Hello World!

#include <stdio.h>
int main (){
printf("************************\n");
printf("*                      *\n");
printf("*     Hello World!     *\n");
printf("*                      *\n");
printf("************************\n");
return 0;
}

三个数中取最大值

#include <stdio.h>
int main(){
  int a,b,c,max;
  printf("please input a,b,c:\n");
  scanf("%d,%d,%d",&a,&b,&c);   //输入的3个数以逗号分隔，如果以空格分隔会出错
  //scanf("%d%d%d",&a,&b,&c);   //可以使用空格分隔
  max=a;
  if(max<b)
    max=b;
  if(max<c)
      max=c;
  printf("The largest number is %d\n",max);
  return 0;
}

两数之和

#include <stdio.h>
int main() {
  int a, b, sum;
  printf("please input two number:\n");
  scanf("%d%d", &a, &b);
  sum = a + b;
  printf("The sum is: %d\n", sum);
  return 0;
}

两个整数比大小

#include <stdio.h>
int main() {
  int a, b, max, min;
  printf("please input two number:\n");
  scanf("%d%d", &a, &b);
  if (a > b) {
    max = a;
    min = b;
    printf("max=%d,min=%d\n", max, min);
  } else if (a < b) {
    max = b;
    min = a;
    printf("max=%d,min=%d\n", max, min);
  } else if (a = b) {
    printf("Two number are equal!\n");
  }
  return 0;
}

求两个整数中的较大者

#include <stdio.h>
//主函数
int main()                             //定义主函数
{                                      //主函数体开始
  int max(int x, int y);               //对被调用函数max的声明
  int a, b, c;                         //定义变量a,b,c
  printf("please input two number:");  //打印输出，提示用户输入两个数字
  scanf("%d%d", &a, &b);               //输入变量a和b的值
  c = max(a, b);          //调用max函数，将得到的值赋给c
  printf("max=%d\n", c);  //输出c的值
  return 0;               //返回函数值为0
}  //主函数体结束
//求两个整数中的较大者的max函数
int max(int x, int y) {  //定义mmax函数，函数值为整型，形式参数x和y为整型
  int z;  // max函数中的声明部分，定义本函数中用到的变量z为整型
  if (x > y) z = x;  //若x>y成立，将x的值赋给变量z
  else z = y;  //否则（即x>y不成立），将y的值赋给变量z
  return (z);  //将z的值作为max的函数值，返回到调用max函数的位置
}

运行时输入a,b,c三个值，输出其中值最大者

#include <stdio.h>
int main() {
  int a, b, c, max;
  printf("please input a,b,c:");
  scanf("%d,%d,%d", &a, &b, &c);
  a > b ? (a > c ? (max = a) : (max = c))
        : (b > c ? (max = b) : (max = c));
              printf("The largest number is %d\n", max);
  return 0;
}

求阶乘n!

#include <stdio.h>
int main() {
  int t = 1, n;
  printf("please input n:");
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    t = t * i;
  }
  printf("The multiplicative of n is :%d", t);
  return 0;
}

求5！

#include <stdio.h>
int main() {
  int i = 2, t = 1;
  while (i <= 5) {
    t = t * i;
    i = i + 1;
  }
  printf("%d\n", t);
  return 0;
}

输出成绩在80分以上的学生的学号和成绩

#include <stdio.h>
int main() {
  int a[10] = {91, 82, 73, 54, 95, 66, 77, 48, 99, 76},
      b[10] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
  printf("Student number and grades of students with scores over 80:\n");
  for (int i = 1; i <= 10; i++) {
    if (a[i] >= 80) printf("%d,%d\n", b[i], a[i]);
  }
  return 0;
}

判断2000-2500间的闰年

#include <stdio.h>
int main() {
  int n;
  printf("Please enter a year from 2000 to 2500:");
  scanf("%d", &n);
  if (n >= 2000 && n <= 2500) {
    if (n % 4 == 0) {
      if (n % 100 == 0) {
        if (n % 400 == 0) {
          printf("%d is a leap year.", n);
        } else
          printf("%d is not a leap year.", n);
      } else {
        printf("%d is a leap year.", n);
      }
    } else {
      printf("%d is not a leap year.", n);
    }
  } else {
    printf("Input error, please re-enter a year from 2000 to 2500:");
  }
  return 0;
}

