PAT甲级 1029

Median

Given an increasing sequence S of N integers, the median is the number at the middle position. For example, the median of S1={11, 12, 13, 14} is 12, and the median of S2={9, 10, 15, 16, 17} is 15. The median of two sequences is defined to be the median of the nondecreasing sequence which contains all the elements of both sequences. For example, the median of S1 and S2 is 13.

Given two increasing sequences of integers, you are asked to find their median.

Input

Each input file contains one test case. Each case occupies 2 lines, each gives the information of a sequence. For each sequence, the first positive integer N (<=1000000) is the size of that sequence. Then N integers follow, separated by a space. It is guaranteed that all the integers are in the range of long int.

Output

For each test case you should output the median of the two given sequences in a line.

Sample Input
4 11 12 13 14
5 9 10 15 16 17
Sample Output
13

---

1. 考点

   1. 二分法

2. 题解

   1. 分为2个二分
   2. 第一个二分，寻找中位数。第二个二分统计小于当前这个数的个数。
   3. 如何当前2个二分找的位置之和等于整个list的一半，即为中位数。

#include<iostream>
#include<stdio.h>
using namespace std;
const int maxn=1000000+10;
typedef long long ll;
ll firstSequence[maxn],secondSequence[maxn];
int input_data(ll num[]){
    int n;
    scanf("%d",&n);
    int i;
    for(i=1;i<=n;i++){
        scanf("%lld",num+i);
    }
    return n;
}
int getNumOfLess(ll value,int n,ll num[]){
    int low=1,high=n,mid;
    while(low<=high){
        mid=(low+high)/2;
        if(value<num[mid]){
            high=mid-1;
        }else{
            low=mid+1;
        }
    }
    return low-1;
}
ll findMedian(ll num1[],int n1,int n2,ll num2[]){
    ll rst=-1;
    int low=1,high=n1,mid;
    int pos;
    while(low<=high){
        mid=(low+high)/2;
        pos=getNumOfLess(num1[mid],n2,num2);
        if(mid+pos==(n1+n2+1)/2){
            rst=num1[mid];
            break;
        }else if(mid+pos<(n1+n2+1)/2){
            low=mid+1;
        }else{
            high=mid-1;
        }
    }
    return rst;
}
int main(){
    freopen("./in","r",stdin);
    int n1,n2;
    ll rst;
    n1=input_data(firstSequence);
    n2=input_data(secondSequence);
    rst=max(findMedian(firstSequence,n1,n2,secondSequence),findMedian(secondSequence,n2,n1,firstSequence));
    printf("%lld",rst);
}