标签：01 08 ribbon length cutting 2021 piece dp first

E - Cut Ribbon

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.

题意: 就是一个长为N的彩带，剪成长度为a,b,c中的一种，色带片的数量应最大
题解: 这个题虽然简单但是，是一个完全背包问题。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int dp[4010];//不能开小 
int main()
{
	int i,j,n,m,t[3];
	scanf("%d %d %d %d",&n,&t[0],&t[1],&t[2]);
	memset(dp,-40,sizeof(dp));//负数要尽量小。 
	dp[0]=0;
	for(i=0;i<3;i++)//3个可以“无限取的”，“完全背包” 
	{
		for(j=t[i];j<=n;j++)//枚举每一个，尽可能的装。 
		{
			dp[j]=max(dp[j],dp[j-t[i]]+1);// 
		}
	}
	printf("%d\n",dp[n]);
	return 0;
}

这个是我感觉讲背包比较好的链接
https://valen.blog.csdn.net/article/details/87878853?spm=1001.2014.3001.5506

标签：01,08,ribbon,length,cutting,2021,piece,dp,first
来源： https://blog.csdn.net/weixin_54383625/article/details/119301390

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