如何檢查輸入數字的次數

Question

我試圖解決我收到的任務表上的一個問題，以幫助我進一步瞭解我的課程中的C++代碼。

的問題是（我引述）：

寫一個程序：

• 要求用戶輸入1和5之間10個號碼爲一個陣列並在屏幕上顯示陣列
• 創建第二個大小爲5的數組，並用零填充
• 計算第一個數組中已輸入多少個1s，2s，... 5s並將此數字存儲在第二個數組中。
• 顯示第二個數組，如下例所示。

問題是如何去檢查數字輸入的次數。我正在考慮一個for循環，但是我寫這個循環的方式根本上是不正確的，所以我發現自己正在努力看到我所遇到的錯誤。也許我錯過了一些簡單的東西？任何幫助都會很棒。

這是我的（可怕的）for循環嘗試，所以你可以看到我的錯誤。

#include <iostream> 
#include <windows.h> 

using namespace std; 

int main() 
{ 
    int input[10]; 
    const int MAX_NO = 5; 
    int COUNT[5] = { 0,0,0,0,0 }; 
    int count = 10; 

    for (int i = 0; i < count; i++) 
    { 
     cout << "Please enter a number for value " << i + 1 << " :"; 
     cin >> input[i]; 

     while (input[i] < 1 || input[i] > 5) 
     { 
      cout << "Error: Enter another number between 1 and 5: "; 
      cin >> input[i]; 
     } 
    } 

    cout << endl << "You entered "; 

    for (int i = 0; i < count; i++) 
    { 
     cout << input[i] << " "; 
    } 

    cout << "\n"; 
    // show how many times 1 number appears 

    for (int i = 1; i <= 5; i++) 
    { 
     if (input[i] == i) 
     { 
      COUNT[i]++; 
     } 
    } 


    for (int i = 0; i < MAX_NO; i++) 
    { 
     cout << i + 1 << " appears " << COUNT[i] 
      << " times in the input" << endl; 
    } 
    cout << endl; 

    system("pause"); 
    return 0; 
} 

Answer 1

將

COUNT[ input[i]-1 ]++; 

在你的第一個循環（驗證後）。一旦你這樣做了，你不需要第二個循環來計算結果。

這從內到外首先得到什麼input[i]，然後用它來修改COUNT陣列中的(input[i]-1)'th位置。如果用戶在循環的第一次運行中輸入4，則i == 0和input[i] == 4。由於數組是從0開始的，因此它將增加COUNT[input[i]-1]，在這種情況下它是COUNT[4-1] == COUNT[3]。

在您的初始循環運行後，1的數量將在COUNT[0]之間，2的數量將會在COUNT[1]等等中。

Answer 2

for (int i = 1; i <= 5; i++) 
{ 
    if (input[i] == i) 
    { 
     COUNT[i]++; 
    } 
} 

讓我們來看看它在做什麼。它從檢查input[1]開始。 （這應該是input[0]，因爲數組索引從0開始）。然後它檢查input[1]是否等於1.如果是，則它增加COUNT[1]。 接下來，它會檢查input[2]。然後它檢查input[2]是否等於2.如果是，則增加COUNT[2]。等到它經歷了input[5]。
你看到這個問題嗎？你只是檢查前5個輸入，只檢查一個值。

Answer 3

#include <iostream> 
#include <windows.h> 
using namespace std; 
int main() 
{ 
//declare a constant values 
int input[10]; 
int count = 10; //all constant MUST be in capital letters 

//second array filled with zeros 
const int MAX_NO = 5; 
int COUNT[5] = { 0, 0, 0, 0, 0 }; 

//ask user for 10 input values 
for (int i = 0; i < count; i++) 
{ 
    cout << "Please enter a number for value " << i + 1 << " :"; 
    cin >> input[i]; 
    //check if input numbers are between 1 and 5 inclusive 
    while (input[i] < 1 || input[i] > 5) 
    { 
     cout << "Error: Enter another number between 1 and 5: "; 
     cin >> input[i]; 
    } 
    /* show how many times 1 number appears. 
    this section should be in the main loop which would enable the program to check how many times a 
    number is entered so that it is stored in the second array. changed i to secondCount because this is the counting index of the second array not the first which you've called i (one of the reason you'd all zero as output when u ran your code)*/ 

    for (int secondCount = 1; secondCount <= MAX_NO; secondCount++) 
    { 
     if (input[i] == secondCount) 
     { 
      COUNT[secondCount-1]+= 1; //use minus 1 from i and increment. += 1 is the same as COUNT++ 
     } 
    } 
} 

//display number entered in the first array 
cout << endl << "You entered "; 

for (int i = 0; i < count; i++) 
{ 
    cout << input[i] << " "; 
} 

cout << "\n"; 

//display how many times a number is entered. 

for (int secondCount = 0; secondCount < MAX_NO; secondCount++) 
{ 
    cout << secondCount + 1 << " appears " << COUNT[secondCount] 
     << " times in the input" << endl; 
} 
cout << endl; 

system("pause"); 
return 0; 
} 

OUTPUT：

Please enter a number for value 1 = 1 
Please enter a number for value 2 = 1 
Please enter a number for value 3 = 1 
Please enter a number for value 4 = 2 
Please enter a number for value 5 = 3 
Please enter a number for value 6 = 2 
Please enter a number for value 7 = 4 
Please enter a number for value 8 = 4 
Please enter a number for value 9 = 3 
Please enter a number for value 10 = 2 

You entered: 1 1 1 2 3 2 4 4 3 2 

1 appears 3 times in the input 
2 appears 3 times in the input 
3 appears 2 times in the input 
4 appears 2 times in the input 
5 appears 0 times in the input