S1E16：拾遗 | 课后测试题及答案

boluohmx

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李寻欢2333

有点难搞

我要进大厂！

坚持

摸你穷

0.
100个
1.
11个
2.
c = 5;
b = 5;
a = 5;
3.
b = 4;
c = 8;
a = 14;
4.
z = x ? x :-x;
5.
A.
if(size > 12)
{
        cost = cost * 1.05;
        flag = 2;
}
bill = cost * flag;
B.
if (ibex > 14)
{
      sheds = 3;
}
  sheds = 2;
  help = 2 * sheds;
C.
没做出来
动动手

ssolstice

1187520285

对答案

李姐万岁

0 会打印100次
1 一次都有不会打印
2 abc
3 a=16，c=10 b=6
4
5 if (size > 12)
{
       cost = cost * 1.05;lag = 2;
}
bill = cost * flag;

while（1）
{scanf("%d", &score);
if (score < 0)
{
         printf("count = %d\n", count);
}
count++;
}

exiler

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Kangboab007520

1

CYihan

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yangyangchirou

salaheiyou

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s1e16

鹹魚居士

DD

87_89_70

0. 10
1. 0
2. a, b, c
3. a =14 , b = 5, c = 9
4. z = x>=0 ? x : -x;
5.
A: if (size > 12){
    cost = cost * 1.05;
    flag = 2;
    bill = cost * flag;
}
bill = cost * flag;
B: if (ibex > 14){
    sheds = 3;
    help = 2 * sheds;
}
sheds = 2;
help = 2 * sheds;
C: while (1){
    if (score < 0){
    printf("count = %d\n", count);
    }
    count++;
    scanf("%d", &score);
}
0. #include <stdio.h>

int main(){
    int year;
    double night, turtle;
    night = turtle = 10000;
    for (year=0; night<=turtle; year++){
        turtle += 1000;
        night *= 1.05;
    }
    printf("%d年后,黑夜的投资额超过小甲鱼!\n", year);
    printf("小甲鱼的投资额是:%.2f\n", turtle);
    printf("黑夜的投资额是:%.2f\n", night);
    return 0;
}
1. #include <stdio.h>

int main(){
    int money = 400;
    int year;
    for (year=0; money>=0; year++){
        money -= 50;
        money *= 1.08;
    }
    printf("%d年后,小甲鱼败光了所有的家产,再次回到一贫如洗...", year);
    return 0;
}
2.

damienrouge

s

ksksasd

kk

kangzhuo

11

牛头熊觉得很棒

0.10个
1.0个
2.a,b,c
3.a = 14,b = 3,c = 10
4.z = x<0? -x:x;
5.if (size > 12)
{
       cost = cost * 1.05;
        flag = 2;
}
else
{
        bill = cost * flag;
}
if (ibex > 14)
{
         sheds = 3;
}
else
{sheds = 2;
help = 2 * sheds;
}

if (ibex > 14)
{
        sheds = 3;
}
else
{sheds = 2;   
help = 2 * sheds;
}
readin: scanf("%d", &score);
if (score < 0)
{
       printf("count = %d\n", count);
}
else
{count++;
  readin;
}
1.#include <stdio.h>

int main() {
        double xjy = 10000, hy = 10000;
        int year = 0;
        do {
                xjy = xjy + (10000 * 10 / 100);
                hy = hy + (hy * 5 / 100);
                year++;
        } while (hy < xjy);

        printf("经过%d年，黑夜手头的 Money 才会超过小甲鱼\n", year);
        printf("小甲鱼的投资额是：%.2f\n", xjy);
        printf("黑夜的投资额是：%.2f\n", hy);

        return 0;
}
2.#include <stdio.h>

int main() {
        int money = 400;//万元,原始资产
        int year = 0;//年份

        do {
                money = money + (money * 8 / 100) - 50;
                year++;
                printf("%d  ", money);
        } while (money > 50);

        printf("%d年之后，小甲鱼败光了所有家产，再次回到一贫如洗......\n", year);

        return 0;
}
2.#include <stdio.h>
#include <math.h>

int main() {
        float m = 0, n = 1, pi;
        float x = 1, y;

        do {
                x += 2, y = 1.0 / x, y = -y;
                n = n + y;
                m = n;
                printf("%.12f  %.8f\n", n, m);
        } while (fabs(n) >= 0.00000001);

        pi = m * 4;

        printf("pi≈%.7f", pi);

        return 0;
}
2.#include <stdio.h>
#include <math.h>

int main() {
        double m = 0, n = 1, pi;
        float x = 1, y;
        int i = 1;

        do {
                x += 2, y = (1.0 / x) * pow(-1, i) ;
                i++;
                if (fabs(y) < 0.00000001) {
                        break;
                }
                n = n + y;
                m = n;
                printf("%.12f  %.8f\n", y, m);
        } while (i);

        pi = m * 4;

        printf("pi≈%.7f", pi);

        return 0;
}
3.

8749224

1

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qiang