hdu1395 2^x mod n = 1

/**********
2^x mod n = 1
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11751    Accepted Submission(s): 3659

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
题目大意：求一个数n，2的x次方对n取余，如果等于1，那么输出x，没有则输出？
难点：2的x次方可能很大，超出int型范围；
关键点：费马定理，剩余定理
解题时间：2014,07,31
解题思路：根据费马定理可知，只要n是一个大于2的奇数，都存在x，再有剩余定理
  保证在算2的x次方的时候不会超出范围
体会：定理知道了还要会用啊
*************/

#include<stdio.h>int main(){int n,i,s;while(~scanf("%d",&n)){s=1;if(n%2==0||n==1)printf("2^? mod %d = 1\n",n);else{for(i=1;;i++){s=s*2;if(s%n==1)break;s=s%n;//剩余定理保证s不超出范围 }printf("2^%d mod %d = 1\n",i,n);}}return 0;}