Median (25)

时间限制 1000 ms 内存限制 65536 KB 代码长度限制 100 KB 判断程序 Standard (来自小小)
题目描述
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.

输入描述:
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.

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

输入例子:
4 11 12 13 14
5 9 10 15 16 17

输出例子:
13

题解:求解中位数，用二指针的思想，设i,j 扫一遍，便可得到。中间位置为（n+m-1）/2

#include <bits/stdc++.h>
#include <stdlib.h>
using namespace std;
#define vi vector<int>
#define pii pair<int,int>
#define x first
#define y second
#define all(x) x.begin(),x.end()
#define pb push_back
#define mp make_pair
#define SZ(x) x.size()
#define rep(i,a,b) for(int i=a;i<b;i++)
#define per(i,a,b) for(int i=b-1;i>=a;i--)
#define pi acos(-1)
#define mod 1000000007
#define inf 1000000007
#define ll long long
#define DBG(x) cerr<<(#x)<<"="<<x<<"\n";
#define N 200010
template <class U,class T> void Max(U &x, T y){if(x<y)x=y;}
template <class U,class T> void Min(U &x, T y){if(x>y)x=y;}
template <class T> void add(int &a,T b){a=(a+b)%mod;}
const int maxn=1000010;
const int INF=0x7ffffff;
int main()
{int n,m,a[maxn],b[maxn],ans;cin>>n;rep(i,0,n) cin>>a[i];cin>>m;rep(i,0,n) cin>>b[i];//a[n]=b[m]=inf;ans=(n+m-1)/2;int i=0,j=0,k=0;while(k<ans){k++;if(a[i]>b[j]){j++;}else{i++;}}if(a[i]<b[j])cout<<a[i]<<endl;else cout<<b[j]<<endl;}

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