Hdoj 1395 2 ^ x mod n = 1 [brute force]

Strategy: we can see that the remainder is not equal to 1 or the number that can be divisible by 2. We only need to judge whether it is an odd number other than 1, search for 2 ^ X (mod (N) in sequence ))? = 1.

Difficulty: if every time it is based on the original × 2, it will time out. At this time, we can use the same remainder theorem.

AC by SWS;

Question link: http://acm.hdu.edu.cn/showproblem.php? PID = 1, 1395

Code:

# Include <stdio. h> int main () {int N; while (scanf ("% d", & n) = 1) {If (n = 1 | n % 2 = 0) {printf ("2 ^? MoD % d = 1 \ n ", n); continue;} else {int sum = 2, I, Dc = 1; for (I = 1; I ++) if (sum % n! = 1) {sum = sum % N * 2; // same remainder theorem sum × 2% N = (sum % N) × (2% N) = (2 <n) => sum % N * 2} elsebreak; printf ("2 ^ % d mod % d = 1 \ n", I, n) ;}} return 0 ;}