結果

問題 No.613 Solitude by the window
ユーザー 👑 hos.lyric
提出日時 2019-04-27 23:12:38
言語 D
(dmd 2.109.1)
結果
AC  
実行時間 2 ms / 2,000 ms
コード長 2,752 bytes
コンパイル時間 1,029 ms
コンパイル使用メモリ 119,160 KB
実行使用メモリ 6,948 KB
最終ジャッジ日時 2024-06-22 01:03:15
合計ジャッジ時間 1,690 ms
ジャッジサーバーID
（参考情報） 		judge2 / judge1
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 * 3
other AC * 21
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ソースコード

diff #
プレゼンテーションモードにする

/*
  x -> x (x + 4) = (x + 2)^2 - 4
  x + 2 -> (x + 2)^2 - 2
  
  z[0] = 4, z[n + 1] = z[n]^2 - 2
  
  z[n] = a^(2^n) + b^(2^n)
  a + b = 4, a b = 1
  a, b = 2 \pm sqrt(3)
  T^2 - 4 T + 1 = 0
  
  (P / 3) = +1  if P == 1 (mod 3)
  (P / 3) = -1  if P == 2 (mod 3)
  (3 / P) = +1  if P == 1, 11 (mod 12)
  (3 / P) = -1  if P == 5, 7 (mod 12)
  
  (0) P = 3
    z[n] = 2 2^(2^n) = 2  if n >= 1
  
  (+1) P == 1, 11 (mod 12)
    a^*: period | P - 1
  
  (-1) P == 5, 7 (mod 12)
    b = a^P (Frobenius aut.)
    a^(P+1) = 1
    a^*: period | 2 (P + 1)
*/
import std.conv, std.functional, std.stdio, std.string;
import std.algorithm, std.array, std.bigint, std.complex, std.container, std.math, std.numeric, std.range, std.regex, std.typecons;
import core.bitop;
class EOFException : Throwable { this() { super("EOF"); } }
string[] tokens;
string readToken() { for (; tokens.empty; ) { if (stdin.eof) { throw new EOFException; } tokens = readln.split; } auto token = tokens.front; tokens
    .popFront; return token; }
int readInt() { return readToken.to!int; }
long readLong() { return readToken.to!long; }
real readReal() { return readToken.to!real; }
bool chmin(T)(ref T t, in T f) { if (t > f) { t = f; return true; } else { return false; } }
bool chmax(T)(ref T t, in T f) { if (t < f) { t = f; return true; } else { return false; } }
int binarySearch(alias pred, T)(in T[] as) { int lo = -1, hi = cast(int)(as.length); for (; lo + 1 < hi; ) { const mid = (lo + hi) >> 1; 
    (unaryFun!pred(as[mid]) ? hi : lo) = mid; } return hi; }
int lowerBound(T)(in T[] as, T val) { return as.binarySearch!(a => (a >= val)); }
int upperBound(T)(in T[] as, T val) { return as.binarySearch!(a => (a > val)); }
long power(long a, long e, long m) {
  long x = a % m, y = 1 % m;
  for (; e; e >>= 1) {
    if (e & 1) {
      y = (y * x) % m;
    }
    x = (x * x) % m;
  }
  return y;
}
void main() {
  try {
    for (; ; ) {
      const N = readLong();
      const P = readLong();
      
      long ans;
      if (P == 3) {
        ans = (N == 0) ? 1 : 2;
      } else {
        // T^2 - 4 T + 1
        long[] mul(long[] a, long[] b) {
          return [(a[0] * b[0] - a[1] * b[1]) % P, (a[0] * b[1] + a[1] * b[0] + 4 * a[1] * b[1]) % P];
        }
        long n = power(2, N, (P % 12 == 1 || P % 12 == 11) ? (P - 1) : (2 * (P + 1)));
        long[] x = [0, 1], y = [1, 0];
        for (long e = n; e; e >>= 1) {
          if (e & 1) {
            y = mul(y, x);
          }
          x = mul(x, x);
        }
        debug {
          writefln("T^%s == %s + %s T (mod T^2 - 4 T + 1)", n, y[0], y[1]);
        }
        ans = (2 * y[0] + 4 * y[1]) % P;
      }
      ans -= 2;
      ans = (ans % P + P) % P;
      writeln(ans);
    }
  } catch (EOFException e) {
  }
}
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