Megacity（二分答案简单题之杨氏代码）

题目描述:

The administration of the Tomsk Region firmly believes that it's time to become a megacity (that is, get population of one million). Instead of improving the demographic situation, they decided to achieve its goal by expanding the boundaries of the city.

The city of Tomsk can be represented as point on the plane with coordinates (0; 0). The city is surrounded with n other locations, the i-th one has coordinates (xi, yi) with the population of ki people. You can widen the city boundaries to a circle of radius r. In such case all locations inside the circle and on its border are included into the city.

Your goal is to write a program that will determine the minimum radius r, to which is necessary to expand the boundaries of Tomsk, so that it becomes a megacity.

输入描述:

The first line of the input contains two integers n and s (1 ≤ n ≤ 103; 1 ≤ s < 106) — the number of locatons around Tomsk city and the population of the city. Then n lines follow. The i-th line contains three integers — the xi and yi coordinate values of the i-th location and the number ki of people in it (1 ≤ ki < 106). Each coordinate is an integer and doesn't exceed 104 in its absolute value.

It is guaranteed that no two locations are at the same point and no location is at point (0; 0).

输出描述:

In the output, print "-1" (without the quotes), if Tomsk won't be able to become a megacity. Otherwise, in the first line print a single real number — the minimum radius of the circle that the city needs to expand to in order to become a megacity.

The answer is considered correct if the absolute or relative error don't exceed 10 - 6.

题意:

题目中一个叫特斯拉的城市想要合并一些城市使得该城市变为特大城市而成为特大城市的条件就是满足人口达到1000000人首先特斯拉城市位于（0，0）这个坐标题目会给一个n和m 其中n代表有n个城市后面跟随n行，其中每一行都会有一个 x y sum 分别代表 (x,y)坐标和该城市的人口数sum; m代表特斯拉城市已经有m个人了要求合并完这些城市到特大城市的的最小半径是多少

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e3+100;
const double mod=1e-7;
int n,m;
struct add{double x,y,sum;
}s[maxn];
double check(double l,double r){double mid;while(r-l>=mod){mid=(l+r)/2;int sum=m;for(int i=0;i<n;i++){if(s[i].x*s[i].x+s[i].y*s[i].y<=mid*mid){sum+=s[i].sum;}}if(sum>=1000000){r=mid;}else{l=mid;}}return mid;
}
int main(){cin>>n>>m;int ans=m;for(int i=0;i<n;i++){cin>>s[i].x>>s[i].y>>s[i].sum;ans+=s[i].sum;}if(ans<1000000){cout<<"-1"<<endl;}else{printf("%.7lf\n",check(0,1000000000));}return 0;
}