题目地址：

https://www.luogu.com.cn/problem/CF189A

题面翻译：
给一长度为$n$的缎带，要求将其剪成若干长度为$a,b,c$的缎带，且缎带数量尽可能多。

输入格式：
输入仅一行，四个正整数$n,a,b,c(n,a,b,c\le 4000)$。

输出格式：
输出仅一行，即缎带数量的最大值。

题目描述：
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.

输入格式：
The first line contains four space-separated integers $n$ , $a$ , $b$ and $c$ $(1\le n,a,b,c\le 4000)$ — 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.

输出格式：
Print a single number — the maximum possible number of ribbon pieces. It is guaranteed that at least one correct ribbon cutting exists.

完全背包问题，设$f[k][n]$为只考虑前$k$种长度的情况下，长$n$的缎带最多能切成多少段，可以按照最后一段切出的长度是否是$a[k]$来分类，如果不是，则最多$f[k-1][n]$；如果是，则最多$f[k][n-a[k]]$，取最大即可。需要注意在递推的时候，一定要保证方案存在。代码如下：

#include <iostream>
#include <cstring>
using namespace std;

const int N = 4010;
int n, a[4];
int f[N];

int main() {
  scanf("%d", &n);
  for (int i = 1; i <= 3; i++) scanf("%d", &a[i]);

  memset(f, -1, sizeof f);
  f[0] = 0;
  for (int i = 1; i <= 3; i++)
    for (int j = a[i]; j <= n; j++)
      // 当方案存在的时候才递推
      if (~f[j - a[i]])
        f[j] = max(f[j], f[j - a[i]] + 1);

  printf("%d\n", f[n]);
}
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时空复杂度$O(n)$。