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：
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 | /* 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：
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 | /* 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：
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 | /* 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：
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 | /* 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：
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 | #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。
11 21 2 31 2 3 41 2 3 4 5
源代码：
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 | #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：
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 | #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：
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 | #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：
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 | #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：
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 | /* 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：
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 | #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: 3434 = 3 + 3134 = 5 + 2934 = 11 + 2334 = 17 + 17
7、用递归的方式颠倒字符串
源代码：
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 | /* 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
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 | /* 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：
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 | #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：
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 | #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：
