UVA 11538Chess Queen

Question：
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 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 (M × N) board, where the values of M and
N came from the input. All output values will fit in 64-bit signed integer.
Sample Input
2 2
100 223
2300 1000
0 0
Sample Output
12
10907100
11514134000
题意大意：两个皇后放在一个n*m的矩形方阵中，两个皇后在同一水平，同一竖直，同一斜线上均会互相攻击，让你输出共有多少中方案使两个皇后互相攻击。
思路：水平的和竖直的情况非常容易考虑，但斜线就不容易想到了。我们可以先算非最长的对角线的请况，sum+=4*i*(i-1);1<=i

#include <iostream>#include <cstdio>#include <cmath>using namespace std;const int Max=1000005;long long  a[Max];int main(){   long long  n,m;    while(1)    {        long long sum=0;        scanf("%lld%lld",&n,&m);        if(n==0&&m==0)            break;        sum+=(n*(n-1)*m+m*(m-1)*n);     //这里算的是水平和竖直的情况        for(long long i=1;i<min(n,m);i++)        sum+=4*i*(i-1);        //这里算的是非最长对角线的请况        long long  t=abs(n-m),h=min(n,m);  //这里算的最长对角线的情况（当n!=m时，最长对角线不只有一条）        sum+=2*(t+1)*h*(h-1);        printf("%lld\n",sum);    }    return 0;}

体会：把问题细化，可能就会变得简单。