杭电1395 利用同余手法提高效率

 

Problem Description
Give a number n, find the minimum x(x>0) that satisfies 2^x mod n = 1.
 

 

Input
One positive integer on each line, the value of n.
 

 

Output
If the minimum x exists, print a line with 2^x mod n = 1.

Print 2^? mod n = 1 otherwise.

You should replace x and n with specific numbers.
 

 

Sample Input
2 5
 

 

Sample Output
2^? mod 2 = 1 2^4 mod 5 = 1
如果不用同余手法处理 往往是会超时的
同余思想就是多个数乘积的余数等于各个数的余的乘积
posted @ 2016-02-29 22:24  猪突猛进!!!  阅读(210)  评论(0编辑  收藏  举报