同余方程 A^x = B (mod C)的解 - BSGS/exBSGS - 洛谷BSGS题单

文章目录

• • 模板
  • 题目（来自洛谷的BSGS题单）
  • • [TJOI2007] 可爱的质数/【模板】BSGS
    • [SDOI2011]计算器
    • 多少个1？
    • [SDOI2013] 随机数生成器
    • 【模板】扩展 BSGS/exBSGS
    • 【XR-1】快乐肥宅
    • [CQOI2018]破解D-H协议
    • CF1106F Lunar New Year and a Recursive Sequence

这波不写详解了，OI-Wiki有详解。

思路其实很简单，先用hash表存一部分计算结果，然后枚举另一半，同时查找是否存在解。

模板

inline ll mul(ll a, ll b, ll p) { return (a * b - (ll)((long double)a / p * b) * p + p) % p; } 

inline ll qpow(ll x, ll m, ll mod, ll res = 1) {
	for ( ; m; m >>= 1, x = mul(x, x, mod)) if (m & 1) res = mul(res, x, mod);
	return res;
}

inline void exgcd(ll a, ll b, ll &g, ll &x, ll &y) {
	if (b) exgcd(b, a % b, g, y, x), y -= a / b * x;
	else x = 1, y = 0, g = a;
}

inline ll inv(ll x, ll m) {
	ll a, b, g; 
	exgcd(x, m, g, a, b);
	return (a % m + m) % m;
}

map<ll, ll> bs;
inline ll BSGS(ll a, ll b, ll p) { // (a, p) = 1
    if (b == 1 || p == 1) return 0;
	ll m = sqrt(p) + 1; bs.clear();
	for (ll k = 0, t = b; k <= m; k ++, t = mul(t, a, p)) bs[t] = k;
	for (ll x = 1, t = qpow(a, m, p), base = t; x <= m; x ++, t = mul(t, base, p)) 
		if (bs.count(t)) return ((x * m - bs[t]) % p + p) % p;
	return -1;
}

inline ll exBSGS(ll a, ll b, ll c) {
	a %= c, b %= c;
	if (b == 1 || c == 1) return 0;
	ll g = __gcd(a, c), k = 0, t = 1, res;
	while (g ^ 1LL) {
		if (b % g) return -1;
		c /= g, b /= g, t = mul(t, (a / g), c), k ++;
		if (t == b) return k;
		g = __gcd(a, c);
	}
	b = b * inv(t, c) % c, res = BSGS(a, b, c);
    return ~res ? res + k : res;
}

题目（来自洛谷的BSGS题单）

洛谷题单链接

然鹅，本蒟蒻会写的题过于模板。。。

[TJOI2007] 可爱的质数/【模板】BSGS

// #pragma GCC optimize("Ofast")
// #pragma GCC target("avx,avx2,fma")

#include <bits/stdc++.h>
using namespace std;
#define fast ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define fin freopen("input.txt", "r", stdin), freopen("output.txt", "w", stdout)
#define PII pair<int, int>
#define PLL pair<ll, ll>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ld nd << 1
#define rd nd << 1 | 1

typedef vector<int> vec;
// typedef long double ld;
typedef long long ll;
typedef unsigned long long ull;
typedef __int128_t i128;

// const int mod = 1e9 + 7;
// const int mod = 998244353;
const int maxn = 1e6 + 50;
const int inf = 0x3f3f3f3f;			// 1e9;
const ll INF = 0x3f3f3f3f3f3f3f3f;	// 1e18;

ll n, b, p;
map<ll, int> bsgs;

inline ll qpow(ll x, ll m, ll mod, ll res = 1) {
	for ( ; m; m >>= 1, x = x * x % mod) if (m & 1) res = res * x % mod;
	return res;
}

int main() {
	scanf("%lld%lld%lld", &p, &b, &n);
	int m = sqrt(p) + 1;
	for (ll k = 0, t = n % p; k <= m; k ++, t = t * b % p) bsgs[t] = k;
	for (ll x = 1, t = qpow(b, m, p), base = t; x <= m; x ++, t = t * base % p) 
		if (bsgs.count(t)) printf("%lld\n", x * m - bsgs[t]), exit(0);	
	return cout << "no solution", 0;
}

