CodeForces 675B Restoring Painting

G - Restoring Painting
Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d
& %I64u
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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： 给出一个3*3的方格，方格内其中有4个数确定，给出数的范围1到n, 问有多少种填法方使方格内所有的2*2方格的和等于左上角的2*2的方格的和。

思路： 列一下方程你会发现，只要确定左上角的数字就行,最后再乘以n,因为中间那个数可以是1到n的任何数。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
int a,b,c,d,i,n;
scanf("%d%d%d%d%d",&n,&a,&b,&c,&d);
__int64 ans=0;
for(i=1;i<=n;i++)
{
int a1,a2,a3;
a1=b-c+i;
a2=a-d+i;
a3=a+b-c-d+i;
if(a1>=1&&a1<=n&&a2>=1&&a2<=n&&a3>=1&&a3<=n)
ans++;
}
printf("%I64d\n",ans*n);
return 0;
}