CQOI2018D1T1:破解D-H协议（BSGS模板）

版权声明：博主是菜鸡，但转的时候还是说一声吧 https://blog.csdn.net/qq_37666409/article/details/81514918

题目传送门：

洛谷P4454

bzoj5296

loj2531

CQOI2018今年到底考了多少模板题啊……

随便根据A或B求出a，b

然后照着题目说的求K就行了

高次同余一眼BSGS

然后map改成hash_map(绝对不是没有c++11编译器用不了unordered_map)

效率就很稳了

注意hash_map是非标准容器，无法直接支持long long，所以要一些黑科技

#include <bits/stdc++.h>
#include <ext/hash_map>
#define LL long long

namespace __gnu_cxx {
    template<> struct hash< std::string > {
        size_t operator()( const std::string& x ) const {
            return hash< const char* >()( x.c_str() );
        }
    };
}

namespace __gnu_cxx {
    template<> struct hash<long long> {
        size_t operator()(long long x) const {
            return (unsigned) x;
        }
    };
}

using namespace std; 
using namespace __gnu_cxx;

template <typename T> inline void read(T &x) {
    int ch = getchar();
    bool fg = false;
    for (x = 0; !isdigit(ch); ch = getchar()) {
        if (ch == '-') {
            fg = true;
        }
    }
    for (; isdigit(ch); ch = getchar()) {
        x = x * 10 + ch - '0';
    }
    if (fg) {
        x = -x;
    }
}

LL cnt;
hash_map<LL, LL> F;

LL qpow(LL a, LL b, LL c) {
    LL ret = 1;
    while(b) {
        if(b & 1) {
            ret *= a;
            ret %= c;
        }
        b >>= 1;
        a = a * a % c;
    }
    return ret;
}

LL BSGS(LL A, LL B, LL C) {
    if(A == 0 && B == 0)  { 
        return 1;
    }
    F.clear();
    LL m = (LL) ceil(sqrt((double)C)); LL con = B % C , D = 1;
    F[B % C] = 0;
    for(LL i = 1; i <= m; i++) {
        con = con * A % C;
        F[con] = i;
    }
    con = qpow(A, m, C);
    for(LL i = 1; i <= m; i++) {
        D = D * con % C;
        if(F[D]) {
            LL t = i * m - F[D];    
            return (t % C + C) % C;    
        }	
    }
    return 0;
}

signed main() {
    LL g, P, A, B;
    int n;
    read(g), read(P);
    read(n);
    while(n --) {
        read(A), read(B);
        printf("%lld\n", qpow(B, BSGS(g, A, P), P));
    }
    return 0;
}