【BZOJ】【3850】ZCC Loves Codefires】的更多相关文章

ZCC Loves Codefires Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 790    Accepted Submission(s): 420 Problem Description Though ZCC has many Fans, ZCC himself is a crazy Fan of a coder, called…
http://acm.hdu.edu.cn/showproblem.php?pid=4882 ZCC Loves Codefires Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 97    Accepted Submission(s): 55 Problem Description Though ZCC has many Fans,…
ZCC Loves Codefires 题目链接: http://acm.hust.edu.cn/vjudge/contest/121349#problem/B Description Though ZCC has many Fans, ZCC himself is a crazy Fan of a coder, called "Memset137". It was on Codefires(CF), an online competitive programming site, th…
 ZCC Loves Codefires Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 223    Accepted Submission(s): 123 Problem Description Though ZCC has many Fans, ZCC himself is a crazy Fan of a coder, ca…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4882 ---------------------------------------------------------------------------------------------------------------------------------------------------------- 欢迎光临天资小屋:http://user.qzone.qq.com/593830943…
贪心 就跟NOIP2012国王游戏差不多,考虑交换相邻两题的位置,对其他题是毫无影响的,然后看两题顺序先后哪个更优.sort即可. WA了一次的原因:虽然ans开的是long long,但是在这一句:ans+=time*a[i].k;时,还是需要在time(int类型)前面加上(LL)进行类型强制转换. /************************************************************** Problem: 3850 User: ProgrammingAp…
http://www.lydsy.com/JudgeOnline/problem.php?id=3850 题意:类似国王游戏....无意义.. #include <cstdio> #include <cstring> #include <cmath> #include <string> #include <iostream> #include <algorithm> #include <queue> #include &l…
Though ZCC has many Fans, ZCC himself is a crazy Fan of a coder, called "Memset137". It was on Codefires(CF), an online competitive programming site, that ZCC knew Memset137, and immediately became his fan. But why? Because Memset137 can solve a…
类似某noip国王游戏. 考虑交换两个题目的顺序,仅会对这两个题目的贡献造成影响. 于是sort,比较时计算两个题目对答案的贡献,较小的放在前面. #include<cstdio> #include<algorithm> using namespace std; typedef long long ll; struct Point{ll T,K;}a[100001]; bool operator < (const Point &a,const Point &b…
基站选址的区间里隐藏着DP优化的机密…… 分析:       不论是做过乘积最大还是石子合并,或者是其他的入门级别的区间DP题目的人呐,大米并认为读题后就能够轻松得出一个简洁明了的Dp转移方程.       由于这道题每个村庄i仅有两种状态:①自己有一个基站②自己不是基站,但是自己的范围S[i]里有基站.基于这样的关系,可以得出一个容易理解的Dp转移方程:       [设f[k][i]表示1~i的村庄中选取k个村庄安放基站,并且第k个村庄就安放在村庄i,使得所有村庄合法的最小花费]      …