UVa 11538

 

 

Problem A
Chess Queen 
Input: Standard Input

Output: Standard Output

You probably know how the game of chess is played and how chess queen operates. Two chess queens are in attacking position when they are on same row, column or diagonal of a chess board. Suppose two such chess queens (one black and the other white) are placed on (2x2) chess board. They can be in attacking positions in 12 ways, these are shown in the picture below:

Given an (NxM) board you will have to decide in how many ways 2 queens can be in attacking position in that.

Input

Input file can contain up to 5000 lines of inputs. Each line contains two non-negative integers which denote the value of M and N (0< M, N£10⁶) respectively.

Input is terminated by a line containing two zeroes. These two zeroes need not be processed.

Output

For each line of input produce one line of output. This line contains an integer which denotes in how many ways two queens can be in attacking position in  an (MxN) board, where the values of M and N came from the input. All output values will fit in 64-bit signed integer.

Sample Input                              Output for Sample Input

2 2 100 223 2300 1000 0 0

12 10907100 11514134000

思路：

计数原理，将同一横向同一纵向统一斜向的相加即可。

用到的数学公式

1^2+2^2+3^2+……n^2= n(n+1)(2n+1)/6

证明方法如下：

代码：

#include<iostream>
#include<string.h>
#include<stdio.h>
using namespace std;


int main()
{
    long long int n,m,n1,sum;
    //freopen("aa.txt","r",stdin);
    while(scanf("%lld%lld",&n,&m)!=EOF)
    {
        sum=n1=0;
        if(n==0&&m==0)
            break;
        sum+=m*n*(n-1+m-1);
        if(n>m)
        swap(n,m);
        n1+=((n-1)*n*(2*n-1)/6-n*(n-1)/2)*4;
        n1+=(m-n+1)*n*(n-1)*2;
        sum+=n1;
        printf("%lld
",sum);
    }
    return 0;
}