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

25

Sample Output

2^? mod 2 = 12^4 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;}}}}