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ACM ICPC 2017 Warmup Contest 9 I

I. Older Brother

Your older brother is an amateur mathematician with lots of experience. However, his memory is very bad. He recently got interested in linear algebra over finite fields, but he does not remember exactly which finite fields exist. For you, this is an easy question: a finite field of order q exists if and only if q is a prime power, that is, q = $p^kpk holds for some prime number pand some integer k ≥ 1. Furthermore, in that case the field is unique (up to isomorphism).$

The conversation with your brother went something like this:

Input

The input consists of one integer q, satisfying 1 ≤ q ≤ $10^9109.$

Output

Output “yes” if there exists a finite field of order q. Otherwise, output “no”.

样例输入1

样例输出1

no

样例输入2

样例输出2

yes

样例输入3

样例输出3

yes

题目来源

ACM ICPC 2017 Warmup Contest 9

题意：问一个数n是否是一个素数p的k次方

思路：用Pollard_rho分解质因数，看一看所有的质因子是否相等。

  1 //2017-10-24
  2 #include <cstdlib>
  3 #include <iostream>
  4 #include <ctime>
  5 
  6 typedef long long LL;
  7 #define MAXN 10000
  8 
  9 using namespace std;
 10 
 11 LL factor[MAXN];
 12 int tot;
 13 const int S=20;
 14 
 15 LL muti_mod(LL a,LL b,LL c){    //返回(a*b) mod c,a,b,c<2^63
 16     a%=c;
 17     b%=c;
 18     LL ret=0;
 19     while (b){
 20         if (b&1){
 21             ret+=a;
 22             if (ret>=c) ret-=c;
 23         }
 24         a<<=1;
 25         if (a>=c) a-=c;
 26         b>>=1;
 27     }
 28     return ret;
 29 }
 30 
 31 LL pow_mod(LL x,LL n,LL mod){  //返回x^n mod c ,非递归版
 32     if (n==1) return x%mod;
 33     int bit[90],k=0;
 34     while (n){
 35         bit[k++]=n&1;
 36         n>>=1;
 37     }
 38     LL ret=1;
 39     for (k=k-1;k>=0;k--){
 40         ret=muti_mod(ret,ret,mod);
 41         if (bit[k]==1) ret=muti_mod(ret,x,mod);
 42     }
 43     return ret;
 44 }
 45 
 46 bool check(LL a,LL n,LL x,LL t){   //以a为基，n-1=x*2^t，检验n是不是合数
 47     LL ret=pow_mod(a,x,n),last=ret;
 48     for (int i=1;i<=t;i++){
 49         ret=muti_mod(ret,ret,n);
 50         if (ret==1 && last!=1 && last!=n-1) return 1;
 51         last=ret;
 52     }
 53     if (ret!=1) return 1;
 54     return 0;
 55 }
 56 
 57 bool Miller_Rabin(LL n){
 58     LL x=n-1,t=0;
 59     while ((x&1)==0) x>>=1,t++;
 60     bool flag=1;
 61     if (t>=1 && (x&1)==1){
 62         for (int k=0;k<S;k++){
 63             LL a=rand()%(n-1)+1;
 64             if (check(a,n,x,t)) {flag=1;break;}
 65             flag=0;
 66         }
 67     }
 68     if (!flag || n==2) return 0;
 69     return 1;
 70 }
 71 
 72 LL gcd(LL a,LL b){
 73     if (a==0) return 1;
 74     if (a<0) return gcd(-a,b);
 75     while (b){
 76         LL t=a%b; a=b; b=t;
 77     }
 78     return a;
 79 }
 80 
 81 //找出任意质因数
 82 LL Pollard_rho(LL x,LL c){
 83     LL i=1,x0=rand()%x,y=x0,k=2;
 84     while (1){
 85         i++;
 86         x0=(muti_mod(x0,x0,x)+c)%x;
 87         LL d=gcd(y-x0,x);
 88         if (d!=1 && d!=x){
 89             return d;
 90         }
 91         if (y==x0) return x;
 92         if (i==k){
 93             y=x0;
 94             k+=k;
 95         }
 96     }
 97 }
 98 
 99 //递归进行质因数分解N
100 void findfac(LL n){
101     if (!Miller_Rabin(n)){
102         factor[tot++] = n;
103         return;
104     }
105     LL p=n;
106     while (p>=n) p=Pollard_rho(p,rand() % (n-1) +1);
107     findfac(p);
108     findfac(n/p);
109 }
110 
111 int main(){
112     int n;
113     while(cin>>n){
114         if(n == 1){
115             cout<<"no"<<endl;
116             continue;
117         }
118         tot = 0;
119         findfac(n);
120         bool ok = 1;
121         for(int i = 1; i < tot; i++)
122           if(factor[i] != factor[i-1]){
123               ok = 0;
124               break;
125           }
126         if(ok)cout<<"yes"<<endl;
127         else cout<<"no"<<endl;
128     }
129     return 0;
130 }

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原文地址：https://www.cnblogs.com/Penn000/p/7723815.html