Describe
a programmer who started working with an annual salary of N million. He hopes to buy a 60-square-meter house in the Zhongguancun Mansion. The current price is 2 million. Assuming that the house price increases by K% per year, and the programmer’s future annual salary remains unchanged, I don’t eat or drink, I don’t have to pay taxes. I save up all my income every year. How many years can I buy this house? (In the first year, the annual salary is N million, and the house price is 2 million)
Enter a
line containing two positive integers N (10 <= N <= 50), K (1 <= K <= 20), separated by a single space.
Output
If the house can be bought in the 20th year or before, an integer M is output, indicating that it needs to be bought in the M year at the earliest, otherwise Impossible is output.
Sample input
50 10
Sample output
8
#include <iostream>
using namespace std;
int main()
{
int n,k,y=20,i;
double fangjia=200;
cin >>n>>k;
for(i=1;i<=y;i++)
{
if(i*n>=fangjia)
{
cout<<i<<endl;
break;
}
fangjia *= (1+1.0*k/100);//注意这里不要忘记家前面的1.0，也可以写成0.01*k
}
if(i==21)
cout<<"Impossible"<<endl;
return 0;
}