message stringlengths 2 44.5k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 276 109k | cluster float64 23 23 | __index_level_0__ int64 552 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n sheets of square paper of the same size. Align the bottom of these papers horizontally and arrange them in several rows. However, adjacent rows must be arranged so that the left side is not lower than the right side. For example, n When = 5, the following 7 ways of arranging are possible.
<image>
We will represent these by the number of square columns in each column. For example, when n = 5, each of them will be represented.
(5) (4, 1) (3, 2) (3, 1, 1) (2, 2, 1) (2, 1, 1, 1) (1, 1, 1, 1, 1)
It is expressed as.
When n is input, create a program that outputs all in lexicographic order. n β€ 30. However, lexicographic order means that two arrangements (a1, a2, ..., as) are arranged (b1, For b2, ..., bt), when a1> b1 or an integer i> 1 exists and a1 = b1, ..., ai-1 = bi-1 and ai> bi holds (a1) , a2, ..., as) are arranged so that they are output before (b1, b2, ..., bt).
The input data consists of one line, with n written on the first line.
In the output, write one line in lexicographic order and insert a line break at the end. The output of (a1, a2, ..., as) is the integer a1, a2, ..., as. Output in this order separated by blanks.
Input example 1
---
Five
Output example 1
Five
4 1
3 2
3 1 1
2 2 1
2 1 1 1
1 1 1 1 1
input
The input consists of multiple datasets. Input ends when n is 0. The number of datasets does not exceed 5.
output
All data sets are output in lexicographic order.
Example
Input
5
5
0
Output
5
4 1
3 2
3 1 1
2 2 1
2 1 1 1
1 1 1 1 1
5
4 1
3 2
3 1 1
2 2 1
2 1 1 1
1 1 1 1 1
Submitted Solution:
```
def generate_square(rest, output=[10000]):
for i in range(rest, 0, -1):
if output[-1] >= i:
yield from generate_square(rest - i, output + [i])
if rest == 0:
yield output
while 1:
x = int(input().strip())
if x == 0:
break
generator = generate_square(x)
for x in generator:
print(" ".join(map(str, x[1:])))
GN-2:python kodairatomonori$
``` | instruction | 0 | 97,771 | 23 | 195,542 |
No | output | 1 | 97,771 | 23 | 195,543 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n sheets of square paper of the same size. Align the bottom of these papers horizontally and arrange them in several rows. However, adjacent rows must be arranged so that the left side is not lower than the right side. For example, n When = 5, the following 7 ways of arranging are possible.
<image>
We will represent these by the number of square columns in each column. For example, when n = 5, each of them will be represented.
(5) (4, 1) (3, 2) (3, 1, 1) (2, 2, 1) (2, 1, 1, 1) (1, 1, 1, 1, 1)
It is expressed as.
When n is input, create a program that outputs all in lexicographic order. n β€ 30. However, lexicographic order means that two arrangements (a1, a2, ..., as) are arranged (b1, For b2, ..., bt), when a1> b1 or an integer i> 1 exists and a1 = b1, ..., ai-1 = bi-1 and ai> bi holds (a1) , a2, ..., as) are arranged so that they are output before (b1, b2, ..., bt).
The input data consists of one line, with n written on the first line.
In the output, write one line in lexicographic order and insert a line break at the end. The output of (a1, a2, ..., as) is the integer a1, a2, ..., as. Output in this order separated by blanks.
Input example 1
---
Five
Output example 1
Five
4 1
3 2
3 1 1
2 2 1
2 1 1 1
1 1 1 1 1
input
The input consists of multiple datasets. Input ends when n is 0. The number of datasets does not exceed 5.
output
All data sets are output in lexicographic order.
Example
Input
5
5
0
Output
5
4 1
3 2
3 1 1
2 2 1
2 1 1 1
1 1 1 1 1
5
4 1
3 2
3 1 1
2 2 1
2 1 1 1
1 1 1 1 1
Submitted Solution:
```
def main():
pass
if __name__ == "__main__":
main()
``` | instruction | 0 | 97,772 | 23 | 195,544 |
No | output | 1 | 97,772 | 23 | 195,545 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n sheets of square paper of the same size. Align the bottom of these papers horizontally and arrange them in several rows. However, adjacent rows must be arranged so that the left side is not lower than the right side. For example, n When = 5, the following 7 ways of arranging are possible.
<image>
We will represent these by the number of square columns in each column. For example, when n = 5, each of them will be represented.
(5) (4, 1) (3, 2) (3, 1, 1) (2, 2, 1) (2, 1, 1, 1) (1, 1, 1, 1, 1)
It is expressed as.
When n is input, create a program that outputs all in lexicographic order. n β€ 30. However, lexicographic order means that two arrangements (a1, a2, ..., as) are arranged (b1, For b2, ..., bt), when a1> b1 or an integer i> 1 exists and a1 = b1, ..., ai-1 = bi-1 and ai> bi holds (a1) , a2, ..., as) are arranged so that they are output before (b1, b2, ..., bt).
The input data consists of one line, with n written on the first line.
In the output, write one line in lexicographic order and insert a line break at the end. The output of (a1, a2, ..., as) is the integer a1, a2, ..., as. Output in this order separated by blanks.
Input example 1
---
Five
Output example 1
Five
4 1
3 2
3 1 1
2 2 1
2 1 1 1
1 1 1 1 1
input
The input consists of multiple datasets. Input ends when n is 0. The number of datasets does not exceed 5.
output
All data sets are output in lexicographic order.
Example
Input
5
5
0
Output
5
4 1
3 2
3 1 1
2 2 1
2 1 1 1
1 1 1 1 1
5
4 1
3 2
3 1 1
2 2 1
2 1 1 1
1 1 1 1 1
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0507
"""
import sys
from sys import stdin
input = stdin.readline
def solve(n):
print(*n)
if max(n) == 1:
return
l = len(n)
for i in range(l-1, -1, -1):
if n[i] != 1:
n[i] -= 1
j = i + 1
while True:
if j >= l:
n.append(1)
break
if n[i] >= n[j]+1:
n[j] += 1
break
j += 1
break
solve(n)
def main(args):
while True:
n = int(input())
if n == 0:
break
solve([n])
if __name__ == '__main__':
main(sys.argv[1:])
``` | instruction | 0 | 97,773 | 23 | 195,546 |
No | output | 1 | 97,773 | 23 | 195,547 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n sheets of square paper of the same size. Align the bottom of these papers horizontally and arrange them in several rows. However, adjacent rows must be arranged so that the left side is not lower than the right side. For example, n When = 5, the following 7 ways of arranging are possible.
<image>
We will represent these by the number of square columns in each column. For example, when n = 5, each of them will be represented.
(5) (4, 1) (3, 2) (3, 1, 1) (2, 2, 1) (2, 1, 1, 1) (1, 1, 1, 1, 1)
It is expressed as.
When n is input, create a program that outputs all in lexicographic order. n β€ 30. However, lexicographic order means that two arrangements (a1, a2, ..., as) are arranged (b1, For b2, ..., bt), when a1> b1 or an integer i> 1 exists and a1 = b1, ..., ai-1 = bi-1 and ai> bi holds (a1) , a2, ..., as) are arranged so that they are output before (b1, b2, ..., bt).
The input data consists of one line, with n written on the first line.
In the output, write one line in lexicographic order and insert a line break at the end. The output of (a1, a2, ..., as) is the integer a1, a2, ..., as. Output in this order separated by blanks.
Input example 1
---
Five
Output example 1
Five
4 1
3 2
3 1 1
2 2 1
2 1 1 1
1 1 1 1 1
input
The input consists of multiple datasets. Input ends when n is 0. The number of datasets does not exceed 5.
output
All data sets are output in lexicographic order.
Example
Input
5
5
0
Output
5
4 1
3 2
3 1 1
2 2 1
2 1 1 1
1 1 1 1 1
5
4 1
3 2
3 1 1
2 2 1
2 1 1 1
1 1 1 1 1
Submitted Solution:
```
def f(n,m,l):
if n == 0:
s = ''
for i in l:
s = s + str(i) + ' '
s.rstrip()
print(s)
return
else:
for i in range(1,min(n,m)+1):
l.append(min(n,m)+1-i)
f(n-min(n,m)-1+i,min(n,m)+1-i,l)
l.pop()
while True:
n = int(input())
if n == 0:
break
else:
f(n,30,[])
``` | instruction | 0 | 97,774 | 23 | 195,548 |
No | output | 1 | 97,774 | 23 | 195,549 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a regular polygon with n vertices labeled from 1 to n in counter-clockwise order. The triangulation of a given polygon is a set of triangles such that each vertex of each triangle is a vertex of the initial polygon, there is no pair of triangles such that their intersection has non-zero area, and the total area of all triangles is equal to the area of the given polygon. The weight of a triangulation is the sum of weigths of triangles it consists of, where the weight of a triagle is denoted as the product of labels of its vertices.
Calculate the minimum weight among all triangulations of the polygon.
Input
The first line contains single integer n (3 β€ n β€ 500) β the number of vertices in the regular polygon.
Output
Print one integer β the minimum weight among all triangulations of the given polygon.
Examples
Input
3
Output
6
Input
4
Output
18
Note
According to Wiki: polygon triangulation is the decomposition of a polygonal area (simple polygon) P into a set of triangles, i. e., finding a set of triangles with pairwise non-intersecting interiors whose union is P.
In the first example the polygon is a triangle, so we don't need to cut it further, so the answer is 1 β
2 β
3 = 6.
In the second example the polygon is a rectangle, so it should be divided into two triangles. It's optimal to cut it using diagonal 1-3 so answer is 1 β
2 β
3 + 1 β
3 β
4 = 6 + 12 = 18. | instruction | 0 | 97,874 | 23 | 195,748 |
Tags: dp, greedy, math
Correct Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
import time
n = int(input())
start = time.time()
ans = 0
for i in range(2, n):
ans += i*(i+1)
print(ans)
finish = time.time()
#print(finish - start)
``` | output | 1 | 97,874 | 23 | 195,749 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a regular polygon with n vertices labeled from 1 to n in counter-clockwise order. The triangulation of a given polygon is a set of triangles such that each vertex of each triangle is a vertex of the initial polygon, there is no pair of triangles such that their intersection has non-zero area, and the total area of all triangles is equal to the area of the given polygon. The weight of a triangulation is the sum of weigths of triangles it consists of, where the weight of a triagle is denoted as the product of labels of its vertices.
Calculate the minimum weight among all triangulations of the polygon.
Input
The first line contains single integer n (3 β€ n β€ 500) β the number of vertices in the regular polygon.
Output
Print one integer β the minimum weight among all triangulations of the given polygon.
Examples
Input
3
Output
6
Input
4
Output
18
Note
According to Wiki: polygon triangulation is the decomposition of a polygonal area (simple polygon) P into a set of triangles, i. e., finding a set of triangles with pairwise non-intersecting interiors whose union is P.
In the first example the polygon is a triangle, so we don't need to cut it further, so the answer is 1 β
2 β
3 = 6.
In the second example the polygon is a rectangle, so it should be divided into two triangles. It's optimal to cut it using diagonal 1-3 so answer is 1 β
2 β
3 + 1 β
3 β
4 = 6 + 12 = 18. | instruction | 0 | 97,875 | 23 | 195,750 |
Tags: dp, greedy, math
Correct Solution:
```
v = int(input())
ans,a,b = 6,3,4
while(b<=v):
ans += (a*b)
a+=1
b+=1
print(ans)
``` | output | 1 | 97,875 | 23 | 195,751 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a regular polygon with n vertices labeled from 1 to n in counter-clockwise order. The triangulation of a given polygon is a set of triangles such that each vertex of each triangle is a vertex of the initial polygon, there is no pair of triangles such that their intersection has non-zero area, and the total area of all triangles is equal to the area of the given polygon. The weight of a triangulation is the sum of weigths of triangles it consists of, where the weight of a triagle is denoted as the product of labels of its vertices.
