Codeforces 189A：Cut Ribbon（完全背包，DP）

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 nnn. He wants to cut the ribbon in a way that fulfils the following two conditions:

After the cutting each ribbon piece should have length aaa, bbb or ccc.
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 nnn, aaa, bbb and ccc (1 ≤ n, a,b, c≤ 4000)(1 ≤ n, a,b, c≤ 4000)(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 aaa, bbb and ccc 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

input

7 5 5 2

output

Note

In the first example Polycarpus can cut the ribbon in such way: the first piece has length 222, the second piece has length 333.

In the second example Polycarpus can cut the ribbon in such way: the first piece has length 555, the second piece has length 222.

Solve

完全背包，看成给出三种商品，恰好能够装满背包的情况

注意初始化的问题，如果把dpdpdp数组全部初始化为−1-1−1的话，需要注意dp[j]=−1&&dp[j−a[i]]=−1dp[j]=-1\&\&dp[j-a[i]]=-1dp[j]=−1&&dp[j−a[i]]=−1的情况。或者就全部初始值给成小于−1-1−1的数

Code

/*************************************************************************

	 > Author: WZY

	 > School: HPU

	 > Created Time:   2019-04-01 18:30:38

************************************************************************/

#include <cmath>

#include <cstdio>

#include <time.h>

#include <cstring>

#include <limits.h>

#include <iostream>

#include <algorithm>

#include <random>

#include <iomanip>

#include <map>

#include <set>

#include <stack>

#include <queue>

#include <vector>

#include <string>

#include <random>

#define ll long long

#define ull unsigned long long

#define lson o<<1

#define rson o<<1|1

#define ms(a,b) memset(a,b,sizeof(a))

#define SE(N) setprecision(N)

#define PSE(N) fixed<<setprecision(N)

#define bug cerr<<"-------------"<<endl

#define debug(...) cerr<<"["<<#__VA_ARGS__":"<<(__VA_ARGS__)<<"]"<<"\n"

#define LEN(A) strlen(A)

const double E=exp(1);

const double eps=1e-9;

const double pi=acos(-1.0);

const int mod=1e9+7;

const int maxn=1e6+10;

const int maxm=1e3+10;

const int moha=19260817;

const int inf=1<<30;

const ll INF=1LL<<60;

using namespace std;

inline void Debug(){cerr<<'\n';}

inline void MIN(int &x,int y) {if(y<x) x=y;}

inline void MAX(int &x,int y) {if(y>x) x=y;}

inline void MIN(ll &x,ll y) {if(y<x) x=y;}

inline void MAX(ll &x,ll y) {if(y>x) x=y;}

template<class FIRST, class... REST>void Debug(FIRST arg, REST... rest){

	cerr<<arg<<"";Debug(rest...);}

int dp[maxn];

int main(int argc, char const *argv[])

{

	ios::sync_with_stdio(false);cin.tie(0);

	cout.precision(20);

	#ifndef ONLINE_JUDGE

	    freopen("in.txt", "r", stdin);

	    freopen("out.txt", "w", stdout);

		srand((unsigned int)time(NULL));

	#endif

	int n;

	int a[4];

	cin>>n>>a[0]>>a[1]>>a[2];

	ms(dp,-1);

	dp[0]=0;

	for(int i=0;i<3;i++)

		for(int j=a[i];j<=n;j++)

			if(dp[j-a[i]]!=-1)

				MAX(dp[j],dp[j-a[i]]+1);

	cout<<dp[n]<<endl;

	#ifndef ONLINE_JUDGE

	    cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s.\n";

	#endif

	return 0;

}

巴特西