【SPOJ】Power Modulo Inverted（拓展BSGS）

【SPOJ】Power Modulo Inverted（拓展BSGS）

题面

洛谷
求最小的(y)

满足

[kequiv x^y(mod z) ]

题解

拓展(BSGS)模板题

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<set>
#include<map>
#include<vector>
#include<queue>
using namespace std;
#define ll long long
#define RG register
inline int read()
{
    RG int x=0,t=1;RG char ch=getchar();
    while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
    if(ch=='-')t=-1,ch=getchar();
    while(ch<='9'&&ch>='0')x=x*10+ch-48,ch=getchar();
    return x*t;
}
const int HashMod=123456;
struct HashTable
{
	struct Line{int u,v,next;}e[100000];
	int h[HashMod],cnt;
	void Add(int u,int v,int w){e[++cnt]=(Line){w,v,h[u]};h[u]=cnt;}
	void clear(){memset(h,0,sizeof(h));cnt=0;}
	void Insert(int x,int i)
	{
		int k=x%HashMod;
		Add(k,i,x);
	}
	int Query(int x)
	{
		for(int i=h[x%HashMod];i;i=e[i].next)
			if(e[i].u==x)return e[i].v;
		return -1;
	}
}Hash;
int fpow(int a,int b,int MOD)
{
	int s=1;
	while(b){if(b&1)s=1ll*s*a%MOD;a=1ll*a*a%MOD;b>>=1;}
	return s;
}
void NoAnswer(){puts("No Solution");}
void ex_BSGS(int y,int z,int p)
{
	if(z==1){puts("0");return;}
	int k=0,a=1;
	while(233)
	{
		int d=__gcd(y,p);if(d==1)break;
		if(z%d){NoAnswer();return;}
		z/=d;p/=d;++k;a=1ll*a*y/d%p;
		if(z==a){printf("%d
",k);return;}
	}
	Hash.clear();
	int m=sqrt(p)+1;
	for(int i=0,t=z;i<m;++i,t=1ll*t*y%p)Hash.Insert(t,i);
	for(int i=1,tt=fpow(y,m,p),t=1ll*a*tt%p;i<=m;++i,t=1ll*t*tt%p)
	{
		int B=Hash.Query(t);if(B==-1)continue;
		printf("%d
",i*m-B+k);return;
	}
	NoAnswer();
}
int main()
{
	int x,z,k;
	while(233)
	{
		x=read();z=read();k=read();
		if(x==0&&z==0&&k==0)break;
		ex_BSGS(x,k,z);
	}
	return 0;
}