[SDOI2011]计算器

全是板子。。。

// #pragma GCC optimize("Ofast")
// #pragma GCC target("avx,avx2,fma")

#include <bits/stdc++.h>
using namespace std;
#define fast ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define fin freopen("input.txt", "r", stdin), freopen("output.txt", "w", stdout)
#define PII pair<int, int>
#define PLL pair<ll, ll>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ld nd << 1
#define rd nd << 1 | 1

typedef vector<int> vec;
// typedef long double ld;
typedef long long ll;
typedef unsigned long long ull;
typedef __int128_t i128;

// const int mod = 1e9 + 7;
// const int mod = 998244353;
const int maxn = 1e6 + 50;
const int inf = 0x3f3f3f3f;			// 1e9;
const ll INF = 0x3f3f3f3f3f3f3f3f;	// 1e18;

int T, op;

inline ll qpow(ll x, ll m, ll mod, ll res = 1) {
	for ( ; m; m >>= 1, x = x * x % mod) if (m & 1) res = res * x % mod;
	return res;
}

inline void exgcd(ll a, ll b, ll &g, ll &x, ll &y) {
	if (b) {
		exgcd(b, a % b, g, x, y);
		swap(x, y), y -= a / b * x;
	} else {
		x = 1, y = 0, g = a;
	}
}

inline ll inv(ll x, ll m) {
	ll a, b, g; 
	exgcd(x, m, g, a, b);
	return (a % m + m) % m;
}

map<ll, ll> bs;
ll bsgs(ll a, ll b, ll c) {
	if (b == 1 || c == 1) return 0;
	ll g = __gcd(a, c), k = 0, t = 1;
	while (g ^ 1LL) {
		if (b % g) return -1;
		c /= g, b /= g, t = t * (a / g) % c, k ++;
		if (t == b) return k;
		g = __gcd(a, c);
	}
	b = b * inv(t, c) % c;

	ll m = sqrt(c) + 1; bs.clear();
	for (ll k = 0, t = b; k <= m; k ++, t = t * a % c) bs[t] = k;
	for (ll x = 1, t = qpow(a, m, c), base = t; x <= m; x ++, t = t * base % c) 
		if (bs.count(t)) return  ((x * m - bs[t]) % c + c) % c + k;
	return -1;
}

int main() {
	ll y, z, p, ans = 1;
	scanf("%d %d", &T, &op);
	switch (op) {
		case 1: {
			while (T -- ) {
				scanf("%lld%lld%lld", &y, &z, &p);
				printf("%lld\n", qpow(y % p, z, p));
			}
		}; break;
		case 2: {
			while (T -- ) {
				scanf("%lld%lld%lld", &y, &z, &p);
				ll g, a, b; exgcd(y, p, g, a, b);
				if (z % g) puts("Orz, I cannot find x!");
				else printf("%lld\n", (a * (z / g) % p + p) % p);
			}
		}; break;
		case 3: {
			while (T -- ) {
				scanf("%lld%lld%lld", &y, &z, &p);
				ans = bsgs(y % p, z % p, p);
				if (ans == -1) puts("Orz, I cannot find x!");
				else printf("%lld\n", ans);
			}
		}; break;
	}
	return 0;
}

多少个1？

等比求个和，变板子了

// #pragma GCC optimize("Ofast")
// #pragma GCC target("avx,avx2,fma")

#include <bits/stdc++.h>
using namespace std;
#define fast ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define fin freopen("input.txt", "r", stdin), freopen("output.txt", "w", stdout)
#define PII pair<int, int>
#define PLL pair<ll, ll>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ld nd << 1
#define rd nd << 1 | 1

typedef vector<int> vec;
// typedef long double ld;
typedef long long ll;
typedef unsigned long long ull;
typedef __int128_t i128;

