PAT A1096.Consecutive Factors

Among all the factors of a positive integer N, there may exist several consecutive numbers. For example, 630 can be factored as 3*5*6*7, where 5, 6, and 7 are the three consecutive numbers. Now given any positive N, you are supposed to find the maximum number of consecutive factors, and list the smallest sequence of the consecutive factors.

Input Specification:

Each input file contains one test case, which gives the integer N (1<N<2³¹).

Output Specification:

For each test case, print in the first line the maximum number of consecutive factors. Then in the second line, print the smallest sequence of the consecutive factors in the format "factor[1]*factor[2]*...*factor[k]", where the factors are listed in increasing order, and 1 is NOT included.

Sample Input:

630

Sample Output:

3
5*6*7

/********************************************************************/

看到这道题目，我的第一反应是又写了一遍生成素数表的代码。

写完回头一看，妈的原来是考虑全部因数，不是只考虑素因数……

还是非常简单的题目，只要逐一遍历就成，需要注意的有俩点

1.数值上限很大，在遍历到数值N的根的时候就可以结束循环了。

2.因为1导致的，如果只遍历到数值N的根时，对于N是素数的情况会出现问题。解决的方式也很简单，这种情况下单独输出1和N的值就行。

附上代码，其中大部分是一开始写的素数表，之后被我注释掉了……

#include <iostream>
#include <cmath>
using namespace std;
/*int * MakePrimeList(int &PrimeCount ,int N)
{
    bool *IsNotPrime = new bool[N/2+1]();
    PrimeCount = 1;
    for(int i=2;i<=N/4;i++)
    {
        for(int j=i;j*i<=N/2;j++)
        {
            if(IsNotPrime[i*j]==false)
            {
                IsNotPrime[i*j]=true;
            }
        }
    }
    int *PrimeList = new long long[N/2+1]();
    for(int i = 2;i<=N/2;i++)
    {
        if(IsNotPrime[i]==false)
        {
            PrimeList[PrimeCount]=i;
            PrimeCount++;
        }
    }
    delete []IsNotPrime;
    return PrimeList;
}*/

int main()
{
    int N = 0;

    //int *PrimeList = new long long[N/2+1]();
    //int *FactorList = new int [N/2+1]();
    cin>>N;

        //int PrimeCount=0,FactorCount = 0;
        //PrimeList = MakePrimeList(PrimeCount,N);

        /*

        for(int i = 1;i<=PrimeCount;i++)
        {
            cout<<PrimeList[i]<<' ';
        }

        cout<<endl;
        //*/
        //int temp = 1;
        /*while(N!=1)
        {
            if(N%PrimeList[temp]==0)
            {
                N/=PrimeList[temp];
                FactorList[++FactorCount]=PrimeList[temp];
                cout<<PrimeList[temp]<<'*';
            }
            else temp++;
        }
        for(int i=1;i<=FactorCount;i++)
        {
            cout<<FactorList[i]<<' ';
        }*/
        int MaxSeq = 0,tempSeq = 1,FirstNum = 2,tempNum = 2,tempProduct = N;
        int sqr = (int)sqrt(1.0*N);
        for(int i=2;i<=sqr;i++)
        {
            if(MaxSeq>=1 && i>sqr) break;

            if(N%i==0)
            {
                tempNum=i;
                tempProduct/=i;
                tempSeq=1;
                while(tempProduct%(++tempNum)==0)
                {
                    tempProduct/=tempNum;
                    tempSeq++;
                }
                if(MaxSeq<tempSeq)
                {
                    FirstNum = i;
                    MaxSeq=tempSeq;
                }
                tempProduct = N;
            }
        }
        if(MaxSeq==0)
            cout<<1<<endl<<N<<endl;  
        else
        {
            cout<<MaxSeq<<endl;
            int i =0;
            for(i=FirstNum;i<FirstNum+MaxSeq-1;i++)
            {
                cout<<i<<'*';
            }
            cout<<i;
        }

    //delete []PrimeList;
    //delete []FactorList;
    return 0;
}