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            試題

            C語言面試的10個經典基礎算法及代碼

            時間:2025-01-04 22:31:44 試題 我要投稿
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            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關注我們應屆畢業生考試網!

            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:

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