Calculate the minimum weight among all triangulations of the polygon.
Input
The first line contains single integer n (3 β€ n β€ 500) β the number of vertices in the regular polygon.
Output
Print one integer β the minimum weight among all triangulations of the given polygon.
Examples
Input
3
Output
6
Input
4
Output
18
Note
According to Wiki: polygon triangulation is the decomposition of a polygonal area (simple polygon) P into a set of triangles, i. e., finding a set of triangles with pairwise non-intersecting interiors whose union is P.
In the first example the polygon is a triangle, so we don't need to cut it further, so the answer is 1 β
2 β
3 = 6.
In the second example the polygon is a rectangle, so it should be divided into two triangles. It's optimal to cut it using diagonal 1-3 so answer is 1 β
2 β
3 + 1 β
3 β
4 = 6 + 12 = 18. | instruction | 0 | 97,876 | 23 | 195,752 |
Tags: dp, greedy, math
Correct Solution:
```
a = int(input())
ans = 0
for i in range(a - 2):
ans += (i + 2) * (i + 3)
print(ans)
``` | output | 1 | 97,876 | 23 | 195,753 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a regular polygon with n vertices labeled from 1 to n in counter-clockwise order. The triangulation of a given polygon is a set of triangles such that each vertex of each triangle is a vertex of the initial polygon, there is no pair of triangles such that their intersection has non-zero area, and the total area of all triangles is equal to the area of the given polygon. The weight of a triangulation is the sum of weigths of triangles it consists of, where the weight of a triagle is denoted as the product of labels of its vertices.
Calculate the minimum weight among all triangulations of the polygon.
Input
The first line contains single integer n (3 β€ n β€ 500) β the number of vertices in the regular polygon.
Output
Print one integer β the minimum weight among all triangulations of the given polygon.
Examples
Input
3
Output
6
Input
4
Output
18
Note
According to Wiki: polygon triangulation is the decomposition of a polygonal area (simple polygon) P into a set of triangles, i. e., finding a set of triangles with pairwise non-intersecting interiors whose union is P.
In the first example the polygon is a triangle, so we don't need to cut it further, so the answer is 1 β
2 β
3 = 6.
In the second example the polygon is a rectangle, so it should be divided into two triangles. It's optimal to cut it using diagonal 1-3 so answer is 1 β
2 β
3 + 1 β
3 β
4 = 6 + 12 = 18. | instruction | 0 | 97,877 | 23 | 195,754 |
Tags: dp, greedy, math
Correct Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
n = int(input())
s = 0
for i in range (3,n+1):
s += i * (i-1)
print(s)
``` | output | 1 | 97,877 | 23 | 195,755 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a regular polygon with n vertices labeled from 1 to n in counter-clockwise order. The triangulation of a given polygon is a set of triangles such that each vertex of each triangle is a vertex of the initial polygon, there is no pair of triangles such that their intersection has non-zero area, and the total area of all triangles is equal to the area of the given polygon. The weight of a triangulation is the sum of weigths of triangles it consists of, where the weight of a triagle is denoted as the product of labels of its vertices.
Calculate the minimum weight among all triangulations of the polygon.
Input
The first line contains single integer n (3 β€ n β€ 500) β the number of vertices in the regular polygon.
Output
Print one integer β the minimum weight among all triangulations of the given polygon.
Examples
Input
3
Output
6
Input
4
Output
18
Note
According to Wiki: polygon triangulation is the decomposition of a polygonal area (simple polygon) P into a set of triangles, i. e., finding a set of triangles with pairwise non-intersecting interiors whose union is P.
In the first example the polygon is a triangle, so we don't need to cut it further, so the answer is 1 β
2 β
3 = 6.
In the second example the polygon is a rectangle, so it should be divided into two triangles. It's optimal to cut it using diagonal 1-3 so answer is 1 β
2 β
3 + 1 β
3 β
4 = 6 + 12 = 18. | instruction | 0 | 97,878 | 23 | 195,756 |
Tags: dp, greedy, math
Correct Solution:
```
n =int(input())
if n >= 3 and n <= 500 :
print((((n*n*n)-n)//3)-2)
``` | output | 1 | 97,878 | 23 | 195,757 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a regular polygon with n vertices labeled from 1 to n in counter-clockwise order. The triangulation of a given polygon is a set of triangles such that each vertex of each triangle is a vertex of the initial polygon, there is no pair of triangles such that their intersection has non-zero area, and the total area of all triangles is equal to the area of the given polygon. The weight of a triangulation is the sum of weigths of triangles it consists of, where the weight of a triagle is denoted as the product of labels of its vertices.
Calculate the minimum weight among all triangulations of the polygon.
Input
The first line contains single integer n (3 β€ n β€ 500) β the number of vertices in the regular polygon.
Output
Print one integer β the minimum weight among all triangulations of the given polygon.
Examples
Input
3
Output
6
Input
4
Output
18
Note
According to Wiki: polygon triangulation is the decomposition of a polygonal area (simple polygon) P into a set of triangles, i. e., finding a set of triangles with pairwise non-intersecting interiors whose union is P.
In the first example the polygon is a triangle, so we don't need to cut it further, so the answer is 1 β
2 β
3 = 6.
In the second example the polygon is a rectangle, so it should be divided into two triangles. It's optimal to cut it using diagonal 1-3 so answer is 1 β
2 β
3 + 1 β
3 β
4 = 6 + 12 = 18. | instruction | 0 | 97,879 | 23 | 195,758 |
Tags: dp, greedy, math
Correct Solution:
```
n = int(input())
i = 2
s = 0
while i < n:
s += i * (i+1)
i += 1
print(s)
``` | output | 1 | 97,879 | 23 | 195,759 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a regular polygon with n vertices labeled from 1 to n in counter-clockwise order. The triangulation of a given polygon is a set of triangles such that each vertex of each triangle is a vertex of the initial polygon, there is no pair of triangles such that their intersection has non-zero area, and the total area of all triangles is equal to the area of the given polygon. The weight of a triangulation is the sum of weigths of triangles it consists of, where the weight of a triagle is denoted as the product of labels of its vertices.
Calculate the minimum weight among all triangulations of the polygon.
Input
The first line contains single integer n (3 β€ n β€ 500) β the number of vertices in the regular polygon.
Output
Print one integer β the minimum weight among all triangulations of the given polygon.
Examples
Input
3
Output
6
Input
4
Output
18
Note
According to Wiki: polygon triangulation is the decomposition of a polygonal area (simple polygon) P into a set of triangles, i. e., finding a set of triangles with pairwise non-intersecting interiors whose union is P.
In the first example the polygon is a triangle, so we don't need to cut it further, so the answer is 1 β
2 β
3 = 6.
In the second example the polygon is a rectangle, so it should be divided into two triangles. It's optimal to cut it using diagonal 1-3 so answer is 1 β
2 β
3 + 1 β
3 β
4 = 6 + 12 = 18. | instruction | 0 | 97,880 | 23 | 195,760 |
Tags: dp, greedy, math
Correct Solution:
```
def nCk(n, k):
if n < k:
return 0
if k == 0:
return 1
if k == 1:
return n if n != 0 else 0
x = n
for i in range(1, k):
x *= (n-i)
y = 1
while k > 0:
y *= k
k -= 1
x //= y
return x
n = int(input())
print(6 * nCk(n - 3, 0) + 12 * nCk(n - 3, 1) + 8 * nCk(n - 3, 2) + 2 * nCk(n - 3, 3))
``` | output | 1 | 97,880 | 23 | 195,761 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a regular polygon with n vertices labeled from 1 to n in counter-clockwise order. The triangulation of a given polygon is a set of triangles such that each vertex of each triangle is a vertex of the initial polygon, there is no pair of triangles such that their intersection has non-zero area, and the total area of all triangles is equal to the area of the given polygon. The weight of a triangulation is the sum of weigths of triangles it consists of, where the weight of a triagle is denoted as the product of labels of its vertices.
Calculate the minimum weight among all triangulations of the polygon.
Input
The first line contains single integer n (3 β€ n β€ 500) β the number of vertices in the regular polygon.
Output
Print one integer β the minimum weight among all triangulations of the given polygon.
Examples
Input
3
Output
6
Input
4
Output
18
Note
According to Wiki: polygon triangulation is the decomposition of a polygonal area (simple polygon) P into a set of triangles, i. e., finding a set of triangles with pairwise non-intersecting interiors whose union is P.
In the first example the polygon is a triangle, so we don't need to cut it further, so the answer is 1 β
2 β
3 = 6.
In the second example the polygon is a rectangle, so it should be divided into two triangles. It's optimal to cut it using diagonal 1-3 so answer is 1 β
2 β
3 + 1 β
3 β
4 = 6 + 12 = 18. | instruction | 0 | 97,881 | 23 | 195,762 |
Tags: dp, greedy, math
Correct Solution:
```
Q = int(input())
s = 0
for i in range(3,Q+1):
s += (i-1)*i
print(s)
``` | output | 1 | 97,881 | 23 | 195,763 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a regular polygon with n vertices labeled from 1 to n in counter-clockwise order. The triangulation of a given polygon is a set of triangles such that each vertex of each triangle is a vertex of the initial polygon, there is no pair of triangles such that their intersection has non-zero area, and the total area of all triangles is equal to the area of the given polygon. The weight of a triangulation is the sum of weigths of triangles it consists of, where the weight of a triagle is denoted as the product of labels of its vertices.
Calculate the minimum weight among all triangulations of the polygon.
Input
The first line contains single integer n (3 β€ n β€ 500) β the number of vertices in the regular polygon.
Output
Print one integer β the minimum weight among all triangulations of the given polygon.
Examples
Input
3
Output
6
Input
4
Output
18
Note
According to Wiki: polygon triangulation is the decomposition of a polygonal area (simple polygon) P into a set of triangles, i. e., finding a set of triangles with pairwise non-intersecting interiors whose union is P.
In the first example the polygon is a triangle, so we don't need to cut it further, so the answer is 1 β
2 β
3 = 6.
In the second example the polygon is a rectangle, so it should be divided into two triangles. It's optimal to cut it using diagonal 1-3 so answer is 1 β
2 β
3 + 1 β
3 β
4 = 6 + 12 = 18.
Submitted Solution:
```
n = int(input())
res = 0
for i in range(n - 2):
res += (i + 2) * (i + 3)
print(res)
``` | instruction | 0 | 97,882 | 23 | 195,764 |
Yes | output | 1 | 97,882 | 23 | 195,765 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a regular polygon with n vertices labeled from 1 to n in counter-clockwise order. The triangulation of a given polygon is a set of triangles such that each vertex of each triangle is a vertex of the initial polygon, there is no pair of triangles such that their intersection has non-zero area, and the total area of all triangles is equal to the area of the given polygon. The weight of a triangulation is the sum of weigths of triangles it consists of, where the weight of a triagle is denoted as the product of labels of its vertices.
Calculate the minimum weight among all triangulations of the polygon.
Input
The first line contains single integer n (3 β€ n β€ 500) β the number of vertices in the regular polygon.
Output
Print one integer β the minimum weight among all triangulations of the given polygon.
