most hugs if you arrive on the third day of the year: 3+1+2+3+1+2=12 hugs.
In the third test case, the numbers of the days are \{1,2,3,4,1,2, 1,2,3, 1, 1,2,3\}. You will get the most hugs if you come on the twelfth day of the year: your friend will hug you 2+3+1+2+3+4=15 times. | instruction | 0 | 48,993 | 4 | 97,986 |
Tags: binary search, brute force, greedy, implementation, two pointers
Correct Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
pr = stdout.write
mod=10**9+7
def ni():
return int(raw_input())
def li():
return map(int,raw_input().split())
def pn(n):
stdout.write(str(n)+'\n')
def pa(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return map(int,stdin.read().split())
range = xrange # not for python 3.0+
# main code
def gs(n,m):
return ((n*(n+1))/2)-((m*(m+1))/2)
n,x=li()
l=li()
l*=2
dp=[0]*(2*n+1)
dp1=[0]*(2*n+1)
for i in range(1,2*n+1):
dp[i]=dp[i-1]+l[i-1]
dp1[i]=dp1[i-1]+((l[i-1]*(l[i-1]+1))/2)
l1,r1=1,n
ans=0
while l1<=n:
l2=l1
r2=r1
while l2<=r2:
#print l1,r1,l2,r2
mid=(l2+r2)/2
if dp[r1]-dp[mid-1]<=x:
ans=max(ans,dp1[r1]-dp1[mid-1])
r2=mid-1
else:
if dp[r1]-dp[mid]<=x:
temp=x-(dp[r1]-dp[mid])
#print temp,l[mid-1],l[mid-1]-temp
ans=max(ans,dp1[r1]-dp1[mid]+gs(l[mid-1],l[mid-1]-temp))
#r2=mid-1
break
else:
l2=mid+1
l1+=1
r1+=1
pn(ans)
``` | output | 1 | 48,993 | 4 | 97,987 |
Provide tags and a correct Python 2 solution for this coding contest problem.
You've been in love with Coronavirus-chan for a long time, but you didn't know where she lived until now. And just now you found out that she lives in a faraway place called Naha.
You immediately decided to take a vacation and visit Coronavirus-chan. Your vacation lasts exactly x days and that's the exact number of days you will spend visiting your friend. You will spend exactly x consecutive (successive) days visiting Coronavirus-chan.
They use a very unusual calendar in Naha: there are n months in a year, i-th month lasts exactly d_i days. Days in the i-th month are numbered from 1 to d_i. There are no leap years in Naha.
The mood of Coronavirus-chan (and, accordingly, her desire to hug you) depends on the number of the day in a month. In particular, you get j hugs if you visit Coronavirus-chan on the j-th day of the month.
You know about this feature of your friend and want to plan your trip to get as many hugs as possible (and then maybe you can win the heart of Coronavirus-chan).
Please note that your trip should not necessarily begin and end in the same year.
Input
The first line of input contains two integers n and x (1 ≤ n ≤ 2 ⋅ 10^5) — the number of months in the year and the number of days you can spend with your friend.
The second line contains n integers d_1, d_2, …, d_n, d_i is the number of days in the i-th month (1 ≤ d_i ≤ 10^6).
It is guaranteed that 1 ≤ x ≤ d_1 + d_2 + … + d_n.
Output
Print one integer — the maximum number of hugs that you can get from Coronavirus-chan during the best vacation in your life.
Examples
Input
3 2
1 3 1
Output
5
Input
3 6
3 3 3
Output
12
Input
5 6
4 2 3 1 3
Output
15
Note
In the first test case, the numbers of the days in a year are (indices of days in a corresponding month) \{1,1,2,3,1\}. Coronavirus-chan will hug you the most if you come on the third day of the year: 2+3=5 hugs.
In the second test case, the numbers of the days are \{1,2,3,1,2,3,1,2,3\}. You will get the most hugs if you arrive on the third day of the year: 3+1+2+3+1+2=12 hugs.
In the third test case, the numbers of the days are \{1,2,3,4,1,2, 1,2,3, 1, 1,2,3\}. You will get the most hugs if you come on the twelfth day of the year: your friend will hug you 2+3+1+2+3+4=15 times. | instruction | 0 | 48,994 | 4 | 97,988 |
Tags: binary search, brute force, greedy, implementation, two pointers
Correct Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
pr = stdout.write
mod=10**9+7
def ni():
return int(raw_input())
def li():
return map(int,raw_input().split())
def pn(n):
stdout.write(str(n)+'\n')
def pa(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return map(int,stdin.read().split())
range = xrange # not for python 3.0+
# main code
def gs(n,m):
return ((n*(n+1))/2)-((m*(m+1))/2)
n,x=li()
l=li()
l*=2
dp=[0]*(2*n+1)
dp1=[0]*(2*n+1)
for i in range(1,2*n+1):
dp[i]=dp[i-1]+l[i-1]
dp1[i]=dp1[i-1]+((l[i-1]*(l[i-1]+1))/2)
l1,r1=1,n
ans=0
while l1<=n:
l2=l1
r2=r1
while l2<=r2:
#print l1,r1,l2,r2
mid=(l2+r2)/2
if dp[r1]-dp[mid-1]<=x:
ans=max(ans,dp1[r1]-dp1[mid-1])
r2=mid-1
else:
if dp[r1]-dp[mid]<=x:
temp=x-(dp[r1]-dp[mid])
#print temp,l[mid-1],l[mid-1]-temp
ans=max(ans,dp1[r1]-dp1[mid]+gs(l[mid-1],l[mid-1]-temp))
r2=mid-1
#break
else:
l2=mid+1
l1+=1
r1+=1
pn(ans)
``` | output | 1 | 48,994 | 4 | 97,989 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've been in love with Coronavirus-chan for a long time, but you didn't know where she lived until now. And just now you found out that she lives in a faraway place called Naha.
You immediately decided to take a vacation and visit Coronavirus-chan. Your vacation lasts exactly x days and that's the exact number of days you will spend visiting your friend. You will spend exactly x consecutive (successive) days visiting Coronavirus-chan.
They use a very unusual calendar in Naha: there are n months in a year, i-th month lasts exactly d_i days. Days in the i-th month are numbered from 1 to d_i. There are no leap years in Naha.
The mood of Coronavirus-chan (and, accordingly, her desire to hug you) depends on the number of the day in a month. In particular, you get j hugs if you visit Coronavirus-chan on the j-th day of the month.
