kiki's game Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 40000/1000 K (Java/Others)Total Submission(s): 5094    Accepted Submission(s): 2985 Problem Description Recently kiki has nothing to do. While she is bored, an idea appears in his mi…
A Simple Nim 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5795 Description Two players take turns picking candies from n heaps,the player who picks the last one will win the game.On each turn they can pick any number of candies which come from the…
Permutation Bo 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5753 Description There are two sequences h1∼hn and c1∼cn. h1∼hn is a permutation of 1∼n. particularly, h0=hn+1=0. We define the expression [condition] is 1 when condition is True,is 0 whe…
SG打表找规律 HDU 5795 题目连接 #include<iostream> #include<cstdio> #include<cmath> #include<algorithm> #include<cstring> using namespace std; #define MAXN 10000 int sg[MAXN],visit[MAXN]; int getsg(int n) { int i,j; ) return sg[n]; mem…
题目 //1,1,2,3,5,8,13,21,34,55…… //斐波纳契数列 #include<math.h> #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; int main() { ]; int len,i,a,b,c; while(gets(str)) { len = strlen(str); a=; b=c=; ;i<len;i++)…
邂逅明下 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1432    Accepted Submission(s): 670 Problem Description 当日遇到月,于是有了明.当我遇到了你,便成了侣.那天,日月相会,我见到了你.而且,大地失去了光辉,你我是否成侣?这注定是个凄美的故事.(以上是废话)小t和所有世俗的人们一…
Exclusive or 题目链接: http://acm.hust.edu.cn/vjudge/contest/121336#problem/J Description Given n, find the value of Note: ♁ denotes bitwise exclusive-or. Input The input consists of several tests. For each tests: A single integer n (2≤n<10^500). Output…
Couple doubi 题目链接: http://acm.hust.edu.cn/vjudge/contest/121334#problem/D Description DouBiXp has a girlfriend named DouBiNan.One day they felt very boring and decided to play some games. The rule of this game is as following. There are k balls on th…
看了解题报告,发现看不懂 QAQ 比较简单的解释是这样的: 可以先暴力下达标,然后会发现当前数 和 上一个数 的差值是一个 固定值, 而且等于当前数与i(第i个数)的商, 于是没有规律的部分暴力解决,有规律的套公式 //#pragma comment(linker, "/STACK:16777216") //for c++ Compiler #include <stdio.h> #include <iostream> #include <cstring&g…
题意: 给出N个小时,分配这些小时去写若干份论文,若用1小时写一份论文,该论文会被引用A次,新写一篇论文的话,全面的论文会被新论文引用一次. 找最大的H,H是指存在H遍论文,而且这些论文各被引用大于H次. 思路: 第一步,我们要想怎么样分配这些时间,毫无疑问,当N个小时写N份论文时为最优方案. 证明如下:  如果一开始的A>数量的话,就一直写新的:(没疑问吧!!!,如果A>数量,你还写旧的不是浪费时间吗???)  然后我们就会到一个分水岭 到达 A=数量(这时还是可以接受的,因为这里的A也算进…