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10个经典的C语言面试基础算法及代码

时间:2015-02-18来源: 作者:点击:
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。

//本文来自安云网

 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。 //本文来自安云网

  C语言实现的代码如下:

//内容来自AnYun.ORG

/* 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;
} //本文来自安云网 

  结果输出: //内容来自安云网

Enter number of terms: 10
Fibonacci Series: 0+1+1+2+3+5+8+13+21+34+ //内容来自AnYun.ORG 

  也可以使用下面的源代码: //内容来自AnYun.ORG

/* 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;
} //内容来自安云网 

  结果输出:

//内容来自安云网

Enter an integer: 200
Fibonacci Series: 0+1+1+2+3+5+8+13+21+34+55+89+144+ //内容来自AnYun.ORG 

 2、回文检查

  源代码:

//内容来自AnYun.ORG

/* 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;
} //内容来自安云网 

  结果输出: //内容来自安云网

Enter an integer: 12321
12321 is a palindrome. 
//内容来自AnYun.ORG

 3、质数检查

  注:1既不是质数也不是合数。

//本文来自安云网

  源代码:

//内容来自安云网

/* 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;
} 

//内容来自AnYun.ORG

  结果输出: //本文来自安云网

Enter a positive integer: 29
29 is a prime number. 
//本文来自安云网

 4、打印金字塔和三角形

  使用 * 建立三角形 //本文来自安云网

*
* *
* * *
* * * *
* * * * * //本文来自安云网 

  源代码: //内容来自安云网

#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;
} //内容来自AnYun.ORG 

  如下图所示使用数字打印半金字塔。

//内容来自安云网

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

//本文来自安云网

  源代码:

//本文来自安云网

#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;
} //内容来自AnYun.ORG 

  用 * 打印半金字塔

//本文来自安云网

* * * * *
* * * *
* * * 
* *
* 
//本文来自安云网

  源代码: //内容来自AnYun.ORG

#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;
} //内容来自AnYun.ORG 

  用 * 打印金字塔 //内容来自AnYun.ORG

        *
      * * *
    * * * * *
  * * * * * * *
* * * * * * * * * //本文来自安云网 

  源代码: //内容来自安云网

#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;
} //本文来自安云网 

  用 * 打印倒金字塔

//本文来自安云网

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

//内容来自安云网

  源代码:

//内容来自安云网

#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;
} //本文来自安云网 

 5、简单的加减乘除计算器

  源代码:

//本文来自安云网

/* 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;
} //内容来自安云网 

  结果输出:

//本文来自安云网

Enter operator either + or - or * or divide : -
Enter two operands: 3.4
8.4
3.4 - 8.4 = -5.0 //内容来自AnYun.ORG 

 6、检查一个数能不能表示成两个质数之和

  源代码: //内容来自AnYun.ORG

#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;
} //内容来自安云网 

  结果输出: //内容来自AnYun.ORG

Enter a positive integer: 34
34 = 3 + 31
34 = 5 + 29
34 = 11 + 23
34 = 17 + 17 //内容来自AnYun.ORG 

 7、用递归的方式颠倒字符串

  源代码: //内容来自AnYun.ORG

/* 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);
    }
} //内容来自安云网 

  结果输出: //本文来自安云网

Enter a sentence: margorp emosewa
awesome program //内容来自AnYun.ORG 

 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;
} 

//内容来自AnYun.ORG

  结果输出: //内容来自AnYun.ORG

10个经典的C语言面试基础算法及代码 //内容来自AnYun.ORG

 9、使用多维数组实现两个矩阵的相加

  源代码:

//本文来自安云网

#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;
} 
//内容来自AnYun.ORG

  结果输出:

//内容来自安云网

10个经典的C语言面试基础算法及代码

//本文来自安云网

 10、矩阵转置

  源代码: //本文来自安云网

#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;
} 

//本文来自安云网

  结果输出: //内容来自安云网

10个经典的C语言面试基础算法及代码 //本文来自安云网

  via: codeceo

//内容来自AnYun.ORG

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