c/c++语言开发共享loj#2531. 「CQOI2018」破解 D-H 协议(BSGS)

题意

sol

搞个bsgs板子出题人也是很棒棒哦

#include<bits/stdc++.h>  #define pair pair<int, int> #define mp(x, y) make_pair(x, y) #define fi first #define se second #define int long long  #define ll long long  #define ull unsigned long long  #define fin(x) {freopen(#x".in","r",stdin);} #define fout(x) {freopen(#x".out","w",stdout);} using namespace std; const int maxn = 1e6 + 10, inf = 1e9 + 10;; int mod; const double eps = 1e-9; template <typename a, typename b> inline bool chmin(a &a, b b){if(a > b) {a = b; return 1;} return 0;} template <typename a, typename b> inline bool chmax(a &a, b b){if(a < b) {a = b; return 1;} return 0;} template <typename a, typename b> inline ll add(a x, b y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;} template <typename a, typename b> inline void add2(a &x, b y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);} template <typename a, typename b> inline ll mul(a x, b y) {return 1ll * x * y % mod;} template <typename a, typename b> inline void mul2(a &x, b y) {x = (1ll * x * y % mod + mod) % mod;} template <typename a> inline void debug(a a){cout << a << 'n';} template <typename a> inline ll sqr(a x){return 1ll * x * x;} template <typename a, typename b> inline ll fp(a a, b p, int md = mod) {int b = 1;while(p) {if(p & 1) b = mul(b, a);a = mul(a, a); p >>= 1;}return b;} template <typename a> a inv(a x) {return fp(x, mod - 2);} 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 g; map<int, int> mp; int solve(int x) {//g^ret = x (mod p)     mp.clear(); int block = ceil(sqrt(mod)), base = fp(g, block);     for(int i = 0, cur = x; i <= block; i++, mul2(cur, g)) mp[cur] = i;     for(int i = 1, cur = base; i <= block; i++, mul2(cur, base)) if(mp[cur]) return i * block - mp[cur];         return 0; } /* int solve(int x) {     int now = 1;     for(int i = 0; i<= mod; i++) {         if(now == x) return i;         mul2(now, g);     }     assert(1 == 2); } */ signed main() {     //freopen("a.in", "r", stdin);     g = read(); mod = read();     int n = read();     while(n--) {         int a = read(), b = read();         cout << fp(g, solve(a) * solve(b)) << 'n';;     }     return 0; }