POJ 1742 dp

最新推荐文章于 2020-07-31 10:59:51 发布

black_miracle

最新推荐文章于 2020-07-31 10:59:51 发布

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Coins

Time Limit: 3000MS		Memory Limit: 30000K

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.

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

楼教主的男人八题中的一题，貌似是最简单的。

题意：给你n种硬币，一个m，接下来n个数是每种硬币的面值，最后n个数是每种硬币的数量，问你用这些硬币可以组成多少种不超过m元的组合。

多重背包，好好学习一下。

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int dp[100005],sum[100005];
int v[105],c[105];
int main(){
	int n,m,i,j;
	while(scanf("%d%d",&n,&m)!=EOF){
		if(n==0&&m==0)break;
		memset(dp,0,sizeof(dp));
		for(i=1;i<=n;i++)scanf("%d",&v[i]);
		for(i=1;i<=n;i++)scanf("%d",&c[i]);
		dp[0]=1;
		int ans=0;
		for(i=1;i<=n;i++){
			memset(sum,0,sizeof(sum));
			for(j=v[i];j<=m;j++){
				if(!dp[j]&&dp[j-v[i]]&&sum[j-v[i]]<c[i]){
					dp[j]=1;
					sum[j]=sum[j-v[i]]+1;
					ans++;
				}
			}
		}
		cout<<ans<<endl;
	}
	return 0;
}