Examples
Input
3
Output
6
Input
4
Output
18
Note
According to Wiki: polygon triangulation is the decomposition of a polygonal area (simple polygon) P into a set of triangles, i. e., finding a set of triangles with pairwise non-intersecting interiors whose union is P.
In the first example the polygon is a triangle, so we don't need to cut it further, so the answer is 1 β
2 β
3 = 6.
In the second example the polygon is a rectangle, so it should be divided into two triangles. It's optimal to cut it using diagonal 1-3 so answer is 1 β
2 β
3 + 1 β
3 β
4 = 6 + 12 = 18.
Submitted Solution:
```
n=int(input());print((n**3-n)//3-2)
``` | instruction | 0 | 97,883 | 23 | 195,766 |
Yes | output | 1 | 97,883 | 23 | 195,767 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a regular polygon with n vertices labeled from 1 to n in counter-clockwise order. The triangulation of a given polygon is a set of triangles such that each vertex of each triangle is a vertex of the initial polygon, there is no pair of triangles such that their intersection has non-zero area, and the total area of all triangles is equal to the area of the given polygon. The weight of a triangulation is the sum of weigths of triangles it consists of, where the weight of a triagle is denoted as the product of labels of its vertices.
Calculate the minimum weight among all triangulations of the polygon.
Input
The first line contains single integer n (3 β€ n β€ 500) β the number of vertices in the regular polygon.
Output
Print one integer β the minimum weight among all triangulations of the given polygon.
Examples
Input
3
Output
6
Input
4
Output
18
Note
According to Wiki: polygon triangulation is the decomposition of a polygonal area (simple polygon) P into a set of triangles, i. e., finding a set of triangles with pairwise non-intersecting interiors whose union is P.
In the first example the polygon is a triangle, so we don't need to cut it further, so the answer is 1 β
2 β
3 = 6.
In the second example the polygon is a rectangle, so it should be divided into two triangles. It's optimal to cut it using diagonal 1-3 so answer is 1 β
2 β
3 + 1 β
3 β
4 = 6 + 12 = 18.
Submitted Solution:
```
# https://codeforces.com/problemset/problem/1140/D
n = int(input())
dp = [0]*(n+1)
dp[3] = 6
for x in range(4, n+1):
dp[x] = dp[x-1] + x*(x-1)
print(dp[n])
``` | instruction | 0 | 97,884 | 23 | 195,768 |
Yes | output | 1 | 97,884 | 23 | 195,769 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a regular polygon with n vertices labeled from 1 to n in counter-clockwise order. The triangulation of a given polygon is a set of triangles such that each vertex of each triangle is a vertex of the initial polygon, there is no pair of triangles such that their intersection has non-zero area, and the total area of all triangles is equal to the area of the given polygon. The weight of a triangulation is the sum of weigths of triangles it consists of, where the weight of a triagle is denoted as the product of labels of its vertices.
Calculate the minimum weight among all triangulations of the polygon.
Input
The first line contains single integer n (3 β€ n β€ 500) β the number of vertices in the regular polygon.
Output
Print one integer β the minimum weight among all triangulations of the given polygon.
Examples
Input
3
Output
6
Input
4
Output
18
Note
According to Wiki: polygon triangulation is the decomposition of a polygonal area (simple polygon) P into a set of triangles, i. e., finding a set of triangles with pairwise non-intersecting interiors whose union is P.
In the first example the polygon is a triangle, so we don't need to cut it further, so the answer is 1 β
2 β
3 = 6.
In the second example the polygon is a rectangle, so it should be divided into two triangles. It's optimal to cut it using diagonal 1-3 so answer is 1 β
2 β
3 + 1 β
3 β
4 = 6 + 12 = 18.
Submitted Solution:
```
n=int(input())
summ=0
for i in range(3,n+1):
summ+=(1*(i-1)*(i))
print(summ)
``` | instruction | 0 | 97,885 | 23 | 195,770 |
Yes | output | 1 | 97,885 | 23 | 195,771 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a regular polygon with n vertices labeled from 1 to n in counter-clockwise order. The triangulation of a given polygon is a set of triangles such that each vertex of each triangle is a vertex of the initial polygon, there is no pair of triangles such that their intersection has non-zero area, and the total area of all triangles is equal to the area of the given polygon. The weight of a triangulation is the sum of weigths of triangles it consists of, where the weight of a triagle is denoted as the product of labels of its vertices.
Calculate the minimum weight among all triangulations of the polygon.
Input
The first line contains single integer n (3 β€ n β€ 500) β the number of vertices in the regular polygon.
Output
Print one integer β the minimum weight among all triangulations of the given polygon.
Examples
Input
3
Output
6
Input
4
Output
18
Note
According to Wiki: polygon triangulation is the decomposition of a polygonal area (simple polygon) P into a set of triangles, i. e., finding a set of triangles with pairwise non-intersecting interiors whose union is P.
In the first example the polygon is a triangle, so we don't need to cut it further, so the answer is 1 β
2 β
3 = 6.
In the second example the polygon is a rectangle, so it should be divided into two triangles. It's optimal to cut it using diagonal 1-3 so answer is 1 β
2 β
3 + 1 β
3 β
4 = 6 + 12 = 18.
Submitted Solution:
```
def solve(a, n):
# print("Here")
q = []
if n==4:
r = a[0]*a[1]*a[2] + a[2]*a[3]*a[0]
return r
if n==3:
r = a[0]*a[1]*a[2]
return r
r = 0
q = [a[0]]
for i in range(0, n-2, 2):
r += a[i]*a[i+1]*a[i+2]
# print(i+1,i+2,i+3)
q.append(a[i+2])
if n%2==0:
r += a[-2]*a[-1]*a[0]
# print(i-1,i,1)
# print(q)
# print('-'*10)
return r + solve(q, len(q))
#-----------------------------------------
t = 1
for _ in range(t):
n = int(input())
# a = list(input())
a = list(range(1, n+1))
res = solve(a, n)
print(res)
# print(*res)
``` | instruction | 0 | 97,886 | 23 | 195,772 |
No | output | 1 | 97,886 | 23 | 195,773 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a regular polygon with n vertices labeled from 1 to n in counter-clockwise order. The triangulation of a given polygon is a set of triangles such that each vertex of each triangle is a vertex of the initial polygon, there is no pair of triangles such that their intersection has non-zero area, and the total area of all triangles is equal to the area of the given polygon. The weight of a triangulation is the sum of weigths of triangles it consists of, where the weight of a triagle is denoted as the product of labels of its vertices.
Calculate the minimum weight among all triangulations of the polygon.
Input
The first line contains single integer n (3 β€ n β€ 500) β the number of vertices in the regular polygon.
Output
Print one integer β the minimum weight among all triangulations of the given polygon.
Examples
Input
3
Output
6
Input
4
Output
18
Note
According to Wiki: polygon triangulation is the decomposition of a polygonal area (simple polygon) P into a set of triangles, i. e., finding a set of triangles with pairwise non-intersecting interiors whose union is P.
In the first example the polygon is a triangle, so we don't need to cut it further, so the answer is 1 β
2 β
3 = 6.
In the second example the polygon is a rectangle, so it should be divided into two triangles. It's optimal to cut it using diagonal 1-3 so answer is 1 β
2 β
3 + 1 β
3 β
4 = 6 + 12 = 18.
Submitted Solution:
```
from math import *
n=int(input())
print((n-1)*n*(2*n-1)//6+(n-1)*(n-2)//2-2)
``` | instruction | 0 | 97,887 | 23 | 195,774 |
No | output | 1 | 97,887 | 23 | 195,775 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a regular polygon with n vertices labeled from 1 to n in counter-clockwise order. The triangulation of a given polygon is a set of triangles such that each vertex of each triangle is a vertex of the initial polygon, there is no pair of triangles such that their intersection has non-zero area, and the total area of all triangles is equal to the area of the given polygon. The weight of a triangulation is the sum of weigths of triangles it consists of, where the weight of a triagle is denoted as the product of labels of its vertices.
Calculate the minimum weight among all triangulations of the polygon.
Input
The first line contains single integer n (3 β€ n β€ 500) β the number of vertices in the regular polygon.
Output
Print one integer β the minimum weight among all triangulations of the given polygon.
Examples
Input
3
Output
6
Input
4
Output
18
Note
According to Wiki: polygon triangulation is the decomposition of a polygonal area (simple polygon) P into a set of triangles, i. e., finding a set of triangles with pairwise non-intersecting interiors whose union is P.
In the first example the polygon is a triangle, so we don't need to cut it further, so the answer is 1 β
2 β
3 = 6.
In the second example the polygon is a rectangle, so it should be divided into two triangles. It's optimal to cut it using diagonal 1-3 so answer is 1 β
2 β
3 + 1 β
3 β
4 = 6 + 12 = 18.
Submitted Solution:
```
from collections import defaultdict
from functools import reduce
from itertools import permutations
mi = lambda: [int(i) for i in input().split()]
flat = lambda l: reduce(lambda a, b: a + b, l)
n = mi()[0]
if n == 3:
print(6)
elif n == 4:
print(18)
else:
u = [*range(2, n + 2)]
a = [u.pop(-1)]
d = 0
while u:
a.append(u.pop(d))
if u:
a.insert(0, u.pop(d))
d = -1 if d == 0 else 0
# print(a)
r = 0
for i in range(len(a)):
r += a[i - 1] * a[i]
print(r)
# # # a = [2, 6, 3, 5, 4]
# a = [2, 3, 4, 5, 6, 7, 8]
# m = 1000000000000000
#
# for p in permutations(a, len(a)):
# r = 0
#
# for i in range(len(p)):
# r += p[i - 1] * p[i]
#
# # if r == 72:
# print(r, p)
# m = min(m, r)
#
# print(m)
``` | instruction | 0 | 97,888 | 23 | 195,776 |
No | output | 1 | 97,888 | 23 | 195,777 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a regular polygon with n vertices labeled from 1 to n in counter-clockwise order. The triangulation of a given polygon is a set of triangles such that each vertex of each triangle is a vertex of the initial polygon, there is no pair of triangles such that their intersection has non-zero area, and the total area of all triangles is equal to the area of the given polygon. The weight of a triangulation is the sum of weigths of triangles it consists of, where the weight of a triagle is denoted as the product of labels of its vertices.
Calculate the minimum weight among all triangulations of the polygon.
Input
The first line contains single integer n (3 β€ n β€ 500) β the number of vertices in the regular polygon.
Output
Print one integer β the minimum weight among all triangulations of the given polygon.
Examples
Input
3
Output
6
Input
4
Output
18
Note
According to Wiki: polygon triangulation is the decomposition of a polygonal area (simple polygon) P into a set of triangles, i. e., finding a set of triangles with pairwise non-intersecting interiors whose union is P.
In the first example the polygon is a triangle, so we don't need to cut it further, so the answer is 1 β
2 β
3 = 6.
In the second example the polygon is a rectangle, so it should be divided into two triangles. It's optimal to cut it using diagonal 1-3 so answer is 1 β
2 β
3 + 1 β
3 β
4 = 6 + 12 = 18.
Submitted Solution:
```
n=int(input())
su=0
for i in range(2,n-1):
su=su+i*i+1
print(su)
``` | instruction | 0 | 97,889 | 23 | 195,778 |
No | output | 1 | 97,889 | 23 | 195,779 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Pete and Bob invented a new interesting game. Bob takes a sheet of paper and locates a Cartesian coordinate system on it as follows: point (0, 0) is located in the bottom-left corner, Ox axis is directed right, Oy axis is directed up. Pete gives Bob requests of three types:
* add x y β on the sheet of paper Bob marks a point with coordinates (x, y). For each request of this type it's guaranteed that point (x, y) is not yet marked on Bob's sheet at the time of the request.
