linear_algebra_matrix/test/linalg_longlong_matmul.test.cpp

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Last update: 2022-11-15 00:34:03+09:00
Problem: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=ITP1_7_D

Depends on

Code

#include "../linalg_longlong.hpp"
#include <iostream>
using namespace std;
#define PROBLEM "http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=ITP1_7_D"
using lint = long long;

int main() {
    int N, M, L;
    cin >> N >> M >> L;
    vector<vector<lint>> A(N, vector<lint>(M)), B(M, vector<lint>(L));
    for (auto &v : A) {
        for (auto &x : v) cin >> x;
    }
    for (auto &v : B) {
        for (auto &x : v) cin >> x;
    }
    auto C = matmul(A, B, lint(1e13));
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < L - 1; j++) printf("%lld ", C[i][j]);
        printf("%lld\n", C[i].back());
    }
}

#line 2 "number/bare_mod_algebra.hpp"
#include <algorithm>
#include <cassert>
#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;
    if (b != 0) {
        d = extgcd(b, a % b, y, x), y -= (a / b) * x;
    } else {
        x = 1, y = 0;
    }
    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) {
    Int x, y;
    extgcd<Int>(a, m, x, y);
    x %= m;
    return x + (x < 0) * m;
}

// Require: 1 <= b
// return: (g, x) s.t. g = gcd(a, b), xa = g MOD b, 0 <= x < b/g
template <class Int> /* constexpr */ std::pair<Int, Int> inv_gcd(Int a, Int b) {
    a %= b;
    if (a < 0) a += b;
    if (a == 0) return {b, 0};
    Int s = b, t = a, m0 = 0, m1 = 1;
    while (t) {
        Int u = s / t;
        s -= t * u, m0 -= m1 * u;
        auto tmp = s;
        s = t, t = tmp, tmp = m0, m0 = m1, m1 = tmp;
    }
    if (m0 < 0) m0 += b / s;
    return {s, m0};
}

template <class Int>
/* constexpr */ std::pair<Int, Int> crt(const std::vector<Int> &r, const std::vector<Int> &m) {
    assert(r.size() == m.size());
    int n = int(r.size());
    // Contracts: 0 <= r0 < m0
    Int r0 = 0, m0 = 1;
    for (int i = 0; i < n; i++) {
        assert(1 <= m[i]);
        Int r1 = r[i] % m[i], m1 = m[i];
        if (r1 < 0) r1 += m1;
        if (m0 < m1) {
            std::swap(r0, r1);
            std::swap(m0, m1);
        }
        if (m0 % m1 == 0) {
            if (r0 % m1 != r1) return {0, 0};
            continue;
        }
        Int g, im;
        std::tie(g, im) = inv_gcd<Int>(m0, m1);

        Int u1 = m1 / g;
        if ((r1 - r0) % g) return {0, 0};

        Int x = (r1 - r0) / g % u1 * im % u1;
        r0 += x * m0;
        m0 *= u1;
        if (r0 < 0) r0 += m0;
    }
    return {r0, m0};
}

// 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
// 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) {
    Int r = 0, m = 1;
    assert(A.size() == M.size());
    assert(B.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);
        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);
        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;
}
#line 4 "linear_algebra_matrix/linalg_longlong.hpp"
#include <cstdlib>
#line 6 "linear_algebra_matrix/linalg_longlong.hpp"

// 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
    int H = mtr.size(), W = mtr[0].size(), c = 0;
    for (int h = 0; h < H; h++) {
        if (c == W) break;
        int piv = -1;
        for (int j = h; j < H; j++)
            if (mtr[j][c]) {
                if (piv == -1 or abs(mtr[j][c]) > abs(mtr[piv][c])) piv = j;
            }
        if (piv == -1) {
            c++;
            h--;
            continue;
        }
        std::swap(mtr[piv], mtr[h]);
        if (h != piv) {
            for (int w = 0; w < W; w++) {
                mtr[piv][w] =
                    mtr[piv][w] ? mod - mtr[piv][w] : 0; // To preserve sign of determinant
            }
        }
        lint pivinv = mod_inverse<lint>(mtr[h][c], mod);
        for (int hh = 0; hh < H; hh++) {
            if (hh == h) continue;
            lint coeff = mtr[hh][c] * pivinv % mod;
            for (int w = W - 1; w >= c; w--) {
                mtr[hh][w] = mtr[hh][w] - mtr[h][w] * coeff % mod;
                if (mtr[hh][w] < 0) mtr[hh][w] += mod;
            }
        }
        c++;
    }
    return mtr;
}

template <typename lint>
int rank_gauss_jordan(const std::vector<std::vector<lint>> &mtr) // Rank of Gauss-Jordan eliminated matrix
{
    for (int h = (int)mtr.size() - 1; h >= 0; h--) {
        for (auto v : mtr[h])
            if (v) return h + 1;
    }
    return 0;
}

template <typename lint, typename mdint>
lint mod_determinant(std::vector<std::vector<lint>> mtr, mdint mod) {
    if (mtr.empty()) return 1 % mod;
    assert(mtr.size() == mtr[0].size());
    lint ans = 1;
    mtr = gauss_jordan(mtr, mod);
    for (int i = 0; i < (int)mtr.size(); i++) ans = ans * mtr[i][i] % mod;
    return ans;
}

template <typename lint, typename mdint>
std::vector<std::vector<lint>>
matmul(const std::vector<std::vector<lint>> &A, const std::vector<std::vector<lint>> &B, mdint mod) {
    int H = A.size(), W = B[0].size(), K = B.size();
    std::vector<std::vector<lint>> C(H, std::vector<lint>(W));
    for (int i = 0; i < H; i++) {
        for (int j = 0; j < W; j++) {
            for (int k = 0; k < K; k++) (C[i][j] += A[i][k] * B[k][j]) %= mod;
        }
    }
    return C;
}

template <typename lint, typename mdint>
std::vector<lint>
matmul(const std::vector<std::vector<lint>> &A, const std::vector<lint> &v, mdint mod) {
    std::vector<lint> res(A.size());
    for (int i = 0; i < (int)A.size(); i++) {
        for (int j = 0; j < (int)v.size(); j++) (res[i] += A[i][j] * v[j]) %= mod;
    }
    return res;
}

template <typename lint, typename powint, typename mdint>
std::vector<std::vector<lint>> matpower(std::vector<std::vector<lint>> X, powint n, mdint mod) {
    std::vector<std::vector<lint>> res(X.size(), std::vector<lint>(X.size()));
    for (int i = 0; i < (int)res.size(); i++) res[i][i] = 1;
    while (n) {
        if (n & 1) res = matmul(res, X, mod);
        X = matmul(X, X, mod);
        n >>= 1;
    }
    return res;
}
#line 2 "linear_algebra_matrix/test/linalg_longlong_matmul.test.cpp"
#include <iostream>
using namespace std;
#define PROBLEM "http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=ITP1_7_D"
using lint = long long;

int main() {
    int N, M, L;
    cin >> N >> M >> L;
    vector<vector<lint>> A(N, vector<lint>(M)), B(M, vector<lint>(L));
    for (auto &v : A) {
        for (auto &x : v) cin >> x;
    }
    for (auto &v : B) {
        for (auto &x : v) cin >> x;
    }
    auto C = matmul(A, B, lint(1e13));
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < L - 1; j++) printf("%lld ", C[i][j]);
        printf("%lld\n", C[i].back());
    }
}