Restoring Painting

最新推荐文章于 2021-03-12 20:03:27 发布

转载最新推荐文章于 2021-03-12 20:03:27 发布 · 335 阅读

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本文介绍了一个编程挑战，即根据已知条件恢复一幅3x3的画作，确保每个2x2方格内的数字之和等于左上角2x2方格的总和。通过枚举左上角的数字并检查其它角落数字的有效性来解决问题。

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Restoring Painting

CodeForces - 675B

Vasya works as a watchman in the gallery. Unfortunately, one of the most expensive paintings was stolen while he was on duty. He doesn't want to be fired, so he has to quickly restore the painting. He remembers some facts about it.

The painting is a square 3 × 3, each cell contains a single integer from 1 to n, and different cells may contain either different or equal integers.
The sum of integers in each of four squares 2 × 2 is equal to the sum of integers in the top left square 2 × 2.
Four elements a, b, c and d are known and are located as shown on the picture below.

$\text{[math]}$

Help Vasya find out the number of distinct squares the satisfy all the conditions above. Note, that this number may be equal to 0, meaning Vasya remembers something wrong.

Two squares are considered to be different, if there exists a cell that contains two different integers in different squares.

Input

The first line of the input contains five integers n, a, b, c and d (1 ≤ n ≤ 100 000, 1 ≤ a, b, c, d ≤ n) — maximum possible value of an integer in the cell and four integers that Vasya remembers.

Output

Print one integer — the number of distinct valid squares.

Example

Input

2 1 1 1 2

Output

Input

3 3 1 2 3

Output

Note

Below are all the possible paintings for the first sample. $\text{[math]}$ $\text{[math]}$

In the second sample, only paintings displayed below satisfy all the rules. $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$

思路：观察发现中间那个数无论是几都无所谓，四四方格若相等中间的数都会被抵消。暴力枚举最左上的数，则另三个角的数是固定的，判断这三个数是否符合条件。

#include<stdio.h>
int main()
{
	int a,b,c,d,x,y,z,i;
	long long ans,n;
	while(~scanf("%lld%d%d%d%d",&n,&a,&b,&c,&d))
	{
		ans=0;
		for(i=1;i<=n;i++)
		{
			x=b+i-c;
			y=i+a-d;
			z=b+i+a-c-d;
			if(x>=1&&x<=n&&y>=1&&y<=n&&z>=1&&z<=n)
			ans++;
		}
		printf("%lld\n",ans*n);
		
	}
	return 0;
}