// const int mod = 1e9 + 7;
// const int mod = 998244353;
const int maxn = 1e6 + 50;
const int inf = 0x3f3f3f3f;			// 1e9;
const ll INF = 0x3f3f3f3f3f3f3f3f;	// 1e18;

int T, op;

inline ll mul(ll a, ll b, ll mod, ll res = 0) {
	for ( ; b; b >>= 1, a = (a + a) % mod) if (b & 1) res = (res + a) % mod;
	return res;
}

inline ll qpow(ll x, ll m, ll mod, ll res = 1) {
	for ( ; m; m >>= 1, x = mul(x, x, mod)) if (m & 1) res = mul(res, x, mod);
	return res;
}

inline void exgcd(ll a, ll b, ll &g, ll &x, ll &y) {
	if (b) {
		exgcd(b, a % b, g, x, y);
		swap(x, y), y -= a / b * x;
	} else {
		x = 1, y = 0, g = a;
	}
}

inline ll inv(ll x, ll m) {
	ll a, b, g; 
	exgcd(x, m, g, a, b);
	return (a % m + m) % m;
}

map<ll, ll> bs;
ll bsgs(ll a, ll b, ll c) {
	if (b == 1 || c == 1) return 0;
	/* ll g = __gcd(a, c), k = 0, t = 1;
	while (g ^ 1LL) {
		if (b % g) return -1;
		c /= g, b /= g, t = mul(t, (a / g), c), k ++;
		if (t == b) return k;
		g = __gcd(a, c);
	}
	b = b * inv(t, c) % c; */

	ll m = sqrt(c) + 1; bs.clear();
	for (ll k = 0, t = b; k <= m; k ++, t = mul(t, a, c)) bs[t] = k;
	for (ll x = 1, t = qpow(a, m, c), base = t; x <= m; x ++, t = mul(t, base, c)) 
		if (bs.count(t)) return ((x * m - bs[t]) % c + c) % c;
	return -1;
}

int main() {
	ll k, m, ans; 
	scanf("%lld %lld", &k, &m);
	ans = bsgs(10LL % m, (9LL * k + 1LL) % m, m);
	printf("%lld", ans);
	return 0;
}

[SDOI2013] 随机数生成器

特判，特判，特判!!!

// #pragma GCC optimize("Ofast")
// #pragma GCC target("avx,avx2,fma")

#include <bits/stdc++.h>
using namespace std;
#define fast ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define fin freopen("input.txt", "r", stdin), freopen("output.txt", "w", stdout)
#define PII pair<int, int>
#define PLL pair<ll, ll>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ld nd << 1
#define rd nd << 1 | 1

typedef vector<int> vec;
// typedef long double ld;
typedef long long ll;
typedef unsigned long long ull;
typedef __int128_t i128;

// const int mod = 1e9 + 7;
// const int mod = 998244353;
const int maxn = 1e6 + 50;
const int inf = 0x3f3f3f3f;			// 1e9;
const ll INF = 0x3f3f3f3f3f3f3f3f;	// 1e18;

int T, op;

/* inline ll mul(ll a, ll b, ll mod, ll res = 0) {
	for ( ; b; b >>= 1, a = (a + a) % mod) if (b & 1) res = (res + a) % mod;
	return res;
} */

inline ll mul(ll a, ll b, ll mod) { return a * b % mod; } 

inline ll qpow(ll x, ll m, ll mod, ll res = 1) {
	for ( ; m; m >>= 1, x = mul(x, x, mod)) if (m & 1) res = mul(res, x, mod);
	return res;
}

inline void exgcd(ll a, ll b, ll &g, ll &x, ll &y) {
	if (b) {
		exgcd(b, a % b, g, x, y);
		swap(x, y), y -= a / b * x;
	} else {
		x = 1, y = 0, g = a;
	}
}

inline ll inv(ll x, ll m) {
	ll a, b, g; 
	exgcd(x, m, g, a, b);
	return (a % m + m) % m;
}

map<ll, ll> bs;
ll bsgs(ll a, ll b, ll c) {
	if (b == 1 || c == 1) return 0;
	/* ll g = __gcd(a, c), k = 0, t = 1;
	while (g ^ 1LL) {
		if (b % g) return -1;
		c /= g, b /= g, t = mul(t, (a / g), c), k ++;
		if (t == b) return k;
		g = __gcd(a, c);
	}
	b = b * inv(t, c) % c; */

	ll m = sqrt(c) + 1; bs.clear();
	for (ll k = 0, t = b; k <= m; k ++, t = mul(t, a, c)) bs[t] = k;
	for (ll x = 1, t = qpow(a, m, c), base = t; x <= m; x ++, t = mul(t, base, c)) 
		if (bs.count(t)) return ((x * m - bs[t]) % c + c) % c;
	return -1;
}

