uva 10392 - Factoring Large Numbers

最新推荐文章于 2021-11-25 20:45:14 发布

滑头鬼之亨

最新推荐文章于 2021-11-25 20:45:14 发布

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分类专栏：数论待解决文章标签： numbers cryptography input integer output 算法

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本文介绍了一种简单但效率较低的大数质因数分解算法。该算法通过遍历从2开始的所有整数来尝试分解输入的大数，直到找到所有质因数。尽管这种方法对于非常大的数可能效率不高，但对于理解基本的质因数分解过程很有帮助。

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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'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

90
1234567891
18991325453139
12745267386521023
-1

Sample Output

    2
    3
    3
    5

    1234567891

    3
    3
    13
    179
    271
    1381
    2423

    30971
    411522630413

分解质因数

偷懒的作法时间复杂度极高，下次换个好的算法再做过...

#include<stdio.h>
#include<math.h>
void main()
{long long a,k,i;
 while(scanf("%lld",&a),a!=-1)
 {k=sqrt(a);
  i=2; 
  while ((a!=1)&&(i<=k))
  {while (a%i==0) {a=a/i; printf("    %lld\n",i);}
   ++i;
  }
  if (a!=1) printf("    %lld\n",a);
  printf("\n");
 }
}