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10个经典的C语言面试基础算法及代码
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作者：码农网 – 小峰
编辑：strongerHuang
链接：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。
一、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: 10Fibonacci Series: 0+1+1+2+3+5+8+13+21+34+二、回文检查
/* 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;}Enter an integer: 1232112321 is a palindrome.三、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: 2929 is a prime number.四、打印金字塔和三角形
** ** * ** * * ** * * * *#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;}11 21 2 31 2 3 41 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：
* * * * ** * * ** * * * **#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;}五、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: 3.48.43.4 - 8.4 = -5.0六、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;}
/* Function to check prime number */int prime(int n){ 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 + 17七、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 emosewaawesome program八、实现二进制与十进制之间的相互转换
源代码：
/* 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;}
/* Function to convert decimal to binary.*/int decimal_binary(int n){ int rem, i=1, binary=0; while (n!=0) { rem=n%2; n/=2; binary+=rem*i; i*=10; } return binary;}
/* Function to convert binary to decimal.*/int binary_decimal(int n){ 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：
九、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：
十、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：
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