C语言的10大基础算法

c语言的10大基础算法

 

 

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。也许它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_s("%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_s("%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_s("%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_s("%d",&n);
for(i=2;i<=n/2;++i)
{
if(n%i==0)
{
flag=1;
break;
}
}
if (flag==0)
printf("%d 是质数.",n);
else
printf("%d 不是质数.",n);
return 0;
}
结果输出：

enter a positive integer: 29
29 是质数.

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_s("%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_s("%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_s("%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_s("%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_s("%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_s("%c",&o);
printf("enter two operands: ");
scanf_s("%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_s("%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: 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：

/* 不使用字符串反转用户输入的句子的示例. */

#include <stdio.h>
void reverse();
int main()
{
printf("enter a sentence: ");
reverse();
return 0;
}
void reverse()
{
char c;
scanf_s("%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根据用户输入的数据将二进制数转换成十进制数或十进制数转换成二进制数的程序源代码. */

#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. 输入字母'd'将二进制转换为十进制.\n");
printf("2. 输入字母'b'将十进制转换为二进制.\n");
scanf_s("%c",&c);
if (c =='d' || c == 'd')
{
printf("输入二进制数: ");
scanf_s("%d", &n);
printf("%d in binary = %d in decimal", n, binary_decimal(n));
}
if (c =='b' || c == 'b')
{
printf("输入一个十进制数字: ");
scanf_s("%d", &n);
printf("%d in decimal = %d in binary", n, decimal_binary(n));
}
return 0;
}

int decimal_binary(int n) /* 函数将十进制转换为二进制。*/
{
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) /* 将二进制转换为十进制的函数.*/
{
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_s("%d",&r);
printf("enter number of columns (between 1 and 100): ");
scanf_s("%d",&c);
printf("\n输入第一个矩阵中的元素:\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_s("%d",&a[i][j]);
}

/* storing elements of second matrix entered by user. */

printf("输入第二个矩阵的元素:\n");
for(i=0;i<r;++i)
for(j=0;j<c;++j)
{
printf("enter element a%d%d: ",i+1,j+1);
scanf_s("%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("\n两个矩阵之和为: \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;
}

输出结果：

 

 

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("输入矩阵的行和列: ");
scanf_s("%d %d", &r, &c);

/* 将用户输入的矩阵元素存储在数组a[]中。*/
printf("\n输入矩阵元素:\n");
for(i=0; i<r; ++i)
for(j=0; j<c; ++j)
{
printf("enter elements a%d%d: ",i+1,j+1);
scanf_s("%d",&a[i][j]);
}
/* 显示矩阵 a[][] */
printf("\n输入矩阵: \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");
}

/* 求矩阵a[][]的转置并存储在数组trans[][]中. */
for(i=0; i<r; ++i)
for(j=0; j<c; ++j)
{
trans[j][i]=a[i][j];
}

/* 显示转置，即显示数组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;
}

输出结果：