随笔分类 - 水题
摘要:这题用Java最方便了,直接用Java的正则表达式匹配功能~~~~~~~import java.util.*;import java.util.regex.Matcher;import java.util.regex.Pattern;public class Main { public static void main(String[] args) { Scanner cin = new Scanner(System.in); while(cin.hasNextInt()) { int N = cin.nextInt(); ...
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摘要:水题,K为64的时候单独处理,其余情况就很easy了。/* * hdu2116/win.cpp * Created on: 2012-11-2 * 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:平衡二叉树就可以,我是用的set。/* * hdu1908/win.cpp * Created on: 2012-11-2 * Author : ben */#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <ctime>#include <iostream>#include <algorithm>#include <queue>#include <set>#include <
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摘要:判断图是否连通,连通的话如果每个点的度为偶数就存在欧拉回路,否则就不存在。/* * hdu1878/win.cpp * Created on: 2012-9-7 * Author : ben */#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <ctime>#include <iostream>#include <algorithm>#include <queue>#include <s
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摘要:挺水的题。我的做法是用ans = N*M*K减去不和谐的数目。首先,对每一个paints shoes对,ans -= N,这是没有问题的。这个操作全执行完以后就能记录每个paints跟多少shoes和谐(用paintspair[i]表示)。然后对于每一个clothes paints对,ans -= paintspair[i],这样就避免了重复计数。/* * hdu4451/win.cpp * Created on: 2012-10-29 * Author : ben */#include <cstdio>#include <cstdlib>#include <cs
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摘要:刚看的时候还以为挺难的,试了一下最水的方法,居然过了。就是从N以后往下试,试到smith数停止就行,而且对于smith数的判断也可以简单处理,循环到sqrt(n)即可,唉,这种简单方法我这次现场赛咋没想到呢……/* * hdu1333/win.cpp * Created on: 2012-10-27 * Author : ben */#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <ctime>#include <iost
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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:明明贪心是正确的嘛,不知道为什么官方的解题报告说是状态压缩DP。/* * hdu2001/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&g
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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: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: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:水题,打素数表即可,都用不着筛法。#include <cstdio>#include <cstring>#include <cstdlib>#include <algorithm>#include <cmath>using namespace std;const int MAXI = 200;const int MAXV = 1100;int prime[MAXI], I;int facnum[MAXV];void init() { I = 0; memset(prime, 0, sizeof(prime)); prime[I++]
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摘要:水题,不过还是挺考查对STL的熟悉程度的,我用的set做,居然也用了好长时间。/* * hdu1263/win.cpp * Created on: 2011-10-13 * Author : ben */#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <ctime>#include <iostream>#include <algorithm>#include <queue>#include &
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摘要:现场赛中最水的题,发现这题时比赛已经进行了半个小时,可是我打了半个小时还没有搞定,看着旁边的队伍一个个地升起气球,更加紧张,后面还有一点小问题干脆让海峰写了。今天发现杭电加了现场赛的题目,20分钟就把这题给过了。看来赛场上的心理素质还得训练,不然正式比赛的时候难以发挥出最好的水平~/* * hdu4054/win.cpp * Created on: 2011-10-6 * Author : ben */#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#inc
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摘要:水题,理解题意后不难想到,只要对所有物品排个序,然后依次从小到大,三个三个地拿即可。/* * hdu1678/win.cpp * Created on: 2011-10-1 * Author : ben */#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <algorithm>#include <queue>using namespace std;void work();int main() {#ifndef ONL
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