HDU 2815 扩展baby step giant step 算法

题目大意就是求 a^x = b(mod c) 中的x

用一般的baby step giant step 算法会超时

这里参考的是http://hi.baidu.com/aekdycoin/item/236937318413c680c2cf29d4

map平衡树查找值

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
#include <map>
using namespace std;
#define ll long long

int q_pow(int a , int b , int mod)
{
    ll ans = 1;
    while(b)
    {
        if(b&1) ans =((ll)ans*a)%mod;
        a = ((ll)a*a)%mod;
        b>>=1;
    }
    return ans;
}

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

int ex_gcd(int a , int &x , int b , int &y)
{
    if(b == 0){
        x=1 , y=0;
        return a;
    }
    int ans = ex_gcd(b , x , a%b , y);
    int t = x;
    x=y , y = t-a/b*y;
    return ans;
}

int inv(int a , int b , int mod)
{
    int x , y , d;
    d = ex_gcd(a , x , mod , y);
    int e = (ll)x*b%mod;
    return e<0?e+mod:e;
}

int BabyStep(int A,int B,int C){
    map<int,int> Hash;
    ll buf=1%C,D=buf,K;
    int i,d=0,tmp;
    for(i=0;i<=100;buf=buf*A%C,++i)
        if(buf==B) return i;
    while((tmp=gcd(A,C))!=1)
    {
        if(B%tmp)return -1;
        ++d;
        C/=tmp;
        B/=tmp;
        D=D*A/tmp%C;
    }
    Hash.clear();
    int M=(int)ceil(sqrt((double)C));
    for(buf=1%C,i=0;i<=M;buf=buf*A%C,++i)
        if(!Hash.count((int)buf))Hash[(int)buf]=i;
    for(i=0,K=q_pow((ll)A,M,C);i<=M;D=D*K%C,++i)
    {
        tmp=inv(D,B,C);
        if(tmp>=0&&Hash.count(tmp))return i*M+Hash[tmp]+d;
    }
    return -1;
}
int main()
{
   // freopen("a.in" ,"r" , stdin);
    int k , p , n;
    while(scanf("%d%d%d" , &k , &p , &n) == 3)
    {
        if(n>=p){
            puts("Orz,I can’t find D!");
            continue;
        }
        int ans = BabyStep(k , n , p);
        if(ans == -1) puts("Orz,I can’t find D!");
        else printf("%d
" , ans);
    }
    return 0;
}

 hash表查找值（效率高很多）hash是线性查找：

#include<iostream>
#include<map>
#include<cmath>
#include <cstdio>
using namespace std;
typedef long long LL;
const int maxn = 65535;
struct hash{
    int a,b,next;
}Hash[maxn << 1];
int flg[maxn + 66];
int top,idx;
//hash值插入
void ins(int a,int b){
    int k = b & maxn;
    if(flg[k] != idx){
        flg[k] = idx;
        Hash[k].next = -1;
        Hash[k].a = a;
        Hash[k].b = b;
        return ;
    }
    while(Hash[k].next != -1){
        if(Hash[k].b == b) return ;
        k = Hash[k].next;
    }
    Hash[k].next = ++ top;
    Hash[top].next = -1;
    Hash[top].a = a;
    Hash[top].b = b;
}
//hash值查找
int find(int b){
    int k = b & maxn;
    if(flg[k] != idx) return -1;
    while(k != -1){
        if(Hash[k].b == b) return Hash[k].a;
        k = Hash[k].next;
    }
    return -1;
}
int gcd(int a,int b)
{
    return b?gcd(b,a%b):a;
}
int ext_gcd(int a,int b,int& x,int& y)
{
    int t,ret;
    if (!b){x=1,y=0;return a;}
    ret=ext_gcd(b,a%b,x,y);
    t=x,x=y,y=t-a/b*y;
    return ret;
}
int Inval(int a,int b,int n){
    int x,y,e;
    ext_gcd(a,n,x,y);
    e=(LL)x*b%n;
    return e<0?e+n:e;
}
int pow_mod(LL a,int b,int c)
{
    LL ret=1%c;
    a%=c;
    while(b){
        if(b&1)ret=ret*a%c;
        a=a*a%c;
        b>>=1;
    }
    return ret;
}
int BabyStep(int A,int B,int C){
    top = maxn; ++ idx;
    LL buf=1%C,D=buf,K;
    int i,d=0,tmp;
    for(i=0;i<=100;buf=buf*A%C,++i)
        if(buf==B)return i;
    while((tmp=gcd(A,C))!=1){
        if(B%tmp)return -1;
        ++d;
        C/=tmp;
        B/=tmp;
        D=D*A/tmp%C;
    }
    int M=(int)ceil(sqrt(C+0.5));
    for(buf=1%C,i=0;i<=M;buf=buf*A%C,++i)
        ins(i,buf);
    for(i=0,K=pow_mod((LL)A,M,C);i<=M;D=D*K%C,++i){
        tmp=Inval((int)D,B,C);
        int w ;
        if(tmp>=0&&(w = find(tmp)) != -1) return i*M+w+d;
    }
    return -1;
}
int main(){
    int A,B,C;
    while(scanf("%d%d%d",&A,&C,&B)!=EOF){
        if(B>C){
            puts("Orz,I can’t find D!");
            continue;
        }
        int tmp=BabyStep(A,B,C);
        if(tmp<0)
            puts("Orz,I can’t find D!");
        else printf("%d
",tmp);
    }
    return 0;
}