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10个经典的C语言面试基础算法及代码

时间:2022-12-05 02:15:31 C语言 我要投稿
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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,下面小编为大家整理了10个经典的C语言面试基础算法及代码,希望能帮到大家!

10个经典的C语言面试基础算法及代码

  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

  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

  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

  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

  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

  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。

  1

  1 2

  1 2 3

  1 2 3 4

  1 2 3 4 5

  源代码:

  #include

  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:

  #include

  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:

  #include

  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:

  #include

  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:

  /* Source code to create a simple calculator for addition, subtraction, multiplication and division using switch...case statement in C programming. */

  # include

  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

  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:

  /* Example to reverse a sentence entered by user without using strings. */

  #include

  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

  /* C programming source code to convert either binary to decimal or decimal to binary according to data entered by user. */

  #include

  #include

  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:

  #include

  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:

  #include

  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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