* remove x y β on the sheet of paper Bob erases the previously marked point with coordinates (x, y). For each request of this type it's guaranteed that point (x, y) is already marked on Bob's sheet at the time of the request.
* find x y β on the sheet of paper Bob finds all the marked points, lying strictly above and strictly to the right of point (x, y). Among these points Bob chooses the leftmost one, if it is not unique, he chooses the bottommost one, and gives its coordinates to Pete.
Bob managed to answer the requests, when they were 10, 100 or 1000, but when their amount grew up to 2Β·105, Bob failed to cope. Now he needs a program that will answer all Pete's requests. Help Bob, please!
Input
The first input line contains number n (1 β€ n β€ 2Β·105) β amount of requests. Then there follow n lines β descriptions of the requests. add x y describes the request to add a point, remove x y β the request to erase a point, find x y β the request to find the bottom-left point. All the coordinates in the input file are non-negative and don't exceed 109.
Output
For each request of type find x y output in a separate line the answer to it β coordinates of the bottommost among the leftmost marked points, lying strictly above and to the right of point (x, y). If there are no points strictly above and to the right of point (x, y), output -1.
Examples
Input
7
add 1 1
add 3 4
find 0 0
remove 1 1
find 0 0
add 1 1
find 0 0
Output
1 1
3 4
1 1
Input
13
add 5 5
add 5 6
add 5 7
add 6 5
add 6 6
add 6 7
add 7 5
add 7 6
add 7 7
find 6 6
remove 7 7
find 6 6
find 4 4
Output
7 7
-1
5 5
Submitted Solution:
```
import bisect
n = int(input())
points = list()
x_points = list()
corresponding_y_points = list()
y_points = list()
corresponding_x_points = list()
def insert_point(point):
index = bisect.bisect(points, point)
points.insert(index, point)
index = bisect.bisect(x_points, point[0])
x_points.insert(index, point[0])
corresponding_y_points.insert(index, point[1])
index = bisect.bisect(y_points, point[1])
y_points.insert(index, point[1])
corresponding_x_points.insert(index, point[0])
def find_first_and_second_point(point, x=True):
if x:
first_point = point[0]
second_point = point[1]
else:
first_point = point[1]
second_point = point[0]
return first_point, second_point
def find_index_point_to_remove(ordered_points, corresponding_points, point, x=True):
first_point, second_point = find_first_and_second_point(point, x)
first_index = bisect.bisect_left(ordered_points, first_point)
second_index = first_index
for index in range(first_index, len(ordered_points)):
if ordered_points[index] != ordered_points[first_index]:
break
if corresponding_points[index] == second_point:
second_index = index
break
return second_index
def remove_point(point):
points.remove(point)
del corresponding_y_points[find_index_point_to_remove(x_points, corresponding_y_points, point)]
x_points.remove(current_point[0])
del corresponding_x_points[find_index_point_to_remove(y_points, corresponding_x_points, point, False)]
y_points.remove(current_point[1])
def find_point(ordered_points, corresponding_points, point, x=True):
first_point, second_point = find_first_and_second_point(point, x)
first_index = bisect.bisect_left(ordered_points, first_point + 1)
if first_index >= len(ordered_points):
return -1, -1
found = False
for index in range(first_index, len(ordered_points)):
if corresponding_points[index] > second_point:
first_index = index
found = True
break
if not found:
return -1, -1
min_point = corresponding_points[first_index]
second_index = first_index
for index in range(first_index + 1, len(ordered_points)):
if ordered_points[index] != ordered_points[first_index]:
break
if corresponding_points[index] < min_point:
min_point = corresponding_points[index]
second_index = index
return first_index, second_index
for _ in range(n):
request = input().split()
current_point = (int(request[1]), int(request[2]))
if "add" in request:
insert_point(current_point)
elif "remove" in request:
remove_point(current_point)
else:
first_index_x, first_index_y = find_point(x_points, corresponding_y_points, current_point)
second_index_y, second_index_x = find_point(y_points, corresponding_x_points, current_point, False)
if first_index_x == -1 and second_index_x == -1:
print("-1")
else:
if x_points[first_index_x] > current_point[0] and corresponding_y_points[first_index_y] > current_point[1]:
print(x_points[first_index_x], corresponding_y_points[first_index_y])
elif corresponding_x_points[second_index_x] > current_point[0] and y_points[second_index_y] > current_point[1]:
print(corresponding_x_points[second_index_x], y_points[second_index_y])
else:
print("-1")
``` | instruction | 0 | 98,156 | 23 | 196,312 |
No | output | 1 | 98,156 | 23 | 196,313 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Pete and Bob invented a new interesting game. Bob takes a sheet of paper and locates a Cartesian coordinate system on it as follows: point (0, 0) is located in the bottom-left corner, Ox axis is directed right, Oy axis is directed up. Pete gives Bob requests of three types:
* add x y β on the sheet of paper Bob marks a point with coordinates (x, y). For each request of this type it's guaranteed that point (x, y) is not yet marked on Bob's sheet at the time of the request.
* remove x y β on the sheet of paper Bob erases the previously marked point with coordinates (x, y). For each request of this type it's guaranteed that point (x, y) is already marked on Bob's sheet at the time of the request.
* find x y β on the sheet of paper Bob finds all the marked points, lying strictly above and strictly to the right of point (x, y). Among these points Bob chooses the leftmost one, if it is not unique, he chooses the bottommost one, and gives its coordinates to Pete.
Bob managed to answer the requests, when they were 10, 100 or 1000, but when their amount grew up to 2Β·105, Bob failed to cope. Now he needs a program that will answer all Pete's requests. Help Bob, please!
Input
The first input line contains number n (1 β€ n β€ 2Β·105) β amount of requests. Then there follow n lines β descriptions of the requests. add x y describes the request to add a point, remove x y β the request to erase a point, find x y β the request to find the bottom-left point. All the coordinates in the input file are non-negative and don't exceed 109.
Output
For each request of type find x y output in a separate line the answer to it β coordinates of the bottommost among the leftmost marked points, lying strictly above and to the right of point (x, y). If there are no points strictly above and to the right of point (x, y), output -1.
Examples
Input
7
add 1 1
add 3 4
find 0 0
remove 1 1
find 0 0
add 1 1
find 0 0
Output
1 1
3 4
1 1
Input
13
add 5 5
add 5 6
add 5 7
add 6 5
add 6 6
add 6 7
add 7 5
add 7 6
add 7 7
find 6 6
remove 7 7
find 6 6
find 4 4
Output
7 7
-1
5 5
Submitted Solution:
```
a=[]
n=0
for i in range(int(input())):
b,x,y=input().split()
x=int(x)
y=int(y)
c=n/2
d=round(c/2)
d=min(d,n-1)
d=max(d,0)
while round(c)>0:
if a[d][0]>x and a[d][1]>y:
d+=round(c/2)
d=min(d,n-1)
d=max(d,0)
else:
d-=round(c/2)
d=min(d,n-1)
d=max(d,0)
c/=2
while 0<d and a[d][0]>x and a[d][1]>y:
d-=1
while d<n-1 and (a[d][0]<=x or a[d][1]<=y):
d+=1
if b=='add':
n+=1
if d>=n-1 or (a[d][0]>=x and a[d][1]>=y):
a=a[:d]+[[x,y]]+a[d:]
else:
a=a[:d+1]+[[x,y]]+a[d+1:]
elif b=='remove':
d+=1
d=min(d,n-1)
while a[d][0]!=x and a[d][1]!=y:
d-=1
a=a[:d]+a[d+1:]
n-=1
else:
if d>=n or (d==n-1 and (a[d][0]==x or a[d][1]==y)):
print(-1)
else:
if a[d][0]==x or a[d][1]==y:
print(*a[d+1])
else:
print(*a[d])
``` | instruction | 0 | 98,157 | 23 | 196,314 |
No | output | 1 | 98,157 | 23 | 196,315 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Pete and Bob invented a new interesting game. Bob takes a sheet of paper and locates a Cartesian coordinate system on it as follows: point (0, 0) is located in the bottom-left corner, Ox axis is directed right, Oy axis is directed up. Pete gives Bob requests of three types:
* add x y β on the sheet of paper Bob marks a point with coordinates (x, y). For each request of this type it's guaranteed that point (x, y) is not yet marked on Bob's sheet at the time of the request.
* remove x y β on the sheet of paper Bob erases the previously marked point with coordinates (x, y). For each request of this type it's guaranteed that point (x, y) is already marked on Bob's sheet at the time of the request.
* find x y β on the sheet of paper Bob finds all the marked points, lying strictly above and strictly to the right of point (x, y). Among these points Bob chooses the leftmost one, if it is not unique, he chooses the bottommost one, and gives its coordinates to Pete.
Bob managed to answer the requests, when they were 10, 100 or 1000, but when their amount grew up to 2Β·105, Bob failed to cope. Now he needs a program that will answer all Pete's requests. Help Bob, please!
Input
The first input line contains number n (1 β€ n β€ 2Β·105) β amount of requests. Then there follow n lines β descriptions of the requests. add x y describes the request to add a point, remove x y β the request to erase a point, find x y β the request to find the bottom-left point. All the coordinates in the input file are non-negative and don't exceed 109.
Output
For each request of type find x y output in a separate line the answer to it β coordinates of the bottommost among the leftmost marked points, lying strictly above and to the right of point (x, y). If there are no points strictly above and to the right of point (x, y), output -1.
Examples
Input
7
add 1 1
add 3 4
find 0 0
remove 1 1
find 0 0
add 1 1
find 0 0
Output
1 1
3 4
1 1
Input
13
add 5 5
add 5 6
add 5 7
add 6 5
add 6 6
add 6 7
add 7 5
add 7 6
add 7 7
find 6 6
remove 7 7
find 6 6
find 4 4
Output
7 7
-1
5 5
Submitted Solution:
```
l = int(input())
items = []
points = []
for i in range(l):
item = input().split()
item[1] = int(item[1])
item[2] = int(item[2])
items.append(item)
for item in items:
if item[0] == 'add':
points.append(item[1:])
points.sort(key=lambda point: point[0])
elif item[0] == 'remove':
points.remove(item[1:])
else:
check = 0
for point in points:
if point[0] > item[1] and point[1] > item[2]:
check = 1
print(point[0], point[1])
break;
if (not check):
print(-1)
``` | instruction | 0 | 98,158 | 23 | 196,316 |
No | output | 1 | 98,158 | 23 | 196,317 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Pete and Bob invented a new interesting game. Bob takes a sheet of paper and locates a Cartesian coordinate system on it as follows: point (0, 0) is located in the bottom-left corner, Ox axis is directed right, Oy axis is directed up. Pete gives Bob requests of three types:
* add x y β on the sheet of paper Bob marks a point with coordinates (x, y). For each request of this type it's guaranteed that point (x, y) is not yet marked on Bob's sheet at the time of the request.
* remove x y β on the sheet of paper Bob erases the previously marked point with coordinates (x, y). For each request of this type it's guaranteed that point (x, y) is already marked on Bob's sheet at the time of the request.
* find x y β on the sheet of paper Bob finds all the marked points, lying strictly above and strictly to the right of point (x, y). Among these points Bob chooses the leftmost one, if it is not unique, he chooses the bottommost one, and gives its coordinates to Pete.
Bob managed to answer the requests, when they were 10, 100 or 1000, but when their amount grew up to 2Β·105, Bob failed to cope. Now he needs a program that will answer all Pete's requests. Help Bob, please!
