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PAT 1029
感觉已经没啥心情用JAVA和PAT斗智斗勇了,会内存超限,是不是读进来就内存超限了?
二分做的方法有点bug,又写了一个merge sort的,想着百万数量级应该不会超时吧这都超时也太蛋疼了,结果居然内存爆了...
1 import java.util.*; 2 import java.io.*; 3 4 class FastReader{ 5 BufferedReader reader; 6 StringTokenizer tokenizer; 7 8 public FastReader(InputStream stream){ 9 reader = new BufferedReader(new InputStreamReader(stream), 1 << 10);10 tokenizer = null;11 }12 13 public String next(){14 while (tokenizer == null || !tokenizer.hasMoreTokens()){15 try{16 tokenizer = new StringTokenizer(reader.readLine());17 } catch (Exception e){18 throw new RuntimeException(e);19 }20 }21 22 return tokenizer.nextToken();23 }24 25 public long next_long(){26 return Long.parseLong(next());27 }28 29 public int next_int(){30 return Integer.parseInt(next());31 }32 }33 34 public class Main{35 static long[] seq_a;36 static long[] seq_b;37 38 public static void main(String[] args){39 FastReader reader = new FastReader(System.in);40 int N;41 42 N = reader.next_int();43 seq_a = new long[(int) N];44 for (int i = 0; i < N; i++)45 seq_a[i] = reader.next_long();46 47 N = reader.next_int();48 seq_b = new long[(int) N];49 for (int i = 0; i < N; i++)50 seq_b[i] = reader.next_long();51 52 int total = seq_a.length + seq_b.length;53 int target = ((total & 0x1) != 0) ? total / 2 + 1 : total / 2;54 55 int pa = 0, pb = 0;56 int cnt = 1;57 while (cnt < target){58 if (seq_a[pa] < seq_b[pb])59 pa++;60 else61 pb++;62 63 cnt++;64 }65 66 long ans = Math.min(seq_a[pa], seq_b[pb]);67 System.out.println(ans);68 }69 }
PAT,想说爱你不容易... 既然这么不待见JAVA为什么还给JAVA这么严苛的条件,给C++倒是很宽嘛,不用二分做都能过的时间...
PAT 1029
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