You know about this feature of your friend and want to plan your trip to get as many hugs as possible (and then maybe you can win the heart of Coronavirus-chan).
Please note that your trip should not necessarily begin and end in the same year.
Input
The first line of input contains two integers n and x (1 ≤ n ≤ 2 ⋅ 10^5) — the number of months in the year and the number of days you can spend with your friend.
The second line contains n integers d_1, d_2, …, d_n, d_i is the number of days in the i-th month (1 ≤ d_i ≤ 10^6).
It is guaranteed that 1 ≤ x ≤ d_1 + d_2 + … + d_n.
Output
Print one integer — the maximum number of hugs that you can get from Coronavirus-chan during the best vacation in your life.
Examples
Input
3 2
1 3 1
Output
5
Input
3 6
3 3 3
Output
12
Input
5 6
4 2 3 1 3
Output
15
Note
In the first test case, the numbers of the days in a year are (indices of days in a corresponding month) \{1,1,2,3,1\}. Coronavirus-chan will hug you the most if you come on the third day of the year: 2+3=5 hugs.
In the second test case, the numbers of the days are \{1,2,3,1,2,3,1,2,3\}. You will get the most hugs if you arrive on the third day of the year: 3+1+2+3+1+2=12 hugs.
In the third test case, the numbers of the days are \{1,2,3,4,1,2, 1,2,3, 1, 1,2,3\}. You will get the most hugs if you come on the twelfth day of the year: your friend will hug you 2+3+1+2+3+4=15 times.
Submitted Solution:
```
from bisect import *
from collections import *
from math import gcd,ceil,sqrt,floor,inf
from heapq import *
from itertools import *
from operator import add,mul,sub,xor,truediv,floordiv
from functools import *
#------------------------------------------------------------------------
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#------------------------------------------------------------------------
def RL(): return map(int, sys.stdin.readline().rstrip().split())
def RLL(): return list(map(int, sys.stdin.readline().rstrip().split()))
def N(): return int(input())
#------------------------------------------------------------------------
from types import GeneratorType
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
farr=[1]
ifa=[]
def fact(x,mod=0):
if mod:
while x>=len(farr):
farr.append(farr[-1]*len(farr)%mod)
else:
while x>=len(farr):
farr.append(farr[-1]*len(farr))
return farr[x]
def ifact(x,mod):
global ifa
ifa.append(pow(farr[-1],mod-2,mod))
for i in range(x,0,-1):
ifa.append(ifa[-1]*i%mod)
ifa=ifa[::-1]
def per(i,j,mod=0):
if i<j: return 0
if not mod:
return fact(i)//fact(i-j)
return farr[i]*ifa[i-j]%mod
def com(i,j,mod=0):
if i<j: return 0
if not mod:
return per(i,j)//fact(j)
return per(i,j,mod)*ifa[j]%mod
def catalan(n):
return com(2*n,n)//(n+1)
def isprime(n):
for i in range(2,int(n**0.5)+1):
if n%i==0:
return False
return True
def lowbit(n):
return n&-n
def inverse(a,m):
a%=m
if a<=1: return a
return ((1-inverse(m,a)*m)//a)%m
class BIT:
def __init__(self,arr):
self.arr=arr
self.n=len(arr)-1
def update(self,x,v):
while x<=self.n:
self.arr[x]+=v
x+=x&-x
def query(self,x):
ans=0
while x:
ans+=self.arr[x]
x&=x-1
return ans
'''
class SMT:
def __init__(self,arr):
self.n=len(arr)-1
self.arr=[0]*(self.n<<2)
self.lazy=[0]*(self.n<<2)
def Build(l,r,rt):
if l==r:
self.arr[rt]=arr[l]
return
m=(l+r)>>1
Build(l,m,rt<<1)
Build(m+1,r,rt<<1|1)
self.pushup(rt)
Build(1,self.n,1)
def pushup(self,rt):
self.arr[rt]=self.arr[rt<<1]+self.arr[rt<<1|1]
def pushdown(self,rt,ln,rn):#lr,rn表区间数字数
if self.lazy[rt]:
self.lazy[rt<<1]+=self.lazy[rt]
self.lazy[rt<<1|1]+=self.lazy[rt]
self.arr[rt<<1]+=self.lazy[rt]*ln
self.arr[rt<<1|1]+=self.lazy[rt]*rn
self.lazy[rt]=0
def update(self,L,R,c,l=1,r=None,rt=1):#L,R表示操作区间
if r==None: r=self.n
if L<=l and r<=R:
self.arr[rt]+=c*(r-l+1)
self.lazy[rt]+=c
return
m=(l+r)>>1
self.pushdown(rt,m-l+1,r-m)
if L<=m: self.update(L,R,c,l,m,rt<<1)
if R>m: self.update(L,R,c,m+1,r,rt<<1|1)
self.pushup(rt)
def query(self,L,R,l=1,r=None,rt=1):
if r==None: r=self.n
#print(L,R,l,r,rt)
if L<=l and R>=r:
return self.arr[rt]
m=(l+r)>>1
self.pushdown(rt,m-l+1,r-m)
ans=0
if L<=m: ans+=self.query(L,R,l,m,rt<<1)
if R>m: ans+=self.query(L,R,m+1,r,rt<<1|1)
return ans
'''
class DSU:#容量+路径压缩
def __init__(self,n):
self.c=[-1]*n
def same(self,x,y):
return self.find(x)==self.find(y)
def find(self,x):
if self.c[x]<0:
return x
self.c[x]=self.find(self.c[x])
return self.c[x]
def union(self,u,v):
u,v=self.find(u),self.find(v)
if u==v:
return False
if self.c[u]<self.c[v]:
u,v=v,u
self.c[u]+=self.c[v]
self.c[v]=u
return True
def size(self,x): return -self.c[self.find(x)]
class UFS:#秩+路径
def __init__(self,n):
self.parent=[i for i in range(n)]
self.ranks=[0]*n
def find(self,x):
if x!=self.parent[x]:
self.parent[x]=self.find(self.parent[x])
return self.parent[x]
def union(self,u,v):
pu,pv=self.find(u),self.find(v)
if pu==pv:
return False
if self.ranks[pu]>=self.ranks[pv]:
self.parent[pv]=pu
if self.ranks[pv]==self.ranks[pu]:
self.ranks[pu]+=1
else:
self.parent[pu]=pv
def Prime(n):
c=0
prime=[]
flag=[0]*(n+1)
for i in range(2,n+1):
if not flag[i]:
prime.append(i)
c+=1
for j in range(c):
if i*prime[j]>n: break
flag[i*prime[j]]=prime[j]
if i%prime[j]==0: break
#print(flag)
return flag
def dij(s,graph):
d={}
d[s]=0
heap=[(0,s)]
