message stringlengths 2 44.5k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 42 109k | cluster float64 5 5 | __index_level_0__ int64 84 217k |
|---|---|---|---|---|---|
Provide a correct Python 3 solution for this coding contest problem.
We have two bottles for holding water.
Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.
Bottle 2 contains C milliliters of water.
We will transfer water from Bottle 2 to Bottle 1 as much as possible.
How much amount of water will remain in Bottle 2?
Constraints
* All values in input are integers.
* 1 \leq B \leq A \leq 20
* 1 \leq C \leq 20
Input
Input is given from Standard Input in the following format:
A B C
Output
Print the integer representing the amount of water, in milliliters, that will remain in Bottle 2.
Examples
Input
6 4 3
Output
1
Input
8 3 9
Output
4
Input
12 3 7
Output
0 | instruction | 0 | 1,479 | 5 | 2,958 |
"Correct Solution:
```
a,b,c=map(int,input().split())
print(b+c-a if a<b+c else 0)
``` | output | 1 | 1,479 | 5 | 2,959 |
Provide a correct Python 3 solution for this coding contest problem.
We have two bottles for holding water.
Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.
Bottle 2 contains C milliliters of water.
We will transfer water from Bottle 2 to Bottle 1 as much as possible.
How much amount of water will remain in Bottle 2?
Constraints
* All values in input are integers.
* 1 \leq B \leq A \leq 20
* 1 \leq C \leq 20
Input
Input is given from Standard Input in the following format:
A B C
Output
Print the integer representing the amount of water, in milliliters, that will remain in Bottle 2.
Examples
Input
6 4 3
Output
1
Input
8 3 9
Output
4
Input
12 3 7
Output
0 | instruction | 0 | 1,480 | 5 | 2,960 |
"Correct Solution:
```
A,B,C = [int(v) for v in input().split()]
print(max(C-(A-B),0))
``` | output | 1 | 1,480 | 5 | 2,961 |
Provide a correct Python 3 solution for this coding contest problem.
We have two bottles for holding water.
Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.
Bottle 2 contains C milliliters of water.
We will transfer water from Bottle 2 to Bottle 1 as much as possible.
How much amount of water will remain in Bottle 2?
Constraints
* All values in input are integers.
* 1 \leq B \leq A \leq 20
* 1 \leq C \leq 20
Input
Input is given from Standard Input in the following format:
A B C
Output
Print the integer representing the amount of water, in milliliters, that will remain in Bottle 2.
Examples
Input
6 4 3
Output
1
Input
8 3 9
Output
4
Input
12 3 7
Output
0 | instruction | 0 | 1,481 | 5 | 2,962 |
"Correct Solution:
```
a,b,c=map(int,input().split())
v=b+c-a
print(v if v>0 else 0)
``` | output | 1 | 1,481 | 5 | 2,963 |
Provide a correct Python 3 solution for this coding contest problem.
We have two bottles for holding water.
Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.
Bottle 2 contains C milliliters of water.
We will transfer water from Bottle 2 to Bottle 1 as much as possible.
How much amount of water will remain in Bottle 2?
Constraints
* All values in input are integers.
* 1 \leq B \leq A \leq 20
* 1 \leq C \leq 20
Input
Input is given from Standard Input in the following format:
A B C
Output
Print the integer representing the amount of water, in milliliters, that will remain in Bottle 2.
Examples
Input
6 4 3
Output
1
Input
8 3 9
Output
4
Input
12 3 7
Output
0 | instruction | 0 | 1,482 | 5 | 2,964 |
"Correct Solution:
```
A,B,C=map(int,input().split())
tmp=max(0,C-(A-B))
print(tmp)
``` | output | 1 | 1,482 | 5 | 2,965 |
Provide a correct Python 3 solution for this coding contest problem.
We have two bottles for holding water.
Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.
Bottle 2 contains C milliliters of water.
We will transfer water from Bottle 2 to Bottle 1 as much as possible.
How much amount of water will remain in Bottle 2?
Constraints
* All values in input are integers.
* 1 \leq B \leq A \leq 20
* 1 \leq C \leq 20
Input
Input is given from Standard Input in the following format:
A B C
Output
Print the integer representing the amount of water, in milliliters, that will remain in Bottle 2.
Examples
Input
6 4 3
Output
1
Input
8 3 9
Output
4
Input
12 3 7
Output
0 | instruction | 0 | 1,483 | 5 | 2,966 |
"Correct Solution:
```
A, B, C = map(int, input().split(' '))
print(max(C - (A - B), 0))
``` | output | 1 | 1,483 | 5 | 2,967 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have two bottles for holding water.
Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.
Bottle 2 contains C milliliters of water.
We will transfer water from Bottle 2 to Bottle 1 as much as possible.
How much amount of water will remain in Bottle 2?
Constraints
* All values in input are integers.
* 1 \leq B \leq A \leq 20
* 1 \leq C \leq 20
Input
Input is given from Standard Input in the following format:
A B C
Output
Print the integer representing the amount of water, in milliliters, that will remain in Bottle 2.
Examples
Input
6 4 3
Output
1
Input
8 3 9
Output
4
Input
12 3 7
Output
0
Submitted Solution:
```
a,b,c = list(map(int,input().split(" ")))
print(max(0, c -a+b))
``` | instruction | 0 | 1,484 | 5 | 2,968 |
Yes | output | 1 | 1,484 | 5 | 2,969 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have two bottles for holding water.
Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.
Bottle 2 contains C milliliters of water.
We will transfer water from Bottle 2 to Bottle 1 as much as possible.
How much amount of water will remain in Bottle 2?
Constraints
* All values in input are integers.
* 1 \leq B \leq A \leq 20
* 1 \leq C \leq 20
Input
Input is given from Standard Input in the following format:
A B C
Output
Print the integer representing the amount of water, in milliliters, that will remain in Bottle 2.
Examples
Input
6 4 3
Output
1
Input
8 3 9
Output
4
Input
12 3 7
Output
0
Submitted Solution:
```
a = list(map(int,input().split()))
print(max(0,a[2]-(a[0]-a[1])))
``` | instruction | 0 | 1,485 | 5 | 2,970 |
Yes | output | 1 | 1,485 | 5 | 2,971 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have two bottles for holding water.
Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.
Bottle 2 contains C milliliters of water.
We will transfer water from Bottle 2 to Bottle 1 as much as possible.
How much amount of water will remain in Bottle 2?
Constraints
* All values in input are integers.
* 1 \leq B \leq A \leq 20
* 1 \leq C \leq 20
Input
Input is given from Standard Input in the following format:
A B C
Output
Print the integer representing the amount of water, in milliliters, that will remain in Bottle 2.
Examples
Input
6 4 3
Output
1
Input
8 3 9
Output
4
Input
12 3 7
Output
0
Submitted Solution:
```
a,b,c=map(int,input().split())
print(abs(min(a-b-c,0)))
``` | instruction | 0 | 1,486 | 5 | 2,972 |
Yes | output | 1 | 1,486 | 5 | 2,973 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have two bottles for holding water.
Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.
Bottle 2 contains C milliliters of water.
We will transfer water from Bottle 2 to Bottle 1 as much as possible.
How much amount of water will remain in Bottle 2?
Constraints
* All values in input are integers.
* 1 \leq B \leq A \leq 20
* 1 \leq C \leq 20
Input
Input is given from Standard Input in the following format:
A B C
Output
Print the integer representing the amount of water, in milliliters, that will remain in Bottle 2.
Examples
Input
6 4 3
Output
1
Input
8 3 9
Output
4
Input
12 3 7
Output
0
Submitted Solution:
```
A,B,C=map(int,input().split())
print(C-min(C,A-B))
``` | instruction | 0 | 1,487 | 5 | 2,974 |
Yes | output | 1 | 1,487 | 5 | 2,975 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have two bottles for holding water.
Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.
Bottle 2 contains C milliliters of water.
We will transfer water from Bottle 2 to Bottle 1 as much as possible.
How much amount of water will remain in Bottle 2?
Constraints
* All values in input are integers.
* 1 \leq B \leq A \leq 20
* 1 \leq C \leq 20
Input
Input is given from Standard Input in the following format:
A B C
Output
Print the integer representing the amount of water, in milliliters, that will remain in Bottle 2.
Examples
Input
6 4 3
Output
1
Input
8 3 9
Output
4
Input
12 3 7
Output
0
Submitted Solution:
```
BottleOne = int(input())
BottleTwo = int(input())
BottleThree = int(input())
ans = int((BottleThree+BottleTwo)) - int((BottleOne))
print(int(ans))
``` | instruction | 0 | 1,488 | 5 | 2,976 |
No | output | 1 | 1,488 | 5 | 2,977 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have two bottles for holding water.
Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.
Bottle 2 contains C milliliters of water.
We will transfer water from Bottle 2 to Bottle 1 as much as possible.
How much amount of water will remain in Bottle 2?
Constraints
* All values in input are integers.
* 1 \leq B \leq A \leq 20
* 1 \leq C \leq 20
Input
Input is given from Standard Input in the following format:
A B C
Output
Print the integer representing the amount of water, in milliliters, that will remain in Bottle 2.
Examples
Input
6 4 3
Output
1
Input
8 3 9
Output
4
Input
12 3 7
Output
0
Submitted Solution:
```
# チートシート
a, b, c = map(int, input().split())
tmp = a-b
print(c-tmp)
``` | instruction | 0 | 1,489 | 5 | 2,978 |
No | output | 1 | 1,489 | 5 | 2,979 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have two bottles for holding water.
Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.
Bottle 2 contains C milliliters of water.
We will transfer water from Bottle 2 to Bottle 1 as much as possible.
How much amount of water will remain in Bottle 2?
Constraints
* All values in input are integers.
* 1 \leq B \leq A \leq 20
* 1 \leq C \leq 20
Input
Input is given from Standard Input in the following format:
A B C
Output
Print the integer representing the amount of water, in milliliters, that will remain in Bottle 2.
