杭电 1395 2^x mod n = 1 暴力题

      一直以为有什么高深的算法，，没想到暴力一下就能过。。这太悲剧了。可以用欧拉定理证明其存在性。欧拉定理是这样的，如果a和m互质且a

2^x mod n = 1

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5974    Accepted Submission(s): 1808

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

 

ac代码：

#include 
#include 
using namespace std;
int main(){
	int n;
	while(~scanf("%d",&n)){
		if(n%2&&n>1){
		  int s=1,x;
		  for(x=1;;++x){
		    s=s*2%n;
			if(s==1){
			  printf("2^%d mod %d = 1\n",x,n);break;
			}
		  }
		}
		else{
		  printf("2^? mod %d = 1\n",n);
		}
	}
  return 0;
}