hdu 1029 求出现次数过半的数

题目传送门//res tp hdu

已知必定存在一个元素出现次数超过一半，考虑用栈
若当前元素等于栈顶元素，入栈，反之出栈，并将当前元素入栈
最终的栈顶元素即是所求

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define per(i,a,b) for(int i = (a);i>=(b);--i)
#define fo(i,a,b) for(int i =(a);i<(b);++i)
#define de(x) cout<<#x<<" = "<<x<<endl;
#define endl '\n'
#define ls(p) ((p)<<1)
#define rs(p) (((p)<<1)|1)
using namespace std;
typedef long long ll;
const int mn = 1e6+10;
int n;
ll Top,t;
int cnt;
// stack
int main(){
	while(scanf("%d",&n)!=EOF){
		cnt = 1;
		scanf("%lld",&Top);
		rep(i,2,n){
			scanf("%lld",&t);
			if(t == Top)++cnt;
			else{
				if(cnt == 1)
					Top = t;
				else{
					--cnt;
				}
			}
		}
		printf("%lld\n",Top);
		
		
	}
	
	
	
}