UVA 11538 - Chess Queen
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:
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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 |
Problemsetter: Shahriar Manzoor
Special Thanks to: Mohammad Mahmudur Rahman
m行n列的表格
1、 在同一行 互相攻击的 方案:n*m*(n-1)
2、 在同一列 互相攻击的 方案:n*m*(m-1)
3、在同一斜线 互相攻击的方案:以 m<n 为例 { sum{ for( i=1;i<=n-1;i++) i*(i-1) } *2 + [n-(m-1)]*m*(m-2) } *2 , 大于则交换 m,n
#include <iostream> #include <cstdio> #include <algorithm> using namespace std; unsigned long long n,m,sum; int main(){ while(cin>>m>>n && (m||n)){ sum=m*n*(n-1) + m*n*(m-1); if(n<m) swap(n,m); sum+=m*(m-1)*(m-2)/3*4; sum+=(n+1-m)*m*(m-1)*2; cout<<sum<<endl; } return 0; }