C语言程序设计-现代方法课后编程题作业

2.1编写一个程序，使用printf在屏幕上显示下面的图形

#include 

int main()
{
    printf("          *n");
    printf("         *n");
    printf("        *n");
    printf("       *n");
    printf("*     *n");
    printf(" *   *n");
    printf("   *n");
    
    return 0;
}

2.2编写一个计算机球体体积的程序，其中球体半径为10m，参考公式v=4/3ΠR

#include 

int main()
{
    printf("球体体积：%.2f " , 4.0f/3.0f * 3.14 * 1000);
    return 0;
}

2.3修改上题程序，使用户可以自行录入球衣半径

#include 

int main()
{
    float r ;

    printf("球体半径为：");

    scanf("%f" , &r);

    printf("球体体积为：%.2f" , 4.0f/3.0f * 3.14 * r * r * r);
    
    return 0;
}

2.4编写一个程序，要求用户输入一个美金数量，然后显示出5%税率的相应金额，格式如下Enter an amount : 100. 00 

With tax added: $105.00

​
#include 

int main()
{
    float amount , addeddmoney;

    printf("Enter an amount:");
    scanf("%f", &amount);

    addeddmoney = amount + amount * 0.05;

    printf("With tax added:$%.2fn", addeddmoney );
    
    return 0;
}

​

2.5编程要求用户输入x的值，然后显示如下多项式的值：

3x^5+2x^4-5x^3-x^2+7x-6

#include 

int main()
{
    int x;

    printf("x为");
    scanf("%d", &x);
    printf("值为：%dn", (3*x*x*x*x*x)+(2*x*x*x*x)-(5*x*x*x)-(x*x)+(7*x-6));
    
    return 0;
}

​

​

2.6修改上题，用如下公式对多项式求值：

（（（（3x+2）x-5）x-1）x+7）x-6

#include 

int main()
{
    int x;

    printf("x为");
    scanf("%d", &x);
    printf("值为：%dn", ((((3*x+2)*x-5)*x-1)*x+7)*x-6);

    return 0;
}

2.7编写一个程序，要求用户输入一个美金数量，然后显示出如何用最少的20美元、10美元、5美元和1美元来付款：

Enter a dollar amount: 93
$20 bills: 4
$10 bills: 1
$5 bills: 0
$1 bills: 3
提示：将付款金额除以20，确定20美元的数量，然后从付款金额中减去20美元的总金额。对其他面值的钞票重复这一操作。确保在程序中始终使用整数值，不要用浮点数

#include 

int main()
{
    int money ;

    printf("Enter a dollar amount:");
    scanf("%d" , &money);
    //金额要减去整个数
    printf("$20 bills:%dn", money/20);
    money=money-20*(money/20);

    printf("$10 bills:%dn", money/10);
    money=money-10*(money/10);

    printf("$5 bills:%dn",money/5);
    money=money-5*(money/5);

    printf("$1 bills:%dn",money/1);

    return 0;
}

编程计算第一、第二、第三个月还贷后剩余的贷款金额：
Enter amount of loan: 20000.00
Enter interest rate: 6.0
Enter monthly payment: 386.66

Balance remaining after first payment: $19713.34
Balance remaining after second payment: $19425.25
Balance remaining after third payment: $19135.71
在显示每次还款后的余额时保留两位小数。提示：每个月的贷款余额减去还款金额后，还需要加上贷款余额与月利率的乘积。月利率的计算方法是把用户输入的利率转换成百分数再除以12。

#include 

int main()
{
    float loan , rete , payment ;
    printf("Enter amount of loan:");
    scanf("%f" , &loan);
    
    printf("Enter interest rate:");
    scanf("%f" , &rete);

    printf("Enter monthly payment:");
    scanf("%f" , &payment);

    loan = loan - payment + (loan * (rete / 100 / 12) );
    printf("Balance remaining after first payment:%.2fn" , loan);

    loan = loan - payment + (loan * (rete / 100 / 12) );
    printf("Balance remaining after second payment:%.2fn" , loan);

    loan = loan - payment + (loan * (rete / 100 / 12) );
    printf("Balance remaining after second payment:%.2fn" , loan);

    return 0;
}