1. Single branch selection structure and if statement
#include <stdio.h>
#include <math.h>
int main()
{
double a = 3, b = 4, c = 5, s, area;
if (a + b > c && b + c > a && a + c > b)
{
s = (a + b + c) / 2;
area = sqrt(s * (s - a) * (s - b) * (s - c));
printf("area=%7.2f\n", area);
}
}
2. Double branch selection structure and if statement
#include <stdio.h>
#include <math.h>
int main()
{
double result, n, d;
scanf_s("%lf %lf", &n, &d);
if (d != 0)
{
result = n / d;
printf("%f\n", result);
}
else
{
printf("除数不能为0!\n");
}
}
3. Example 1: A program to find the area of a triangle
#include <stdio.h>
#include <math.h>
int main()
{
double a = 3, b = 4, c = 5, s, area;
if (a + b > c && b + c > a && a + c > b)
{
s = (a + b + c) / 2;
area = sqrt(s * (s - a) * (s - b) * (s - c));
printf("area=%7.2f\n", area);
}
else
{
printf("三边数值不能构成三角形\n");
}
}
4. Example 2: Find the roots of a quadratic equation in one unknown
#include <stdio.h>
#include <math.h>
int main()
{
double a, b, c, x1, x2;
scanf_s("%lf %lf %lf", &a, &b, &c);
if ((b * b - 4 * a * c) >= 0)
{
x1 = (-b + sqrt(b * b - 4 * a * c)) / (2 * a);
x2 = (-b - sqrt(b * b - 4 * a * c)) / (2 * a);
printf("x1=%.2f\n", x1);
printf("x2=%.2f\n", x2);
}
else
{
printf("方程无实根！\n");
}
return 0;
}
运行结果:
-6 3 7
x1=-0.86
x2=1.36
Five. Typical mistakes using if statements
#include <stdio.h>
int main()
{
int a = 0;
if (a == 100);
printf("满分!\n");
}
#include <stdio.h>
int main()
{
int a = 0;
if (a == 100);
printf("满分!\n");
else
printf("再努力!");
}
#include <stdio.h>
int main()
{
int a = 0;
if (a == 100)
printf("满分!\n");
printf("哈哈!\n");
else
printf("再努力!");
}
Six. Practical projects
1. Enter a number to determine whether it can be divisible by 3 or 5. If it is divisible by at least one of these two numbers, print this number, otherwise it will not print.
#include <stdio.h>
int main()
{
int x;
scanf_s("%d", &x);
if (x % 3 == 0 || x % 5 == 0)
{
printf("%d\n", x);
}
}
2. Input integers a and b. If the sum of the two numbers is greater than 100, then output the two numbers and the number in the hundreds place , otherwise output the sum of the two numbers.
#include <stdio.h>
int main()
{
int a, b, sum;
scanf_s("%d %d", &a, &b);
sum = a + b;
if (sum > 100)
{
sum /= 100;
sum %= 10;
printf("两数和百位上的数字为%d\n", sum);
}
else
{
printf("两数和为%d\n", sum);
}
}
3. Compile the program, calculate the value of the piecewise function and output it (x is just an integer)
#include <stdio.h>
int main()
{
int x, y;
scanf_s("%d", &x);
if (x >= 1)
{
y = x - 1;
}
else
{
y = -x + 1;
}
printf("y=%d\n", y);
}
4. Write a program to find the y value (the x value is input by the keyboard and evaluated by the trigonometric formula).
#include <stdio.h>
#include <math.h>
int main()
{
double x, y;
const PI = 3.1415926;
scanf_s("%lf", &x);
x = x * PI / 180;
if (x >= 0)
{
y = (sin(x) + cos(x)) / 2;
}
else
{
y = (sin(x) - cos(x)) / 2;
}
printf("y=%.2f\n", y);
}
5. Calculate weekly wage by piecewise function
#include <stdio.h>
int main()
{
int rate, hour;
double salary;
scanf_s("%d %d", &rate, &hour);
if (hour > 40)
{
salary = 40 * rate + (hour - 40) * rate * 1.5;
}
else
{
salary = rate * hour;
}
printf("%.2f", salary);
}
6. Input hours and minutes, and output in the form of hh:mm. Example: input 14 25, output 14:25, input 8 9, output 08:09
#include <stdio.h>
int main()
{
int h, m;
printf("输入小时和分钟");
scanf_s("%d %d", &h, &m);
printf("%02d:%02d\n", h, m);
}
#include <stdio.h>
int main()
{
int h, m;
printf("输入小时和分钟: ");
scanf_s("%d %d", &h, &m);
if (h < 10)
printf("0");
printf("%d:", h);
if (m < 10)
printf("0");
printf("%d\n", m);
}