B - Gena's Code It's the year 4527 and the tanks game that we all know and love still exists. There also exists Great Gena's code, written in 2016. The problem this code solves is: given the number of tanks that go into the battle from each country,…

B. Gena's Code 题目连接: http://www.codeforces.com/contest/614/problem/B Description It's the year 4527 and the tanks game that we all know and love still exists. There also exists Great Gena's code, written in 2016. The problem this code solves is: give…

It's the year 4527 and the tanks game that we all know and love still exists. There also exists Great Gena's code, written in 2016. The problem this code solves is: given the number of tanks that go into the battle from each country, find their produ…

题目链接:Codeforces 417D Cunning Gena 题目大意:n个小伙伴.m道题目,每一个监视器b花费,给出n个小伙伴的佣金,所须要的监视器数,以及能够完毕的题目序号. 注意,这里仅仅要你拥有的监视器数量大于小伙伴须要的监视器数量就可以. 求最少花费多少金额能够解决全部问题. 解题思路:dp[i],i为一个二进制数.表示完毕这些题目的最小代价,可是这里要注意,由于有个监视器的数量.普通情况下要开一个二维的状态.可是2^20次方有一百万,再多一维的数组会超内存,所以我的做法是将每一…

#include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <math.h> using namespace std; int flag; +]; int zero; +]; bool Perfect() { int len=strlen(s); ; i<len; i++) ; ]!=; ; } int main() { f…

这道题提醒我两点: 1.break时一定要检查清楚 2.字符串直接赋值一定要注意结束符,最好能用strcpy 以上是debug的惨痛教训 #include <iostream> #include <string> #include <cstring> #include <cstdlib> #include <cstdio> #include <cmath> #include <algorithm> #include <…

题目链接 D. Cunning Gena time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output A boy named Gena really wants to get to the "Russian Code Cup" finals, or at least get a t-shirt. But the offered p…

告诉你若干个(<=100)武器的花费以及武器能消灭的怪物编号,问消灭所有怪物(<=100)的最小花费...当然每个武器可以无限次使用,不然这题就太水了╮(╯▽╰)╭ 这题当时比赛的时候连题都还没看就结束了....赛后一看,果断是重复覆盖... 不过之后一直没敲...然后今天算是补回来吧,同时也把好久以前学的DLX复习一下... DLX的话,双向十字链表...具体的话,百度Google什么的dancing links... 一开始敲的时候还是挺顺利的,因为之前做过几次重复覆盖都是直接拿精确覆盖的…

题意:有n个空心物品,每个物品有外部体积outi和内部体积ini,如果ini>outj,那么j就可以套在i里面.现在我们要选出n个物品的一个子集,这个子集内的k个物品全部套在一起,且剩下的物品都无法添加到这个子集中(没有空间塞进去). 定义浪费的空间为子集中空心的部分,即ini1+(ini2−outi1)+(ini3−outi2)+⋯+(inik−outik−1)ini1+(ini2−outi1)+(ini3−outi2)+⋯+(inik−outik−1).求浪费空间最少的子集个数. 解法:第一…

A. Link/Cut Tree time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Programmer Rostislav got seriously interested in the Link/Cut Tree data structure, which is based on Splay trees. Specifica…