Uva 11538 - Chess Queen
http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2533
|
11538 - Chess Queen Time limit: 2.000 seconds |
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 (2 × 2) chess board. They can be in attacking positions in 12 ways, these are shown in the picture below:

Figure: in a (2 × 2) chessboard 2 queens can be in attacking position in 12 ways Given an (N × M) 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 (M × N) board, where the values of M and N came from the input. All output values will fit in 64-bit signed integer.
Sample Input
2 2
100 223
2300 1000
0 0
Sample Output
12
10907100
11514134000
分析;
公式: n*m*(m+n-2)+2*n*(n-1)*(3*m-n-1)/3 。
AC代码:
// UVa11538 Chess Queen
#include<iostream>
#include<algorithm>
using namespace std;
int main() {
unsigned long long n, m; // 最大可以保存2^64-1>1.8*10^19
while(cin >> n >> m) {
if(!n && !m) break;
if(n > m) swap(n, m); // 这样就避免了对n<=m和n>m两种情况分类讨论
cout << n*m*(m+n-)+*n*(n-)*(*m-n-)/ << endl;
}
return ;
}
吐槽一下:uva 上面看不了自己的代码,只能自己把每个代码都保存下来咯,,,Orz
Uva 11538 - Chess Queen的更多相关文章
- uva 11538 Chess Queen<计数>
链接:http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&am ...
- 组合数学 UVa 11538 Chess Queen
Problem A Chess Queen Input: Standard Input Output: Standard Output You probably know how the game o ...
- 【基本计数方法---加法原理和乘法原理】UVa 11538 - Chess Queen
题目链接 题意:给出m行n列的棋盘,当两皇后在同行同列或同对角线上时可以互相攻击,问共有多少种攻击方式. 分析:首先可以利用加法原理分情况讨论:①两皇后在同一行:②两皇后在同一列:③两皇后在同一对角线 ...
- 【组合计数】UVA - 11538 - Chess Queen
考虑把皇后放在同一横排或者统一纵列,答案为nm(m-1)和nm(n-1),显然. 考虑同一对角线的情况不妨设,n<=m,对角线从左到右依次为1,2,3,...,n-1,n,n,n,...,n(m ...
- UVa 11538 Chess Queen (排列组合计数)
题意:给定一个n*m的棋盘,那么问你放两个皇后相互攻击的方式有多少种. 析:皇后攻击,肯定是行,列和对角线,那么我们可以分别来求,行和列其实都差不多,n*A(m, 2) + m*A(n, 2), 这是 ...
- UVa11538 A Chess Queen
A Chess Queen Problem A Chess Queen Input: Standard Input Output: Standard Output You probably know ...
- 【计数原理】【UVA11538】 Chess Queen
传送门 Description 给你一个n*m的棋盘,在棋盘上放置一黑一白两个皇后,求两个皇后能够互相攻击的方案个数 Input 多组数据,每组数据包括: 一行,为n和m 输入结束标志为n=m=0. ...
- 【策略】UVa 278 - Chess
Chess Almost everyone knows the problem of putting eight queens on an chessboard such that no Quee ...
- UVA11538 - Chess Queen(数学组合)
题目链接:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem ...
随机推荐
- GIT 在本地保存账户和密码
原文链接:http://www.jianshu.com/p/908591004f3b 解决方法,在本地的工程文件夹的.git下打开config文件 添加: [credential] helper = ...
- 设计模式:职责链模式(Chain Of Responsibility)
定 义:使多个对象都有机会处理请求,从而避免请求的发送者和接受者之间的耦合关系.将这些对象连成一条链,并沿着这条链传递请求,直到有一个对象处理它为止. 结构图: 处理请求类: //抽象处理类 abs ...
- 设计模式:工厂方法模式(Factory Method)
定义:定义一个用于创建对象的接口,让子类决定实例化哪一个类. 工厂方法使一个类的实例化延迟到其子类. 结构图: 示例: HTML代码: <html xmlns="http://www. ...
- 转:ASP.NET MVC3 Model验证总结
http://www.wyjexplorer.cn/Post/2012/8/3/model-validation-in-aspnet-mvc3 ASP.NET MVC3中的Model是自验证的,这是通 ...
- Silverlight4-安装顺序(VS2010)
1.vs2010 2. Silverlight4_Tools 3.Silverlight_Developer 4.Microsoft Expression Blend Preview for Silv ...
- CS6破解
1) 序列号这里为大家生成了两个,可以通过软件验证:1325-0949-2080-9819-3777-32301325-0160-5283-9851-2671-8951 2) 破解补丁安装时会用到,请 ...
- 根据 字数 确定 UI控件高度
//字体 textLabel.font = [UIFont systemFontOfSize:13]; CGFloat labelWidth = [UIScreen mainScreen].bound ...
- 在bash shell中使用getfattr查看文件扩展属性
getfattr用法 用于获取文件扩展属性,返回一系列键值对,参考Linux Man Page. 常用OPTIONS -n name, --name=name Dump the value of th ...
- jquery在线手册
开发时用到jquery,有几个函数想不起来怎么用,找了一下jquery在线手册. 记录一下,下回有需要再看看. 链接:http://www.chenfahui.cn/jq/
- 修改seacherbar 取消按钮属性
继承UISearchBar. 重载setShowCancelButton. [super xxx]; ... for (UIView *subview in self.subview) { if ...