Problem F: Factoring Large Numbers

One of the central ideas behind much cryptography is that factoringlarge numbers is computationally intensive. In this context one mightuse a 100 digit number that was a product of two 50 digit primenumbers. Even with the fastest projected computers this factorizationwill take hundreds of years.

You don't have those computers available, but if you are clever youcan still factor fairly large numbers.

Input

The input will be a sequence of integer values, one per line,terminated by a negative number. The numbers will fit in gcc'slong long int datatype.You may assume that there will beat most one factor more than 1000000.

Output

Each positive number from the input must be factored and all factors(other than 1) printed out. The factors must be printed in ascendingorder with 4 leading spaces preceding a left justified number, andfollowed by a single blank line.

Sample Input

90

1234567891

18991325453139

12745267386521023

-1

Sample Output

    2

    3

    3

    5

    1234567891

    3

    3

    13

    179

    271

    1381

    2423

    30971

    411522630413

题意:将给出的N分解成几个素数的乘式, 但不包括1

做法: 从2开始遍历到N的开方数, 每一个因子都拿去重复的判断是否能对其整除, 若能, 输出因子i, 更新N, 如此做还能提高效率~

AC代码:

#include<stdio.h>

#include<math.h>

int main() {

	long long int N;

	while(scanf("%lld", &N) != EOF) {

		if(N < 0)

			break;

		for(int i = 2; i <= sqrt(N); i++) {

			while(N % i == 0) {

				printf("    %d\n", i);

				N = N / i;

			}

		}

		if(N > 1)

			printf("    %lld\n", N);

		printf("\n");

	}

	return 0;

}