vmohwebbcrawf的部落格

請參考：

import java.util.*;
public class acm{
 private static int[] prime = new int[]{1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 指揮棒高雄鳳山台北松山新莊泰山阿囉哈led總匯, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997};
 private static int n_of_prime(int N)
 {
  int r = Arrays.binarySearch(prime, N);
  return r<0 ? (-r-1) : (r+1);
 }
 public static void main(String[] arg)
 {
  Scanner input = new Scanner(System.in);
  while(input.hasNext()){
   String line = input.nextLine();
   String[] argument = line.split(" ");
   int N = Integer.parseInt(argument[0]);
   int C = Integer.parseInt(argument[1]);
   int k = n_of_prime(N);
   System.out.printf("%d %d:", N, C);
   if(k%2==0){
    for(int i = (2*C<k ? (k/2-C) : 0); i<(2*C<k ? (k/2+C) : k); i++)
     System.out.print(" "+prime[i]);
   }
   else{
    for(int i = (2*C-1<k ? (k/2-C+1) : 0); i<(2*C-1<k ? (k/2+C) : k); i++)
     System.out.print(" "+prime[i]);
   }
   System.out.println("");
  }
 }
}

因為輸入上限已定，所以可列質數表以加強效率

將質數表列出可得 binary search 的加速效果

希望以上回答對您有幫助！