2^x mod n = 1 【杭电-HDOJ-1395】 附题

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/*
2^x mod n = 1
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11800    Accepted Submission(s): 3673
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

*/

#include<stdio.h>int main(){    int i,j,n,s;    while(~scanf("%d",&n)){      if(n==1||n%2==0)          printf("2^? mod %d = 1\n",n);      else{          __int64  s=1;          int i;          for(i=1;;i++){              s*=2;              s%=n;        //控制s的值在一定范围内,乘2的同时对n取余               if(s==1){    //  if(s%n==1)                  break;               }           }         printf("2^%d mod %d = 1\n",i,n);      }   } return 0;}



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