Problem F: Factoring Large Numbers
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's
long 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
90 1234567891 18991325453139 12745267386521023 -1
Sample Output
2
3
3
5
1234567891
3
3
13
179
271
1381
2423
30971
411522630413
题目大意:
求出一个数所有的质因子
解析:
一个数如果能被前面的质因子完全整除,那么新的数就不会被含有这个质因子的数整除。
#include<stdio.h>
#include<math.h>
int main() {
long long num;
while(scanf("%lld",&num) != EOF) {
if(num < 0)
break;
for(long long i=2;i<=sqrt(num);i++) {
while(num % i ==0) {
num /= i;
printf(" %lld\n",i);
}
}
printf(" %lld\n\n",num);
}
return 0;
}
本文详细阐述了如何使用高效算法来分解具有100位数的数字,这些数字由两个50位的质数相乘而成。通过迭代检查从2开始的每个整数是否为该数的因子,并将因子按升序打印,直至分解完成。
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