Examples
Input
6 4 3
Output
1
Input
8 3 9
Output
4
Input
12 3 7
Output
0
Submitted Solution:
```
A, B, C = [int(i) for i in input().split(" ")]
print(C-A+B if C-A+B else 0)
``` | instruction | 0 | 1,490 | 5 | 2,980 |
No | output | 1 | 1,490 | 5 | 2,981 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have two bottles for holding water.
Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.
Bottle 2 contains C milliliters of water.
We will transfer water from Bottle 2 to Bottle 1 as much as possible.
How much amount of water will remain in Bottle 2?
Constraints
* All values in input are integers.
* 1 \leq B \leq A \leq 20
* 1 \leq C \leq 20
Input
Input is given from Standard Input in the following format:
A B C
Output
Print the integer representing the amount of water, in milliliters, that will remain in Bottle 2.
Examples
Input
6 4 3
Output
1
Input
8 3 9
Output
4
Input
12 3 7
Output
0
Submitted Solution:
```
def main():
n = int(input())
h = list(map(int, input().split()))
pre = 0
flag = True
for i in range(n):
if h[i] - 1 >= pre:
h[i] -= -1
if h[i] < pre:
flag = False
break
pre = h[i]
if flag:
print('Yes')
else:
print('No')
if __name__ == '__main__':
main()
``` | instruction | 0 | 1,491 | 5 | 2,982 |
No | output | 1 | 1,491 | 5 | 2,983 |
Provide a correct Python 3 solution for this coding contest problem.
In programming, hexadecimal notation is often used.
In hexadecimal notation, besides the ten digits 0, 1, ..., 9, the six letters `A`, `B`, `C`, `D`, `E` and `F` are used to represent the values 10, 11, 12, 13, 14 and 15, respectively.
In this problem, you are given two letters X and Y. Each X and Y is `A`, `B`, `C`, `D`, `E` or `F`.
When X and Y are seen as hexadecimal numbers, which is larger?
Constraints
* Each X and Y is `A`, `B`, `C`, `D`, `E` or `F`.
Input
Input is given from Standard Input in the following format:
X Y
Output
If X is smaller, print `<`; if Y is smaller, print `>`; if they are equal, print `=`.
Examples
Input
A B
Output
<
Input
E C
Output
>
Input
F F
Output
= | instruction | 0 | 1,537 | 5 | 3,074 |
"Correct Solution:
```
x, y = input().split();print("=" if x==y else "<>"[x>y::2])
``` | output | 1 | 1,537 | 5 | 3,075 |
Provide a correct Python 3 solution for this coding contest problem.
In programming, hexadecimal notation is often used.
In hexadecimal notation, besides the ten digits 0, 1, ..., 9, the six letters `A`, `B`, `C`, `D`, `E` and `F` are used to represent the values 10, 11, 12, 13, 14 and 15, respectively.
In this problem, you are given two letters X and Y. Each X and Y is `A`, `B`, `C`, `D`, `E` or `F`.
When X and Y are seen as hexadecimal numbers, which is larger?
Constraints
* Each X and Y is `A`, `B`, `C`, `D`, `E` or `F`.
Input
Input is given from Standard Input in the following format:
X Y
Output
If X is smaller, print `<`; if Y is smaller, print `>`; if they are equal, print `=`.
Examples
Input
A B
Output
<
Input
E C
Output
>
Input
F F
Output
= | instruction | 0 | 1,538 | 5 | 3,076 |
"Correct Solution:
```
a,b = input().split()
if a>b:
print(">")
if a == b:
print("=")
if a<b:
print("<")
``` | output | 1 | 1,538 | 5 | 3,077 |
Provide a correct Python 3 solution for this coding contest problem.
In programming, hexadecimal notation is often used.
In hexadecimal notation, besides the ten digits 0, 1, ..., 9, the six letters `A`, `B`, `C`, `D`, `E` and `F` are used to represent the values 10, 11, 12, 13, 14 and 15, respectively.
In this problem, you are given two letters X and Y. Each X and Y is `A`, `B`, `C`, `D`, `E` or `F`.
When X and Y are seen as hexadecimal numbers, which is larger?
Constraints
* Each X and Y is `A`, `B`, `C`, `D`, `E` or `F`.
Input
Input is given from Standard Input in the following format:
X Y
Output
If X is smaller, print `<`; if Y is smaller, print `>`; if they are equal, print `=`.
Examples
Input
A B
Output
<
Input
E C
Output
>
Input
F F
Output
= | instruction | 0 | 1,539 | 5 | 3,078 |
"Correct Solution:
```
a,b=map(str,input().split())
print('>'if a>b else'<'if a<b else'=')
``` | output | 1 | 1,539 | 5 | 3,079 |
Provide a correct Python 3 solution for this coding contest problem.
In programming, hexadecimal notation is often used.
In hexadecimal notation, besides the ten digits 0, 1, ..., 9, the six letters `A`, `B`, `C`, `D`, `E` and `F` are used to represent the values 10, 11, 12, 13, 14 and 15, respectively.
In this problem, you are given two letters X and Y. Each X and Y is `A`, `B`, `C`, `D`, `E` or `F`.
When X and Y are seen as hexadecimal numbers, which is larger?
Constraints
* Each X and Y is `A`, `B`, `C`, `D`, `E` or `F`.
Input
Input is given from Standard Input in the following format:
X Y
Output
If X is smaller, print `<`; if Y is smaller, print `>`; if they are equal, print `=`.
Examples
Input
A B
Output
<
Input
E C
Output
>
Input
F F
Output
= | instruction | 0 | 1,540 | 5 | 3,080 |
"Correct Solution:
```
x,y=input().split();print("=><"[(x>y)-(x<y)])
``` | output | 1 | 1,540 | 5 | 3,081 |
Provide a correct Python 3 solution for this coding contest problem.
In programming, hexadecimal notation is often used.
In hexadecimal notation, besides the ten digits 0, 1, ..., 9, the six letters `A`, `B`, `C`, `D`, `E` and `F` are used to represent the values 10, 11, 12, 13, 14 and 15, respectively.
In this problem, you are given two letters X and Y. Each X and Y is `A`, `B`, `C`, `D`, `E` or `F`.
When X and Y are seen as hexadecimal numbers, which is larger?
Constraints
* Each X and Y is `A`, `B`, `C`, `D`, `E` or `F`.
Input
Input is given from Standard Input in the following format:
X Y
Output
If X is smaller, print `<`; if Y is smaller, print `>`; if they are equal, print `=`.
Examples
Input
A B
Output
<
Input
E C
Output
>
Input
F F
Output
= | instruction | 0 | 1,542 | 5 | 3,084 |
"Correct Solution:
```
a,b=input().split();print('>=<'[(a<b)+(a<=b)])
``` | output | 1 | 1,542 | 5 | 3,085 |
Provide a correct Python 3 solution for this coding contest problem.
In programming, hexadecimal notation is often used.
In hexadecimal notation, besides the ten digits 0, 1, ..., 9, the six letters `A`, `B`, `C`, `D`, `E` and `F` are used to represent the values 10, 11, 12, 13, 14 and 15, respectively.
In this problem, you are given two letters X and Y. Each X and Y is `A`, `B`, `C`, `D`, `E` or `F`.
When X and Y are seen as hexadecimal numbers, which is larger?
Constraints
* Each X and Y is `A`, `B`, `C`, `D`, `E` or `F`.
Input
Input is given from Standard Input in the following format:
X Y
Output
If X is smaller, print `<`; if Y is smaller, print `>`; if they are equal, print `=`.
Examples
Input
A B
Output
<
Input
E C
Output
>
Input
F F
Output
= | instruction | 0 | 1,544 | 5 | 3,088 |
"Correct Solution:
```
x, y = input().split()
print("<" if x < y else (">" if x > y else "="))
``` | output | 1 | 1,544 | 5 | 3,089 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In programming, hexadecimal notation is often used.
In hexadecimal notation, besides the ten digits 0, 1, ..., 9, the six letters `A`, `B`, `C`, `D`, `E` and `F` are used to represent the values 10, 11, 12, 13, 14 and 15, respectively.
In this problem, you are given two letters X and Y. Each X and Y is `A`, `B`, `C`, `D`, `E` or `F`.
When X and Y are seen as hexadecimal numbers, which is larger?
Constraints
* Each X and Y is `A`, `B`, `C`, `D`, `E` or `F`.
Input
Input is given from Standard Input in the following format:
X Y
Output
If X is smaller, print `<`; if Y is smaller, print `>`; if they are equal, print `=`.
Examples
Input
A B
Output
<
Input
E C
Output
>
Input
F F
Output
=
Submitted Solution:
```
x,y=input().split()
if x>y:
print(">")
elif y>x:
print("<")
else:
print("=")
``` | instruction | 0 | 1,545 | 5 | 3,090 |
Yes | output | 1 | 1,545 | 5 | 3,091 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In programming, hexadecimal notation is often used.
In hexadecimal notation, besides the ten digits 0, 1, ..., 9, the six letters `A`, `B`, `C`, `D`, `E` and `F` are used to represent the values 10, 11, 12, 13, 14 and 15, respectively.
In this problem, you are given two letters X and Y. Each X and Y is `A`, `B`, `C`, `D`, `E` or `F`.
When X and Y are seen as hexadecimal numbers, which is larger?
Constraints
* Each X and Y is `A`, `B`, `C`, `D`, `E` or `F`.
Input
Input is given from Standard Input in the following format:
X Y
Output
If X is smaller, print `<`; if Y is smaller, print `>`; if they are equal, print `=`.
Examples
Input
A B
Output
<
Input
E C
Output
>
Input
F F
Output
=
Submitted Solution:
```
x,y=input().split()
if x<y:
print("<")
elif x>y:
print(">")
else:
print("=")
``` | instruction | 0 | 1,547 | 5 | 3,094 |
Yes | output | 1 | 1,547 | 5 | 3,095 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In programming, hexadecimal notation is often used.
