Merge pull request #422 from hitonanode/refactor-bare-mod-algebra

refactor bare_mod_algebra and so on

Author: hitonanode

linear_algebra_matrix/linalg_longlong.hpp

@@ -4,7 +4,6 @@
 #include <cstdlib>
 #include <vector>
 
-// CUT begin
 template <typename lint, typename mdint>
 std::vector<std::vector<lint>> gauss_jordan(std::vector<std::vector<lint>> mtr, mdint mod) {
     // Gauss-Jordan elimination 行基本変形のみを用いるガウス消去法
@@ -28,7 +27,7 @@ std::vector<std::vector<lint>> gauss_jordan(std::vector<std::vector<lint>> mtr,
                     mtr[piv][w] ? mod - mtr[piv][w] : 0; // To preserve sign of determinant
             }
         }
-        lint pivinv = mod_inverse<lint>(mtr[h][c], mod);
+        lint pivinv = inv_mod<lint>(mtr[h][c], mod);
         for (int hh = 0; hh < H; hh++) {
             if (hh == h) continue;
             lint coeff = mtr[hh][c] * pivinv % mod;

number/bare_mod_algebra.hpp

@@ -1,11 +1,11 @@
 #pragma once
 #include <algorithm>
 #include <cassert>
+#include <numeric>
 #include <tuple>
 #include <utility>
 #include <vector>
 
-// CUT begin
 // Solve ax+by=gcd(a, b)
 template <class Int> Int extgcd(Int a, Int b, Int &x, Int &y) {
     Int d = a;
@@ -16,9 +16,10 @@ template <class Int> Int extgcd(Int a, Int b, Int &x, Int &y) {
     }
     return d;
 }
-// Calculate a^(-1) (MOD m) s if gcd(a, m) == 1
-// Calculate x s.t. ax == gcd(a, m) MOD m
-template <class Int> Int mod_inverse(Int a, Int m) {
+
+// Calculate a^(-1) (MOD m) if gcd(a, m) == 1
+// Calculate x s.t. ax == gcd(a, m) MOD m and 0 <= x < m
+template <class Int> Int inv_mod(Int a, Int m) {
     Int x, y;
     extgcd<Int>(a, m, x, y);
     x %= m;
@@ -76,37 +77,25 @@ template <class Int>
 
 // 蟻本 P.262
 // 中国剰余定理を利用して，色々な素数で割った余りから元の値を復元
-// 連立線形合同式 A * x = B mod M の解
+// 連立線形合同式 A_i x = R_i mod M_i の解
 // Requirement: M[i] > 0
 // Output: x = first MOD second (if solution exists), (0, 0) (otherwise)
 template <class Int>
 std::pair<Int, Int>
-linear_congruence(const std::vector<Int> &A, const std::vector<Int> &B, const std::vector<Int> &M) {
+linear_congruence(const std::vector<Int> &A, const std::vector<Int> &R, const std::vector<Int> &M) {
     Int r = 0, m = 1;
     assert(A.size() == M.size());
-    assert(B.size() == M.size());
+    assert(R.size() == M.size());
     for (int i = 0; i < (int)A.size(); i++) {
         assert(M[i] > 0);
         const Int ai = A[i] % M[i];
-        Int a = ai * m, b = B[i] - ai * r, d = std::__gcd(M[i], a);
+        Int a = ai * m, b = R[i] - ai * r, d = std::gcd(M[i], a);
         if (b % d != 0) {
             return std::make_pair(0, 0); // 解なし
         }
-        Int t = b / d * mod_inverse<Int>(a / d, M[i] / d) % (M[i] / d);
+        Int t = b / d * inv_mod<Int>(a / d, M[i] / d) % (M[i] / d);
         r += m * t;
         m *= M[i] / d;
     }
     return std::make_pair((r < 0 ? r + m : r), m);
 }
-
-template <class Int = int, class Long = long long> Int pow_mod(Int x, long long n, Int md) {
-    static_assert(sizeof(Int) * 2 <= sizeof(Long), "Watch out for overflow");
-    if (md == 1) return 0;
-    Int ans = 1;
-    while (n > 0) {
-        if (n & 1) ans = (Long)ans * x % md;
-        x = (Long)x * x % md;
-        n >>= 1;
-    }
-    return ans;
-}

