1096 Consecutive Factors （20 分)

2019-04-16

1096 Consecutive Factors （20 分)

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:

Sample Output:

1
2

3
5*6*7

作者: CHEN, Yue

单位: 浙江大学

时间限制: 400 ms

内存限制: 64 MB

代码长度限制: 16 KB

题目大意

给出一个数，分解因数。求因数可以组成连续最长序列的最小值。

分析

一开始拿到题没什么思路。网上查了一下，暴力解决。i从2开始，遍历至根号n，计算从i开始连续因数的长度。如果是长度为0，说明是质数，直接输出本身。

代码

#include <iostream>
#include <vector>
#include <stack>
#include <algorithm>
#include <deque>
#include <map>
#include <set>
#include <cstring>
#include <cmath>
#include <sstream>

using namespace std;
int n,cnt,start,maxcnt=-1;
int main() {
    cin>>n;
    int f=sqrt(n);
    int flag=0;
    for(int i=2;i<=f;i++){
        int x=n;
        cnt=0;
        for(int j=i;j<=f;j++){
            if(x%j==0){
                cnt++;
                x/=j;
                if(cnt>maxcnt){
                    maxcnt=cnt;
                    start=i;
                }
            } else
                break;
        }
    }
    if(maxcnt==-1)
    {
        cout<<"1"<<endl;
        cout<<n;
    }else {
        cout<<maxcnt<<endl;
        cout<<start;
        for(int i=start+1;i<start+maxcnt;i++)
            cout<<"*"<<i;
    }
    return 0;
}