In hexadecimal notation, besides the ten digits 0, 1, ..., 9, the six letters `A`, `B`, `C`, `D`, `E` and `F` are used to represent the values 10, 11, 12, 13, 14 and 15, respectively.
In this problem, you are given two letters X and Y. Each X and Y is `A`, `B`, `C`, `D`, `E` or `F`.
When X and Y are seen as hexadecimal numbers, which is larger?
Constraints
* Each X and Y is `A`, `B`, `C`, `D`, `E` or `F`.
Input
Input is given from Standard Input in the following format:
X Y
Output
If X is smaller, print `<`; if Y is smaller, print `>`; if they are equal, print `=`.
Examples
Input
A B
Output
<
Input
E C
Output
>
Input
F F
Output
=
Submitted Solution:
```
X, Y = input().split()
print('>' if X>Y else '<' if X<Y else '=')
``` | instruction | 0 | 1,548 | 5 | 3,096 |
Yes | output | 1 | 1,548 | 5 | 3,097 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In programming, hexadecimal notation is often used.
In hexadecimal notation, besides the ten digits 0, 1, ..., 9, the six letters `A`, `B`, `C`, `D`, `E` and `F` are used to represent the values 10, 11, 12, 13, 14 and 15, respectively.
In this problem, you are given two letters X and Y. Each X and Y is `A`, `B`, `C`, `D`, `E` or `F`.
When X and Y are seen as hexadecimal numbers, which is larger?
Constraints
* Each X and Y is `A`, `B`, `C`, `D`, `E` or `F`.
Input
Input is given from Standard Input in the following format:
X Y
Output
If X is smaller, print `<`; if Y is smaller, print `>`; if they are equal, print `=`.
Examples
Input
A B
Output
<
Input
E C
Output
>
Input
F F
Output
=
Submitted Solution:
```
I=input()
s=I.split(" ")
#print(s)
c=sorted(s)
#print(c)
if s==c and s[0]!=s[1]:
print("<")
if s!=c and s[0]!=s[1]:
print(">")
else:
print("=")
``` | instruction | 0 | 1,549 | 5 | 3,098 |
No | output | 1 | 1,549 | 5 | 3,099 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In programming, hexadecimal notation is often used.
In hexadecimal notation, besides the ten digits 0, 1, ..., 9, the six letters `A`, `B`, `C`, `D`, `E` and `F` are used to represent the values 10, 11, 12, 13, 14 and 15, respectively.
In this problem, you are given two letters X and Y. Each X and Y is `A`, `B`, `C`, `D`, `E` or `F`.
When X and Y are seen as hexadecimal numbers, which is larger?
Constraints
* Each X and Y is `A`, `B`, `C`, `D`, `E` or `F`.
Input
Input is given from Standard Input in the following format:
X Y
Output
If X is smaller, print `<`; if Y is smaller, print `>`; if they are equal, print `=`.
Examples
Input
A B
Output
<
Input
E C
Output
>
Input
F F
Output
=
Submitted Solution:
```
x,y = map(int, input().split())
a=0
b=0
if x=="A":
a=10
elif x=="B":
a=11
elif x=="C":
a=12
elif x=="D":
a=13:
elif x=="E":
a=14
elif x=="F":
a=15
if y=="A":
b=10
elif y=="B":
b=11
elif x=="C":
b=12
elif x=="D":
b=13
elif x=="E":
b=14
elif x=="F":
b=15
if a<b:
print("<")
elif a>b:
print(">")
elif a==b:
print("=")
``` | instruction | 0 | 1,551 | 5 | 3,102 |
No | output | 1 | 1,551 | 5 | 3,103 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In programming, hexadecimal notation is often used.
In hexadecimal notation, besides the ten digits 0, 1, ..., 9, the six letters `A`, `B`, `C`, `D`, `E` and `F` are used to represent the values 10, 11, 12, 13, 14 and 15, respectively.
In this problem, you are given two letters X and Y. Each X and Y is `A`, `B`, `C`, `D`, `E` or `F`.
When X and Y are seen as hexadecimal numbers, which is larger?
Constraints
* Each X and Y is `A`, `B`, `C`, `D`, `E` or `F`.
Input
Input is given from Standard Input in the following format:
X Y
Output
If X is smaller, print `<`; if Y is smaller, print `>`; if they are equal, print `=`.
Examples
Input
A B
Output
<
Input
E C
Output
>
Input
F F
Output
=
Submitted Solution:
```
import sys
import math
# 入力
a, b, c = map(int, input().split())
# 計算
if a < (b + 2*c):
print("a >= b + 2*cとなるようにしてください")
sys.exit()
elif a == (b + 2*c):
print("1")
else:
max = math.floor((a + c) / (b + 2*c))
print(max)
``` | instruction | 0 | 1,552 | 5 | 3,104 |
No | output | 1 | 1,552 | 5 | 3,105 |
Provide a correct Python 3 solution for this coding contest problem.
You are given an integer sequence of length N, a = {a_1, a_2, …, a_N}, and an integer K.
a has N(N+1)/2 non-empty contiguous subsequences, {a_l, a_{l+1}, …, a_r} (1 ≤ l ≤ r ≤ N). Among them, how many have an arithmetic mean that is greater than or equal to K?
Constraints
* All input values are integers.
* 1 ≤ N ≤ 2 \times 10^5
* 1 ≤ K ≤ 10^9
* 1 ≤ a_i ≤ 10^9
Input
Input is given from Standard Input in the following format:
N K
a_1
a_2
:
a_N
Output
Print the number of the non-empty contiguous subsequences with an arithmetic mean that is greater than or equal to K.
Examples
Input
3 6
7
5
7
Output
5
Input
1 2
1
Output
0
Input
7 26
10
20
30
40
30
20
10
Output
13 | instruction | 0 | 1,553 | 5 | 3,106 |
"Correct Solution:
```
class Bit:
def __init__(self, n):
self.size = n
self.tree = [0] * (n + 1)
def sum(self, i):
s = 0
while i > 0:
s += self.tree[i]
i -= i & -i
return s
def add(self, i, x):
while i <= self.size:
self.tree[i] += x
i += i & -i
def main():
import sys
read = sys.stdin.read
N, K = (int(i) for i in input().split())
A = [int(s) for s in read().rstrip().split('\n')]
from itertools import accumulate
S = list(accumulate([0] + A))
B = [s - K*i for i, s in enumerate(S)]
c = {b: i+1 for i, b in enumerate(sorted(set(B)))}
f = Bit(N+1)
ans = 0
for j in range(N+1):
ans += f.sum(c[B[j]])
f.add(c[B[j]], 1)
print(ans)
if __name__ == '__main__':
main()
``` | output | 1 | 1,553 | 5 | 3,107 |
Provide a correct Python 3 solution for this coding contest problem.
You are given an integer sequence of length N, a = {a_1, a_2, …, a_N}, and an integer K.
a has N(N+1)/2 non-empty contiguous subsequences, {a_l, a_{l+1}, …, a_r} (1 ≤ l ≤ r ≤ N). Among them, how many have an arithmetic mean that is greater than or equal to K?
Constraints
* All input values are integers.
* 1 ≤ N ≤ 2 \times 10^5
* 1 ≤ K ≤ 10^9
* 1 ≤ a_i ≤ 10^9
Input
Input is given from Standard Input in the following format:
N K
a_1
a_2
:
a_N
Output
Print the number of the non-empty contiguous subsequences with an arithmetic mean that is greater than or equal to K.
Examples
Input
3 6
7
5
7
Output
5
Input
1 2
1
Output
0
Input
7 26
10
20
30
40
30
20
10
Output
13 | instruction | 0 | 1,554 | 5 | 3,108 |
"Correct Solution:
```
from bisect import bisect_left
import sys
input = sys.stdin.readline
class BinaryIndexedTree():
def __init__(self, N):
self.N = N
self.arr = [0] * (N+1)
def query(self, i):
ret = 0
i += 1
while i:
ret += self.arr[i]
lsb = i & (-i)
i -= lsb
return ret
def add(self, i, v):
i += 1
while i < self.N+1:
lsb = i & (-i)
self.arr[i] += v
i += lsb
def search(self, x):
lo = -1
hi = self.N
while lo < hi:
mid = (lo+hi)//2
if self.query(mid) < x:
lo = mid + 1
else:
hi = mid
return lo
def main():
N, K = map(int, input().split())
A = [int(input()) for _ in range(N)]
B = [0] * (N+1)
for i in range(N):
B[i+1] = B[i] + A[i] - K
sortedB = sorted(B)
C = [0] * (N+1)
bit = BinaryIndexedTree(N+1)
ans = 0
for i in range(N+1):
x = bisect_left(sortedB, B[i])
C[i] = x
ans += bit.query(x)
bit.add(C[i], 1)
print(ans)
if __name__ == "__main__":
main()
``` | output | 1 | 1,554 | 5 | 3,109 |
Provide a correct Python 3 solution for this coding contest problem.
You are given an integer sequence of length N, a = {a_1, a_2, …, a_N}, and an integer K.
a has N(N+1)/2 non-empty contiguous subsequences, {a_l, a_{l+1}, …, a_r} (1 ≤ l ≤ r ≤ N). Among them, how many have an arithmetic mean that is greater than or equal to K?
Constraints
* All input values are integers.
* 1 ≤ N ≤ 2 \times 10^5
* 1 ≤ K ≤ 10^9
* 1 ≤ a_i ≤ 10^9
Input
Input is given from Standard Input in the following format:
N K
a_1
a_2
:
a_N
Output
Print the number of the non-empty contiguous subsequences with an arithmetic mean that is greater than or equal to K.
