phlaoi
发表于 2021-11-29 22:08:15
1
风亦
发表于 2021-11-30 23:12:14
1010
a b c
b = 3,c = 8, a = (3,4,8,14)
小明要人带
发表于 2021-12-1 09:19:30
1
Donja
发表于 2021-12-1 19:30:34
查看参考答案
jack_straw_man
发表于 2021-12-1 21:28:53
0. 10
1. 0如果++i 就是 10个
2. a, b, c
3. a= 14; b = 5; c = 9;
4. z = a>0?a:-a;
5.
A.
if (size > 12)
{
cost = cost * 1.05;
flag = 2;
}
bill = cost * flag;
B.
if (ibex > 14)
{
sheds = 2;
sheds = 3;
}
help = 2 * sheds;
C.
while(1)
{
scanf("%d", &score);
if (score < 0)
{
break;
}
count++;
}
printf("count = %d\n", count);
l14564
发表于 2021-12-2 01:31:10
0.110
1.i++ 相当于会先用i=0进行赋值再给i加1?(a++) 是先将变量 a 的值(5)做为整个表达式的值返回,再将 a 自增 1(类似于 a = a + 1)?i++先返回数值再进行自增?++i先进行自增再返回数值
2.a bc?
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;
B.
if (ibex > 14)
{
sheds = 3;
}
sheds = 2;
help = 2 * sheds;
C.
do
{
scanf("%d", &score);
}
while (count++,score > 0);
printf("count = %d\n", count);
Churling
发表于 2021-12-2 10:55:36
3
sznar
发表于 2021-12-2 20:03:31
1
很酷1212
发表于 2021-12-2 20:25:30
1
鱼鱼鱼小
发表于 2021-12-3 10:33:04
000
yong_ze_
发表于 2021-12-3 13:24:30
答案
张文耀860
发表于 2021-12-3 16:21:04
联想知道
Rachel·梨
发表于 2021-12-4 11:11:11
{:10_257:}
学c的小同鞋
发表于 2021-12-4 11:32:17
完成
zzl-jy
发表于 2021-12-4 22:04:03
{:5_90:}
salv
发表于 2021-12-5 10:32:43
{:5_97:}
kg120
发表于 2021-12-5 11:39:03
最后一题也可以拓展为分别计算小、中、老兔子的数量
xj-608111
发表于 2021-12-6 08:58:56
1
V-Mn
发表于 2021-12-6 11:07:24
0.
100
1.
10
2.
a,b,c
3.
b=5 c=9 a=14
4.
z=(x>0?x:(-x))
5.
if (size > 12)
{cost = cost * 1.05;}
bill = cost * flag;
B.
if (ibex > 14)
{
sheds = 3;
}
sheds = 2;
help = 2 * sheds;
C.
readin: scanf("%d", &score);
if (score < 0)
{
printf("count = %d\n", count);
}
count++;
readin: scanf("%d", &score);
0.
#include<stdio.h>
#include<iostream>
using namespace std;
int main()
{
int n,m1=10000,m0=10000;
for(n=0;m1>=m0;n++)
{
m1+=10000*0.1;
m0*=1.05;
}
cout<<n<<'\n'<<m1<<'\n'<<m0;
return 0;
}
1.
#include<stdio.h>
#include<iostream>
using namespace std;
int main()
{
int n,a=4000000;
for(n=0;a>=500000;n++)
{
a=(a-500000)*1.08;
}
cout<<n;
return 0;
}
2.
#include<stdio.h>
int main()
{
double PI=0.0,n,k=1;
for(n=1;1/n>0.00000001;n+=2)
{
PI=PI+k*(1/n);
k=-k;
}
printf("PI=%.7f",4*PI);
return 0;
}
3.
找不出函数关系式
妮妮666
发表于 2021-12-6 17:22:30
6