PAT 1096 Consecutive Factors （暴力，最大连续乘积）

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

给你一个数N，该数可以表示成 c* i *（i+1）*（i+2）*... , 问你连续序列最长是多少

直接暴力搞，先判断一下素数....

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
bool prime(int num)
{
if(num==2) return true;
int i,n=sqrt(num);
for(i=2;i<=n;i++) {
if(num%i==0) return false;
}
return true;
}
int main()
{
int t,n,m,i,j,s,len;
scanf("%d",&n);
m=sqrt(n);
if(prime(n)) printf("1\n%d\n",n);
else {
s=0;len=0;
for(i=2;i<=m;i++) {
t=1;
for(j=0;;j++) {
t*=(i+j);
if(n%t) break;
}
if(j>len) {
len=j;
s=i;
}
}
printf("%d\n",len);
for(i=s;i<s+len;i++) {
if(i!=s) printf("*");
printf("%d",i);
}
printf("\n");
}
return 0;
}