HDU Coins(dp)

Coins

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13459    Accepted Submission(s): 5383

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

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

 

Sample Output

84

 

Source
题意：n是有几种金币，m是总的价格，a数组是金币的价值，c数组是对应金币的数量，求解这些金币可以组成的小于m的价值。

题解：

对于每个硬币而言:
1、ai*ci>=m，就是完全背包。
2、ai*ci<m，就是多重背包，处理方法就是二进制转换为0-1背包来处理。

 

#include <stdio.h>#include <iostream>#include <string.h>using namespace std;const int inf = -0X3f3f3f3f;int a[100100], c[100100];int dp[100100];int m;void slove(int p, int w)//p价值，w数量{    int i, j;    if(w*p >= m)    {        for(int j = p; j <= m; j++)        {            dp[j] = max(dp[j], dp[j-p] + p);        }    }    else    {        int k = 1;        while(k <= w)        {            for(i = m; i >= k*p; i--)            {                dp[i] = max(dp[i], dp[i-k*p]+k*p);            }               w -= k;            k*=2;        }        for(i = m; i >= w*p; i--)        {            dp[i] = max(dp[i], dp[i-w*p]+w*p);        }    }}int main(){    int n, i, j;    while(~scanf("%d%d", &n, &m)&&(n||m))    {        for(i = 1; i <= n; i++)        {            scanf("%d", &a[i]);        }        for(i = 1; i <= n; i++)        {            scanf("%d", &c[i]);        }        memset(dp, inf, sizeof(dp));        dp[0] = 0;        for(i = 1; i <= n; i++)        {            slove(a[i], c[i]);        }        int sum = 0;        for(i = 1; i <= m; i++)        {            if(dp[i]>0)                sum++;        }        printf("%d\n", sum);    }}