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    C语言的10大基础算法

    作者:翻斗街扛把子胡图图 时间:2021-07-06 17:44

    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 + %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:

    /* 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: \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);
    /* 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;
    }

    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站长博客网站的支持!
    如果你觉得本文对你有帮助,欢迎转载,烦请注明出处,谢谢!

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