Sgu 114

114. Telecasting station

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

 开始还自己一个一个的枚举想3分每个区间段，找最优，后来发现是线性的，看是大于还是小于0去两边，结果发现没加上取左边的，每次都是按取右边的在算，所以还是有问题，不过记得之前看过白书说中位点最优！- -

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 citizens displeasure 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

---

Author	: Michael R. Mirzayanov
Resource	: PhTL #1 Training Contests
Date	: Fall 2001

#pragma comment(linker,"/STACK:102400000,102400000")
#include <cstdio>
#include <vector>
#include <cmath>
#include <queue>
#include <set>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define INF 0x7fffffff 
#define mod 1000000007
#define ll long long
#define maxn 100025
#define pi acos(-1.0)
int n, m;
struct node{
    ll x, c;
}a[maxn];
bool cmp(node a, node b){
    return a.x < b.x;
}
int main(){
    scanf("%d", &n);
    ll s = 0,cnt=0;
    for (int i = 0; i < n;i++){
        scanf("%lld%lld", &a[i].x, &a[i].c);
        cnt += a[i].c;
    }
    sort(a, a + n, cmp);
    int p = 0;
    double ans = 0;
    for (int i = 0; i < n; i++){
        s += a[i].c;
        if (s >= cnt / 2){p = i; break;}
    }
    ans = a[p].x;
    if (cnt % 2 == 2&&s==cnt/2)ans = ((double)(a[p].x + a[p + 1].x))/2;
    printf("%.5lf\n", ans);
    return 0;
}