奶头挺立呻吟高潮av全片,成人试看120秒体验区,性欧美极品v,A片高潮抽搐揉捏奶头视频

試題

C語言面試的10個經典基礎算法及代碼

時間:2024-09-16 07:36:40 試題 我要投稿
  • 相關推薦

C語言面試的10個經典基礎算法及代碼

  算法是一個程序和軟件的靈魂,下面百分網小編為大家搜索整理了關于C語言面試的10個經典基礎算法及代碼,歡迎參考學習,希望對大家有所幫助!想了解更多相關信息請持續(xù)關注我們應屆畢業(yè)生考試網!

C語言面試的10個經典基礎算法及代碼

  1、計算Fibonacci數列

  Fibonacci數列又稱斐波那契數列,又稱黃金分割數列,指的是這樣一個數列:1、1、2、3、5、8、13、21。

  C語言實現的代碼如下:

  /* 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

  {

  display=t1+t2;

  t1=t2;

  t2=display;

  ++count;

  printf("%d+",display);

  }

  return 0;

  }

  結果輸出:

  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碼:

  /* 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

  {

  printf("%d+",display);

  t1=t2;

  t2=display;

  display=t1+t2;

  }

  return 0;

  }

  結果輸出:

  Enter an integer: 200

  Fibonacci Series: 0+1+1+2+3+5+8+13+21+34+55+89+144+

  2、回文檢查

  源代碼:

  /* 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;

  }

  結果輸出:

  Enter an integer: 12321

  12321 is a palindrome.

  3、質數檢查

  注:1既不是質數也不是合數。

  源代碼:

  /* 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;

  }

  結果輸出:

  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

  *

  * *

  * * *

  * * * *

  * * * * *

  源代碼:

  #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("\n");

  }

  return 0;

  }

  如下圖所示使用數字打印半金字塔。

  1

  1 2

  1 2 3

  1 2 3 4

  1 2 3 4 5

  源代碼:

  #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("\n");

  }

  return 0;

  }

  用 * 打印半金字塔

  * * * * *

  * * * *

  * * *

  * *

  *

  源代碼:

  #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("\n");

  }

  return 0;

  }

  用 * 打印金字塔

  *

  * * *

  * * * * *

  * * * * * * *

  * * * * * * * * *

  源代碼:

  #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("\n");

  }

  return 0;

  }

  用 * 打印倒金字塔

  * * * * * * * * *

  * * * * * * *

  * * * * *

  * * *

  *

  源代碼:

  #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

  printf(" ");

  for(j=i;j<=2*i-1;++j)

  printf("* ");

  for(j=0;j

  printf("* ");

  printf("\n");

  }

  return 0;

  }

  5、簡單的加減乘除計算器

  源代碼:

  /* 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;

  }

  結果輸出:

  Enter operator either + or - or * or divide : -

  Enter two operands: 3.4

  8.4

  3.4 - 8.4 = -5.0

  6、檢查一個數能不能表示成兩個質數之和

  源代碼:

  #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", 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;

  }

  結果輸出:

  Enter a positive integer: 34

  34 = 3 + 31

  34 = 5 + 29

  34 = 11 + 23

  34 = 17 + 17

  7、用遞歸的方式顛倒字符串

  源代碼:

  /* 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 != '\n')

  {

  Reverse();

  printf("%c",c);

  }

  }

  結果輸出:

  Enter a sentence: margorp emosewa

  awesome program

  8、實現二進制與十進制之間的相互轉換

  /* 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:\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;

  }

  結果輸出:

  9、使用多維數組實現兩個矩陣的相加

  源代碼:

  #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("\nEnter elements of 1st matrix:\n");

  /* Storing elements of first matrix entered by user. */

  for(i=0;i

  for(j=0;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

  for(j=0;j

  {

  printf("Enter element a%d%d: ",i+1,j+1);

  scanf("%d",&b[i][j]);

  }

  /*Adding Two matrices */

  for(i=0;i

  for(j=0;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

  for(j=0;j

  {

  printf("%d ",sum[i][j]);

  if(j==c-1)

  printf("\n\n");

  }

  return 0;

  }

  結果輸出:

  10、矩陣轉置

  源代碼:

  #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("\nEnter elements of matrix:\n");

  for(i=0; i

  for(j=0; 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

  for(j=0; 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

  for(j=0; j

  {

  trans[j][i]=a[i][j];

  }

  /* Displaying the transpose,i.e, Displaying array trans[][]. */

  printf("\nTranspose of Matrix:\n");

  for(i=0; i

  for(j=0; j

  {

  printf("%d ",trans[i][j]);

  if(j==r-1)

  printf("\n\n");

  }

  return 0;

  }

  結果輸出:

【C語言面試的10個經典基礎算法及代碼】相關文章:

10個經典的C語言面試基礎算法及代碼10-06

C語言快速排序算法及代碼06-25

C語言選擇排序算法及實例代碼07-25

C語言插入排序算法及實例代碼07-02

C語言奇偶排序算法詳解及實例代碼10-30

C語言基礎算法案例(精選)08-21

C語言字符串快速壓縮算法代碼09-05

C語言二分查找(折半查找)算法及代碼08-07

桶排序算法的理解及C語言版代碼示例07-11

C語言精簡代碼10-03

主站蜘蛛池模板: 崇信县| 通州区| 永顺县| 西和县| 区。| 嘉定区| 白玉县| 麻城市| 麻阳| 齐齐哈尔市| 巫山县| 息烽县| 昌黎县| 集安市| 亳州市| 安徽省| 苍山县| 宁阳县| 庐江县| 清丰县| 乐平市| 嘉兴市| 泰顺县| 迁安市| 腾冲县| 德格县| 东台市| 山阳县| 邻水| 郸城县| 福鼎市| 高淳县| 潼关县| 六枝特区| 雷波县| 弋阳县| 嘉荫县| 油尖旺区| 湾仔区| 堆龙德庆县| 岳普湖县|