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…

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…

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.…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3177 \(describe\): 有一个山洞,山洞的容积最大为\(v\).现在你有\(n\)个物品,这些物品在往山洞里搬和放在山洞所需要占用山洞的体积是两个不同的值\(B\),\(A\).你可以理解为在搬运这个物品进洞时需要的容积为一个\(B\),放下物品后的容积是一个\(A\).在任何时刻搬运物品都不允许超过山洞的最大容积.试求能不能把所有物品搬进去 题解: 这个题正解是贪心...没错... 题目…

以防万一,题目原文和链接均附在文末.那么先是题目分析: [一句话题意] “就是统计一篇文章里不同单词的总数”(已经是一句话了..) [题目分析] 明显需要去重,上set,因为按行分析,又没有EOLN用,于是上istringstream. [算法流程] 读一行塞一行干一行爱一行.....发这篇的目的其实是备忘istringstream的用法的.这道题没难点. #include <iostream> #include <sstream> #include <string>…