cf 830 D1. Balance (Easy version)

题目
题意: 维护一个集合。给定两种操作，1.插入一个集合中没有的数x；2.查找不在集合中的最小的k的倍数。
思路: 拿一个set维护集合中的数，另外拿map维护每个k上一次枚举到哪里，因为mex随着集合的增加只增不减的。这样复杂度也就n+n/2+n/3…≈nlnn
代码:

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5+10;
int n,m,k,T;
set<ll> sa;
map<ll,ll> mp;
void solve()
{char ch; ll x;cin>>ch>>x;if(ch=='+'){sa.insert(x);}else{for(ll i=mp[x]==0?x:mp[x];;i+=x){if(!sa.count(i)){mp[x] = i;cout<<i<<"\n";return ;}}}
}
signed main(void)
{ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);cin>>T;while(T--)solve();return 0;
}

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