10个经典的C语言面试基础算法及代码作者：码农网 – 小峰

原文地址：http://www.codeceo.com/article/10-c-interview-algorithm.html

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：

  1. /* Displaying Fibonacci sequence up to nth term where n is entered by user. */#include <stdio.h>int main()
    {
  2.   int count, n, t1=0, t2=1, display=0;
  3.   printf("Enter number of terms: ");
  4.   scanf("%d",&n);
  5.   printf("Fibonacci Series: %d+%d+", t1, t2); /* Displaying first two terms */
  6.   count=2;    /* count=2 because first two terms are already displayed. */
  7.   while (count<n)
  8.   {
  9.       display=t1+t2;
  10.       t1=t2;
  11.       t2=display;
  12.       ++count;
  13.       printf("%d+",display);
  14.   }
  15.   return 0;
    }

结果输出：

  1. 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：

  1. /* Displaying Fibonacci series up to certain number entered by user. */
  3. #include <stdio.h>
    int main()
    {
  4.   int t1=0, t2=1, display=0, num;
  5.   printf("Enter an integer: ");
  6.   scanf("%d",&num);
  7.   printf("Fibonacci Series: %d+%d+", t1, t2); /* Displaying first two terms */
  8.   display=t1+t2;
  9.   while(display<num)
  10.   {
  11.       printf("%d+",display);
  12.       t1=t2;
  13.       t2=display;
  14.       display=t1+t2;
  15.   }
  16.   return 0;
    }

结果输出：

  1. 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：

  1. /* C program to check whether a number is palindrome or not */
  3. #include <stdio.h>
    int main()
    {
  4.   int n, reverse=0, rem,temp;
  5.   printf("Enter an integer: ");
  6.   scanf("%d", &n);
  7.   temp=n;
  8.   while(temp!=0)
  9.   {
  10.      rem=temp%10;
  11.      reverse=reverse*10+rem;
  12.      temp/=10;
  13.   }  
    /* Checking if number entered by user and it's reverse number is equal. */
  14.   if(reverse==n)
  15.       printf("%d is a palindrome.",n);
  16.   else
  17.       printf("%d is not a palindrome.",n);
  18.   return 0;
    }

结果输出：

  1. 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：

  1. /* C program to check whether a number is prime or not. */#include <stdio.h>int main(){
  2.   int n, i, flag=0;
  3.   printf("Enter a positive integer: ");
  4.   scanf("%d",&n);
  5.   for(i=2;i<=n/2;++i)
  6.   {
  7.       if(n%i==0)
  8.       {
  9.           flag=1;
  10.           break;
  11.       }
  12.   }
  13.   if (flag==0)
  14.       printf("%d is a prime number.",n);
  15.   else
  16.       printf("%d is not a prime number.",n);
  17.   return 0;
    }

结果输出：

  1. 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

  1. *
    * *
    * * *
    * * * *
    * * * * *

源代码：

  1. #include <stdio.h>
    int main()
    {
  2.     int i,j,rows;
  3.     printf("Enter the number of rows: ");
  4.     scanf("%d",&rows);
  5.     for(i=1;i<=rows;++i)
  6.     {
  7.         for(j=1;j<=i;++j)
  8.         {
  9.            printf("* ");
  10.         }
  11.         printf("\n");
  12.     }
  13.     return 0;
    }

如下图所示使用数字打印半金字塔。

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

源代码：

  1. #include <stdio.h>
    int main()
    {
  2.     int i,j,rows;
  3.     printf("Enter the number of rows: ");
  4.     scanf("%d",&rows);
  5.     for(i=1;i<=rows;++i)
  6.     {
  7.         for(j=1;j<=i;++j)
  8.         {
  9.            printf("%d ",j);
  10.         }
  11.         printf("\n");
  12.     }
  13.     return 0;
    }

用 * 打印半金字塔

  1. * * * * *
    * * * *
    * * *
    * *
    *

源代码：

  1. #include <stdio.h>
    int main()
    {
  2.     int i,j,rows;
  3.     printf("Enter the number of rows: ");
  4.     scanf("%d",&rows);
  5.     for(i=rows;i>=1;--i)
  6.     {
  7.         for(j=1;j<=i;++j)
  8.         {
  9.            printf("* ");
  10.         }
  11.     printf("\n");
  12.     }
  13.     return 0;
    }

