题目大意：维护一种数据结构，支持以下操作：
1 x 若x不存在,插入x2 x 若x存在,删除x3 输出当前最小值,若不存在输出-14 输出当前最大值,若不存在输出-15 x 输出x的前驱,若不存在输出-16 x 输出x的后继,若不存在输出-17 x 若x存在,输出1,否则输出-1
这题卡Treap，，要写线段树
ZKW大法好啊 可惜我这个沙茶又写挂了……
#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#define M 2100100using namespace std;int n,m,q,tree[M];void Add(int x){x+=q;if(tree[x])return ;for(;x;x>>=1)tree[x]++;}void Del(int x){x+=q;if(!tree[x])return ;for(;x;x>>=1)tree[x]–;}int Get_Min(){int x;if(!tree[1])return 0;for(x=1;x<=q;x=tree[x<<1]?x<<1:x<<1|1);return x-q;}int Get_Max(){int x;if(!tree[1])return 0;for(x=1;x<=q;x=tree[x<<1|1]?x<<1|1:x<<1);return x-q;}int Get_Pred(int x){for(x+=q;x^1;x>>=1)if(x&1&&tree[x>>1]>tree[x])break;if(x==1)return 0;for(x^=1;x<=q;x=tree[x<<1|1]?x<<1|1:x<<1);return x-q;}int Get_Secc(int x){for(x+=q;x^1;x>>=1)if(~x&1&&tree[x>>1]>tree[x])break;if(x==1)return 0;for(x^=1;x<=q;x=tree[x<<1]?x<<1:x<<1|1);return x-q;}int Exsist(int x){return tree[x+q]?1:-1;}int main(){//freopen("3685.in","r",stdin);//freopen("3685.out","w",stdout);int i,p,x;cin>>n>>m;for(q=1;q<=n;q<<=1);for(i=1;i<=m;i++){scanf("%d",&p);if(p==1){scanf("%d",&x);Add(x+1);}if(p==2){scanf("%d",&x);Del(x+1);}if(p==3)printf("%d\n",Get_Min()-1);if(p==4)printf("%d\n",Get_Max()-1);if(p==5){scanf("%d",&x);printf("%d\n",Get_Pred(x+1)-1);}if(p==6){scanf("%d",&x);printf("%d\n",Get_Secc(x+1)-1);}if(p==7){scanf("%d",&x);printf("%d\n",Exsist(x+1));}}}
一起吃早餐，午餐，晚餐。或许吃得不好，
