Codeforces Round #353 (Div. 2) B. Restoring Painting_vasya works as a watchman in the gallery-优快云博客

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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.

   Examples 
 
     input 
   
2 1 1 1 2

     output 
   
2

     input 
   
3 3 1 2 3

     output 
   
6

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

题意：给一个3*3的矩阵

$\text{[math]}$ 题意：矩阵中可以填1-n共n个数字，已知a,b,c,d的值，要求每个2*2的矩阵数值之和都和左上角2*2矩阵相同，问一共有多少种填法

思路：一共有4个2*2的矩阵，且都包含最中间的那个方框，所以它可以放1-n任意一个数，可以不管它；

假设左上角放的是x，根据可以得到三个方程；

1<=x+a-c<=n,所以 c-b+1<=x<=c-b+n;

1<=x+a-d<=n,所以d-a+1<=x<=d-a+n;

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

同时1<=x<=n;

所以x最小可以取cnt1=max(1,c-b+1,d-a+1,c+d-a-b+1);

最大可以取cnt2=min(n,c-b+n,d-a+n,c+d-a-b+n);

如果cnt2<cnt1,没有可行解，答案是0；否则ans=(cnt2-cnt1+1)*n;

要注意的是ans可能会很大，要用long long；

贴上AC代码：

#include<stdio.h>
#include<algorithm>
using namespace std;
int main()
{
    int n,a,b,c,d,i,cnt1,cnt2;long long ans;
    scanf("%d%d%d%d%d",&n,&a,&b,&c,&d);
    cnt1=max(1,max(max(c-b+1,d-a+1),c+d-a-b+1));
    cnt2=min(n,min(min(c-b+n,d-a+n),c+d-a-b+n));
    if(cnt2>=cnt1)
    ans=(long long)(cnt2-cnt1+1)*n;
    else ans=0;
    printf("%lld\n",ans);
}