C語(yǔ)言的10大基礎(chǔ)算法

更新時(shí)間：2019年09月04日 13:12:16 作者：翻斗街扛把子胡圖圖

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。這篇文章主要介紹了C語(yǔ)言的10大基礎(chǔ)算法,需要的朋友可以參考下

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)列又稱斐波那契數(shù)列，又稱黃金分割數(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 <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;
}

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 <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;
}

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 <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;
}

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 <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(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 <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(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 <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 * 打印金字塔

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

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;
}

用 * 打印倒金字塔

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

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;
}

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 <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;
}

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 <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;
}

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 <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(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 <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;
}

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 <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;
}

10、矩陣轉(zhuǎn)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;
}

總結(jié)

以上所述是小編給大家介紹的C語(yǔ)言的10大基礎(chǔ)算法,希望對(duì)大家有所幫助，如果大家有任何疑問(wèn)請(qǐng)給我留言，小編會(huì)及時(shí)回復(fù)大家的。在此也非常感謝大家對(duì)腳本之家網(wǎng)站的支持！
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