• 10个经典的C语言面试基础算法及代码

    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：

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    /* 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：

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    /* 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：

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    /* 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：

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    /* 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：

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    #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(" ");
        }
        return 0;
    }

    　　如下图所示使用数字打印半金字塔。

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

    　　源代码：

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    #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(" ");
        }
        return 0;
    }

    　　用 * 打印半金字塔

    * * * * *
* * * *
* * * 
* *
*

    　　源代码：

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    #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(" ");
        }
        return 0;
    }

    　　用 * 打印金字塔

            *
      * * *
    * * * * *
  * * * * * * *
* * * * * * * * *

    　　源代码：

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    #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(" ");
        }
        return 0;
    }

    　　用 * 打印倒金字塔

    * * * * * * * * *
  * * * * * * *
    * * * * *
      * * *
        *

    　　源代码：

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    #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(" ");
        }
        return 0;
    }

    　5、简单的加减乘除计算器

    　　源代码：

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    /* 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：

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    #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, 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：

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    /* 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 != ' ')
        {
            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

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    /* 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: ");
       printf("1. Enter alphabet 'd' to convert binary to decimal. ");
       printf("2. Enter alphabet 'b' to convert decimal to binary. ");
       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：

    　9、使用多维数组实现两个矩阵的相加

    　　源代码：

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    #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(" Enter elements of 1st matrix: ");
      
    /* 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: ");
        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(" Sum of two matrix is: ");
        for(i=0;i<r;++i)
           for(j=0;j<c;++j)
           {
               printf("%d   ",sum[i][j]);
               if(j==c-1)
                   printf(" ");
           }
      
        return 0;
    }

    　　结果输出：

    　10、矩阵转置

    　　源代码：

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    #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(" Enter elements of matrix: ");
        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(" Entered Matrix: ");
        for(i=0; i<r; ++i)
        for(j=0; j<c; ++j)
        {
            printf("%d  ",a[i][j]);
            if(j==c-1)
                printf(" ");
        }
      
    /* 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(" Transpose of Matrix: ");
        for(i=0; i<c; ++i)
        for(j=0; j<r; ++j)
        {
            printf("%d  ",trans[i][j]);
            if(j==r-1)
                printf(" ");
        }
        return 0;
    }

    　　结果输出：
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  • 原文地址：https://www.cnblogs.com/qxzy/p/4234825.html
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