CF动态规划一--Cut Ribbon--完全背包

桑榆207

于 2019-07-21 22:06:24 发布

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分类专栏： ACM动态规划

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本文介绍了一种算法，用于解决将固定长度的丝带切割成长度为a、b或c的小段，使得小段数量最大的问题。通过动态规划方法实现了这一目标，并给出了具体的代码实现。

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A. Cut Ribbon

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Polycarpus has a ribbon, its length is n. He wants to cut the ribbon in a way that fulfils the following two conditions:

After the cutting each ribbon piece should have length a, b or c.
After the cutting the number of ribbon pieces should be maximum.

Help Polycarpus and find the number of ribbon pieces after the required cutting.

Input

The first line contains four space-separated integers n, a, b and c (1 ≤ n, a, b, c ≤ 4000) — the length of the original ribbon and the acceptable lengths of the ribbon pieces after the cutting, correspondingly. The numbers a, b and c can coincide.

Output

Print a single number — the maximum possible number of ribbon pieces. It is guaranteed that at least one correct ribbon cutting exists.

Examples

input

Copy

5 5 3 2

output

Copy

input

Copy

7 5 5 2

output

Copy

Note

In the first example Polycarpus can cut the ribbon in such way: the first piece has length 2, the second piece has length 3.

In the second example Polycarpus can cut the ribbon in such way: the first piece has length 5, the second piece has length 2.

#include<bits/stdc++.h>
using namespace std;
#define maxn 5000
int a[maxn];
int dp[maxn];
int main()
{
    int n;
    scanf("%d%d%d%d",&n,&a[0],&a[1],&a[2]);
    memset(dp,-9999,sizeof(dp));
    //将n分成长度为a,b,c的几份，使得份数最大
    dp[0]=0;
    for(int i=0;i<3;i++)
    {
        for(int j=a[i];j<=n;j++)
        {
            dp[j]=max(dp[j],dp[j-a[i]]+1);
        }
    }
    printf("%d\n",dp[n]);
    return 0;
}