按照题意模拟判断即可

#include <iostream>
using namespace std;
int main()
{
    int k,n,m;
    cin>>k>>n>>m;
    double ans = n;
    int fg = 0;
    if( m <= k)
    {
    	fg = 1;
    	if(n <= 100)
    	{
    		ans=n*1.0*0.9;
		}
		else if(n > 100 && n<=1000)
		{
			ans = n*1.0*0.8;
		}
		else ans = n*1.0*0.7;
	}
	printf("%d %.2lf",fg,ans);
    return 0;
}