Codeforces 189 A. Cut Ribbon(DP 恰装满的完全背包问题)

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

2

input

Copy

7 5 5 2

output

Copy

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.

题意：一块长为n的布，现在要把它剪开，只能剪成a, b, c这样的小块布，求最多能剪成多少块。。。

分析：就是一个需要刚好装满的完全背包问题，只有三种商品a, b, c，能取无限件物品，每件物品价值是1，求最大价值。。。。

代码：

#include<iostream>
#include<string>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define ll long long
using namespace std;
int INF = 1e9+7;
const int maxn = 4e4+7;
int n, a[3], dp[maxn];

int main()
{
    scanf("%d%d%d%d", &n, &a[0], &a[1], &a[2]);
    dp[0] = 0;
    for(int i = 1; i <= n; i++) dp[i] = -INF;   //刚好装满的背包，dp[0] = 0, dp[1..n] = -INF
    for(int i = 1; i <= n; i++)
         for(int j = 0; j < 3; j++)
            if(i >= a[j]) dp[i] = max(dp[i], dp[i - a[j]] + 1);
    printf("%d\n",dp[n]);
    return 0;
}