SGU 114

114. Telecasting station

time limit per test: 0.25 sec. 
memory limit per test: 4096 KB

 

Every city in Berland is situated on Ox axis. The government of the country decided to build new telecasting station. After many experiments Berland scientists came to a conclusion that in any city citizensdispleasure is equal to product of citizens amount in it by distance between city and TV-station. Find such point on Ox axis for station so that sum of displeasures of all cities is minimal.

 

Input

Input begins from line with integer positive number N (0<N<15000) – amount of cities in Berland. Following N pairs (X, P) describes cities (0<X, P<50000), where X is a coordinate of city and P is an amount of citizens. All numbers separated by whitespace(s).

 

Output

Write the best position for TV-station with accuracy 10^-5.

 

Sample Input

4
1 3
2 1
5 2
6 2

Sample Output

3.00000

把权值看成有多少个人，求中位数。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>

using namespace std;

#define maxn 15005

struct node  {

        int x,p;
};

int n;
node s[maxn];

bool cmp(node a,node b) {
        return a.x < b.x;
}

int main() {
        //freopen("sw.in","r",stdin);

        scanf("%d",&n);

        int sum = 0;
        for(int i = 1; i <= n; ++i) {
                scanf("%d%d",&s[i].x,&s[i].p);
                sum += s[i].p;
        }

        sort(s + 1,s + n + 1,cmp);

        int now = 0;
        double m1,m2;
        for(int i = 1; i <= n; ++i) {
                now += s[i].p;
                if(now >= sum / 2) {
                        m1 = s[i].x;
                        if(now >= sum / 2 + 1) m2 = s[i].x;
                        else m2 = s[i + 1].x;
                        break;
                }

        }

        if(sum % 2) printf("%.5f\n",m2);
        else printf("%.5f\n",(m1 + m2) / 2);

        return 0;




}