C语言的10大基础算法

2020-01-28 14:18:20王振洲

4、打印金字塔和三角形

 使用 * 建立三角形

*
* *
* * *
* * * *
* * * * *

源代码:


#include <stdio.h>
int main()
{
int i,j,rows;
printf("Enter the number of rows: ");
scanf("%d",&rows);
for(i=1;i<=rows;++i)
{
for(j=1;j<=i;++j)
{
printf("* ");
}
printf("n");
}
return 0;
}

如下图所示使用数字打印半金字塔。

1
1 2
1 2 3
1 2 3 4
1 2 3 4 5

源代码:


#include <stdio.h>
int main()
{
int i,j,rows;
printf("Enter the number of rows: ");
scanf("%d",&rows);
for(i=1;i<=rows;++i)
{
for(j=1;j<=i;++j)
{
printf("%d ",j);
}
printf("n");
}
return 0;
}

用 * 打印半金字塔

* * * * *
* * * *
* * *
* *
*

源代码:


#include <stdio.h>
int main()
{
int i,j,rows;
printf("Enter the number of rows: ");
scanf("%d",&rows);
for(i=rows;i>=1;--i)
{
for(j=1;j<=i;++j)
{
printf("* ");
}
printf("n");
}
return 0;
}

用 * 打印金字塔

*
* * *
* * * * *
* * * * * * *
* * * * * * * * *

源代码:


#include <stdio.h>
int main()
{
int i,space,rows,k=0;
printf("Enter the number of rows: ");
scanf("%d",&rows);
for(i=1;i<=rows;++i)
{
for(space=1;space<=rows-i;++space)
{
printf(" ");
}
while(k!=2*i-1)
{
printf("* ");
++k;
}
k=0;
printf("n");
}
return 0;
}

用 * 打印倒金字塔

* * * * * * * * *
* * * * * * *
* * * * *
* * *
*

源代码:


#include<stdio.h>
int main()
{
int rows,i,j,space;
printf("Enter number of rows: ");
scanf("%d",&rows);
for(i=rows;i>=1;--i)
{
for(space=0;space<rows-i;++space)
printf(" ");
for(j=i;j<=2*i-1;++j)
printf("* ");
for(j=0;j<i-1;++j)
printf("* ");
printf("n");
}
return 0;
}

5、简单的加减乘除计算器

源代码:


/* Source code to create a simple calculator for addition, subtraction, multiplication and division using switch...case statement in C programming. */

# include <stdio.h>
int main()
{
char o;
float num1,num2;
printf("Enter operator either + or - or * or divide : ");
scanf("%c",&o);
printf("Enter two operands: ");
scanf("%f%f",&num1,&num2);
switch(o) {
case '+':
printf("%.1f + %.1f = %.1f",num1, num2, num1+num2);
break;
case '-':
printf("%.1f - %.1f = %.1f",num1, num2, num1-num2);
break;
case '*':
printf("%.1f * %.1f = %.1f",num1, num2, num1*num2);
break;
case '/':
printf("%.1f / %.1f = %.1f",num1, num2, num1/num2);
break;
default:
/* If operator is other than +, -, * or /, error message is shown */
printf("Error! operator is not correct");
break;
}
return 0;
}

结果输出:

Enter operator either + or - or * or divide : -
Enter two operands: 3.4