2^x mod n = 1（欧拉定理，欧拉函数，快速幂乘）

2^x mod n = 1

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9231    Accepted Submission(s): 2837

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.

 

思路：

1. 当n为偶数时，bn + 1（b为整数）是奇数，而2^x是偶数，故 2^x mod n = 1不可能成立；

2. 当n等于1时，不能成立

3. 当n为非1的奇数时，n和2互质，由欧拉定理：若a，n为正整数，且两者互素，则a^phi(n) mod n = 1，其中phi(n)是n的欧拉函数。知2^phi(n) mod n = 1.因此phi(n)必是符合要求的x，但phi(n)未必是最小的，遍历小于其的正整数，逐一试验即可，计算2^x mod n时用快速幂乘。

 

AC Code：

#include <iostream>
#include <vector>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;

//计算n的欧拉函数
int Eular(int n)
{
    int res = 1, i;
    for (i = 2; i * i <= n; i++){
        if (n % i == 0){
            n /= i;
            res *= (i - 1);
            while (n % i == 0){
                n /= i;
                res *= i;
            }
        }
    }
    if (n > 1) res *= (n - 1);
    return res;
}

//快速幂乘计算2^b % n
int myPow(int b, int n)  
{
    if(b == 0) return 1;
    long long c = myPow(b >> 1, n);
    c = (c * c) % n;
    if(b & 1) c = (2 * c) % n;
    return c;
}

int main()
{
    int n, x;
    bool ok;
    while(scanf("%d", &n) != EOF){
        ok = 0;
        if((n & 1) && (n - 1)){
            ok = 1;
            int phi = Eular(n);
            for(x = 1; x < phi; x++){
                if(myPow(x, n) == 1) break;
            }
        }
        if(ok) printf("2^%d mod %d = 1
", x, n);
        else printf("2^%? mod %d = 1
", n);
    }
    return 0;
}