HDU 2844 Coin 多重背包

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HDU 2844 Coin 多重背包

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…

hdu 2844 Coins (多重背包+二进制优化)

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…

HDu -2844 Coins多重背包

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…

HDU - 2844 Coins(多重背包+完全背包)

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…

HDU 2844 Coins (多重背包计数空间换时间)

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…

hdu 2844 coins(多重背包二进制拆分法)

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…