[C language] question set of IX

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Question 41 → Sum of each digit of the number 『 recursive 』

Enter an unsigned number , Just use the input function scanf() Input , But note that this is unsigned input, so we need to add... Before the data type unsigned( Unsigned type ) Can only be .

Calculate the sum of each digit of the number , Suppose that the number we enter is 19, So it's equivalent to 9+1=10, such .

Use recursion to realize , I've already said all the previous questions , Then let's talk about .

summary ： A function that calls itself within its body is called a recursive call , This function is called a recursive function . Executing a recursive function will call itself repeatedly , Each call enters a new layer , When the innermost function is executed , Exit layer by layer from the inside to the outside .

Then recursion also has restrictions , Otherwise it will cause death , This is equivalent to an endless cycle .

There are restrictions , When this restriction is met , Recursion will not continue . After each recursive call, it gets closer and closer to this limit .

Of course , This problem is also very easy to solve without recursion , The key point is just two lines of code, as follows ↓

sum += n % 10;// Find a bit sum = sum + n % 10
n = n / 10; // Erase a bit

Question 42 → Bubble sort

Topic content → Create a function Bubble_Sort() Implement bubble sorting .

Compare adjacent elements . If the first one is bigger than the second one , Just swap them .

Do the same for each pair of adjacent elements , From the beginning of the first couple to the end of the last couple . At this point , The last element should be the largest number .

 

 

Repeat the above steps for all elements , Except for the last one .

Keep repeating the above steps for fewer and fewer elements each time , Until there's no pair of numbers to compare . 

Bubble sort is to move small elements forward or to move large elements back . Comparison is the comparison of two adjacent elements , The exchange also happens between these two elements . therefore , If two elements are equal , No more exchange ; If two equal elements are not adjacent , So even if we exchange the two in front of each other to make the two adjacent , It won't be exchanged at this time , So the order of the same elements hasn't changed , So bubble sorting is a stable algorithm .

Bubble sorting based sorting algorithm , It is also an algorithm that we must firmly grasp .

Question 43 → Study Group

  

Topic content → Classmate Zhang San has a study group with ⑤ personal , Everyone has ③ The test results of this course , Seek the average score of the whole component subject and the total average score of each subject , Please use C Language programming help Zhang San realize it .

Input description : Three lines of input , Input the Chinese scores of five people respectively , Math scores 、 English scores .

Output description : Two line output , The average score of one subject and the total average score of three subjects .

The solution is as follows ↓

The main purpose of this topic is to examine the use of arrays ( One dimensional array 、 Two dimensional array ), Be careful : When we finish calculating the total score of the current subject , And it carries out the average score of the current subject . Remember to give it a clear score in the current subject 0 The operation of .

Question 44 → Multiplication of positive integers

Topic content → This topic requires the use of recursion to achieve not using  *  Operator , Realize the multiplication of two positive integers .

In fact, we have talked about recursion many times in the previous questions , Here, let's talk about the limitation of recursion ↓

Every recursive function should only make a limited number of recursive calls , Otherwise it will enter a dead end , Never quit , Such a procedure is meaningless .

There are restrictions , When this restriction is met , Recursion will not continue .

After each recursive call, it gets closer and closer to this limit .

In fact, recursion doesn't have to be too complicated , Keep recursion simple , Understand the characteristics of the problem. In fact, it is better to use recursion in this way . This is also the best way I recommend beginners to understand recursion . Like this topic, i jiode If it is a function argument, it must have two parameters , First of all, the problem requires that we cannot use multiplication and are recursive , Then we have addition and subtraction , Then you can think about how to solve this problem recursively ,4 * 4 = 16 = (4 + 4 * (4-1)) For example, can you make it a recursive feature to solve . notes : The constraints of recursion are very easy here , because 0x Any number has to 0 Then we can set up  a  or  b  The two bit constraints .

Question 45 → Array element exchange

Topic content → Will array  arr1  Numbers and arrays of elements  arr2  Exchange of elements .

int arr1[5] = { 1, 3, 5, 7, 9 };

int arr2[5] = { 2, 4, 6, 8, 10 };

Exchange these two sets of values , Get the value exchanged below . As shown below ↓

int arr1[5] = { 2, 4, 6, 8, 10 };int arr2[5] = { 1, 3, 5, 7, 9 };

Output content : To print out each element , Elements before switching and circuits after switching .

If you know how to exchange two variables , Then you can easily solve this problem . It just adds the knowledge points of one-dimensional array . It's a little more difficult than ordinary exchange. It's just a loss , So hurry to realize it .

Question 41 の Code

#define _CRT_SECURE_NO_WARNINGS 1
#include<stdio.h>
int print(unsigned int n)
{
// The reason for this is n>9 Because n<10 No matter what number we input, it will be the final sum of the sum
if (n > 9)
{
// Take the mold and get a bit , Divide by ten .
return print(n / 10) + n % 10;
}
else
{
return n;
}
}
int main(void)
{
unsigned int num = 0;
printf(&quot; Please enter a number →&quot;);
scanf(&quot;%d&quot;, &num);
int ret = print(num);
printf(&quot;ret = %d\n&quot;, ret);
return 0;
}

