山东省第五届ACM省赛题——Weighted Median（模拟）

题目描述
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?

输入
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).

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

示例输入
7
10 35 5 10 15 5 20
10 35 5 10 15 5 20
示例输出
20

本来看到n是10^7就想着是不是有什么高效的算法，结果一看别人做的才知道可以直接模拟过程，从1到n循环，结果700多MS过了。。。

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define MAXN 10000010
using namespace std;
struct Number
{
    int x,w;
};
Number num[MAXN];
bool cmp(Number a,Number b)
{
    return a.xlong long sum,first,last,ans;
int main()
{
    double S;
    int n,m,i;
    while(~scanf("%d",&n))
    {
        sum=0;
        for(i=1; i<=n; ++i)
            scanf("%d",&num[i].x);
        for(i=1; i<=n; ++i)
        {
            scanf("%d",&num[i].w);
            sum+=num[i].w;
        }
        sort(num+1,num+1+n,cmp);
        S=sum*0.5;
        first=last=0;
        for(i=1;i<=n-1;++i)
        {
            first+=num[i].w;
            last=sum-first-num[i+1].w;
            if(first1].x;
                break;
            }
        }
        printf("%lld\n",ans);
    }
    return 0;
}