用 $\texttt{map}$：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <map>
using namespace std;

typedef long long ll;

map<ll, ll> hash;
ll p, g, t;

ll quickpow(ll a, ll b, ll p) {			//快速求出a^b%p 
	ll res = 1ll;
	a %= p;
	for(; b; b >>= 1, a = a * a % p)
		if(b & 1ll)	res = res * a % p;
	return res;
}
ll bsgs(ll a, ll b, ll p) {				//求出a^x=（同余于）b(mod p)的最小的x 
	hash.clear();
	b %= p;
	ll t = (ll)sqrt(p) + 1;
	for(ll j = 0; j < t; ++j) {
		ll val = b * quickpow(a, j, p) % p;
		hash[val] = j;
	}
	a = quickpow(a, t, p);
	if(!a)	return !b ? 1 : -1;
	for(ll i = 0; i <= t; ++i) {
		ll val = quickpow(a, i, p) % p;
		int j = hash.find(val) == hash.end() ? -1 : hash[val];
		if(j >= 0 && i * t - j >= 0)	return i * t - j;
	}
	return -1;
}
inline ll read() {
	ll f = 1ll, x = 0ll;
	char c = getchar();
	while(c < '0' || c > '9') {
		if(c == '-')	f = -1ll;
		c = getchar();
	}
	while(c >= '0' && c <= '9') {
		x = x * 10ll + c - '0';
		c = getchar();
	}
	return x * f;
}

int main() {
	g = read(), p = read(), t = read();
	while(t--) {
		ll A = read(), B = read();
		ll a = bsgs(g, A, p);
		printf("%lld\n", quickpow(B, a, p));
	}
	return 0;
}

用这个代码交就40分6TLE

用 unordered_map：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <map>
#include <unordered_map>
using namespace std;

typedef long long ll;

unordered_map<ll, ll> cjytql;
ll p, g, t;

ll quickpow(ll a, ll b, ll p) {			//快速求出a^b%p 
	ll res = 1ll;
	a %= p;
	for(; b; b >>= 1, a = a * a % p)
		if(b & 1ll)	res = res * a % p;
	return res;
}
ll bsgs(ll a, ll b, ll p) {				//求出a^x=（同余于）b(mod p)的最小的x 
	cjytql.clear();
	b %= p;
	ll t = (ll)sqrt(p) + 1;
	for(ll j = 0; j < t; ++j) {
		ll val = b * quickpow(a, j, p) % p;
		cjytql[val] = j;
	}
	a = quickpow(a, t, p);
	if(!a)	return !b ? 1 : -1;
	for(ll i = 0; i <= t; ++i) {
		ll val = quickpow(a, i, p) % p;
		int j = cjytql.find(val) == cjytql.end() ? -1 : cjytql[val];
		if(j >= 0 && i * t - j >= 0)	return i * t - j;
	}
	return -1;
}
inline ll read() {
	ll f = 1ll, x = 0ll;
	char c = getchar();
	while(c < '0' || c > '9') {
		if(c == '-')	f = -1ll;
		c = getchar();
	}
	while(c >= '0' && c <= '9') {
		x = x * 10ll + c - '0';
		c = getchar();
	}
	return x * f;
}

int main() {
	g = read(), p = read(), t = read();
	while(t--) {
		ll A = read(), B = read();
		ll a = bsgs(g, A, p);
		printf("%lld\n", quickpow(B, a, p));
	}
	return 0;
}

用这个代码交就AC了？

有哪位大佬能够讲一下是怎么回事嘛qwq