Examples
Input
3 6
7
5
7
Output
5
Input
1 2
1
Output
0
Input
7 26
10
20
30
40
30
20
10
Output
13 | instruction | 0 | 1,555 | 5 | 3,110 |
"Correct Solution:
```
from collections import defaultdict, deque, Counter
from heapq import heappush, heappop, heapify
from bisect import bisect_right, bisect_left
import random
from itertools import permutations, accumulate, combinations, product
import sys
import string
from bisect import bisect_left, bisect_right
from math import factorial, ceil, floor, gamma, log
from operator import mul
from functools import reduce
sys.setrecursionlimit(2147483647)
INF = 1 << 100
def LI(): return list(map(int, sys.stdin.buffer.readline().split()))
def I(): return int(sys.stdin.buffer.readline())
def LS(): return sys.stdin.buffer.readline().rstrip().decode('utf-8').split()
def S(): return sys.stdin.buffer.readline().rstrip().decode('utf-8')
def IR(n): return [I() for i in range(n)]
def LIR(n): return [LI() for i in range(n)]
def SR(n): return [S() for i in range(n)]
def LSR(n): return [LS() for i in range(n)]
def SRL(n): return [list(S()) for i in range(n)]
def MSRL(n): return [[int(j) for j in list(S())] for i in range(n)]
mod = 1000000007
class BIT:
def __init__(self, size):
self.bit = [0] * size
self.size = size
self.total = 0
def add(self, i, w):
x = i + 1
while x <= self.size:
self.bit[x - 1] += w
x += x & -x
return
def sum(self, i):
res = 0
x = i + 1
while x:
res += self.bit[x - 1]
x -= x & -x
return res
n, k = LI()
A = IR(n)
A = [a - k for a in A]
A = [0] + list(accumulate(A))
D = {}
for i, v in enumerate(sorted(set(A))):
D[v] = i
for i in range(n + 1):
A[i] = D[A[i]]
bit = BIT(max(A) + 1)
c = 0
for a in A:
c += bit.sum(a)
bit.add(a, 1)
print(c)
``` | output | 1 | 1,555 | 5 | 3,111 |
Provide a correct Python 3 solution for this coding contest problem.
You are given an integer sequence of length N, a = {a_1, a_2, …, a_N}, and an integer K.
a has N(N+1)/2 non-empty contiguous subsequences, {a_l, a_{l+1}, …, a_r} (1 ≤ l ≤ r ≤ N). Among them, how many have an arithmetic mean that is greater than or equal to K?
Constraints
* All input values are integers.
* 1 ≤ N ≤ 2 \times 10^5
* 1 ≤ K ≤ 10^9
* 1 ≤ a_i ≤ 10^9
Input
Input is given from Standard Input in the following format:
N K
a_1
a_2
:
a_N
Output
Print the number of the non-empty contiguous subsequences with an arithmetic mean that is greater than or equal to K.
Examples
Input
3 6
7
5
7
Output
5
Input
1 2
1
Output
0
Input
7 26
10
20
30
40
30
20
10
Output
13 | instruction | 0 | 1,557 | 5 | 3,114 |
"Correct Solution:
```
from itertools import accumulate
from bisect import bisect_right
N, K = map(int, input().split())
A = [-19] + [int(input()) for i in range(N)]
diff = [a - K for a in A]
diff = list(accumulate(diff))
class BinaryIndexedTree:
def __init__(self, n):
self.size = n
self.bit = [0] * (n + 1)
def sum(self, i):
s = 0
while i > 0:
s += self.bit[i]
i -= (i & -i)
return s
def add(self, i, x):
while i <= self.size:
self.bit[i] += x
i += (i & -i)
def reset(self):
self.bit = [0] * (self.size + 1)
BIT = BinaryIndexedTree(N + 1)
temp = sorted(diff)
order = [bisect_right(temp, d) for d in diff]
ans = 0
for x in order:
ans += BIT.sum(x)
BIT.add(x, 1)
print(ans)
``` | output | 1 | 1,557 | 5 | 3,115 |
Provide a correct Python 3 solution for this coding contest problem.
You are given an integer sequence of length N, a = {a_1, a_2, …, a_N}, and an integer K.
a has N(N+1)/2 non-empty contiguous subsequences, {a_l, a_{l+1}, …, a_r} (1 ≤ l ≤ r ≤ N). Among them, how many have an arithmetic mean that is greater than or equal to K?
Constraints
* All input values are integers.
* 1 ≤ N ≤ 2 \times 10^5
* 1 ≤ K ≤ 10^9
* 1 ≤ a_i ≤ 10^9
Input
Input is given from Standard Input in the following format:
N K
a_1
a_2
:
a_N
Output
Print the number of the non-empty contiguous subsequences with an arithmetic mean that is greater than or equal to K.
Examples
Input
3 6
7
5
7
Output
5
Input
1 2
1
Output
0
Input
7 26
10
20
30
40
30
20
10
Output
13 | instruction | 0 | 1,558 | 5 | 3,116 |
"Correct Solution:
```
from bisect import bisect_left
class Bit:
def __init__(self, n):
self.size = n
self.tree = [0] * (n+1)
def sum(self, i): # [0, i)
s = 0
while i > 0:
s += self.tree[i]
i -= i & -i
return s
def add(self, i, x): # i > 0
assert i > 0
while i <= self.size:
self.tree[i] += x
i += i & -i
n, k = map(int, input().split())
a = [int(input()) for _ in range(n)]
d = [0] * (n+1)
d2 = {0}
for i in range(n):
d[i+1] = d[i] + a[i] - k
d2.add(d[i+1])
d2 = sorted(d2)
bit = Bit(len(d2))
ans = 0
for i in range(n+1):
d2_i = bisect_left(d2, d[i])
ans += bit.sum(d2_i+1)
bit.add(d2_i+1, 1)
print(ans)
``` | output | 1 | 1,558 | 5 | 3,117 |
Provide a correct Python 3 solution for this coding contest problem.
You are given an integer sequence of length N, a = {a_1, a_2, …, a_N}, and an integer K.
a has N(N+1)/2 non-empty contiguous subsequences, {a_l, a_{l+1}, …, a_r} (1 ≤ l ≤ r ≤ N). Among them, how many have an arithmetic mean that is greater than or equal to K?
Constraints
* All input values are integers.
* 1 ≤ N ≤ 2 \times 10^5
* 1 ≤ K ≤ 10^9
* 1 ≤ a_i ≤ 10^9
Input
Input is given from Standard Input in the following format:
N K
a_1
a_2
:
a_N
Output
Print the number of the non-empty contiguous subsequences with an arithmetic mean that is greater than or equal to K.
Examples
Input
3 6
7
5
7
Output
5
Input
1 2
1
Output
0
Input
7 26
10
20
30
40
30
20
10
Output
13 | instruction | 0 | 1,559 | 5 | 3,118 |
"Correct Solution:
```
import sys
from copy import deepcopy
F = sys.stdin
N, K = map(int, F.readline().strip("\n").split())
A = [None] * (N + 1)
A[0] = 0
A_order = {0}
for a in range(N):
A[a+1] = int(F.readline().strip("\n")) - K + A[a]
A_order |= {A[a+1]}
maxA = len(A_order)
A_order = list(A_order)
A_order.sort()
compressed_A = dict()
for i in range(maxA):
compressed_A[A_order[i]] = i
bit = [0] * (maxA + 1)
total_inversion = 0
for i in range(N+1):
b = compressed_A[A[i]] + 1
while b <= maxA + 1:
bit[b - 1] += 1
b += b & -b
#lower sum
lower = compressed_A[A[i]] + 1
lower_sum = 0
while lower > 0:
lower_sum += bit[lower - 1]
lower -= lower & -lower
total_inversion += i + 1 - lower_sum
print(N * (N + 1) // 2 - total_inversion)
``` | output | 1 | 1,559 | 5 | 3,119 |
Provide a correct Python 3 solution for this coding contest problem.
You are given an integer sequence of length N, a = {a_1, a_2, …, a_N}, and an integer K.
a has N(N+1)/2 non-empty contiguous subsequences, {a_l, a_{l+1}, …, a_r} (1 ≤ l ≤ r ≤ N). Among them, how many have an arithmetic mean that is greater than or equal to K?
Constraints
* All input values are integers.
* 1 ≤ N ≤ 2 \times 10^5
* 1 ≤ K ≤ 10^9
* 1 ≤ a_i ≤ 10^9
Input
Input is given from Standard Input in the following format:
N K
a_1
a_2
:
a_N
Output
Print the number of the non-empty contiguous subsequences with an arithmetic mean that is greater than or equal to K.
Examples
Input
3 6
7
5
7
Output
5
Input
1 2
1
Output
0
Input
7 26
10
20
30
40
30
20
10
Output
13 | instruction | 0 | 1,560 | 5 | 3,120 |
"Correct Solution:
```
n,k=map(int,input().split())
b=[0]*(n+1)
for i in range(n):
b[i+1]=b[i]+int(input())-k
a=sorted(set(b))
s={a[i-1]:i for i in range(1,len(a)+1)}
c=[]
for i in b:
c.append(s[i])
b=[0]*(len(a)+1)
def f(x):
c=0
while x>0:
c+=b[x]
x-=x&-x
return c
def g(x):
while x<len(a)+1:
b[x]+=1
x+=x&-x
e=0
for i in range(n+1):
e+=f(c[i])
g(c[i])
print(e)
``` | output | 1 | 1,560 | 5 | 3,121 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer sequence of length N, a = {a_1, a_2, …, a_N}, and an integer K.
a has N(N+1)/2 non-empty contiguous subsequences, {a_l, a_{l+1}, …, a_r} (1 ≤ l ≤ r ≤ N). Among them, how many have an arithmetic mean that is greater than or equal to K?
Constraints
* All input values are integers.
