1. Fill in the blanks
1、26
Two, short answer questions
1. Enter 10 scores from the keyboard (between 0 and 100), and after removing the highest and lowest scores, find the average score, and keep 1 decimal place to display the result. (Do not have any prompt text, and the result is a double-precision number without a carriage return)
Method 1: Use an array
/*
时间：2021-3-30
作者：童话
环境：Win 10 、 Visual Studio 2019
*/
double scores[10]; // 储存分数的数组
double max; // 分数最大值的变量
double min; // 分数最小值的变量
double sum; // 分数求和变量
int i; // 循环变量
/* 循环输入分数 */
for ( i = 0; i < 10; i++)
{
scanf("%lf", &scores[i]);
}
/* 初始化最值为数组第一个元素 */
max = min = scores[0];
for ( i = 1; i < 10; i++)
{
if (scores[i] > max)
{
max = scores[i];
}
if (scores[i] < min)
{
min = scores[i];
}
}
/* 初始化和为0 */
sum = 0;
for (i = 0; i < 10; i++)
{
sum += scores[i];
}
/* 累加的和减去最值 */
sum -= (min + max);
/*输出 %.1f 表示输出一位小数*/
printf("%.1f", sum / 8);
Method Two:
/*
时间：2021-3-30
作者：童话
环境：Win 10 、 Visual Studio 2019
*/
double score; // 分数
double max = 0; // 最大值，初始化为分数范围的下限
double min = 100; // 最小值，初始化为分数范围的上限
double sum = 0; // 分数的和，初始化为0
int i; // 循环变量
for (i = 0; i < 10; i++)
{
/* 输入 */
scanf("%lf", &score);
/* 判断最值 */
if (score > max)
{
max = score;
}
if (score < min)
{
min = score;
}
/* 累加求和 */
sum += score;
}
/* 输出 */
printf("%.1f", (sum - max - min) / 8);
3. Procedural questions
1. Please program to find all the daffodil numbers and sum them. The daffodil number is a three-digit number. The cube of the hundreds digit, the cube of the tens digit, and the cube of the single digit must add up to the number itself. For example: 153=1 1 1+5 5 5+3 3 3 is the number of narcissus. (Display all daffodil numbers and their sum in one line from small to large, separate the numbers with a space, do not prompt text and enter)
/*
时间：2021-3-30
作者：童话
环境：Win 10 、 Visual Studio 2019
*/
int i; /* 循环变量 */
int a; /* 个位 */
int b; /* 十位 */
int c; /* 百位 */
int sum = 0; /* 和 */
for (i = 100; i <= 999; i++)
{
a = i % 10;
b = i / 10 % 10;
c = i / 100;
if (i == a * a * a + b * b * b + c * c * c)
{
sum += i;
printf("%d ", i);
}
}
printf("%d", sum);
2、
Given that the minimum salary of a company’s employees is 500, the relationship between the profit (a positive integer) of the project received in a certain month and the profit commission is as follows (measurement unit: yuan)
profit≤1000 no commission
1000<profit≤2000 commission 10%;
2000 ＜profit≤5000 commission 15%;
5000＜profit≤10000 commission 20%;
10000＜profit commission 25%.
It is required to input the project profit of a certain employee for a certain month, and output the actual salary of the employee. (Double-precision real number calculation, the result shows that keep 2 decimal places without carriage return)
Choose one of the two, recommend the second
/*
时间：2021-3-30
作者：童话
环境：Win 10 、 Visual Studio 2019
*/
int profit;
double com = 0;
scanf("%d", &profit);
if (profit <= 1000)
com = 0;
else if (profit > 1000 && profit <= 2000)
com = 0.1;
else if (profit > 2000 && profit <= 5000)
com = 0.15;
else if (profit > 5000 && profit <= 10000)
com = 0.2;
else if (profit > 10000)
com = 0.25;
printf("%.2f", 500 + profit * com);
/*
时间：2021-3-30
作者：童话的女朋友
环境：Win 10 、 学习通
*/
int profit,r;
double x;
scanf("%d",&profit);
r=(profit-1)/1000;
switch(r)
{
case 0:x=500;break;
case 1:x=500+profit*0.1;break;
case 2:
case 3:
case 4:x=500+profit*0.15;break;
case 5:
case 6:
case 7:
case 8:
case 9:x=500+profit*0.2;break;
default:x=500+profit*0.25;break;
}
printf("%.2f",x);
3、
Enter the integer n, calculate the expression
and display the result. (The result is a double-precision number, display 10 digits after the decimal point, do not enter)
/*
时间：2021-3-30
作者：童话
环境：Win 10 、 Visual Studio 2019
*/
int n;
int i, j;
long long t; // int 和 long 不通过
double e = 1;
scanf("%d", &n);
for (i = 1; i <= n; i++)
{
t = 1;
for (j = 1; j <= i; j++)
{
t *= j;
}
e += (1.0 / t);
}
printf("%.10f", e);