算法是一個程序和軟體的靈魂，作為一名優秀的程式設計師，只有對一些基礎的算法有著全面的掌握，才會在設計程序和編寫代碼的過程中顯得得心應手。本文包括了經典的Fibonacci數列、簡易計算器、回文檢查、質數檢查等算法。
1、計算Fibonacci數列
Fibonacci數列又稱斐波那契數列，又稱黃金分割數列，指的是這樣一個數列：1、1、2、3、5、8、13、21。
C語言實現的代碼如下：
Enter number of terms: 10Fibonacci Series: 0+1+1+2+3+5+8+13+21+34+
也可以使用下面的原始碼：/* 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); display=t1+t2; while(display<num) { printf("%d+",display); t1=t2; t2=display; display=t1+t2; } return 0;}結果輸出：
Enter an integer: 200Fibonacci Series: 0+1+1+2+3+5+8+13+21+34+55+89+144+2、回文檢查
原始碼：
/* 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; } if(reverse==n) printf("%d is a palindrome.",n); else printf("%d is not a palindrome.",n); return 0;}結果輸出：
Enter an integer: 1232112321 is a palindrome.3、質數檢查
註：1既不是質數也不是合數。
原始碼：
/* 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;}結果輸出：
Enter a positive integer: 2929 is a prime number.4、列印金字塔和三角形
使用 * 建立三角形
*
* *
* * *
* * * *
* * * * *
原始碼：
#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;}如下圖所示使用數字列印半金字塔。
11 21 2 31 2 3 41 2 3 4 5原始碼：
#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;}用 * 列印半金字塔
* * * * ** * * ** * * * **原始碼：
#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 * 列印金字塔
* * * * * * * * * * * * * * * ** * * * * * * * *原始碼：
#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 * 列印倒金字塔
* * * * * * * * * * * * * * * * * * * * * * * * *原始碼：
#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、簡單的加減乘除計算器
原始碼：
/* 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: printf("Error! operator is not correct"); break; } return 0;}結果輸出：
Enter operator either + or - or * or divide : -Enter two operands: 3.48.43.4 - 8.4 = -5.06、檢查一個數能不能表示成兩個質數之和
原始碼：
#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) { int i, flag=1; for(i=2; i<=n/2; ++i) if(n%i==0) flag=0; return flag;}結果輸出：
Enter a positive integer: 3434 = 3 + 3134 = 5 + 2934 = 11 + 2334 = 17 + 177、用遞歸的方式顛倒字符串
原始碼：
/* 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); }}結果輸出：
Enter a sentence: margorp emosewaawesome program8、實現二進位與十進位之間的相互轉換
/* 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) { 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) { int decimal=0, i=0, rem; while (n!=0) { rem = n%10; n/=10; decimal += rem*pow(2,i); ++i; } return decimal;}結果輸出：
9、使用多維數組實現兩個矩陣的相加
原始碼：
#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");
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]); }
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]); }
for(i=0;i<r;++i) for(j=0;j<c;++j) sum[i][j]=a[i][j]+b[i][j];
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;}結果輸出：
10、矩陣轉置
原始碼：
#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);
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]); } 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"); }
for(i=0; i<r; ++i) for(j=0; j<c; ++j) { trans[j][i]=a[i][j]; }
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;}結果輸出：