这个带有数组的伪代码的逻辑是否正确？

Question

我在将这个作业问题转换为伪代码时遇到了一些麻烦。我想知道逻辑是否正确。

问题：

编写一个程序，打印以下元素：1、3、5、7、9、11、13、15、17、19，直到用户输入的值。程序每行打印的数字不应超过 10 个。打印的最后一行可能小于 10。除了第一个数字和最后一个数字外，应该有一个“，”分隔所有数字。每行的最后一个数字应该有一个“。”在它之后。在退出之前，它还应该打印所有元素的总和：计算项目的总和并输入结果。例如，如果用户输入 76 你的输出应该如下 您的程序的示例运行应如下所示：

输入您的号码（>= 1 和

1、3、5、7、9、11、13、15、17、19。 21,23,25,27,29,31,33,35,37,39。 41,43, 45, 47, 49, 51, 53, 55, 57, 59。 61,63, 65, 67, 69, 71, 73, 75。

以上元素之和1444。

我的尝试：

Algorithm ElementsSequenceSeries

// declare variables
Var num, count, odd, sum, arrayOne, arrayTwo, arrayThree, arrayFour, arrayFive

// initialize variables
Count = 0; sum = 0; odd=0; arrayOne = 0; arrayTwo = 0; arrayThree = 0; arrayFour = 0; arrayFive = 0;

<BeginAlg>

Print (Input 1<= num <= 100); 

Count = count + 1;
Odd = (count*2)–1; // create a list of odd numbers up to a user-defined value less than 100

If (num > odd)
  Print(Odd);

    // For the indices 0 to 8, place a comma after each element of the array.
    For (num <=20 and num> odd) and (arrayOne[9]= {0,1,2,3,4,5,6,7,8})
        Print (arrayOne[]= {0,1,2,3,4,5,6,7,8} + “,”);
    EndFor

    // For the index 9, place a fullstop after it.
    For (num <=20 && num> odd) and (array One[]= 9)
        Print arrayOne[]= 9 + “.”)
    EndFor


    For (num <=40 and num> odd) and (arrayTwo[]= {0,1,2,3,4,5,6,7,8})
        Print (arrayTwo[]= {0,1,2,3,4,5,6,7,8} + “,”);
    EndFor

    For (num <=40 and num> odd) and (arrayTwo[]= 9)
        Print arrayTwo[]= 9 + “.”)
    EndFor

    For (num <=60 and num> odd) and (arrayThree[]={0,1,2,3,4,5,6,7,8})
        Print (arrayThree[]= {0,1,2,3,4,5,6,7,8} + “,”);
    EndFor

    For (num <=60 and num> odd) and (arrayThree[]= 9)
        Print arrayThree[]= 9 + “.”)
    EndFor


    For (num <=80 and num> odd) and (arrayFour[]={0,1,2,3,4,5,6,7,8})
        Print (arrayFour []= {0,1,2,3,4,5,6,7,8} + “,”);
    EndFor

    For (num <=80 and num> odd) and (arrayFour []= 9)
        Print arrayFour []= 9 + “.”)
    EndFor


    For (num <=100 and num> odd) and (arrayFive[]={0,1,2,3,4,5,6,7,8})
        Print (arrayFive []= {0,1,2,3,4,5,6,7,8} + “,”);
    EndFor

    For (num <=100 and num> odd) and (arrayFive []= length)
        Print arrayFive []= length + “.”);
    EndFor

EndIf

Sum = arrayOne, arrayTwo, arrayThree, arrayFour, arrayFive;
Print(sum);

<EndAlg>

Answer 1

您提出的算法使用可以节省的内存，因为我们需要打印所有奇数，所以可以使用四个变量来解决问题，一个用于跟踪总和，一个用于换行，一个用于用户输入和一个作为循环变量。

Variables:

sum   :int   // keeps track of the sum
count :int   // keeps track of the Integer printed in a line.
i     :int   // looping variable.
n     :int   // Given as input
Algorithm:

/* BEGIN ALGORITHM */
n=input();
count=0;
sum=0
for(i=1:i<n:i=i+2){
                  print("i,");
                  count=count+1;
                  if(count>9){ 
                               print("/n"); // change line;
                               count=0;
                               }
                   sum=sum+i;
                   }
print("n."); 
print("Sum is (sum+n)");
/* END ALGORITHM */