Codeforces Round #353 (Div. 2) B. Restoring Painting

Description

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.

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.

Sample Input

Input

2 1 1 1 2

Output

2

Input

3 3 1 2 3

Output

6

题意：如图给出a,b,c,d;现在向空的方格内填数，使每个2*2的矩形的和等于左上角2*2矩形,以及方格内最大能填的数。次题关键是找出几个等量关系。

另左上角为x1,右上角为x2,左下角为x3,右下角为x4.中间的数i（1->n)

则有：

x1+a+b+i=x2+a+c+i;

x1+a+b+i=b+d+x3+i;

x1+a+b+i=x4+c+d+1;

推出：

x2=x1+b-c;x3=x1+a-d;x4=x1+a+b-c-d;

然后x2,x3,x4都大于1小于n;

同时x1>1&&x1<n;

推导得出

 1<=x1+a-c<=n,所以 c-b+1<=x1<=c-b+n;

  1<=x1+a-d<=n,所以d-a+1<=x1<=d-a+n;

  1<=a+b+x1-c-d<=n,所以c+d-a-b+1<=x1<=c+d-a-b+n;

    所以x1最小可以取ans1=max(1,c-b+1,d-a+1,c+d-a-b+1);

              最大可以取ans2=min(n,c-b+n,d-a+n,c+d-a-b+n);

答案即为（ans2-ans1)*n;

代码如下：

#include<stdio.h>#include<algorithm>using namespace std;int min(int a,int b){if(a>b)return b;elsereturn a;}int max(int a,int b){if(a>b)return a;elsereturn b;}int main(){int n,a,b,c,d;int ans1,ans2;long long ans;while(scanf("%d",&n)!=EOF){scanf("%d%d%d%d",&a,&b,&c,&d);ans1=max(1,max(max(c-b+1,d-a+1),c+d-a-b+1));          ans2=min(n,min(min(c-b+n,d-a+n),c+d-a-b+n));                   if(ans2>=ans1)           ans=(long long)(ans2-ans1+1)*n;          else ans=0;           printf("%lld\n",ans);  }}