結果
| 問題 | 
                            No.8037 Restricted Lucas (Hard)
                             | 
                    
| コンテスト | |
| ユーザー | 
                             | 
                    
| 提出日時 | 2018-04-02 00:16:15 | 
| 言語 | D  (dmd 2.109.1)  | 
                    
| 結果 | 
                             
                                WA
                                 
                             
                            
                         | 
                    
| 実行時間 | - | 
| コード長 | 2,321 bytes | 
| コンパイル時間 | 1,313 ms | 
| コンパイル使用メモリ | 148,596 KB | 
| 実行使用メモリ | 6,944 KB | 
| 最終ジャッジ日時 | 2024-06-13 00:09:32 | 
| 合計ジャッジ時間 | 1,942 ms | 
| 
                            ジャッジサーバーID (参考情報)  | 
                        judge4 / judge5 | 
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| ファイルパターン | 結果 | 
|---|---|
| sample | WA * 1 | 
| other | WA * 6 | 
ソースコード
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_mul = (ten + ten + ten + ten + two).to!char.to!string;
immutable string op_mod = (ten + ten + ten + seven).to!char.to!string;
immutable long MOD = ten ^^ nine + seven;
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] += mul(m_one[i][k], m_two[k][j]);
                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;
    while (T--) {
        long N = readln.chomp.to!long;
        if (N == one) {
            one.writeln;
        } else if (N == two) {
            three.writeln;
        } else {
            long a = fib(N + one);
            long b = fib(N - three);
            long c = mod(a - b, MOD);
            long d = mod(c + MOD, MOD);
            d.writeln;
        }
    }
}