* 1 ≤ N ≤ 2 \times 10^5
* 1 ≤ K ≤ 10^9
* 1 ≤ a_i ≤ 10^9
Input
Input is given from Standard Input in the following format:
N K
a_1
a_2
:
a_N
Output
Print the number of the non-empty contiguous subsequences with an arithmetic mean that is greater than or equal to K.
Examples
Input
3 6
7
5
7
Output
5
Input
1 2
1
Output
0
Input
7 26
10
20
30
40
30
20
10
Output
13
Submitted Solution:
```
import sys
input = sys.stdin.readline
n, k = [int(item) for item in input().split()]
a = [0]
for i in range(n):
a.append(int(input()) - k)
for i in range(1, n+1):
a[i] += a[i - 1]
a_set = sorted(list(set(a)))
comp = dict()
for i, item in enumerate(a_set):
comp[item] = i
for i, item in enumerate(a):
a[i] = comp[item]
n = len(a_set) + 5
bit = [0] * (n + 1)
# Add w to ax
def bit_add(x, w):
while x <= n:
bit[x] += w
x += x & -x
# Sum a1 to ax
def bit_sum(x):
ret = 0
while x > 0:
ret += bit[x]
x -= x & -x
return ret
ans = 0
for item in a:
ans += bit_sum(item + 1)
bit_add(item + 1, 1)
print(ans)
``` | instruction | 0 | 1,561 | 5 | 3,122 |
Yes | output | 1 | 1,561 | 5 | 3,123 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer sequence of length N, a = {a_1, a_2, …, a_N}, and an integer K.
a has N(N+1)/2 non-empty contiguous subsequences, {a_l, a_{l+1}, …, a_r} (1 ≤ l ≤ r ≤ N). Among them, how many have an arithmetic mean that is greater than or equal to K?
Constraints
* All input values are integers.
* 1 ≤ N ≤ 2 \times 10^5
* 1 ≤ K ≤ 10^9
* 1 ≤ a_i ≤ 10^9
Input
Input is given from Standard Input in the following format:
N K
a_1
a_2
:
a_N
Output
Print the number of the non-empty contiguous subsequences with an arithmetic mean that is greater than or equal to K.
Examples
Input
3 6
7
5
7
Output
5
Input
1 2
1
Output
0
Input
7 26
10
20
30
40
30
20
10
Output
13
Submitted Solution:
```
import bisect
from itertools import accumulate
# python template for atcoder1
import sys
sys.setrecursionlimit(10**9)
input = sys.stdin.readline
def solve():
N, K = map(int, input().split())
L = [int(input()) for _ in range(N)]
L_acc = [0]+list(accumulate(L))
for i in range(N+1):
L_acc[i] -= (i)*K
L_sort = list(sorted(set(L_acc)))
L_comp = [-1]*(N+1)
for i in range(N+1):
key = L_acc[i]
ind = bisect.bisect_left(L_sort, key)
L_comp[i] = ind
# BIT
bit = [0]*(N+1)
def sum_bit(i):
s = 0
while i > 0:
s += bit[i]
i -= i & (-i)
return s
def add(i, x):
while i <= N:
bit[i] += x
i += i & (-i)
ans = 0
for i, l in enumerate(L_comp):
if l == 0:
ans += N-i
continue
ans += sum_bit(l)
add(l, 1)
print(ans)
solve()
``` | instruction | 0 | 1,562 | 5 | 3,124 |
Yes | output | 1 | 1,562 | 5 | 3,125 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer sequence of length N, a = {a_1, a_2, …, a_N}, and an integer K.
a has N(N+1)/2 non-empty contiguous subsequences, {a_l, a_{l+1}, …, a_r} (1 ≤ l ≤ r ≤ N). Among them, how many have an arithmetic mean that is greater than or equal to K?
Constraints
* All input values are integers.
* 1 ≤ N ≤ 2 \times 10^5
* 1 ≤ K ≤ 10^9
* 1 ≤ a_i ≤ 10^9
Input
Input is given from Standard Input in the following format:
N K
a_1
a_2
:
a_N
Output
Print the number of the non-empty contiguous subsequences with an arithmetic mean that is greater than or equal to K.
Examples
Input
3 6
7
5
7
Output
5
Input
1 2
1
Output
0
Input
7 26
10
20
30
40
30
20
10
Output
13
Submitted Solution:
```
import sys
input = sys.stdin.readline
class BIT1():
"""
Binary Indexed Tree (1-indexed)
"""
def __init__(self, n):
self.n = n
self.bit = [0] * (self.n + 1)
self.data = [0] * (self.n + 1)
def add(self, idx, x):
# add x to idx-th element
# idx: 1-indexed
self.data[idx] += x
while idx <= self.n:
self.bit[idx] += x
idx += (idx & (-idx))
def sum(self, idx):
# get sum of [1, idx]
# idx: 1-indexed
s = 0
while idx:
s += self.bit[idx]
idx -= (idx & (-idx))
return s
n, k = map(int, input().split())
a = [int(input()) for _ in range(n)]
s = [0] * (n + 1)
for i in range(n):
s[i + 1] = s[i] + a[i]
t = [s[i] - k * i for i in range(n + 1)]
# 座標圧縮
zipped = {}
for i, xi in enumerate(sorted(set(t))):
zipped[xi] = i + 1
m = len(zipped)
t2 = [0] * (n + 1)
for i in range(n + 1):
t2[i] = zipped[t[i]]
ans = 0
bit = BIT1(m)
for i in range(n + 1):
ans += bit.sum(t2[i])
bit.add(t2[i], 1)
print(ans)
``` | instruction | 0 | 1,563 | 5 | 3,126 |
Yes | output | 1 | 1,563 | 5 | 3,127 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer sequence of length N, a = {a_1, a_2, …, a_N}, and an integer K.
a has N(N+1)/2 non-empty contiguous subsequences, {a_l, a_{l+1}, …, a_r} (1 ≤ l ≤ r ≤ N). Among them, how many have an arithmetic mean that is greater than or equal to K?
Constraints
* All input values are integers.
* 1 ≤ N ≤ 2 \times 10^5
* 1 ≤ K ≤ 10^9
* 1 ≤ a_i ≤ 10^9
Input
Input is given from Standard Input in the following format:
N K
a_1
a_2
:
a_N
Output
Print the number of the non-empty contiguous subsequences with an arithmetic mean that is greater than or equal to K.
Examples
Input
3 6
7
5
7
Output
5
Input
1 2
1
Output
0
Input
7 26
10
20
30
40
30
20
10
Output
13
Submitted Solution:
```
from collections import defaultdict
n, k = map(int, input().split())
a = [int(input()) - k for _ in range(n)]
s = [0] * (n + 1)
for i in range(n):
s[i + 1] = s[i] + a[i]
mp = defaultdict(int)
b = sorted(list(set(s)))
num = 1
for c in b:
mp[c] = num
num += 1
m = len(mp)
d = [0] * (m + 1)
def add(i):
while i <= m:
d[i] += 1
i += i & -i
def qqq(i):
res = 0
while i > 0:
res += d[i]
i -= i & -i
return res
ans = 0
for i in range(n + 1):
ans += qqq(mp[s[i]])
add(mp[s[i]])
print(ans)
``` | instruction | 0 | 1,564 | 5 | 3,128 |
Yes | output | 1 | 1,564 | 5 | 3,129 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer sequence of length N, a = {a_1, a_2, …, a_N}, and an integer K.
a has N(N+1)/2 non-empty contiguous subsequences, {a_l, a_{l+1}, …, a_r} (1 ≤ l ≤ r ≤ N). Among them, how many have an arithmetic mean that is greater than or equal to K?
Constraints
* All input values are integers.
* 1 ≤ N ≤ 2 \times 10^5
* 1 ≤ K ≤ 10^9
* 1 ≤ a_i ≤ 10^9
Input
Input is given from Standard Input in the following format:
N K
a_1
a_2
:
a_N
Output
Print the number of the non-empty contiguous subsequences with an arithmetic mean that is greater than or equal to K.
Examples
Input
3 6
7
5
7
Output
5
Input
1 2
1
Output
0
Input
7 26
10
20
30
40
30
20
10
Output
13
Submitted Solution:
```
n, k = map(int, input().rstrip("\n").split(" "))
from collections import defaultdict
h = []
total = 0
prev_counter = defaultdict(int)
for i in range(n):
value = int(input())
counter = defaultdict(int)
diff = value - k
if diff >= 0:
total += 1
counter[diff] = 1
for key, count in prev_counter.items():
if (key + diff) >= 0:
total += count
counter[key + diff] += count
prev_counter = counter
print(total )
``` | instruction | 0 | 1,565 | 5 | 3,130 |
No | output | 1 | 1,565 | 5 | 3,131 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer sequence of length N, a = {a_1, a_2, …, a_N}, and an integer K.
a has N(N+1)/2 non-empty contiguous subsequences, {a_l, a_{l+1}, …, a_r} (1 ≤ l ≤ r ≤ N). Among them, how many have an arithmetic mean that is greater than or equal to K?
Constraints
* All input values are integers.
* 1 ≤ N ≤ 2 \times 10^5
* 1 ≤ K ≤ 10^9
* 1 ≤ a_i ≤ 10^9
Input
Input is given from Standard Input in the following format:
N K
a_1
a_2
:
a_N
Output
Print the number of the non-empty contiguous subsequences with an arithmetic mean that is greater than or equal to K.
