demo_3_20

#define _CRT_SECURE_NO_WARNINGS 1
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>

int main()
{
    int n = 0;
    double a, single_num = 0.0,total_sum=0.0;
    //从标准输入获取数字
    scanf("%d %lf", &n, &a);
    for (int i = 0; i < n; i++)
    {
        single_num += a*pow(10, i);
        total_sum += single_num;
    }
    printf("%lf\n", total_sum);
    system("pause");
    return 0;
}

int main()
{
    //求1-20的阶乘
    double total_sum = 0.0;
    //遍历获取[1,20]当中的每一个数字
    for (int i = 1; i <= 20; i++)
    {
        //对每一个数字进行阶乘
        double single_num = 1.0;
        for (int j = i; j > 0; j--)
        {
            single_num *= j;
        }
        //对每个数字阶乘的结果进行求和
        total_sum += single_num;
    }
    printf("%lf\n", total_sum);
    system("pause");
    return 0;
}

int main()
{
    double total_sum = 0.0, sum1 = 0.0, sum2 = 0.0, sum3 = 0.0;
    //[1-100]
    for (int i = 1; i <= 100; i++)
    {
        //累加
        sum1 += i;
        //[1-50]
        if (i <= 50)
        {
            //累加
            sum2 += i*i;
        }
        if (i <= 10)
        {
            sum3 = 1.0 / i;
        }
    }
    total_sum = sum1 + sum2 + sum3;
    printf("%.2lf\n", total_sum);
    system("pause");
    return 0;
}

int main()
{
    int a, b, c;
    //获取[100,999]的数字
    for (int i = 100; i <= 999; i++)
    {
        //获取三位数字中的每一位
        a = i / 100;
        b = i / 10 % 10;
        c = i % 10;
        //进行判断
        if (a*a*a + b*b*b + c*c*c == i)
        {
            printf("%d ", i);
        }
    }
    system("pause");
    return 0;
}

int main()
{
    for (int data = 2; data <= 1000; data++)
    {
        int factor_sum = 0;
        for (int factor = 1; factor <= data / 2; factor++)
        {
            if (data%factor == 0)
            {
                factor_sum += factor;
            }
        }
        //判断
        if (factor_sum == data)
        {
            printf("%d its factors are ", data);
            for (int factor = 1; factor < data /2; factor++)
            {
                if (data%factor == 0)
                {
                    printf("%d,", factor);
                }
            }
            printf("\n");
        }
    }
    system("pause");
    return 0;
}

int main()
{
    //a代表分子，b代表分母，total_sum代表分式之和
    double a = 2.0, b = 1.0, total_sum = 0.0;
    for (int i = 0; i < 20; i++)
    {
        total_sum = a / b;
        //更新分子和分母
        double tmp = a;
        a = a + b;
        b = tmp;
    }
    printf("%lf\n", total_sum);
    system("pause");
    return 0;
}

int main()
{
    //定义高度
    double total_meter = 100.0;
    //定义小球经过的米数
    double ball_total_sum = 0.0;
    for (int i = 0; i < 10; i++)
    {
        //下落+回弹
        ball_total_sum += total_meter;
        //回弹距离等于高度的一半
        total_meter /= 2;
        ball_total_sum += total_meter;
    }
    //第十次回弹的距离
    ball_total_sum -= total_meter;
    printf("小球经过%lf米，第十次回弹的距离为%lf\n", ball_total_sum, total_meter);
    system("pause");
    return 0;
}

int main()
{
    int day = 9;
    int cur_day_count = 1;
    int prev_day_count;
    while (day > 0)
    {
        prev_day_count = (cur_day_count + 1) * 2;
        //更新
        cur_day_count = prev_day_count;
        day--;
    }
    printf("第一天获取桃子的数量%d\n", prev_day_count);
    system("pause");
    return 0;
}

int main()
{
    //从标准输入当中获取a的值
    float a, x0, x1;
    scanf("%f", &a);
    //计算x0和x1的值
    x0 = a / 2;
    x1 = (x0 + a / x0) / 2;
    //判断是否小于10^-5次方
    while (fabs(x0 - x1) >= 1e-5)
    {
        //更新x0和x1的值
        x0 = x1;
        x1 = (x0 + a / x0) / 2;
    }
    printf("[%f]的平方根为[%f]\n", a, x1);
    system("pause");
    return 0;
}