输出2000-2500间的闰年

#include <stdio.h>
int main() {
  int i=0;     //定义变量i来计算2000-2500年闰年的数量
  printf("The leap years in 2000-2500 are:");
  for(int n = 2000; n <= 2500; n++ ) {
    if (n % 4 == 0) {
      if (n % 100 == 0) {
        if (n % 400 == 0) {
          printf("'%d'", n);
          ++i;
        }
      } else {
        printf("'%d'", n);
        ++i;
      }
    }
  }
  printf("\nThere are %d leap years in 2000-2500.",i);
  return 0;
}

#include <stdio.h>
int main() {
  int year;
  printf("Enter year:");
  scanf("%d", &year);
  if(year > 2000 && year <2500)
  (year % 4 == 0 && year % 100 != 0) || year % 400 == 0 ? printf("yes") : printf("no");
  else printf("error");
  return 0;
}

求多项式1-1/2+1/3-1/4+1/5-1/6+···+1/99-1/100

求： $\displaystyle {1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} +\cdots +\frac{1}{97} - \frac{1}{98} + \frac{1}{99} - \frac{1}{100}}$ 的值。

#include <stdio.h>
int main() {
  int sign = 1;
  double deno = 2.0, sum = 1.0, term;  //定义deno，sum，term为双精度型变量
  while (deno <= 100) {
    sign = -sign;
    term = sign / deno;
    sum = sum + term;
    deno = deno + 1;
  }
  printf("%f\n", sum);
  return 0;
}

温度转换：华氏法（℉）转换为摄氏法（℃）

#include <stdio.h>
int main() {
  float f, c;     //定义华氏度f和摄氏度c为单精度浮点型变量
  printf("f=");
  scanf("%f", &f);
  c = (5.0 / 9) * (f - 32);   //两变量转换公式为c=(5/9)*(f-32)
  printf("f=%f\nc=%f", f, c);
  return 0;
}

编辑器中各数据类型在内存中占用的存储单元长度

#include <stdio.h>
int main() {
  int sizeof_int, sizeof_short, sizeof_long, sizeof_long_long_int, 
  sizeof_char, sizeof_float, sizeof_double;
  sizeof_int = sizeof(int);
  sizeof_short = sizeof(short);
  sizeof_long = sizeof(long);
  sizeof_long_long_int = sizeof(long long int);
  sizeof_char = sizeof(char);
  sizeof_float = sizeof(float);
  sizeof_double = sizeof(double);
  printf(
      "int=%d\nshort int=%d\nlong int=%d\nlong long int=%d\n"
      "char=%d\nfloat=%d\ndouble=%d\n",
      sizeof_int, sizeof_short, sizeof_long, sizeof_long_long_int, 
      sizeof_char, sizeof_float, sizeof_double);
  return 0;
}

无符号位整形数据用“%u”格式输出

#include <stdio.h>
int main() {
unsigned int a;    //定义a为无符号整型变量
  a=2147483648;    //指定用无符号十进制数的格式输出
  printf("%u",a);
}

字符常量有符号无符号

#include <stdio.h>
int main() {
  unsigned char a = 128;   //无符号取值范围为0~255
  signed char b = 128;     //有符号取值范围为-128~127，b=228时溢出
  printf("%c = %d\n", a, a);
  printf("%c = %d\n", b, b);
}

大小写字母转换

#include <stdio.h>
int main() {
  char letter, big, small;
  printf("Enter a letter:");
  scanf("%c", &letter);
  if (letter > 64 && letter < 91) {   //大写字母ASCII码范围65~90
    big = letter;
    small = big + 32;
  } else if (letter > 96 && letter < 123) {   //小写字母ASCII码范围97~122
    small = letter;
    big = small - 32;
  } else {
    printf("Input error");
  }
  printf("big=%c\nsmall=%c", big, small);
  return 0;
}

