思路：排个序然后扫一遍即可
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
#define LL long long
const int maxn = 100000+100;
LL a[maxn];
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
for(int i = 1;i<=n;i++)
scanf("%lld",&a[i]);
sort(a+1,a+1+n);
for(int i = 1;i<=n;i+=3)
{
if(a[i]==a[i+1]&&a[i+1]==a[i+2])
continue;
else
{
printf("%lld\n",a[i]);
break;
}
}
}
}
时间限制 1000 ms 内存限制 65536 KB
题目描述
Given an array with N
输入格式
Several test cases are given, terminated by EOF.
Each test case consists of two lines. The first line gives the length of array N(1≤N≤105), and the other line describes the Nelements. All elements are ranged in [0,263−1].
输出格式
Output the answer for each test case, one per line.
输入样例
4
1 1 1 3
10
1 2 3 1 2 3 1 2 3 4
输出样例
3
4
