谢谢小甲鱼
0.90
1.
10
2.a,b,c
3
b=4
c=
4.
z=(x>=0 ? x : -x)
5.
A:if (size > 12)
{
cost = cost * 1.05;
flag = 2;
break;
}
bill = cost * flag;
B
if (ibex > 14)
{
sheds = 3;
break;
}
sheds = 2;
help = 2 * sheds;
C
readin: scanf("%d", &score);
if (score < 0)
{
goto stage2;
}
count++;
goto readin;
stage2: printf("count = %d\n", count);
看看
11
看看
9 314
#include<stdio.h>
int main()
{
int f=10000,count=0;
float sum1=10000,sum2=10000;
while(sum2<=sum1)
{
sum1=sum1+f*0.1;
sum2=1.05*sum2;
count++;
}
printf("%d年后，黑夜的投资额超过小甲鱼！\n",count);
printf("小甲鱼的投资额是：%.2f\n",sum1);
printf("黑夜的投资额是：%.2f\n",sum2);
return 0;
}
#include<stdio.h>
int main()
{
int a=4000000,count=0;
while(a>0)
{
a-=500000;
a=1.08*a;
count++;
}
printf("%d年之后，小甲鱼败光了所有的家产，再次回到一贫如洗。",count);
return 0;
}
#include<stdio.h>
#include<math.h>
int main()
{
double a=1,sum=4;
int count=1;
while(1)
{
a=1/(1+(float)count*2);
if(a>=pow(10,-8))
{
if((int)count%2!=0)
{
a=-a;
sum+=4*a;
a=-a;
count++;
}
else
{
sum+=4*a;
count++;
}
}
else
{
break;
}
}
printf("pi保留7位是%.7lf",sum);
return 0;
}
0，81
1,0
2,5
3，
{:10_277:}
0:10个A
1:11个B
2：a,b,c
3:a=14,b=9,c=5
4:
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:
while(1)
{
scanf("%d", &score);
if (score < 0)
{
printf("count = %d\n", count);
break;
}
count++;
}
动动手：
#include<stdio.h>
int main()
{
float x=10000,b=10000;
int count=0;
while(x>=b)
{
x+=1000;
b=b*1.05;
count++;
}
printf("%d年后，黑夜的投资额超越小甲鱼！\n小甲鱼的投资额是：%.2f.\n黑夜的投资额是:%.2f.\n",count,x,b);
}
1：
#include<stdio.h>
int main()
{
float x=4000000;
int count=0;
while(x>0)
{
x=(x-500000)*1.08;
count++;
}
printf("%d年后，小甲鱼败光所有家产，再次回到一贫如洗。。。。。\n",count);
}
2:
#include<stdio.h>
#include<math.h>
int main()
{
double PI=1.0,i=1.0,num=1.0,I;
int n=0;
while(fabs(num)>=pow(10,-8))
{
if(i>0)
{
i=-(i+2);
}
else
{
i=-(i-2);
}
num=1/i;
PI+=num;
}
PI=4*PI;
printf("PI的值为%.7f\n",PI);
}
11
I LOVEV
33444444444555555555666666666666667777888899999999999999999999999000000000000000011
0 10
1 0
2 abc
3 14 5 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
do
{
scanf("%d", &score);
if(score<0)
{
printf("count = %d\n", count);
}
count++;
}while(score>=0);
0. 100個
1. 10
2. a, b, c = 5, 5, 5;
3. a = 14, b = 5, c = 9
4. z = x > o? x : y
5. A:
if (size > 12) {
cost *= 1.05; break;}
bill = cost * flag;
B:
if (ibex > 14) {
sheds = 3; }
help = 2 * sheds;
C:
while 1 {
scanf("%d", &score);
if (score < 0) {
printf("count = %d\n", count); break; }}
count++;
1111111111111
测试题：
0.10次
1.0次 开始时 while = 0;//为假就不执行了？
2. a, b, c ; lvalue 是用于识别或定位存储位置的标识符
3.
b=3,c=b+1+5,a = c+1 + b+1;
a == 14; (9+1 + 3+1)
b == 5; (3+1+1)
c == 4; (3+1)
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:
if (score < 0)
{
printf("count = %d\n", count);
}
count++;
scanf("%d", &score);
动动手：
0.
#include <stdio.h>
#include <math.h>
int main()
{
double i, xjy, hy;
for(i=0;hy <= xjy;i++)
{
xjy = hy = 10000;
xjy += (xjy* 0.1)*i ;//单利率:利息=本金x年利率x年数
hy = hy*pow((1 + 0.05) ,i);//复利率:F=P*(1+i)^n
}
printf("%.f年之后,黑夜的投资额超过小甲鱼!\n",i);
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()
{
double yuan,year;
yuan = 40000000;
for (year=0;yuan >= 0;year++)
{
yuan -=5000000;
yuan += yuan * 0.08;
}
printf("%.f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 i=1.0, n=1.0 ,s=1,Pi=0;
for (;fabs(n) >= 10^(-8); i += 2)
{
Pi += n;
s= -s;
n = s/i;
}
Pi = 4*Pi;
printf("Pi = %.7f",Pi);
return 0;
}
3.
#include <stdio.h>
int main()
{
int sum, t=2;
for(int year=720;year >= 0;year-=60)
{
sum = t;
for (int mother=2;mother >= 0;mother--)
{
t +=2;
}
t *= 2;
}
printf("%d",sum);
return 0;
}
57332只兔兔？
1
1