結果

問題 No.916 Encounter On A Tree
コンテスト
ユーザー hos.lyric
提出日時 2019-10-25 21:44:58
言語 D
(dmd 2.109.1)
結果
WA  
実行時間 -
コード長 4,118 bytes
コンパイル時間 1,609 ms
コンパイル使用メモリ 153,996 KB
実行使用メモリ 20,172 KB
最終ジャッジ日時 2024-06-22 02:54:52
合計ジャッジ時間 5,467 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge5
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 * 5
other AC * 31 WA * 25
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 # 

import std.conv, std.functional, std.range, std.stdio, std.string;
import std.algorithm, std.array, std.bigint, std.complex, std.container, std.math, std.numeric, 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)); }

struct ModInt(int M_) {
  alias M = M_;
  int x;
  this(ModInt a) { x = a.x; }
  this(long a) { x = cast(int)(a % M); if (x < 0) x += M; }
  ref ModInt opAssign(long a) { return (this = ModInt(a)); }
  ref ModInt opOpAssign(string op)(ModInt a) {
    static if (op == "+") { x += a.x; if (x >= M) x -= M; }
    else static if (op == "-") { x -= a.x; if (x < 0) x += M; }
    else static if (op == "*") { x = cast(int)((cast(long)(x) * a.x) % M); }
    else static assert(false);
    return this;
  }
  ref ModInt opOpAssign(string op)(long a) {
    static if (op == "^^") {
      ModInt t2 = this, te = ModInt(1);
      for (long e = a; e; e >>= 1) {
        if (e & 1) te *= t2;
        t2 *= t2;
      }
      x = cast(int)(te.x);
      return this;
    } else return mixin("this " ~ op ~ "= ModInt(a)");
  }
  ModInt inv() const {
    int a = x, b = M, y = 1, z = 0, t;
    for (; ; ) {
      t = a / b; a -= t * b;
      if (a == 0) {
        assert(b == 1 || b == -1);
        return ModInt(b * z);
      }
      y -= t * z;
      t = b / a; b -= t * a;
      if (b == 0) {
        assert(a == 1 || a == -1);
        return ModInt(a * y);
      }
      z -= t * y;
    }
  }
  ModInt opUnary(string op)() const if (op == "-") { return ModInt(-x); }
  ModInt opBinary(string op, T)(T a) const { return mixin("ModInt(this) " ~ op ~ "= a"); }
  ModInt opBinaryRight(string op)(long a) const { return mixin("ModInt(a) " ~ op ~ "= this"); }
  string toString() const { return x.to!string; }
}

enum MO = 10L^^9 + 7;
alias Mint = ModInt!MO;

enum LIM = 2^^20 + 10;
Mint[] inv, fac, invFac;
void prepare() {
  inv = new Mint[LIM];
  fac = new Mint[LIM];
  invFac = new Mint[LIM];
  inv[1] = 1;
  foreach (i; 2 .. LIM) {
    inv[i] = -(Mint.M / i) * inv[cast(size_t)(Mint.M % i)];
  }
  fac[0] = invFac[0] = 1;
  foreach (i; 1 .. LIM) {
    fac[i] = fac[i - 1] * i;
    invFac[i] = invFac[i - 1] * inv[i];
  }
}
Mint binom(long n, long k) {
  if (0 <= k && k <= n) {
    assert(n < LIM);
    return fac[cast(size_t)(n)] * invFac[cast(size_t)(k)] * invFac[cast(size_t)(n - k)];
  } else {
    return Mint(0);
  }
}


void main() {
  prepare();
  
  try {
    for (; ; ) {
      const D = readInt();
      const L = readInt();
      const R = readInt();
      const K = readInt();
      
      Mint ans = 1;
      foreach (i; 0 .. D) {
        ans *= fac[1 << i];
      }
      
      const A = bsr(L);
      const B = bsr(R);
      debug {
        writefln("D = %s, A = %s, B = %s, K = %s", D, A, B, K);
      }
      if (A == B) {
        if (K % 2 == 0 && 1 <= K / 2 && K / 2 <= A) {
          ans *= inv[(1 << A) - 1];
          ans *= 1 << (K / 2 - 1);
        } else {
          ans = 0;
        }
      } else {
        if ((K - (R - L)) % 2 == 0 && R - L <= K && K <= R + L) {
          ans *= inv[1 << A];
          ans *= 1 << max((K - (R - L)) / 2 - 1, 0);
        } else {
          ans = 0;
        }
      }
      writeln(ans);
    }
  } catch (EOFException e) {
  }
}
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