hdu2844—Coins(多重背包)
来源:互联网 发布:台电x16 plus淘宝 编辑:程序博客网 时间:2024/05/29 11:30
题目链接:传送门
Coins
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 15108 Accepted Submission(s): 5994
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(without change) and he known the price would not more than m.But he didn't know the exact price of the watch.
You are to write a program which reads n,m,A1,A2,A3...An and C1,C2,C3...Cn corresponding to the number of Tony's coins of value A1,A2,A3...An then calculate how many prices(form 1 to m) Tony can pay use these coins.
You are to write a program which reads n,m,A1,A2,A3...An and C1,C2,C3...Cn corresponding to the number of Tony's coins of value A1,A2,A3...An then calculate how many prices(form 1 to m) Tony can pay use these coins.
Input
The input contains several test cases. The first line of each test case contains two integers n(1 ≤ n ≤ 100),m(m ≤ 100000).The second line contains 2n integers, denoting A1,A2,A3...An,C1,C2,C3...Cn (1 ≤ Ai ≤ 100000,1 ≤ Ci ≤ 1000). The last test case is followed by two zeros.
Output
For each test case output the answer on a single line.
Sample Input
3 101 2 4 2 1 12 51 4 2 10 0
Sample Output
84
Source
2009 Multi-University Training Contest 3 - Host by WHU
解题思路:多重背包模板题
#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <string>#include <stack>#include <queue>using namespace std;typedef long long ll;typedef pair<int,int>PA;const int N = 120;const int M = 100119;const int INF = 0x3fffffff;int cost[N],num[N],dp[M];int main(){ int n,m; while( ~scanf("%d%d",&n,&m)&&n&&m ){ memset( dp , 0 , sizeof(dp) ); for( int i = 0 ; i < n ; ++i ){ scanf("%d",&cost[i]); } for( int i = 0 ; i < n ; ++i ){ scanf("%d",&num[i]); } for( int i = 0 ; i < n ; ++i ){ //当物品数量乘花费>=m,用完全背包解 if( cost[i]*num[i] >= m ){ for( int j = cost[i] ; j <= m ; ++j ){ dp[j] = max( dp[j] , dp[j-cost[i]]+cost[i] ); } continue; } int rec = num[i],k = 1; //将物品数量分解成二进制数用01背包解 //当时k<rec写成k<num[i]runtime了m次,心累 while( k < rec ){ for( int j = m ; j >= k*cost[i] ; --j ){ dp[j] = max( dp[j] , dp[j-k*cost[i]]+cost[i]*k ); } rec = rec-k; k = k*2; } for( int j = m ; j >= rec*cost[i] ; --j ){ dp[j] = max( dp[j] , dp[j-rec*cost[i]]+cost[i]*rec ); } } int ans = 0; for( int i = 1 ; i <= m ; ++i ){ if( dp[i] == i ) ++ans; } printf("%d\n",ans); } return 0;}
阅读全文
0 0
- hdu2844—Coins(多重背包)
- hdu2844 Coins(多重背包)
- HDU2844-Coins(多重背包)
- HDU2844 Coins(多重背包)
- HDU2844:Coins(多重背包)
- hdu2844 Coins 多重背包
- hdu2844 Coins --多重背包
- hdu2844 Coins(多重背包+二进制优化)
- hdu2844 Coins(多重背包模板题)
- hdu2844 Coins 多重背包+完全背包
- 动态规划:HDU2844-Coins(多重背包的二进制优化)
- hdu2844 Coins (多重背包+二进制优化)
- 动态规划--多重背包--hdu2844 coins
- hdu2844 完全背包+多重背包(经典)
- Coins(多重背包)
- Coins(多重背包)
- hdu2844 多重背包
- hdu2844(多重背包)
- 迭代器
- meanshift算法通俗讲解
- spring 3.0 应用springmvc 构造RESTful URL 详细讲解
- 找出1~n之间的所有素数
- 数据挖掘RapidMiner工具使用----产品介绍与安装过程
- hdu2844—Coins(多重背包)
- Spring1 Bean实例化
- angularjs指令中的compile与link函数详解
- spring2 bean作用域 和 生命周期
- 各种常用的排序算法实现对数组的排序——整理总结(代码实现)
- JVM的内存模型
- $GLOBALS['HTTP_RAW_POST_DATA'] 和$_POST的区别
- java 中获取各种路径
- Spark Streaming