Codeforces Round #353 (Div. 2) B. Restoring Painting(方块内放数,问有多少种不同的放法)_vasya works as a watchman in the gallery. unfortun-CSDN博客

本文链接：https://blog.csdn.net/acm_fighting/article/details/51437269

探讨一种算法：给定3x3绘画的部分数值，寻找满足特定条件的绘画组合总数。通过枚举和计算验证，实现快速找出所有可能方案。

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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

题意:有一个3*3的正方形,你已经知道了a[1][2],a[2][1],a[2][3],a[3][2]的值,每个方块中可以放置1-n的值,然后使得四个2*2的正方形的和都相同,问有多少种不同的放法。

思路:

我们可以枚举其中的一个方块的值,然后可以求出另外方块的值,判断这样是否合法,

又因为中间方块的值是任意1-n,所以只要把上一步算出的合法数*n便是答案。

By zidaoziyan, contest: Codeforces Round #353 (Div. 2), problem: (B) Restoring Painting, Accepted, #
 #include<bits/stdc++.h>
using namespace std;

int main(){
    int n,a,b,c,d;
    scanf("%d%d%d%d%d",&n,&a,&b,&c,&d);
    int ans=0;
    for(int i=1;i<=n;i++){
        int x4=a+i-d;
        int x5=x4+b-c;
        int x2=d+x5-a;
        if(x4<=0||x4>n||x5<=0||x5>n||x2<=0||x2>n)
            continue;
        ans++;
    }
    printf("%lld\n",1LL*ans*n);
}