PAT 1096 甲级 Consecutive Factors (20分)

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本文介绍了一种算法，用于找出正整数N中最大的连续因子序列，并列出最小的连续因子序列。通过两次循环实现，首先从2到N的平方根范围内寻找可能的连续因子，再在满足条件的序列中找到最长的连续因子个数。

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
567

这题要求一定数学功底，逻辑想不出来就别强求了…
利用两次循环。
第一次循环从i = 2 开始，一直到N的开平方为止。因为开平方之后的计算都不成立。
内部的第二次循环要求连续 j 都是N的因子，只要N%j != 0 即求下一个j++。否则退出循环，外层i++。
第二次循环期间要求计算满足条件 j 的个数。
在第一次循环结束末尾，判断MAX和满足条件的pos(PS：就是i) 是否需要更换。
输出结果。

#include<iostream>
#include<vector>
#include<algorithm>
#include<math.h>

using namespace std;

int main(){
	int N;
	cin>>N;
	int i  = 2,j = 2;
	int sqr = sqrt(1.0*N);
	int count = 0;
	int n = N;
	int max = 0;
	int start = N;
	while(j <= sqr){
		i = j ;
		while(i <= n){
			if(N%i == 0){
				N /= i;
				count++;
				i++;
			}else{
				break;
			}
		}
		if(max < count){
			max = count;
			start = j;
		}
		N = n ;
		j++;
		count = 0;
	}
	if(max){
		cout<<max<<endl;
		for(int i = 0;i < max;i++){
			if(!i){
				cout<<start;
			}else{
				cout<<"*"<<start + i;
			}			
		}
	}else{
		cout<<1<<endl;
		cout<<n;
	}
	
	
	return 0;
}