結果

問題 No.8037 Restricted Lucas (Hard)
ユーザー nebukuro09
提出日時 2018-04-02 00:39:42
言語 D
(dmd 2.109.1)
結果
AC  
実行時間 2 ms / 2,000 ms
コード長 2,683 bytes
コンパイル時間 1,569 ms
コンパイル使用メモリ 148,384 KB
実行使用メモリ 6,944 KB
最終ジャッジ日時 2024-06-13 00:09:35
合計ジャッジ時間 2,040 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge3
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 1
other AC * 6
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

import std.stdio, std.array, std.string, std.conv, std.algorithm;
import std.typecons, std.range, std.random, std.math, std.container;
import std.numeric, std.bigint, core.bitop, std.bitmanip;
immutable long zero = false;
immutable long one = true;
immutable long two = one << one;
immutable long three = two | one;
immutable long four = two << one;
immutable long five = four | one;
immutable long six = four | two;
immutable long seven = four | three;
immutable long eight = four << one;
immutable long nine = eight | one;
immutable long ten = eight | two;
immutable string str_A = ((one << six) | one).to!char.to!string;
immutable string str_B = ((one << six) | two).to!char.to!string;
immutable string op_add = ((one << five) | eight | two | one).to!char.to!string;
immutable string op_sub = ((one << five) | eight | four | one).to!char.to!string;
immutable string op_mul = ((one << five) | eight | two).to!char.to!string;
immutable string op_mod = ((one << five) | four | one).to!char.to!string;
immutable long MOD = ten ^^ nine  | seven;
long add(long A, long B) {
    return mixin(str_A ~ op_add ~ str_B);
}
long sub(long A, long B) {
    return mixin(str_A ~ op_sub ~ str_B);
}
long mul(long A, long B) {
    A = mod(A, MOD);
    B = mod(B, MOD);
    return mixin(str_A ~ op_mul ~ str_B);
}
long mod(long A, long B) {
    return mixin(str_A ~ op_mod ~ str_B);
}
long[][] matmul(int N, long[][] m_one, long[][] m_two) {
    auto ret = new long[][](N, N);
    foreach (i; zero..N) {
        foreach (j; zero..N) {
            ret[i][j] = zero;
            foreach (k; zero..N) {
                ret[i][j] = add(ret[i][j], mod(mul(m_one[i][k], m_two[k][j]), MOD));
                ret[i][j] = mod(ret[i][j], MOD);
            }
        }
    }
    return ret;
}
long[][] matpow(int N, long K, long[][] mat) {
    auto ret = new long[][](N, N);
    foreach (i; zero..N)
        foreach (j; zero..N)
            ret[i][j] = i == j ? one : zero;
    while (K > zero) {
        if ((K & one) == one)
            ret = matmul(N, ret, mat);
        mat = matmul(N, mat, mat);
        K >>= one;
    }
    return ret;
}
long fib(long n) {
    long[][] m = [[one, one], [one, zero]];
    return matpow(two, n, m)[zero][zero];
}
void main() {
    int T = readln.chomp.to!int;
    foreach (_; zero..T) {
        long N = readln.chomp.to!long;
        if (N == one) {
            one.writeln;
        } else if (N == two) {
            three.writeln;
        } else {
            long a = fib(add(N, one));
            long b = fib(sub(N, three));
            long c = mod(sub(a, b), MOD);
            long d = mod(add(c, MOD), MOD);
            d.writeln;
        }
    }
}
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