TOJ 1203.Factoring Large Numbers

题目链接 : http://acm.tju.edu.cn/toj/showp1203.html

One of the central idea 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’s long long int datatype.

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.

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大数求因子的题，本来以为会超时，结果也没有，通过这道题坚信了自己平常使用的求因子、质数的方法还是蛮快的嘛。水题上代码不多说。

#include <stdio.h>using namespace std;int main(){    long long num;    while(~scanf("%lld",&num) && num!=-1){        for(long long i=2;i*i<=num;i++)            while(!(num%i)){                num/=i;                printf("    %lld\n",i);            }        if(num>1)            printf("    %lld\n",num);        printf("\n");    }}