Input
The first input line contains number n (1 β€ n β€ 2Β·105) β amount of requests. Then there follow n lines β descriptions of the requests. add x y describes the request to add a point, remove x y β the request to erase a point, find x y β the request to find the bottom-left point. All the coordinates in the input file are non-negative and don't exceed 109.
Output
For each request of type find x y output in a separate line the answer to it β coordinates of the bottommost among the leftmost marked points, lying strictly above and to the right of point (x, y). If there are no points strictly above and to the right of point (x, y), output -1.
Examples
Input
7
add 1 1
add 3 4
find 0 0
remove 1 1
find 0 0
add 1 1
find 0 0
Output
1 1
3 4
1 1
Input
13
add 5 5
add 5 6
add 5 7
add 6 5
add 6 6
add 6 7
add 7 5
add 7 6
add 7 7
find 6 6
remove 7 7
find 6 6
find 4 4
Output
7 7
-1
5 5
Submitted Solution:
```
import bisect
n = int(input())
points = list()
x_points = list()
corresponding_y_points = list()
for _ in range(n):
request = input().split()
current_point = (int(request[1]), int(request[2]))
if "add" in request:
index = bisect.bisect(points, current_point)
points.insert(index, current_point)
index = bisect.bisect(x_points, current_point[0])
x_points.insert(index, current_point[0])
corresponding_y_points.insert(index, current_point[1])
elif "remove" in request:
points.remove(current_point)
index_x = bisect.bisect_left(x_points, current_point[0])
index_y = index_x
for index in range(index_x, len(x_points)):
if x_points[index] != x_points[index_x]:
break
if corresponding_y_points[index] == current_point[1]:
index_y = index
break
del corresponding_y_points[index_y]
x_points.remove(current_point[0])
else:
try:
index_x = bisect.bisect_left(x_points, current_point[0] + 1)
for index in range(index_x, len(x_points)):
if corresponding_y_points[index] > current_point[1]:
index_x = index
break
y_min = corresponding_y_points[index_x]
index_y = index_x
for index in range(index_x, len(x_points)):
if x_points[index] != x_points[index_x]:
break
if corresponding_y_points[index] < y_min:
y_min = corresponding_y_points[index]
index_y = index
if x_points[index_x] > current_point[0] and corresponding_y_points[index_y] > current_point[1]:
print(x_points[index_x], corresponding_y_points[index_y])
else:
print("-1")
except Exception:
print("-1")
``` | instruction | 0 | 98,159 | 23 | 196,318 |
No | output | 1 | 98,159 | 23 | 196,319 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Dima has two sequences of points with integer coordinates: sequence (a1, 1), (a2, 2), ..., (an, n) and sequence (b1, 1), (b2, 2), ..., (bn, n).
Now Dima wants to count the number of distinct sequences of points of length 2Β·n that can be assembled from these sequences, such that the x-coordinates of points in the assembled sequence will not decrease. Help him with that. Note that each element of the initial sequences should be used exactly once in the assembled sequence.
Dima considers two assembled sequences (p1, q1), (p2, q2), ..., (p2Β·n, q2Β·n) and (x1, y1), (x2, y2), ..., (x2Β·n, y2Β·n) distinct, if there is such i (1 β€ i β€ 2Β·n), that (pi, qi) β (xi, yi).
As the answer can be rather large, print the remainder from dividing the answer by number m.
Input
The first line contains integer n (1 β€ n β€ 105). The second line contains n integers a1, a2, ..., an (1 β€ ai β€ 109). The third line contains n integers b1, b2, ..., bn (1 β€ bi β€ 109). The numbers in the lines are separated by spaces.
The last line contains integer m (2 β€ m β€ 109 + 7).
Output
In the single line print the remainder after dividing the answer to the problem by number m.
Examples
Input
1
1
2
7
Output
1
Input
2
1 2
2 3
11
Output
2
Note
In the first sample you can get only one sequence: (1, 1), (2, 1).
In the second sample you can get such sequences : (1, 1), (2, 2), (2, 1), (3, 2); (1, 1), (2, 1), (2, 2), (3, 2). Thus, the answer is 2.
Submitted Solution:
```
from math import sqrt,ceil,gcd
from collections import defaultdict
def modInverse(b,m):
g = gcd(b, m)
if (g != 1):
# print("Inverse doesn't exist")
return -1
else:
# If b and m are relatively prime,
# then modulo inverse is b^(m-2) mode m
return pow(b, m - 2, m)
def solve():
n = int(input())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
m = int(input())
dp = [1]*(10**5+1000)
for i in range(1,10**5+1000):
dp[i] = dp[i-1]*i
dp[i]%=m
l1 = []
l2 = []
for i in range(n):
l1.append([a[i],i+1])
l2.append([b[i],i+1])
z = l1+l2
z = sorted(z, key=lambda x: (x[0], x[1]))
hash = defaultdict(list)
for a,b in z:
hash[a].append(b)
ans = 1
print(z)
for i in hash:
z1 = dp[len(hash[i])]
j = 0
z2 = 1
while j<len(hash[i]):
k = hash[i][j]
cnt = 0
while j<len(hash[i]) and hash[i][j] == k:
cnt+=1
j+=1
# print(i,cnt)
z2 *= modInverse(dp[cnt],m)
# print(i,z2)
ans*=z1*z2
ans%=m
print(ans)
# print(z)
# t = int(input())
# for _ in range(t):
solve()
``` | instruction | 0 | 98,167 | 23 | 196,334 |
No | output | 1 | 98,167 | 23 | 196,335 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Dima has two sequences of points with integer coordinates: sequence (a1, 1), (a2, 2), ..., (an, n) and sequence (b1, 1), (b2, 2), ..., (bn, n).
Now Dima wants to count the number of distinct sequences of points of length 2Β·n that can be assembled from these sequences, such that the x-coordinates of points in the assembled sequence will not decrease. Help him with that. Note that each element of the initial sequences should be used exactly once in the assembled sequence.
Dima considers two assembled sequences (p1, q1), (p2, q2), ..., (p2Β·n, q2Β·n) and (x1, y1), (x2, y2), ..., (x2Β·n, y2Β·n) distinct, if there is such i (1 β€ i β€ 2Β·n), that (pi, qi) β (xi, yi).
As the answer can be rather large, print the remainder from dividing the answer by number m.
Input
The first line contains integer n (1 β€ n β€ 105). The second line contains n integers a1, a2, ..., an (1 β€ ai β€ 109). The third line contains n integers b1, b2, ..., bn (1 β€ bi β€ 109). The numbers in the lines are separated by spaces.
The last line contains integer m (2 β€ m β€ 109 + 7).
Output
In the single line print the remainder after dividing the answer to the problem by number m.
Examples
Input
1
1
2
7
Output
1
Input
2
1 2
2 3
11
Output
2
Note
In the first sample you can get only one sequence: (1, 1), (2, 1).
In the second sample you can get such sequences : (1, 1), (2, 2), (2, 1), (3, 2); (1, 1), (2, 1), (2, 2), (3, 2). Thus, the answer is 2.
Submitted Solution:
```
from math import sqrt,ceil
from collections import defaultdict
def solve():
n = int(input())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
m = int(input())
l1 = []
l2 = []
for i in range(n):
l1.append([a[i],i+1])
l2.append([b[i],i+1])
z = l1+l2
z.sort()
hash = defaultdict(list)
for a,b in z:
hash[a].append(b)
ans = 1
for i in hash:
ans*=len(hash[i])
ans%=m
print(ans)
# t = int(input())
# for _ in range(t):
solve()
``` | instruction | 0 | 98,168 | 23 | 196,336 |
No | output | 1 | 98,168 | 23 | 196,337 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Dima has two sequences of points with integer coordinates: sequence (a1, 1), (a2, 2), ..., (an, n) and sequence (b1, 1), (b2, 2), ..., (bn, n).
Now Dima wants to count the number of distinct sequences of points of length 2Β·n that can be assembled from these sequences, such that the x-coordinates of points in the assembled sequence will not decrease. Help him with that. Note that each element of the initial sequences should be used exactly once in the assembled sequence.
Dima considers two assembled sequences (p1, q1), (p2, q2), ..., (p2Β·n, q2Β·n) and (x1, y1), (x2, y2), ..., (x2Β·n, y2Β·n) distinct, if there is such i (1 β€ i β€ 2Β·n), that (pi, qi) β (xi, yi).
As the answer can be rather large, print the remainder from dividing the answer by number m.
Input
The first line contains integer n (1 β€ n β€ 105). The second line contains n integers a1, a2, ..., an (1 β€ ai β€ 109). The third line contains n integers b1, b2, ..., bn (1 β€ bi β€ 109). The numbers in the lines are separated by spaces.
The last line contains integer m (2 β€ m β€ 109 + 7).
Output
In the single line print the remainder after dividing the answer to the problem by number m.
Examples
Input
1
1
2
7
Output
1
Input
2
1 2
2 3
11
Output
2
Note
In the first sample you can get only one sequence: (1, 1), (2, 1).
In the second sample you can get such sequences : (1, 1), (2, 2), (2, 1), (3, 2); (1, 1), (2, 1), (2, 2), (3, 2). Thus, the answer is 2.
Submitted Solution:
```
f = lambda: map(int, input().split())
n = int(input())
c, d = {}, {}
for x, y in zip(f(), f()):
c[x] = c.get(x, 0) + 1
c[y] = c.get(y, 0) + 1
if x == y: d[x] = d.get(x, 0) + 2
s, m = 1, int(input())
for k, v in c.items():
u = d.get(k, 0)
for i in range(v - u, v, 2): s = s * (i * i + i) // 2 % m
for i in range(v - u): s = s * (i + 1) % m
print(s)
``` | instruction | 0 | 98,169 | 23 | 196,338 |
No | output | 1 | 98,169 | 23 | 196,339 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Dima has two sequences of points with integer coordinates: sequence (a1, 1), (a2, 2), ..., (an, n) and sequence (b1, 1), (b2, 2), ..., (bn, n).
Now Dima wants to count the number of distinct sequences of points of length 2Β·n that can be assembled from these sequences, such that the x-coordinates of points in the assembled sequence will not decrease. Help him with that. Note that each element of the initial sequences should be used exactly once in the assembled sequence.
Dima considers two assembled sequences (p1, q1), (p2, q2), ..., (p2Β·n, q2Β·n) and (x1, y1), (x2, y2), ..., (x2Β·n, y2Β·n) distinct, if there is such i (1 β€ i β€ 2Β·n), that (pi, qi) β (xi, yi).
As the answer can be rather large, print the remainder from dividing the answer by number m.
Input
The first line contains integer n (1 β€ n β€ 105). The second line contains n integers a1, a2, ..., an (1 β€ ai β€ 109). The third line contains n integers b1, b2, ..., bn (1 β€ bi β€ 109). The numbers in the lines are separated by spaces.
The last line contains integer m (2 β€ m β€ 109 + 7).
Output
In the single line print the remainder after dividing the answer to the problem by number m.
Examples
Input
1
1
2
7
Output
1
Input
2
1 2
2 3
11
Output
2
Note
In the first sample you can get only one sequence: (1, 1), (2, 1).
In the second sample you can get such sequences : (1, 1), (2, 2), (2, 1), (3, 2); (1, 1), (2, 1), (2, 2), (3, 2). Thus, the answer is 2.
