bzoj 2242 计算器

数论裸题，只要会算法即可。

膜拜了bsgs，注意以下几个点

1.y要先模p，特判y=0，z=0情况

2.y^x不要用快速幂求，累乘即可

3.小步的时候不要忘*z

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>

//#define md
#define ll long long
#define inf (int) 1e9
#define eps 1e-8
#define N 1000010
using namespace std;

struct ha
{
	int a[N],pos[N];
	int mxn;
	void clear() { memset(a,-1,sizeof(a)); mxn=1000000;}
	void insert(int x,int ps)
	{
		int now=x%mxn;
		while (a[now]!=-1) {now++; if (now>mxn) now=1; } 
		a[now]=x; pos[now]=ps;
	}
	int find(int x)
	{
		int now=x%mxn;
		while (a[now]!=-1) { if (a[now]==x) return pos[now]; now++; if (now>mxn) now=1; }
		return -1;
	}
}hash;

int mi(ll a,int b,int md)
{
	ll ans=1;
	while (b)
	{
		if (b&1) ans=ans*a%md;
		a=a*a%md; b>>=1;
	}
	return (int)ans;
}
int ni(ll a,int md) { return mi(a,md-2,md); }
int gcd(int a,int b) { return b==0?a:gcd(b,a%b);}
void exgcd(int a,int b,ll &x,ll &y)
{
	if (b==0) { x=1; y=0; return; }
	exgcd(b,a%b,x,y);
	ll t=y; y=x-(a/b)*y; x=t;
}
	
int main()
{
	int tt,opt;
	scanf("%d%d",&tt,&opt);
	while (tt--)
	{
		switch (opt)
		{
			case 1:
				{
					ll a; int b,md;
					scanf("%lld%d%d",&a,&b,&md);
					printf("%d\n",mi(a,b,md));
					break;
				}
			case 2:
				{
					int a,b,c;
					scanf("%d%d%d",&a,&c,&b);
					int t=gcd(a,b);
					if (c%t) { printf("Orz, I cannot find x!\n"); continue;}
					a/=t; b/=t; c/=t;
					ll x,y;
					exgcd(a,b,x,y);
					x=x*c%b; while (x<0) x+=b;
					printf("%lld\n",x);
					break;
				}
			case 3:
				{
					ll y,z; int md;
					scanf("%lld%lld%d",&y,&z,&md);
					y%=md;
					if (!y&&!z) { printf("1\n"); continue;}
					if (!y) { printf("Orz, I cannot find x!\n"); continue;}
					int m=sqrt(md-1),ans=-1;
					ll ji=z,nic=ni(y,md);
					hash.clear();
					for (int i=0;i<m;i++) { hash.insert((int)ji,i); ji=ji*nic%md;}
					ji=1; ll c=mi(y,m,md);
					for (int i=0;i<m;i++)
					{
						int x=hash.find(ji);
						if (x!=-1) { ans=i*m+x; break; }
						ji=ji*c%md;
					}
					if (ans==-1) printf("Orz, I cannot find x!\n"); else printf("%d\n",ans);
					break;
				}
		}
	}
	return 0;
}