本文共 3859 字，大约阅读时间需要 12 分钟。

1、斐波拉契数列

#include 
   
    using namespace std;int main(){    int n1 = 1, n2 = 1, display;    int count;    cin>>count;    cout<
    
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2、回文数判断

#include 
   
    using namespace std;int main(){    int num, rem, reserve = 0, tmp;    cin>>num;    tmp = num;    while(tmp != 0) {        reserve = reserve*10 + (rem = tmp%10);        tmp = tmp/10;    }    if(num == reserve) {        cout<<"yes"<
   

3、判读质数

#include 
   
    using namespace std;int main(){    /*    2\3\5\7\11    */    int num;    bool flag = false;    cin>>num;    if(num == 1) cout<<"1既不是质数也不是合数!"<
   

4、打印三角形

#include
   
    /* * 三角型 */int main(){    int rows,i,j,space;    printf("Enter number of rows: ");    scanf("%d",&rows);    //i = rows    for(i=rows;i>=1;--i){            //rows - i        for(space=0;space
   

5、简单计算器

# include 
   
    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;}
   

6、判断一个数能不能表示成两哥质数的和

#include 
   
    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;}
   

7、递归的方式颠倒字符串

#include 
   
    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);    }}
   

8、十进制 与 二进制的转换

#include 
   
    #include 
    
     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;        //进位        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;}
    
   

9、数组转置

#include 
   
    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
   

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