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实验一
2020-11-06 22:11:00 【树大数媒201王婧】
1.缩写程序,输出以下信息: **********¥¥ ¥¥¥ This is my first C program! **********¥¥¥¥¥
#include<stdlib.h>
int main(void)
{
printf("**********¥¥\n");/*wj*/
printf("¥¥¥\n");
printf("This i my first C program!\n");
printf("**********¥¥¥¥¥\n");
system("pause");
return 0;
}```
2.输入圆柱的半径r 和高h,计算并输出其体积。
```#include<stdio.h>
#include<stdlib.h>
#define PI 3.1415926
int main(void)
{
float r, h;
float v;
printf("请输入圆柱体的半径r:");
scanf("%f", &r);
printf("请输入圆柱体的高h:");
scanf("%f", &h);
v = PI * r * r * h;
printf("圆柱体的体积v为:%f", v);
system("pause");
return 0;
}```
3.输入三个数到变量a,b,c 中,求它们的平均值。
```#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
#include<stdlib.h>
int main(void)
{
int a, b, c;/*number*/
float aver;
printf("请输入a:");
scanf("%d", &a);
printf("请输入b:");
scanf("%d", &b);
printf("请输入c:");
scanf("%d", &c);
aver = (a + b + c) / 3.0;
printf("平均值为:%f", aver);
system("pause");
return 0;
}
4.输入秒数,将它按小时.分钟.秒的形式来输出。例如输入24680,则输出6 小时51
分20 秒.
```#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
#include<stdlib.h>
int main(void)
{
int s;
int hour, minute, second;/*单位 */
printf("请输入秒数:");
scanf("%d", &s);
hour = s/3600;
minute = (s % 3600) / 60;
second = s % 60;
printf("%d秒等于%d小时%d分%d秒", s, hour, minute, second);
system("pause");
return 0;
}```
5.
(1)编写一个计算球体体积的程序,其中球体半径为10m(注意分数的写法)
(2)修改上题中的程序,使用户可以自行输入球体的半径。
```#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
#include<stdlib.h>
#define PI 3.1415926
int main(void)
{
float r;/*半径*/
float v;/*体积*/
printf("请输入球体的半径r:");
scanf("%f", &r);
v = 4.00/ 3.00* PI * r * r * r;
printf("球体的体积v为:%f", v);
system("pause");
return 0;
}```
6.编写一个程序,使用printf 在屏幕上显示下面的图形:
#define _CRT_SECURE_NO_WARNINGS #include<stdio.h> #include<stdlib.h> int main(void) { printf(" *\n"); printf(" *\n"); printf(" *\n"); printf(" * *\n"); printf(" * *\n"); printf(" *\n"); system("pause"); return 0; }``` 7.编写一个程序,要求用户输入一个美元变量,然后显示出增加5%税率后的相应金 额,格式如下所示 Enter an amount: 100.00 With tax added:$105.00
int main(void)
{
float money;/*钱*/
float moneyWithTax;
printf("Enter an amount:");
scanf("%f", &money);
moneyWithTax = money * (1 + 0.05);
printf("With tax added:$%.2f",moneyWithTax);
system("pause");
return 0;
}```
8.(1)编程要求用户输入x 的值,然后显示如下多项式的值:
3x5+2x4-5x3-x2+7x-6
```#include<stdio.h>
#include<stdlib.h>
int main(void)
{
float x, y;/*数*/
printf("请输入x的值:");
scanf("%f", &x);
y = 3 * x * x * x * x * x + 2 * x * x * x * x - 5 * x * x * x - x * x + 7 * x - 6;
printf("多项式3x5+2x4-5x3-x2+7x-6的值为:%f", y);
system("pause");
return 0;
}```
(2)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<stdio.h>
#include<stdlib.h>
int main(void)
{
float x, y;/*数*/
printf("请输入x的值:");
scanf("%f", &x);
y = ((((3 * x + 2) * x - 5) * x - 1) * x + 7) * x - 6;
printf("多项式((((3x+2)x-5)x-1)x+7)x-6的值为:%f", y);
system("pause");
return 0;
}```
9.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,确保在程序中使用整数值,不要使用浮点数)
```#include<stdio.h>
#include<stdlib.h>
int main(void)
{
int money, twenty, ten, five, one;/*12345*/
printf("Enter a dollar amount:");
scanf_s("%d", &money);
twenty = money / 20;
money = money - twenty * 20;
ten = money / 10;
money = money - ten * 10;
five = money / 5;
money = money - 5 * five;
one = money;
printf("$20 bills:%d\n", twenty);
printf("$10 bills:%d\n", ten);
printf("$5 bills:%d\n", five);
printf("$1 bills:%d\n", one);
system("pause");
return 0;
}```
版权声明
本文为[树大数媒201王婧]所创,转载请带上原文链接,感谢
https://my.oschina.net/u/4774086/blog/4706902
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