杭电 HDU ACM 1395 2^x mod n = 1

时间：2015-04-05 09:06:40 收藏：0 阅读：104

2^x mod n = 1

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13610 Accepted Submission(s): 4208

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

Author

MA, Xiao

对x递增，每步取模，ls数组记录是否此余数出现过，如果出现过，那么接下来必将循环，所以出现重复即跳出即可。

#include<iostream>
using namespace std;
int main()
{
	int n,i,j;
	while(cin>>n)
	{
		int ls[10000];int sum=1;
		memset(ls,0,sizeof(ls));
		for(i=1; ;i++)
		{
			sum*=2;
			sum%=n;
			
			if(sum==1)
			{
				cout<<"2^"<<i<<" mod "<<n<<" = 1"<<endl;
				break;
			}
				
			 if(ls[sum])
			 {
				 cout<<"2^? mod "<<n<<" = "<<"1"<<endl;
				 break;
			 }
			 ls[sum]=1;
		}
	}
	return 0;
}