poj 3243 Clever Y 高次方程

  1  Accepted    8508K     579MS   C++    2237B/**
 hash的强大，，还是高次方程，不过要求n不一定是素数
 **/
 #include <iostream>
 #include <cstdio>
 #include <cmath>
 #include <cstring>
 #include <algorithm>
 using namespace std;
 long long a,b,n;
 const int maxn = ;
 bool Hash[maxn];
 long long val[maxn];
 long long idx[maxn];

 long long gcd(long long a,long long b){
     if(b==)
         return a;
     return gcd(b,a%b);
 }

 void ex_gcd(long long a,long long b,long long &x,long long &y){
     if(b==){
         x=;
         y=;
         return ;
     }
     ex_gcd(b,a%b,x,y);
     long long tmp= x-(a/b)*y;
     x = y;
     y = tmp;
     return ;
 }

 void Insert(long long id,long long num){
     long long k = num%maxn;
     while(Hash[k]&&val[k]!=num){
         k++;
         if(k==maxn) k = k-maxn;
     }
     if(!Hash[k]){
         Hash[k] = true;
         val[k] = num;
         idx[k] = id;
     }
     return;
 }

 long long found(long long num){
     long long k = num%maxn;
     while(Hash[k]&&val[k]!=num){
         k++;
         if(k==maxn) k-=maxn;
     }
     if(Hash[k]){
         return idx[k];
     }
     return -;
 }

 long long baby_step(long long a,long long b,long long n){
     long long temp =;
     long long i;
     for(i=;i<=;i++){
         if(temp==b%n) return i;
         temp = temp*a%n;
     }
     long long tmp,d =,cnt=;
     memset(Hash,false,sizeof(Hash));
     memset(val,-,sizeof(val));
     memset(idx,-,sizeof(idx));

     while((tmp=gcd(a,n))!=){
         if(b%tmp)
             return -;
         cnt++;
         n = n/tmp;
         b = b/tmp;
         d =d*a/tmp%n;
     }
     long long cur =;
     long long m = ceil(sqrt(n+0.5));
     for(i=;i<m;i++){
         Insert(i,cur);
         cur = cur*a%n;
     }
     long long x,y;
     for(i=;i<m;i++){
         ex_gcd(d,n,x,y);
         x = x*b%n;
         x = (x%n+n)%n;
         long long k = found(x);
         if(k!=-)
             return i*m+k+cnt;
         d = d*cur%n;
     }
     return -;
 }

 int main()
 {
     while(scanf("%I64d%I64d%I64d",&a,&n,&b)==){
         if(a==&&b==&&n==)
             break;
         long long res = baby_step(a,b,n);
         if(res==-){
             printf("No Solution\n");
         }else{
             printf("%I64d\n",res);
         }
     }
     return ;
 }