dp第四题

#include <string.h>
#include <stdio.h>
#include <iostream>
using namespace std;
int dp[4050];
int main()
{
    int n,a,b,c;
    while(cin >> n >> a >> b >> c)
    {
        memset(dp,-1,sizeof(dp));
        dp[a]=1;dp[b]=1;dp[c]=1;
        for(int i=1;i<=n;i++)
        {
            if(i-a>0 && dp[i-a]!=-1 && dp[i-a]+1>dp[i]) dp[i]=dp[i-a]+1;
            if(i-b>0 && dp[i-b]!=-1 && dp[i-b]+1>dp[i]) dp[i]=dp[i-b]+1;
            if(i-c>0 && dp[i-c]!=-1 && dp[i-c]+1>dp[i]) dp[i]=dp[i-c]+1;
        }
//        for(int i=1;i<=n;i++)
//        {
//            cout << dp[i] << " ";
//        }
        cout << dp[n] << endl;
    }
    return 0;
}

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.

某晚codeforces上的2房A题。

状态:dp[i]代表结到长度为i时，所能结的最大段数。

状态转移方程:dp[i]=dp[i-a]+1;

考虑三个情况:

1.减出界;

2.初始值为-1,不进行处理,因为无法形成一段的剪法，再加一块，也无法形成正确的剪法;

3.dp[i]<dp[i-a]+1,求最大的段数,所以dp[i]要比dp[i-1]存的段数多。