The trouble of Xiaoqian

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2277 Accepted Submission(s): 805

Problem Description In the country of ALPC , Xiaoqian is a very famous mathematician. She is immersed in calculate, and she want to use the minimum number of coins in every shopping. (The numbers of the shopping include the coins she gave the store and the store backed to her.)
And now , Xiaoqian wants to buy T (1 ≤ T ≤ 10,000) cents of supplies. The currency system has N (1 ≤ N ≤ 100) different coins, with values V1, V2, ..., VN (1 ≤ Vi ≤ 120). Xiaoqian is carrying C1 coins of value V1, C2 coins of value V2, ...., and CN coins of value VN (0 ≤ Ci ≤ 10,000). The shopkeeper has an unlimited supply of all the coins, and always makes change in the most efficient manner .But Xiaoqian is a low-pitched girl , she wouldn’t like giving out more than 20000 once. Input There are several test cases in the input.
Line 1: Two space-separated integers: N and T.
Line 2: N space-separated integers, respectively V1, V2, ..., VN coins (V1, ...VN)
Line 3: N space-separated integers, respectively C1, C2, ..., CN
The end of the input is a double 0. Output Output one line for each test case like this ”Case X: Y” : X presents the Xth test case and Y presents the minimum number of coins . If it is impossible to pay and receive exact change, output -1. Sample Input 3 70 5 25 50 5 2 1 0 0 Sample Output Case 1: 3

 1 #include<iostream>
 2 #include<stdio.h>
 3 #include<algorithm>
 4 #include<string.h>
 5 #include<string>
 6 #define INF  9999999
 7 #include<algorithm>
 8 using namespace
 
 std;
 9 int dp1[20005],dp2[20005];
10 int val[150],num[150];
11 int min(int x,int y)
12 {
13     return x>y? y:x;
14 }
15 int main()
16 {
17     int n,m;
18     int T=1;
19     while(cin>>n>>m)
20     {
21         if(n==0&&m==0)  break;
22         for(int i=1;i<=n;i++)
23             cin>>val[i];
 
24         for(int i=1;i<=n;i++)
25             cin>>num[i];
26         for(int i=1;i<=20000;i++)
27             dp1[i]=dp2[i]=INF;
28         dp1[0]=dp2[0]=0;
29         for(int i=1;i<=n;i++)//多重背包
30         {
31             for(int k=1,flag=0;;k*=2)
32             {
33                 if(k*2>num[i])
34                 {
35                     k=num[i]-k+1;
36                     flag=1;
37                 }
38                 for(int j=20000;j>=k*val[i];j--)
39                 {
40                     dp1[j]=min(dp1[j],dp1[j-k*val[i]]+k);
41                 }
42                 if(flag)
43                     break;
44             }
45         }
46         for(int i=1;i<=n;i++)//完全背包,這裏有兩種寫法，一種是這種常規的優化了的，還有一種適合新手，雖然復雜度會高些，但是新手理解起來會更容易。
47         {
48             for(int j=val[i];j<=20000;j++)49                 dp2[j]=min(dp2[j],dp2[j-val[i]]+1);
50         }
　　　　　　　/*
　　　　　　　 for(int i=1;i<=n;i++)
　　　　　　　{

　　　　　　　　　　for(int k=1;k*val[i]<20000;k*=2)//類比於多重背包
            　　　{
               　　　　 for(int j=20000;j>=k*val[i];j--)
               　　 　　{
                    　　　　dp2[j]=min(dp2[j],dp2[j-k*val[i]]+k);
                　　　　}　　
           　　 　　}
      　　　　 }

　　　　　　　　　　*/

51         int lss=INF;
52         for(int i=m;i<=20000;i++)
53         {
54             lss=min(lss,dp1[i]+dp2[i-m]);
55         }
56         if(lss>20000)
57             cout<<"Case "<<T++<<": "<<-1<<endl;
58         else
59             cout<<"Case "<<T++<<": "<<lss<<endl;
60     }
61     return 0;
62 }

HDU-3591 混合背包