Codeforces 890B

B. Vlad and Cafes
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Vlad likes to eat in cafes very much. During his life, he has visited cafes n times. Unfortunately, Vlad started to feel that his last visits are not any different from each other. To fix that Vlad had a small research.

First of all, Vlad assigned individual indices to all cafes. Then, he wrote down indices of cafes he visited in a row, in order of visiting them. Now, Vlad wants to find such a cafe that his last visit to that cafe was before his last visits to every other cafe. In other words, he wants to find such a cafe that he hasn't been there for as long as possible. Help Vlad to find that cafe.

Input
In first line there is one integer n (1 ≤ n ≤ 2·105) — number of cafes indices written by Vlad.

In second line, n numbers a1, a2, ..., an (0 ≤ ai ≤ 2·105) are written — indices of cafes in order of being visited by Vlad. Vlad could visit some cafes more than once. Note that in numeration, some indices could be omitted.

Output
Print one integer — index of the cafe that Vlad hasn't visited for as long as possible.

Examples
input
5
1 3 2 1 2
output
3
input
6
2 1 2 2 4 1
output
2
Note
In first test, there are three cafes, and the last visits to cafes with indices 1 and 2 were after the last visit to cafe with index 3; so this cafe is the answer.

In second test case, there are also three cafes, but with indices 1, 2 and 4. Cafes with indices 1 and 4 were visited after the last visit of cafe with index 2, so the answer is 2. Note that Vlad could omit some numbers while numerating the cafes.

 思路：

利用Set添加元素，从后到前删除元素，size==1,停止，输出set中的begin元素

代码：

#include <bits/stdc++.h>
using namespace std;
int ans[200005];
set<int> s;
int main() {
    ios::sync_with_stdio(false);
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;++i) {
        scanf("%d",&ans[i]);
        s.insert(ans[i]);
    }
    for(int i=n;i>=1;--i) {
        if(s.size()==1)
            break;
        if(s.find(ans[i])!=s.end()) {
            s.erase(ans[i]);
        }
    }
    printf("%d
",*s.begin());
    return 0;
}