uva 10392 Factoring Large Numbers

Problem :

One of the central ideas behind much cryptography is that factoring large numbers is computationally intensive. In this context one might use a 100 digit number that was a product of two 50 digit prime numbers. Even with the fastest projected computers this factorization will take hundreds of years.

You don't have those computers available, but if you are clever you can 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 be at 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 ascending order with 4 leading spaces preceding a left justified number, and followed by a single blank line.

Sample Input

9012345678911899132545313912745267386521023-1

Sample Output

    2    3    3    5    1234567891    3    3    13    179    271    1381    2423    30971    411522630413

题目大意：将一个数分解成最小的因子。

解题思路：想不到什么好办法 , 就用最笨的办法 , 然后过了 , 不要学我.

#include<stdio.h>#include<math.h>int main() {long long N;while (scanf("%lld", &N) != EOF, N > 0) {long long i = 0;for (i = 2; i < sqrt(N); i++) {    //上限到根号N,不然会超时while (N % i == 0) {printf("    %lld\n", i);N = N / i;}}if (i = N) {printf("    %lld\n", N);}printf("\n");}return 0;}