算法是一個(gè)程序和軟件的靈魂，作為一名優(yōu)秀的程序員，只有對(duì)一些基礎(chǔ)的算法有著全面的掌握，才會(huì)在設(shè)計(jì)程序和編寫(xiě)代碼的過(guò)程中顯得得心應(yīng)手。本文是近百個(gè)C語(yǔ)言算法系列的第二篇，包括了經(jīng)典的Fibonacci數(shù)列、簡(jiǎn)易計(jì)算器、回文檢查、質(zhì)數(shù)檢查等算法。也許他們能在你的畢業(yè)設(shè)計(jì)或者面試中派上用場(chǎng)。

1、計(jì)算Fibonacci數(shù)列

Fibonacci數(shù)列又稱(chēng)斐波那契數(shù)列，又稱(chēng)黃金分割數(shù)列，指的是這樣一個(gè)數(shù)列：1、1、2、3、5、8、13、21。

C語(yǔ)言實(shí)現(xiàn)的代碼如下：

/* 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
	

結(jié)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 
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
	

結(jié)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 
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;
}

結(jié)果輸出：

Enter an integer: 12321
12321 is a palindrome.

3、質(zhì)數(shù)檢查

注：1既不是質(zhì)數(shù)也不是合數(shù)。

源代碼：

/* 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;
}

結(jié)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

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

源代碼：

#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("* ");
        }
        printf("\n");
    }
    return 0;
}

如下圖所示使用數(shù)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 
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

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

源代碼：

#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

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

源代碼：

#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

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

源代碼：

#include
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
	

5、簡(jiǎn)單的加減乘除計(jì)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 
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;
}

結(jié)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(gè)數(shù)能不能表示成兩個(gè)質(zhì)數(shù)之和

源代碼：

#include 
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;
}

結(jié)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 
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(jié)果輸出：

Enter a sentence: margorp emosewa
awesome program

8、實(shí)現(xiàn)二進(jìn)制與十進(jìn)制之間的相互轉(zhuǎn)換

/* 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;
}

結(jié)果輸出：

9、使用多維數(shù)組實(shí)現(xiàn)兩個(gè)矩陣的相加

源代碼：

#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
	

結(jié)果輸出：

10、矩陣轉(zhuǎn)置

源代碼：

#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
	

結(jié)果輸出：