CodeForces 189A-Cut Ribbon

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本文探讨了如何使用动态规划解决切割特定长度丝带的问题，目标是在切割后得到尽可能多的长度为a、b或c的丝带片段。通过输入丝带原始长度n及可接受的片段长度a、b、c，算法输出最优的片段数量。实例展示了解决过程，并提供了AC代码作为参考。

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

小结：果然，dp我不是一般的弱，真的真的还需要多多锻炼.....

以下是AC代码：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#define mnum (~0U>>2)
int n,dp[4100];
void Init()
{
//printf("%d\n\n",-mnum);
for(int i=0;i<=n;i++)
dp[i]=-mnum;
}
int main()
{
int a,b,c;
while(~scanf("%d%d%d%d",&n,&a,&b,&c))
{
Init();
//printf("%d\n",dp[0]);
dp[0]=0;
for(int i=1;i<=n;i++)
{
if(i-a>=0)
dp[i]=dp[i-a]+1>dp[i]?dp[i-a]+1:dp[i];
if(i-b>=0)
dp[i]=dp[i-b]+1>dp[i]?dp[i-b]+1:dp[i];
if(i-c>=0)
dp[i]=dp[i-c]+1>dp[i]?dp[i-c]+1:dp[i];
}
printf("%d\n",dp[n]);
}
return 0;
}