2014
02-28

# Uva 11538-Chess Queen[组合数学]

## Chess Queen

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 (2×2) chess board. They can be in attacking positions in 12 ways, these are shown in the picture below:

 Figure: in a (2×2) 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 2100 223 2300 1000 0 0 1210907100 11514134000

【题意】

1.在一行或一列：n*m（m-1），m*n*（n-1）

2.在对角线，假设n<m,则各对角线长度：1，2，3……n-1,n,n,……n,n-1,n-2,……1.

n-m+1个n。假设长度为l，则该对角线的方案数：l*(l-1)。将所有的加起来即可。

//============================================================================
// Name        : uva-11538.cpp
// Author      : coder
// Version     :
// Description : Hello World in C++, Ansi-style
//============================================================================
#include <iostream>
#include <stdio.h>
using namespace std;

typedef long long ll;
ll ans,m,n;
int main() {
//freopen("in.txt", "r", stdin);
while(scanf("%lld%lld",&m,&n) && m && n){
if(m > n) swap(m,n);
ans = 0;
ans += m*(m-1)*n + n*(n-1)*m; //横向和竖向
ll x=0;//x为斜向
//此处可用公式优化。
for(ll i=2; i<m; i++){
x += i*(i-1)*2;
}
x += (n-m+1) * m * (m-1);
ans += x*2;//两个方向
printf("%lld\n",ans);
}
return 0;
}

#include<iostream>
#include<cstdio>
using namespace std;

int main()
{
long long int n,m;
while(scanf("%lld%lld",&n,&m)!=EOF&&n&&m)
{
if(n>m)
swap(n,m);
long long int ans=(n-1)*n*m  +   (m-1)*m*n  +   2*n*(n-1)*(m-n+1);
ans+=2*2*(n-2)*(n-1)*(n)/3;
printf("%lld\n",ans);
}
return 0;
}

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