经常遇到的10大C语言基础算法（珍藏版源码）
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 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+
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: 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 <stdio.h>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 <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: 3.48.43.4 - 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 <stdio.h>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 <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
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：
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：
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：
文章来源：综合网络，如有侵权，联系删除。
赞 (0)