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discrete_logarithm
(math/discrete_logarithm.hpp)

• View this file on GitHub
• Last update: 2025-07-01 03:22:56+09:00
• Include: #include "math/discrete_logarithm.hpp"

discrete_logarithm

使い方

discrete_logarithm(ll x, ll y, ll m) : $x^k \pmod{m}$ となる最小の $k$ を返す。無い場合は $-1$ が返る。 $O(\sqrt{M})$

参考

OI wiki

Depends on

• floor K 乗根 (math/iroot.hpp)
• math/power.hpp

Verified with

• verify/yosupo/discrete_logarithm.test.cpp

Code

#pragma once

#include "../math/iroot.hpp"
#include "../math/power.hpp"

#include<algorithm>
#include<cassert>
#include<cmath>
#include<map>
#include<numeric>
#include<vector>

long long __modinv(long long a, long long m){
	long long b=m, u=1, v=0;
	while(b){
		long long t = a/b;
		a -= t*b; std::swap(a, b);
		u -= t*v; std::swap(u, v);
	}
	u %= m;
	if(u < 0)u += m;
	return u;
}

long long discrete_logarithm(long long x, long long y, long long m) {
	assert(x < m && y < m);
	if(m == 1)return 0;
	if(y == 1)return 0;
	if(x == 0){
		if(y == 1)return 0;
		else if(y == 0)return 1;
		else return -1;
	}

	if(std::gcd(x, m) != 1){
		long long d = std::gcd(x, m);
		if(y%d)return -1;
		y /= d;
		m /= d;
		long long ret = discrete_logarithm(x%m, (y*__modinv(x/d, m))%m, m);
		if(ret == -1)return -1;
		else return ret+1;
	}

	long long sq = iroot(m);
	if(sq < 3)sq = 3;
	std::vector<long long> _b(sq);
	for(int i = 0;i < sq;i++)_b[i] = (i == 0 ? 1 : (_b[i-1]*x)%m);

	std::map<long long, long long> b;
	for(int i = sq-1;i >= 0;i--)b[_b[i]] = i;

	long long inv = __modinv((_b.back()*x)%m, m);

	for(int i = 0;i < sq;i++){
		long long num = (y*power(inv, i, m))%m;
		if(b.contains(num)){
			return i*sq + b[num];
		}
	}
	return -1;
};

#line 2 "math/discrete_logarithm.hpp"

#line 2 "math/iroot.hpp"

#include<cmath>
#include<limits>

unsigned long long iroot(unsigned long long n, int k=2){
	constexpr unsigned long long LIM = std::numeric_limits<unsigned long long>::max();
	if(n <= 1 || k == 1){
		return n;
	}
	if(k >= 64){
		return 1;
	}
	if(k == 2){
		return sqrtl(n);
	}

	if(n == LIM)n--;

	auto safe_mul = [&](unsigned long long &x, unsigned long long &y) -> void {
		if(x <= LIM / y){
			x *= y;
		}else{
			x = LIM;
		}
	};

	auto power = [&](unsigned long long a, int b) -> unsigned long long {
		unsigned long long ret = 1;
		while(b){
			if(b & 1)safe_mul(ret, a);
			safe_mul(a, a);
			b >>= 1;
		}
		return ret;
	};

	unsigned long long ret = (k == 3 ? cbrt(n)-1 : std::pow(n, std::nextafter(1.0/double(k), 0.0)));
	while(power(ret+1, k) <= n)ret++;
	return ret;
}
#line 2 "math/power.hpp"

#include <type_traits>

template<typename T>
concept NotPrimitiveInt =
	!(std::is_same_v<T, int> ||
		std::is_same_v<T, long> ||
		std::is_same_v<T, long long> ||
		std::is_same_v<T, unsigned> ||
		std::is_same_v<T, unsigned long> ||
		std::is_same_v<T, unsigned long long>);

template<NotPrimitiveInt T>
T power(T n, long long k) {
	T ret = 1;
	while(k > 0) {
		if(k & 1)ret *= n;
		n = n*n;
		k >>= 1;
	}
	return ret;
}

long long power(long long n, long long k, long long p) {
	long long ret = 1;
	while(k > 0){
		if(k & 1)ret = ret*n % p;
		n = n*n % p;
		k >>= 1;
	}
	return ret;
}
#line 5 "math/discrete_logarithm.hpp"

#include<algorithm>
#include<cassert>
#line 9 "math/discrete_logarithm.hpp"
#include<map>
#include<numeric>
#include<vector>

long long __modinv(long long a, long long m){
	long long b=m, u=1, v=0;
	while(b){
		long long t = a/b;
		a -= t*b; std::swap(a, b);
		u -= t*v; std::swap(u, v);
	}
	u %= m;
	if(u < 0)u += m;
	return u;
}

long long discrete_logarithm(long long x, long long y, long long m) {
	assert(x < m && y < m);
	if(m == 1)return 0;
	if(y == 1)return 0;
	if(x == 0){
		if(y == 1)return 0;
		else if(y == 0)return 1;
		else return -1;
	}

	if(std::gcd(x, m) != 1){
		long long d = std::gcd(x, m);
		if(y%d)return -1;
		y /= d;
		m /= d;
		long long ret = discrete_logarithm(x%m, (y*__modinv(x/d, m))%m, m);
		if(ret == -1)return -1;
		else return ret+1;
	}

	long long sq = iroot(m);
	if(sq < 3)sq = 3;
	std::vector<long long> _b(sq);
	for(int i = 0;i < sq;i++)_b[i] = (i == 0 ? 1 : (_b[i-1]*x)%m);

	std::map<long long, long long> b;
	for(int i = sq-1;i >= 0;i--)b[_b[i]] = i;

	long long inv = __modinv((_b.back()*x)%m, m);

	for(int i = 0;i < sq;i++){
		long long num = (y*power(inv, i, m))%m;
		if(b.contains(num)){
			return i*sq + b[num];
		}
	}
	return -1;
};

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