seen=set()
while heap:
dis,u=heappop(heap)
if u in seen:
continue
for v in graph[u]:
if v not in d or d[v]>d[u]+graph[u][v]:
d[v]=d[u]+graph[u][v]
heappush(heap,(d[v],v))
return d
def GP(it): return [[ch,len(list(g))] for ch,g in groupby(it)]
class DLN:
def __init__(self,val):
self.val=val
self.pre=None
self.next=None
def nb(i,j):
for ni,nj in [[i+1,j],[i-1,j],[i,j-1],[i,j+1]]:
if 0<=ni<n and 0<=nj<m:
yield ni,nj
@bootstrap
def gdfs(r,p):
if len(g[r])==1 and p!=-1:
yield None
for ch in g[r]:
if ch!=p:
yield gdfs(ch,r)
yield None
t=1
for i in range(t):
n,x=RL()
d=RLL()
d=d+d
pred=[0]
preh=[0]
for i in range(2*n):
pred.append(d[i]+pred[-1])
preh.append(d[i]*(d[i]+1)//2+preh[-1])
ans=0
for i in range(2*n):
if pred[i+1]>=x:
l=0
r=i
while l<r:
mid=(l+r+1)//2
if pred[i+1]-pred[mid]>=x:
l=mid
else:
r=mid-1
tmp=preh[i+1]-preh[l]
mis=pred[i+1]-pred[l]-x
tmp-=mis*(mis+1)//2
ans=max(ans,tmp)
print(ans)
'''
sys.setrecursionlimit(200000)
import threading
threading.stack_size(10**8)
t=threading.Thread(target=main)
t.start()
t.join()
'''
'''
sys.setrecursionlimit(200000)
import threading
threading.stack_size(10**8)
t=threading.Thread(target=main)
t.start()
t.join()
'''
``` | instruction | 0 | 48,995 | 4 | 97,990 |
Yes | output | 1 | 48,995 | 4 | 97,991 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've been in love with Coronavirus-chan for a long time, but you didn't know where she lived until now. And just now you found out that she lives in a faraway place called Naha.
You immediately decided to take a vacation and visit Coronavirus-chan. Your vacation lasts exactly x days and that's the exact number of days you will spend visiting your friend. You will spend exactly x consecutive (successive) days visiting Coronavirus-chan.
They use a very unusual calendar in Naha: there are n months in a year, i-th month lasts exactly d_i days. Days in the i-th month are numbered from 1 to d_i. There are no leap years in Naha.
The mood of Coronavirus-chan (and, accordingly, her desire to hug you) depends on the number of the day in a month. In particular, you get j hugs if you visit Coronavirus-chan on the j-th day of the month.
You know about this feature of your friend and want to plan your trip to get as many hugs as possible (and then maybe you can win the heart of Coronavirus-chan).
Please note that your trip should not necessarily begin and end in the same year.
Input
The first line of input contains two integers n and x (1 ≤ n ≤ 2 ⋅ 10^5) — the number of months in the year and the number of days you can spend with your friend.
The second line contains n integers d_1, d_2, …, d_n, d_i is the number of days in the i-th month (1 ≤ d_i ≤ 10^6).
It is guaranteed that 1 ≤ x ≤ d_1 + d_2 + … + d_n.
Output
Print one integer — the maximum number of hugs that you can get from Coronavirus-chan during the best vacation in your life.
Examples
Input
3 2
1 3 1
Output
5
Input
3 6
3 3 3
Output
12
Input
5 6
4 2 3 1 3
Output
15
Note
In the first test case, the numbers of the days in a year are (indices of days in a corresponding month) \{1,1,2,3,1\}. Coronavirus-chan will hug you the most if you come on the third day of the year: 2+3=5 hugs.
In the second test case, the numbers of the days are \{1,2,3,1,2,3,1,2,3\}. You will get the most hugs if you arrive on the third day of the year: 3+1+2+3+1+2=12 hugs.
In the third test case, the numbers of the days are \{1,2,3,4,1,2, 1,2,3, 1, 1,2,3\}. You will get the most hugs if you come on the twelfth day of the year: your friend will hug you 2+3+1+2+3+4=15 times.
Submitted Solution:
```
from bisect import bisect_left as bl
from bisect import bisect_right as br
from heapq import heappush,heappop
import math
from collections import *
from functools import reduce,cmp_to_key,lru_cache
from itertools import accumulate
import io, os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
# import sys
# input = sys.stdin.readline
M = mod = 10**9 + 7
def factors(n):return sorted(set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))
def inv_mod(n):return pow(n, mod - 2, mod)
def li():return [int(i) for i in input().rstrip().split()]
def st():return str(input().rstrip())[2:-1]
def val():return int(input().rstrip())
def li2():return [str(i)[2:-1] for i in input().rstrip().split()]
def li3():return [int(i) for i in st()]
n, x = li()
l = li()
l = l + l
startindices = [0 for i in l]
i = j = ans = curr = days = 0
while j < 2*n:
if days + l[j] <= x:
days += l[j]
curr += (l[j] * (l[j] + 1))//2
else:
while l[j] + days - (l[i] - startindices[i] ) > x:
left = startindices[i]
days -= l[i] - left
curr -= (l[i] * (l[i] + 1))//2 - (left * (left + 1))//2
i += 1
deletetill = l[j] + days - x
left = startindices[i]
curr -= (left + deletetill) *( left + deletetill + 1)//2 - (left * (left + 1))//2
startindices[i] += deletetill
days = x
curr += l[j] * (l[j] + 1)//2
ans = max(ans,curr)
j += 1
print(ans)
``` | instruction | 0 | 48,996 | 4 | 97,992 |
Yes | output | 1 | 48,996 | 4 | 97,993 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've been in love with Coronavirus-chan for a long time, but you didn't know where she lived until now. And just now you found out that she lives in a faraway place called Naha.
