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 <stdio.h>#include <math.h>#include <string.h>#include <stdlib.h>#include <iostream>#include <sstream>#include <algorithm>#include <set>#include <queue>#include <stack>#include <map>#include <bitset>#define MAXN 10000010using namespace std;struct Number{    int x,w;};Number num[MAXN];bool cmp(Number a,Number b){    return a.x<b.x;}long 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(first<S&&last<=S)            {                ans=num[i+1].x;                break;            }        }        printf("%lld\n",ans);    }    return 0;}