int main() {
	ll p, a, b, x1, t, ans;
	scanf("%d", &T);
	while (T -- ) {
		scanf("%lld%lld%lld%lld%lld", &p, &a, &b, &x1, &t);

		if (x1 == t) {
			puts("1");
			continue;
		}

		if (a == 0LL) {
			if (b == t) puts("2");
			else puts("-1");
			continue;
		}

		if (a == 1LL) {
			if (b == 0) puts("-1");
			else {
				ans = qpow(b, p - 2, p) * ((t - x1) % p + p) % p;
				printf("%lld\n", ans + 1);
			}
			continue;
		}

		ll u = (t * (a - 1) + b) % p;
		ll d = (x1 * (a - 1) + b) % p;
		ans = bsgs(a, u * qpow(d, p - 2, p) % p, p);
		if (ans == -1) puts("-1");
		else printf("%lld\n", ans + 1);
	}
	return 0;
}

【模板】扩展 BSGS/exBSGS

// #pragma GCC optimize("Ofast")
// #pragma GCC target("avx,avx2,fma")

#include <bits/stdc++.h>
using namespace std;
#define fast ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define fin freopen("input.txt", "r", stdin), freopen("output.txt", "w", stdout)
#define PII pair<int, int>
#define PLL pair<ll, ll>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ld nd << 1
#define rd nd << 1 | 1

typedef vector<int> vec;
// typedef long double ld;
typedef long long ll;
typedef unsigned long long ull;
// typedef __int128_t i128;

// const int mod = 1e9 + 7;
// const int mod = 998244353;
const int maxn = 1e6 + 50;
const int inf = 0x3f3f3f3f;			// 1e9;
const ll INF = 0x3f3f3f3f3f3f3f3f;	// 1e18;

int T;

inline ll lcm(ll a, ll b) { return a / __gcd(a, b) * b; }

/* inline ll mul(ll a, ll b, ll mod, ll res = 0) {
	for ( ; b; b >>= 1, a = (a + a) % mod) if (b & 1) res = (res + a) % mod;
	return res;
} */

inline ll mul(ll a, ll b, ll mod) { return a * b % mod; } 

inline ll qpow(ll x, ll m, ll mod, ll res = 1) {
	for ( ; m; m >>= 1, x = mul(x, x, mod)) if (m & 1) res = mul(res, x, mod);
	return res;
}

inline void exgcd(ll a, ll b, ll &g, ll &x, ll &y) {
	if (b) exgcd(b, a % b, g, y, x), y -= a / b * x;
	else x = 1, y = 0, g = a;
}

inline ll inv(ll x, ll m) {
	ll a, b, g; 
	exgcd(x, m, g, a, b);
	return (a % m + m) % m;
}

map<ll, ll> bs;
inline ll BSGS(ll a, ll b, ll p) { // (a, p) = 1
    if (b == 1 || p == 1) return 0;
	ll m = sqrt(p) + 1; bs.clear();
	for (ll k = 0, t = b; k <= m; k ++, t = mul(t, a, p)) bs[t] = k;
	for (ll x = 1, t = qpow(a, m, p), base = t; x <= m; x ++, t = mul(t, base, p)) 
		if (bs.count(t)) return ((x * m - bs[t]) % p + p) % p;
	return -1;
}

inline ll exBSGS(ll a, ll b, ll c) {
	a %= c, b %= c;
	if (b == 1 || c == 1) return 0;
	ll g = __gcd(a, c), k = 0, t = 1, res;
	while (g ^ 1LL) {
		if (b % g) return -1;
		c /= g, b /= g, t = mul(t, (a / g), c), k ++;
		if (t == b) return k;
		g = __gcd(a, c);
	}
	b = b * inv(t, c) % c, res = BSGS(a, b, c);
    return ~res ? res + k : res;
}

ll a, b, p, ans;

int main() {
	while (~scanf("%lld%lld%lld", &a, &p, &b) && (a || b || p)) {
		if (~(ans = exBSGS(a, b, p))) printf("%lld\n", ans);
		else puts("No Solution");
	}
	return 0;
}

