Uva 11538 Chess Queen

Description

 

 

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

 

思路：

在同一行：n*m*(m-1)

在同一列：m*n*(n-1)

对角线：2*(2*(A(2,2)+A(3,2)+A(4,2)+…+A(m-1,2))+A(m,2)*(n-m+1))

n小于m时交换，这样就避免分类

因为两条对角线对称，所以直接乘以2

对角线上面的个数：2,3,4,…m-1,m…m,m-1,…4,3,2.

其中m的个数为n-m+1个...。。

但是后来发现有直接计算对角线所有情况的公式：

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

n-m+1个m。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 2*(3n - m - 1)*m(m - 1)/3.

时间直接从一千多毫秒到30毫秒。还有题目的数据有些坑，m ,n应该不止题目给出的10^6。应该超出10^9。所以m,n要定义为long long 型的。

//这个是没有直接用计算对角线公式所有情况的。

#include <iostream>
#include <cmath>
#include <cstdio>
using namespace std;
int main()
{
    long long  n,m;
    while(cin>>n>>m && n && m)
    {
        if(n < m)
        {
            int temp = m;
            m = n;
            n = temp;
        }
        long long  total = 0;
        total = m*(m-1)*n + n*(n-1)*m;
        long long sum = 0;
        for(long long i = 2; i <= m-1; i++)
        {
            sum += i*(i-1);
        }
        sum = sum*2 + (n-m+1)*m*(m-1);
        sum = sum*2;
        total = total + sum;
        cout<<total<<endl;
    }
    return 0;
}

 //用对角线公式

#include <iostream>
#include <cmath>
#include <cstdio>
using namespace std;
int main()
{
    long long  n,m;
    while(cin>>n>>m && n && m)
    {

        if(n < m)
        {
            int temp = m;
            m = n;
            n = temp;
        }
        long long  total = 0;
        total = m*(m-1)*n + n*(n-1)*m + 2*(3*n - m - 1)*m*(m - 1)/3 ;
        cout<<total<<endl;

    }
    return 0;
}