算法是一個程序和軟件的靈魂，作為一名優秀的程序員，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。本文是 近百個C語言算法系列的第二篇，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。也許他們能在你的畢業設計或者面試中派上用 場。

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：

/ 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;
}</pre> 
  
結果輸出：

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); / Displaying first two terms /
  display=t1+t2;
  while(display<num)
  {
      printf("%d+",display);
      t1=t2;
      t2=display;
      display=t1+t2;
  }
  return 0;
}</pre> 
  
結果輸出：

Enter an integer: 200
Fibonacci Series: 0+1+1+2+3+5+8+13+21+34+55+89+144+
 
  
2、回文檢查

源代碼：
/ 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=reverse10+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;
}</pre> 
  
結果輸出：

Enter an integer: 12321
12321 is a palindrome.
 
  
3、質數檢查

注：1既不是質數也不是合數。
源代碼：
/ 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;
}</pre> 
  
結果輸出：

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
*


*







*</pre> 
源代碼：





#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;
}</pre>
如下圖所示使用數字打印半金字塔。
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5

源代碼：
#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("%d ",j);
  }
  printf("\n");

}
  return 0;
}</pre>
用 * 打印半金字塔
    *






</pre>
源代碼：
#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;
}</pre>
用 * 打印金字塔
        *


* * *











*</pre> 
源代碼：













#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;
}</pre>
用 * 打印倒金字塔
        *




* * *
  *</pre> 

源代碼：
#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;
}</pre>
5、簡單的加減乘除計算器
源代碼：
/ 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, num1num2);
            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;
}</pre> 
  
結果輸出：

Enter operator either + or - or * or divide : -
Enter two operands: 3.4
8.4
3.4 - 8.4 = -5.0
 
  
6、檢查一個數能不能表示成兩個質數之和

源代碼：
#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)      / Function to check prime number /
{
    int i, flag=1;
    for(i=2; i<=n/2; ++i)
       if(n%i==0)
          flag=0;
    return flag;
}</pre> 
  
結果輸出：

Enter a positive integer: 34
34 = 3 + 31
34 = 5 + 29
34 = 11 + 23
34 = 17 + 17
 
  
7、用遞歸的方式顛倒字符串

源代碼：
/ 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);
    }
}</pre> 
  
結果輸出：

Enter a sentence: margorp emosewa
awesome program
 
  
8、實現二進制與十進制之間的相互轉換

/ 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)  / Function to convert decimal to binary./
{
    int rem, i=1, binary=0;
    while (n!=0)
    {
        rem=n%2;
        n/=2;
        binary+=remi;
        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;
}</pre> 
  
結果輸出：

9、使用多維數組實現兩個矩陣的相加
源代碼：
#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");

/ 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;

}</pre> 
  
結果輸出：

10、矩陣轉置
源代碼：
#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);

/ 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;
}</pre> 
  
結果輸出：



 

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