Codeforces Round #119 (Div. 2)A. Cut Ribbon

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.

这题n^2枚举可以解决，但是也可以用dp O(n)啊

dp得到的数。

dp[i]代表 组成i 需要的最多ribbon块数。

/* ***********************************************
Author        :guanjun
Created Time  :2016/10/7 18:22:02
File Name     :cf119a.cpp
************************************************ */
#include <bits/stdc++.h>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 10010
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << 61;
const double eps=1e-5;
using namespace std;
priority_queue<int,vector<int>,greater<int> >pq;
struct Node{
    int x,y;
};
struct cmp{
    bool operator()(Node a,Node b){
        if(a.x==b.x) return a.y> b.y;
        return a.x>b.x;
    }
};

bool cmp(int a,int b){
    return a>b;
}
int dp[4010];
int main()
{
    #ifndef ONLINE_JUDGE
    //freopen("in.txt","r",stdin);
    #endif
    //freopen("out.txt","w",stdout);
    int n,a,b,c;
    while(cin>>n>>a>>b>>c){
        cle(dp);
        dp[a]=dp[b]=dp[c]=1;
        for(int i=1;i<=n;i++){
            if(i>a&&dp[i-a])dp[i]=max(dp[i],dp[i-a]+1);
            if(i>b&&dp[i-b])dp[i]=max(dp[i],dp[i-b]+1);
            if(i>c&&dp[i-c])dp[i]=max(dp[i],dp[i-c]+1);
        }
        cout<<dp[n]<<endl;
    }
    return 0;
}