P2485 [SDOI2011]计算器 取模三大基础模板

手敲了这题，差不多入门同余了

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define ls (o<<1)
#define rs (o<<1|1)
#define pb push_back
//#define a(i,j) a[(i)*(m+2)+(j)]  //m是矩阵的列数
const double PI= acos(-1.0);
const int M = 1e5+7;
/*
int head[M],cnt;
void init(){cnt=0,memset(head,0,sizeof(head));}
struct EDGE{int to,nxt,val;}ee[M*2];
void add(int x,int y,int z){ee[++cnt].nxt=head[x],ee[cnt].to=y,ee[cnt].val=z,head[x]=cnt;}
*/
ll qpow(ll a,ll b,ll p)
{
	ll ans=1;
	while(b)
	{
		if(b&1)ans=ans*a%p;
		a=a*a%p;
		b/=2;
	}
	return ans;
}
ll bsgs(ll a,ll b,ll p)// a^x \equiv  b (mod p)
{
	map<ll,ll>hash;hash.clear();
	b%=p;
	ll t=(ll)sqrt(p)+1;
	for(int j=0;j<t;j++)
	{
		ll val=(ll)b*qpow(a,j,p)%p;//b*a^j
		hash[val]=j;
	}
	a=qpow(a,t,p);//a^t
	if(a==0)return b==0?1:-1;//0 % b =0  只有b是0才有解 
	for(ll i=0;i<=t;i++)
	{
		ll val=qpow(a,i,p);
		ll j= hash.find(val)==hash.end() ? -1 : hash[val];
		if(j>=0&&i*t-j>=0)return i*t-j;//最非负整数解 
	}
	return -1;
}
ll exgcd(ll a,ll b,ll &x,ll &y)
{
	if(b==0)
	{
		x=1,y=0;
		return a;
	}
	ll d=exgcd(b,a%b,x,y);
	ll z=x;
	x=y,y=z-y*(a/b);
	return d;
}
int main()
{
	ios::sync_with_stdio(false);
  	cin.tie(0);
  	ll a,b,p;
  	int t,q;
  	cin>>t>>q;
  	while(t--)
  	{
  		cin>>a>>b>>p;
  		if(q==1)cout<<qpow(a,b,p)<<endl;
  		if(q==2)
  		{
  			ll x,y;
  			// a*x+p*y=b 
  			ll tp=exgcd(a,p,x,y);
  			//cout<<tp<<"  -- "<<x<<endl;
  			x*=b/tp;//一个特解 
  			ll m=p/tp;
  		//	cout<<m<<endl;
  			if(b%tp!=0)cout<<"Orz, I cannot find x!"<<endl;
  			else cout<<(x%m+m)%m<<endl;
		}
		if(q==3)
		{
			int tp=bsgs(a,b,p);
			if(tp>=0)cout<<tp<<endl;
			else cout<<"Orz, I cannot find x!\n";
		}
	}
	  	
	return 0;
}