Weighted Median

For n elements x1, x2, ..., xn with positive integer weights w1, w2, ..., wn. The weighted median is the element xk satisfying 
 and   , S indicates   

Can you compute the weighted median in O(n) worst-case?

 

Input

There are several test cases. For each case, the first line contains one integer n(1 ≤  n ≤ 10^7) — the number of elements in the sequence. The following line contains n integer numbers xi (0 ≤ xi ≤ 10^9). The last line contains n integer numbers wi (0 < wi < 10^9).

 

Output

One line for each case, print a single integer number— the weighted median of the sequence.

 

Sample Input

7
10 35 5 10 15 5 20
10 35 5 10 15 5 20

Sample Output

20

Hint

The S which indicates the sum of all weights may be exceed a 32-bit integer. If S is 5,  equals 2.5.

 

这是一道山东省省赛的题，英语不好的我看了以后一脸懵逼，最后花了好长时间才看懂，原来很是简单

题目意思：
给一些数x和它们对应的权值w，按照如图所示公式，s是所有权值w的总和。
求一个xk，使得满足前两个公式。

解题思路：
用结构体保存元素值和权值，先升序排列，累计xk之前的权值和，直到该权值大于s。

#include<stdio.h>
#include<algorithm>
using namespace std;
struct message{
int x;
int w;
}a[10000010];
int my_cmp(message a,message b)
{
    if(a.x<b.x)
        return 1;
    else
        return 0;
}
int main()
{
    int n,i;
    double s,y;
    while(scanf("%d",&n)!=EOF)
    {
        s=0;
        for(i=0;i<n;i++)
        scanf("%d",&a[i].x);
        for(i=0;i<n;i++)
        {
            scanf("%d",&a[i].w);
            s=s+a[i].w;
        }
        s=s/2.0;
        y=0;
        sort(a,a+n,my_cmp);
        for(i=0;i<n;i++)
        {
            y=y+a[i].w;
            if(y>s)
            {
                printf("%d
",a[i].x);
                break;
            }
        }
    }
    return 0;
}