How many---hdu2609(最小表示)】的更多相关文章

Give you n ( n < 10000) necklaces ,the length of necklace will not large than 100,tell me How many kinds of necklaces total have.(if two necklaces can equal by rotating ,we say the two necklaces are some). For example 0110 express a necklace, you can…
#include <iostream> #include <algorithm> #include <string.h> #include <cstdio> #include <vector> using namespace std; ; struct elem { ]; int len; bool operator <(const elem &rhs)const{ ); } bool operator == (const elem…
题目链接:https://vjudge.net/problem/HDU-2609 How many Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3272    Accepted Submission(s): 1457 Problem Description Give you n ( n < 10000) necklaces ,the…
How many Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4210    Accepted Submission(s): 1936 Problem Description Give you n ( n < 10000) necklaces ,the length of necklace will not large than 10…
Give you n ( n < 10000) necklaces ,the length of necklace will not large than 100,tell meHow many kinds of necklaces total have.(if two necklaces can equal by rotating ,we say the two necklaces are some).For example 0110 express a necklace, you can r…
题意:有n个环形字符串,一个环形字符串移动会形成不能的字符串,我们把它们看作同一串字符串,求有多少个不同的字符串....... 思路:用最小表示发将一个环形串的最小字典序找出来,然后让这个环形串按照这个顺序来组成一个新的串,其他串都这样处理,然后去重,输出结果就是了....... #include<iostream> #include<stdio.h> #include<string.h> #include<algorithm> using namespac…
Give you n ( n < 10000) necklaces ,the length of necklace will not large than 100,tell me How many kinds of necklaces total have.(if two necklaces can equal by rotating ,we say the two necklaces are some). For example 0110 express a necklace, you can…
1391: [Ceoi2008]order Time Limit: 10 Sec  Memory Limit: 64 MBSubmit: 1509  Solved: 460[Submit][Status][Discuss] Description 有N个工作,M种机器,每种机器你可以租或者买过来. 每个工作包括若干道工序,每道工序需要某种机器来完成,你可以通过购买或租用机器来完成. 现在给出这些参数,求最大利润 Input 第一行给出 N,M(1<=N<=1200,1<=M<=12…
在讲解PriorityQueue之前,需要先熟悉一个有序数据结构:最小堆. 最小堆是一种经过排序的完全二叉树,其中任一非终端节点数值均不大于其左孩子和右孩子节点的值. 可以得出结论,如果一棵二叉树满足最小堆的要求,那么,堆顶(根节点)也就是整个序列的最小元素. 最小堆的例子如下图所示:…
#define WM_SHOWTASK (WM_USER + 1) void CTestDlg::OnSysCommand(UINT nID, LPARAM lParam) { if ((nID & 0xFFF0) == IDM_ABOUTBOX) { CAboutDlg dlgAbout; dlgAbout.DoModal(); } else { //增加关闭时最小化到托盘的函数 if (nID == SC_CLOSE) //SC_MINIMIZE { ToTray(); return; }…