HDOJ 2^x mod n = 1 1395

2^x mod n = 1

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13630    Accepted Submission(s): 4213

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

25

 

Sample Output

2^? mod 2 = 12^4 mod 5 = 1

 

Author

MA, Xiao

 

Source

ZOJ Monthly, February 2003

 

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#include<stdio.h>int F_pow(int x,int n,int modn){int res=1;while(n){    if(n&1) res=res*x%modn;    x=x*x%modn;    n>>=1;}    return res;}int main(){int F_pow(int x,int n,int modn);int n;while(scanf("%d",&n)!=EOF){int i=0;if(!(n&1)||n<=1){printf("2^? mod %d = 1\n",n);continue;}int num=0;for(i=1;;i++){num=F_pow(2,i,n);if(num==1) break;}printf("2^%d mod %d = 1\n",i,n);}return 0;}