经常遇到的10大C语言基础算法（珍藏版源码）

时间：2020-10-15 11:20:33

关键字： C语言嵌入式

[导读]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。

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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 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;}

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("%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;}

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 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 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.

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",&rows);//微信搜索公众号【C语言中文社区】关注回复C语言，免费领取200G学习资料 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

* * * * ** * * ** * * * **

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: "); 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=rows;i>=1;--i) { for(space=0;space printf(" "); for(j=i;j<=2*i-1;++j) printf("* "); for(j=0;j-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 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

6、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;//微信搜索公众号【C语言中文社区】关注回复C语言，免费领取200G学习资料 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: 3434 = 3 + 3134 = 5 + 2934 = 11 + 2334 = 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 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

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 #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;}

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,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;}

结果输出：

10、矩阵转置

源代码：

#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;}

结果输出：

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来源 | 知乎

| 整理文章为传播相关技术，版权归原作者所有 |

| 如有侵权，请联系删除 |

   
    
           
       
        
         
          
           
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