ZOJ 1489 HDU1395 2^x mod n = 1 数学

2^x mod n = 1

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Time Limit:
2 Seconds      Memory Limit:65536 KB

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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",n);
		else{
		  while(t%n!=1){
		    t*=2;
	        k++;
	        t%=n;
		  }
		  printf("2^%d mod %d = 1\n",k,n);
	    }
	}
	return 0;
}