与嵌套循环相关的难题

Question

对于给定的输入N,所包含的语句执行了多少次？

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;

任何人都可以找到一个简单的方法或公式来做到这一般.请解释.

Answer 1

• 首先,我编写了一个C生成总和的代码:

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");
}

• 简单编译并生成结果N=1 to N=10:

$ 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How this works?用一些图表探索:

  对于N=1:

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

那是(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

那是(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N 三个循环的总和是:

(1)+(和0f 1到2)+ ... +(1到(N-2)之和)+(1到(N-1)之和)+(1到N之和)

或者我们可以把它写成:

=>(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)

你可以在这里参考简化计算:( 我在这里问)

[ 你的答案 ]

= ( ((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,该算法的复杂性为O(n ³)

编辑:

以下循环也具有相同的计数,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;