Coins

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 20860    Accepted Submission(s): 8198

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

 

多重揹包的模板題

 1 #include <iostream>
 2 #include <map>
 3 #include <math.h>
 4 #include <algorithm>
 5 #include <vector>
 6
 
 #include <cstdlib>
 7 #include <cstdio>
 8 #include <cstring>
 9 #include <set>
10 using namespace std;
11 int n,m,a[105],c[105],dp[100005];
12 void comdp(int w,int v)
13 {
14     int i;
15     for(i=w; i<=m; i++)
16         dp[i]=max(dp[i],dp[i-w]+v);
17 }
18 void zeroone(int w,int v)
19 {
20     int i;
21     for(i=m; i>=w; i--)
22         dp[i]=max(dp[i],dp[i-w]+v);
23 }
24 void multidp(int w,int v,int cnt)//此時開始多重揹包，dp[i]表示揹包中重量為i時所包含的最大價值
25 {
26     if(cnt*w>=m)//此時相當於物品數量無限進行完全揹包
27     {
28         comdp(w,v);
29         return;
30     }
31     int k=1;//否則進行01揹包轉化，具體由程式碼下數學定理可得
32     while(k<=cnt)
33     {
34         zeroone(k*w,k*v);
35         cnt-=k;
36         k*=2;
37     }
38     zeroone(cnt*w,cnt*v);
39     return ;
40 }
41 int main()
42 {
43     while(~scanf("%d %d",&n,&m))
44     {
45         if(n==0&&m==0)break;
46         for(int i=1;i<=n;i++)scanf("%d",&a[i]);
47         for(int i=1;i<=n;i++)scanf("%d",&c[i]);
48         memset(dp,0,sizeof(dp));
49         int ans=0;
50         for(int i=1;i<=n;i++)
51         {
52             multidp(a[i],a[i],c[i]);
53         }
54         for(int i=1;i<=m;i++)
55         {
56             if(dp[i]==i)ans++;
57         }
58         printf("%d\n",ans);
59     }
60     return 0;
61 }

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