ZOJ1049

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1049

假设当前已侵蚀的半径为R，一年后为r，则会有(π*r*r-π*R*R)/2 == 50，可以得到r*r=100/π+R*R。

#include<iostream>
#include<cmath>

using namespace std;

int main()
{
    int n,years;
    float x,y,R,r,d;
    const float c = 100/3.1415926;

    cin>>n;
    for (int i=1; i<=n; i++)
    {
        cin>>x>>y;
        d = (x*x+y*y)*3.1415926/100;
        years = (int)d;
        if (d-years>0) years++;
        cout<<"Property "<<i<<": This property will begin eroding in year "<<years<<"."<<endl;
    }
    cout<<"END OF OUTPUT."<<endl;

    return 0;
}

当R>=sqrt(x*x+y*y)时，即R*R>=x*x+y*y时，房子就会被淹掉。此时R*R是100/π的整数倍，假设为k倍，那么k>=(x*x+y*y)*π/100,k取(x*x+y*y)*π/100上取整。