D - Cut Ribbon （dp）

最新推荐文章于 2022-10-23 18:09:04 发布

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最新推荐文章于 2022-10-23 18:09:04 发布

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本文介绍了一道关于如何将固定长度的绸缎按照指定长度进行切割的问题，目标是在遵循给定长度约束的情况下，尽可能多地切割出绸缎段落。文章详细解析了解决方案的思路，并提供了一个使用动态规划算法实现的具体代码示例。

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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

Input

7 5 5 2

Output

题目解析：

题目说给出给出一个绸缎的长度n。然后，是三个数，a,b,c。就是表示绸缎可以剪成a,b,c的长度。求，最多可以剪多少段？

思路解析：

刚看到题的时候，把or看成and了，靠。一开始用母函数写了一个。wrong了。之后才知道是一道dp。我们可以把n看成背包容积，把a,b,c看成不同的花费。则状态转移方程式为dp[i] = max(dp[i],dp[i-x]+1)(i-x>=0)表示当长度为i的时候考虑放x的性价比是多少。所以，我们可以用一个for就可以解决该问题了。

#include <stdio.h>
#include <string.h>

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;
}