{:10_257:}
0.10个
1.10个
2.
3.
4.
5.利用标志位
1
aa
k看看
努力努力再努力
0
goto answer
/*
//投资问题
#include <stdio.h>
#include <math.h>
#define DAN 0.1
#define FUHE 0.05
int main()
{
float _jiayu = 10000, _heiye = 10000;
int year;
for(year = 1; _jiayu >= _heiye; year++)
{
_jiayu = 10000 * (1 + DAN * year);
_heiye = 10000 * pow((FUHE + 1), year);
}
printf("%d年后\n", year);
printf("小甲鱼:%.2f\n", _jiayu);
printf("黑夜:%.2f\n", _heiye);
return 0;
} //自己写的*/
/*
//破产
#include <stdio.h>
#define BONUS 0.08
int main()
{
int year;
float yuan;
for(yuan=4000000,year=0;yuan>=0;year++)
{
yuan = yuan-500000;
yuan = yuan * (1 + BONUS);
}
printf("%d年之后破产\n", year);
return 0;
}//自己写的*/
/*
//求PI
#include <stdio.h>
#include <math.h>
int main()
{
double pi = 0,p = 0, pn;
int n;
pn = (double)(1/(2*n-1));
for(n = 1;fabs(pn) >= pow(10,-8);n++)
{
p = p + pn * pow(-1,n-1);
}
pi = 4 * p;
printf("PI = %.7lf\n", pi);
return 0;
}//自己写的*/
0.81
1.9
2.a=5
3.b=3,c=9,a=14
4.c=x>0?x:-x;
5.if(size>12)
{
cost*=1.05;
flag=2;
}
else
{
bill=cost*flag;
}
if(ibex>14)
{
shed=3;
}
shed=2;
help=2*sheds;
动动手
0.#include<stdio.h>
#include<math.h>
int main()
{
int i,n=1;
double r1=0.1,r2=0.05,x,y;
i=10000;
for(n=1;;n++)
{
x=i*(1+n*r1);
y=i*(pow(1+r2,n));
if(x<y)
{
printf("%d年后,黑夜的投资额超过小甲鱼\n",n);
printf("小甲鱼的投资额是:%.2lf\n",x);
printf("黑夜的投资额是:%d\n",y);
break;
}
}
return 0;
}
1.#include<stdio.h>
#include<math.h>
int main()
{
int year,i,j;
double r1=0.08;
i=4e6;
for(year=1;;year++)
{
i-=5e5;
i*=(1+r1);
if(i<=0)
{
printf("%d年之后,小甲鱼败光了所有的家产,再次回到了一贫如洗...\n",year);
break;
}
}
return 0;
}
3.
#include<stdio.h>
int main()
{
int i;
int f1=1,f2=1,f3;
for(i=4;i<=24;i++)
{
f3=f1+f2;
f1=f2;
f2=f3;
}
printf("两年后的兔子数为%d",f3);
return 0;
}
1
{:5_90:}
小甲鱼赛高
。
1
1
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dsads
1
朕想知道