Submitted Solution:
```
n=int(input())
a=sorted(list(map(int,input().split()))+list(map(int,input().split())))
m=int(input())
b=[]
c=1
for i in range(1,n+1):
c*=i
c%=m
b.append(c)
d=0
e=1
for i in range(n):
if a[i]!=a[i-1]:
e*=b[i-d-1]
e%=m
d=i
e*=b[n-d-1]
print(e%m)
``` | instruction | 0 | 98,170 | 23 | 196,340 |
No | output | 1 | 98,170 | 23 | 196,341 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya is a Greencode wildlife preservation society proponent. One day he found an empty field nobody owned, divided it into n Γ m squares and decided to plant a forest there. Vasya will plant nm trees of all different heights from 1 to nm. For his forest to look more natural he wants any two trees growing in the side neighbouring squares to have the absolute value of difference in heights to be strictly more than 1. Help Vasya: make the plan of the forest planting for which this condition is fulfilled.
Input
The first line contains two space-separated integers n and m (1 β€ n, m β€ 100) β the number of rows and columns on Vasya's field
Output
If there's no solution, print -1. Otherwise, print n lines containing m numbers each β the trees' planting plan. In every square of the plan the height of a tree that should be planted on this square should be written. If there are several solutions to that problem, print any of them.
Examples
Input
2 3
Output
3 6 2
5 1 4
Input
2 1
Output
-1 | instruction | 0 | 98,222 | 23 | 196,444 |
Tags: constructive algorithms
Correct Solution:
```
# https://codeforces.com/problemset/problem/45/J
def solve(n, m):
if n*m == 1:
return [[1]]
if n*m <= 3 or (n == 2 and m == 2):
return -1
a = [[0] * m for _ in range(n)]
base1 = 1
base2 = (n*m) // 2 + 1
for i in range(n):
for j in range(m):
if (i+j) % 2 == 0:
a[i][j] = base2
base2+=1
else:
a[i][j] = base1
base1+=1
return a
def pr(a, n, m):
if a == -1:
print(a)
else:
for _ in range(n):
print(' '.join([str(x) for x in a[_]]))
n, m = map(int, input().split())
pr(solve(n, m), n, m)
``` | output | 1 | 98,222 | 23 | 196,445 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya is a Greencode wildlife preservation society proponent. One day he found an empty field nobody owned, divided it into n Γ m squares and decided to plant a forest there. Vasya will plant nm trees of all different heights from 1 to nm. For his forest to look more natural he wants any two trees growing in the side neighbouring squares to have the absolute value of difference in heights to be strictly more than 1. Help Vasya: make the plan of the forest planting for which this condition is fulfilled.
Input
The first line contains two space-separated integers n and m (1 β€ n, m β€ 100) β the number of rows and columns on Vasya's field
Output
If there's no solution, print -1. Otherwise, print n lines containing m numbers each β the trees' planting plan. In every square of the plan the height of a tree that should be planted on this square should be written. If there are several solutions to that problem, print any of them.
Examples
Input
2 3
Output
3 6 2
5 1 4
Input
2 1
Output
-1
Submitted Solution:
```
# https://codeforces.com/problemset/problem/45/J
def solve(n, m):
if n*m == 1:
return [1]
if n*m < 5:
return [-1]
if n == 2 and m == 3:
return [[1,5,3], [4,2,6]]
a = [[0] * m for _ in range(n)]
base1 = 1
base2 = (n*m+1) // 2 + 1
for i in range(n):
for j in range(m):
if (i+j) % 2 == 0:
a[i][j] = base1
base1+=1
else:
a[i][j] = base2
base2+=1
return a
def pr(a, n, m):
if len(a) == 1:
print(a[0])
else:
for _ in range(n):
print(' '.join([str(x) for x in a[_]]))
n, m = map(int, input().split())
pr(solve(n, m), n, m)
``` | instruction | 0 | 98,227 | 23 | 196,454 |
No | output | 1 | 98,227 | 23 | 196,455 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya is a Greencode wildlife preservation society proponent. One day he found an empty field nobody owned, divided it into n Γ m squares and decided to plant a forest there. Vasya will plant nm trees of all different heights from 1 to nm. For his forest to look more natural he wants any two trees growing in the side neighbouring squares to have the absolute value of difference in heights to be strictly more than 1. Help Vasya: make the plan of the forest planting for which this condition is fulfilled.
Input
The first line contains two space-separated integers n and m (1 β€ n, m β€ 100) β the number of rows and columns on Vasya's field
Output
If there's no solution, print -1. Otherwise, print n lines containing m numbers each β the trees' planting plan. In every square of the plan the height of a tree that should be planted on this square should be written. If there are several solutions to that problem, print any of them.
Examples
Input
2 3
Output
3 6 2
5 1 4
Input
2 1
Output
-1
Submitted Solution:
```
# Legends Always Come Up with Solution
# Author: Manvir Singh
import os
import sys
from io import BytesIO, IOBase
def main():
n,m=map(int,input().split())
if n==1 and m==1:
print(1)
elif n<=2 and m<=2:
print(-1)
else:
a=[]
z=1
for i in range(n):
a.append([])
for j in range(m):
a[-1].append(z)
z+=1
for j in range(1,m,2):
for i in range(n-1):
a[i][j],a[i+1][j]=a[i+1][j],a[i][j]
for i in a:
print(*i)
# 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")
if __name__ == "__main__":
main()
``` | instruction | 0 | 98,229 | 23 | 196,458 |
No | output | 1 | 98,229 | 23 | 196,459 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya is a Greencode wildlife preservation society proponent. One day he found an empty field nobody owned, divided it into n Γ m squares and decided to plant a forest there. Vasya will plant nm trees of all different heights from 1 to nm. For his forest to look more natural he wants any two trees growing in the side neighbouring squares to have the absolute value of difference in heights to be strictly more than 1. Help Vasya: make the plan of the forest planting for which this condition is fulfilled.
Input
The first line contains two space-separated integers n and m (1 β€ n, m β€ 100) β the number of rows and columns on Vasya's field
Output
If there's no solution, print -1. Otherwise, print n lines containing m numbers each β the trees' planting plan. In every square of the plan the height of a tree that should be planted on this square should be written. If there are several solutions to that problem, print any of them.
Examples
Input
2 3
Output
3 6 2
5 1 4
Input
2 1
Output
-1
Submitted Solution:
```
# Legends Always Come Up with Solution
# Author: Manvir Singh
import os
import sys
from io import BytesIO, IOBase
def main():
n,m=map(int,input().split())
if n==1 and m==1:
print(1)
elif n<2 or m<2 or (n==2 and m==2):
print(-1)
else:
a=[]
z=1
for i in range(n):
a.append([])
for j in range(m):
a[-1].append(z)
z+=1
for j in range(1,m,2):
for i in range(n-1):
a[i][j],a[i+1][j]=a[i+1][j],a[i][j]
for i in a:
print(*i)
# 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")
if __name__ == "__main__":
main()
``` | instruction | 0 | 98,230 | 23 | 196,460 |
No | output | 1 | 98,230 | 23 | 196,461 |
Provide a correct Python 3 solution for this coding contest problem.
Your task is to develop a tiny little part of spreadsheet software.
Write a program which adds up columns and rows of given table as shown in the following figure:
<image>
Input
The input consists of several datasets. Each dataset consists of:
n (the size of row and column of the given table)
1st row of the table
2nd row of the table
:
:
nth row of the table
The input ends with a line consisting of a single 0.
Output
For each dataset, print the table with sums of rows and columns. Each item of the table should be aligned to the right with a margin for five digits. Please see the sample output for details.
Example
Input
4
52 96 15 20
86 22 35 45
45 78 54 36
16 86 74 55
4
52 96 15 20
86 22 35 45
45 78 54 36
16 86 74 55
0
Output
52 96 15 20 183
86 22 35 45 188
45 78 54 36 213
16 86 74 55 231
199 282 178 156 815
52 96 15 20 183
86 22 35 45 188
45 78 54 36 213
16 86 74 55 231
199 282 178 156 815 | instruction | 0 | 98,513 | 23 | 197,026 |
"Correct Solution:
```
while True:
n=int(input())
if n==0:
break
ts=[0]*n
ys=0
a=[]
for i in range(n):
b=list(map(int,input().split()))
s=sum(b)
a.append(b+[s])
for i in range(n):
ts[i]+=b[i]
ys+=s
t=sum(ts)
for i in range(n):
for j in range(n+1):
print(str(a[i][j]).rjust(5),end='')
print()
for i in range(n):
print(str(ts[i]).rjust(5),end='')
print(str(t).rjust(5))
``` | output | 1 | 98,513 | 23 | 197,027 |
Provide a correct Python 3 solution for this coding contest problem.
Your task is to develop a tiny little part of spreadsheet software.
Write a program which adds up columns and rows of given table as shown in the following figure:
<image>
Input
The input consists of several datasets. Each dataset consists of:
n (the size of row and column of the given table)
1st row of the table
2nd row of the table
:
:
nth row of the table
The input ends with a line consisting of a single 0.
Output
For each dataset, print the table with sums of rows and columns. Each item of the table should be aligned to the right with a margin for five digits. Please see the sample output for details.
Example
Input
4
52 96 15 20
86 22 35 45
45 78 54 36
16 86 74 55
4
52 96 15 20
86 22 35 45
45 78 54 36
16 86 74 55
0
Output
52 96 15 20 183
86 22 35 45 188
45 78 54 36 213
16 86 74 55 231
199 282 178 156 815
52 96 15 20 183
86 22 35 45 188
45 78 54 36 213
16 86 74 55 231
199 282 178 156 815 | instruction | 0 | 98,514 | 23 | 197,028 |
"Correct Solution:
```
while True:
n=int(input())
if not n: break
a=[list(map(int,input().split())) for _ in range(n)]
t=[[0 for _ in range(n+1)] for _ in range(n+1)]
for i in range(n):
for j in range(n):
t[i][j]=a[i][j]
t[i][n]+=a[i][j]
t[n][j]+=a[i][j]
t[n][n]+=a[i][j]
for i in range(n+1):
print("".join(map("{0:>5}".format,t[i])))
``` | output | 1 | 98,514 | 23 | 197,029 |
Provide a correct Python 3 solution for this coding contest problem.
Your task is to develop a tiny little part of spreadsheet software.
Write a program which adds up columns and rows of given table as shown in the following figure:
<image>
Input
The input consists of several datasets. Each dataset consists of:
n (the size of row and column of the given table)
1st row of the table
2nd row of the table
:
:
nth row of the table
The input ends with a line consisting of a single 0.
Output
For each dataset, print the table with sums of rows and columns. Each item of the table should be aligned to the right with a margin for five digits. Please see the sample output for details.
Example
Input
4
52 96 15 20
86 22 35 45
45 78 54 36
16 86 74 55
4
52 96 15 20
86 22 35 45
45 78 54 36
16 86 74 55
0
Output
52 96 15 20 183
86 22 35 45 188
45 78 54 36 213
16 86 74 55 231
199 282 178 156 815
52 96 15 20 183
86 22 35 45 188
45 78 54 36 213
16 86 74 55 231
199 282 178 156 815 | instruction | 0 | 98,515 | 23 | 197,030 |
"Correct Solution:
```
while True:
n = int(input())
if n == 0:
break
rows = [list(map(int, input().split())) for _ in range(n)]
for row in rows:
print("".join(map("{0:>5}".format, row)), end="")
print("{:>5}".format(sum(row)))
columns = list(zip(*rows))
print("".join(map("{0:>5}".format, list(sum(column) for column in columns))), end="")
print("{0:>5}".format(sum(sum(row) for row in rows)))
``` | output | 1 | 98,515 | 23 | 197,031 |
Provide a correct Python 3 solution for this coding contest problem.
