903256108
发表于 2020-2-24 18:29:36
a
cq75558012
发表于 2020-2-24 19:50:12
朕想知道!
wuyo
发表于 2020-2-25 13:49:43
答案
dcx1230000
发表于 2020-2-25 15:39:24
0.100
1.11
2.a
3.
akilameruna
发表于 2020-2-25 16:31:25
{:10_277:}
2842172730
发表于 2020-2-25 20:19:06
1
xuhaikuan2015
发表于 2020-2-25 21:13:02
查看答案
莱布妮子接班人
发表于 2020-2-26 10:34:59
1
tixunfeifu
发表于 2020-2-26 11:43:00
学习打卡
初来乍到的小白
发表于 2020-2-26 11:49:22
{:10_256:}
自学小白菜
发表于 2020-2-26 15:45:03
0.10次
1.0次吧,不太确定
2.a
3.b = 4;c = 10 ; a = 4
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(scanf("%d",&score)<0)
{
printf("count=%d\n",count);
count++;
}
动手题:
0.
#include <stdio.h>
int main()
{
int count=0;
float i,j;
i =10000.00;
j =10000.00;
for(i,j;j<=i;)
{
i = 10000*0.1+i;
j = j *1.05;
count++;
}
printf("%d年后,黑夜的投资超过小甲鱼!\n",count);
printf("小甲鱼的投资是%.2f\n",i);
printf("黑夜的投资是%.2f\n",j);
return 0;
}
好的,这道题还简单
1.
#include <stdio.h>
int main()
{
float i = 400.00;
int count = 0;
for(i;i>0;)
{
i = i+ i * 0.08-50;
count++;
}
printf("%d年后,小甲鱼再次一贫如洗。哈哈哈\n",count);
return 0;
}
2.
不太懂这道题的思路
3.写不出来,思路很明确,但是写for里面的时候卡住了
徐德民
发表于 2020-2-26 20:08:39
1
王玉炫
发表于 2020-2-26 21:04:32
1
D语言玩家
发表于 2020-2-26 23:55:16
已完成
1315357277
发表于 2020-2-27 10:02:22
ccc
xyf0410
发表于 2020-2-27 11:37:27
{:10_279:}
sssuhl
发表于 2020-2-27 12:06:10
6666666666666
牙牙乐
发表于 2020-2-27 14:52:26
010
1 11
2 a= 5
b = a
c= b
3
4z =
今秋的雨很凉
发表于 2020-2-27 18:41:46
0.100
1.无数
2.a,b,c
3.c=3,a=14,b=9
4.int x,z;
if(x >= 0)
{
z = x;
}
else
{
z = -x
}
5.if(size > 12)
{
cost = cost * 1.05;
flag = 2;
}
else
{
blii = cost * flag;
}
if(ibex > 14)
{
sheds = 3;
}
else
{
sheds = 2;
help = 2 * sheds;
}
readin:scanf("%d", &score);
while(score >= 0)
{
count++;
scanf("%d", &score);
}
printf("count = %d\n", count);
#include<stdio.h>
int main()
{
double xjy=10000;
double hy=10000;
int year;
for(year=1; xjy >= hy; year++)
{
xjy = 10000 + 10000 * year * 0.1;
hy = hy + hy * 0.05;
}
year--;
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:%.2lf\n",xjy);
printf("黑夜的投资额是: %.2lf\n",hy);
return 0;
}
1.#include<stdio.h>
int main()
{
int jc=4000000;
int year;
for(year=0; jc > 0; year++)
{
jc = jc * (1 + 0.08);
jc = jc - 500000;
}
year--;
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);
}
2.#include<stdio.h>
#include<math.h>
int main()
{
double sum;
double j;
int i;
for(sum = 1.0,i = 1,j=0;sum < 1000; sum = sum + 2,i++)
{
j = j + (pow(-1,i - 1) * (1 / sum));
}
printf("Π/4≈%.7lf\n",j);
}
3.2的24次方
战术咕咕
发表于 2020-2-27 19:42:16
对照答案