MN_Guan
发表于 2020-3-7 22:05:17
0:10
1:运行前感觉是10次,运行后是0次,因为i++执行时i=0,不进入while循环
2:a、b、c
3:a=15 b=5c=10
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;
else
sheds=2;
help = 2 * sheds;
------------------------
C:
do
{
scanf("%d",&score);
count++;
}
while(!(score<0));
printf("count = %d\n",count);
--------------------------
动动手:
0:
#include<stdio.h>
int main()
{
int i=0;
float FishC=10000;
float night=10000;
while(night<=FishC)
{
FishC+=10000*0.1;
night+=night*0.05;
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",FishC);
printf("黑夜的投资额是%.2f\n",night);
return 0;
}
1:
#include<stdio.h>
int main()
{
float total=400;
int i=1;
while((total-=50)>0)
{
total+=total*0.08;
i++;
}
printf("%d年之后,小甲鱼败光了了所有的家产,再次回到一贫如洗......\n",i);
return 0;
}
2:
判断条件降了一阶,运行时间太长了,这是1e-7的结果:3.1415966
#include<stdio.h>
#include<math.h>
int main()
{
// float numerator=1;//分子
float approximate_PI=0;//近似值
int i=0;
float denominator=1;//分母
float limit=pow(10,-7);
float fraction;//分数
while((fraction=1/denominator)>=limit)
{
approximate_PI+=pow(-1,i++)*fraction;
denominator+=2;
printf("%.9f %d\n",fraction,i);
}
printf("PI的近似值为:%.7f\n",4*approximate_PI);
return 0;
}
3:
//the answer is 75025
#include<stdio.h>
int main()
{
int year0=1,year1=0,year2=0;
for(int i=1;i<=24;i++)
{
year2+=year1;
year1=year0;
year0=year2;
}
printf("两年之后可以繁殖%d对兔子。\n",year0+year1+year2);
return 0;
}
千秋寒夏
发表于 2020-3-7 23:18:20
浮点数
1394116864
发表于 2020-3-8 09:56:14
123
瞌睡小子
发表于 2020-3-8 10:43:42
拾遗
1106824646
发表于 2020-3-8 13:32:52
sadfgh
啊川
发表于 2020-3-8 15:01:27
0.
huanghuiyv
发表于 2020-3-8 15:33:29
无条件支持楼主!
yuanguohua
发表于 2020-3-8 17:00:12
1
大明天子
发表于 2020-3-8 17:12:29
割白菜喂猪 发表于 2017-1-5 23:13
地方
做
金咕咕
发表于 2020-3-9 12:47:53
瞅一下
我来学习de
发表于 2020-3-9 15:42:12
#include<stdio.h>
int main()
{
int x,h,n,temp;
int j;
while(x>h)
{
for(n=1;;n++)
{
temp=10000*0.1*n;
x=temp+10000;
for(j=1;j<n;j++)
{
if(j==1)
{
h=(1+0.05)*10000;
}
else
{
h=(1+0.05)*h;
}
}
}
}
printf("%d后,黑的资产超过小",n);
printf("小的资产是%d\n黑的资产是%d",x,h);
return 0;
}
#include<stdio.h>
int main()
{
int n=1;
int money=400;
while(money)
{
if(n==1)
{
money=400*0.08;
}
else
{
money=money*0.08;
}
money-=50;
n++;
}
printf("%d后,资产用光",n+1);
return 0;
}
humanR
发表于 2020-3-9 17:56:46
1
dhzj159
发表于 2020-3-9 18:44:17
ss's
無用。
发表于 2020-3-9 20:40:45
我来检查答案了{:5_109:}
jiangyifei
发表于 2020-3-9 22:48:17
测试题答案:
0.10个
1.11个
2.a,b,c
3.a=14,b=5,c=9
4.
#include <stdio.h>
int main()
{
float x;
floata z;
printf("请输入一个数:");
scanf("%f",&x);
z=x>0 ? x:x*(-1);
printf("%f的绝对值是:%f\n",x,z);
return 0;
}
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.readin: scanf("%d", &score);
if (score < 0)
{
printf("count = %d\n", count);
}
count++;
动动手答案:
0.
suqiang357
发表于 2020-3-10 00:01:07
1
三台子刘德华
发表于 2020-3-10 15:02:00
滴滴答滴答
Wenty
发表于 2020-3-10 16:16:00
{:10_291:}
幸运巴别塔
发表于 2020-3-10 17:02:08
00
lh1592788511
发表于 2020-3-10 17:41:48
ljk