10个经典的C语言面试基础算法及代码作者:码农网 – 小峰
原文地址: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。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 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: 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 <stdio.h>
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:
2、回文检查Enter an integer: 200Fibonacci Series: 0+1+1+2+3+5+8+13+21+34+55+89+144+
源代码:
/* 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;
}
结果输出:
3、质数检查Enter an integer: 1232112321 is a palindrome.
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:
4、打印金字塔和三角形Enter a positive integer: 29
29 is a prime number.
使用 * 建立三角形
*
* *
* * *
* * * *
* * * * *
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("* "); } 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 <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(" "); } return 0;
}
用 * 打印半金字塔
* * * * *
* * * *
* * *
* *
*
源代码:
#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(" "); } 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(" "); } return 0;
}
用 * 打印倒金字塔
* * * * * * * * * * * * * * * * * * * * * * * * *
源代码:
5、简单的加减乘除计算器#include<stdio.h>
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;
}
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:
6、检查一个数能不能表示成两个质数之和Enter operator either + or - or * or divide : -
Enter two operands: 3.4
8.43.4 - 8.4 = -5.0
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, 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:
7、用递归的方式颠倒字符串Enter a positive integer: 34
34 = 3 + 31
34 = 5 + 29
34 = 11 + 23
34 = 17 + 17
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 != ' ') { Reverse(); printf("%c",c); }
}
结果输出:
8、实现二进制与十进制之间的相互转换Enter a sentence: margorp emosewa awesome program
/* 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: "); 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%