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

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

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

         }

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

         return ;

 }