算法是一个程序和软件的灵魂，作为一名优秀的程序员，只有对一些基础的算法有着全面的掌握，才会在设计程序和编写代码的过程中显得得心应手。本文是近百个C语言算法系列的第二篇，包括了经典的Fibonacci数列、简易计算器、回文检查、质数检查等算法。这些东西在你找工作面试的时候都可能成为你的加分项。

1、计算Fibonacci数列

Fibonacci数列又称斐波那契数列，又称黄金分割数列，指的是这样一个数列：1、1、2、3、5、8、13、21。

C语言实现的代码如下：

/* Displaying Fibonacci sequence up to nth term where n is entered by user. */

#include <stdio.h>

int main()

{

int count, n, t1=0, t2=1, display=0;

printf("Enter number of terms: ");

scanf("%d",&n);

printf("Fibonacci Series: %d+%d+", t1, t2); /* Displaying first two terms */

count=2; /* count=2 because first two terms are already displayed. */

while (count<n)

{

display=t1+t2;

t1=t2;

t2=display;

++count;

printf("%d+",display);

}

return 0;

}

结果输出：

Enter number of terms: 10

Fibonacci Series: 0+1+1+2+3+5+8+13+21+34+

也可以使用下面的源代码：

/* Displaying Fibonacci series up to certain number entered by user. */

#include <stdio.h>

int main()

{

int t1=0, t2=1, display=0, num;

printf("Enter an integer: ");

scanf("%d",&num);

printf("Fibonacci Series: %d+%d+", t1, t2); /* Displaying first two terms */

display=t1+t2;

while(display<num)

{

printf("%d+",display);

t1=t2;

t2=display;

display=t1+t2;

}

return 0;

}

结果输出：

Enter an integer: 200

Fibonacci Series: 0+1+1+2+3+5+8+13+21+34+55+89+144+

2、回文检查

源代码：

/* C program to check whether a number is palindrome or not */

#include <stdio.h>

int main()

{

int n, reverse=0, rem,temp;

printf("Enter an integer: ");

scanf("%d", &n);

temp=n;

while(temp!=0)

{

rem=temp%10;

reverse=reverse*10+rem;

temp/=10;

}

/* Checking if number entered by user and it's reverse number is equal. */

if(reverse==n)

printf("%d is a palindrome.",n);

else

printf("%d is not a palindrome.",n);

return 0;

}

结果输出：

Enter an integer: 12321

12321 is a palindrome.

3、质数检查

注：1既不是质数也不是合数。

源代码：

/* C program to check whether a number is prime or not. */

#include <stdio.h>

int main()

{

int n, i, flag=0;

printf("Enter a positive integer: ");

scanf("%d",&n);

for(i=2;i<=n/2;++i)

{

if(n%i==0)

{

flag=1;

break;

}

}

if (flag==0)

printf("%d is a prime number.",n);

else

printf("%d is not a prime number.",n);

return 0;

}

结果输出：

Enter a positive integer: 29

29 is a prime number.

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

8.4

3.4 - 8.4 = -5.0

6、检查一个数能不能表示成两个质数之和

源代码：

#include <stdio.h>

int prime(int n);

int main()

{

int n, i, flag=0;

printf("Enter a positive integer: ");

scanf("%d",&n);

for(i=2; i<=n/2; ++i)

{

if (prime(i)!=0)

{

if ( prime(n-i)!=0)

{

printf("%d = %d + %d\n", n, i, n-i);

flag=1;

}

}

}

if (flag==0)

printf("%d can't be expressed as sum of two prime numbers.",n);

return 0;

}

int prime(int n) /* Function to check prime number */

{

int i, flag=1;

for(i=2; i<=n/2; ++i)

if(n%i==0)

flag=0;

return flag;

}

结果输出：

Enter a positive integer: 34

34 = 3 + 31

34 = 5 + 29

34 = 11 + 23

34 = 17 + 17

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