BestCoder-Round#33】的更多相关文章

主题链接:pid=4858">http://acm.hdu.edu.cn/showproblem.php?pid=4858 项目管理 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 760    Accepted Submission(s): 262 Problem Description 我们建造了一个大项目! 这个项目有n个节…
King's Game  Accepts: 249  Submissions: 671  Time Limit: 2000/1000 MS (Java/Others)  Memory Limit: 65536/65536 K (Java/Others) 问题描述 为了铭记历史,国王准备在阅兵的间隙玩约瑟夫游戏.它召来了 n(1\le n\le 5000)n(1≤n≤5000) 个士兵,逆时针围成一个圈,依次标号 1, 2, 3 ... n1,2,3...n. 第一轮第一个人从 11 开始报数,报…
King's Phone  Accepts: 310  Submissions: 2980  Time Limit: 2000/1000 MS (Java/Others)  Memory Limit: 65536/65536 K (Java/Others) 问题描述 阅兵式上,国王见到了很多新奇东西,包括一台安卓手机.他很快对手机的图形解锁产生了兴趣. 解锁界面是一个 3 \times 33×3 的正方形点阵,第一行的三个点标号 1, 2, 31,2,3,第二行的三个点标号 4, 5, 64,5…
Rikka with Phi  Accepts: 5  Submissions: 66  Time Limit: 16000/8000 MS (Java/Others)  Memory Limit: 131072/131072 K (Java/Others) Problem Description Rikka and Yuta are interested in Phi function (which is known as Euler's totient function). Yuta giv…
1.BestCoder Round #89 2.总结:4个题,只能做A.B,全都靠hack上分.. 01  HDU 5944   水 1.题意:一个字符串,求有多少组字符y,r,x的下标能组成等比数列. 2.总结:有个坑,y,r,x顺序组公比q>1,也可反着来x,r,y顺序组. #include<iostream> #include<cstring> #include<cmath> #include<queue> #include<algorit…
BestCoder Round #90 本次至少暴露出三个知识点爆炸.... A. zz题 按题意copy  Init函数 然后统计就ok B. 博弈 题  不懂  推了半天的SG.....  结果这个题.... C 数据结构题   我写了半个小时分块   然后发现     改的是颜色.... 我的天  炸炸炸 D. 没看懂题目要干啥.....  官方题解要搞死小圆…
BestCoder Round #7 Start Time : 2014-08-31 19:00:00    End Time : 2014-08-31 21:00:00Contest Type : Register Public   Contest Status : Ended Current Server Time : 2014-08-31 21:12:12 Solved Pro.ID Title Ratio(Accepted / Submitted)   1001 Little Pony…
Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 354    Accepted Submission(s): 100 Problem Description ZYB has a tree with N nodes,now he wants you to solve the numbers of nodes distanced no m…
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 175    Accepted Submission(s): 74 Problem Description ZYB has a premutation P,but he only remeber the reverse log of each prefix of the premutat…
题目传送门 /* 设一个b[]来保存每一个a[]的质因数的id,从后往前每一次更新质因数的id, 若没有,默认加0,nlogn复杂度: 我用暴力竟然水过去了:) */ #include <cstdio> #include <iostream> #include <cstring> #include <string> #include <algorithm> using namespace std; ; const int INF = 0x3f3f…