Consecutive Factors (20)

题目描述
Among all the factors of a positive integer N, there may exist several consecutive numbers. For example, 630 can be factored as 356*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.

输入描述:
Each input file contains one test case, which gives the integer N (131).

输出描述:
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.

输入例子:
630

输出例子:
3
567

思路:
这道题挺有意思，最先我是想要将找到所有因素然后维护最大，后面发现想错了，因为输出并没有让我们求完所有因素，这道题只需要找到最大连续因素。数n的大小是int，大概在12!以内，又因为我们枚举时其实到sqrt(n)就可以了，所以时间复杂度大致在11*sqrt(n)，直接暴力维护最大值就可以了

#include <iostream>
#include <cmath>
using namespace std;
int main(){
    long long n;
    cin>>n;
    long long sqr=(long long)sqrt(1.0*n);
    long long first=0,len=0;               //len为最长连续整数的长度，first是其中第一个
    for(long long i=2;i<=sqr;i++){         //遍历第一个整数
        long long temp=1,j;                //temp为当前连续整数的乘积
        for(j=i;;j++){
            temp*=j;
            if(n%temp!=0) break;           //j不断加1直到temp不能整除n
        }
        if(j-i>len){                       //更新最长长度与第一个整数
            len=j-i;
            first=i;
        }
    }
    if(len==0) cout<<1<<endl<<n;          //根号n范围内都无解，只能输出n本身（素数）
    else{
        cout<<len<<endl;
        for(long long i=0;i<len;i++){
            cout<<first+i;
            if(i<len-1) cout<<"*";
        }
    }
    return 0;
}