結果

問題 No.1241 Eternal Tours
ユーザー 👑 hos.lyrichos.lyric
提出日時 2020-09-06 15:14:44
言語 Java
(openjdk 23)
結果
AC  
実行時間 923 ms / 6,000 ms
コード長 3,606 bytes
コンパイル時間 3,141 ms
コンパイル使用メモリ 80,832 KB
実行使用メモリ 84,008 KB
最終ジャッジ日時 2024-11-29 11:05:30
合計ジャッジ時間 20,919 ms
ジャッジサーバーID
（参考情報） 		judge5 / judge4
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 * 4
other AC * 40
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 static java.lang.Math.*;
import static java.math.BigInteger.*;
import static java.util.Arrays.*;
import static java.util.Collections.*;
import java.math.*;
import java.util.*;
import java.io.*;

public class Main {
  public static void main(String[] args) { new Main().run(); }
  Scanner in = new Scanner(System.in);
  void _out(Object...os) { System.out.println(deepToString(os)); }
  void _err(Object...os) { System.err.println(deepToString(os)); }
  
  final long MO = 998244353;
  final long G = 3;
  
  void run() {
    for (; in.hasNext(); ) {
      int X = in.nextInt();
      int Y = in.nextInt();
      long T = in.nextLong();
      int A = in.nextInt();
      int B = in.nextInt();
      int C = in.nextInt();
      int D = in.nextInt();
      
      int m = 1 << (X + 1);
      int n = 1 << (Y + 1);
      
      long[] gms = new long[m];
      long[] gns = new long[n];
      gms[0] = 1;
      gns[0] = 1;
      gms[1] = power(G, (MO - 1) / m);
      gns[1] = power(G, (MO - 1) / n);
      for (int i = 2; i < m; ++i) {
        gms[i] = (gms[i - 1] * gms[1]) % MO;
      }
      for (int i = 2; i < n; ++i) {
        gns[i] = (gns[i - 1] * gns[1]) % MO;
      }
      
      long[][] f = new long[m][n];
      f[0][0] = f[1][0] = f[0][1] = f[m - 1][0] = f[0][n - 1] = 1;
      
      for (int x = 0; x < m; ++x) {
        fft(n, gns, f[x]);
      }
      for (int y = 0; y < n; ++y) {
        long[] work = new long[m];
        for (int x = 0; x < m; ++x) {
          work[x] = f[x][y];
        }
        fft(m, gms, work);
        for (int x = 0; x < m; ++x) {
          f[x][y] = work[x];
        }
      }
      
      long TT = 1 + (T - 1) % (MO - 1);
      for (int x = 0; x < m; ++x) for (int y = 0; y < n; ++y) {
        f[x][y] = power(f[x][y], TT);
      }
      
      for (int i = 1; i < m - i; ++i) {
        long t = gms[i];
        gms[i] = gms[m - i];
        gms[m - i] = t;
      }
      for (int i = 1; i < n - i; ++i) {
        long t = gns[i];
        gns[i] = gns[n - i];
        gns[n - i] = t;
      }
      for (int x = 0; x < m; ++x) {
        fft(n, gns, f[x]);
      }
      for (int y = 0; y < n; ++y) {
        long[] work = new long[m];
        for (int x = 0; x < m; ++x) {
          work[x] = f[x][y];
        }
        fft(m, gms, work);
        for (int x = 0; x < m; ++x) {
          f[x][y] = work[x];
        }
      }
      long invMN = power((1L * m * n) % MO, MO - 2);
      for (int x = 0; x < m; ++x) for (int y = 0; y < n; ++y) {
        f[x][y] = (f[x][y] * invMN) % MO;
      }
      
      long ans = 0;
      for (int s : new int[]{+1, -1}) for (int t : new int[]{+1, -1}) {
        int dx = (s * C - A) & (m - 1);
        int dy = (t * D - B) & (n - 1);
        ans += s * t * f[dx][dy];
      }
      ans = (ans % MO + MO) % MO;
      System.out.println(ans);
    }
  }
  
  long power(long a, long e) {
    if (e == 0) {
      return 1 % MO;
    } else {
      long b = power(a, e / 2);
      b = (b * b) % MO;
      if (e % 2 != 0) {
        b = (b * a) % MO;
      }
      return b;
    }
  }
  
  void fft(int n, long[] gs, long[] xs) {
    for (int l = n, shift = 0; (l /= 2) >= 1; ++shift) {
      for (int i = 0; i < l; ++i) {
        for (int j = i; j < n; j += l * 2) {
          long t = (xs[j] - xs[j + l]) % MO;
          xs[j] = (xs[j] + xs[j + l]) % MO;
          xs[j + l] = (gs[i << shift] * t) % MO;
        }
      }
    }
    for (int i = 0, j = 1; j < n; ++j) {
      for (int k = n / 2; k > (i ^= k); k /= 2) {}
      if (j < i) {
        long t = xs[i];
        xs[i] = xs[j];
        xs[j] = t;
      }
    }
  }
}
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