POJ1742:Coins(多重背包)

最新推荐文章于 2020-07-31 10:59:51 发布

原创最新推荐文章于 2020-07-31 10:59:51 发布 · 5.4k 阅读

1 ·

CC 4.0 BY-SA版权

文章标签：

#背包 #POJ

背包专栏收录该内容

49 篇文章

订阅专栏

本文探讨了在银陆使用不同价值的银币支付不超过特定价格的商品时，可以支付的不同价格的数量。通过输入银币的价值和数量，以及商品的价格上限，计算出符合条件的支付方案总数。

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

这道题与杭电的一道题的题意是相同的，之所以拿出来是因为杭电那道题的代码在这里过不了，所以贴一个新的代码

#include <stdio.h>
#include <algorithm>
#include <string.h>

int dp[100005];
int sum[100005];
int v[105],c[105];

int main()
{
    int i,j,k,n,m;
    while(~scanf("%d%d",&n,&m),n+m)
    {
        for(i = 1;i<=n;i++)
        scanf("%d",&v[i]);
        for(i = 1;i<=n;i++)
        scanf("%d",&c[i]);
        memset(dp,0,sizeof(dp));
        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++;
                }
            }
        }
        printf("%d\n",ans);
    }

    return 0;
}