Codeforces 189A：Cut Ribbon（完全背包，DP）

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最新推荐文章于 2024-02-15 14:54:16 发布

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分类专栏： Codeforces DP

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本文介绍了一个典型的完全背包问题案例，通过帮助角色Polycarpus切割丝带的方式，最大化切割后的丝带段数量，每段长度限定为预设的几种可能。文章提供了完整的代码实现，展示了如何使用动态规划解决该问题，并特别关注了初始化问题及其解决方案。

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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 $c$ $(1 \leq n, a, b, c \leq 4000)$ — 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

input

7 5 5 2

output

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 $2$ .

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 $d p$ 数组全部初始化为 $- 1$ 的话，需要注意 $dp[j]=−1&&dp[j−a[i]]=−1dp[j]=-1\&\&dp[j-a[i]]=-1$ 的情况。或者就全部初始值给成小于 $- 1$ 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;
}