SDAU练习三1020

Problem T

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 9   Accepted Submission(s) : 2

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.<br><br>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

 

题目大意：

现在有好多硬币，面值分别为   a1.a2....ai，给出每个硬币的数量，求可以组成多少个不同的价值。

思路：

这个是多重背包的问题。

关于多重背包求解当这件物品的数量*这件物品的体积  >=    背包容量   那么这件这就相当于完全背包，完全背包的定义：每件物品无限可用。

如果不够的话，那么就用  0  1  来解，把  1,2,3...k  件物品当成一件物品来看。

感想：

一个典型的问题，思路可以借鉴，水平有限，，代码写不好。，

AC代码：

1. #include <cstdlib>  
2. #include <cstring>  
3. #include <cstdio>  
4. #include <iostream>  
5. using namespace std;  
6. int val[900005];  
7. int num[900005];  
8. int dp[900005];  
9. int v;  
10. void ZeroOnePack(int cost,int weight)  
11. {  
12.     for(int i=v;i>=cost;i--)  
13.        if(dp[i-cost]+weight>dp[i]) dp[i]=dp[i-cost]+weight;  
14. }  
15. void CompletePack(int cost,int weight)  
16. {  
17.     for(int i=cost;i<=v;i++)  
18.         if(dp[i-cost]+weight>dp[i]) dp[i]=dp[i-cost]+weight;  
19. }  
20. void MultiplePack(int cost ,int weight,int amount)  
21. {  
22.     if(cost*amount>=v) CompletePack(cost,weight);  
23.     else  
24.     {  
25.         for(int k=1;k<amount;)  
26.         {  
27.             ZeroOnePack(k*cost,k*weight);  
28.             amount-=k;  
29.             k<<=1;  
30.         }  
31.         ZeroOnePack(amount*cost,amount*weight);  
32.     }  
33. }  
34. int main()  
35. {  
36.     //freopen("r.txt","r",stdin);  
37.     int i,n,m,j;  
38.   
39.     while(~scanf("%d%d",&n,&v))  
40.     {  
41.         if(n==0&&v==0) break;  
42.   
43.         for(i=0;i<n;i++)  
44.         {  
45.             scanf("%d",&val[i]);  
46.         }  
47.         for(i=0;i<n;i++)  
48.         {  
49.             scanf("%d",&num[i]);  
50.         }  
51.   
52.         memset(dp,0,sizeof(dp));  
53.   
54.         for(i=0;i<n;i++)  
55.         {  
56.             MultiplePack(val[i],val[i],num[i]);  
57.         }  
58.         int temp=0,cou=0;  
59.         for(i=0;i<=v;i++)  
60.         {  
61.             if(temp!=dp[i])  
62.             {  
63.                 temp=dp[i];  
64.                 cou++;  
65.             }  
66.         }  
67.         cout<<cou<<endl;  
68.     }  
69.     return 0;  
70. }