c语言 循环 函数相关练习

 

#include <stdio.h>

//1.求任意10个整数的最大数
int get_max_1(int arr[])
{
    int i = 0;
    int max = arr[0];
    for(i=1; i<10; i++)
    {
        if(max<arr[i])
        {
            max=arr[i];
        }
    }
    return max;
}

int get_max_2(int* arr)
{
    int i = 0;
    int max = *arr;
    for(i=1; i<10; i++)
    {
        if(max < *(arr+i))
        {
            max = *(arr+i);
        }
    }
    return max;
}

//2.求1/1-1/2+1/3-1/4+1/5...+1/99-1/100的值
double caculation(int n)
{
    double sum = 0.0;
    int i = 0;
    int flag = 1;
    for(i=1; i<=n; i++)
    {
        sum += flag*(1.0/i);
        flag = -flag;
    }
    return sum;
}

//3.实现九九乘法表
void multiplication_table()
{
    int i = 0;
    int j = 0;
    for(i=1;i<=9;i++) // line
    {
        for(j=1;j<=i;j++)
        {
            printf("%d*%d=%-2d ",i,j,i*j);
        }
        printf("\n");
    }
}


//4.编写程序数一下1-100的所有数中，出现多少个数字9？
int count_9()
{
    int counter = 0;
    int i = 0;
    for(i=1;i<=100;i++)
    {
        if(i%10 == 9)
        {
            counter++;
        }
        else if (i/10==9)
        {
            counter++;
        }
    }
    return counter;

}

//5. revert the adcdef string: abcdef-->fedcba
int my_strlen(char* arr)
{
    int count = 0;
    while(*arr != '\0')
    {
        count++;
        arr++;
    }
    return count;
}

//method-1
void revert_string_1(char* arr)
{
    int left = 0;
    int right = my_strlen(arr) - 1;
    while(left<right)
    {
        char temp = *(arr+left);
        *(arr+left) = *(arr+right);
        *(arr+right) = temp;
        left++;
        right--;
    }
}

//method-2: 递归实现
void revert_string_2(char * arr)
{
    char temp = *arr;
    int right = my_strlen(arr)-1;
    *arr = *(arr+right);
    *(arr+right) = '\0';
    if(my_strlen(arr)>1)
    {
        revert_string_2(arr+1);
    }
    *(arr+right) = temp;
}

//6.编写一个函数实现n的k次方，用递归实现
// k=0, n**k = 1
// k>0, n**k = n*Pow(n, k-1)
// k<0, n**k = 1.0/n*Pow(n, k-1)
double Pow(int n, int k)
{
    if(k==0)
    {
        return 1.0;
    }
    else if(k>0)
    {
        return 1.0*n*Pow(n, k-1);
    }
    else if (k<0)
    {
        return 1.0/Pow(n, -k);
    }
}

// 7.给定一个非负整数，求该整数的各位之和；
int my_sum(int num)
{
    if(num>9)
    {
        return my_sum(num/10)+ my_sum(num%10);
    }
    else 
    {
        return num;
    }
}



void main()
{
    //1.
    int my_arr1[10]={-1,2,4,5,19,45,34,64,100,99};
    int max_val_1 = get_max_1(my_arr1);
    int max_val_2 = get_max_2(my_arr1);
    printf("The max value is %d\n", max_val_1);
    printf("The max value is %d\n", max_val_2);

    //2.
    double sum = caculation(100);
    printf("1/1-1/2+1/3...-1/100=%lf\n", sum);

    //3.
    multiplication_table();

    //4.
    int count = count_9();
    printf("\ncounter of 9 in 1-100 is %d\n", count);

    //5.
    char arr[] = "abcdef";
    revert_string_2(arr);
    printf("%s\n", arr);

    //6.
    int n = 0;
    int k = 0;
    scanf("%d%d", &n, &k);
    double sum_pow = Pow(n, k);
    printf("%d 的%d次方:%lf\n", n, k, sum_pow);

    //7.
    int num = 123456;
    int num_sum = my_sum(num);
    printf("The sum of num is %d\n", num_sum);

}