Preface ： The code is for reference only , The school compiler is too low-level to fine tune
Probably not BUG When I finished writing the code, I was already VS2019 I've run on it
When reading the code, be sure to understand the comments
I wish you a happy study ！ Happy everyday ！

experiment 1　VC++6.0 Development environment and sequential structure programming

1. Enter a Celsius temperature , Output Fahrenheit temperature required . The formula for converting Celsius temperature to Fahrenheit temperature is f=9/5*c+32.

#include <stdio.h>
int main()
{
    
	double c = 0;
	double f = 0;
	scanf("%lf", &c);
	f = 9.0 / 5 * c + 32;// It can't be used here 9/5  The compiler will default to integer division 
	printf("%lf", f);
	return 0;
}

2. Enter two integers , Output their sum 、 The square of sum 、 The sum of the squares .

#include <stdio.h>
int main()
{
    
	int a = 0;
	int b = 0;
	int sum1 = 0;
	int sum2 = 0;
	int sum3 = 0;// First define variables and several containers for storing and 
	scanf("%d", &a);
	scanf("%d", &b);
	sum1 = a + b;
	sum2 = (a + b) * (a + b);
	sum3 = a * a + b * b;
	printf("%d\n",sum1);
	printf("%d\n", sum2);
	printf("%d\n", sum3);
	return 0;
}

3、 2006 year , A football team won 98 game , lost 55 game . Use this information to write a C Program , Calculate and show the team in 2006 Percentage of wins during the year .

#include <stdio.h>
int main()
{
    
	double win = 98;
	double lose = 55;
	double total = 0;
	total = win + lose;
	double rate = win / total;
	printf("%lf%%", rate*100);
	return 0;
}

experiment 2 Choose structural programming

1． Enter several characters , Count the number of numeric characters respectively 、 The number of English letters .

 #include <stdio.h>
int main()
{
    
    int i = 0;
    int arr[100] = {
    0};
    int count = 0;
    int num1 = 0;
    int num2 = 0;
    printf(" Please enter the content :\n");
    while((arr[i]=getchar())!='\n')// Space as end sign 
    {
    
        i++;
        count++;
    }
// Enter the string 
    for(i=0; i<count; i++)
    {
    

        if(arr[i]<='z' && arr[i]>='A')
        {
    
            num1++;
        }
        if(arr[i]>='0'&&arr[i]<='9')
        {
    
            num2++;
       }

    }
    printf(" This string of characters has %d Letters , Yes %d A digital .\n",num1,num2);
    return 0;
}

2． Programming to solve ax2+bx+c=0 The root of the , among a、b、c Input by keyboard .

#include <stdio.h>
#include<math.h>
int main()
{
    
	double a = 0;
	double b = 0;
	double c = 0;
	double d1 = 0;
	double d2 = 0;
	printf(" Please enter three coefficients :");
	scanf("%lf %lf %lf", & a, &b, &c);
	if (a == 0)
	{
    
		printf(" The root of the equation is ：%lf", -b / c);
	}
	else if (a != 0 && ( b * b - 4 * a * c ) >0)
	{
    
		d1 = (-b + sqrt(b * b - 4 * a * c)) / (2 * a);
		d2 = (-b - sqrt(b * b - 4 * a * c)) / (2 * a);
		printf(" The root of the equation is ：%lf and %lf", d1, d2);
	}
	else
	{
    
		printf(" This system of equations has no solution ！");
	}

	return 0;
}

3． Know the profit of the project received by the employees of a company in a month profit( Integers ) The relationship with profit commission is as follows ( The unit of measurement is yuan )：

profit≤1000				 No commission 
1000 < profit≤2000			 Royalty 10 %
2000 < profit≤5000			 Royalty 15 %
5000 < profit≤10000		 Royalty 20 %
10000 < profit				 Royalty 25 %
 Enter the number of profits , Calculate the profit commission .

#include <stdio.h>
int main()
{
    
	double commission = 0;
	int profit = 0;
	printf(" Please enter the number of profits ：");
	scanf("%d", &profit);
	if (profit <= 1000)
	{
    
		profit = 0;
	}
	else if (1000 < profit && profit <= 2000)
	{
    
		commission = profit * 0.1;
	}
	else if (2000 < profit && profit <= 5000)
	{
    
		commission = profit * 0.15 ;
	}
	else if (5000 < profit && profit <= 10000)
	{
    
		commission = profit * 0.2;
	}
	else if (10000 < profit)
	{
    
		commission = profit * 0.25 ;
	}
	printf(" Put forward to be %lf", commission);
	return 0;
}

experiment 3　 Cyclic structure programming （ One ）

⒈ Make a program , Please Fibonacci (Fibonacci) Sequence ：1,1,2,3,5,8,………. Please output before 20 term . The sequence satisfies the relation ： Fn = Fn - 1 + Fn - 2.

#include <stdio.h>
int main()
{
    
    int arr[20] = {
     1,1,2,0, };
    int i = 0;
    printf("%d ,%d ",arr[0],arr[1]);
    for (i = 3; i <= 19; i++)
    {
    
        arr[i] = arr[i - 1] + arr[i - 2];
        printf("%d ", arr[i]);
    }
    return 0;
}

⒉ Enter two positive integers m and n, Find the greatest common divisor and the least common multiple .

#include <stdio.h>
int main()
{
    
    int m = 0;
    int n = 0;
    int max = 0;// This is used to store the larger of the two numbers 
    int i = 0;
    scanf("%d %d", &m, &n);
    if (m > n)
    {
    
        max = m;
    }
    else
    {
    
        max = n;
    }
    for (i = max; i > 2; i--)
    {
    
        if (m % i == 0 && n % i == 0)
        {
    
            printf(" The greatest common divisor is %d\n", i);
            break;
        }
    }
    for (i = max;; max++)
    {
    
        if (max % m == 0 && max % n == 0)
        {
    
            printf(" The least common multiple is %d\n", max);
            break;
        }
    }
    return 0;
}

⒊ seek Sn=a+aa+aaa+……+aa…a It's worth , Its a Chinese stands for 1 To 9 A number in . for example ：a representative 2, Then ask 2 + 22 + 222 + 2222 + 22222（ here n = 5）,a and n Input by keyboard .

#include<stdio.h>
#include<math.h> 
int main()
{
    
 int n = 0;
 int num = 0;
 int a[10] = {
    0};
 int sum = 0;
 while (scanf("%d", &n) != EOF)
	{
    
		for (int i = 0; i < n; i++)
		{
    
			num = num + 2 * pow(10, i);
			a[i] = num;
		}

		for (int i = 0; i < n; i++)
		{
    
			sum = sum + a[i];

		}

		printf("%d\n", sum);
		sum = 0;// The next set of data is updated to the initial state , namely sum by 0,Num=0 
		Num = 0;
	}

	return 0;
}

⒋ Enter a positive integer from the keyboard n, Calculate the sum of the bits of the number and output . for example , The number of inputs is 5246, The calculation ：5 + 2 + 4 + 6 = 17 And the output .

#include <stdio.h>
int main()
{
    
	int n = 0;
	int ret = 0;
	int sum = 0;
	scanf("%d", &n);
	while (n > 0)
	{
    
		ret = n % 10;
		sum += ret;
		n = n / 10;
	}
	printf("%d", sum);
	return 0;
}