【XR-1】快乐肥宅

BSGS求出特解（可能无解或者为唯一解），再用CRT合并答案即可（可能会爆long long）。

好毒瘤，枯了，先咕了。

[CQOI2018]破解D-H协议

// #pragma GCC optimize("Ofast")
// #pragma GCC target("avx,avx2,fma")

#include <bits/stdc++.h>
using namespace std;
#define fast ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define fin freopen("input.txt", "r", stdin), freopen("output.txt", "w", stdout)
#define PII pair<int, int>
#define PLL pair<ll, ll>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ld nd << 1
#define rd nd << 1 | 1

typedef vector<int> vec;
// typedef long double ld;
typedef long long ll;
typedef unsigned long long ull;
// typedef __int128_t i128;

// const int mod = 1e9 + 7;
// const int mod = 998244353;
const int maxn = 1e6 + 50;
const int inf = 0x3f3f3f3f;			// 1e9;
const ll INF = 0x3f3f3f3f3f3f3f3f;	// 1e18;

int T;

inline ll lcm(ll a, ll b) { return a / __gcd(a, b) * b; }

/* inline ll mul(ll a, ll b, ll mod, ll res = 0) {
	for ( ; b; b >>= 1, a = (a + a) % mod) if (b & 1) res = (res + a) % mod;
	return res;
}  */

inline ll mul(ll a, ll b, ll p) { return (a * b - (ll)((long double)a / p * b) * p + p) % p; } 

inline ll qpow(ll x, ll m, ll mod, ll res = 1) {
	for ( ; m; m >>= 1, x = mul(x, x, mod)) if (m & 1) res = mul(res, x, mod);
	return res;
}

inline void exgcd(ll a, ll b, ll &g, ll &x, ll &y) {
	if (b) exgcd(b, a % b, g, y, x), y -= a / b * x;
	else x = 1, y = 0, g = a;
}

inline ll inv(ll x, ll m) {
	ll a, b, g; 
	exgcd(x, m, g, a, b);
	return (a % m + m) % m;
}

map<ll, ll> bs;
inline ll BSGS(ll a, ll b, ll p) { // (a, p) = 1
    if (b == 1 || p == 1) return 0;
	ll m = sqrt(p) + 1; bs.clear();
	for (ll k = 0, t = b; k <= m; k ++, t = mul(t, a, p)) bs[t] = k;
	for (ll x = 1, t = qpow(a, m, p), base = t; x <= m; x ++, t = mul(t, base, p)) 
		if (bs.count(t)) return ((x * m - bs[t]) % p + p) % p;
	return -1;
}

inline ll exBSGS(ll a, ll b, ll c) {
	a %= c, b %= c;
	if (b == 1 || c == 1) return 0;
	ll g = __gcd(a, c), k = 0, t = 1, res;
	while (g ^ 1LL) {
		if (b % g) return -1;
		c /= g, b /= g, t = mul(t, (a / g), c), k ++;
		if (t == b) return k;
		g = __gcd(a, c);
	}
	b = b * inv(t, c) % c, res = BSGS(a, b, c);
    return ~res ? res + k : res;
}

ll g, p, A, B, K;

int main() {
	scanf("%lld%lld%d", &g, &p, &T);
	while (T -- ) {
		scanf("%lld%lld", &A, &B);
		ll a = exBSGS(g, A, p), b = exBSGS(g, B, p);
		printf("%lld\n", qpow(qpow(g, a, p), b, p));
	}
	return 0;
}

CF1106F Lunar New Year and a Recursive Sequence

用离散对数将表达式转换成一个线性方程，显然先求原根，再矩阵快速幂就好了。

// #pragma GCC optimize("Ofast")
// #pragma GCC target("avx,avx2,fma")

#include <bits/stdc++.h>
using namespace std;
#define fast ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define fin freopen("input.txt", "r", stdin), freopen("output.txt", "w", stdout)
#define PII pair<int, int>
#define PLL pair<ll, ll>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ld nd << 1
#define rd nd << 1 | 1

typedef vector<int> vec;
// typedef long double ld;
typedef long long ll;
typedef unsigned long long ull;
// typedef __int128_t i128;

