D - Cut Ribbon （dp）

D -Cut Ribbon

Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Description

Polycarpus has a ribbon, its length isn. He wants to cut the ribbon in a way that fulfils the following two conditions:

After the cutting each ribbon piece should have lengtha, b orc. 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 integersn, 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 Input

Input

5 5 3 2

Output

2

Input

7 5 5 2

Output

2

題目解析：

題目說給出給出一個綢緞的長度n。然後，是三個數，a,b,c。就是表示綢緞可以剪成a,b,c的長度。求，最多可以剪多少段？

思路解析：

剛看到題的時候，把or看成and了，靠。一開始用母函數寫了一個。wrong了。之後才知道是一道dp。我們可以把n看成背包容積，把a,b,c看成不同的花費。則狀態轉移方程式為dp[i] = max(dp[i],dp[i-x]+1)(i-x>=0)表示當長度為i的時候考慮放x的性價比是多少。所以，我們可以用一個for就可以解決該問題了。

#include 
#include 

const int INF = ~0U >>2;
const int N = 4e3 + 5;
int n,dp[N];

void Init()
{
    for(int i = 0;i <= n;++i)
      dp[i] = -INF;
}
int main()
{
    int a,b,c;
    while(~scanf("%d%d%d%d",&n,&a,&b,&c))
    {
        Init();dp[0] = 0;//
        for(int i = 1;i <= n;++i)
        {
            if(i - a>=0){
               if(dp[i] < dp[i-a]+1)
                  dp[i] = dp[i-a]+1;
            }
            if(i - b >=0){
               if(dp[i] < dp[i-b]+1)
                 dp[i] = dp[i-b]+1;
            }
            if(i - c >=0){
               if(dp[i] < dp[i-c]+1)
                 dp[i] = dp[i-c]+1;
            }
        }
        printf("%d\n",dp[n]);
    }
    return 0;
}