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: 10
Fibonacci 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: 200
Fibonacci 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: 12321
12321 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: 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("* ");
 }
 printf("n");
 }
 return 0;
}

如下图所示使用数字打印半金字塔。

1
1 2
1 2 3
1 2 3 4
1 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: 3.4
8.4
3.4 - 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;
 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 + %dn", 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: 34
34 = 3 + 31
34 = 5 + 29
34 = 11 + 23
34 = 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 emosewa
awesome 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: nn");
 for(i=0;i<r;++i)
 for(j=0;j<c;++j)
 {
 printf("%d ",sum[i][j]);
 if(j==c-1)
 printf("nn");
 }
 return 0;
}

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("nn");
 }
/* 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("nn");
 }
 return 0;
}