DessertTime

There are multiple test cases. The first line of input contains an integer T (1 ≤ T ≤ 10), indicating the number of test cases. For each test case:

The first line contains an integer N (3 ≤ N ≤ $10^5$ ), indicating the number of plates.

The second line contains N integers, $a_1$ , $a_2$ ,……, $a_n$ (1 ≤ $a_i$ ≤ $10^9$ ), indicating the number of cakes on i-th plate. It is guaranteed every number appears more than once.

C++14(g++5.4) 解法, 执行用时: 113ms, 内存消耗: 10304K, 提交时间: 2020-10-05 21:32:41

#include<cstdio>
#include<algorithm>
using namespace std;
int n,a[100005],ans[4],len[4],now,flag,pre;
int main() {
	int t;
	scanf("%d",&t);
	while(t--) {
		scanf("%d",&n);
		for(int i=1; i<=n; i++)scanf("%d",&a[i]);
		sort(a+1,a+n+1);
		int cnt=1;
		for(int i=n-1;i>=0;--i){
			if(a[i]==a[i+1])cnt++;
			else {
				if(cnt&1){
					if(i)printf("%d\n",a[i+1]);
					else printf("-1\n");
					goto  JUMP; 
				}
				cnt=1;
			}
		}
		printf("%d\n",a[1]);
		JUMP:;
	}
}

C++(clang++ 11.0.1) 解法, 执行用时: 161ms, 内存消耗: 816K, 提交时间: 2022-12-07 20:38:49

#include<bits/stdc++.h>
using namespace std;
const int N=3e5+5;
bool cmp(int a,int b){
	return a>b;
}
int a[N];
int main(){
	int t,n;
	cin>>t;
	while(t--){
		cin>>n;
		int ans=-1,f=0;
		for(int i=0;i<n;i++){
			scanf("%d",&a[i]);
		}
		sort(a,a+n,cmp);
		int cnt=1;
		for(int i=1;i<n;i++){
			if(a[i]!=a[i-1]){
				if(cnt&1){
					ans=a[i-1];
					break;
				}
				cnt=1;
			}else cnt++;
		}
		if(ans==-1&&(!(cnt&1))) ans=a[n-1];
		cout<<ans<<endl;
	}
	return 0;
}

C++(g++ 7.5.0) 解法, 执行用时: 182ms, 内存消耗: 1344K, 提交时间: 2023-01-15 17:20:52

#include<bits/stdc++.h>
#define maxn 100007
int A[maxn];
int main(){
	int t;
	scanf("%d",&t);
	while(t--){
		int n,m,q;
		scanf("%d",&n);
		for(int i=0;i<n;i++){
			scanf("%d",&A[i]);	
		}
		A[n]=0;
		std::sort(A,A+n);
		q=A[0];
		int x,i=n-1,f=1;
		while(i){
			while(i&&A[i--]==A[i]);
		    x=n-i-1;
		    if((x%2&&i)||(i==0&&n%2==0)){
			printf("%d\n",A[i+1]);
			f=0;break;
			}
		}
		if(f) printf("-1\n");
	}
}

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