HDU 1395 2^x mod n = 1（快速幂取模）

2^x mod n = 1

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15722    Accepted Submission(s): 4871

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

Source

ZOJ Monthly, February 2003

AC代码：

 

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<cstdlib>
#include<iomanip>
#include<algorithm>
#include<time.h>
typedef long long LL;
using namespace std;
int q_mod(long long a,int b,int c)  //快速幂取模 
{
	long long ans=1;
	while(b>0)
	{
		if(b&1)  //b&1相当于b%2==1 
			ans=(ans*a)%c;
			 
		b>>=1;  //右移 
		a=(a*a)%c;
	}
	return ans;
}
int main()
{
	int x,n;
	while(cin>>n)
	{
		if(!(n&1)||n<=1)
			cout<<"2^? mod "<<n<<" = 1"<<endl;
		else
		{
			for(int i=1;;i++)
				if(q_mod(2,i,n)==1)
				{			
					cout<<"2^"<<i<<" mod "<<n<<" = 1"<<endl;
					break;
				}
		}

	}
}