求一个数学难题的算法

Question

F0=1000
F1=F0*a-30;
F2=F1*a-30;
F3=F2*a-30;
... ....
Fn-1=Fn-2*a-30
Fn=Fn-1*a-30

Fn=0;
输入n;
输出a

Answer 1

 #include <stdio.h>

int n;
double a;

double fun(int tInputValue)
{
    if(tInputValue == 0)
        return 1000.0;
    else
        return fun(tInputValue-1)*a-30;
}

int main()
{
    scanf("%d", &n);

    for(a =0; a < 1000000; a=a+0.01)
    {
        if( 30-fun(n-1)*a < 1e-06 )
        {
            printf("%lf\n", a);
            break;
        }
    }
}