Your task is to develop a tiny little part of spreadsheet software.
Write a program which adds up columns and rows of given table as shown in the following figure:
<image>
Input
The input consists of several datasets. Each dataset consists of:
n (the size of row and column of the given table)
1st row of the table
2nd row of the table
:
:
nth row of the table
The input ends with a line consisting of a single 0.
Output
For each dataset, print the table with sums of rows and columns. Each item of the table should be aligned to the right with a margin for five digits. Please see the sample output for details.
Example
Input
4
52 96 15 20
86 22 35 45
45 78 54 36
16 86 74 55
4
52 96 15 20
86 22 35 45
45 78 54 36
16 86 74 55
0
Output
52 96 15 20 183
86 22 35 45 188
45 78 54 36 213
16 86 74 55 231
199 282 178 156 815
52 96 15 20 183
86 22 35 45 188
45 78 54 36 213
16 86 74 55 231
199 282 178 156 815 | instruction | 0 | 98,516 | 23 | 197,032 |
"Correct Solution:
```
from copy import deepcopy
def main():
while True:
n = int(input())
if n == 0:
break
tate = [0 for i in range(n)]
for i in range(n):
num = 0
m = map(int, input().split())
hoge = deepcopy(m)
tate = [x + y for (x, y) in zip(tate, hoge)]
for j in m:
print("{:5}".format(j), end='')
num += j
print("{:5}".format(num))
num = 0
for j in tate:
print("{:5}".format(j), end='')
num += j
print("{:5}".format(num))
if __name__ == '__main__':
main()
``` | output | 1 | 98,516 | 23 | 197,033 |
Provide a correct Python 3 solution for this coding contest problem.
Your task is to develop a tiny little part of spreadsheet software.
Write a program which adds up columns and rows of given table as shown in the following figure:
<image>
Input
The input consists of several datasets. Each dataset consists of:
n (the size of row and column of the given table)
1st row of the table
2nd row of the table
:
:
nth row of the table
The input ends with a line consisting of a single 0.
Output
For each dataset, print the table with sums of rows and columns. Each item of the table should be aligned to the right with a margin for five digits. Please see the sample output for details.
Example
Input
4
52 96 15 20
86 22 35 45
45 78 54 36
16 86 74 55
4
52 96 15 20
86 22 35 45
45 78 54 36
16 86 74 55
0
Output
52 96 15 20 183
86 22 35 45 188
45 78 54 36 213
16 86 74 55 231
199 282 178 156 815
52 96 15 20 183
86 22 35 45 188
45 78 54 36 213
16 86 74 55 231
199 282 178 156 815 | instruction | 0 | 98,517 | 23 | 197,034 |
"Correct Solution:
```
while True:
n = int(input())
if n == 0:
break
a = []
for i in range(n):
a.append([int(num) for num in input().split()])
for i in range(n):
sum = 0
for j in range(n):
sum += a[i][j]
a[i].append(sum)
a.append([])
for i in range(n + 1):
sum = 0
for j in range(n):
sum += a[j][i]
a[n].append(sum)
for i in range(n + 1):
for j in range(n + 1):
print('{0:>5}'.format(a[i][j]), end="")
print()
``` | output | 1 | 98,517 | 23 | 197,035 |
Provide a correct Python 3 solution for this coding contest problem.
Your task is to develop a tiny little part of spreadsheet software.
Write a program which adds up columns and rows of given table as shown in the following figure:
<image>
Input
The input consists of several datasets. Each dataset consists of:
n (the size of row and column of the given table)
1st row of the table
2nd row of the table
:
:
nth row of the table
The input ends with a line consisting of a single 0.
Output
For each dataset, print the table with sums of rows and columns. Each item of the table should be aligned to the right with a margin for five digits. Please see the sample output for details.
Example
Input
4
52 96 15 20
86 22 35 45
45 78 54 36
16 86 74 55
4
52 96 15 20
86 22 35 45
45 78 54 36
16 86 74 55
0
Output
52 96 15 20 183
86 22 35 45 188
45 78 54 36 213
16 86 74 55 231
199 282 178 156 815
52 96 15 20 183
86 22 35 45 188
45 78 54 36 213
16 86 74 55 231
199 282 178 156 815 | instruction | 0 | 98,518 | 23 | 197,036 |
"Correct Solution:
```
while True:
n=int(input())
if n==0:
break
i=0
mat=[]
#γͺγΉγγδΈγ€γ«
while i<n:
tmp_mat=list(map(int, input().split()))
mat.extend(tmp_mat)
i+=1
total=0
tmp_last=[0 for l in range(n+1)]
for x in range(len(mat)):
total+=mat[x]
tmp_last[x%n]+=mat[x]
print(str(mat[x]).rjust(5),end="")
if x%n==n-1:
tmp_last[n]+=total
print(str(total).rjust(5))
total=0
for y in range(n+1):
print(str(tmp_last[y]).rjust(5),end="")
print("")
``` | output | 1 | 98,518 | 23 | 197,037 |
Provide a correct Python 3 solution for this coding contest problem.
Your task is to develop a tiny little part of spreadsheet software.
Write a program which adds up columns and rows of given table as shown in the following figure:
<image>
Input
The input consists of several datasets. Each dataset consists of:
n (the size of row and column of the given table)
1st row of the table
2nd row of the table
:
:
nth row of the table
The input ends with a line consisting of a single 0.
Output
For each dataset, print the table with sums of rows and columns. Each item of the table should be aligned to the right with a margin for five digits. Please see the sample output for details.
Example
Input
4
52 96 15 20
86 22 35 45
45 78 54 36
16 86 74 55
4
52 96 15 20
86 22 35 45
45 78 54 36
16 86 74 55
0
Output
52 96 15 20 183
86 22 35 45 188
45 78 54 36 213
16 86 74 55 231
199 282 178 156 815
52 96 15 20 183
86 22 35 45 188
45 78 54 36 213
16 86 74 55 231
199 282 178 156 815 | instruction | 0 | 98,519 | 23 | 197,038 |
"Correct Solution:
```
def main():
while True:
N = int(input())
if N == 0:
break
A = [
list(map(int, input().split(" ")))
for _ in range(N)
]
tot = [sum(col[r] for col in A) for r in range(N)]
A.append(tot)
for col in A:
col.append(sum(col))
for col in A:
print("".join("{:5}".format(x) for x in col))
if __name__ == "__main__":
main()
``` | output | 1 | 98,519 | 23 | 197,039 |
Provide a correct Python 3 solution for this coding contest problem.
Your task is to develop a tiny little part of spreadsheet software.
Write a program which adds up columns and rows of given table as shown in the following figure:
<image>
Input
The input consists of several datasets. Each dataset consists of:
n (the size of row and column of the given table)
1st row of the table
2nd row of the table
:
:
nth row of the table
The input ends with a line consisting of a single 0.
Output
For each dataset, print the table with sums of rows and columns. Each item of the table should be aligned to the right with a margin for five digits. Please see the sample output for details.
Example
Input
4
52 96 15 20
86 22 35 45
45 78 54 36
16 86 74 55
4
52 96 15 20
86 22 35 45
45 78 54 36
16 86 74 55
0
Output
52 96 15 20 183
86 22 35 45 188
45 78 54 36 213
16 86 74 55 231
199 282 178 156 815
52 96 15 20 183
86 22 35 45 188
45 78 54 36 213
16 86 74 55 231
199 282 178 156 815 | instruction | 0 | 98,520 | 23 | 197,040 |
"Correct Solution:
```
def output_res(row):
print(''.join(map(lambda num: str(num).rjust(5), row)), str(sum(row)).rjust(5), sep='')
def main():
while True:
n = int(input())
if n == 0:
break
bottom_total = [0 for _ in range(n)]
for _ in range(n):
each_line = list(map(int, input().split(' ')))
output_res(each_line)
bottom_total = [x + y for x, y in zip(bottom_total, each_line)]
output_res(bottom_total)
if __name__ == '__main__':
main()
``` | output | 1 | 98,520 | 23 | 197,041 |
Provide a correct Python 3 solution for this coding contest problem.
Hint
* One grid may be filled more than once
* Even if it can be represented by one line segment such as '1', it may be represented by two or more line segments.
Sample Input 1
Formula for Sample Input 1.
Sample Input 2
Sample Input 2 formula. Another character may get inside the smallest rectangle that covers one character, such as the "-7" part.
Constraints
The input satisfies the following conditions.
* 1 β€ N β€ 150
* 0 β€ Xi1, Yi1, Xi2, Yi2 β€ 200 (1 β€ i β€ N)
* Xi1 = Xi2 or Yi1 = Yi2 (1 β€ i β€ N)
* Numbers appearing in the middle of the calculation and in the result fit in a signed 32-bit integer
Input
N
X11 Y11 X12 Y12
X21 Y21 X22 Y22
::
XN1 YN1 XN2 YN2
The first line is given one integer N that represents the number of line segments. Next, the line segment information is given in N lines. Four integers Xi1, Yi1, Xi2, Yi2 are given on the i-th line of the line segment information, separated by blanks. Represents the X and Y coordinates of the first endpoint of the line segment, and the X and Y coordinates of the second endpoint, respectively.
Output
Output the calculation result of each case on one line.
Examples
Input
4
1 1 1 5
3 3 5 3
4 2 4 4
7 1 7 5
Output
2
Input
23
1 1 3 1
3 1 3 3
3 3 1 3
1 3 1 5
1 5 3 5
5 2 7 2
7 2 7 4
7 4 5 4
5 4 5 6
5 6 7 6
11 4 13 4
13 0 15 0
15 0 15 4
18 2 18 2
21 5 23 5
21 5 21 7
21 7 23 7
23 7 23 9
23 9 21 9
24 0 26 0
24 0 24 4
24 4 26 4
26 0 26 4
Output
-328 | instruction | 0 | 98,546 | 23 | 197,092 |
"Correct Solution:
```
dic = {((0, 0), (0, 1), (0, 2), (0, 3), (0, 4),
(1, 0), (1, 4),
(2, 0), (2, 1), (2, 2), (2, 3), (2, 4)):"0",
((0, 0), (0, 1), (0, 2), (0, 3), (0, 4)):"1",
((0, 0), (0, 2), (0, 3), (0, 4),
(1, 0), (1, 2), (1, 4),
(2, 0), (2, 1), (2, 2), (2, 4)):"2",
((0, 0), (0, 2), (0, 4),
(1, 0), (1, 2), (1, 4),
(2, 0), (2, 1), (2, 2), (2, 3), (2, 4)):"3",
((0, 0), (0, 1), (0, 2),
(1, 2),
(2, 0), (2, 1), (2, 2), (2, 3), (2, 4)):"4",
((0, 0), (0, 1), (0, 2), (0, 4),
(1, 0), (1, 2), (1, 4),
(2, 0), (2, 2), (2, 3), (2, 4)):"5",
((0, 0), (0, 1), (0, 2), (0, 3), (0, 4),
(1, 0), (1, 2), (1, 4),
(2, 0), (2, 2), (2, 3), (2, 4)):"6",
((0, 0),
(1, 0),
(2, 0), (2, 1), (2, 2), (2, 3), (2, 4)):"7",
((0, 0), (0, 1), (0, 2), (0, 3), (0, 4),
(1, 0), (1, 2), (1, 4),
(2, 0), (2, 1), (2, 2), (2, 3), (2, 4)):"8",
((0, 0), (0, 1), (0, 2), (0, 4),
(1, 0), (1, 2), (1, 4),
(2, 0), (2, 1), (2, 2), (2, 3), (2, 4)):"9",
((0, 0),
(1, -1), (1, 0), (1, 1),
(2, 0)):"+",
((0, 0),
(1, 0),
(2, 0)):"-",
((0, 0), ):"*"}
n = int(input())
mp = [[0] * 201 for _ in range(201)]
for _ in range(n):
x1, y1, x2, y2 = map(int, input().split())
if x1 == x2:
y1, y2 = min(y1, y2), max(y1, y2)
for y in range(y1, y2 + 1):
mp[y][x1] = 1
elif y1 == y2:
x1, x2 = min(x1, x2), max(x1, x2)
for x in range(x1, x2 + 1):
mp[y1][x] = 1
checked = [[False] * 201 for _ in range(201)]
vec = ((1, 0), (0, -1), (-1, 0), (0, 1))
def search(x, y, ix, iy, stack):
for dx, dy in vec:
if not 0 <= y + dy < 201 or not 0 <= x + dx < 201:continue
if mp[y + dy][x + dx] == 1 and not checked[y + dy][x + dx]:
checked[y + dy][x + dx] = True
stack.append((x + dx - ix, y + dy - iy))
search(x + dx, y + dy, ix, iy, stack)
s = ""
for x in range(201):
for y in range(201):
if mp[y][x] == 1 and not checked[y][x]:
checked[y][x] = True
stack = [(0, 0)]
search(x, y, x, y, stack)
stack.sort()
s += dic[tuple(stack)]
print(eval(s))
``` | output | 1 | 98,546 | 23 | 197,093 |
Provide tags and a correct Python 3 solution for this coding contest problem.
