当前位置：网站首页>Introduction to C language (6)

Introduction to C language (6)

2022-07-18 07:08:00 【InfoQ】

1. Character type judgment

Title Description

Enter a character from the keyboard , Determine whether the character is capitalized 、 Lowercase letters 、 Numeric characters or other characters . Output corresponding prompt information respectively .

Input

Enter a character .

Output

If the character is a capital letter , The output “upper”; If it's lowercase , The output “lower”; If it is a numeric character , The output “digit”; For other characters , The output “other”.（ The output does not contain double quotes ）.

The sample input  

E

Sample output  

upper

#include<stdio.h>
#include<ctype.h>
int main()
{
 char ch;
 scanf(&quot;%c&quot;,&ch);
 if(islower(ch))
 printf(&quot;lower\n&quot;);
 else if(isupper(ch))
 printf(&quot;upper\n&quot;);
 else if(isdigit(ch))
 printf(&quot;digit\n&quot;);
 else
 printf(&quot;other\n&quot;);
 return 0;
 
}

2. Judge the number of daffodils

Title Description

Spring is the season of flowers , Narcissus is one of the most charming representatives , There's a narcissus number in math , He defines it like this ：“ Narcissistic number ” A three digit number , The cube sum of its digits is equal to itself , such as ：153=1

.  Now ask for a three digit number , Judge whether the number is daffodil number , If it is , Output “yes”, Otherwise output “no” 

Input

Enter a positive three digit integer .

Output

Output “yes” or “no”.

The sample input  

153

Sample output  

yes

#include <stdio.h>
#include <stdlib.h>
 
int main()
{ int m,a,b,c;
 scanf (&quot;%d&quot;,&m);
 a=m/100;
 b=(m/10)%10;
 c=m%10;
 if(m==a*a*a+b*b*b+c*c*c){
 printf(&quot;yes&quot;);
 }
 else{
 printf(&quot;no&quot;);
 }
 return 0;
}

3.I love Leap year ！

Title Description

  According to a year , Determine if it's a leap year .

Input

  Enter as an integer , Represents a year .

Output

If it's a leap year , Output "Yes", Otherwise output "No". The output is on a single line .

The sample input  

2012

Sample output  

Yes

#include <stdio.h>
#include <stdlib.h>
 
int main()
 
{
 
 int y;
 scanf(&quot;%d&quot;,&y);
 
 if(y%400==0||(y%4==0&&y%100!=0)){
 
 printf(&quot;Yes\n&quot;);}
 else
 {
 printf(&quot;No\n&quot;);
 }
 
 return 0;
}

4. Triangle determination

Title Description

Here are three positive integers , Judge whether using these three integers as the side length can form a triangle .

Input

The input is three int The positive integer of the range , There is a space between .

Output

If you can form a triangle , Output "Yes", Otherwise output "No"

The sample input  

3 4 5

Sample output  

Yes

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int main()
 
{
 int a,b,c;
 scanf(&quot;%d %d %d&quot;,&a,&b,&c);
if(a+b>c&&a+c>b&&b+c>a)
{
 printf(&quot;Yes&quot;);
}else
 printf(&quot;No&quot;);
 
 return 0;
}

5. Judge right triangle

Title Description

Enter three positive integers , Judge whether using these three integers as the side length can form a right triangle .

Input

Enter three positive integers .

Output

Whether it can form a right triangle . If you can output ：yes. If not , Output ：no.

The sample input  

6 8 10

Sample output  

yes

#include <stdio.h>
#include <stdlib.h>
int main()
 
{
 int a,b,c;
 scanf(&quot;%d%d%d&quot;,&a,&b,&c);
if(a*a+b*b==c*c||a*a+c*c==b*b||b*b+c*c==a*a)
{
 printf(&quot;yes&quot;);
}else
 printf(&quot;no&quot;);
 return 0;
}

原网站

版权声明
本文为[InfoQ]所创，转载请带上原文链接，感谢
https://yzsam.com/2022/199/202207151654459440.html