UVa - 11538 - Chess Queen
先上题目
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
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2 2 100 223 2300 1000 0 0 |
12 10907100 11514134000 |
图片可能看不到，可以去UVa上面找这题看一下。
题意就是在一个n*m的棋盘里面放两只王后一黑一白，求出她们会互相攻击的摆法的数目。
这里先分析她们的互相攻击关系有3种：同一行(a)，同一列(b)，对角线(c)。然后对她们进行分类计算。
求出a=n*m*(m-1);
b=m*n*(n-1);
c=2*(2*n*(n-1)*(3*m-n-1))/3;
其中c是对角线的情况。这里需要用推出公式，由Ni=1,2,3···，(n-m+1)个n，···3,2,1可得出其中一只王后在棋盘上的每一条斜线上面方法，而另一只王后则在第i种情况下有(Ni)-1种情况ΣNi*(Ni-1)。同时有两个方向的斜线要考虑，所以要乘以2。
这里还要需要注意乘法溢出的问题，最好使用unsigned long long。由于输入输出的麻烦，以后做排列组合的问题最后用cin和cout来完成。
还有一个问题需要注意的就是这里要假设n<=m因为c的计算公式是在这个条件下进行的。所以在计算前要判断一下两者大小，并且确定是否要交换。
上代码
1 #include <stdio.h> 2 #include <string.h> 3 #include <iostream> 4 5 using namespace std; 6 7 int main() 8 { 9 unsigned long long n,m; 10 while((cin>>n>>m),(n || m)) 11 { 12 unsigned long long a,b,c,d,sum; 13 if(n>m) 14 { 15 n=n^m; 16 m=n^m; 17 n=n^m; 18 } 19 a=n*m*(m-1); 20 b=m*n*(n-1); 21 c=2*n*(n-1); 22 d=(3*m-n-1); 23 if(c%3) d/=3; 24 else c/=3; 25 c=c*d; 26 sum=a+b+c; 27 cout<<a+b+c<<endl; 28 29 } 30 return 0; 31 } 32 33 34 /* 35 36 2 2 37 100 223 38 2300 1000 39 0 0 40 41 42 12 43 10907100 44 11514134000 45 46 */
