ACM: uva 11538 - Chess Queen

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 (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 andN (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.

 

SampleInput                             

2 2

100 223

2300 1000

0 0

Output for Sample Input

12

10907100

11514134000

 

题意: chess中的皇后问题, 在一个n*m的范围内, 两个皇后能够相互攻击的摆放方式.

 

解题思路:

      1. 计数问题, 有三种相对摆放方式: 水平, 竖直, 对角线. 根据加法原理即可, 并且没有交集.

         水平和竖直是一样的, 只要n*m矩形旋转90度. 所以结果是: n*m*(m-1)+n*m*(n-1);

      2. 对角线复杂些, 先来确定对角线的长度: 1,2,3,...,n-2,n-1,n,n,n,...,n,n,n-1,n-2,...,2,1;

         其中n的个数是m-n+1 (其中假设m>n);

         结果: 2*(2*∑i*(i-1) + (m-n+1)*n*(n-1))  其中累加的范围是(1<=i<=n-1);

         化简得: 2*n*(n-1)*(3*m-n-1)/3

      3. 综上所述: n*m*(n+m-2)+2*n*(n-1)*(3*m-n-1)/3

 

代码:

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;

ll n, m;

int main()
{
 while(scanf("%lld %lld", &n, &m) != EOF)
 {
  if(n == 0 && m == 0) break;

  if(n > m) swap(n, m);

  printf("%lld\n", n*m*(n+m-2)+2*n*(n-1)*(3*m-n-1)/3);
 }

 return 0;
}