You immediately decided to take a vacation and visit Coronavirus-chan. Your vacation lasts exactly x days and that's the exact number of days you will spend visiting your friend. You will spend exactly x consecutive (successive) days visiting Coronavirus-chan.
They use a very unusual calendar in Naha: there are n months in a year, i-th month lasts exactly d_i days. Days in the i-th month are numbered from 1 to d_i. There are no leap years in Naha.
The mood of Coronavirus-chan (and, accordingly, her desire to hug you) depends on the number of the day in a month. In particular, you get j hugs if you visit Coronavirus-chan on the j-th day of the month.
You know about this feature of your friend and want to plan your trip to get as many hugs as possible (and then maybe you can win the heart of Coronavirus-chan).
Please note that your trip should not necessarily begin and end in the same year.
Input
The first line of input contains two integers n and x (1 ≤ n ≤ 2 ⋅ 10^5) — the number of months in the year and the number of days you can spend with your friend.
The second line contains n integers d_1, d_2, …, d_n, d_i is the number of days in the i-th month (1 ≤ d_i ≤ 10^6).
It is guaranteed that 1 ≤ x ≤ d_1 + d_2 + … + d_n.
Output
Print one integer — the maximum number of hugs that you can get from Coronavirus-chan during the best vacation in your life.
Examples
Input
3 2
1 3 1
Output
5
Input
3 6
3 3 3
Output
12
Input
5 6
4 2 3 1 3
Output
15
Note
In the first test case, the numbers of the days in a year are (indices of days in a corresponding month) \{1,1,2,3,1\}. Coronavirus-chan will hug you the most if you come on the third day of the year: 2+3=5 hugs.
In the second test case, the numbers of the days are \{1,2,3,1,2,3,1,2,3\}. You will get the most hugs if you arrive on the third day of the year: 3+1+2+3+1+2=12 hugs.
In the third test case, the numbers of the days are \{1,2,3,4,1,2, 1,2,3, 1, 1,2,3\}. You will get the most hugs if you come on the twelfth day of the year: your friend will hug you 2+3+1+2+3+4=15 times.
Submitted Solution:
```
n,x = map(int,input().split())
arr = list(map(int,input().split()))
i=j=su=hu=an=0
su += arr[j%n]
hu += su*(su+1)//2
while i != n:
while su < x:
j += 1
su += arr[j%n]
hu += arr[j%n]*(arr[j%n]+1)//2
if arr[i]>su-x:
hu1 = hu
y = su-x
hu1 -= y*(y+1)//2
an = max(hu1,an)
su -= arr[i]
hu -= arr[i]*(arr[i]+1)//2
i += 1
print(an)
``` | instruction | 0 | 48,997 | 4 | 97,994 |
Yes | output | 1 | 48,997 | 4 | 97,995 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've been in love with Coronavirus-chan for a long time, but you didn't know where she lived until now. And just now you found out that she lives in a faraway place called Naha.
You immediately decided to take a vacation and visit Coronavirus-chan. Your vacation lasts exactly x days and that's the exact number of days you will spend visiting your friend. You will spend exactly x consecutive (successive) days visiting Coronavirus-chan.
They use a very unusual calendar in Naha: there are n months in a year, i-th month lasts exactly d_i days. Days in the i-th month are numbered from 1 to d_i. There are no leap years in Naha.
The mood of Coronavirus-chan (and, accordingly, her desire to hug you) depends on the number of the day in a month. In particular, you get j hugs if you visit Coronavirus-chan on the j-th day of the month.
You know about this feature of your friend and want to plan your trip to get as many hugs as possible (and then maybe you can win the heart of Coronavirus-chan).
Please note that your trip should not necessarily begin and end in the same year.
Input
The first line of input contains two integers n and x (1 ≤ n ≤ 2 ⋅ 10^5) — the number of months in the year and the number of days you can spend with your friend.
The second line contains n integers d_1, d_2, …, d_n, d_i is the number of days in the i-th month (1 ≤ d_i ≤ 10^6).
It is guaranteed that 1 ≤ x ≤ d_1 + d_2 + … + d_n.
Output
Print one integer — the maximum number of hugs that you can get from Coronavirus-chan during the best vacation in your life.
Examples
Input
3 2
1 3 1
Output
5
Input
3 6
3 3 3
Output
12
Input
5 6
4 2 3 1 3
Output
15
Note
In the first test case, the numbers of the days in a year are (indices of days in a corresponding month) \{1,1,2,3,1\}. Coronavirus-chan will hug you the most if you come on the third day of the year: 2+3=5 hugs.
In the second test case, the numbers of the days are \{1,2,3,1,2,3,1,2,3\}. You will get the most hugs if you arrive on the third day of the year: 3+1+2+3+1+2=12 hugs.
In the third test case, the numbers of the days are \{1,2,3,4,1,2, 1,2,3, 1, 1,2,3\}. You will get the most hugs if you come on the twelfth day of the year: your friend will hug you 2+3+1+2+3+4=15 times.
Submitted Solution:
```
from sys import stdin, gettrace
if not gettrace():
def input():
return next(stdin)[:-1]
# def input():
# return stdin.buffer.readline()
def main():
n,x = map(int, input().split())
dd = [int(a) for a in input().split()]
ddh = []
ddd = [0]
for i in range(n):
ddh.append((dd[i]*(dd[i]+1))//2)
for i in range(2*n):
ddd.append(dd[i%n]+ddd[-1])
best = 0
hugs = 0
l = 1
r = 1
while ddd[r] < x:
hugs += ddh[(r-1)%n]
r += 1
while r < n*2:
hugs += ddh[(r-1)%n]
firstday = ddd[r] - x
while firstday >= ddd[l]:
hugs -= ddh[(l-1)%n]
l += 1
lostdays = dd[(l-1)%n] - (ddd[l] - firstday)
h = hugs - (lostdays*(lostdays+1))//2
best = max(best, h)
r+=1
print(best)
if __name__ == "__main__":
main()
``` | instruction | 0 | 48,998 | 4 | 97,996 |
Yes | output | 1 | 48,998 | 4 | 97,997 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've been in love with Coronavirus-chan for a long time, but you didn't know where she lived until now. And just now you found out that she lives in a faraway place called Naha.
You immediately decided to take a vacation and visit Coronavirus-chan. Your vacation lasts exactly x days and that's the exact number of days you will spend visiting your friend. You will spend exactly x consecutive (successive) days visiting Coronavirus-chan.
