loj#2531. 「CQOI2018」破解 D-H 协议(BSGS)
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2022-04-08 19:13:54
题意 "题目链接" Sol 搞个BSGS板子出题人也是很棒棒哦 cpp include define Pair pair define MP(x, y) make_pair(x, y) define fi first define se second define int long long def ......
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; }
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