Running results  

Please enter a number →1234

ret = 10

Question 42 の Code

#define _CRT_SECURE_NO_WARNINGS 1
#include<stdio.h>

void Bubble_Sort(int arr[],int sz)
{
int i = 0;
// Determine the number of times to sort
for (i = 0; i < sz - 1; i++)
{
// Number of times per exchange , Number of first exchanges n-1, successively ......
int j = 0;
// Prevent invalid loops , That is, when our array is bubble sorted .
int flag = 1;
 // Determine the number of exchanges .
for (j = 0; j < sz - 1 - i; j++)
{
// Compare the first number with the second number , Exchange when the first number is greater than the second number .
if (arr[j] > arr[j + 1])
{
// Create temporary variables , swapping ！
int change;
change = arr[j];
arr[j] = arr[j+1];
arr[j+1] = change;
flag = 0;
}
}
if (flag == 1)
{
break;
}
}
}
int main(void)
{
int i = 0;
// Reverse order of array
int arr[] = { 9, 8, 7, 6, 5, 4, 3, 2, 1, 0 };
//sz Is the size of the total elements of the array  
int sz = sizeof(arr) / sizeof(arr[0]);
// Call function
Bubble_Sort(arr,sz);
for (i = 0; i < sz; i++)
{
printf(&quot;%d &quot;, arr[i]);
}
return 0;
}

Running results  

0 1 2 3 4 5 6 7 8 9 

Question 43 の Code

#define _CRT_SECURE_NO_WARNINGS 1
#include <stdio.h>
int main()
{
int i = 0;
int j = 0;// Loop initialization subscript
int arr[3][5] = { 0 };// That's ok - subject , Column - Student .
int sum = 0; // The total score of the current subject
int average = 0; // Total average score
int v[3]; // Average number of subjects
printf(&quot; Please enter the grades of students in each subject :\n&quot;);
for (i = 0; i < 3; i++)
{
printf(&quot;\n A subject counts into 5 Second grade \n&quot;);
if (i == 0)
printf(&quot; Mathematics :&quot;);
if (i == 1)
printf(&quot; Chinese subject :&quot;);
if (i == 2)
printf(&quot; English subjects :&quot;);
for (j = 0; j < 5; j++)
{
scanf(&quot;%d&quot;, &arr[i][j]); // Enter each student's grade in each subject
sum += arr[i][j]; // Calculate the total score of the current subject (sum)
}
v[i] = sum / 5; //  Average score of current account , Divide the total score by 5
sum = 0;  //  Clear the total score of the current subject 0
}
printf(&quot;\n Math scores  = %d\n Chinese achievement  = %d\n English scores  = %d\n&quot;, v[0], v[1], v[2]);
average = v[0] + v[1] + v[2];
printf(&quot; average : %d\n&quot;, average / 3);
return 0;
}

Running results  

Please enter the grades of students in each subject :

A subject counts into 5 Second grade mathematics :80 80 80 80 80 A subject counts into 5 Sub grade Chinese subject :90 90 90 90 90 A subject counts into 5 The second grade is English :100 100 100 100 100 Math scores  = 80 Chinese achievement  = 90 English scores  = 100 average : 90

Question 44 の Code

#define _CRT_SECURE_NO_WARNINGS 1
#include <stdio.h>
int Multiply(int a, int b)
{
if (b == 0)
{
return 0;
}
return a + Multiply(a, b - 1);// a + a * b
}
int main(void)
{
int i = 0;
int j = 0;
printf(&quot; Please enter two numbers :&quot;);
scanf(&quot;%d %d&quot;, &i, &j);
Multiply(i, j);
printf(&quot;%d\n&quot;, Multiply(i, j));
return 0;
}

Running results  

Please enter two numbers ：4  5 

20

 

Question 45 の Code

#include<stdio.h>
#define number 5
int main(void)
{
int i = 0;
int arr1[number] = { 1, 3, 5, 7, 9 };
int arr2[number] = { 2, 4, 6, 8, 10 };
 for (i = 0; i < number; i++)
 {
 printf(&quot; No exchange of previous values :arr1[%d] = %d\n&quot;,i + 1, arr1[i]);
 }
for (i = 0; i < number; i++)
{
printf(&quot; No exchange of previous values :arr2[%d] = %d\n&quot;, i + 1, arr2[i]);
}
printf(&quot;\n&quot;);
int sz = sizeof(arr1) / sizeof(arr1[0]);
for (i = 0; i < sz; i++)
{
int tmp;
tmp = arr1[i];
arr1[i] = arr2[i];
arr2[i] = tmp;
printf(&quot;arr1[%d] = %-2d &quot;, i + 1, arr1[i]);
printf(&quot;arr2[%d] = %-2d\n&quot;, i + 1, arr2[i]);
}
return 0;
}

Running results  

 

No exchange of previous values :arr1[1] = 1 No exchange of previous values :arr1[2] = 3 No exchange of previous values :arr1[3] = 5 No exchange of previous values :arr1[4] = 7 No exchange of previous values :arr1[5] = 9 No exchange of previous values :arr2[1] = 2 No exchange of previous values :arr2[2] = 4 No exchange of previous values :arr2[3] = 6 No exchange of previous values :arr2[4] = 8 No exchange of previous values :arr2[5] = 10arr1[1] = 2   arr2[1] = 1 arr1[2] = 4   arr2[2] = 3 arr1[3] = 6   arr2[3] = 5 arr1[4] = 8   arr2[4] = 7 arr1[5] = 10  arr2[5] = 9 

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