They use a very unusual calendar in Naha: there are n months in a year, i-th month lasts exactly d_i days. Days in the i-th month are numbered from 1 to d_i. There are no leap years in Naha.
The mood of Coronavirus-chan (and, accordingly, her desire to hug you) depends on the number of the day in a month. In particular, you get j hugs if you visit Coronavirus-chan on the j-th day of the month.
You know about this feature of your friend and want to plan your trip to get as many hugs as possible (and then maybe you can win the heart of Coronavirus-chan).
Please note that your trip should not necessarily begin and end in the same year.
Input
The first line of input contains two integers n and x (1 ≤ n ≤ 2 ⋅ 10^5) — the number of months in the year and the number of days you can spend with your friend.
The second line contains n integers d_1, d_2, …, d_n, d_i is the number of days in the i-th month (1 ≤ d_i ≤ 10^6).
It is guaranteed that 1 ≤ x ≤ d_1 + d_2 + … + d_n.
Output
Print one integer — the maximum number of hugs that you can get from Coronavirus-chan during the best vacation in your life.
Examples
Input
3 2
1 3 1
Output
5
Input
3 6
3 3 3
Output
12
Input
5 6
4 2 3 1 3
Output
15
Note
In the first test case, the numbers of the days in a year are (indices of days in a corresponding month) \{1,1,2,3,1\}. Coronavirus-chan will hug you the most if you come on the third day of the year: 2+3=5 hugs.
In the second test case, the numbers of the days are \{1,2,3,1,2,3,1,2,3\}. You will get the most hugs if you arrive on the third day of the year: 3+1+2+3+1+2=12 hugs.
In the third test case, the numbers of the days are \{1,2,3,4,1,2, 1,2,3, 1, 1,2,3\}. You will get the most hugs if you come on the twelfth day of the year: your friend will hug you 2+3+1+2+3+4=15 times.
Submitted Solution:
```
import sys
def II():
return int(sys.stdin.readline())
def LI():
return list(map(int, sys.stdin.readline().split()))
def MI():
return map(int, sys.stdin.readline().split())
def SI():
return sys.stdin.readline().strip()
n,x = MI()
d = LI()
l = [0]
for i in range(1,10**6+1):
l.append(l[-1]+i)
m = max(d)
if x<=m:
print(l[m]-l[m-x])
else:
count = l[m]
x-=m
tt = []
for i in range(n):
if d[i] == m:
tt.append(i)
ans = 0
for i in (tt):
cl = count
cr = count
le = i-1
r = (i+1)%n
while(x>d[le]):
cl+=l[d[le]]
x-=d[le]
le-=1
if le<0:
le+=n
cl+=l[d[le]]-l[d[le]-x]
while(x>d[r]):
cl+=l[d[r]]
x-=d[r]
r+=1
if r>=n:
r-=n
cr+=l[x]
ans = max(cl,cr)
print(ans)
``` | instruction | 0 | 48,999 | 4 | 97,998 |
No | output | 1 | 48,999 | 4 | 97,999 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've been in love with Coronavirus-chan for a long time, but you didn't know where she lived until now. And just now you found out that she lives in a faraway place called Naha.
You immediately decided to take a vacation and visit Coronavirus-chan. Your vacation lasts exactly x days and that's the exact number of days you will spend visiting your friend. You will spend exactly x consecutive (successive) days visiting Coronavirus-chan.
They use a very unusual calendar in Naha: there are n months in a year, i-th month lasts exactly d_i days. Days in the i-th month are numbered from 1 to d_i. There are no leap years in Naha.
The mood of Coronavirus-chan (and, accordingly, her desire to hug you) depends on the number of the day in a month. In particular, you get j hugs if you visit Coronavirus-chan on the j-th day of the month.
You know about this feature of your friend and want to plan your trip to get as many hugs as possible (and then maybe you can win the heart of Coronavirus-chan).
Please note that your trip should not necessarily begin and end in the same year.
Input
The first line of input contains two integers n and x (1 ≤ n ≤ 2 ⋅ 10^5) — the number of months in the year and the number of days you can spend with your friend.
The second line contains n integers d_1, d_2, …, d_n, d_i is the number of days in the i-th month (1 ≤ d_i ≤ 10^6).
It is guaranteed that 1 ≤ x ≤ d_1 + d_2 + … + d_n.
Output
Print one integer — the maximum number of hugs that you can get from Coronavirus-chan during the best vacation in your life.
Examples
Input
3 2
1 3 1
Output
5
Input
3 6
3 3 3
Output
12
Input
5 6
4 2 3 1 3
Output
15
Note
In the first test case, the numbers of the days in a year are (indices of days in a corresponding month) \{1,1,2,3,1\}. Coronavirus-chan will hug you the most if you come on the third day of the year: 2+3=5 hugs.
In the second test case, the numbers of the days are \{1,2,3,1,2,3,1,2,3\}. You will get the most hugs if you arrive on the third day of the year: 3+1+2+3+1+2=12 hugs.
In the third test case, the numbers of the days are \{1,2,3,4,1,2, 1,2,3, 1, 1,2,3\}. You will get the most hugs if you come on the twelfth day of the year: your friend will hug you 2+3+1+2+3+4=15 times.
Submitted Solution:
```
n, x=map(int, input().split())
d=list(map(int, input().split()))
td=d+d
s=[0]*len(td)
s[0]=int((d[0]+1)*d[0]/2)
for i in range(1, len(td)):
s[i]=int((td[i]+1)*td[i]/2)
td[i]+=td[i-1]
res=0
key=0
val=0
for i in range(1, len(td)):
val+=s[i]
while td[i]-td[key+1]>=x:
val-=s[key+1]
key+=1
cnt=td[i]-td[key]-x
if cnt<0:
cnt=0
res=max(res, int(val-cnt*(cnt+1)/2))
#print("%s %s %s --- %s" %(val, key, cnt, res))
print(res)
``` | instruction | 0 | 49,000 | 4 | 98,000 |
No | output | 1 | 49,000 | 4 | 98,001 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've been in love with Coronavirus-chan for a long time, but you didn't know where she lived until now. And just now you found out that she lives in a faraway place called Naha.
You immediately decided to take a vacation and visit Coronavirus-chan. Your vacation lasts exactly x days and that's the exact number of days you will spend visiting your friend. You will spend exactly x consecutive (successive) days visiting Coronavirus-chan.
