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 = 499991;
bool Hash[maxn];
long long val[maxn];
long long idx[maxn];

long long gcd(long long a,long long b){
    if(b==0)
        return a;
    return gcd(b,a%b);
}

void ex_gcd(long long a,long long b,long long &x,long long &y){
    if(b==0){
        x=1;
        y=0;
        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 -1;
}

long long baby_step(long long a,long long b,long long n){
    long long temp =1;
    long long i;
    for(i=0;i<=100;i++){
        if(temp==b%n) return i;
        temp = temp*a%n;
    }
    long long tmp,d =1,cnt=0;
    memset(Hash,false,sizeof(Hash));
    memset(val,-1,sizeof(val));
    memset(idx,-1,sizeof(idx));

    while((tmp=gcd(a,n))!=1){
        if(b%tmp)
            return -1;
        cnt++;
        n = n/tmp;
        b = b/tmp;
        d =d*a/tmp%n;
    }
    long long cur =1;
    long long m = ceil(sqrt(n+0.5));
    for(i=0;i<m;i++){
        Insert(i,cur);
        cur = cur*a%n;
    }
    long long x,y;
    for(i=0;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!=-1)
            return i*m+k+cnt;
        d = d*cur%n;
    }
    return -1;
}

int main()
{
    while(scanf("%I64d%I64d%I64d",&a,&n,&b)==3){
        if(a==0&&b==0&&n==0)
            break;
        long long res = baby_step(a,b,n);
        if(res==-1){
            printf("No Solution
");
        }else{
            printf("%I64d
",res);
        }
    }
    return 0;
}