麻烦嵌套的循环运行时间 [英] Trouble with nested for-loop running time

问题描述

我一直在思考过这个问题了几天，现在和我挂了计算时代的第二个嵌套的循环运行的次数。我相信，我有正确的公式，用于确定运行时间对于其他两个for循环，但该第三人有我挂断了​​电话。我的第一个循环运行N-1次。等式确定的倍环路＃2是运行的数量; 1至n-1的总和。如果有人可以帮助我了解如何找到时间循环＃3的运行数量将大大AP preciated。

 的for（int i = 1; I＆LT; = N-1;我++）{
        对于（INT J = I + 1; J＆LT; = N; J ++）{
            对于（INT K = 1; K＆LT; = j的; k ++）{
            }
        }
    }
 

解决方案

第三循环运行C倍：

  C = SUM（SUM（SUM（1，K = 1 .. j）中，J = I + 1。N），I = 1 ... N-1）
  =萨姆（萨姆（J，J = i + 1的。N）中，i = 1 ... N-1）的
  = 2 + 3 + 4 + ... + N
      + 3 + 4 + ... + N
    ...
                    + N
  = 2 * 1 + 3 * 2 + 4 * 3 + 5 * 4 + ... + N *（N-1）
  =（1 * 1 + 1）+（2 * 2 + 2）+（3 * 3 + 3）+ ... +（（N-1）*（N-1）+ N-1）的
  =（1 ^ 2 + 2 ^ 2 + ......第（n-1）^ 2）+（1 + 2 + 3 + ... +（N-1））
  =（N-1）* N *（2 * N-1）/ 6 +（N-1）* n / 2个
  =（N-1）* N *（2 * N + 2）/ 6
  =为O（n ^ 3）
 

在这里，我使用的公式：

  1 ^ 2 + 2 ^ 2 + ... + M ^ 2 = M *（M + 1）×（2×M + 1）/ 6
 

和

  1 + 2 + ... + M = M *（M + 1）/ 2
 

I have been thinking over this problem for a few days now and am hung up on calculating the number of times the second nested for-loop will run. I believe that I have the correct formula for determining the running time for the other two for-loops, but this third one has me hung up. I have the first loop running n-1 times. The equation to determine the number of times loop #2 runs is; The summation of 1 to n-1. If anyone could help me understand how to find the number of times loop #3 runs it would be greatly appreciated.

    for ( int i=1; i<=n-1; i++ ) {
        for ( int j=i+1; j<=n; j++ ) {
            for ( int k=1; k<=j; k++ ) {
            }
        }
    }

解决方案

The third loop runs C times:

C = Sum( Sum ( Sum ( 1 , k = 1 .. j ) , j = i+1 .. n ) , i = 1 .. n-1 )
  = Sum( Sum (           j            , j = i+1 .. n ) , i = 1 .. n-1 )
  = 2 + 3 + 4 + ... + n
      + 3 + 4 + ... + n
    ...
                    + n
  = 2*1 + 3*2 + 4*3 + 5*4 + ... + n*(n-1)
  = (1*1 + 1) + (2*2 + 2) + (3*3 + 3) + ... + ((n-1)*(n-1) + n-1)
  = (1^2 + 2^2 + ... (n-1)^2) + (1 + 2 + 3 + ... + (n-1))
  = (n-1)*n*(2*n-1)/6 + (n-1)*n/2 
  = (n-1)*n*(2*n+2)/6
  = O(n^3)

Here I used the formulas:

1^2 + 2^2 + ... + m^2 = m*(m+1)*(2*m+1)/6

and

1 + 2 + ... + m = m*(m+1)/2

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