K - *Restoring Painting

最新推荐文章于 2017-09-05 20:03:16 发布

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最新推荐文章于 2017-09-05 20:03:16 发布

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分类专栏：算法练习文章标签： ACM hpu 算法练习思维

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帮助Vasya恢复被盗的3x3方形画作，已知四个角落及中心点的部分数值，需填充1到n之间的整数，使得每个2x2区域的和相等。计算所有可能的不同组合。

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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. $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$

题解：给一个3x3的方格，四个角和中间的数未知，填上从1到n的数使满足四个角2x2的和相等，易知和与中间的数无关。故只需对一个角的数暴力

然后判断其他的是否合法；最后乘以n;注意爆仓

#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long LL;
int main() {
	int n,a,b,c,d;
	while(scanf("%d %d %d %d %d",&n,&a,&b,&c,&d)!=EOF) {
		LL res1=max(max(1,1-a+d),max(1-b+c,1-a-b+c+d));
		LL res2=min(min(n,n-a+d),min(n-b+c,n-a-b+c+d));
		printf("%lld\n",res2<res1?0:(res2-res1+1)*n);
	}
	return 0;
}
/*
	 x1+a+b			x1
	=x2+a+c		x2=x1+b-c
	=x3+b+d		x3=x1+a-d
	=x4+c+d		x4=x1+a+b-c-d
				1<= x <=n
*/