输入的字母统一以小写形式输出

#include <stdio.h>
int main() {
  char ch;
  printf("Enter a letter:");
  scanf("%c", &ch);
  if (ch > 'A' && ch < 'Z' || ch > 'a' && ch < 'z') {
    ch = (ch > 'A' && ch < 'Z') ? (ch + 32) : ch;
    printf("%c", ch);
  } else
    printf("error");
  return 0;
}

求已知三角形三边长，求三角形面积

已知三角形面积公式： $area=\sqrt{s(s-a)(s-b)(s-c)}$
其中， $s=\frac{a+b+c}{2}$ ，a，b，c为三角形的边长。

#include <math.h>  //调用数学函数库，后面计算用sqrt函数开根号
#include <stdio.h>
int main() {
  double a, b, c, s, area;
  printf("Enter the triangle side length a,b,c:");
  scanf("%lf%lf%lf", &a, &b, &c);
  if (a + b > c && a + c > b && b + c > b) {
    s = (a + b + c) / 2;
    area = sqrt(s * (s - a) * (s - b) * (s - c));
    printf("a=%f\tb=%f\tc=%f\narea=%f\n", a, b, c, area);
  } else {
    printf("error");
  }
  return 0;
}

求一元二次方程 $ax^2+bx+c=0$ 的根。a,b,c由键盘输入，设 $b^2-4ac≥0$ .

已知，当 $b^2-4ac≥0$ 时，一元二次方程 $ax^2+bx+c=0$ 有根。
实数根： $x_1=\frac{-b+\sqrt{b^2-4ac}}{2a},x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$
令 $p=\frac{-b}{2a},q=\frac{\sqrt{b^2-4ac}}{2a}$ ,则 $x_1=p+q,x_2=p-q$ .

#include <math.h>  //调用数学函数库，后面计算用sqrt函数开根号
#include <stdio.h>
int main() {
  double a, b, c, disc, x1, x2, p, q;
  printf("Please enter a, b, c:");
  scanf("%lf%lf%lf", &a, &b, &c);
  disc = b * b - 4 * a * c;
  if (disc >= 0) {
    p = -b / (2.0 * a);
    q = sqrt(disc) / (2.0 * a);
    x1 = p + q;
    x2 = p - q;
    printf("x1=%7.2f\nx2=%7.2f", x1, x2);  //%7.2f指定输出数据占7列，其中小数部分占2列
  } else {
    printf("The equation has no real root.");  //当disc<0时，方程无实根
  }
  return 0;
}

对输入的三个数按从小到大的顺序排列

#include <stdio.h>
int main() {
  int a, b, c, t;
  printf("Enter three number:");
  scanf("%d%d%d", &a, &b, &c);
  if (a > b) {
    t = a;
    a = b;
    b = t;
  }
  if (a > c) {
    t = a;
    a = c;
    c = t;
  }
  if (b > c) {
    t = b;
    b = c;
    c = t;
  }
  printf("%d<%d<%d", a, b, c);
  return 0;
}

阶跃函数

$y=\begin{cases}-1& \text{（x<0）}\\0 & \text{（x=0）}\\1& \text{（x>0）}\end{cases}$ 输入一个x值，输出相应y值。

#include <stdio.h>
int main() {
  float x;
  int y;
  printf("Enter x:");
  scanf("%f", &x);
  if (x < 0) {
    y = -1;
  } else if (x == 0) {
    y = 0;
  } else {
    y = 1;
  }
  printf("y=%d", y);
  return 0;
}

#include <stdio.h>
int main() {
  float x;
  int y;
  printf("Enter x:");
  scanf("%f", &x);
  y = (x < 0) ? -1 : ((x == 0) ? 0 : 1);
  printf("y=%d", y);
  return 0;
}

成绩等级

#include <stdio.h>
int main() {
  char grade;
  scanf("%c",&grade);     //输入等级
  printf("Your score:");
  switch(grade){
    case'A':printf("85~100\n");break;
    case'B':printf("70~84\n");break;
    case'C':printf("60~69\n");break;
    case'D':printf("<60\n");break;
    default:printf("enter data error!\n");
  }
  return 0;
}