HDU 1395 2^x mod n = 1 violence ~ ~ Best to learn Euler theorem ~ ~ ~

2^X mod n = 1Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 13341 Accepted Submission (s): 4143


Problem descriptiongive a number n, find the minimum x (x>0) that satisfies 2^x mod n = 1.

Inputone positive integer on each line, the value of N.

Outputif 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

Authorma, Xiao Euler theorem Baidu Encyclopedia mode operation Nature as a acmer, or take a good look at the above Baidu Encyclopedia, very useful

According to modulo p multiplication inverse: For the integer A, p if there is an integer b, satisfies a*b mod p=1 is called B is a modulo p multiplication inverse element.

A necessary and sufficient condition of the multiplication inverse of the existence modulo P is gcd (a,p) = 1, which makes A=2^x,b=1,p=n

If present x uses 2^x mod n=1 then gcd (2^x,n) =1

(1) Because the value of x is greater than 0. The 2^x factor is only one 2, so when n is even, gcd (2^x,n) =2k (k=1,2,3 ...), i.e. no x at this time, makes 2^x mod n=1.

(2) When n is odd gcd (2^x,n) = 1, an x is required to make the 2^x mod n=1.

(3) As the result of any modulo 1 is 0, so when n=1, regardless of x value, 2^x mod n=0.

Combining the above (1), (2), (3), when the value of n is 1 or even, there is no X to make the 2^x mod n=1, and in other cases there must be an X to make 2^x mod n = 1.

Code:

#include <stdio.h>int main () {int n, while (~scanf ("%d", &n)) {if (N==1 | | n%2==0) {printf ("2^? MoD%d = 1\n ", n);} Else{int j = 1, Mi=2;while (true) {mi%= n, if (mi = = 1) {printf ("2^%d mod%d = 1\n", j,n); Mi *= 2; ++j;} }}return 0;}

with June

HDU 1395 2^x mod n = 1 violence ~ ~ Best to learn Euler theorem ~ ~ ~