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经典C语言面试基础算法
浏览数:27 

来源: 码农网
原文: 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:
/* 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:

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、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: