HDU 1029 Ignatius and the Princess IV 水题

思路：排序后的$a[(n+1)/2]$即是答案。

http://acm.hdu.edu.cn/showproblem.php?pid=1029

/*********************************************
    Problem : HDU 1029
    Author  : NMfloat
    InkTime (c) NM . All Rights Reserved .
********************************************/

#include <map>
#include <set>
#include <queue>
#include <cmath>
#include <ctime>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>

#define rep(i,a,b)  for(int i = a ; i <= b ; i ++)
#define rrep(i,a,b) for(int i = b ; i >= a ; i --)
#define repE(p,u) for(Edge * p = G[u].first ; p ; p = p -> next)
#define cls(a,x)   memset(a,x,sizeof(a))
#define eps 1e-8

using namespace std;

const int MOD = 1e9+7;
const int INF = 0x3f3f3f3f;
const int MAXN = 1e6+5;
const int MAXE = 2e5+5;

typedef long long LL;
typedef unsigned long long ULL;

int T,n,m,k;

int A[MAXN];

void input() {
    rep(i,1,n) scanf("%d",&A[i]);
}

void solve() {
    sort(A+1,A+n+1);
    printf("%d\n",A[(n+1)/2]);
}

int main(void) {
    //freopen("a.in","r",stdin);
    //scanf("%d",&T); while(T--) {
    //while(~scanf("%d %d",&n,&m)) {
    while(~scanf("%d",&n)) {
        input();
        solve();
    }
    return 0;
}