They use a very unusual calendar in Naha: there are n months in a year, i-th month lasts exactly d_i days. Days in the i-th month are numbered from 1 to d_i. There are no leap years in Naha.
The mood of Coronavirus-chan (and, accordingly, her desire to hug you) depends on the number of the day in a month. In particular, you get j hugs if you visit Coronavirus-chan on the j-th day of the month.
You know about this feature of your friend and want to plan your trip to get as many hugs as possible (and then maybe you can win the heart of Coronavirus-chan).
Please note that your trip should not necessarily begin and end in the same year.
Input
The first line of input contains two integers n and x (1 ≤ n ≤ 2 ⋅ 10^5) — the number of months in the year and the number of days you can spend with your friend.
The second line contains n integers d_1, d_2, …, d_n, d_i is the number of days in the i-th month (1 ≤ d_i ≤ 10^6).
It is guaranteed that 1 ≤ x ≤ d_1 + d_2 + … + d_n.
Output
Print one integer — the maximum number of hugs that you can get from Coronavirus-chan during the best vacation in your life.
Examples
Input
3 2
1 3 1
Output
5
Input
3 6
3 3 3
Output
12
Input
5 6
4 2 3 1 3
Output
15
Note
In the first test case, the numbers of the days in a year are (indices of days in a corresponding month) \{1,1,2,3,1\}. Coronavirus-chan will hug you the most if you come on the third day of the year: 2+3=5 hugs.
In the second test case, the numbers of the days are \{1,2,3,1,2,3,1,2,3\}. You will get the most hugs if you arrive on the third day of the year: 3+1+2+3+1+2=12 hugs.
In the third test case, the numbers of the days are \{1,2,3,4,1,2, 1,2,3, 1, 1,2,3\}. You will get the most hugs if you come on the twelfth day of the year: your friend will hug you 2+3+1+2+3+4=15 times.
Submitted Solution:
```
import os
import sys
if os.path.exists('/mnt/c/Users/Square/square/codeforces'):
f = iter(open('D.txt').readlines())
def input():
return next(f)
# input = lambda: sys.stdin.readline().strip()
else:
input = lambda: sys.stdin.readline().strip()
fprint = lambda *args: print(*args, flush=True)
n, x = map(int, input().split())
A = list(map(int, input().split()))
A = A + A
def f(x):
return (x*(x+1))//2
S = [A[0]]
H = [f(A[0])]
for i in A[1:]:
S.append(S[-1] + i)
H.append(H[-1] + f(i))
n = len(A)
res1 = 0
pos2 = 0
for pos1, a in enumerate(A[:-1]):
while pos2 < n and S[pos2] - S[pos1] < x - 1:
pos2 += 1
I = S[pos2-1] - S[pos1]
t = x - 1 - I
if pos2 >= n or t > S[pos2]:
break
dH = H[pos2-1]-H[pos1]
# print('pos1 =', pos1, 'pos2 =', pos2, 't =', t, 'I =', I, 'dH =', dH, 'res =', a + dH + f(t))
res1 = max(a + dH + f(t), res1)
# print(S)
res2 = 0
pos1 = 0
for pos2, a in list(enumerate(A)):
while S[pos2] - S[pos1] > x:
pos1 += 1
I = S[pos2] - S[pos1]
t = x - I
dH = H[pos2]-H[pos1]
# print('pos1 =', pos1, 'pos2 =', pos2, 't =', t, 'A[pos1] =', A[pos1], 'I =', I, 'dH =', dH, 'res =', dH + f(A[pos1]) - f(A[pos1] - t))
if t > A[pos1]:
continue
res2 = max(dH + f(A[pos1]) - f(A[pos1] - t), res2)
res = max(res1, res2)
print(res)
``` | instruction | 0 | 49,001 | 4 | 98,002 |
No | output | 1 | 49,001 | 4 | 98,003 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've been in love with Coronavirus-chan for a long time, but you didn't know where she lived until now. And just now you found out that she lives in a faraway place called Naha.
You immediately decided to take a vacation and visit Coronavirus-chan. Your vacation lasts exactly x days and that's the exact number of days you will spend visiting your friend. You will spend exactly x consecutive (successive) days visiting Coronavirus-chan.
They use a very unusual calendar in Naha: there are n months in a year, i-th month lasts exactly d_i days. Days in the i-th month are numbered from 1 to d_i. There are no leap years in Naha.
The mood of Coronavirus-chan (and, accordingly, her desire to hug you) depends on the number of the day in a month. In particular, you get j hugs if you visit Coronavirus-chan on the j-th day of the month.
You know about this feature of your friend and want to plan your trip to get as many hugs as possible (and then maybe you can win the heart of Coronavirus-chan).
Please note that your trip should not necessarily begin and end in the same year.
Input
The first line of input contains two integers n and x (1 ≤ n ≤ 2 ⋅ 10^5) — the number of months in the year and the number of days you can spend with your friend.
The second line contains n integers d_1, d_2, …, d_n, d_i is the number of days in the i-th month (1 ≤ d_i ≤ 10^6).
It is guaranteed that 1 ≤ x ≤ d_1 + d_2 + … + d_n.
Output
Print one integer — the maximum number of hugs that you can get from Coronavirus-chan during the best vacation in your life.
Examples
Input
3 2
1 3 1
Output
5
Input
3 6
3 3 3
Output
12
Input
5 6
4 2 3 1 3
Output
15
Note
In the first test case, the numbers of the days in a year are (indices of days in a corresponding month) \{1,1,2,3,1\}. Coronavirus-chan will hug you the most if you come on the third day of the year: 2+3=5 hugs.
In the second test case, the numbers of the days are \{1,2,3,1,2,3,1,2,3\}. You will get the most hugs if you arrive on the third day of the year: 3+1+2+3+1+2=12 hugs.
In the third test case, the numbers of the days are \{1,2,3,4,1,2, 1,2,3, 1, 1,2,3\}. You will get the most hugs if you come on the twelfth day of the year: your friend will hug you 2+3+1+2+3+4=15 times.
