思路：三种操作，增加一个数，减少一个数，询问大于某个数的第k个数
第一种操作就是普通的update，第二个询问前先求query(x)-query(x-1)是否为0，为0就是没有这个元素，第三个操作大于某个数的第K个数相当于求第query(x)+k个数，那么二分一下就可以了
#include<bits/stdc++.h> using namespace std; const int maxn = 100000+50; int c[maxn]; int lowbit(int x){return x&(-x);} void update(int i,int v) { while(i<=maxn) { c[i]+=v; i+=lowbit(i); } } int query(int i) { int ans = 0; while(i) { ans+=c[i]; i-=lowbit(i); } return ans; } int find(int k) { int ans = maxn; int l = 1,r=maxn; while(l<=r) { int mid = (l+r)>>1; if(query(mid)>=k) ans=mid,r=mid-1; else l=mid+1; } return ans; } int main() { int q; while(scanf("%d",&q)!=EOF) { memset(c,0,sizeof(c)); while(q--) { int op,x; scanf("%d%d",&op,&x); if(op==0) update(x,1); else if(op==1) { if(query(x)-query(x-1)==0) puts("No Elment!"); else update(x,-1); } else { int k; scanf("%d",&k); if(query(maxn)-query(x) < k) puts("Not Find!"); else printf("%d\n",find(k+query(x))); } } } }
Description
For the k-th number, we all should be very familiar with it. Of course,to kiki it is also simple. Now Kiki meets a very similar problem, kiki wants to design a container, the container is to support the three operations. Push: Push a given element e to container Pop: Pop element of a given e from container Query: Given two elements a and k, query the kth larger number which greater than a in container; Although Kiki is very intelligent, she can not think of how to do it, can you help her to solve this problem?Input
Input some groups of test data ,each test data the first number is an integer m (1 <= m <100000), means that the number of operation to do. The next m lines, each line will be an integer p at the beginning, p which has three values: If p is 0, then there will be an integer e (0 <e <100000), means press element e into Container. If p is 1, then there will be an integer e (0 <e <100000), indicated that delete the element e from the container If p is 2, then there will be two integers a and k (0 <a <100000, 0 <k <10000),means the inquiries, the element is greater than a, and the k-th larger number.Output
For each deletion, if you want to delete the element which does not exist, the output "No Elment!". For each query, output the suitable answers in line .if the number does not exist, the output "Not Find!".Sample Input
5 0 5 1 2 0 6 2 3 2 2 8 1 7 0 2 0 2 0 4 2 1 1 2 1 2 2 1 3 2 1 4Sample Output
No Elment! 6 Not Find! 2 2 4 Not Find!