結果

問題 No.990 N×Mマス計算(Kの倍数)
ユーザー tenten
提出日時 2022-08-30 12:48:17
言語 Java
(openjdk 23)
結果
AC  
実行時間 634 ms / 2,000 ms
コード長 2,796 bytes
コンパイル時間 3,159 ms
コンパイル使用メモリ 78,780 KB
実行使用メモリ 60,848 KB
最終ジャッジ日時 2024-11-07 07:59:58
合計ジャッジ時間 8,838 ms
ジャッジサーバーID
(参考情報)
judge2 / 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 * 2
other AC * 19
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 java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) throws Exception {
Scanner sc = new Scanner();
int n = sc.nextInt();
int m = sc.nextInt();
int k = sc.nextInt();
if (sc.next().charAt(0) == '+') {
HashMap<Integer, Long> bCounts = new HashMap<>();
for (int i = 0; i < m; i++) {
int x = sc.nextInt() % k;
bCounts.put(x, bCounts.getOrDefault(x, 0L) + 1);
}
HashMap<Integer, Long> aCounts = new HashMap<>();
for (int i = 0; i < n; i++) {
int x = sc.nextInt() % k;
aCounts.put(x, aCounts.getOrDefault(x, 0L) + 1);
}
long ans = 0;
for (int x : bCounts.keySet()) {
if (x == 0) {
ans += aCounts.getOrDefault(x, 0L) * bCounts.get(x);
} else {
ans += aCounts.getOrDefault(k - x, 0L) * bCounts.get(x);
}
}
System.out.println(ans);
} else {
HashMap<Integer, Long> bCounts = new HashMap<>();
for (int i = 0; i < m; i++) {
int x = getGCD(k, sc.nextInt());
bCounts.put(x, bCounts.getOrDefault(x, 0L) + 1);
}
HashMap<Integer, Long> aCounts = new HashMap<>();
for (int i = 0; i < n; i++) {
int x = getGCD(k, sc.nextInt());
aCounts.put(x, aCounts.getOrDefault(x, 0L) + 1);
}
long ans = 0;
for (int x : bCounts.keySet()) {
for (int y : aCounts.keySet()) {
if (y % (k / x) == 0) {
ans += bCounts.get(x) * aCounts.get(y);
}
}
}
System.out.println(ans);
}
}
static int getGCD(int x, int y) {
if (y == 0) {
return x;
} else {
return getGCD(y, x % y);
}
}
}
class Scanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
StringBuilder sb = new StringBuilder();
public Scanner() throws Exception {
}
public int nextInt() throws Exception {
return Integer.parseInt(next());
}
public long nextLong() throws Exception {
return Long.parseLong(next());
}
public double nextDouble() throws Exception {
return Double.parseDouble(next());
}
public String next() throws Exception {
while (!st.hasMoreTokens()) {
st = new StringTokenizer(br.readLine());
}
return st.nextToken();
}
}
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