S1E16：拾遗 | 课后测试题及答案

yinda_peng

answer

小六御文

1

阿彬

0.10个A
1.0个B
2.?
3.?
4.

12323sd

1

W021024

去

Brick_Porter

#include <stdio.h>
#include <math.h>

void exercise0();
void exercise1();
void exercise2();
void exercise3();

int main()
{
    exercise0();
    exercise1();
    exercise2();
    exercise3();
    return 0;
}

void exercise0()
{
    int i;
    float profit_a, profit_b, principal, rate_a, rate_b;

    i = 0;
    profit_a = profit_b = principal = 10000;
    rate_a = 0.1;
    rate_b = 0.05;
    while (!(profit_b > profit_a))
    {
        profit_a += principal * rate_a;
        profit_b += profit_b * rate_b;
        i++;
    }
    printf("%d年后，黑夜的投资额超过小甲鱼！\n", i);
    printf("小甲鱼的投资额是：%.2f\n", profit_a);
    printf("黑夜的投资额是：%.2f\n", profit_b);
}

void exercise1()
{
    int year = 0, expense = 50, principal = 400;
    double rate = 0.08;
    while (principal > 0)
    {
        principal -= expense;
        principal += principal * rate;
        year++;
    }
    printf("%d年之后，小甲鱼败光了所有的家产，再次回到一贫如洗……\n", year);
}

void exercise2()
{
    int mark, base;
    double pi, item;

    mark = base = 1;
    pi = 0;
    do
    {
        item = mark / (double)base;
        pi += item;
        mark = -mark;
        base += 2;
    } while (fabs(item) > 1e-8);
    printf("pi = %.7lf\n", 4 * pi);
}

int bunny(int i)
{
    if (i < 3)
        return 2;
    return bunny(i - 1) + bunny(i - 2);
}

void exercise3()
{
    printf("%ld\n", bunny(24));
}

18335681882

0. 100
1. 10
2.
3. ++c + ++b
4. x < 0 ? z = -x : z = x;
5.A.  if(size > 12) cost *= 1.05;else bill = cost * flag
B. if(ibex > 14) sheds = 3; else: {sheds - 3; help = 2 * sheds;}
C. do{scanf("%d",&score); count++}while(score < 0);printf("count=%d\n",count)

FGNHBH

。

dogdoggodlike

2333

bianhua_j

1

MuxiAcio

T.T

二拾三

晚风拾夕

答案

Yvonne-

1

裤头哥

qweqweqweqweqwe

liunian233

lz11115

16

杰Sunny

0. 100次
1. 11次
2. a
3. a = 14, b = 3, c = 9
4. z = x > 0 ? x:(-x)

Dengsanshou

P16：拾遗
3.
#include <stdio.h>
#include <math.h>
int main()
{
        int n;//months
        int pair;//the number of rabits
        int i;
        int n1;
        int n2;
        printf("请输入繁殖月数：");
        scanf_s("%d", &n);
        if (n <= 2)
        {
                pair = 1;
                n1 = 1;
                n2 = 1;
        }
        n1 = 1;
        n2 = 1;
        for (i = 3; i <= n; i++)
        {
                pair = n1 + n2;
                n1 = n2;
                n2 = pair;
               
        }
        printf("%d月后，兔子的总对数为：%d\n", n, pair);
        return 0;
}