Ladder match L1-006 factor consecutive title

Simple simulation

Integer decomposed into a continuous factors.

Input formats:

In a given input row positive integer  N .

Output formats:

First, the number of the first output line of the longest continuous factors; in line 2, and then press the  因子1*因子2*……*因子k format of the output of the minimum contiguous factor sequence, wherein the ascending order output factor, not including one.

Sample input:

630

 

Sample output:

3
5*6*7


Violence done, simply enumerate what, keep the maximum length 
must be open ll wow! 
Code:

 1 #include <stdio.h>
 2 #include <iostream>
 3 #include <algorithm>
 4 #include <string.h>
 5 #include <queue>
 6 #include <map>
 7 #include <math.h>
 8 #include <set>
 9 #include <vector>
10 #include <stack>
11 #define ll long long
12 #define mod 998244353
13 const int maxn=2e5+5;
14 const int nowmi = 0x3f3f3f3f;
15 const int N=1000010;
16 using namespace std;
17 
18 ll isprime(ll x)
19 {
20     if(x==1)
21         return 0;
22     for(ll i=2;i*i<=x;i++)
23         if(x%i==0)
24             return 0;
25     return 1;
26 }
27 int main(){
28 
29     ll n;
30      CIN >> n-;
 31 is      IF (isPrime (n-))
 32      {
 33 is          COUT << . 1 << endl << n-;
 34 is          return  0 ;
 35      }
 36      LL len = 0 , S, ANS; // maximum length, the product , continuous factors most front of a 
37 [      for (LL I = 2 ; I * I <= n-; I ++ ) {
 38 is          IF (n-% I == 0 ) {
 39              S = I;
 40              LL J;
 41 is              for (J = + I . 1 ; J * J <= n-; J ++ ) {
42                 s*=j;
43                 if(n%s!=0)
44                     break;
45             }
46             if(j-i>len){//更新长度
47                 len=j-i;
48                 ans=i;
49             }
50         }
51     }
52     cout<<len<<endl;
53     for(ll i=ans;i<=ans+len-1;i++){
54         if(i!=ans)
55             cout<<"*";
56         cout<<i;
57     }
58     return 0;
59 }