BSGS模板（a^x==b(mod c)）

a^x==b(mod c) 可以求满足的最小自然数x 

#include <iostream>
#include <algorithm>
#include <cmath>
#include <tr1/unordered_map>
using namespace std::tr1;
using namespace std;
void exgcd(int a,int &x,int b,int &y){//ax+by=1
    if(!b){
        x=1;y=0;
        return ;
    }
    exgcd(b,y,a%b,x);
    y-=a/b*x;
}
int inverse(int x,int y){//x^(-1)(mod y) <=> x*x^(-1)+y*k=1
    int inv_x,k;
    exgcd(x,inv_x,y,k);
    return (inv_x%y+y)%y;
}
int BSGS(int a,int b,int c){//a^x=b(mod c)
    //特判答案<=100的情况
    for(int x=0,pow_a_x=1%c;x<=100;++x){
        if(pow_a_x==b)return x;
        pow_a_x=(long long)pow_a_x*a%c;
    }
    //通过预处理使得a,c互质
    int base_count=0,D=1;
    while(1){
        int d=__gcd(a,c);
        if(d==1)break;
        if(b%d)return -1;
        b/=d;c/=d;
        D=(long long)D*(a/d)%c;
        ++base_count;
    }
    b=(long long)b*inverse(D,c)%c;
    //解a^(x-base_count)=b(mod c)
    int n=sqrt(c);
    unordered_map<int,int>hash_table;
    int pow_a_j=1;
    for (int j = 1; j <= n;++j){
        pow_a_j=(long long)pow_a_j*a%c;
        hash_table[(long long)pow_a_j*b%c]=j;
    }
    int pow_a_n=pow_a_j,pow_a_in=1,max_i=(c+n-1)/n;
    for (int i = 1; i <= max_i;++i){
        pow_a_in=(long long)pow_a_in*pow_a_n%c;
        if(hash_table.count(pow_a_in)) return i*n-hash_table[pow_a_in]+base_count;
    }
    return -1;
}
int main(){

    return 0;
}