UVa11538

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 (N × M) board you will have to decide in how many ways 2 queens can be in attacking
position in that.
Input
Input ?le 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 (M × N) board, where the values of M and
N came from the input. All output values will ?t in 64-bit signed integer.
Sample Input
2 2
100 223
2300 1000
0 0
Sample Output
12
10907100
11514134000

题意：

       给一块nxm的国际象棋棋盘，在上面摆放两个不同色皇后，问使得这两个皇后可以相互攻击的摆放方案数。

分析：

       分类讨论，设A(n,m)是同一行放置两个皇后的方案数。那么A(n,m)=n*m*(m-1)。

设B(n,m) 是同一列放置两个皇后的方案数，B(n,m)=A(m,n)=n*m*(n-1)。D(n,m)是同一对角线放置两个皇后的方案数。不妨设n<=m，所有/向的对角线从左至右依次长度为1、2、…、n-1、n、n、…(一共m-n+1个n)、n-1、n-2、…、1。所以D(n,m)=2(2sigma(i from 1 to n-1)i(i-1)+(m-n+1)n(n-1))。总方案数为A(n,m)+B(n,m)+D(n,m)。

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>                //用于cin/cout，因为这样可以与平台无关的读写64bit整数，更为方便 
 4 #include <algorithm>            //为了使用swap 
 5 #define ULL unsigned long long         //最大可以保存到2的64次方减1，如果是long long则是到2的63次方减1 
 6 using namespace std; 
 7 
 8 int main()
 9 {
10     ULL n, m;
11     while(cin >> n >> m)
12     {
13         if(!n && !m) break;
14         if(n > m) swap(n, m);        //避免分情况讨论 
15         cout << n * m * (m + n - 2) + 2 * n * (n-1) *(3 * m - n - 1) / 3 << endl;
16     }
17     return 0;
18 }