【背包专题】D

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.

InputThe 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.OutputFor 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：问能够将输入的钱币组合为1~m之间的数的种类总数，比如样例2：m=5，钱币可以组合为：1,2,4,5.所以总的种类数为4.
思路：没有求最值，而是求可到达状态数。参考《背包九讲》里的第九讲。定义dp[i]表示是否可以从这些硬币中选取一些硬币，使得面值之和为i，dp[i]的取值为1表示可以，0表示不可以。
　　状态转移方程为dp[i] = dp[i]||dp[i-value[i]]||...dp[i-num[i]*value[i]]
处理每种硬币时，每个状态要转移的次数和num[i]有关，num[i]较大时，时间复杂度比较高，我们可以采用二进制数的方法优化，1,2,4...2^n，只要是小于num[i]的数，我们都可以用这些数字组合而来，将每一组硬币
面值相加作为一个硬币来处理，就可以转为01背包的问题来求解。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define N 100010

int n,m;
int dp[N],w[10010],value[110],num[110];

int main()
{
    int i,j,ans,x;
    while(scanf("%d%d",&n,&m),n+m)
    {
        memset(dp,0,sizeof(dp));
        for(i = 1; i <= n; i ++)
            scanf("%d",&value[i]);//硬币价值 
        for(i = 1; i <= n; i ++)
            scanf("%d",&num[i]);//硬币数量 
        ans = 0;    
        for(i = 1; i <= n; i ++)
        {
            for(j = 1;num[i] > 0 ; j*=2)//二进制划分硬币数量 
            {
                x = min(j,num[i]);
                w[++ans] = x*value[i];
                num[i] -= x;
            }
        }
        dp[0] = 1;//初始状态为可到达状态 
        for(i = 1; i <= ans; i ++)
        {
            for(j = m; j >= w[i];j--)
            {
                if(dp[j-w[i]])//如果状态可以到达 
                    dp[j] = 1;
            }
        }
        ans = 0;
        for(i = 1; i <= m; i ++)
            if(dp[i])
                ++ans;
        printf("%d
",ans);
    }
    return 0;
}