Submitted Solution:
```
n,x = map(int, input().split())
a= list(map(int, input().split()))
ar = [i for j in a for i in range(1,j+1) ]
#print(ar)
ans=-1
if len(ar)-x == 0:
print(sum(ar))
exit()
for i in range(0,len(ar)-x):
s=0
s = sum(ar[i:i+x])
if s>=ans:
ans = s
k = ar[len(ar)-x+1:] + ar[:x-1]
for i in range(0,len(k)-x):
s=0
s = sum(k[i:i+x])
if s>=ans:
ans = s
print(ans)
``` | instruction | 0 | 49,002 | 4 | 98,004 |
No | output | 1 | 49,002 | 4 | 98,005 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasiliy spent his vacation in a sanatorium, came back and found that he completely forgot details of his vacation!
Every day there was a breakfast, a dinner and a supper in a dining room of the sanatorium (of course, in this order). The only thing that Vasiliy has now is a card from the dining room contaning notes how many times he had a breakfast, a dinner and a supper (thus, the card contains three integers). Vasiliy could sometimes have missed some meal, for example, he could have had a breakfast and a supper, but a dinner, or, probably, at some days he haven't been at the dining room at all.
Vasiliy doesn't remember what was the time of the day when he arrived to sanatorium (before breakfast, before dinner, before supper or after supper), and the time when he left it (before breakfast, before dinner, before supper or after supper). So he considers any of these options. After Vasiliy arrived to the sanatorium, he was there all the time until he left. Please note, that it's possible that Vasiliy left the sanatorium on the same day he arrived.
According to the notes in the card, help Vasiliy determine the minimum number of meals in the dining room that he could have missed. We shouldn't count as missed meals on the arrival day before Vasiliy's arrival and meals on the departure day after he left.
Input
The only line contains three integers b, d and s (0 ≤ b, d, s ≤ 1018, b + d + s ≥ 1) — the number of breakfasts, dinners and suppers which Vasiliy had during his vacation in the sanatorium.
Output
Print single integer — the minimum possible number of meals which Vasiliy could have missed during his vacation.
Examples
Input
3 2 1
Output
1
Input
1 0 0
Output
0
Input
1 1 1
Output
0
Input
1000000000000000000 0 1000000000000000000
Output
999999999999999999
Note
In the first sample, Vasiliy could have missed one supper, for example, in case he have arrived before breakfast, have been in the sanatorium for two days (including the day of arrival) and then have left after breakfast on the third day.
In the second sample, Vasiliy could have arrived before breakfast, have had it, and immediately have left the sanatorium, not missing any meal.
In the third sample, Vasiliy could have been in the sanatorium for one day, not missing any meal. | instruction | 0 | 49,318 | 4 | 98,636 |
Tags: binary search, constructive algorithms, greedy, implementation, math
Correct Solution:
```
b, d, s = map(int, input().split())
days = max(b, d, s)
print(max(0, max(days - b - 1, 0)+ max(days - d - 1, 0)+ max(days - s - 1, 0)))
``` | output | 1 | 49,318 | 4 | 98,637 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasiliy spent his vacation in a sanatorium, came back and found that he completely forgot details of his vacation!
Every day there was a breakfast, a dinner and a supper in a dining room of the sanatorium (of course, in this order). The only thing that Vasiliy has now is a card from the dining room contaning notes how many times he had a breakfast, a dinner and a supper (thus, the card contains three integers). Vasiliy could sometimes have missed some meal, for example, he could have had a breakfast and a supper, but a dinner, or, probably, at some days he haven't been at the dining room at all.
Vasiliy doesn't remember what was the time of the day when he arrived to sanatorium (before breakfast, before dinner, before supper or after supper), and the time when he left it (before breakfast, before dinner, before supper or after supper). So he considers any of these options. After Vasiliy arrived to the sanatorium, he was there all the time until he left. Please note, that it's possible that Vasiliy left the sanatorium on the same day he arrived.
According to the notes in the card, help Vasiliy determine the minimum number of meals in the dining room that he could have missed. We shouldn't count as missed meals on the arrival day before Vasiliy's arrival and meals on the departure day after he left.
Input
The only line contains three integers b, d and s (0 ≤ b, d, s ≤ 1018, b + d + s ≥ 1) — the number of breakfasts, dinners and suppers which Vasiliy had during his vacation in the sanatorium.
Output
Print single integer — the minimum possible number of meals which Vasiliy could have missed during his vacation.
Examples
Input
3 2 1
Output
1
Input
1 0 0
Output
0
Input
1 1 1
Output
0
Input
1000000000000000000 0 1000000000000000000
Output
999999999999999999
Note
In the first sample, Vasiliy could have missed one supper, for example, in case he have arrived before breakfast, have been in the sanatorium for two days (including the day of arrival) and then have left after breakfast on the third day.
In the second sample, Vasiliy could have arrived before breakfast, have had it, and immediately have left the sanatorium, not missing any meal.
In the third sample, Vasiliy could have been in the sanatorium for one day, not missing any meal. | instruction | 0 | 49,319 | 4 | 98,638 |
Tags: binary search, constructive algorithms, greedy, implementation, math
Correct Solution:
```
b, d, s = map(int, input().split())
def avg(x, y, z):
A = [x, y, z]
A.sort()
return A[1]
print(max(max(b, d, s) - min(b, d, s) - 1, 0) + max(max(b, d, s) - avg(b, d, s) - 1, 0))
``` | output | 1 | 49,319 | 4 | 98,639 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasiliy spent his vacation in a sanatorium, came back and found that he completely forgot details of his vacation!
Every day there was a breakfast, a dinner and a supper in a dining room of the sanatorium (of course, in this order). The only thing that Vasiliy has now is a card from the dining room contaning notes how many times he had a breakfast, a dinner and a supper (thus, the card contains three integers). Vasiliy could sometimes have missed some meal, for example, he could have had a breakfast and a supper, but a dinner, or, probably, at some days he haven't been at the dining room at all.
Vasiliy doesn't remember what was the time of the day when he arrived to sanatorium (before breakfast, before dinner, before supper or after supper), and the time when he left it (before breakfast, before dinner, before supper or after supper). So he considers any of these options. After Vasiliy arrived to the sanatorium, he was there all the time until he left. Please note, that it's possible that Vasiliy left the sanatorium on the same day he arrived.
According to the notes in the card, help Vasiliy determine the minimum number of meals in the dining room that he could have missed. We shouldn't count as missed meals on the arrival day before Vasiliy's arrival and meals on the departure day after he left.
Input
The only line contains three integers b, d and s (0 ≤ b, d, s ≤ 1018, b + d + s ≥ 1) — the number of breakfasts, dinners and suppers which Vasiliy had during his vacation in the sanatorium.
