Cut Ribbon



Cut Ribbon

time limit per test

2 seconds

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 ≤ 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.

Sample test(s)

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.

一个绳子，可以分成a,b,c长度，最多分成多少段？

一开始的暴力方法，全分成a，b,c的段数,然后暴力判断，然后超时- -。接着就是剪枝，贪心的想法，并且缩小b，c段数，即确定a之后再求b,c的段数

如果有，最大段数就是这个。然而有反例，比如53,10,11,23.那么在3个10，一个23就会退出，实际上2个10,3个11才是最优解。因此在判断b，c段的时候不可以跳出，而是用max来找。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int max(int a, int b)
{
 return a > b ? a : b;
}
int main()
{
 int i, j, m, n, a, b, c, ans, pre1, pre2, pre3, k, temp, leap = 0, temp1;
 ans = 0;
 cin >> n >> a >> b >> c;
 if (a > b){ temp = a; a = b; b = temp; }
 if (b > c){ temp = b; b = c; c = temp; }
 if (a > b){ temp = a; a = b; b = temp; }
 pre1 = n / a;
 if (n%a == 0)
  ans = n / a;
 else
 {
  for (i = pre1; i >= 0; i--)
  {
   temp = n - a*i;
   if (temp%b == 0)
   {
    ans = max(ans, temp / b + i);
   }
   else{
    pre2 = temp / b;
    for (j = pre2; j >= 0; j--)
    {
     temp1 = temp - b*j;
     if (temp1%c == 0)
     {
      ans = max(ans, i + j + temp1 / c);
      break;
     }
    }
   }
  }
 }
 cout << ans << endl;
 return 0;
}

2.网上用的是dp法，看的我眼都要瞎了。离散化dp，然后dp[i+a]=max(dp[i+a],dp[i]+1);

然后最后输出dp[n]，还需多练。还有就是从前向后找需要注意数组大小的问题，或者加一个判断if(i+a<=n)

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
int dp[100000];
int max(int a, int b)
{
 return a>b ? a : b;
}
int main()
{
 int a, b, c, n, i, j;
 cin >> n >> a >> b >> c;
 memset(dp, 0, sizeof(dp));
 dp[a] = dp[b] = dp[c] = 1;
 for (i = 1; i <= n; i++)
 {
  if (dp[i])
  {
   dp[i + a] = max(dp[i + a], dp[i] + 1);
   dp[i + b] = max(dp[i + b], dp[i] + 1);
   dp[i + c] = max(dp[i + c], dp[i] + 1);
  }
 }
 cout << dp[n] << endl;
 return 0;
}