1.4.38 3-sum的初级算法的实现。通过实验评估以下ThreeSum内循环的实现性能：
for(int i=0;i<N;i++)
  for(int j=0;j<N;j++)
    for(int k=0;k<N;k++)
       if(i<j && j<k)
          if(a[i]+a[j]+a[k]==0)
             cnt++;
为此实现另一个版本的DoublingTest,计算该程序和ThreeSum的运行时间的比例。
答：

public class E1d4d38
{
    public static int countOfThreeSum(int[] a)
    {
        int N=a.length;
        int cnt=0;
        for (int i=0;i<N;i++)
            for (int j=i+1;j<N;j++)
               for(int k=j+1;k<N;k++)
                     if(a[i]+a[j]+a[k]==0)
                        cnt++;
        return cnt;
    }//end countOfThreeSum
   
   public static int countOfBasicThreeSum(int[] a)
    {
        int N=a.length;
        int cnt=0;
        for (int i=0;i<N;i++)
            for (int j=0;j<N;j++)
               for(int k=0;k<N;k++)
                   if(i<j && j<k)
                     if(a[i]+a[j]+a[k]==0)
                        cnt++;
        return cnt;
    }//end countOfBasicThreeSum
  
 
    public static void timeTrial(int N)
    {
        int MAX=1000000;
        int[] a=new int[N];
        for(int i=0;i<N;i++)
            a[i]=StdRandom.uniform(-MAX,MAX);
      
        //
        Stopwatch timerBasicThreeSum=new Stopwatch();
        int cntBasicThreeSum=countOfBasicThreeSum(a);
        double timeBasicThreeSum= timerBasicThreeSum.elapsedTime();
         //
        Stopwatch timerThreeSum=new Stopwatch();
        int cntThreeSum=countOfThreeSum(a);
        double timeThreeSum= timerThreeSum.elapsedTime();
        //
        StdOut.printf("N=%7d BasicTime=%7.1f  ThreeSumTime=%7.1f  Rate=%7.1f \n",N,timeBasicThreeSum,timeThreeSum,timeBasicThreeSum/timeThreeSum);
    }
   
    public static void main(String[] args)
    {
        for (int N=250;N<Integer.MAX_VALUE/2;N+=N)
             {
                 timeTrial(N);
             }
    }//end main
}