Codefroces 675 B. Restoring Painting(数学推公式)

Posted 2020-07-16 ITAK

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B. Restoring Painting

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

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.

In the second sample, only paintings displayed below satisfy all the rules.

题目大意：

就是给定一个3*3的矩阵，然后保证每一个2*2的矩阵的和是相等的，有四个数是已知的，如上图，给定一个数n（n<10^5）每个数都是<=n的同时>=1的已知的数 a, b, c, d，然后让你求的是有多少种方法可以让每一个2*2的矩阵的和是相等的。

解题思路：
其实我们现在先将剩余位置的方格中，填入相应的数字（未知数用字母代替就行）

问号那个地方从左往右，从上往下分别是e, f, g, h, i，那么我们可以列出方程：

b+e =  f+c

b+h = c+i

a+e = h+d

a+f =  d+i

通过上述方程我们可以发现，g（也就是中间那个数）填什么都可以，所以我们求出方法数之后只需要乘以n就行了。我们可以通过枚举一个未知数e，e从1到n那么剩下的未知数也可以用e和已知的数来表示了，我们只需要判断这些数是不是在1-n这个范围内就行了最后乘以n（最后会爆int）

上代码：

#include <iostream>

using namespace std;
typedef long long LL;
int main()
{
    LL a, b, c, d, n, e, f, i, h;
    while(cin>>n>>a>>b>>c>>d)
    {
        LL sum = 0;
        for(LL e=1; e<=n; e++)
        {
            f = b+e-c;
            h = a+e-d;
            i = b-c+h;
            if( (h>=1&&h<=n) && (f>=1&&f<=n) && (i>=1&&i<=n) )
                sum++;
        }
        cout<<sum*n<<endl;
    }
    return 0;
}