用 * 打印金字塔

  1.         *
  2.       * * *
  3.     * * * * *
  4.   * * * * * * *
    * * * * * * * * *

源代码：

  1. #include <stdio.h>
    int main()
    {
  2.     int i,space,rows,k=0;
  3.     printf("Enter the number of rows: ");
  4.     scanf("%d",&rows);
  5.     for(i=1;i<=rows;++i)
  6.     {
  7.         for(space=1;space<=rows-i;++space)
  8.         {
  9.            printf("  ");
  10.         }
  11.         while(k!=2*i-1)
  12.         {
  13.            printf("* ");
  14.            ++k;
  15.         }
  16.         k=0;
  17.         printf("\n");
  18.     }
  19.     return 0;
    }

用 * 打印倒金字塔

  1. * * * * * * * * *
  2.   * * * * * * *
  3.     * * * * *
  4.       * * *
  5.         *

源代码：

  1. #include<stdio.h>
    int main()
    {
  2.     int rows,i,j,space;
  3.     printf("Enter number of rows: ");
  4.     scanf("%d",&rows);
  5.     for(i=rows;i>=1;--i)
  6.     {
  7.         for(space=0;space<rows-i;++space)
  8.            printf("  ");
  9.         for(j=i;j<=2*i-1;++j)
  10.           printf("* ");
  11.         for(j=0;j<i-1;++j)
  12.             printf("* ");
  13.         printf("\n");
  14.     }
  15.     return 0;
    }

5、简单的加减乘除计算器

源代码：

  1. /* 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()
    {
  2.     char o;
  3.     float num1,num2;
  4.     printf("Enter operator either + or - or * or divide : ");
  5.     scanf("%c",&o);
  6.     printf("Enter two operands: ");
  7.     scanf("%f%f",&num1,&num2);
  8.     switch(o) {
  9.         case '+':
  10.             printf("%.1f + %.1f = %.1f",num1, num2, num1+num2);
  11.             break;
  12.         case '-':
  13.             printf("%.1f - %.1f = %.1f",num1, num2, num1-num2);
  14.             break;
  15.         case '*':
  16.             printf("%.1f * %.1f = %.1f",num1, num2, num1*num2);
  17.             break;
  18.         case '/':
  19.             printf("%.1f / %.1f = %.1f",num1, num2, num1/num2);
  20.             break;
  21.         default:
  22.             /* If operator is other than +, -, * or /, error message is shown */
  23.             printf("Error! operator is not correct");
  24.             break;
  25.     }
  26.     return 0;
    }

结果输出：

  1. Enter operator either + or - or * or divide : -
    Enter two operands: 3.4
    8.43.4 - 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：

  1. #include <stdio.h>
    int prime(int n);
    int main()
    {
  2.     int n, i, flag=0;
  3.     printf("Enter a positive integer: ");
  4.     scanf("%d",&n);
  5.     for(i=2; i<=n/2; ++i)
  6.     {
  7.         if (prime(i)!=0)
  8.         {
  9.             if ( prime(n-i)!=0)
  10.             {
  11.                 printf("%d = %d + %d\n", n, i, n-i);
  12.                 flag=1;
  13.             }
  15.         }
  16.     }
  17.     if (flag==0)
  18.       printf("%d can't be expressed as sum of two prime numbers.",n);
  19.     return 0;}int prime(int n)      /* Function to check prime number */{
  20.     int i, flag=1;
  21.     for(i=2; i<=n/2; ++i)
  22.        if(n%i==0)
  23.           flag=0;
  24.     return flag;
    }

结果输出：

  1. 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：

  1. /* Example to reverse a sentence entered by user without using strings. */
    #include <stdio.h>
    void Reverse();
    int main()
    {
  2.     printf("Enter a sentence: ");
  3.     Reverse();
  4.     return 0;
    }
    void Reverse()
    {
  5.     char c;
  6.     scanf("%c",&c);
  7.     if( c != '\n')
  8.     {
  9.         Reverse();
  10.         printf("%c",c);
  11.     }
    }

