UVa11538 A Chess Queen

A Chess Queen

 

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:

 

 

 

Figure: in a (2x2) chessboard 2 queens can be in attacking position in 12 ways

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

 

Problemsetter: Shahriar Manzoor

Special Thanks to: Mohammad Mahmudur Rahman

 

 

 

 

/*
分三种情况：
同行 
同列
同斜
*/

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<queue>
#include<set>
#include<vector>
#include<bitset>
using namespace std;
typedef long long ll;

const int M=1000010;

int get(){
    char c;
    int res=0;
    while(c=getchar(),!isdigit(c));
    do{
        res=(res<<3)+(res<<1)+(c-'0');
    }while(c=getchar(),isdigit(c));
    return res;
}




int main()
{
    ll ans1,ans2,ans3,ans,n,m;
    while(~scanf("%lld%lld",&n,&m))
    {
        if(n==0&m==0)break;
        if(n>m)swap(n,m);
        ans1=n*m*(m-1);//同行 
        ans2=n*m*(n-1);//同列
        ans3=2*n*(n-1)*(3*m-n-1)/3;//同斜
        ans=ans1+ans2+ans3;
        printf("%lld\n",ans);         
    }
    return 0;
}