hdu 3591  The trouble of Xiaoqian 题意:xiaoqi要买一个T元的东西,当前的货币有N种,xiaoqi对于每种货币有Ci个:题中定义了最小数量即xiaoqi拿去买东西的钱的张数加上店家找的零钱的张数(店家每种货币有无限多张,且找零是按照最小的数量找零的):问xiaoqi买元东西的最小数量? 多重背包+完全背包: 思路:这个最小数量是拿去买东西的张数和找零的张数之和,其实我们只需要将这两个步骤分开,开算出能买T元东西的前i最少f[i]张,这里涉及到容量问题:容量只…

HDU 3591 The trouble of Xiaoqian(多重背包+全然背包) pid=3591">http://acm.hdu.edu.cn/showproblem.php? pid=3591 题意: 有一个具有n种货币的货币系统, 每种货币的面值为val[i]. 如今小杰手上拿着num[1],num[2],-num[n]个第1种,第2种-第n种货币去买价值为T(T<=20000)的商品, 他给售货员总价值>=T的货币,然后售货员(可能,假设小杰给的钱>T,那肯…

题目大意 有 \(N\) 种不同面值的硬币,分别给出每种硬币的面值 \(v_i\) 和数量 \(c_i\).同时,售货员每种硬币数量都是无限的,用来找零. 要买价格为 \(T\) 的商品,求在交易中最少使用的硬币的个数(指的是交易中给售货员的硬币个数与找回的硬币个数之和). 个数最多不能超过 \(20000\),如果不能实现,输出 \(-1\):否则输出此次交易中使用的最少的硬币个数. 样例 有 \(3\) 种硬币,面值分别为 \(5, 25 50\),个数分别为 \(5, 2, 1\),要买…

The trouble of Xiaoqian Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2166    Accepted Submission(s): 773 Problem Description In the country of ALPC , Xiaoqian is a very famous mathematician.…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=3591 The trouble of Xiaoqian Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2798    Accepted Submission(s): 972 Problem Description In the countr…

The trouble of Xiaoqian Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1472 Accepted Submission(s): 502 Problem Description In the country of ALPC , Xiaoqian is a very famous mathematician. She i…

题目: The trouble of Xiaoqian Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1997    Accepted Submission(s): 711 Problem Description In the country of ALPC , Xiaoqian is a very famous mathematici…

The trouble of Xiaoqian Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1076    Accepted Submission(s): 355 Problem Description In the country of ALPC , Xiaoqian is a very famous mathematician.…

//给出Xiaoqian的钱币的价值和其身上有的每种钱的个数 //商家的每种钱的个数是无穷,xiaoqian一次最多付20000 //问如何付钱交易中钱币的个数最少 //Xiaoqian是多重背包 //商家是全然背包 #include<cstdio> #include<cstring> #include<iostream> using namespace std ; const int maxn = 20010 ; const int inf = 0x3f3f3f3f…

题目:click here 题意: 给定5组数据,每组数据选择一个数,看是否能找到5个数的和为零. 分析: 千万不要~~T~~ 普通线性查找: #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; typedef long long ll; const int INF = 0x3f3f3f3f; ; int t, n…