为什么说算法是一个程序和软件的灵魂

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。也许他们能在你的毕业设计或者面试中派上用场。

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 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	

结果输出：

Enter number of terms: 10Fibonacci 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 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	

结果输出：

Enter an integer: 200Fibonacci 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 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: 1232112321 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 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: 2929 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 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。

11 21 2 31 2 3 41 2 3 4 5

源代码：

#include 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 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 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：

#includeint main(){    int rows,i,j,space;    printf("Enter number of rows: ");    scanf("%d",&rows);    for(i=rows;i>=1;--i)    {        for(space=0;space	

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 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;}