// const int mod = 1e9 + 7;
const int mod = 998244353;
const int maxn = 1e6 + 50;
const int inf = 0x3f3f3f3f;			// 1e9;
const ll INF = 0x3f3f3f3f3f3f3f3f;	// 1e18;

int T;

inline ll lcm(ll a, ll b) { return a / __gcd(a, b) * b; }

/* inline ll mul(ll a, ll b, ll mod, ll res = 0) {
	for ( ; b; b >>= 1, a = (a + a) % mod) if (b & 1) res = (res + a) % mod;
	return res;
}  */

inline ll mul(ll a, ll b, ll p) { return (a * b - (ll)((long double)a / p * b) * p + p) % p; } 

inline ll qpow(ll x, ll m, ll mod, ll res = 1) {
	for ( ; m; m >>= 1, x = mul(x, x, mod)) if (m & 1) res = mul(res, x, mod);
	return res;
}

inline void exgcd(ll a, ll b, ll &g, ll &x, ll &y) {
	if (b) exgcd(b, a % b, g, y, x), y -= a / b * x;
	else x = 1, y = 0, g = a;
}

inline ll inv(ll x, ll m) {
	ll a, b, g; 
	exgcd(x, m, g, a, b);
	return (a % m + m) % m;
}

map<ll, ll> bs;
inline ll BSGS(ll a, ll b, ll p) { // (a, p) = 1
    if (b == 1 || p == 1) return 0;
	ll m = sqrt(p) + 1; bs.clear();
	for (ll k = 0, t = b; k <= m; k ++, t = mul(t, a, p)) bs[t] = k;
	for (ll x = 1, t = qpow(a, m, p), base = t; x <= m; x ++, t = mul(t, base, p)) 
		if (bs.count(t)) return ((x * m - bs[t]) % p + p) % p;
	return -1;
}

inline ll exBSGS(ll a, ll b, ll c) {
	a %= c, b %= c;
	if (b == 1 || c == 1) return 0;
	ll g = __gcd(a, c), k = 0, t = 1, res;
	while (g ^ 1LL) {
		if (b % g) return -1;
		c /= g, b /= g, t = mul(t, (a / g), c), k ++;
		if (t == b) return k;
		g = __gcd(a, c);
	}
	b = b * inv(t, c) % c, res = BSGS(a, b, c);
    return ~res ? res + k : res;
}

int k;
ll n, m;
ll base[105][105], ans[105][105], tmp[105][105], a[105];

inline void mat_mul(int op, ll p) {
	switch(op) {
		case 1: {
			memset(tmp, 0, sizeof tmp);
			for (int i = 1; i <= k; i++) 
				for (int j = 1; j <= k; j++) 
					for (int o = 1; o <= k; o++) 
						tmp[i][j] = (tmp[i][j] + base[i][o] * base[o][j] % p) % p;
			memcpy(base, tmp, sizeof tmp);
		}; break;
	 	case 2: {
			memset(tmp, 0, sizeof tmp);
			for (int i = 1; i <= k; i++) 
				for (int j = 1; j <= k; j++) 
					for (int o = 1; o <= k; o++) 
						tmp[i][j] = (tmp[i][j] + ans[i][o] * base[o][j] % p) % p;
			memcpy(ans, tmp, sizeof tmp);
		};break;
	}
}

void mat_pow(ll ind, ll p) {
	while (ind) {
		if (ind & 1) mat_mul(2, p);
		mat_mul(1, p), ind >>= 1;
	}
}

int main() {
	scanf("%d", &k);
	for (int i = 1; i <= k; i++) scanf("%lld", &base[1][i]), ans[i][i] = 1LL;
	for (int i = 2; i <= k; i++) base[i][i - 1] = 1LL;
	scanf("%lld %lld", &n, &m);
	mat_pow(n - k, mod - 1);
	ll t = exBSGS(3, m, mod) % (mod - 1);
	// cout << t << '\n';
	ll g, x, y; exgcd(ans[1][1], mod - 1, g, x, y);
	if (t % g) puts("-1");
	else printf("%lld", qpow(3, mul(t / g, (x % (mod - 1) + mod - 1) % (mod - 1), mod - 1), mod));
	return 0;
}

标签：洛谷,int,res,exBSGS,return,mod,BSGS,ll,define
来源： https://blog.csdn.net/qq_41743740/article/details/118946068