/*
2^x mod n = 1
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11800 Accepted Submission(s): 3673
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
*/
#include<stdio.h>
int main(){
int i,j,n,s;
while(~scanf("%d",&n)){
if(n==1||n%2==0) printf("2^? mod %d = 1\n",n);
else{
__int64 s=1;
int i;
for(i=1;;i++){
s*=2;
s%=n; //控制s的值在一定范围内，乘2的同时对n取余
if(s==1){ // if(s%n==1)
break;
}
}
printf("2^%d mod %d = 1\n",i,n);
}
}
return 0;
}
2^x mod n = 1 【杭电-HDOJ-1395】 附题,布布扣,bubuko.com
2^x mod n = 1 【杭电-HDOJ-1395】 附题
原文：http://blog.csdn.net/holyang_1013197377/article/details/38321333