Examples
Input
3 6
7
5
7
Output
5
Input
1 2
1
Output
0
Input
7 26
10
20
30
40
30
20
10
Output
13
Submitted Solution:
```
import sys,bisect,math
input = sys.stdin.readline
# i が与えられたとき a1~aiまでの和を計算する( O(log n) )
def bitsum(bit,i):
s = 0
while i > 0:
s += bit[i]
i -= i&(-i)
return s
# i と x が与えられたとき ai += x とする ( O(log n) )
def bitadd(bit,i,x):
while i <= n:
bit[i] += x
i += i&(-i)
return bit
n,k = map(int,input().split())
a = [int(input()) for i in range(n)]
#b[j+1]-b[i]:a[i]~a[j]
b = [0]
for i in range(n):
b.append(b[-1]+a[i])
#c[i]:b[i]-k*i
c = [0]*(n+1)
for i in range(n+1):
c[i] = b[i]-k*i
d = sorted(c)
for i in range(n+1):
k = bisect.bisect_left(d,c[i])
d[k] = d[k]
c[i] = k+1
#print(c)
n_ = int(math.log2(n+1)) + 1
bit = [0] * (2**n_+1)
ass = 0
for j in range(n+1):
ass += j-bitsum(bit,c[j])
#bitsum(bit,a[j]): j-1番目までの中でa[j]以下のもの
bit = bitadd(bit,c[j],1)
print(n*(n+1)//2-ass)
``` | instruction | 0 | 1,566 | 5 | 3,132 |
No | output | 1 | 1,566 | 5 | 3,133 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer sequence of length N, a = {a_1, a_2, …, a_N}, and an integer K.
a has N(N+1)/2 non-empty contiguous subsequences, {a_l, a_{l+1}, …, a_r} (1 ≤ l ≤ r ≤ N). Among them, how many have an arithmetic mean that is greater than or equal to K?
Constraints
* All input values are integers.
* 1 ≤ N ≤ 2 \times 10^5
* 1 ≤ K ≤ 10^9
* 1 ≤ a_i ≤ 10^9
Input
Input is given from Standard Input in the following format:
N K
a_1
a_2
:
a_N
Output
Print the number of the non-empty contiguous subsequences with an arithmetic mean that is greater than or equal to K.
Examples
Input
3 6
7
5
7
Output
5
Input
1 2
1
Output
0
Input
7 26
10
20
30
40
30
20
10
Output
13
Submitted Solution:
```
class Bit:
def __init__(self, n):
self.size = n
self.tree = [0] * (n + 1)
def sum(self, i):
s = 0
while i > 0:
s += self.tree[i]
i -= i & -i
return s
def add(self, i, x):
while i <= self.size:
self.tree[i] += x
i += i & -i
N, K = map(int, input().split())
ac = [0] * (N + 1)
eps = 10 ** -6
place = {0: 0}
for i in range(N):
a = int(input()) - K
ac[i+1] = ac[i] + a + eps * (i+1)
place[ac[i+1]] = i+1
ac.sort()
ac_cmpr = [0] * (N+1)
for i in range(N+1):
ac_cmpr[place[ac[i]]] = i + 1
bit = Bit(N+1)
bit.add(ac_cmpr[0], 1)
ans = 0
for i in range(1, N+1):
ans += bit.sum(ac_cmpr[i])
bit.add(ac_cmpr[i], 1)
print(ans)
``` | instruction | 0 | 1,567 | 5 | 3,134 |
No | output | 1 | 1,567 | 5 | 3,135 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer sequence of length N, a = {a_1, a_2, …, a_N}, and an integer K.
a has N(N+1)/2 non-empty contiguous subsequences, {a_l, a_{l+1}, …, a_r} (1 ≤ l ≤ r ≤ N). Among them, how many have an arithmetic mean that is greater than or equal to K?
Constraints
* All input values are integers.
* 1 ≤ N ≤ 2 \times 10^5
* 1 ≤ K ≤ 10^9
* 1 ≤ a_i ≤ 10^9
Input
Input is given from Standard Input in the following format:
N K
a_1
a_2
:
a_N
Output
Print the number of the non-empty contiguous subsequences with an arithmetic mean that is greater than or equal to K.
Examples
Input
3 6
7
5
7
Output
5
Input
1 2
1
Output
0
Input
7 26
10
20
30
40
30
20
10
Output
13
Submitted Solution:
```
import sys
input=sys.stdin.readline
N,K=map(int,input().split())
A=[0]
for i in range(N):
a=int(input())
A.append(a-K)
B=[A[0]]
for i in range(1,N+1):
B.append(B[-1]+A[i])
#print(A)
#print(B)
ans=0
for i in range(1,N+1):
for j in range(i,N+1):
if B[j]-B[i-1]>=0:
ans+=1
print(ans)
``` | instruction | 0 | 1,568 | 5 | 3,136 |
No | output | 1 | 1,568 | 5 | 3,137 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which manipulates a sequence A = {a1, a2, . . . , an} with the following operations:
* add(i, x): add x to ai.
* getSum(s, t): print the sum of as, as+1,...,at.
Note that the initial values of ai (i = 1, 2, . . . , n) are 0.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* If comi is 0, then 1 ≤ xi ≤ n, 0 ≤ yi ≤ 1000.
* If comi is 1, then 1 ≤ xi ≤ n, 1 ≤ yi ≤ n.
Input
n q
com1 x1 y1
com2 x2 y2
...
comq xq yq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, q queries are given where com represents the type of queries. '0' denotes add(xi, yi) and '1' denotes getSum(xi, yi).
Output
For each getSum operation, print the sum in a line.
Example
Input
3 5
0 1 1
0 2 2
0 3 3
1 1 2
1 2 2
Output
3
2 | instruction | 0 | 1,635 | 5 | 3,270 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
"""
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=DSL_2_B&lang=jp
"""
import sys
from sys import stdin
input = stdin.readline
class BIT(object):
def __init__(self, n):
self.n = n
self.bit = [0] * (n + 1)
def sum(self, i):
s = 0
# ?????????????????????????????¢?????????????¶?????????????
while i > 0:
s += self.bit[i]
# i &= (i - 1)
i -= (i & -i)
return s
def add(self, i, x):
while i <= self.n:
self.bit[i] += x
i += (i & -i)
def find(self, s, t):
t_sum = self.sum(t)
s_sum = self.sum(s - 1)
return t_sum - s_sum
def main(args):
n, q = map(int, input().split(' '))
rq = BIT(n+1)
for _ in range(q):
com, x, y = map(int, input().split(' '))
if com == 0:
rq.add(x, y)
elif com == 1:
res = rq.find(x, y)
print(res)
if __name__ == '__main__':
main(sys.argv[1:])
``` | output | 1 | 1,635 | 5 | 3,271 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which manipulates a sequence A = {a1, a2, . . . , an} with the following operations:
* add(i, x): add x to ai.
* getSum(s, t): print the sum of as, as+1,...,at.
Note that the initial values of ai (i = 1, 2, . . . , n) are 0.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* If comi is 0, then 1 ≤ xi ≤ n, 0 ≤ yi ≤ 1000.
* If comi is 1, then 1 ≤ xi ≤ n, 1 ≤ yi ≤ n.
Input
n q
com1 x1 y1
com2 x2 y2
...
comq xq yq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, q queries are given where com represents the type of queries. '0' denotes add(xi, yi) and '1' denotes getSum(xi, yi).
Output
For each getSum operation, print the sum in a line.
Example
Input
3 5
0 1 1
0 2 2
0 3 3
1 1 2
1 2 2
Output
3
2 | instruction | 0 | 1,636 | 5 | 3,272 |
"Correct Solution:
```
#####segfunc#####
def segfunc(x, y):
return x + y
#################
#####ide_ele#####
ide_ele = 0
#################
class SegTree:
"""
init(init_val, ide_ele): 配列init_valで初期化 O(N)
update(k, x): k番目の値をxに更新 O(N)
query(l, r): 区間[l, r)をsegfuncしたものを返す O(logN)
"""
def __init__(self, init_val, segfunc, ide_ele):
"""
init_val: 配列の初期値
segfunc: 区間にしたい操作
ide_ele: 単位元
n: 要素数
num: n以上の最小の2のべき乗
tree: セグメント木(1-index)
"""
n = len(init_val)
self.segfunc = segfunc
self.ide_ele = ide_ele
self.num = 1 << (n - 1).bit_length()
self.tree = [ide_ele] * 2 * self.num
# 配列の値を葉にセット
for i in range(n):
self.tree[self.num + i] = init_val[i]
# 構築していく
for i in range(self.num - 1, 0, -1):
self.tree[i] = segfunc(self.tree[2 * i], self.tree[2 * i + 1])
def update(self, k, x):
"""
k番目の値をxに更新
k: index(0-index)
x: update value
"""
k += self.num
self.tree[k] = x
while k > 1:
self.tree[k >> 1] = segfunc(self.tree[k], self.tree[k ^ 1])
k >>= 1
def query(self, l, r):
"""
[l, r)のsegfuncしたものを得る
l: index(0-index)
r: index(0-index)
"""
res = self.ide_ele
l += self.num
r += self.num
while l < r:
if l & 1:
res = segfunc(res, self.tree[l])
l += 1
if r & 1:
res = segfunc(res, self.tree[r - 1])
l >>= 1
r >>= 1
return res
n, q = [int(x) for x in input().split()]
seg = SegTree([0]*n, segfunc, ide_ele)
for _ in range(q):
com, x, y = [int(x) for x in input().split()]
if com:
print(seg.query(x - 1, y))
else:
x -= 1
y += seg.tree[seg.num + x]
seg.update(x, y)
``` | output | 1 | 1,636 | 5 | 3,273 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which manipulates a sequence A = {a1, a2, . . . , an} with the following operations:
* add(i, x): add x to ai.
* getSum(s, t): print the sum of as, as+1,...,at.
Note that the initial values of ai (i = 1, 2, . . . , n) are 0.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* If comi is 0, then 1 ≤ xi ≤ n, 0 ≤ yi ≤ 1000.
* If comi is 1, then 1 ≤ xi ≤ n, 1 ≤ yi ≤ n.
Input
n q
com1 x1 y1
com2 x2 y2
...
comq xq yq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, q queries are given where com represents the type of queries. '0' denotes add(xi, yi) and '1' denotes getSum(xi, yi).
