HDU 1395

 

2^x mod n = 1

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16883    Accepted Submission(s): 5262

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 n;
    while(~scanf("%d",&n)&&n){
        if(n==1||n%2==0){
            printf("2^? mod %d = 1\n",n);
            continue;
        }
        else{
            int x=1,mut=2;
            mut%=n;
            while(mut!=1){
                x++;
                mut=mut*2%n;
            }
            printf("2^%d mod %d = 1\n",x,n);
        }
    }
    return 0;
}

开始时Time Limit Exceeded

改进1.n==1情况  2.循环式取余，减少运算量