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…

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…

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…

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) Problem Description Whuacmers use coins.They have coins of value A1,A2,A3...An Silve…

Coins Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6279    Accepted Submission(s): 2561 Problem Description Whuacmers use coins.They have coins of value A1,A2,A3...An Silverland dollar. One d…

作为一个oier,以及大学acm党背包是必不可少的一部分.好久没做背包类动规了.久违地练习下-.- dd__engi的背包九讲:http://love-oriented.com/pack/ 鸣谢http://blog.csdn.net/eagle_or_snail/article/details/50987044,这里有大部分比较有趣的dp练手题. hud 2602 01背包板子题 #include<cstdio> #include<iostream> #include<cs…

HDOJ(HDU).2844 Coins (DP 多重背包+二进制优化) 题意分析 先把每种硬币按照二进制拆分好,然后做01背包即可.需要注意的是本题只需要求解可以凑出几种金钱的价格,而不需要输出种数.因此用0表示不可以,1表示可以.最后对dp数组扫描一遍即可. 代码总览 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #define nmax 100…

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…

题意是给你几个数,再给你这几个数的可以用的个数,然后随机找几个数来累加, 让我算可以累加得到的数的种数! 解题思路:先将背包初始化为-1,再用多重背包计算,最后检索,若bb[i]==i,则说明i这个数是可以得到的!一个循环计算可以达到的数的个数,最后输出就好了! #include<stdio.h> #define max(a,b) a>b?a:b int bb[500000]; int vv; void shun(int cost,int weight) { int i; for(i=c…

点我看题目 题意 :Whuacmers有n种硬币,分别是面值为A1,A2,.....,An,每一种面值的硬币的数量分别是C1,C2,......,Cn,Whuacmers想买钱包,但是想给人家刚好的钱,不喜欢再找钱那么麻烦,但是他不知道钱包的具体钱数,只知道不会超过m,所以问你手里的钱能表示多少个不超过m的钱数. 思路 :多重背包.不知道的去搜背包九讲,几乎都一样. 我想说这个小哥儿真逗 #include <stdio.h> #include <string.h> #include…

<题目连接> 题目大意: 一位同学想要买手表,他有n种硬币,每种硬币已知有num[i]个.已知手表的价钱最多m元,问她用这些钱能够凑出多少种价格来买手表. 解题分析: 很明显,这是一道多重背包的问题,下面是用二进制拆分的多重背包的万能模板. #include <bits/stdc++.h> using namespace std; const int INF = 1e9; ],cnt[],dp[)]; void OneZeroPack(int m,int v,int value){…

题意:给定n种硬币,每种价值是a,数量是c,让你求不大于给定V的不同的价值数,就是说让你用这些硬币来组成多少种不同的价格,并且价格不大于V. 析:一看就应该知道是一个动态规划的背包问题,只不过是变形,那我们就统计不大于V的不同价格数,也容易实现, 对于多重背包我们是把它转化为01背包和完全背包来解决的. 代码如下: #include <cstdio> #include <iostream> #include <cstring> using namespace std;…

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…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1963 //多重背包 #include <cstdio> #include <cstring> #include <iostream> using namespace std; + ; #define N 15 long long dp[maxn], ans; int c[N], w[N], V; void Pack(int C, int W) { for(int i = C…

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…

Coins Time Limit: / MS (Java/Others) Memory Limit: / K (Java/Others) Total Submission(s): Accepted Submission(s): Problem Description Whuacmers use coins.They have coins of value A1,A2,A3...An Silverland dollar. One day Hibix opened purse and found t…

Big Event in HDU Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 23728    Accepted Submission(s): 8363 Problem Description Nowadays, we all know that Computer College is the biggest department…

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…

http://acm.hdu.edu.cn/showproblem.php?pid=1059 Dividing Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 29901    Accepted Submission(s): 8501 Problem Description Marsha and Bill own a collection…

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…

题目大意:某个人有n种硬币,每种硬币价值为v,数量为c,问在总价值不超过m的条件下,最多有多少种组合方式. 题目思路: 1.对于某种硬币 如果v*c 大于 m,就意味着无论取多少枚硬币,只要总价值不大于m就取不完该种硬币--完全背包. 2.如果某种硬币,如果v*c 不大于m,就意味着这是多重背包,因此可以用二进制方法优化一下. 3.对于不大于m的任意数字j,dp[j]=0代表无法组合成j,dp[j]=1 代表可以组合成j,dp[j]=dp[j]|dp[j-v[i]*k](k为选择第i种硬币的数量…

Dividing Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 20635    Accepted Submission(s): 5813 Problem Description Marsha and Bill own a collection of marbles. They want to split the collection…

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[…

题目链接:https://vjudge.net/contest/228640#problem/F 转载于:http://www.voidcn.com/article/p-mxcorksq-gh.html 题目大意: 给定N种钱,每种面值Ai,数量Ci,问可凑出的总价值中有多少种在(0,M]范围之间 #include <iostream> #include <cstring> #include <algorithm> using namespace std; ], c[]…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=2191 悼念512汶川大地震遇难同胞——珍惜现在,感恩生活 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 35770    Accepted Submission(s): 15088 Problem Description 急!灾区的食物依…

题目链接:http://poj.org/problem?id=1742 http://acm.hdu.edu.cn/showproblem.php?pid=2844 题目分类:动态规划 代码: #include<stdio.h> #include<string.h> #include<math.h> #include<algorithm> using namespace std; int n,m; ]; struct P { int a,b; }point[…

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…