While playing with geometric figures Alex has accidentally invented a concept of a n-th order rhombus in a cell grid.
A 1-st order rhombus is just a square 1 Γ 1 (i.e just a cell).
A n-th order rhombus for all n β₯ 2 one obtains from a n-1-th order rhombus adding all cells which have a common side with it to it (look at the picture to understand it better).
<image>
Alex asks you to compute the number of cells in a n-th order rhombus.
Input
The first and only input line contains integer n (1 β€ n β€ 100) β order of a rhombus whose numbers of cells should be computed.
Output
Print exactly one integer β the number of cells in a n-th order rhombus.
Examples
Input
1
Output
1
Input
2
Output
5
Input
3
Output
13
Note
Images of rhombus corresponding to the examples are given in the statement. | instruction | 0 | 98,677 | 23 | 197,354 |
Tags: dp, implementation, math
Correct Solution:
```
n = int(input())
if n == 1:
print(1)
else:
print( int((2*n-1)**2 - 4*(n*(n-1)/2) ))
``` | output | 1 | 98,677 | 23 | 197,355 |
Provide tags and a correct Python 3 solution for this coding contest problem.
While playing with geometric figures Alex has accidentally invented a concept of a n-th order rhombus in a cell grid.
A 1-st order rhombus is just a square 1 Γ 1 (i.e just a cell).
A n-th order rhombus for all n β₯ 2 one obtains from a n-1-th order rhombus adding all cells which have a common side with it to it (look at the picture to understand it better).
<image>
Alex asks you to compute the number of cells in a n-th order rhombus.
Input
The first and only input line contains integer n (1 β€ n β€ 100) β order of a rhombus whose numbers of cells should be computed.
Output
Print exactly one integer β the number of cells in a n-th order rhombus.
Examples
Input
1
Output
1
Input
2
Output
5
Input
3
Output
13
Note
Images of rhombus corresponding to the examples are given in the statement. | instruction | 0 | 98,678 | 23 | 197,356 |
Tags: dp, implementation, math
Correct Solution:
```
# your code goes here
import math
n=int(input())
sum=1
for i in range(2,n+1):
sum+=4*(i-1)
print(sum)
``` | output | 1 | 98,678 | 23 | 197,357 |
Provide tags and a correct Python 3 solution for this coding contest problem.
While playing with geometric figures Alex has accidentally invented a concept of a n-th order rhombus in a cell grid.
A 1-st order rhombus is just a square 1 Γ 1 (i.e just a cell).
A n-th order rhombus for all n β₯ 2 one obtains from a n-1-th order rhombus adding all cells which have a common side with it to it (look at the picture to understand it better).
<image>
Alex asks you to compute the number of cells in a n-th order rhombus.
Input
The first and only input line contains integer n (1 β€ n β€ 100) β order of a rhombus whose numbers of cells should be computed.
Output
Print exactly one integer β the number of cells in a n-th order rhombus.
Examples
Input
1
Output
1
Input
2
Output
5
Input
3
Output
13
Note
Images of rhombus corresponding to the examples are given in the statement. | instruction | 0 | 98,679 | 23 | 197,358 |
Tags: dp, implementation, math
Correct Solution:
```
def ans(n):
if n==1:
return 1
else:
return ((n-1)*4) + ans(n-1)
n = int(input())
print(ans(n))
``` | output | 1 | 98,679 | 23 | 197,359 |
Provide tags and a correct Python 3 solution for this coding contest problem.
While playing with geometric figures Alex has accidentally invented a concept of a n-th order rhombus in a cell grid.
A 1-st order rhombus is just a square 1 Γ 1 (i.e just a cell).
A n-th order rhombus for all n β₯ 2 one obtains from a n-1-th order rhombus adding all cells which have a common side with it to it (look at the picture to understand it better).
<image>
Alex asks you to compute the number of cells in a n-th order rhombus.
Input
The first and only input line contains integer n (1 β€ n β€ 100) β order of a rhombus whose numbers of cells should be computed.
Output
Print exactly one integer β the number of cells in a n-th order rhombus.
Examples
Input
1
Output
1
Input
2
Output
5
Input
3
Output
13
Note
Images of rhombus corresponding to the examples are given in the statement. | instruction | 0 | 98,680 | 23 | 197,360 |
Tags: dp, implementation, math
Correct Solution:
```
n = int(input())
print(2 * n**2 - 2 * n + 1)
################################################################
``` | output | 1 | 98,680 | 23 | 197,361 |
Provide tags and a correct Python 3 solution for this coding contest problem.
While playing with geometric figures Alex has accidentally invented a concept of a n-th order rhombus in a cell grid.
A 1-st order rhombus is just a square 1 Γ 1 (i.e just a cell).
A n-th order rhombus for all n β₯ 2 one obtains from a n-1-th order rhombus adding all cells which have a common side with it to it (look at the picture to understand it better).
<image>
Alex asks you to compute the number of cells in a n-th order rhombus.
Input
The first and only input line contains integer n (1 β€ n β€ 100) β order of a rhombus whose numbers of cells should be computed.
Output
Print exactly one integer β the number of cells in a n-th order rhombus.
Examples
Input
1
Output
1
Input
2
Output
5
Input
3
Output
13
Note
Images of rhombus corresponding to the examples are given in the statement. | instruction | 0 | 98,681 | 23 | 197,362 |
Tags: dp, implementation, math
Correct Solution:
```
n=int(input())
a=[]
l=1;
n=2*(n**2)-2*n+1
print(n)
``` | output | 1 | 98,681 | 23 | 197,363 |
Provide tags and a correct Python 3 solution for this coding contest problem.
While playing with geometric figures Alex has accidentally invented a concept of a n-th order rhombus in a cell grid.
A 1-st order rhombus is just a square 1 Γ 1 (i.e just a cell).
A n-th order rhombus for all n β₯ 2 one obtains from a n-1-th order rhombus adding all cells which have a common side with it to it (look at the picture to understand it better).
<image>
Alex asks you to compute the number of cells in a n-th order rhombus.
Input
The first and only input line contains integer n (1 β€ n β€ 100) β order of a rhombus whose numbers of cells should be computed.
Output
Print exactly one integer β the number of cells in a n-th order rhombus.
Examples
Input
1
Output
1
Input
2
Output
5
Input
3
Output
13
Note
Images of rhombus corresponding to the examples are given in the statement. | instruction | 0 | 98,682 | 23 | 197,364 |
Tags: dp, implementation, math
Correct Solution:
```
ar=[0 for i in range(0,101)]
def res():
ar[0]=1
s=1
for i in range(1,101):
ar[i]=ar[i-1]+4*s
s+=1
n=int(input())
res()
print(ar[n-1])
``` | output | 1 | 98,682 | 23 | 197,365 |
Provide tags and a correct Python 3 solution for this coding contest problem.
While playing with geometric figures Alex has accidentally invented a concept of a n-th order rhombus in a cell grid.
A 1-st order rhombus is just a square 1 Γ 1 (i.e just a cell).
A n-th order rhombus for all n β₯ 2 one obtains from a n-1-th order rhombus adding all cells which have a common side with it to it (look at the picture to understand it better).
<image>
Alex asks you to compute the number of cells in a n-th order rhombus.
Input
The first and only input line contains integer n (1 β€ n β€ 100) β order of a rhombus whose numbers of cells should be computed.
Output
Print exactly one integer β the number of cells in a n-th order rhombus.
Examples
Input
1
Output
1
Input
2
Output
5
Input
3
Output
13
Note
Images of rhombus corresponding to the examples are given in the statement. | instruction | 0 | 98,683 | 23 | 197,366 |
Tags: dp, implementation, math
Correct Solution:
```
def solve(n):
if n == 1: return 1
if n == 2: return 5
return solve(n-1) + 4*(n-1)
def main():
n = int(input())
print(solve(n))
main()
``` | output | 1 | 98,683 | 23 | 197,367 |
Provide tags and a correct Python 3 solution for this coding contest problem.
While playing with geometric figures Alex has accidentally invented a concept of a n-th order rhombus in a cell grid.
A 1-st order rhombus is just a square 1 Γ 1 (i.e just a cell).
A n-th order rhombus for all n β₯ 2 one obtains from a n-1-th order rhombus adding all cells which have a common side with it to it (look at the picture to understand it better).
<image>
Alex asks you to compute the number of cells in a n-th order rhombus.
Input
The first and only input line contains integer n (1 β€ n β€ 100) β order of a rhombus whose numbers of cells should be computed.
Output
Print exactly one integer β the number of cells in a n-th order rhombus.
Examples
Input
1
Output
1
Input
2
Output
5
Input
3
Output
13
Note
Images of rhombus corresponding to the examples are given in the statement. | instruction | 0 | 98,684 | 23 | 197,368 |
Tags: dp, implementation, math
Correct Solution:
```
n=int(input())
a=2*n*(n-1)+1
print(a)
``` | output | 1 | 98,684 | 23 | 197,369 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
While playing with geometric figures Alex has accidentally invented a concept of a n-th order rhombus in a cell grid.
A 1-st order rhombus is just a square 1 Γ 1 (i.e just a cell).
A n-th order rhombus for all n β₯ 2 one obtains from a n-1-th order rhombus adding all cells which have a common side with it to it (look at the picture to understand it better).
<image>
Alex asks you to compute the number of cells in a n-th order rhombus.
Input
The first and only input line contains integer n (1 β€ n β€ 100) β order of a rhombus whose numbers of cells should be computed.
Output
Print exactly one integer β the number of cells in a n-th order rhombus.
Examples
Input
1
Output
1
Input
2
Output
5
Input
3
Output
13
Note
Images of rhombus corresponding to the examples are given in the statement.
Submitted Solution:
```
n = int(input())
sum = 1
for i in range(1, n):
sum = sum + 4 * i
print(sum)
``` | instruction | 0 | 98,685 | 23 | 197,370 |
Yes | output | 1 | 98,685 | 23 | 197,371 |
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