本系列仅学习交流使用。
第一题
示例代码：
#include<stdio.h>
int main(void) {
printf_s(" *\n");
printf_s(" *\n");
printf_s(" *\n");
printf_s(" * *\n");
printf_s(" * *\n");
printf_s(" *\n");
return 0;
}注意点：换行符“\n”
第二、三题
首先看一下4/3在代码运行后输出值：
#include<stdio.h>
int main(void)
{
printf("4 / 3 = %d\n", 4 / 3);
printf("4.0f / 3.0f = %f", 4.0f / 3.0f);
return 0;
}输出为1.
分析：计算时强制做了一个取整的操作，实际上为浮点数。
示例代码：
#include<stdio.h>
#define pi 3.14
#define r 10.0
int main(void)
{
float vol1 = 4.0f / 3.0f * pi * r * r * r;
printf("第二题计算得到体积为：%f\n", vol1);
float a = 1.0;
printf("输入半径：");
scanf_s("%f", &a);
float vol = 4.0f / 3.0f * pi * a * a * a;
printf("第三题计算得到体积为：%f", vol);
return 0;
}终端输入，得到输出
第四题
示例代码：
#include<stdio.h>
int main()
{
printf("Enter an amount: ");
float a = 1.0;
scanf_s("%f", &a);
float tax_a = a * (1 + 0.05);
printf("With tax added: $%.2f", tax_a);
return 0;
}输出
第五题
示例代码：
#include<stdio.h>
int main()
{
float x = 1.0;
printf("请赋予x初始值：");
scanf_s("%f", &x);
float result = 3 * x * x * x * x * x + 2 * x * x * x * x + 7 * x - 5 * x * x * x - x * x - 6;
printf("%.3f", result);
return 0;
}输入输出：
第六题
示例代码：
#include<stdio.h>
int main()
{
float x = 1.0;
printf("请赋予x初始值：");
scanf_s("%f", &x);
float result = ((((3*x+2)*x-5)*x-1)*x+7)*x-6;
printf("%.3f", result);
return 0;
}输出：
第七题
题目分析：这个题的意思就是，每次都付20，一直到付20付多了的那次，选用10元；然后付10，10最多也就一次，结合题目中的解释应该不难理解。
示例代码：
#include<stdio.h>
int main()
{
printf("Enter a dollar amount: ");
int amount = 0, twenty_amount = 0, five_amount = 0, one_amount = 0, ten_amount = 0;
scanf_s("%d", &amount);
twenty_amount = amount / 20;
ten_amount = (amount - twenty_amount * 20)/10;
five_amount = (amount - twenty_amount * 20 - ten_amount * 10) / 5;
one_amount = amount - twenty_amount * 20 - ten_amount * 10 - five_amount * 5;
printf("\n");
printf("$20 bills: %d\n", twenty_amount);
printf("$10 bills: %d\n", ten_amount);
printf(" $5 bills: %d\n", five_amount);
printf(" $1 bills: %d\n", one_amount);
return 0;
}输出：
分析：巧妙利用"/"在进行整数间的除法运算时是向下取整的。
第八题
示例代码：
#include<stdio.h>
int main()
{
printf("Enter amount of loan: ");
float loan = 0.0f;
scanf_s("%f", &loan);
printf("Enter interest rate: ");
float rate = 0.0f;
scanf_s("%f", &rate);
printf("Enter monthly payment: ");
float payment = 0.0f;
scanf_s("%f", &payment);
float first_month, second_month, third_month;
printf("\n");
//first_month = (loan - payment) * (1 + ((rate / 100.0f) / 12.0f));
first_month = (loan - payment) + loan * ((rate / 100.0f) / 12.0f);
second_month = (first_month - payment) + first_month * ((rate / 100.0f) / 12.0f);
third_month = (second_month - payment) + second_month * ((rate / 100.0f) / 12.0f);
printf("Balance remaining after first payment: %.2f\n", first_month);
printf("Balance remaining after second payment: %.2f\n", second_month);
printf("Balance remaining after third payment: %.2f\n", third_month);
return 0;
}输出：
分析，这个代码实现上不难，但是我在第一次审题的时候没写对，错误请看代码注释部分处，输出结果不对反过来再去看题，原来是文字游戏。。。。
在代码实现第八题的过程中，发现有一部分可以抽象出来，单独写一个函数的，这样可以使代码看起来没那么臃肿，下面给出我的另一个代码实现：
#include<stdio.h>
float money(float, float, float);
int main()
{
float loan = 0.0f;
float payment = 0.0f;
float rate = 0.0f;
printf("Enter amount of loan: ");
scanf_s("%f", &loan);
printf("Enter interest rate: ");
scanf_s("%f", &rate);
printf("Enter monthly payment: ");
scanf_s("%f", &payment);
float first_month, second_month, third_month;
printf("\n");
first_month = money(loan, rate, payment);
second_month = money(first_month, rate, payment);
third_month = money(second_month, rate, payment);
printf("Balance remaining after first payment: %.2f\n", first_month);
printf("Balance remaining after second payment: %.2f\n", second_month);
printf("Balance remaining after third payment: %.2f\n", third_month);
return 0;
}
float money(float balance, float rate, float payment)
{
float num_balance = (balance - payment) + balance * ((rate / 100.0f) / 12.0f);
return num_balance;
}
输出：
重复操作的东西都可以进行抽象，以一种简单的形式去实现，上述代码可以继续精简，这里就不给出啦，后续可以自己试一下。