2^X mod n = 1

Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 13345 Accepted Submission (s): 4146

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 Input25

Sample output2^? MoD 2 = 12^4 MoD 5 = 1

Authorma, Xiao

#include <cstdio> #include <cmath>int powermod (int a,int b,int c)//fast power {    int ans=1;    if (a%c==0) return 0;    a=a%c;    while (b)    {        if (b&1)            ans=ans*a%c;        a=a*a%c;        b>>=1;    }    return ans;} int main () {    int i,n;    The odd number except 11 has the result, the even number must have no result while    (~SCANF ("%d", &n))    {        if (n%2==0| | N==1)//2^x dual number to find the remainder result is even, not 1   1 when the result does not exist       {printf ("2^? MoD%d = 1\n ", n); continue;}            for (i=1;; i++)//For 2^X mod N, when 1<=i<=n can get all the remainder results            if (powermod (2,i,n) ==1)            {                printf ("2^%d mod%d = 1\ n ", i,n);                Break;}}}    

　　

2^X mod n = 1