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 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.
输入
Each input file contains one test case, which gives the integer N ( 1 < N < 2 3 1 ) (1<N<2^31) (1<N<231).
输出
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.
思路
要连续，则最大也是从根号N，所以遍历判断即可。
代码
#include<iostream>
#include<cstdio>
#include<stdlib.h>
using namespace std;
int mmax = 0;
int dfs(int all, int cur)
{
int l = 0;
for (int i = cur; all % i == 0; i++)
{
all /= i;
l++;
}
return l;
}
int main()
{
int N;
int num = 0;
int start = 0;
cin >> N;
for (int i = 2; N/i >=i; i++)
{
num = dfs(N, i);
if (mmax < num)
{
mmax = num;
start = i;
}
}
if (mmax == 0)
{
printf("1\n");
}
else
printf("%d\n", mmax);
if (mmax >= 1)
{
printf("%d", start);
for (int i = 1; i <= mmax - 1; i++)
{
printf("*%d", start + i);
}
}
else {
printf("%d", N);
}
cout << endl;
}