Factorization Prime Factor Algorithm

 

# Include <stdio. h> <br/> # include <math. h> <br/> int main () <br/>{< br/> int n, I; <br/> while (scanf ("% d", & n) & n) <br/>{< br/> for (I = 2; I <= sqrt (n); I ++) <br/> if (n % I = 0 & (n/= I) <br/> printf ("% d", I --); <br/> printf ("% d/n", n); <br/>}< br/> return 0; <br/>}< br/>

 

 

Someone asked me how the program works.

 

Decompose n into prime factors

 

General method: Each I from 2 to sqrt (n) is divided by n. If I can be divisible, it is determined whether I is a prime number, if yes, I is a qualitative factor of n, and then n = n/I, And then I is returned to 2, and then the quality factor of n is searched.

 

My optimization: the general idea remains unchanged and some pruning is performed. First, I try to divide each I from 2 to sqrt (n) by n. If I can divide n, so we don't need to judge I, I must be the quality factor of n, and n = n/I, because n may have multiple identical quality factors. To avoid omission, we only need to set I --, when the next loop is skipped, It is offset by I ++. The value of I remains unchanged because it ranges from 2 ~ Each number of I has already determined whether it can divide n. Therefore, it is not necessary to roll back I to 2. The value of I remains unchanged when I jumps to the next loop, in the end, n will be converted into a prime number, which is also a prime factor. Therefore, writing a program is the code at the beginning. As for efficiency, I think it is still relatively high ..