SGU114-Telecasting station

114. Telecasting station

time limit per test: 0.5 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. FollowingN pairs (X, P) describes cities (0<X, P<50000), whereX 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

41 32 15 26 2

Sample Output

3.00000

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题意是说，有个国家的城市之间要建个公交（我英语不好、翻译可能会跑偏、也可能是电视塔、也可能是别的、我也不知道是啥），然后有一个不满意度，咱们自己想想也知道，肯定都喜欢把地铁站设在自己家门口。然后这个不满意度=城市的人数*这个站的位置。。。然后的然后、我也不会做这个题，直接上理论：数学家说，这是一个带权中位数问题，具体内容见http://baike.baidu.com/link?url=NZbFTsJCCn_jsx8W24aLu6XGEVdpDH0hpO_SIzeCF7vFi_Nz-xV4pe7tzuJBpQsQSzoa719FmHYb3lr7QKV3q_   这个讲的很明白了。然后，下面的代码就自然手到擒来了！！！

 

#include<iostream>#include<string.h>#include<stdio.h>#include<ctype.h>#include<algorithm>#include<stack>#include<queue>#include<set>#include<math.h>#include<vector>#include<map>#include<deque>#include<list>using namespace std;struct N{    double x;    int p;} a[15009];int b[15009];int cmp(N a,N b){    return a.x<b.x;}int main(){    int n;    while(scanf("%d",&n)!=EOF)    {        memset(b,0,sizeof(b));        for(int i=0; i<n; i++)            scanf("%d%d",&a[i].x,&a[i].p);        sort(a,a+n,cmp);        int s=0;        for(int i=0; i<n; i++)        {            s+=a[i].p;            b[i]=s;        }        int mid=s/2,w;        for(int i=0; i<n; i++)        {            if(b[i]>=mid)            {                w=i;                break;            }        }        printf("%d\n",a[w].x);    }    return 0;}