Time limit 1000 ms Memory limit 30000 kB description Charlie is a driver of Advanced Cargo Movement, Ltd. Charlie drives a lot and so he often buys coffee at coffee vending machines at motorests. Charlie hates change. That is basically the setup of y…
http://poj.org/problem?id=1787   描述 Charlie is a driver of Advanced Cargo Movement, Ltd. Charlie drives a lot and so he often buys coffee at coffee vending machines at motorests. Charlie hates change. That is basically the setup of your next task. Yo…
Charlie's Change Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 3720   Accepted: 1125 Description Charlie is a driver of Advanced Cargo Movement, Ltd. Charlie drives a lot and so he often buys coffee at coffee vending machines at motore…
题意: 就是找出来一个字典序最小的硬币集合,且这个硬币集合里面所有硬币的值的和等于题目中的M 题解: 01背包加一下记录路径,如果1硬币不止一个,那我们也不采用多重背包的方式,把每一个1硬币当成一个独立的单位来进行01背包dp 但是我们知道背包dp的路径可能不止一条,而我们要从中得到字典序最小的序列,我的代码中两次不同的排序会得到最大/小字典序 降序 == 最小字典序 升序 == 最大字典序 1 #include<iostream> 2 #include<queue> 3 #inc…
Charlie is a driver of Advanced Cargo Movement, Ltd. Charlie drives a lot and so he often buys coffee at coffee vending machines at motorests. Charlie hates change. That is basically the setup of your next task. Your program will be given numbers and…
http://poj.org/problem?id=1787 Charlie's Change Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 4512   Accepted: 1425 Description Charlie is a driver of Advanced Cargo Movement, Ltd. Charlie drives a lot and so he often buys coffee at co…
这题有点多重背包的感觉,但还是用完全背包解决,dp[j]表示凑到j元钱时的最大硬币数,pre[j]是前驱,used[j]是凑到j时第i种硬币的用量 △回溯答案时i-pre[i]就是硬币价值 #include<iostream> #include<cstdio> #include<cstring> using namespace std; ]={,,,},num[],ans[]; ],pre[],used[];//记录回溯状态,记录每个状态用了多少硬币 int main(…
True Liars Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 2713   Accepted: 868 Description After having drifted about in a small boat for a couple of days, Akira Crusoe Maeda was finally cast ashore on a foggy island. Though he was exha…
链接:https://www.nowcoder.com/acm/contest/141/A来源:牛客网 Eddy was a contestant participating in ACM ICPC contests. ACM is short for Algorithm, Coding, Math. Since in the ACM contest, the most important knowledge is about algorithm, followed by coding(impl…
https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=565 记录路径可以用一个二维数组,记录改变时的量.然后从后往前可以推得所有的值. #include <iostream> #include <string> #include <cstring> #include <cstdlib> #incl…