动态规划 多重背包 Coins

        People in Silverland use coins.They have coins of value A1,A2,A3...An Silverland dollar.One day Tony opened his money-box 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

1.注意此处的手表面额就相当于是背包。

2.num[j-a[i]]数组就表示，要凑出面额j所需的金币i已经用的个数。

3.红色字体处表示：j面值之前没有凑出来过,并且j-p[i]面值的凑出来过，这样就可以增加一张p[i]凑成新面值.类似于经典背包问题：j就表示背包的容量，p[i]就表示重量一样

#include<iostream>
#include<cstdio>
using namespace std;
int dp[100005];//表示1-m内的金额是否凑出过 
int a[105],c[105];//a是拥有的金币的金额，c是拥有的每种金币的个数 
int num[100005];//用来标记某凑某面值时，某金币已经使用的个数 
int main()
{
    int i,j,k,n,m,cnt;
    while(scanf("%d%d",&n,&m))
    {
    if(n==0&&m==0)
      break;
     for(i=0;i<n;i++)
        scanf("%d",&a[i]);
     for(i=0;i<n;i++)
        scanf("%d",&c[i]); 
     for(i=1;i<=m;i++) 
  dp[i]=0;
     dp[0]=1;
     cnt=0;
     for(i=0;i<n;i++)
     {
       for(j=0;j<=m;j++) 
   num[j]=0;
       for(j=a[i];j<=m;j++)
       {
        //cout<<"num[j-p[i]]="<<num[j-p[i]]<<endl;
         if(!dp[j]&&dp[j-a[i]]&&num[j-a[i]]<c[i])//j面值之前没有凑出来过,并且j-p[i]面值的凑出来过，这样就可以增加一张p[i]凑成新面值.j就表示背包的容量，p[i]就表示重量一样  
         {
           num[j]=num[j-a[i]]+1;
           //cout<<"i="<<i<<" "<<"j="<<j<<" "<<"num[j]="<<num[j]<<endl; 
           dp[j]=1;
           cnt++;
         }
       }
     }
     printf("%d\n",cnt);
  }
  return 0;
}