【BZOJ2242】计算器

题解：
1.快速幂
2.exgcd/逆元
3.bsgs模板题

//by sdfzchy#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<map>#include<cmath>using namespace std;typedef long long LL;const int inf=(1<<30);int n,m;inline int in(){    char ch=getchar();    int f=1,tmp=0;    while(ch<'0'||ch>'9') {if(ch=='-') f=-1;ch=getchar();}    while(ch>='0'&&ch<='9') {tmp=(tmp<<1)+(tmp<<3)+(ch-'0');ch=getchar();}    return tmp*f;}LL x,y,p,k;LL ksm(LL a,LL b){    LL ret=1;    while(b)    {        if(b&1) ret=ret*a%p;        a=a*a%p;        b>>=1;      }       return ret;}LL inv(LL a){    if(a%p==0) return -1;    return ksm(a,p-2);}LL ans;LL bsgs(LL x,LL y){    map<LL,LL> ma;    x%=p,y%=p;    if(!x&&!y) return 1;    if(y==1) return 0;    if(x%p==0) return -1;    m=sqrt(p+0.5);    LL o=y%p;    for(int i=0;i<m;i++)    {        if(!ma.count(o)) ma[o]=i;        o=o*x%p;    }    LL oo=ksm(x,m); o=1;    for(int i=1;(i-1)*(i-1)<=p;i++)    {        o=o*oo%p;        if(ma.count(o)) return ((i*m-ma[o])%p+p)%p;    }    return -1;} int main(){    int T=in();k=in();    while(T--)    {        ans=0;        scanf("%lld%lld%lld",&x,&y,&p);        if(k==1) ans=ksm(x%p,y);         else if(k==2)        {            ans=inv(x);            if(ans!=-1) ans=ans*y%p;        }        else ans=bsgs(x,y);        if(ans==-1) puts("Orz, I cannot find x!");        else printf("%lld\n",ans);    }    return 0;}