Coins
| Time Limit: 3000MS | Memory Limit: 30000K | |
| Total Submissions: 28448 | Accepted: 9645 |
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.
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.
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
Source
代码：1297MS
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
#define M 105
#define N 100005
int wight[M],cost[M],dp[N],user[N];
int main()
{
int i,j,n,m,ans;
while(scanf("%d%d",&n,&m)!=EOF &&n && m)
{
for(i=1;i<=n;i++) cin>>wight[i];
for(i=1;i<=n;i++) cin>>cost[i];
memset(dp,0,sizeof(dp));
dp[0]=1;ans=0;
for(i=1;i<=n;i++)
{
memset(user,0,sizeof(user)); //这里user表示的是这样的钱币在到达某种状态时，用了多少。j表示要到达的状态。也就是钱的总额。
for(j=wight[i];j<=m;j++)
{ //假设该钱币总额状态没到过，而前一个状态到过，也就是说，仅仅要加一个wight[i]表示的钱币就能到到该状态，
//而到达前一状态还剩这类钱币。就表示该状态能够到达。 if(!dp[j] && dp[j-wight[i]] && user[j-wight[i]]+1<=cost[i])
{
dp[j]=1;
user[j]=user[j-wight[i]]+1; //用一个钱币到达了该状态，这一个钱币当然要加上。 ans++; //能到达的钱币总额状态加一。 }
}
}
cout<<ans<<endl;
}
return 0;
}
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