poj 1742coins（最佳化的多重背包）

Description

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

3 101 2 4 2 1 12 51 4 2 10 0

Sample Output

84

題目的意思：

  第一行輸入，n,m分別表示n種硬幣，m表示總錢數。

  第二行輸入n個硬幣的價值，和n個硬幣的數量。

  輸出這些硬幣能表示的所有在m之內的硬幣種數。

代碼：

 
#include<iostream>using namespace std;int dp[100005];int p[105],c[105];int num[100005];int main(){    int i,j,k,n,m,cnt;    while(scanf("%d%d",&n,&m),n+m)    {     for(i=0;i<n;i++)        scanf("%d",&p[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=p[i];j<=m;j++)       {         if(!dp[j]&&dp[j-p[i]]&&num[j-p[i]]<c[i])         {           num[j]=num[j-p[i]]+1;           dp[j]=1;           cnt++;         }       }     }     printf("%d\n",cnt);  }  return 0;}