HDU 1395 2^x mod n = 1 （欧拉函数）

Posted 2020-10-09 hinata_hajime

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

 

 

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分析：

欧拉定理 1、初等数论中的欧拉定理：　　对于互质的整数a和n，有a^φ(n) ≡ 1 (mod n)

φ(n)为n的欧拉函数

如果我们设m为φ(n)，那么m可能不是最小的解，但是最小的解必然是m的因数

所以我们只需要判定m的因数即可

代码如下：

#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);//先进行除法是为了防止中间数据的溢出
             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;
}