BZOJ 2242: [SDOI2011]计算器 [快速幂 BSGS]

2242: [SDOI2011]计算器

题意：求\(a^b \mod p,\ ax \equiv b \mod p,\ a^x \equiv b \mod p\)，p是质数

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这种裸题我竟然WA了好多次

第三个注意判断a和b整除p的情况

#pragma GCC optimize ("O2")
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <map>
using namespace std;
typedef long long ll;
#define fir first
#define sec second
inline int read() {char c=getchar(); int x=0, f=1;while(c<'0' || c>'9') {if(c=='-')f=-1; c=getchar();}while(c>='0' && c<='9') {x=x*10+c-'0'; c=getchar();}return x*f;
}int a, b, p;
ll Pow(ll a, int b, int p) {a%=p; ll ans=1;for(; b; b>>=1, a=a*a%p)if(b&1) ans=ans*a%p;return ans;
}
ll inv(int a, int p) {if(a%p==0) return -1;return Pow(a, p-2, p);
}
map<int, int> ma;
ll ind(int a, int b, int p) {a%=p; b%=p;ma.clear();ll e=1; int m=sqrt(p)+0.5;for(int i=0; i<m; i++) {if(!ma.count(e)) ma[e]=i;e=e*a%p;}e=Pow(e, p-2, p);for(int i=0; i<m; i++) {if(ma.count(b)) return i*m + ma[b];b=b*e%p;}return -1;
}
int main() {//freopen("in","r",stdin);freopen("calc.in","r",stdin);freopen("calc.out","w",stdout);int T=read(), type=read();while(T--) {a=read(); b=read(); p=read();ll ans;if(type==1) ans=Pow(a, b, p);else if(type==2) {ans=inv(a, p);if(ans!=-1) ans=ans*b%p;}else ans=ind(a, b, p);if(ans==-1) puts("Orz, I cannot find x!");else printf("%lld\n", ans);}
}