UVA11538 Chess Queen

最新推荐文章于 2019-02-15 06:16:35 发布

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最新推荐文章于 2019-02-15 06:16:35 发布

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原文链接：http://www.cnblogs.com/huibixiaoxing/p/8294739.html

本文探讨了一个n×m棋盘上黑白两皇后相互攻击的方式计数问题，给出了详细的数学推导过程，并提供了一段C++代码实现。

Chess Queen

题目大意：给定一个n × m的棋盘，问有多少种方法放置黑、白两皇后使其相互攻击

同一行/列：nm * (n + m - 2)

不妨令m > n

对角线：不难发现对角线长度为1,2,3...n-1,n,n,n,.....,n,(m - n + 1)个n,n - 1,....,3,2,1

对角线方案 = 2 * Σ(i = 1 to n - 1)i(i - 1) + (m - n + 1) * n * (n - 1)

Σ(i = 1 to n - 1)i(i - 1) = Σ(i = 1 to n - 1)i^2 - Σ(i = 1 to n - 1)i = (n - 1) * n * (2n - 1) / 6 - n (n - 1)/2

(Σ(i = 1 to n)i^2 = n * (n + 1) * (2 * n + 1) / 6)

带入有对角线方案 = 2n * (n - 1) * (3 * m - n - 1)

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <algorithm>
 6 #include <queue>
 7 #include <vector>
 8 #define min(a, b) ((a) < (b) ? (a) : (b))
 9 #define max(a, b) ((a) > (b) ? (a) : (b))
10 #define abs(a) ((a) < 0 ? (-1 * (a)) : (a))
11 inline void swap(long long &a, long long &b)
12 {
13     long long tmp = a;a = b;b = tmp;
14 }
15 inline void read(long long &x)
16 {
17     x = 0;char ch = getchar(), c = ch;
18     while(ch < '0' || ch > '9') c = ch, ch = getchar();
19     while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
20     if(c == '-') x = -x;
21 }
22 
23 const long long INF = 0x3f3f3f3f;
24 long long n,m; 
25 int main()
26 {
27     while(scanf("%lld %lld", &n, &m) != EOF && n * n + m * m)
28     {
29         if(n > m) swap(n, m);
30         printf("%lld\n", n * m * (n - 1) + n * m * (m - 1) + 2 * n * (n - 1) * (3 * m - n - 1) / 3);
31     }
32     return 0;
33 }