number/bare_mod_algebra.md

@@ -0,0 +1,47 @@
+---
+title: Modular arithmetic utilities （C++ の基本型整数に対する拡張 GCD・中国剰余定理・連立線形合同式などの実装）
+documentation_of: ./bare_mod_algebra.hpp
+---
+
+modint を使わない `int` などの整数型上での剰余演算ユーティリティ集．拡張ユークリッドの互除法，モジュラ逆元，中国剰余定理 (CRT)，連立線形合同式の求解などを提供する．
+
+## 使用方法
+
+### `extgcd(a, b, x, y)`
+
+$ax + by = \gcd(a, b)$ を満たす整数 $x, y$ を求め，$\gcd(a, b)$ を返す．
+
+```cpp
+long long x, y;
+long long g = extgcd(12LL, 8LL, x, y); // g = 4
+```
+
+### `inv_mod(a, m)`
+
+$a$ の $m$ を法とする逆元を返す．$\gcd(a, m) = 1$ であることが必要．返り値は $[0, m)$ の範囲．
+
+```cpp
+long long inv = inv_mod(3LL, 7LL); // inv = 5 (3*5 = 15 ≡ 1 mod 7)
+```
+
+### `inv_gcd(a, b)`
+
+$g = \gcd(a, b)$ および $xa \equiv g \pmod{b}$，$0 \le x < b/g$ を満たす $(g, x)$ を返す．$b \ge 1$ が必要．
+
+### `crt(r, m)`
+
+中国剰余定理．$x \equiv r_i \pmod{m_i}$ $(i = 0, \ldots, n-1)$ を同時に満たす $x$ を求める．解が存在すれば $(x \bmod \mathrm{lcm}, \mathrm{lcm})$ を，存在しなければ $(0, 0)$ を返す．
+
+```cpp
+vector<long long> r = {2, 3}, m = {3, 5};
+auto [x, lcm] = crt(r, m); // x = 8, lcm = 15
+```
+
+### `linear_congruence(A, R, M)`
+
+連立線形合同式 $A_i x \equiv R_i \pmod{M_i}$ の解を求める．解が存在すれば $(x \bmod m, m)$ を，存在しなければ $(0, 0)$ を返す．
+
+```cpp
+vector<long long> A = {1, 1}, R = {2, 3}, M = {3, 5};
+auto [x, m] = linear_congruence(A, R, M); // x ≡ 8 mod 15
+```

number/factorize.hpp

@@ -1,8 +1,7 @@
 #pragma once
-#include "../random/xorshift.hpp"
 #include <algorithm>
-#include <array>
 #include <cassert>
+#include <cstdint>
 #include <numeric>
 #include <vector>
 
@@ -28,7 +27,7 @@ inline int get_id(long long n) {
 // Miller-Rabin primality test
 // https://ja.wikipedia.org/wiki/%E3%83%9F%E3%83%A9%E3%83%BC%E2%80%93%E3%83%A9%E3%83%93%E3%83%B3%E7%B4%A0%E6%95%B0%E5%88%A4%E5%AE%9A%E6%B3%95
 // Complexity: O(lg n) per query
-struct {
+inline struct {
     long long modpow(__int128 x, __int128 n, long long mod) noexcept {
         __int128 ret = 1;
         for (x %= mod; n; x = x * x % mod, n >>= 1) ret = (n & 1) ? ret * x % mod : ret;
@@ -53,7 +52,19 @@ struct {
     }
 } is_prime;
 
-struct {
+inline struct {
+    static uint32_t xorshift() noexcept {
+        static uint32_t x = 123456789;
+        static uint32_t y = 362436069;
+        static uint32_t z = 521288629;
+        static uint32_t w = 88675123;
+        uint32_t t = x ^ (x << 11);
+        x = y;
+        y = z;
+        z = w;
+        return w = (w ^ (w >> 19)) ^ (t ^ (t >> 8));
+    }
+
     // Pollard's rho algorithm: find factor greater than 1
     long long find_factor(long long n) {
         assert(n > 1);
@@ -63,7 +74,7 @@ struct {
         auto f = [&](__int128 x) -> long long { return (x * x + c) % n; };
 
         for (int t = 1;; t++) {
-            for (c = 0; c == 0 or c + 2 == n;) c = rand_int() % n;
+            for (c = 0; c == 0 or c + 2 == n;) c = xorshift() % n;
             long long x0 = t, m = std::max(n >> 3, 1LL), x, ys, y = x0, r = 1, g, q = 1;
             do {
                 x = y;
@@ -73,15 +84,15 @@ struct {
                     ys = y;
                     for (int i = std::min(m, r - k); i--;)
                         y = f(y), q = __int128(q) * std::abs(x - y) % n;
-                    g = std::__gcd<long long>(q, n);
+                    g = std::gcd<long long>(q, n);
                     k += m;
                 } while (k < r and g <= 1);
                 r <<= 1;
             } while (g <= 1);
             if (g == n) {
                 do {
                     ys = f(ys);
-                    g = std::__gcd(std::abs(x - ys), n);
+                    g = std::gcd(std::abs(x - ys), n);
                 } while (g <= 1);
             }
             if (g != n) return g;

