HDU 1395 2^x mod n = 1

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本文介绍了一个算法问题，即给定一个正整数n，如何找到满足2^x模n等于1的最小正整数x。通过一个C++代码示例，展示了如何解决这一问题，特别关注于当n为奇数时的特殊情况。

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

爆搜

代码如下：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#define ll long long
#define inf 2147483647
using namespace std;
int main(){
    int n;
    while(scanf("%d",&n)!=EOF){
        int ans=2,t=1;
        if(n%2==0||n==1){
            printf("2^? mod %d = 1\n",n);
        }else{
            while(ans!=1){
                ans*=2;
                ans%=n;
                t++;
            }
            printf("2^%d mod %d = 1\n",t,n);
        }
    }
    return 0;
}