NOI模拟（5.3） CQOI2018D1T1 破解D-H协议

破解D-H协议

题目背景：

5.3 模拟 CQOI2018D1T1  

分析：BSGS

 

显然我们要做的就是对于一个，求出对应的x，然后再发现数据范围是int以内，那么显然直接BSGS就可以了，并且，因为p是质数那么显然gcd(A, p) = 1那么直接上非扩展的就可以了。复杂度O(Tn^1/2)

Source:

/*
	created by scarlyw
*/
#include <cstdio>
#include <string>
#include <algorithm>
#include <cstring>
#include <iostream>
#include <cmath>
#include <cctype>
#include <vector>
#include <set>
#include <queue>

inline char read() {
	static const int IN_LEN = 1024 * 1024;
	static char buf[IN_LEN], *s, *t;
	if (s == t) {
		t = (s = buf) + fread(buf, 1, IN_LEN, stdin);
		if (s == t) return -1;
	}
	return *s++;
}

///*
template<class T>
inline void R(T &x) {
	static char c;
	static bool iosig;
	for (c = read(), iosig = false; !isdigit(c); c = read()) {
		if (c == -1) return ;
		if (c == '-') iosig = true;	
	}
	for (x = 0; isdigit(c); c = read()) 
		x = ((x << 2) + x << 1) + (c ^ '0');
	if (iosig) x = -x;
}
//*/

const int OUT_LEN = 1024 * 1024;
char obuf[OUT_LEN], *oh = obuf;
inline void write_char(char c) {
	if (oh == obuf + OUT_LEN) fwrite(obuf, 1, OUT_LEN, stdout), oh = obuf;
	*oh++ = c;
}


template<class T>
inline void W(T x) {
	static int buf[30], cnt;
	if (x == 0) write_char('0');
	else {
		if (x < 0) write_char('-'), x = -x;
		for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 + 48;
		while (cnt) write_char(buf[cnt--]);
	}
}

inline void flush() {
	fwrite(obuf, 1, oh - obuf, stdout);
}

/*
template<class T>
inline void R(T &x) {
	static char c;
	static bool iosig;
	for (c = getchar(), iosig = false; !isdigit(c); c = getchar())
		if (c == '-') iosig = true;	
	for (x = 0; isdigit(c); c = getchar()) 
		x = ((x << 2) + x << 1) + (c ^ '0');
	if (iosig) x = -x;
}
//*/

long long g, mod, t, x, y;
const int h_mod = 999979;

inline long long mod_pow(long long a, long long b) {
	long long ans = 1;
	for (; b; b >>= 1, a = a * a % mod)
		if (b & 1) ans = ans * a % mod;
	return ans;
}

struct node {
	long long val, key;
	node() {}
	node(long long val, long long key) : val(val), key(key) {}
} ;

std::vector<node> h[h_mod];
std::vector<int> stack;
bool vis[h_mod];

inline void insert(long long pos, long long i) {
	int p = pos % h_mod;
	if (vis[p] == false) stack.push_back(p), vis[p] = true;
	h[p].push_back(node(pos, i));
}

inline long long query(long long x) {
	int p = x % h_mod;
	for (int i = 0; i < h[p].size(); ++i) {
		node *e = &h[p][i];
		if (e->val == x) return e->key;
	}
	return -1;
}

inline long long solve(long long x, long long g, long long mod) {
	long long c = sqrt(mod), p = 1;
	for (int i = 0; i < stack.size(); ++i) 
		h[stack[i]].clear(), vis[stack[i]] = false;
	stack.clear();
	for (long long i = 0; i < c; ++i, p = p * g % mod) {
		if (p == x) return i;
		insert(p * x % mod, i);
	}
	long long cur = 1, t;
	for (long long i = c; i - c + 1 <= mod - 1; i += c) {
		cur = cur * p % mod, t = query(cur);
		if (t != -1) return i - t;
	}
}

int main() {
	freopen("crack.in", "r", stdin);
	freopen("crack.out", "w", stdout);
	R(g), R(mod), R(t);
	while (t--)
		R(x), R(y), std::cout << mod_pow(y, solve(x, g, mod)) << '\n';
	return 0;
}
/*
3 31
3
16 27
21 3
26 9
*/