题意：
设计一个容器，支持插入x，若与容器中的值最小相差为k，则自动忽略。删除操作，把与x相差为k的值都从容器中删除。查询操作，问容器中有没有和x相差为k的数值。
思路：
一个stl中的set 加上cmp结构体重载就搞定了。神奇。
#include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <string> #include <vector> #include <map> #include <set> #include <queue> #include <list> #include <cstdlib> #include <iterator> #include <cmath> #include <iomanip> #include <bitset> #include <cctype> using namespace std; //#pragma GCC optimize(3) //#pragma comment(linker, "/STACK:102400000,102400000") //c++ #define lson (l , mid , rt << 1) #define rson (mid + 1 , r , rt << 1 | 1) #define debug(x) cerr << #x << " = " << x << " "; #define pb push_back #define pq priority_queue typedef long long ll; typedef unsigned long long ull; typedef pair<ll ,ll > pll; typedef pair<int ,int > pii; typedef pair<int ,pii> p3; //priority_queue<int> q;//这是一个大根堆q //priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q #define fi first #define se second //#define endl ' ' #define OKC ios::sync_with_stdio(false);cin.tie(0) #define FT(A,B,C) for(int A=B;A <= C;++A) //用来压行 #define REP(i , j , k) for(int i = j ; i < k ; ++i) //priority_queue<int ,vector<int>, greater<int> >que; const ll mos = 0x7FFFFFFFLL; //2147483647 const ll nmos = 0x80000000LL; //-2147483648 const int inf = 0x3f3f3f3f; const ll inff = 0x3f3f3f3f3f3f3f3fLL; //18 const double PI=acos(-1.0); template<typename T> inline T read(T&x){ x=0;int f=0;char ch=getchar(); while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar(); while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); return x=f?-x:x; } // #define _DEBUG; //*// #ifdef _DEBUG freopen("input", "r", stdin); // freopen("output.txt", "w", stdout); #endif /*-----------------------show time----------------------*/ int n,k; char str[10]; struct cmp { bool operator()(int a,int b)const { if(abs(a-b) <= k)return false; return a < b; } }; set<int,cmp>s; int main(){ scanf("%d%d", &n, &k); for(int i=1; i<=n; i++){ int x; scanf("%s%d", str, &x); if(str[0] == 'a'){ if(s.find(x) == s.end()) s.insert(x); } else if(str[0] == 'q'){ if(s.find(x) == s.end()) puts("No"); else puts("Yes"); } else { while(s.find(x) != s.end()){ s.erase(x); } } } return 0; }