Zoj problem set-1489 2 ^ x mod n = 1

Time Limit: 2 seconds memory limit: 65536 KB

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 shoshould 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

This question seems to have few descriptions. I like this question. I am not very good at English ....

The question is very clear, that is, to divide the X power of 2 by the remainder of N to the X value of 1. When I first saw this question, I thought it would be a good way to use the table-driven method, so I thought I would first create a table of 2 to the power of 32, then, you only need to traverse the table every time you obtain the result.Code:

 
# Include <iostream>
 
 
Using NamespaceSTD;
 
 
Const IntNum = 32;
 
Int* Pow2 =New Int[Num];
 
 
VoidInitial2pow ()
 
 
{
 
 
* Pow2 = 1;
 
For(IntI = 1; I <num; I ++)
 
 
* (Pow2 + I) = 2*(pow2 + I-1 ));
 
 
}
 
 
IntMain ()
 
 
{
 
 
 
Initial2pow ();
 
 
IntN;
 
 
While(CIN> N)
 
 
{
 
If(N % 2 = 0 | n <2)
 
 
{
 
 
Cout <"2 ^? MoD"<N <"= 1"<Endl;
 
 
Continue;
 
}
 
 
BoolFind =False;
 
 
For(IntI = 1; I <num; I ++)
 
 
{
 
If(* (Pow2 + I) % N = 1)
 
 
{
 
 
Cout <"2 ^"<I <"MoD"<N <"= 1"<Endl;
 
 
Find =True;
 
Break;
 
 
}
 
 
}
 
 
If(! Find)
 
 
{
 
Cout <"2 ^? MoD"<N <"= 1"<Endl;
 
 
}
 
 
}
 
 
}
 
 
I feel that the code is concise and can be submitted, but the submitted code gets wa! At this time, I realized that it may be due to precision issues, and the 32-bit int may not be enough. So I used 64-bit long to replace Int. After resubmitting, I still get wa! At this moment, I found that this question is still not that simple! At least it should not be simply a table driver to solve this problem. You still need some mathematical skills. First, I wrote the computing process on top of the PEN:
 
A/B = C .............. D. Multiply the two sides of the equation by 2 to get 2a/B = 2C ....... 2d. Now we can see that the remainder of 2a % B is the same as that of 2D % B!
 
 
If 2D/B = C0 ..... D0, then 2a/B = 2C + C0 ....... D0, with these mathematical inferences, the corrected code is as follows:
 
 
# Include <iostream>
 
 
Using NamespaceSTD;
 
 
IntMain ()
 
 
{
 
IntN;
 
 
While(CIN> N)
 
 
{
 
 
If(N % 2 = 0 | n <2)
 
 
{
 
Cout <"2 ^? MoD"<N <"= 1"<Endl;
 
 
Continue;
 
 
}
 
 
BoolFind =False;
 
IntD = 1;
 
 
For(IntI = 1; I ++)
 
 
{
 
 
D * = 2;
 
 
If(D % N = 1)
 
{
 
 
Cout <"2 ^"<I <"MoD"<N <"= 1"<Endl;
 
 
Find =True;
 
 
Break;
 
}
 
 
D = D % N;
 
 
}
 
 
If(! Find)
 
 
{
 
Cout <"2 ^? MoD"<N <"= 1"<Endl;
 
 
}
 
 
}
 
 
}
 
 
AC successful!