结果输出：

  1. Enter a sentence: margorp emosewa
  2. awesome program

8、实现二进制与十进制之间的相互转换

  1. /* C programming source code to convert either binary to decimal or decimal to binary according to data entered by user. */
  3. #include <stdio.h>
    #include <math.h>
    int binary_decimal(int n);
    int decimal_binary(int n);
    int main()
    {
  4.    int n;
  5.    char c;
  6.    printf("Instructions:\n");
  7.    printf("1. Enter alphabet 'd' to convert binary to decimal.\n");
  8.    printf("2. Enter alphabet 'b' to convert decimal to binary.\n");
  9.    scanf("%c",&c);
  10.    if (c =='d' || c == 'D')
  11.    {
  12.        printf("Enter a binary number: ");
  13.        scanf("%d", &n);
  14.        printf("%d in binary = %d in decimal", n, binary_decimal(n));
  15.    }
  16.    if (c =='b' || c == 'B')
  17.    {
  18.        printf("Enter a decimal number: ");
  19.        scanf("%d", &n);
  20.        printf("%d in decimal = %d in binary", n, decimal_binary(n));
  21.    }
  22.    return 0;
    }
  24. int decimal_binary(int n)  /* Function to convert decimal to binary.*/
    {
  25.     int rem, i=1, binary=0;
  26.     while (n!=0)
  27.     {
  28.         rem=n%2;
  29.         n/=2;
  30.         binary+=rem*i;
  31.         i*=10;
  32.     }
  33.     return binary;
    }
  35. int binary_decimal(int n) /* Function to convert binary to decimal.*/
    {
  36.     int decimal=0, i=0, rem;
  37.     while (n!=0)
  38.     {
  39.         rem = n%10;
  40.         n/=10;
  41.         decimal += rem*pow(2,i);
  42.         ++i;
  43.     }
  44.     return decimal;
    }

结果输出：

aaarticlea/png;base64,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" alt="" data-w="497" data-ratio="0.2454728370221328" data-type="png" data-src="http://mmbiz.qpic.cn/mmbiz/gPKgoIgeYlPWqZXNUicmTuIGYHsFEPzb7glwZpILBx9ia5y9IlSVIXST40bibmdRurllvBcAiaviaUaEhaeNcqGzOvQ/0?wx_fmt=png" data-s="300,640" />

9、使用多维数组实现两个矩阵的相加

源代码：

  1. #include <stdio.h>
    int main()
    {
  2.     int r,c,a[100][100],b[100][100],sum[100][100],i,j;
  3.     printf("Enter number of rows (between 1 and 100): ");
  4.     scanf("%d",&r);
  5.     printf("Enter number of columns (between 1 and 100): ");
  6.     scanf("%d",&c);
  7.     printf("\nEnter elements of 1st matrix:\n");/* Storing elements of first matrix entered by user. */
  9.     for(i=0;i<r;++i)
  10.        for(j=0;j<c;++j)
  11.        {
  12.            printf("Enter element a%d%d: ",i+1,j+1);
  13.            scanf("%d",&a[i][j]);
  14.        }/* Storing elements of second matrix entered by user. */
  16.     printf("Enter elements of 2nd matrix:\n");
  17.     for(i=0;i<r;++i)
  18.        for(j=0;j<c;++j)
  19.        {
  20.            printf("Enter element a%d%d: ",i+1,j+1);
  21.            scanf("%d",&b[i][j]);
  22.        }/*Adding Two matrices */
  24.    for(i=0;i<r;++i)
  25.        for(j=0;j<c;++j)
  26.            sum[i][j]=a[i][j]+b[i][j];/* Displaying the resultant sum matrix. */
  28.     printf("\nSum of two matrix is: \n\n");
  29.     for(i=0;i<r;++i)
  30.        for(j=0;j<c;++j)
  31.        {
  32.            printf("%d   ",sum[i][j]);
  33.            if(j==c-1)
  34.                printf("\n\n");
  35.        }
  37.     return 0;
    }