Output
Print single integer — the minimum possible number of meals which Vasiliy could have missed during his vacation.
Examples
Input
3 2 1
Output
1
Input
1 0 0
Output
0
Input
1 1 1
Output
0
Input
1000000000000000000 0 1000000000000000000
Output
999999999999999999
Note
In the first sample, Vasiliy could have missed one supper, for example, in case he have arrived before breakfast, have been in the sanatorium for two days (including the day of arrival) and then have left after breakfast on the third day.
In the second sample, Vasiliy could have arrived before breakfast, have had it, and immediately have left the sanatorium, not missing any meal.
In the third sample, Vasiliy could have been in the sanatorium for one day, not missing any meal. | instruction | 0 | 49,320 | 4 | 98,640 |
Tags: binary search, constructive algorithms, greedy, implementation, math
Correct Solution:
```
from sys import stdin
b, d, s = map(int, stdin.readline().split())
def calculate(last_meal, b, d, s):
if last_meal == 'b':
if b > d and b > s:
return (b - d - 1) + (b - s - 1)
else:
new_b = max(d, s) + 1
return (new_b - d - 1) + (new_b - s - 1) + new_b - b
elif last_meal == 'd':
days = max(b, d, s + 1)
return days - b + days - d + days - 1 - s
else:
days = max(b, d, s)
return days - b + days - d + days - s
print(min(calculate(l, *days) for l in ('b', 'd', 's') for days in ((b, d, s), (d, s, b), (s, b, d))))
``` | output | 1 | 49,320 | 4 | 98,641 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasiliy spent his vacation in a sanatorium, came back and found that he completely forgot details of his vacation!
Every day there was a breakfast, a dinner and a supper in a dining room of the sanatorium (of course, in this order). The only thing that Vasiliy has now is a card from the dining room contaning notes how many times he had a breakfast, a dinner and a supper (thus, the card contains three integers). Vasiliy could sometimes have missed some meal, for example, he could have had a breakfast and a supper, but a dinner, or, probably, at some days he haven't been at the dining room at all.
Vasiliy doesn't remember what was the time of the day when he arrived to sanatorium (before breakfast, before dinner, before supper or after supper), and the time when he left it (before breakfast, before dinner, before supper or after supper). So he considers any of these options. After Vasiliy arrived to the sanatorium, he was there all the time until he left. Please note, that it's possible that Vasiliy left the sanatorium on the same day he arrived.
According to the notes in the card, help Vasiliy determine the minimum number of meals in the dining room that he could have missed. We shouldn't count as missed meals on the arrival day before Vasiliy's arrival and meals on the departure day after he left.
Input
The only line contains three integers b, d and s (0 ≤ b, d, s ≤ 1018, b + d + s ≥ 1) — the number of breakfasts, dinners and suppers which Vasiliy had during his vacation in the sanatorium.
Output
Print single integer — the minimum possible number of meals which Vasiliy could have missed during his vacation.
Examples
Input
3 2 1
Output
1
Input
1 0 0
Output
0
Input
1 1 1
Output
0
Input
1000000000000000000 0 1000000000000000000
Output
999999999999999999
Note
In the first sample, Vasiliy could have missed one supper, for example, in case he have arrived before breakfast, have been in the sanatorium for two days (including the day of arrival) and then have left after breakfast on the third day.
In the second sample, Vasiliy could have arrived before breakfast, have had it, and immediately have left the sanatorium, not missing any meal.
In the third sample, Vasiliy could have been in the sanatorium for one day, not missing any meal. | instruction | 0 | 49,321 | 4 | 98,642 |
Tags: binary search, constructive algorithms, greedy, implementation, math
Correct Solution:
```
con = list(map(int, input().split()))
k = 0
con.sort()
for i in range(2):
if (con[2] - 1) > con[i]:
k += ((con[2] - 1) - con[i])
print(k)
``` | output | 1 | 49,321 | 4 | 98,643 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasiliy spent his vacation in a sanatorium, came back and found that he completely forgot details of his vacation!
Every day there was a breakfast, a dinner and a supper in a dining room of the sanatorium (of course, in this order). The only thing that Vasiliy has now is a card from the dining room contaning notes how many times he had a breakfast, a dinner and a supper (thus, the card contains three integers). Vasiliy could sometimes have missed some meal, for example, he could have had a breakfast and a supper, but a dinner, or, probably, at some days he haven't been at the dining room at all.
Vasiliy doesn't remember what was the time of the day when he arrived to sanatorium (before breakfast, before dinner, before supper or after supper), and the time when he left it (before breakfast, before dinner, before supper or after supper). So he considers any of these options. After Vasiliy arrived to the sanatorium, he was there all the time until he left. Please note, that it's possible that Vasiliy left the sanatorium on the same day he arrived.
According to the notes in the card, help Vasiliy determine the minimum number of meals in the dining room that he could have missed. We shouldn't count as missed meals on the arrival day before Vasiliy's arrival and meals on the departure day after he left.
Input
The only line contains three integers b, d and s (0 ≤ b, d, s ≤ 1018, b + d + s ≥ 1) — the number of breakfasts, dinners and suppers which Vasiliy had during his vacation in the sanatorium.
Output
Print single integer — the minimum possible number of meals which Vasiliy could have missed during his vacation.
Examples
Input
3 2 1
Output
1
Input
1 0 0
Output
0
Input
1 1 1
Output
0
Input
1000000000000000000 0 1000000000000000000
Output
999999999999999999
Note
In the first sample, Vasiliy could have missed one supper, for example, in case he have arrived before breakfast, have been in the sanatorium for two days (including the day of arrival) and then have left after breakfast on the third day.
In the second sample, Vasiliy could have arrived before breakfast, have had it, and immediately have left the sanatorium, not missing any meal.
In the third sample, Vasiliy could have been in the sanatorium for one day, not missing any meal. | instruction | 0 | 49,322 | 4 | 98,644 |
Tags: binary search, constructive algorithms, greedy, implementation, math
Correct Solution:
```
t = list(map(int,input().split()))
a= max(t)
b= min(t)
if a-b<=1:
print(0)
else:
t.sort()
ans=(a-b-1)
if t[1]+1<t[-1]:
ans+= t[-1]-t[1]-1
print(ans)
``` | output | 1 | 49,322 | 4 | 98,645 |
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