0
ejwjnd
0. 10个
1. 9个
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;
}
else
{
bill = cost * flag;
}
B. if (ibex > 14)
{
sheds = 3;
}
else
{
sheds = 2;
}
help = 2 * sheds;
C.
while(count > 0)
{
readin: scanf("%d", &score);
}
printf("count = %d\n", count);
动动手看图片
{:5_108:}
{:10_266:}{:10_266:}{:10_266:}{:10_266:}{:10_266:}{:10_266:}
1
动手题:兔子那道题,小甲鱼代码输出的应该是可以繁殖的兔子吧,
但题里说的是“两年之后可以繁殖多少对兔子”,我觉得就是一共有多少对兔子,那就应该包括可以繁殖和不可以繁殖的兔子
上代码#include <stdio.h>
int main() {
int month = 0;
int couple = 1;
int baby_0 = 0;
int baby_1 = 0;
while (month < 24) {
couple += baby_1;
baby_1 = baby_0;
baby_0 = couple;
month++;
}
int sum;
sum = couple + baby_0 + baby_1;
printf("%d\n", sum);
来对答案了
1
000
90
11
13
#include<stdio.h>
int main() {
int x, h;
int i=0;//用来计算年份
x = h = 10000;
while (x >= h) {
i++;
x = 10000 * 0.1 + x;
h = h * (1 + 0.05);
}
printf("%d年后黑夜的投资额度超越小甲鱼!\n", i);
printf("小甲鱼的投资额为:%d\n", x);
printf("黑夜的投资额为:%d\n", h);
return 0;
}
有些方法我是真的想不到
v
{:5_108:}
1
最后一题难死
加油
算个利息算的我人要傻了
朕想知道
J