07 2012 档案
摘要:/* * hdu1869/win.cpp * Created on: 2012-7-31 * Author : ben */#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <ctime>#include <iostream>#include <algorithm>#include <queue>#include <set>#include <stack>#include
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摘要:按题目描述的数据量,暴力是不可能过的,居然过了,太假了。。。/* * hdu1238/win.cpp * Created on: 2012-7-31 * Author : ben */#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <ctime>#include <iostream>#include <algorithm>#include <queue>#include <set>
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摘要:数据量很少,暴搜就可以过了/* * hdu1181/win.cpp * Created on: 2012-7-30 * Author : ben */#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <ctime>#include <iostream>#include <algorithm>#include <queue>#include <set>#include <map&
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摘要:简单题,暴搜就可以过了。/* * hdu4152/win.cpp * Created on: 2012-7-28 * Author : ben */#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <ctime>#include <iostream>#include <algorithm>#include <queue>#include <set>#include <map&g
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摘要:比较赤裸的KMP,将主串延长一倍即可达到旋转匹配的效果。/* * hdu2203/win.cpp * Created on: 2012-7-28 * Author : ben */#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <algorithm>#include <iostream>#include <queue>using namespace std;const int MAX_PAR_LEN =
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摘要: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
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摘要: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
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摘要: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
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摘要: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
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摘要:维护一个单调栈,如果当前的高度小于栈顶的,则弹栈,直到当前的高度大于栈顶,将当前的入栈。要注意高度为0的情况,开始忘考虑了,WA了一次。/* * hdu1008/win.cpp * Created on: 2012-7-24 * Author : ben */#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <ctime>#include <iostream>#include <algorithm>#inc
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摘要:一看就可以用母函数做,不过好久没练DP了,所以还是用DP做了。用dp[i][j]表示前i种石头排出j个出来的种数,当考虑第i种石头石,枚举其使用的个数即可。WA了好几次,是整数相乘精度的问题,使用long long就过了。/* * hdu1004/win.cpp * Created on: 2012-7-24 * Author : ben */#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <ctime>#include <
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摘要:罗马数字转阿拉伯数字,还只限12以内的。。。还有更水的赛题么。。/* * hdu1012/win.cpp * Created on: 2012-7-24 * Author : ben */#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <ctime>#include <iostream>#include <algorithm>#include <queue>#include <set&g
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摘要:挺水的,边长肯定是最大的箱子的边长加上次大的箱子的边长/* * hdu1003/win.cpp * Created on: 2012-7-24 * Author : ben */#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <ctime>#include <iostream>#include <algorithm>#include <queue>#include <set>#in
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摘要:水题,不说什么了。/* * hdu1001/win.cpp * Created on: 2012-7-24 * Author : ben */#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <ctime>#include <iostream>#include <algorithm>#include <queue>#include <set>#include <map>#
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摘要:其实挺水的,就是找出相邻的点中斜率绝对值最大的。因为没细看题目,忘了是“绝对值”,WA了两次,杯具。。。/* * hdu1594/win.cpp * Created on: 2012-7-23 * Author : ben */#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <ctime>#include <iostream>#include <algorithm>#include <queue&g
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摘要:其实这题我不会,看了解题报告才做出来。突破点就是0068要找与elnil同角速度的同心圆。在同心圆上,2人的角速度是一样的;而在同心圆内,0068的角速度一直大于elnil的角速度,所以0068可以与elnil在大圆的一条直径上并且使距离达到最大程度。一旦0068到达同心圆上,便直线前进,以保证最快到岸,而此时elnil要走的距离依然是PI*R;/* * hdu1593/win.cpp * Created on: 2012-7-23 * Author : ben */#include <cstdio>#include <cstdlib>#include <cstr
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摘要:很简单的并查集,不过第一次交居然超内存,把height数组改为short int就过了~~/* * hdu1856/win.cpp * Created on: 2012-7-18 * Author : ben */#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <iostream>#include <algorithm>#include <string>using namespace std;const
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摘要:这题严格来说我不会,不过看完题第一反应就是直接看第一个数是如何变化的,然后我就打了个小函数,打完以后交上去准备继续测试找规律,结果一看居然直接就AC了。。。太假了。。#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>using namespace std;int getM(int N) { int i = 1; int ret = 1; while(i != N + 1) { ret++; if(i <= N) { i = i * 2...