int main()
{
    //x0代表xn，x1代表xn+1，f代表数，f1代表导数
    double x0, x1,f,f1;
    //计算n+1的值
    x1 = 1.5;
    do
    {
        x0 = x1;
        f = ((2 * x0 - 4)*x0 + 3)*x0 - 6;
        f1 = (6 * x0 - 8)*x0 + 3;
        x1 = x0 - f / f1;
    } while (fabs(x1-x0)>=1e-5);
    printf("方程1.5附近的根:%lf\n", x1);
    system("pause");
    return 0;
}


int main()
{
    double left = -10, right = 10, mid;
    double tmp = 10;
    while (fabs(tmp) > 1e-5)
    {
        mid = (left + right) / 2;
        tmp = ((2 * mid - 4)*mid + 3)*mid-6;
        if (tmp > 0)
        {
            right = mid;
        }
        else if (tmp < 0)
        {
            left = mid;
        }
    }
    printf("方程在(-10,10)之间的根为:%lf\n", mid);
    system("pause");
    return 0;
}


int main()
{
    //打印上半部分-4行
    for (int i = 0; i < 4; i++)
    {
        //打印空格
        for (int j = 3 - j; j>0; j--)
        {
            printf(" ");
        }
        //打印星星
        for (int j = 2 * i + 1; j > 0; j--)
        {
            printf("*");
        }
        printf("\n");
    }
    //下半部分-3行
    for (int i = 0; i < 3; i++)
    {
        //打印空格
        for (int j = i + 1; j>0; j--)
        {
            printf(" ");
        }
        //打印星星
        for (int j = 7 - 2 * (i + 1); j > 0; j--)
        {
            printf("*");
        }
        printf("\n");
    }
    system("pause");
    return 0;
}


int main()
{
    //列举A的所有对战对象
    for (int A = 'X'; A <= 'Z'; A++)
    {
        //列举B的所有对战对象
        for (int B = 'X'; A <= 'Z'; B++)
        {
            //列举C的所有对战对象
            for (int C = 'X'; A <= 'Z'; C++)
            {
                if (A != 'X'&&C != 'X'&&C != 'Z'&&A != B&&A != C&&B != C)
                {
                    printf("A对战%c,B对战%c,C对战%c\n", A, B, C);
                }
            }
        }
    }
    
    system("pause");
    return 0;
}

int main()
{
    int a = 0, n;
    for (n = 0; n < 3; n++)
    {
        switch (a)
        {
        case 0:
        case 1:a += 1;
        case 2:a += 2; break;
        case 3:a += 3;
        default:a += 4;
        }
        printf("%d\n", a);
    }
    system("pause");
    return 0;
}

int fun(int (*s)[4], int n, int k)
{
    int m, i;
    m = s[0][k];
    for (i = 0; i < n; i++)
    {
        if (s[i][k]>m) m = s[i][k];
    }
    return m;
}
int main()
{
    int a[4][4] = { { 1, 2, 3, 4 }, { 11, 12, 13, 14 }, { 21, 22, 23, 24 }, { 31, 32, 33, 34 } };
    printf("%d\n", fun(a, 4, 0));
    system("pause");
    return 0;
}


int main()
{
    int s, t, A = 10;
    double B = 6;
    s = sizeof(A);
    t = sizeof(B);
    printf("%d,%d\n", s, t);
    system("pause");
    return 0;
}

double f(double x)
{
    return x*x + 1;
}
int main()
{
    double a = 0;
    int i;
    for (i = 0; i < 30; i += 10)
    {
        a += f((double)i);
        printf("%5.0f\n", a);
    }
    system("pause");
    return 0;
}


int main()
{
    int x;
    scanf("%d", &x);
    if (x <= 3);
    else if (x != 10) printf("%d\n", x);
    system("pause");
    return 0;
}

int main()
{
    char a = 'H';
    a = (a >= 'A'&&a <= 'Z') ? (a - 'A' + 'a') : a;
    printf("%c\n", a);
    system("pause");
    return 0;
}

#define SUB(a) (a)-(a)
int main()
{
    int a = 2, b = 3, c = 5, d;
    d = SUB(a + b)*c;
    printf("%d\n", d);
    system("pause");
    return 0;
}