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…
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…
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…
Q: 既是多重背包, 还是找零问题, 怎么处理? A: 题意理解有误, 店主支付的硬币没有限制, 不占额度, 所以此题不比 1252 难多少 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 change…
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 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…
Description People in Silverland use coins.They have coins of value A1,A2,A3...An Silverland dollar.One day Tony opened his money-box and found there were some coins.He decided to buy a very nice watch in a nearby shop. He wanted to pay the exact pri…
前几天刚回到家却发现家里没网线 && 路由器都被带走了,无奈之下只好铤而走险尝试蹭隔壁家的WiFi,不试不知道,一试吓一跳,用个手机软件简简单单就连上了,然后在浏览器输入192.168.1.1就能看到他的路由器的一切信息,包括密码,然后打开笔记本……好了,废话不多说,能连上网后第一时间当然是继续和队友之前约好的训练了. 今天翻看到之前落下的一道混合背包题目,然后在草稿本上慢慢地写递推方程,把一些细节细心地写好…(本来不用太费时间的,可是在家嘛,一会儿妈走来要我教她玩手机,一会儿有一个亲戚朋…
多重背包+完全背包. 买家:多重背包:售货员:完全背包: 开两个数组,分别计算出买家,售货员每个面额的最少张数. 最重要的是上界的处理:上界为maxw*maxw+m(maxw最大面额的纸币). (网上的证明)证明如下: 如果买家的付款数大于了maxw*maxw+m,即付硬币的数目大于了maxw,根据鸽笼原理,至少有两个的和对maxw取模的值相等,也就是说,这部分硬币能够用更少的maxw来代替.证毕. 其实我真心没看懂这个证明.不过我们可以猜想T一定不会太大,我开了10*T然后过了. 这一题学习了…
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…
1.hdu 2126 Buy the souvenirs 题意:给出若干个纪念品的价格,求在能购买的纪念品的数目最大的情况下的购买方案. 思路:01背包+记录方案. #include<iostream> #include<cstdio> #include<cstring> using namespace std; ; ; int cnt[maxw]; int dp[maxw]; int p[maxn]; int main() { int t; scanf("%…
[题意]:已知整个交易系统有N (1 ≤ N ≤ 100)种不同的货币,分别价值V1,V2,V3.......VN(1 ≤ Vi ≤ 120),FJ分别有C1,C2,C3.....CN(0 ≤ Ci ≤10,000)张相应价值货币.FJ只能用仅有的货币去买价值T(1 ≤T≤10,000)分的东西,而老板有无数的货币可以找给他.求FJ给老板的货币数+老板找给FJ的货币数的最小值. Input * Line 1: Two space-separated integers: N and T. * Li…
<span style="color:#3333ff;">/* ------------------------------------------------------------------------------------------------ author : Grant Yuan time : 2014.7.19 aldorithm: 二维背包+全然背包 ----------------------------------------------------…
http://poj.org/problem?id=3260   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…
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…
题意:FJ身上有各种硬币,但是要买m元的东西,想用最少的硬币个数去买,且找回的硬币数量也是最少(老板会按照最少的量自动找钱),即掏出的硬币和收到的硬币个数最少. 思路:老板会自动找钱,且按最少的找,硬币数量也不限,那么可以用完全背包得出组成每个数目的硬币最少数量.而FJ带的钱是有限的,那么必须用多重背包,因为掏出的钱必须大于m,那么我们所要的是大于等于m钱的硬币个数,但是FJ带的钱可能很多,超过m的很多倍都可能,那么肯定要有个背包容量上限,网上说的根据抽屉原理是m+max*max,这里的max指…
http://poj.org/problem?id=3260 这个题目有点小难,我开始没什么头绪,感觉很乱. 后来看了题解,感觉豁然开朗. 题目大意:就是这个人去买东西,东西的价格是T,这个人拥有的纸币和数量.让你求这个人买东西的纸币量加上老板找给他的纸币量最少是多少. 这个老板用于这个人拥有的纸币种类,数量是无限. 思路: 思路就是这个人看成多重背包,老板看成完全背包,f1[i] 表示这个人花了 i 的钱用的最少的纸币.f2[i] 表示老板凑出 i 的钱用的最少的纸币. #include <c…
Coins Time Limit: 3000MS   Memory Limit: 30000K Total Submissions: 34814   Accepted: 11828 Description People in Silverland use coins.They have coins of value A1,A2,A3...An Silverland dollar.One day Tony opened his money-box and found there were some…
Problem Description Whuacmers use coins.They have coins of value A1,A2,A3...An Silverland dollar. One day Hibix opened purse and found there were some coins. He decided to buy a very nice watch in a nearby shop. He wanted to pay the exact price(witho…
Coins Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 12330    Accepted Submission(s): 4922 Problem Description Whuacmers use coins.They have coins of value A1,A2,A3...An Silverland dollar. One…
这道题是典型的多重背包的题目,也是最基础的多重背包的题目 题目大意:给定n和m, 其中n为有多少中钱币, m为背包的容量,让你求出在1 - m 之间有多少种价钱的组合,由于这道题价值和重量相等,所以就是dp[i] = i, 其中dp[i]表示当前背包容量为i 的时候背包能装的价值. 题目思路: 模板 二进制优化 话说那个二进制真的很奇妙,只需要2的1次方 到 2的k-1次方, 到最后在加上一项当前项的个数 - 2 的k次方 + 1,也就是这些系数分别为1; 2; 22 .....2k-1;Mi…
链接:http://acm.hdu.edu.cn/showproblem.php?pid=2844 思路:多重背包 , dp[i] ,容量为i的背包最多能凑到多少容量,如果dp[i] = i,那么代表这个数能凑出来,ans+1: 实现代码: #include<bits/stdc++.h> using namespace std; ; int lis[M],dp[M],a[M],c[M]; int main() { int n,m,idx; while(cin>>n>>m…
Problem Description Whuacmers use coins.They have coins of value A1,A2,A3...An Silverland dollar. One day Hibix opened purse and found there were some coins. He decided to buy a very nice watch in a nearby shop. He wanted to pay the exact price(witho…
题意 给n个币的价值和其数量,问能组合成\(1-m\)中多少个不同的值. 分析 对\(c[i]*a[i]>=m\)的币,相当于完全背包:\(c[i]*a[i]<m\)的币则是多重背包,考虑用二进制优化解决.最后扫一遍\(dp[i]\)统计答案. import java.util.*; import java.math.*; public class Main{ static int MAXN = 100005; static int []dp = new int[MAXN]; static i…
Coins Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 8999    Accepted Submission(s): 3623 Problem Description Whuacmers use coins.They have coins of value A1,A2,A3...An Silverland dollar. One…
Coins                                                                             Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description Whuacmers use coins.They have coins of value A1,A2,A3...An Silve…
Vjudge传送门 $Sol$ 首先发现这是一个多重背包,所以可以用多重背包的一般解法(直接拆分法,二进制拆分法...) 但事实是会TLE,只能另寻出路 本题仅关注“可行性”(面值能否拼成)而不是“最优性”,这是一个特殊之处. 从这里找优化 在“最优性”的问题中,$f[j]$从$f[j]$或$f[j-a[i]]$中转移而来:而在这样的“可行性”问题中,其实只要$f[j]$可行,我们就可以不用考虑$f[j-a[i]$了,也可以反过来说. 于是我们可以考虑一种贪心策略,设$used[j]$表示$f[…