本篇主要介绍了C语言学习入门精通--C语言必备基础算法（附完整代码）   ，通过具体的内容展现，希望对c语言开发的学习有一定的帮助。

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。

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

7、用递归的方式颠倒字符串

源代码：

/* Example to reverse a sentence entered by user without using strings. */

#include <stdio.h>

void Reverse();

int main()

{

    printf("Enter a sentence: ");

    Reverse();

    return 0;

}

void Reverse()

{

    char c;

    scanf("%c",&c);

    if( c != '\n')

    {

        Reverse();

        printf("%c",c);

    }

}

结果输出：

Enter a sentence: margorp emosewa

awesome program

8、实现二进制与十进制之间的相互转换

源代码：

/* C programming source code to convert either binary to decimal or decimal to binary according to data entered by user. */

#include <stdio.h>

#include <math.h>

int binary_decimal(int n);

int decimal_binary(int n);

int main()

{

   int n;

   char c;

   printf("Instructions:\n");

   printf("1. Enter alphabet 'd' to convert binary to decimal.\n");

   printf("2. Enter alphabet 'b' to convert decimal to binary.\n");

   scanf("%c",&c);

   if (c =='d' || c == 'D')

   {

       printf("Enter a binary number: ");

       scanf("%d", &n);

       printf("%d in binary = %d in decimal", n, binary_decimal(n));

   }

   if (c =='b' || c == 'B')

   {

       printf("Enter a decimal number: ");

       scanf("%d", &n);

       printf("%d in decimal = %d in binary", n, decimal_binary(n));

   }

   return 0;

}

int decimal_binary(int n)  /* Function to convert decimal to binary.*/

{

    int rem, i=1, binary=0;

    while (n!=0)

    {

        rem=n%2;

        n/=2;

        binary+=rem*i;

        i*=10;

    }

    return binary;

}

int binary_decimal(int n) /* Function to convert binary to decimal.*/

{

    int decimal=0, i=0, rem;

    while (n!=0)

    {

        rem = n%10;

        n/=10;

        decimal += rem*pow(2,i);

        ++i;

    }

    return decimal;

}

9、使用多维数组实现两个矩阵的相加

源代码：

#include <stdio.h>

int main(){

    int r,c,a[100][100],b[100][100],sum[100][100],i,j;

    printf("Enter number of rows (between 1 and 100): ");

    scanf("%d",&r);

    printf("Enter number of columns (between 1 and 100): ");

    scanf("%d",&c);

    printf("\nEnter elements of 1st matrix:\n");

/* Storing elements of first matrix entered by user. */

    for(i=0;i<r;++i)

       for(j=0;j<c;++j)

       {

           printf("Enter element a%d%d: ",i+1,j+1);

           scanf("%d",&a[i][j]);

       }

/* Storing elements of second matrix entered by user. */

    printf("Enter elements of 2nd matrix:\n");

    for(i=0;i<r;++i)

       for(j=0;j<c;++j)

       {

           printf("Enter element a%d%d: ",i+1,j+1);

           scanf("%d",&b[i][j]);

       }

/*Adding Two matrices */

   for(i=0;i<r;++i)

       for(j=0;j<c;++j)

           sum[i][j]=a[i][j]+b[i][j];

/* Displaying the resultant sum matrix. */

    printf("\nSum of two matrix is: \n\n");

    for(i=0;i<r;++i)

       for(j=0;j<c;++j)

       {

           printf("%d   ",sum[i][j]);

           if(j==c-1)

               printf("\n\n");

       }

    return 0;

}

10、矩阵转置

源代码：

#include <stdio.h>

int main()

{

    int a[10][10], trans[10][10], r, c, i, j;

    printf("Enter rows and column of matrix: ");

    scanf("%d %d", &r, &c);

/* Storing element of matrix entered by user in array a[][]. */

    printf("\nEnter elements of matrix:\n");

    for(i=0; i<r; ++i)

    for(j=0; j<c; ++j)

    {

        printf("Enter elements a%d%d: ",i+1,j+1);

        scanf("%d",&a[i][j]);

    }

/* Displaying the matrix a[][] */

    printf("\nEntered Matrix: \n");

    for(i=0; i<r; ++i)

    for(j=0; j<c; ++j)

    {

        printf("%d  ",a[i][j]);

        if(j==c-1)

            printf("\n\n");

    }

/* Finding transpose of matrix a[][] and storing it in array trans[][]. */

    for(i=0; i<r; ++i)

    for(j=0; j<c; ++j)

    {

       trans[j][i]=a[i][j];

    }

/* Displaying the transpose,i.e, Displaying array trans[][]. */

    printf("\nTranspose of Matrix:\n");

    for(i=0; i<c; ++i)

    for(j=0; j<r; ++j)

    {

        printf("%d  ",trans[i][j]);

        if(j==r-1)

            printf("\n\n");

    }

    return 0;

}

结果输出：

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