這個僞代碼的邏輯是否正確？

Question

我在將這個作業問題轉換爲僞代碼時遇到了一些麻煩。我想知道這個邏輯是否正確。

問題：

編寫一個程序，打印如下內容：1，3，5，7，9，11，13，15，17，19，在這之前用戶輸入的值。該程序不應該打印每行超過10個數字。打印的最後一行可能小於10.除了第一個數字和最後一個數字之外，應該有一個「，」將所有數字分開。每行上的最後一個數字後面應該有一個「。」。在退出之前，還應打印所有元素的總和：計算項目總數並輸入結果。例如，如果用戶輸入76的輸出應該如下 程序應儘可能的樣品運行如下：

輸入數（> = 1和< = 100）：76 你的序列是

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

上述元素的總和1444。

我嘗試：

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

由您提出的算法使用了可以保存下來，因爲我們需要打印的所有奇數的問題，可以用四個變量，一個用於跟蹤總和來解決內存，一個用於改變行，一個用於用戶輸入，另一個用作循環變量。

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 */