codeforces 189A

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

5 5 3 2

output

2

input

7 5 5 2

output

2

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.

题目的大致意思是输入饰带的长度n和三个需求的长度a，b和c，将饰带分成a，b或c的长度，最多能分几个。与背包问题很类似，可以做一个表格进行分析，以第一个输入为例

	0	1	2	3	4	5
5	0	0	0	0	0	1
3	0	0	0	1	0	1
2	0	0	1	1	0	2

 

只有完全分解饰带的才会记录在数据中，如果a，b和c只能用一次，中间的递归过程为

for (i = 0;i < number;i++)
        for (j = length;j >= 0;j--)
            if (length >= l[i] && val[j] < val[j - l[i]] + 1)
                val[j] = val[j - l[i]] + 1;

如果可以重复使用，中间的递归过程为

for (i = 0;i < 3;i++)
            for (j = a[i];j <= len;j++)
                if (val[j] < val[j - a[i]] + 1)
                    val[j] = val[j - a[i]] + 1;

两个代码唯一区别就是从len开始还是从前面开始，从len开始当前的长度不会影响结果，从前面开始当前的长度会影响结果。

#include<iostream>
using namespace std;
const int INF = -999999;
int main()
{
    int len, a[5], i, j, ans, val[4003];
    while (scanf("%d%d%d%d", &len, &a[0], &a[1], &a[2]) != EOF)
    {
        for (i = 1;i <= len;i++)
            val[i] = INF;
        val[0] = 0;
        for (i = 0;i < 3;i++)
            for (j = a[i];j <= len;j++)
                if (val[j] < val[j - a[i]] + 1)
                    val[j] = val[j - a[i]] + 1;
        ans = val[len];
        cout << ans << endl;
    }
    return 0;
}

过渡版本为：

#include<stdio.h>
#include<string.h>
double ribbon(int number, int *l, int length)
{
    int i, j, val[4003];
    memset(val, 0, sizeof(val));
    for (i = 0;i < number;i++)
        for (j = length;j >= 0;j--)
            if (length >= l[i] && val[j] < val[j - l[i]] + 1)
                val[j] = val[j - l[i]] + 1;
    return val[length];
}
int main()
{
    int n, len[5];
    scanf("%d", &n);
    for (int i = 0;i < 3;i++)
        scanf("%d",&len[i]);
    int ans;
    ans = ribbon(3, len, n);
    printf("%d\n",ans);
    return 0;
}