Give a number n, find the minimum x 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：
给你正整数n，求最小的x使得2^x mod n = 1。
1：n=1无解，%后为0。
2：n为偶数无解，（偶%偶）==偶！=1。
3：n为奇数一定有解，对于乘法逆元：在a mod n的操作下，a存在乘法逆元当且仅当a与n互质。（a为偶数，n为奇数，互质）
http://blog.csdn.net/adjky/article/details/70341771 ，大师的代码先阁起，看得懂了再来看。
#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
int n;
int main()
{
while(cin>>n)
{
int k=1,t=2;
if(n==1||n%2==0) printf("2^? mod %d = 1
",n);
else{
while(t%n!=1){
t*=2;
k++;
t%=n;
}
printf("2^%d mod %d = 1
",k,n);
}
}
return 0;
}