poj 2417 Discrete Logging(A^x=B(mod c),普通baby_step)

#include <stdio.h>
#include <iostream>
#include <map>
#include <set>
#include <list>
#include <stack>
#include <vector>
#include <math.h>
#include <string.h>
#include <queue>
#include <string>
#include <stdlib.h>
#include <algorithm>
#define LL long long
#define _LL __int64
#define eps 1e-12
#define PI acos(-1.0)

using namespace std;
const int maxn = 499991;

bool hash[maxn+10];
int idx[maxn+10];
LL val[maxn+10];
//插入哈希表
void insert(int id, LL vv)
{
int v = vv % maxn;
while(hash[v] && val[v] != vv)
{
v++;
if(v == maxn)
v -= maxn;
}
if(!hash[v])
{
hash[v] = true;
idx[v] = id;
val[v] = vv;
}
}
//查找vv相应的jj，A^jj = vv
int found(LL vv)
{
int v = vv%maxn;
while(hash[v] && val[v] != vv)
{
v++;
if(v == maxn)
v -= maxn;
}
if(hash[v] == false)
return -1;
return idx[v];
}

void extend_gcd(LL a, LL b, LL &x, LL &y)
{
if(b == 0)
{
x = 1;
y = 0;
return;
}
extend_gcd(b,a%b,x,y);
LL t = x;
x = y;
y = t-a/b*y;
}
/*
A^x = B(mod C)
令x = i*M+j, 当中M = ceil(sqrt(C*1.0))，(0 <= i,j < M)
那么原式变为A^M^i*A^j = B(mod c)
先枚举j(0~M-1)，将A^j%C存入hash表中
令D = A^M%C，X = A^j,那么D^i*X = B(mod C)
枚举i(0~M-1)求得D^i设为DD。DD*X = B(mod C)
DD，C已知，由于C是素数，gcd(DD,C) = 1,依据扩展欧几里得知在[0,C-1]内有唯一一个解X。
然后在hash表中查找X相应的jj。即A^jj = X。

那么x = i*M+jj，若找不到jj无解。
*/
LL baby_step(LL A, LL B, LL C)
{
memset(hash,false,sizeof(hash));
memset(idx,-1,sizeof(idx));
memset(val,-1,sizeof(val));

LL M = ceil(sqrt(C*1.0));
//将A^j存入hash表中
LL D = 1;
for(int j = 0; j < M; j++)
{
insert(j,D);
D = D*A%C;
}
//D = A^M%C,res = D^i,求方程res*X = B(mod C)中的X，去找X相应的jj，那么x=i*M+jj.
LL res = 1,x,y;
for(int i = 0; i < M; i++)
{
extend_gcd(res,C,x,y);
x = x*B;
x = (x%C+C)%C;
int jj = found(x);
if(jj != -1)
{
return (LL)i*M+jj;
}
res = res*D%C;
}
return -1;
}

int main()
{
LL A,B,C;
while(~scanf("%lld %lld %lld",&C,&A,&B))
{
LL res = baby_step(A,B,C);
if(res == -1)
printf("no solution\n");
else
printf("%lld\n",res);
}
return 0;
}