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查看参考答案
1
// 假设小甲鱼和黑夜手上均有 10000 元,
// 小甲鱼以 10% 的单利息投资,
// 黑夜则以每年 5% 的复合利息投资。
// 计算需要多少年黑夜手头的 Money 才会超过小甲鱼?
#include<stdio.h>
#include<math.h>
int main()
{
float dan,fu=10000;
int year=1;
while(fu<dan)
{
dan=pow((10000*(1+0.1)),year);
fu=(1+0.05);
fu+=fu*0.05;
year++;
}
printf("%d",year);
return 0;
}
回
1
100
0
l-value???
a =14 b = 4 c= 14
z=(a>0?a:-a)
if(size>12)
{
cost = cost*1.05;
flag = 2;
}
bill = coat*flag;
if(ibex>14)
{
sheds= 3;
}
help = 2*sheds;
if(score<0)
{
printf("count = %d\n", count);
}
count++;
scanf("%d,&score);
#include <stdio.h>
#include <stdlib.h>
int jia(int a,int year)
{
a = a+a*0.1*year;
return a;
}
int hei(int b,int year)
{
int i;
for(i=0;i<year;i++)
{
b = (b+b*0.05);
}
return b;
}
int main()
{
int a = 10000;
int b = 10000;
int year;
for(year=1;a>=b;year++)
{
a = jia(a,year);
b = hei(b,year);
}
printf("%d",year);
return 0;
}
#include <stdio.h>
#include <stdlib.h>
int main()
{
int lose(int money,int year);
int money = 400;
int year;
for(year = 1;money>0;year++)
{
money = lose(money,year);
}
printf("%dyear",year);
return 0;
}
int lose(int money,int year)
{
money = (money+money*0.08) - 50;
return money;
}
1
/
ooo
看看
0. 90个‘A’
1.10个‘B’
2.c=5;
b=c;
a=b;
3.b=4
c=9
a=14
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: scanf("%d", &score);
if (score < 0)
{
printf("count = %d\n", count);
}
count++;
scanf("%d", &score);
1
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1
1
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lalala
1
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