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：
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 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：
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 <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: 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("\n");
}
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("\n");
}
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("\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;
}
用 * 打印倒金字塔
* * * * * * * * *
* * * * * * *
* * * * *
* * *
*
源代码：
#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("\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. */
# 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: 3.4
8.43.4 - 8.4 = -5.0
6、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;}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: 34
34 = 3 + 31
34 = 5 + 29
34 = 11 + 23
34 = 17 + 17
7、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 emosewa
awesome program
8、实现二进制与十进制之间的相互转换
/* 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;
} 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;
}
结果输出：
aaarticlea/png;base64,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" alt="" data-w="497" data-ratio="0.2454728370221328" data-type="png" data-src="http://mmbiz.qpic.cn/mmbiz/gPKgoIgeYlPWqZXNUicmTuIGYHsFEPzb7glwZpILBx9ia5y9IlSVIXST40bibmdRurllvBcAiaviaUaEhaeNcqGzOvQ/0?wx_fmt=png" data-s="300,640" />
9、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：
aaarticlea/png;base64,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" alt="" data-w="388" data-ratio="0.7963917525773195" data-type="png" data-src="http://mmbiz.qpic.cn/mmbiz/gPKgoIgeYlPWqZXNUicmTuIGYHsFEPzb7kMV6bOvpia02CwnnG0tqPtraT2s1Gn3NXYcafd600yL8iaQicB4pyPj9g/0?wx_fmt=png" data-s="300,640" />
10、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：
aaarticlea/png;base64,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" alt="" data-w="404" data-ratio="0.9381188118811881" data-type="png" data-src="http://mmbiz.qpic.cn/mmbiz/gPKgoIgeYlPWqZXNUicmTuIGYHsFEPzb7xLmLTEMP5JJMQd6pkAN8nOib9Xkz9NlLaJnWMlKtt200QECjkOkk5tQ/0?wx_fmt=png" />