HDU 1395

讲解转载自：杭电Discuss

1）当n为1，无解。

2）n为偶数2^x显然为偶数，而1为奇数，2^n和1不可能关于n同余，x无解。

3）n为奇数时（n与2互素），由费尔马定理知当x=n-1为一解（但不一定是最小），此时暴力即可。值得注意的是，暴力时为了减小运算量，可以先取摸，再乘2，即代码中的i=(i%n)*2。否则会TLE。

代码（G++）:

#include <cstdlib>
#include <iostream>

using namespace std;

int main(int argc, char *argv[])
{
int n,x,t;
while(cin>>n)
{
if(0==n%2||1==n)  cout<<"2^? mod "<<n<<" = 1"<<endl;
else{
t=2;
for(x=1;x<50000;x++)
{
if(1==t%n)
{
cout<<"2^"<<x<<" mod "<<n<<" = 1"<<endl;
break;
}else t=t%n*2;
}
}
}
system("PAUSE");
return EXIT_SUCCESS;
}

附上原题：

2^x mod n = 1

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)



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