洛谷P2485 [SDOI2011]计算器（exgcd+BSGS）

传送门

一题更比三题强

1操作直接裸的快速幂

2操作用exgcd求出最小正整数解

3操作用BSGS硬上

然后没有然后了

 //minamoto
 #include<cstdio>
 #include<map>
 #include<cmath>
 #include<iostream>
 #define ll long long
 using namespace std;
 #define getc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
 char buf[<<],*p1=buf,*p2=buf;
 inline ll read(){
     #define num ch-'0'
     char ch;bool flag=;ll res;
     while(!isdigit(ch=getc()))
     (ch=='-')&&(flag=true);
     for(res=num;isdigit(ch=getc());res=res*+num);
     (flag)&&(res=-res);
     #undef num
     return res;
 }
 inline ll ksm(ll a,ll b,ll p){
     ll res=;
     while(b){
         if(b&) (res*=a)%=p;
         (a*=a)%=p,b>>=;
     }
     return res;
 }
 ll exgcd(ll a,ll b,ll &x,ll &y){
     if(!b){x=,y=;return a;}
     ll d=exgcd(b,a%b,x,y);
     ll t=x;x=y,y=t-y*(a/b);
     return d;
 }
 map<ll,ll> mp;
 ll BSGS(ll a,ll b,ll p){
     mp.clear();
     b%=p;int t=sqrt((double)p)+,tot=;
     for(int j=;j<t;++j){
         int val=b*tot%p;
         mp[val]=j,tot=tot*a%p;
     }
     if(!tot) return b==?:-;
     a=tot,tot=;
     for(int i=;i<=t;++i){
         int j=mp.find(tot)==mp.end()?-:mp[tot];
         if(j>=&&i*t-j>=) return i*t-j;
         tot=tot*a%p;
     }
     return -;
 }
 ll t,k,z,y,p;
 void solve1(){
     while(t--){
         y=read(),z=read(),p=read();
         printf("%lld\n",ksm(y,z,p));
     }
 }
 void solve2(){
     while(t--){
         y=read(),z=read(),p=read();
         ll x=,yy=,d=;
         d=exgcd(y,p,x,yy);
         if(z%d!=){puts("Orz, I cannot find x!");}
         else{
             ll tmp=abs(p/d);
             x=(((x*z)/d)%tmp+tmp)%tmp;
             printf("%lld\n",x);
         }
     }
 }
 void solve3(){
     while(t--){
         y=read(),z=read(),p=read();
         ll ans=BSGS(y,z,p);
         if(ans!=-) printf("%lld\n",ans);
         else puts("Orz, I cannot find x!");
     }
 }
 int main(){
 //    freopen("testdata.in","r",stdin);
     t=read(),k=read();
     switch(k){
         case :solve1();break;
         case :solve2();break;
         case :solve3();break;
     }
     return ;
 }