10个经典的C语言面试基础算法及代码

1、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。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

2、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 (countn) display=t1+t2;

3、 t1=t2; t2=display; +count; printf(%d+,display); return 0;结果输出：Enter number of terms: 10Fibonacci 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(%

4、d,&num); printf(Fibonacci Series: %d+%d+, t1, t2); /* Displaying first two terms */ display=t1+t2; while(displaynum) printf(%d+,display); t1=t2; t2=display; display=t1+t2; return 0;结果输出：Enter an integer: 200Fibonacci 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

5、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 its reverse number is equal. */ if(reverse=n) printf(%d is a palin

6、drome.,n); else printf(%d is not a palindrome.,n); return 0;结果输出：Enter an integer: 1232112321 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) i

7、f(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: 2929 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,&ro

8、ws); 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。11 21 2 31 2 3 41 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(n); return 0;用 * 打印半金字塔* * * * * * * *

9、* * * *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(n); 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:

10、 ); 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：#includeint main() int rows,i,j,space; printf(Enter number of rows: ); scanf(%d,&rows); for(i=ro

11、ws;i=1;-i) for(space=0;spacerows-i;+space) printf( ); for(j=i;j=2*i-1;+j) printf(* ); for(j=0;ji-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. */

12、# 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-num

13、2); 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 eithe

14、r + or - or * or divide : -Enter two operands: 3.48.43.4 - 8.4 = -5.06、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 + %dn, n, i, n-i); flag=1; if (

15、flag=0) printf(%d cant 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: 3434 = 3 + 3134 = 5 + 2934 = 11 + 2334 = 17 + 177、用递归的方式颠倒字符串源代码：/* Exam

16、ple 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); 结果输出：Enter a sentence: margorp emosewaawesome program8、实现二进制与十进制之间的相互转换/* C

17、 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);

18、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 binar

19、y, 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

20、) 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,a100100,b100100,sum100100,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 e

21、lements of 1st matrix:n); /* Storing elements of first matrix entered by user. */ for(i=0;ir;+i) for(j=0;jc;+j) printf(Enter element a%d%d: ,i+1,j+1); scanf(%d,&aij); /* Storing elements of second matrix entered by user. */ printf(Enter elements of 2nd matrix:n); for(i=0;ir;+i) for(j=0;jc;+j) printf(Enter element a%d%d: ,i+1,j+1); scanf(%d,&bij); /*Adding Two matrices */ for(i=0;ir;+i) for(j=0;jc;+j) sumij=aij+bij; /* Displaying the resultant sum matrix. */ printf(nSum of two matrix is: nn); for(i=0;ir;+i) for(j=0;jc;+j) printf(%d ,sumij); if(j=c-1) printf(nn); return 0;结果输出：10、矩阵转置源代码：#i