链接：

题意：

AFei has many cards. Each card has a number written on it. Now he wants to takes some out of his card and puts them in a box. And he wants to know whether the card with the number x was in the box. So he has the following two operations：

• 0 x （It means to put a card with the number x in the box.）
• 1 x   (It means to query if there is a card with the number x in the box.

思路：

map超时。。离散化，离线处理。

代码：

#include 
      
        using namespace std; typedef long long LL;const int MAXN = 1e6 + 10;const int MOD = 1e9 + 7;int n, m, k, t;struct Node{    int v, pos;    bool operator < (const Node& that) const    {        return this->v < that.v;    }}node[MAXN];int a[MAXN];int vis[MAXN];int op[MAXN]; int main(){    scanf("%d", &n);    for (int i = 1;i <= n;i++)    {        scanf("%d%d", &op[i], &node[i].v);        node[i].pos = i;    }    sort(node+1, node+1+n);    int cnt = 1;    a[node[1].pos] = cnt;    for (int i = 2;i <= n;i++)    {        if (node[i].v == node[i-1].v)            a[node[i].pos] = cnt;        else            a[node[i].pos] = ++cnt;    }    for (int i = 1;i <= n;i++)    {        if (op[i] == 0)            vis[a[i]] = 1;        else            if (vis[a[i]] == 1)                printf("yes\n");            else                printf("no\n");    }     return 0;}