LightOJ1010---Knights in Chessboard (规律题)
来源:互联网 发布:域名国外备案 编辑:程序博客网 时间:2024/05/16 01:10
Given an m x n chessboard where you want to place chess knights. You have to find the number of maximum knights that can be placed in the chessboard such that no two knights attack each other.
Those who are not familiar with chess knights, note that a chess knight can attack 8 positions in the board as shown in the picture below.
Input
Input starts with an integer T (≤ 41000), denoting the number of test cases.
Each case contains two integers m, n (1 ≤ m, n ≤ 200). Here m and n corresponds to the number of rows and the number of columns of the board respectively.
Output
For each case, print the case number and maximum number of knights that can be placed in the board considering the above restrictions.
Sample Input
Output for Sample Input
3
8 8
3 7
4 10
Case 1: 32
Case 2: 11
Case 3: 20
Problem Setter: Jane Alam Jan
规律题
如果
答案是
如果n和m都是奇数且都大于2
答案就是一半的行放
如果n m 都小于等于2
n(m)为1,答案就是m(n)
m(n)有一个为2,那么我们可以考虑可以分出多少个2*2的格子
设有x个,答案是
剩下来可能有一个2*1的,这时如果x是奇数,不能放,如果x是偶数可以放
/************************************************************************* > File Name: LightOJ1010.cpp > Author: ALex > Mail: zchao1995@gmail.com > Created Time: 2015年06月08日 星期一 14时24分24秒 ************************************************************************/#include <functional>#include <algorithm>#include <iostream>#include <fstream>#include <cstring>#include <cstdio>#include <cmath>#include <cstdlib>#include <queue>#include <stack>#include <map>#include <bitset>#include <set>#include <vector>using namespace std;const double pi = acos(-1.0);const int inf = 0x3f3f3f3f;const double eps = 1e-15;typedef long long LL;typedef pair <int, int> PLL;int main() { int t, icase = 1; scanf("%d", &t); while (t--) { int n, m; scanf("%d%d", &n, &m); int ans = 0; if (n >= 3 && m >= 3) { if (n * m % 2 == 0) { ans = n * m / 2; } else { int s = m / 2 + 1; ans += (n / 2 + 1) * s; ans += (n / 2) * (s - 1); } } else { if (n == 1) { ans = m; } else if (m == 1) { ans = n; } else { if (m == 2) { swap(n, m); } if (m <= 3) { ans = 4; } else { int cnt = m / 2; ans = (cnt / 2 + cnt % 2) * 4; if (m % 2 && cnt % 2 == 0) { ans += 2; } } } } printf("Case %d: %d\n", icase++, ans); } return 0;}
- LightOJ1010---Knights in Chessboard (规律题)
- lightoj1010 Knights in Chessboard(找规律)
- Lightoj1010——Knights in Chessboard(找规律)
- light1010 - Knights in Chessboard【找规律】
- 1010 - Knights in Chessboard(找规律)
- lightoj 1010-Knights in Chessboard (规律)
- LightOJ-1010-Knights in Chessboard [规律]
- LightOJ 1010 Knights in Chessboard(数学规律)
- lightoj 1010 - Knights in Chessboard (找规律思维)
- lightoj 1010 - Knights in Chessboard(找规律)
- Knights in Chessboard
- lightoj-1010-Knights in Chessboard
- LightOJ 1010 Knights in Chessboard
- lightoj1010【规律】
- Light oj 1010 - Knights in Chessboard
- lightoj 1010 - Knights in Chessboard 【数学思维】
- LightOJ 1010 Knights in Chessboard <贪心思维>
- Light OJ 1010 - Knights in Chessboard【思维】
- struts2使用session 的三种方式
- 32位和64位系统区别及int字节数
- cgi php-cgi,PHP底层原理
- opencv之莫名其妙的条件宏ICV_DEF_FIND_STUMP_THRESHOLD_SQ解释~
- Ubuntu Server下建立VPN服务器的方法
- LightOJ1010---Knights in Chessboard (规律题)
- C# 判断系统空闲(键盘、鼠标不操作一段时间)
- 引用的意义与本质
- jQuery表单验证插件jQueryValidationEngine使用
- Access violation reading location 0xD15965C4
- 第十四周项目4(1)-处理C++源代码的程序
- 在web.xml中为Struts2配置拦截器
- [Java]多线程之生产者消费者优化版
- json数据生成xml文档