loops - 一个与嵌套循环相关的谜题

Question

对于给定的输入 N，所包含的语句执行多少次？

for i in 1 … N loop
  for j in 1 … i loop
    for k in 1 … j loop
      sum = sum + i ;
    end loop;
  end loop;
end loop;

任何人都可以想出一个简单的方法或公式来做到这一点。请解释。

Answer 1

首先，我写了一个C代码来生成sum:

int main(){
  int i =0, k =0, j =0, n =0;
  int N =0; 
  int sum =0;
  N =10;
  for (n=1; n <= N; n++){
  // unindented code here
  sum =0;
  for (i=1; i<=n; i++)
      for (j=1; j<=i; j++)
          for (k=1; k<=j; k++)
              sum++;

  printf("\n N=%d  sum = %d",n, sum); 
  }
  printf("\n");
}

N=1 to N=10 的简单编译和生成结果:

$ gcc sum.c $ ./a.out

 N=1  sum = 1
 N=2  sum = 4
 N=3  sum = 10
 N=4  sum = 20
 N=5  sum = 35
 N=6  sum = 56
 N=7  sum = 84
 N=8  sum = 120
 N=9  sum = 165
 N=10  sum = 220

然后，尝试用一些图表探索 How this works?:

对于, N=1 :

i<=N,     (i=1)       
            |
j<=i,     (j=1)       
            |
k<=j,     (K=1)       
            |
sum=0.    sum++       ---> sum = 1

即 (1) = 1

对于 N=2 :

i<=N,     (i=1)-------(i=2)
            |     |-----|-----|
j<=i,     (j=1) (j=1)      (j=2)
            |     |     |----|----|
k<=j,     (K=1) (K=1) (K=1)    (K=2)               
            |     |     |        |    
sum=0,    sum++  sum++ sum++   sum++  --> sum = 4

即 (1) + (1 + 2) = 4

对于 N=3 :

i<=N,     (i=1)-------(i=2)--------------------(i=3)
            |     |-----|-----|        |---------|-------------|
j<=i,     (j=1) (j=1)      (j=2)     (j=1)      (j=2)        (j=3)
            |     |     |----|----|    |     |----|----|    |-----|-----|
k<=j,     (K=1) (K=1) (K=1)    (K=2) (K=1) (K=1)    (K=2) (K=1) (K=2) (K=3)
            |     |     |        |     |     |        |     |     |        |
sum=0,    sum++  sum++ sum++  sum++ sum++  sum++    sum++  sum++ sum++  sum++
            --> sum = 10

即 (1) + (1 + 2) + ( 1 + 2 + 3 ) = 10

N = 1, (1)    = 1                                           

N = 2, (1) + (1 + 2)    = 4

N = 3, (1) + (1 + 2) + (1 + 2 + 3)  = 10

N = 4, (1) + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4)  = 20

N = 5, (1) + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4) + (1 + 2 + 3 + 4 + 5)  = 35

最后，我可以理解三个循环中 N 的总和是:

(1) + (sum 0f 1 to 2) + ... + (sum of 1 to (N-2)) + (sum of 1 to (N-1) ) + (sum of 1 to N)

或者我们可以写成:

=> (1) + (1 + 2) + ...+ (1 + 2 +....+ i) + ... + (1 + 2 + ....+ N-1) + (1) + 2 + ....+ N)

=> ( N * 1 ) + ( (N-1) * 2) + ( (N-2) * 3) +...+ ( (N -i+1) * i ) +... + ( 1 * N)

您可以引用这里进行简化计算: (I asked HERE )


[ 你的回答 ]
= ( ((N) * (N+1) * (N+2)) / 6 )
而且，我认为它是正确的。我检查如下:

N = 1,    (1 * 2 * 3)/6  = 1

N = 2,    (2 * 3 * 4)/6 = 4

N = 3,    (3 * 4 * 5)/6 = 6

N = 4,    (4 * 5 * 6)/6 = 10

N = 5,    (5 * 6 * 7)/6 = 35   

此外，该算法的复杂度为 O(n3)

编辑 :

下面的循环也有相同的计数，即 = ( ((N) * (N+1) * (N+2)) / 6 )

for i in 1 … N loop
  for j in i … N loop
    for k in j … N loop
      sum = sum + i ;
    end loop;
  end loop;
end loop;