C语言的10大基础算法

时间:2021-05-22

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:

#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); count=2; while (count<n) {display=t1+t2;t1=t2;t2=display;++count;printf("%d+",display);}return 0;}

结果输出:

Enter number of terms: 10Fibonacci Series: 0+1+1+2+3+5+8+13+21+34+

也可以使用下面的源代码:

#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); 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:

#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;} if(reverse==n) printf("%d is a palindrome.",n);elseprintf("%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:

#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);elseprintf("%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:

# 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 + %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) {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:

#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

#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) {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) {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");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]);}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]);}for(i=0;i<r;++i)for(j=0;j<c;++j)sum[i][j]=a[i][j]+b[i][j];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);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]);}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");}for(i=0; i<r; ++i)for(j=0; j<c; ++j){trans[j][i]=a[i][j];}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;}

总结

以上所述是小编给大家介绍的C语言的10大基础算法,希望对大家有所帮助,如果大家有任何疑问请给我留言,小编会及时回复大家的。在此也非常感谢大家对网站的支持!
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