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摘要:比较简单,直接贴代码/* * hdu2152/win.cpp * Created on: 2012-7-18 * Author : ben */#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <ctime>#include <iostream>#include <algorithm>#include <queue>#include <set>#include <map>
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摘要:不知道大牛们有没有简洁的方法,反正我是一步一步慢慢推公式做出来的。居然推了一下午啊,囧。。。真是老了不中用了。。。其实推完后再整理整理,发现还是挺简单的,其实就是几个数列而已数列①1, 2, 3, 4, 5, 6, 7, 8, 。。。数列②1, 3, 6, 10, 15, 21, 28, 36。。。数列③1, 4, 10, 20, 35, 56, 。。。数列④1, 5, 15, 35, 70数列⑤3, 10, 21, 36,。。。数列⑥1, 6, 15, 28, 。。。我主要是把问题进行分解,三角形个数=正放三角形个数+倒放三角形个数正放三角形个数=边长为1的正放三角形个数+边长为2的正放三角
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摘要:还是比较简单的,看题目的数据范围,貌似简单判断素数的方法(复杂度sqrt(p))也能过,不过手上有素数测试的模板,就直接用了。秒杀。。。/* * hdu1905/win.cpp * Created on: 2012-7-12 * Author : ben */#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <ctime>#include <iostream>#include <algorithm>#inc
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摘要:开始题目看岔了,以为还挺复杂的,打了半天才发现题目是说对2008^N求约数和再计算,不是对N,于是就很简单了。。。/* * hdu1852/win.cpp * Created on: 2012-7-10 * Author : ben */#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <ctime>#include <iostream>#include <algorithm>#include <qu
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摘要:主要就是推公式了,我推了老半天,囧啊。。。还要注意一点就是内切的情况,开始没考虑,WA了一次。/* * hdu1798/win.cpp * Created on: 2012-7-9 * Author : ben */#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <ctime>#include <iostream>#include <algorithm>#include <queue>#inc
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摘要:/* * hdu1787/win.cpp * Created on: 2012-7-9 * Author : ben */#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <ctime>#include <iostream>#include <algorithm>#include <queue>#include <set>#include <map>#include &l
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摘要:这题不错,开始看题目,以为是道水题,花几分钟打了个暴搜,果断runtime error,栈溢出。于是马上找规律,发现可以转换成矩阵乘法和快速幂解决,不过由于审题不清,搞错了a0...a9的顺序,调了老半天才发现。。。。/* * hdu1757/win.cpp * Created on: 2012-7-9 * Author : ben */#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <algorithm>#include &l
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摘要:贪心就可以了,题目公式排版混乱,半天才读懂题。。。/* * hdu1725/win.cpp * Created on: 2012-7-6 * Author : ben */#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <ctime>#include <iostream>#include <algorithm>#include <queue>#include <set>#inclu
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摘要:其实是很简单的题,用定积分得到公式就行了,不过大学都毕业了我居然还会求定积分,忍不住想自夸一下^_^。。。/* * hdu1724/win.cpp * Created on: 2012-7-6 * Author : ben */#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <ctime>#include <iostream>#include <algorithm>#include <queue&g
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摘要:赤裸裸的,直接贴代码。。不过注意题目要求X非负,所以最后还要再处理一下。。/* * hdu2669/win.cpp * Created on: 2012-7-6 * Author : ben */#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <ctime>#include <iostream>#include <algorithm>#include <queue>#include <s
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