hdu 2844 Coins

Coins

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 7939    Accepted Submission(s): 3241

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

 

一般的解法会超时，用一个use数组，记录个数字访问的次数。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <set>#include <queue>using namespace std;#define maxn 100005#define inf 0x7ffffffstruct Node{    int v;    int w;};int use[maxn];bool vis[maxn];Node node[105];int n,m;int main(){    while(scanf("%d%d",&n,&m)!= EOF && (n||m)){        for(int i = 0; i < n; i++){            scanf("%d",&node[i].v);        }        for(int i = 0; i < n; i++){            scanf("%d",&node[i].w);        }       memset(vis,false,sizeof(vis));        vis[0] = true;        int cnt = 0;        for(int i = 0; i < n; i++){            memset(use,0,sizeof(use));            for(int j = node[i].v ; j <= m; j++){                if(!vis[j] && vis[j - node[i].v]){                    if(use[j - node[i].v] < node[i].w){                        use[j] = use[j - node[i].v] +1;                        vis[j] = true;                        cnt++;                    }                }            }        }        printf("%d\n",cnt);    }    return 0;}