10个经典的C语言面试基础算法及代码

jopen 5年前

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 <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

/* 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;  }

结果输出:

10个经典的C语言面试基础算法及代码

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;  }

结果输出:

10个经典的C语言面试基础算法及代码

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:

10个经典的C语言面试基础算法及代码

本文链接:http://www.codeceo.com/article/10-c-interview-algorithm.html
本文作者:码农网 – 小峰