zjfc1080-----2^x mod n = 1

Code
#include<stdio.h>
int process(int i,int n)
{
    long t=2,d=1;
    while(i){
        if(i%2==1)
            d=d*t%n;
        t=t*t%n;
        i/=2;
    }
    return d%n;
}
int main()
{
    int n,i;
    while(scanf("%d",&n)!=EOF){
        if(n%2==0||n==1){
            printf("2^? mod %d = 1\n",n);
            continue;}
        else{
            i=1;
            while(i){
                if(process(i,n)==1){
                    printf("2^%d mod %d = 1\n",i,n);
                    break;
                }
                i++;
            }
        }

    }
}

当n为偶数或者1时，x无解

当n为奇数时，有费马小定理可得，取x=n-1

当n为合数时，n=p1*p2*p3... pi为素数。

所以可以取x=(p1-1)*(p2-1)*(p3-1)      同理可得2^x mod n = 1

转载于:https://www.cnblogs.com/pandy/archive/2009/04/30/1447295.html