Output
For each getSum operation, print the sum in a line.
Example
Input
3 5
0 1 1
0 2 2
0 3 3
1 1 2
1 2 2
Output
3
2 | instruction | 0 | 1,637 | 5 | 3,274 |
"Correct Solution:
```
import math
class BinaryIndexedTree:
def __init__(self, size):
self.N = size
self.tree = [0] * (self.N + 1)
def add(self, idx, val):
while idx <= self.N:
self.tree[idx] += val
idx += (idx & -idx)
def sum_to(self, idx):
result = 0
while idx > 0:
result += self.tree[idx]
idx -= (idx & -idx)
return result
n, q = map(int, input().split())
bit = BinaryIndexedTree(n)
for _ in range(q):
com, n1, n2 = map(int, input().split())
if com == 0:
bit.add(n1, n2)
else:
print(bit.sum_to(n2) - bit.sum_to(n1-1))
``` | output | 1 | 1,637 | 5 | 3,275 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which manipulates a sequence A = {a1, a2, . . . , an} with the following operations:
* add(i, x): add x to ai.
* getSum(s, t): print the sum of as, as+1,...,at.
Note that the initial values of ai (i = 1, 2, . . . , n) are 0.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* If comi is 0, then 1 ≤ xi ≤ n, 0 ≤ yi ≤ 1000.
* If comi is 1, then 1 ≤ xi ≤ n, 1 ≤ yi ≤ n.
Input
n q
com1 x1 y1
com2 x2 y2
...
comq xq yq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, q queries are given where com represents the type of queries. '0' denotes add(xi, yi) and '1' denotes getSum(xi, yi).
Output
For each getSum operation, print the sum in a line.
Example
Input
3 5
0 1 1
0 2 2
0 3 3
1 1 2
1 2 2
Output
3
2 | instruction | 0 | 1,638 | 5 | 3,276 |
"Correct Solution:
```
# operator.add(a, b)
# operator.mul(a, b)
"""
https://komiyam.hatenadiary.org/entry/20131202/1385992406
大雑把に言って、平衡二分探索木からinsert,delete,split,mergeなどができないよう制限したのがsegment treeで、
segment treeの区間[L,R)に対するクエリをL=0に制限したのがbinary indexed treeだと見なすことができます。
"""
class SegmentTree():
"""
update, get を提供するSegmentTree
Attributes
----------
__n : int
葉の数。2 ^ i - 1
__dot :
Segment function
__e: int
単位元
__node: list
Segment Tree
"""
def __init__(self, A, dot, e):
"""
Parameters
----------
A : list
対象の配列
dot :
Segment function
e : int
単位元
"""
n = 2 ** (len(A) - 1).bit_length()
self.__n = n
self.__dot = dot
self.__e = e
self.__node = [e] * (2 * n)
for i in range(len(A)):
self.__node[i + n] = A[i]
for i in range(n - 1, 0, -1):
self.__node[i] = self.__dot(self.__node[2 * i], self.__node[2 * i + 1])
def update(self, i, c):
i += self.__n
node = self.__node
node[i] = c
while i > 1:
i //= 2
node[i] = self.__dot(node[2 * i], node[2 * i + 1])
def get(self, l, r):
vl, vr = self.__e, self.__e
l += self.__n
r += self.__n
while (l < r):
if l & 1:
vl = self.__dot(vl, self.__node[l])
l += 1
l //= 2
if r & 1:
r -= 1
vr = self.__dot(vr, self.__node[r])
r //= 2
return self.__dot(vl, vr)
# http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=DSL_2_A
def AOJ_DSL_2_A():
n, q = map(int, input().split())
e = (1 << 31) - 1
A = [e] * n
seg = SegmentTree(A, min, e)
for i in range(q):
t, x, y = map(int, input().split())
if t == 0:
seg.update(x, y)
else:
print(seg.get(x, y + 1))
# http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=DSL_2_B
def AOJ_DSL_2_B():
from operator import add
import sys
n, q = map(int, input().split())
e = 0
A = [e] * n
seg = SegmentTree(A, add, e)
query = sys.stdin.readlines()
for i in range(q):
t, x, y = map(int, query[i].split())
x -= 1
if t == 0:
A[x] += y
seg.update(x, A[x])
else:
y -= 1
print(seg.get(x, y + 1))
def ABC125_C():
from fractions import gcd
N = int(input())
A = list(map(int, input().split()))
seg = SegmentTree(A, gcd, 0)
ans = 0
for i in range(N):
ans = max(ans, gcd(seg.get(0, i), seg.get(i + 1, N)))
print(ans)
# AOJ_DSL_2_A()
AOJ_DSL_2_B()
# ABC125_C()
``` | output | 1 | 1,638 | 5 | 3,277 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which manipulates a sequence A = {a1, a2, . . . , an} with the following operations:
* add(i, x): add x to ai.
* getSum(s, t): print the sum of as, as+1,...,at.
Note that the initial values of ai (i = 1, 2, . . . , n) are 0.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* If comi is 0, then 1 ≤ xi ≤ n, 0 ≤ yi ≤ 1000.
* If comi is 1, then 1 ≤ xi ≤ n, 1 ≤ yi ≤ n.
Input
n q
com1 x1 y1
com2 x2 y2
...
comq xq yq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, q queries are given where com represents the type of queries. '0' denotes add(xi, yi) and '1' denotes getSum(xi, yi).
Output
For each getSum operation, print the sum in a line.
Example
Input
3 5
0 1 1
0 2 2
0 3 3
1 1 2
1 2 2
Output
3
2 | instruction | 0 | 1,639 | 5 | 3,278 |
"Correct Solution:
```
import sys
input = sys.stdin.readline
class SegmentTree:
def __init__(self, num, ide_ele=0):
tmp = 1
while tmp < num:
tmp *= 2
self.n = tmp
self.ide_ele = ide_ele
self.l = [ide_ele] * (2 * self.n - 1)
def func(self, x, y):
return x + y
# return min(x, y)
def update(self, i, val):
i = self.n - 1 + i
self.l[i] += val
while i > 0:
i = (i-1)//2
self.l[i] = self.func(self.l[i*2+1], self.l[i*2+2])
def query(self, a, b, k, l, r):
# print(a, b, k, l, r)
if r <= a or b <= l:
return self.ide_ele
if a <= l and r <= b:
return self.l[k]
vl = self.query(a, b, k*2+1, l, (l+r)//2)
vr = self.query(a, b, k*2+2, (l+r)//2, r)
return self.func(vl, vr)
def main():
N, Q = map(int, input().split())
segt = SegmentTree(N)
for _ in range(Q):
c, x, y = map(int, input().split())
if c == 0:
segt.update(x-1, y)
else:
print(segt.query(x-1, y, 0, 0, segt.n))
# print(segt.l)
if __name__ == "__main__":
main()
``` | output | 1 | 1,639 | 5 | 3,279 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which manipulates a sequence A = {a1, a2, . . . , an} with the following operations:
* add(i, x): add x to ai.
* getSum(s, t): print the sum of as, as+1,...,at.
Note that the initial values of ai (i = 1, 2, . . . , n) are 0.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* If comi is 0, then 1 ≤ xi ≤ n, 0 ≤ yi ≤ 1000.
* If comi is 1, then 1 ≤ xi ≤ n, 1 ≤ yi ≤ n.
Input
n q
com1 x1 y1
com2 x2 y2
...
comq xq yq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, q queries are given where com represents the type of queries. '0' denotes add(xi, yi) and '1' denotes getSum(xi, yi).
Output
For each getSum operation, print the sum in a line.
Example
Input
3 5
0 1 1
0 2 2
0 3 3
1 1 2
1 2 2
Output
3
2 | instruction | 0 | 1,640 | 5 | 3,280 |
"Correct Solution:
```
class SegmentTree:
'''
SegmentTree(arr, e)
arr : 配列,
e : 単位元\n
segment_tree[i] = x で更新\n
segment_tree[a,b] で区間[a,b)の計算結果を得る(a<bでなければ単位元が返ってくる)
'''
def __init__(self, arr, e):
self.e = e
self.len = len(arr)
# 最下層の長さnの初期化(扱う配列の長さ以上の2冪のうち最も小さいもの)
n = 2 ** (len(bin(self.len + 1)) - 2)
self.n = n
# nodeの初期化
node = [e] * (2 * n)
node[n: n + self.len] = arr
for i in range(n - 1, 0, -1):
node[i] = node[2 * i] + node[2 * i + 1]
self.node = node
def __setitem__(self, i, val):
# 最下層からはじめる。それより上にはn-1個のノードがある
i += self.n
self.node[i] = val
# 最上層に辿り着くまで
while i > 1:
# 親に移動
i //= 2
# 2つの子に対する計算結果
self.node[i] = self.node[2 * i] + self.node[2 * i + 1]
def __getitem__(self, i):
return self.node[self.n + i]
def query(self, a, b):
a = self.trim(a, self.len)
b = self.trim(b, self.len)
# [a,b)が欲しい
# k=0, [l=0,r=n)から始める
# [l,r)が[a,b)の中に入るまで[l,mid), [mid,r)に分割して再帰
def get(k, l, r):
# そもそも[l,r)が[a,b)に入ってない時は計算に影響を与えない単位元を返す
if r <= a or b <= l:
return self.e
if a <= l and r <= b:
return self.node[k]
m = (l + r) // 2
return get(2 * k, l, m) + get(2 * k + 1, m, r)
return get(1, 0, self.n)
def getlist(self):
return self.node[self.n:]
def trim(self, i, n):
i = max(i, -n)
i = min(i, n)
if i < 0:
i += n
return i
# https://onlinejudge.u-aizu.ac.jp/courses/library/3/DSL/2/DSL_2_B
n, q = map(int, input().split())
A = [0] * n
st = SegmentTree(A, 0)
for i in range(q):
com, x, y = map(int, input().split())
x -= 1
if com == 0:
st[x] += y
else:
y -= 1
# print(st[x: y + 1])
print(st.query(x, y + 1))
``` | output | 1 | 1,640 | 5 | 3,281 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which manipulates a sequence A = {a1, a2, . . . , an} with the following operations:
* add(i, x): add x to ai.
