HDU Coins

Coins

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 84 Accepted Submission(s): 52

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.

 

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 10
1 2 4 2 1 1
2 5
1 4 2 1
0 0

 

Sample Output

8
4

 

P03: 多重背包问题

题目

有N种物品和一个容量为V的背包。第i种物品最多有n[i]件可用，每件费用是c[i]，价值是w[i]。求解将哪些物品装入背包可使这些物品的费用

总和不超过背包容量，且价值总和最大。

基本算法

这题目和完全背包问题很类似。基本的方程只需将完全背包问题的方程略微一改即可，因为对于第i种物品有n[i]+1种策略：取0件，

取1件……取n[i]件。令f[i][v]表示前i种物品恰放入一个容量为v的背包的最大权值，则有状态转移方程：

f[i][v]=max{f[i-1][v-k*c[i]]+k*w[i]|0<=k<=n[i]}

复杂度是O(V*Σn[i])。

#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;

int main()
{
    int n,m,A[101],C[101],f[100001],num,count;
    while(scanf("%d%d",&n,&m),n,m)
    {
        for(int i = 0; i < n ; i++)
        scanf("%d",&A[i]);
        for(int i = 0; i < n ; i++)
        scanf("%d",&C[i]);
        memset(f,0,sizeof(f));//标记如果能组成m这种面值的f[m]为1，否则为0。
        f[0] = 1;
        for(int i = 0; i < n ; i++)
        for(int j = 0;j < A[i];j++)//针对每种硬币,只能组成由面值为0--A[i]-1与K*A[i]的加和组成。1<=k<=c[i]
        {
            count = C[i]; //记录使用的次数
            for(int k = j+A[i] ; k <= m;k+=A[i])//
            if(f[k]==1)count = C[i];//如果这种面值的价格不用A[i]这种硬币即可组成，那么这种硬币的数量可以恢复原始数量即一次也没用过
            else if(count>0&&f[k-A[i]]==1)
            {
                f[k] = 1;
                count--;
            }
        }
        num = 0;//记录数目，得到可以组成的金额数目。
        for(int i = 1; i <= m; i++)
        if(f[i]==1)num++;
        printf("%d\n",num);
    }
    return 0;
}