杭電1395

2^x mod n = 1
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 11658    Accepted Submission(s): 3634
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
 
 
 
考え方は:オラの定理
aとmの相互作用であり、a
mが奇数であり、mが1に等しくなければよい
次のコードがあります.
 #include
int main()
{
 int n;
 while(~scanf("%d",&n))
 {
  if(n%2&&n>1)
  {
   int i,s=1;
   for(i=1;;i++)/奇数は必ず解けると判断したのでこのような暴力で必ず飛び出す
   {
    s=s*2%n;
    if(s==1)
    {
     printf("2^%d mod %d = 1",i,n);
     break;
    }
   }
  }
  else
  printf("2^? mod %d = 1",n);
 }
}