为什么单个线程的Java程序有这么多线程？

我有一个具有13个线程的Java程序，尽管其中只有一个线程的CPU使用率为99％，并且已经运行了约24小时。其他人的CPU使用率为0.0％，并且显示到TIME+的任何值，从0:00.0到0:12.82都有3:51.48。该程序旨在成为单线程程序，所以我想知道为什么还有其他线程？

他们在做什么，为什么他们的CPU使用率和TIME +这么少？

更新：我有一个旧的Java程序（第一个程序-不要判断我！），它是单线程的，并且显示相同类型的线程用法...

import java.io.*;

class xdriver {
  static int N = 100;
  static double pi = 3.141592653589793;
  static double one = 1.0;
  static double two = 2.0;

  public static void main(String[] args) {
    //System.out.println("Program has started successfully\n");

    if( args.length == 1) {
      // assume that args[0] is an integer
      N = Integer.parseInt(args[0]);
    }   

    // maybe we can get user input later on this ...
    int nr = N;
    int nt = N;
    int np = 2*N;

    double dr = 1.0/(double)(nr-1);
    double dt = pi/(double)(nt-1);
    double dp = (two*pi)/(double)(np-1);

    System.out.format("nn --> %d\n", nr*nt*np);

    if(nr*nt*np < 0) {
      System.out.format("ERROR: nr*nt*np = %d(long) which is %d(int)\n", (long)( (long)nr*(long)nt*(long)np), nr*nt*np);
      System.exit(1);
    }   

    // inserted to artificially blow up RAM
    double[][] dels = new double [nr*nt*np][3];

    double[] rs = new double[nr];
    double[] ts = new double[nt];
    double[] ps = new double[np];

    for(int ir = 0; ir < nr; ir++) {
      rs[ir] = dr*(double)(ir);
    }   
    for(int it = 0; it < nt; it++) {
      ts[it] = dt*(double)(it);
    }   
    for(int ip = 0; ip < np; ip++) {
      ps[ip] = dp*(double)(ip);
    }   

    double C = (4.0/3.0)*pi;
    C = one/C;

    double fint = 0.0;
    int ii = 0;
    for(int ir = 0; ir < nr; ir++) {
      double r = rs[ir];
      double r2dr = r*r*dr;
      for(int it = 0; it < nt; it++) {
        double t = ts[it];
        double sint = Math.sin(t);
        for(int ip = 0; ip < np; ip++) {
          fint += C*r2dr*sint*dt*dp;

          dels[ii][0] = dr; 
          dels[ii][1] = dt; 
          dels[ii][2] = dp; 
        }   
      }   
    }   

    System.out.format("N ........ %d\n", N); 
    System.out.format("fint ..... %15.10f\n", fint);
    System.out.format("err ...... %15.10f\n", Math.abs(1.0-fint));
  }
}