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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): 17913 Accepted Submission(s): 5609

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

Author MA, Xiao

Source ZOJ Monthly, February 2003

Recommend Ignatius.L | We have carefully selected several similar problems for you: 2462 2421 2521 1695 2466 分析： 歐拉定理 1、初等數論中的歐拉定理：　　對於互質的整數a和n，有a^φ(n) ≡ 1 (mod n) φ(n)為n的歐拉函數 如果我們設m為φ(n)，那麽m可能不是最小的解，但是最小的解必然是m的因數 所以我們只需要判定m的因數即可 代碼如下：

#include <cstdio>
#include 
 
<iostream>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long ll;
ll euler(ll n){ //返回euler(n)
     ll res=n,a=n;
     for(ll i=2;i*i<=a;i++){
         if(a%i==0){
             res=res/i*(i-1);//先進行除法是為了防止中間數據的溢出
             while(a%i==0
 
) a/=i;
         }
     }
     if(a>1) res=res/a*(a-1);
     return res;
}
ll mi(ll b,ll n)
{
  ll ans=1;
  ll a=2;
  while(b>0)
  {
      if(b&1)ans=ans*a%n;
      b=b/2;
      a=(a*a)%n;
  }
  return ans;
}
int main()
{
     ll n,h;
    while(scanf("%lld",&n)!=EOF){
        vector<ll>V;
      if(n%2==0||n==1)
      printf("2^? mod %lld = 1\n",n);
     else
    {
      ll  m=euler(n);
       for(ll i=1;i*i<=m;i++)
       {
          if(m%i==0){
             V.push_back(i);
             if(i!=m/i)
              V.push_back(m/i);
          }
       }
       sort(V.begin(),V.end());
       for(ll i=0;i<V.size();i++)
       {
        if(mi(V[i],n)==1){
        printf("2^%lld mod %lld = 1\n",V[i],n);
        break;
        }
       }
    }
    }
    return 0;
}

HDU 1395 2^x mod n = 1 （歐拉函數）