Coins - HDU 2844 背包dp

01背包问题与二进制优化算法详解

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本文深入探讨了01背包问题的基本概念及其应用，并详细介绍了通过二进制优化方法解决该问题的策略。通过实例分析，展示了如何有效计算在限定条件下能够组成的面额价格数量。

Coins

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6962 Accepted Submission(s): 2829

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

题意：01背包，问能够组成多少种面值的价格。

思路：二进制优化。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n,m,dp[100010],val[110],num[110],ans;
int main()
{ int i,j,k,p;
  while(~scanf("%d%d",&n,&m) && n+m )
  { for(i=1;i<=n;i++)
     scanf("%d",&val[i]);
    for(i=1;i<=n;i++)
     scanf("%d",&num[i]);
    memset(dp,0,sizeof(dp));
    dp[0]=1;
    for(i=1;i<=n;i++)
    { k=1;
      while(num[i]>=k)
      { p=val[i]*k;
        for(j=m;j>=p;j--)
         if(dp[j-p]==1)
          dp[j]=1;
        num[i]-=k;
        k*=2;
      }
      if(num[i])
      { p=val[i]*num[i];
        for(j=m;j>=p;j--)
         if(dp[j-p]==1)
          dp[j]=1;
      }
    }
    ans=0;
    for(i=1;i<=m;i++)
     ans+=dp[i];
    printf("%d\n",ans);
  }
}