Integration of Polynomial （周赛1）

最新推荐文章于 2023-01-07 06:30:00 发布

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最新推荐文章于 2023-01-07 06:30:00 发布

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本文链接：https://blog.youkuaiyun.com/ZHC_AC/article/details/47360353

本文深入探讨了复杂代码逻辑的解析方法，从代码的核心内容出发，详细介绍了如何通过分析关键信息来理解代码逻辑，并提供了一系列在实际编程中应用这些解析技巧的实例。

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Description

Suppose there are a polynomial which has n nonzero terms, please print the integration polynomial of the given polynomial.

The polynomial will be given in the following way, and you should print the result in the same way:

k[1] e[1] k[2] e[2] ... k[n] e[n]

where k[i] and e[i] respectively represent the coefficients and exponents of nonzero terms, and satisfiese[1] < e[2] < ... < e[n].

Note:

Suppose that the constant term of the integration polynomial is 0.
If one coefficient of the integration polynomial is an integer, print it directly.
If one coefficient of the integration polynomial is not an integer, please print it by using fractiona/b which satisfies that a is coprime to b.

Input

There are multiple cases.

For each case, the first line contains one integer n, representing the number of nonzero terms.

The second line contains 2*n integers, representing k[1], e[1], k[2], e[2], ..., k[n], e[n]。

1 ≤ n ≤ 1000

-1000 ≤ k[i] ≤ 1000, k[i] != 0, 1 ≤ i ≤ n

0 ≤ e[i] ≤ 1000, 1 ≤ i ≤ n

Output

Print the integration polynomial in one line with the same format as the input.

Notice that no extra space is allowed at the end of each line.

Sample Input

3
1 0 3 2 2 4

Sample Output

1 1 1 3 2/5 5

定积分

= =

#include <stdio.h>
#include <math.h>

int GY(int x,int y)
{
    int t,r;
    if(x<y)
    {
        t=x;x=y;y=t;
    }

    while(y!=0)
    {r=x%y;x=y;y=r;}

    return x;
}
int main()
{

        int n,x;
        int k[2000],e[2000];
        while(~scanf("%d",&n))
        {
            int i;
            int key;
            for(i=0;i<n;i++)
            {
                scanf("%d %d",&k[i],&e[i]);
                e[i]=e[i]+1;
            }

        for(i=0;i<n;i++)
        {
            if(i<n-1)
            {
                  if(k[i]<0)
                  {
                        x=-1*k[i];

                        if(x%e[i]==0)
                            printf("%d %d ",k[i]/e[i],e[i]);

                        else
                        {
                            key=GY(x,e[i]);
                            printf("%d/%d %d ",k[i]/key,e[i]/key,e[i]);
                         }
                  }

                  else
                  {
                        x=k[i];
                        if(x%e[i]==0)
                            printf("%d %d ",k[i]/e[i],e[i]);

                        else
                        {
                            key=GY(x,e[i]);
                            printf("%d/%d %d ",k[i]/key,e[i]/key,e[i]);
                        }
                  }
            }

            else
            {
                  if(k[i]<0)
                  {
                        x=-1*k[i];
                        if(x%e[i]==0)
                            printf("%d %d\n",k[i]/e[i],e[i]);

                        else
                        {
                            key=GY(x,e[i]);
                            printf("%d/%d %d\n",k[i]/key,e[i]/key,e[i]);
                         }
                  }

                  else
                  {
                        x=k[i];
                        if(x%e[i]==0)
                            printf("%d %d\n",k[i]/e[i],e[i]);

                        else
                        {
                            key=GY(x,e[i]);
                            printf("%d/%d %d\n",k[i]/key,e[i]/key,e[i]);
                        }
                  }
            }
        }
    }


    return 0;
}