number/pow_mod.hpp

@@ -0,0 +1,21 @@
+#pragma once
+#include <cassert>
+#include <type_traits>
+
+template <class Int> Int pow_mod(Int x, long long n, Int md) {
+    using Long =
+        std::conditional_t<std::is_same_v<Int, int>, long long,
+                           std::conditional_t<std::is_same_v<Int, long long>, __int128, void>>;
+    assert(n >= 0 and md > 0);
+    if (md == 1) return 0;
+    if (n == 0) return 1;
+
+    x = (x % md + md) % md;
+    Int ans = 1;
+    while (n > 0) {
+        if (n & 1) ans = (Long)ans * x % md;
+        x = (Long)x * x % md;
+        n >>= 1;
+    }
+    return ans;
+}

number/pow_mod.md

@@ -0,0 +1,17 @@
+---
+title: Modular exponentiation （べき乗 mod）
+documentation_of: ./pow_mod.hpp
+---
+
+整数 $x$, 非負整数 $n$, 正整数 $m$ に対し，$x^n \bmod m$ を $O(\log n)$ で計算する．繰り返し二乗法による実装．`Int` が `int` のとき内部で `long long`，`long long` のとき `__int128` を用いてオーバーフローを回避する．
+
+## 使用方法
+
+```cpp
+int a = pow_mod(3, 100, 1000000007);       // Int = int
+long long b = pow_mod(3LL, 100LL, (long long)1e18 + 9); // Int = long long
+```
+
+- `x`: 底．
+- `n`: 指数（$n \ge 0$）．
+- `md`: 法（$m \ge 1$）．$m = 1$ のとき常に $0$ を返す．

number/primitive_root.hpp

@@ -1,6 +1,6 @@
 #pragma once
-#include "bare_mod_algebra.hpp"
 #include "factorize.hpp"
+#include "pow_mod.hpp"
 
 // Find smallest primitive root for given number n （最小の原始根探索）
 // n must be 2 / 4 / p^k / 2p^k (p: odd prime, k > 0)
@@ -25,10 +25,10 @@ inline long long find_smallest_primitive_root(long long n) {
 
     for (long long g = 1; g < n; g++) {
         if (std::gcd(n, g) != 1) continue;
-        if (pow_mod<long long, __int128>(g, phi, n) != 1) return -1;
+        if (pow_mod(g, phi, n) != 1) return -1;
         bool ok = true;
         for (auto pp : fac) {
-            if (pow_mod<long long, __int128>(g, phi / pp, n) == 1) {
+            if (pow_mod(g, phi / pp, n) == 1) {
                 ok = false;
                 break;
             }

number/test/miller-rabin.test.cpp

@@ -1,5 +1,5 @@
 #include "../factorize.hpp"
-#define PROBLEM "https://yukicoder.me/problems/no/3030"
+#define PROBLEM "https://yukicoder.me/problems/no/8030"
 #include <iostream>
 
 int main() {

number/test/sieve.stress.test.cpp

@@ -1,10 +1,10 @@
 #define PROBLEM "https://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=ITP1_1_A" // DUMMY
 #include "../sieve.hpp"
 #include "../modint_runtime.hpp"
-#include <algorithm>
 #include <cassert>
 #include <cstdio>
 #include <iostream>
+#include <numeric>
 #include <vector>
 using namespace std;
 
@@ -15,7 +15,7 @@ struct Case {
 
 int euler_phi(int x) {
     int ret = 0;
-    for (int d = 1; d <= x; d++) ret += (std::__gcd(d, x) == 1);
+    for (int d = 1; d <= x; d++) ret += (std::gcd(d, x) == 1);
     return ret;
 }
 

rational/rational_number.hpp

@@ -8,7 +8,7 @@ template <class Int, bool AutoReduce = false> struct Rational {
     Int num, den; // den >= 0
 
     static constexpr Int my_gcd(Int a, Int b) {
-        // return __gcd(a, b);
+        // return gcd(a, b);
         if (a < 0) a = -a;
         if (b < 0) b = -b;
         while (a and b) {