结果输出：
aaarticlea/png;base64,iVBORw0KGgoAAAANSUhEUgAAAAEAAAABCAIAAACQd1PeAAAAGXRFWHRTb2Z0d2FyZQBBZG9iZSBJbWFnZVJlYWR5ccllPAAAAyBpVFh0WE1MOmNvbS5hZG9iZS54bXAAAAAAADw/eHBhY2tldCBiZWdpbj0i77u/IiBpZD0iVzVNME1wQ2VoaUh6cmVTek5UY3prYzlkIj8+IDx4OnhtcG1ldGEgeG1sbnM6eD0iYWRvYmU6bnM6bWV0YS8iIHg6eG1wdGs9IkFkb2JlIFhNUCBDb3JlIDUuMC1jMDYwIDYxLjEzNDc3NywgMjAxMC8wMi8xMi0xNzozMjowMCAgICAgICAgIj4gPHJkZjpSREYgeG1sbnM6cmRmPSJodHRwOi8vd3d3LnczLm9yZy8xOTk5LzAyLzIyLXJkZi1zeW50YXgtbnMjIj4gPHJkZjpEZXNjcmlwdGlvbiByZGY6YWJvdXQ9IiIgeG1sbnM6eG1wPSJodHRwOi8vbnMuYWRvYmUuY29tL3hhcC8xLjAvIiB4bWxuczp4bXBNTT0iaHR0cDovL25zLmFkb2JlLmNvbS94YXAvMS4wL21tLyIgeG1sbnM6c3RSZWY9Imh0dHA6Ly9ucy5hZG9iZS5jb20veGFwLzEuMC9zVHlwZS9SZXNvdXJjZVJlZiMiIHhtcDpDcmVhdG9yVG9vbD0iQWRvYmUgUGhvdG9zaG9wIENTNSBXaW5kb3dzIiB4bXBNTTpJbnN0YW5jZUlEPSJ4bXAuaWlkOkJDQzA1MTVGNkE2MjExRTRBRjEzODVCM0Q0NEVFMjFBIiB4bXBNTTpEb2N1bWVudElEPSJ4bXAuZGlkOkJDQzA1MTYwNkE2MjExRTRBRjEzODVCM0Q0NEVFMjFBIj4gPHhtcE1NOkRlcml2ZWRGcm9tIHN0UmVmOmluc3RhbmNlSUQ9InhtcC5paWQ6QkNDMDUxNUQ2QTYyMTFFNEFGMTM4NUIzRDQ0RUUyMUEiIHN0UmVmOmRvY3VtZW50SUQ9InhtcC5kaWQ6QkNDMDUxNUU2QTYyMTFFNEFGMTM4NUIzRDQ0RUUyMUEiLz4gPC9yZGY6RGVzY3JpcHRpb24+IDwvcmRmOlJERj4gPC94OnhtcG1ldGE+IDw/eHBhY2tldCBlbmQ9InIiPz6p+a6fAAAAD0lEQVR42mJ89/Y1QIABAAWXAsgVS/hWAAAAAElFTkSuQmCC" alt="" data-w="388" data-ratio="0.7963917525773195" data-type="png" data-src="http://mmbiz.qpic.cn/mmbiz/gPKgoIgeYlPWqZXNUicmTuIGYHsFEPzb7kMV6bOvpia02CwnnG0tqPtraT2s1Gn3NXYcafd600yL8iaQicB4pyPj9g/0?wx_fmt=png" data-s="300,640" />

10、矩阵转置

源代码：

  1. #include <stdio.h>
    int main()
    {
  2.     int a[10][10], trans[10][10], r, c, i, j;
  3.     printf("Enter rows and column of matrix: ");
  4.     scanf("%d %d", &r, &c);/* Storing element of matrix entered by user in array a[][]. */
  5.     printf("\nEnter elements of matrix:\n");
  6.     for(i=0; i<r; ++i)
  7.     for(j=0; j<c; ++j)
  8.     {
  9.         printf("Enter elements a%d%d: ",i+1,j+1);
  10.         scanf("%d",&a[i][j]);
  11.     }/* Displaying the matrix a[][] */
  12.     printf("\nEntered Matrix: \n");
  13.     for(i=0; i<r; ++i)
  14.     for(j=0; j<c; ++j)
  15.     {
  16.         printf("%d  ",a[i][j]);
  17.         if(j==c-1)
  18.             printf("\n\n");
  19.     }/* Finding transpose of matrix a[][] and storing it in array trans[][]. */
  20.     for(i=0; i<r; ++i)
  21.     for(j=0; j<c; ++j)
  22.     {
  23.        trans[j][i]=a[i][j];
  24.     }/* Displaying the transpose,i.e, Displaying array trans[][]. */
  25.     printf("\nTranspose of Matrix:\n");
  26.     for(i=0; i<c; ++i)
  27.     for(j=0; j<r; ++j)
  28.     {
  29.         printf("%d  ",trans[i][j]);
  30.         if(j==r-1)
  31.             printf("\n\n");
  32.     }
  33.     return 0;
    }

结果输出：
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