* getSum(s, t): print the sum of as, as+1,...,at.
Note that the initial values of ai (i = 1, 2, . . . , n) are 0.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* If comi is 0, then 1 ≤ xi ≤ n, 0 ≤ yi ≤ 1000.
* If comi is 1, then 1 ≤ xi ≤ n, 1 ≤ yi ≤ n.
Input
n q
com1 x1 y1
com2 x2 y2
...
comq xq yq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, q queries are given where com represents the type of queries. '0' denotes add(xi, yi) and '1' denotes getSum(xi, yi).
Output
For each getSum operation, print the sum in a line.
Example
Input
3 5
0 1 1
0 2 2
0 3 3
1 1 2
1 2 2
Output
3
2 | instruction | 0 | 1,641 | 5 | 3,282 |
"Correct Solution:
```
# Range Sum Query
# https://onlinejudge.u-aizu.ac.jp/problems/DSL_2_B
"""
Binary Indexed Tree (Fenwick Tree)
References:
http://hos.ac/slides/20140319_bit.pdf
Verified:
https://judge.yosupo.jp/problem/point_add_range_sum
https://atcoder.jp/contests/jsc2019-qual/tasks/jsc2019_qual_b
"""
from typing import Optional, Sequence
class BinaryIndexedTree:
__slots__ = ["_size", "_tree", "_is_zero_indexed"]
def __init__(self, size: int, initial_values: Optional[Sequence[int]] = None, is_zero_indexed: bool = True):
self._size = size
self._tree = [0] * (self._size + 1)
self._is_zero_indexed = is_zero_indexed
if initial_values:
self._build(initial_values)
def _build(self, initial_values: Sequence[int]):
for i, a in enumerate(initial_values):
self.add(i, a)
def add(self, index: int, value: int) -> None:
"""Add value to tree[index], O(logN)."""
if self._is_zero_indexed:
index += 1
while index <= self._size:
self._tree[index] += value
index += index & -index
def sum(self, index: int) -> int:
"""Return the sum of [1, index], O(logN)."""
ret = 0
while index > 0:
ret += self._tree[index]
index -= index & -index
return ret
def range_sum(self, left: int, right: int) -> int:
"""Return the range sum of [left, right], O(logN)."""
if not self._is_zero_indexed:
left -= 1
return self.sum(right) - self.sum(left)
def main():
N, Q, *queries = map(int, open(0).read().split())
tree = BinaryIndexedTree(N, is_zero_indexed=False)
result = []
for com, x, y in zip(*[iter(queries)] * 3):
if com:
result.append(tree.range_sum(x, y))
else:
tree.add(x, y)
print(*result, sep="\n")
if __name__ == "__main__":
main()
``` | output | 1 | 1,641 | 5 | 3,283 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which manipulates a sequence A = {a1, a2, . . . , an} with the following operations:
* add(i, x): add x to ai.
* getSum(s, t): print the sum of as, as+1,...,at.
Note that the initial values of ai (i = 1, 2, . . . , n) are 0.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* If comi is 0, then 1 ≤ xi ≤ n, 0 ≤ yi ≤ 1000.
* If comi is 1, then 1 ≤ xi ≤ n, 1 ≤ yi ≤ n.
Input
n q
com1 x1 y1
com2 x2 y2
...
comq xq yq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, q queries are given where com represents the type of queries. '0' denotes add(xi, yi) and '1' denotes getSum(xi, yi).
Output
For each getSum operation, print the sum in a line.
Example
Input
3 5
0 1 1
0 2 2
0 3 3
1 1 2
1 2 2
Output
3
2
Submitted Solution:
```
class RSQ:
def __init__(self,n_):
self.n_ = n_
self.n = 1
while self.n < n_: self.n*=2
self.st = [0]*(2*self.n-1)
def update(self,k,x):
k+=(self.n-1)
self.st[k] += x
while k >0:
k = (k-1)//2
self.st[k] = self.st[2*k+1]+self.st[2*k+2]
def search(self,a,b,k,l,r):
if r<=a or b<=l :return 0
if a<=l and r<=b:return self.st[k]
L = self.search(a,b,k*2+1,l,(l+r)//2)
R = self.search(a,b,k*2+2,(l+r)//2,r)
return L+R
def query(self,a,b):
return self.search(a,b,0,0,self.n)
def main():
n,q = map(int,input().split())
ST = RSQ(n)
for _ in range(q):
a,b,c = map(int,input().split())
if a == 0:ST.update(b-1,c)
else :print (ST.query(b-1,c))
if __name__ == '__main__':
main()
``` | instruction | 0 | 1,643 | 5 | 3,286 |
Yes | output | 1 | 1,643 | 5 | 3,287 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which manipulates a sequence A = {a1, a2, . . . , an} with the following operations:
* add(i, x): add x to ai.
* getSum(s, t): print the sum of as, as+1,...,at.
Note that the initial values of ai (i = 1, 2, . . . , n) are 0.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* If comi is 0, then 1 ≤ xi ≤ n, 0 ≤ yi ≤ 1000.
* If comi is 1, then 1 ≤ xi ≤ n, 1 ≤ yi ≤ n.
Input
n q
com1 x1 y1
com2 x2 y2
...
comq xq yq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, q queries are given where com represents the type of queries. '0' denotes add(xi, yi) and '1' denotes getSum(xi, yi).
Output
For each getSum operation, print the sum in a line.
Example
Input
3 5
0 1 1
0 2 2
0 3 3
1 1 2
1 2 2
Output
3
2
Submitted Solution:
```
class SegmentTree:
def __init__(self, N_, element, function_=(lambda l, r: l + r)):
"""
:param N_: セグ木のサイズ
:param element: 単位元
:param function: 子の要素に対して行われる計算,デフォルトだと加算
"""
self.N = 1
while self.N < N_:
self.N *= 2
self.element = element
self.dat = [element for _ in range(self.N * 2)]
self.function = function_
def update(self, k, a):
"""k番目の値(0-indexed)をaに変更"""
k += self.N - 1
self.dat[k] = a
while k > 0:
k = (k - 1) // 2
self.dat[k] = self.function(self.dat[2 * k + 1], self.dat[2 * k + 2])
def query(self, begin, end, k=None, l=None, r=None):
if k is None: return self.query(begin, end, 0, 0, self.N)
if r <= begin or end <= l: return self.element
if begin <= l and r <= end: return self.dat[k]
vl = self.query(begin, end, k * 2 + 1, l, (l + r) // 2)
vr = self.query(begin, end, k * 2 + 2, (l + r) // 2, r)
return self.function(vl, vr)
n, q = map(int, input().split())
n += 1
treeone = SegmentTree(n, 0, (lambda l, r: l + r))
while q > 0:
q -= 1
tmp = list(map(int, input().split()))
if tmp[0] is 0:
treeone.update(tmp[1],treeone.query(tmp[1], tmp[1] + 1) + tmp[2])
else:
print(treeone.query(tmp[1], tmp[2] + 1))
``` | instruction | 0 | 1,644 | 5 | 3,288 |
Yes | output | 1 | 1,644 | 5 | 3,289 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which manipulates a sequence A = {a1, a2, . . . , an} with the following operations:
* add(i, x): add x to ai.
* getSum(s, t): print the sum of as, as+1,...,at.
Note that the initial values of ai (i = 1, 2, . . . , n) are 0.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* If comi is 0, then 1 ≤ xi ≤ n, 0 ≤ yi ≤ 1000.
* If comi is 1, then 1 ≤ xi ≤ n, 1 ≤ yi ≤ n.
Input
n q
com1 x1 y1
com2 x2 y2
...
comq xq yq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, q queries are given where com represents the type of queries. '0' denotes add(xi, yi) and '1' denotes getSum(xi, yi).
Output
For each getSum operation, print the sum in a line.
Example
Input
3 5
0 1 1
0 2 2
0 3 3
1 1 2
1 2 2
Output
3
2
Submitted Solution:
```
def _max(a,b):
return max(a,b)
def _min(a,b):
return min(a,b)
def _sum(a,b):
return a+b
class SegmentTree:
def __init__(self, a):
sz = len(a)
self.n = int(1)
while self.n < sz:
self.n = int(self.n*2)
self.func = _sum
self.init = 0
self.val = [0]*(self.n*2 - 1)
self.val[self.n-1:self.n-1+sz] = a
for i in range(self.n-2, -1, -1):
self.val[i] = self.func(self.val[2*i+1], self.val[2*i+2])
def update(self, pos, x):
# 0-indexed
k = int(pos + self.n-1)
self.val[k] += x
while k > 0:
k = int(k-1) // 2
self.val[k] = self.func(self.val[2*k+1], self.val[2*k+2])
def query(self, a, b, k=0, l=0, r=-1):
# 0-indexed
# [a,b)に対してquery
if r < 0:
r = self.n
if r <= a or b <= l:
return self.init
if a <= l and r <= b:
return self.val[k]
lv = self.query(a, b, 2*k+1, l, (l+r)//2)
rv = self.query(a, b, 2*k+2, (l+r)//2, r)
return self.func(lv, rv)
def get(self, pos):
# 0-indexed
return self.val[pos + self.n-1]
n,q = map(int, input().split())
a = [0]*n
st = SegmentTree(a)
for _ in range(q):
t,x,y = map(int, input().split())
x -= 1
if t == 0:
st.update(x,y)
else:
ans = st.query(x,y)
print(int(ans))
``` | instruction | 0 | 1,645 | 5 | 3,290 |
Yes | output | 1 | 1,645 | 5 | 3,291 |
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