stl（set或map）

https://nanti.jisuanke.com/t/41384

There are nnn points in an array with index from 111 to nnn, and there are two operations to those points.

1: 1 x1 \ x1 x marking the point xxx is not available

2: 2 x2 \ x2 x query for the index of the first available point after that point (including xxx itself) .

Input

nqn\quad qnq

z1x1z_1 \quad x_1z1​x1​

⋮\vdots⋮

zqxqz_q\quad x_qzq​xq​

qqq is the number of queries, zzz is the type of operations, and xxx is the index of operations. 1≤x<n<1091≤x<n<10^91≤x<n<109， 1≤q<1061 \leq q<10^6 1≤q<106 and zzz is 111 or 222

Output

Output the answer for each query.

样例输入

5 3
1 2
2 2
2 1

样例输出

3
1

题意：给出1-n，q个操作
第一个操作（1）：将某个值标记为不可用；
第二个操作（2）：输出该值以后的第一个有用值。
神奇的编译器：c++14超时，c++11就过辽。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <set>
using namespace std;

#define mem(a) memset(a,0,sizeof(a))
#define ll long long
const int maxn = ;
int n,q,a,b;
set<int>s;
int solve()
{
    while(b <= n)
    {
        if(s.find(b) != s.end())
            b++;
        else return b;
    }
    return -;
}

int main()
{
    scanf("%d%d",&n,&q);
    for(int i = ; i < q; i++)
    {
        scanf("%d%d",&a,&b);
        if(a == )
        {
            s.insert(b);
        }
        else
        {
            printf("%d\n",solve());
            // 也可以s.count判断是否存在。
            /*while(s.count(b))
            {
                b++;
            }
            printf("%d\n" , b);*/

        }
    }
    return ;
}

map会超时要用unordered_map

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <set>
#include <unordered_map>
using namespace std;

#define mem(a) memset(a,0,sizeof(a))
#define ll long long
const int maxn = ;
int n,q,a,b;

int main()
{
    unordered_map<int , int>mp;
    scanf("%d%d",&n,&q);
    for(int i = ; i < q; i++)
    {
        scanf("%d%d",&a,&b);
        if(a == )
        {
            mp[b] = - ;
        }
        else
        {
            while(mp[b] == -) b++ ;
            printf("%d\n" , b);
        }
    }
    return ;
}