2^x mod n = 1
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
题意：输入一个数，计算2的几次方能除以它取余等于1。输出格式如上。
这道题终点在数值上，也许x有很大，久指数爆炸了，所以不能直接求2的多少次方，要乘一次取一次余。
#include<stdio.h>
int main()
{
int k,e,n;
while(~scanf("%d",&n))
{
printf("2^");
if(n%2==0||n==1)
{
printf("?");
goto end;
}
e=0;
k=1;
while(++e)
{
k<<=1;
k%=n;
if(k==1)
{
printf("%d",e);
goto end;
}
}
end:
printf(" mod %d = 1\n",n);
}
}
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