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这题看完感觉就是DP,然后我当然就用记忆化搜索打了,打着打着,觉得想法不一定对,但打完交了一遍,MLE,改小数组再交,就过了,但其实还没有完全理解透。算是混过的。相当于DP的方法,dp[i][j]表示前i根筷子组成j组的最小值。我觉得关键的问题在于要先反着排序,然后如果筷子数刚好是人数的3倍时,最小的两根必定会在同一组中,且是一组中的较小的两根。/* * hdu1500/win.cpp * Created on: 2012-7-27 * Author : ben */#include <cstdio>#include <cstdlib>#include <cstr 阅读全文
posted @ 2012-07-27 23:11
moonbay
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这题几年前就看过,一直没看懂啊。。。今天翻解题报告,才知道就是求n/1、n/2、n/3、...、n/(n-1)、n/n的和,并且以分数的形式表示,还算简单。/* * hdu1099/win.cpp * Created on: 2012-7-27 * Author : ben */#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <ctime>#include <iostream>#include <algorit 阅读全文
posted @ 2012-07-27 13:34
moonbay
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就是找一个区间,其被覆盖次数最多,输出即可。。/* * hdu1050/win.cpp * Created on: 2012-7-27 * Author : ben */#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <ctime>#include <iostream>#include <algorithm>#include <queue>#include <set>#includ 阅读全文
posted @ 2012-07-27 11:10
moonbay
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水题,直接贴代码/* * hdu1009/win.cpp * Created on: 2012-7-27 * Author : ben */#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <ctime>#include <iostream>#include <algorithm>#include <queue>#include <set>#include <map>#i 阅读全文
posted @ 2012-07-27 10:31
moonbay
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