Description Farmer John has gone to town to buy some farm supplies. Being a very efficient man, he always pays for his goods in such a way that the smallest number of coins changes hands, i.e., the number of coins he uses to pay plus the number of co…

The Fewest Coins DescriptionFarmer John has gone to town to buy some farm supplies. Being a very efficient man, he always pays for his goods in such a way that the smallest number of coins changes hands, i.e., the number of coins he uses to pay plus…

POJ 3260 The Fewest Coins(多重背包+全然背包) http://poj.org/problem?id=3260 题意: John要去买价值为m的商品. 如今的货币系统有n种货币,相应面值为val[1],val[2]-val[n]. 然后他身上每种货币有num[i]个. John必须付给售货员>=m的金钱, 然后售货员会用最少的货币数量找钱给John. 问你John的交易过程中, 他给售货员的货币数目+售货员找钱给他的货币数目 的和最小值是多少? 分析: 本题与POJ 12…

支付对应的是多重背包问题,找零对应完全背包问题. 难点在于找上限T+maxv*maxv,可以用鸽笼原理证明,实在想不到就开一个尽量大的数组. 1 #include <map> 2 #include <set> 3 #include <cmath> 4 #include <queue> 5 #include <cstdio> 6 #include <vector> 7 #include <climits> 8 #includ…

题目描述 Farmer John has gone to town to buy some farm supplies. Being a very efficient man, he always pays for his goods in such a way that the smallest number of coins changes hands, i.e., the number of coins he uses to pay plus the number of coins he…

Description Farmer John has gone to town to buy some farm supplies. Being a very efficient man, he always pays for his goods in such a way that the smallest number of coins changes hands, i.e., the number of coins he uses to pay plus the number of co…

题目代号:POJ 3260 题目链接:http://poj.org/problem?id=3260 The Fewest Coins Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 6715 Accepted: 2072 Description Farmer John has gone to town to buy some farm supplies. Being a very efficient man, he alway…

The Fewest Coins Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 6299   Accepted: 1922 Description Farmer John has gone to town to buy some farm supplies. Being a very efficient man, he always pays for his goods in such a way that the sm…

The Fewest Coins POJ - 3260 完全背包+多重背包.基本思路是先通过背包分开求出"付出"指定数量钱和"找"指定数量钱时用的硬币数量最小值,然后枚举找的钱,那么付出的钱也随之确定,对于每个枚举出的找的钱可以得到一个答案,那么枚举所有可能的找的钱取答案的最大值即可. 这里有一个对于找钱上限的证明.如果不知道,也可以随便搞一个(比如以下用的10000,注意空间,试过5000可以过). 错误记录: 4.多重背包优化中,把除以2写成<<=2…

题意:给你n组物品和自己有的价值s,每组有l个物品和有一种类型: 0:此组中最少选择一个 1:此组中最多选择一个 2:此组随便选 每种物品有两个值:是需要价值ci,可获得乐趣gi 问在满足条件的情况下,可以得到的最大的乐趣是多少,如果不能满足条件就输出-1 题解:二维01背包 dp[i][j]:前i组物品我们拥有j的价值时最大可获得的乐趣 0:我们需要先把dp[i]所有赋值为负无穷,这样就只能最少选一个才能改变负无穷 1:我们不需要:dp[i][j-ci]+gi(在此组中再选一个),这样就一定最…