《C语言程序设计:现代方法(第2版)(K.N.King 著)》学习笔记七:C语言基本概念(6)

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编程题


编程题 1
编程题 2
[我的答案]

  1. 程序代码如下:
#include <stdio.h>
int main(void)
{
    printf("       *\n");
    printf("      *\n");
    printf("     *\n");
    printf("*   *\n");
    printf(" * *\n");
    printf("  *\n");

    return 0;
}
  1. 程序代码如下:
#include <stdio.h>
#define PI 3.14f
int main(void)
{
    printf("%f\n", (4.0f / 3.0f) * PI * 10.0f * 10.0f * 10.0f );

    return 0;
}
  1. 程序代码如下:
#include <stdio.h>
#define PI 3.14f
int main(void)
{
    float r;

    scanf("%f", &r);
    printf("%f\n", (4.0f / 3.0f) * PI * r * r * r );

    return 0;
}
  1. 程序代码如下:
#include <stdio.h>
int main(void)
{
    float amount;

    printf("Enter an amount: ");
    scanf("%f", &amount);
    printf("With tax added: %.2f\n", 1.05f * amount);

    return 0;
}
  1. 程序代码如下:
#include <stdio.h>
int main(void)
{
    float x;

    scanf("%f", &x);
    printf("%f", (3 * x * x * x * x * x) + (2 * x * x * x * x) - (5 * x * x * x) - (x * x) + (7 * x) - 6);

    return 0;
}
  1. 程序代码如下:
#include <stdio.h>
int main(void)
{
    float x;

    scanf("%f", &x);
    printf("%f", ((((3 * x + 2) * x - 5) * x - 1) * x + 7) * x - 6);

    return 0;
}
  1. 程序代码如下:
#include <stdio.h>
int main(void)
{
    int twenty, ten, five, one, amount;

    printf("Enter a dollar amount: ");
    scanf("%d", &amount);

    twenty = amount / 20;
    amount = amount - twenty * 20;
    ten = amount / 10;
    amount = amount - ten * 10;
    five = amount / 5;
    amount = amount - five * 5;
    one = amount;

    printf("$20 bills: %d\n", twenty);
    printf("$10 bills: %d\n", ten);
    printf(" $5 bills: %d\n", five);
    printf(" $1 bills: %d\n", one);

    return 0;
}
  1. 程序代码如下:
#include <stdio.h>
int main(void)
{
    float loan, rate, payment;

    printf("Enter amount of loan: ");
    scanf("%f", &loan);
    printf("Enter interest rate: ");
    scanf("%f", &rate);
    printf("Enter monthly payment: ");
    scanf("%f", &payment);

    loan = loan * (1.0f + 0.01f * rate / 12.0f);
    loan = loan - payment;
    printf("\nBalance remaining after first payment: $%.2f\n", loan);
    loan = loan * (1.0f + 0.01f * rate / 12.0f);
    loan = loan - payment;
    printf("Balance remaining after second payment: $%.2f\n", loan);
    loan = loan * (1.0f + 0.01f * rate / 12.0f);
    loan = loan - payment;
    printf("Balance remaining after third payment: $%.2f\n", loan);

    return 0;
}

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