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?
InputInput 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.
OutputFor 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 4
Sample Output
No Elment! 6 Not Find! 2 2 4 Not Find!
题解：题目有3种操作，0 ： 插入一个数； 1 ：删除一个数(如果不存在，就输出“No Elment!”), 2 : 查询大于 a 的第K小的数，没有就输出"Not Find!" ；
用可持久化线段树，看代码即可明白几乎是模版；
1 #include<bits/stdc++.h> 2 using namespace std; 3 const int maxn=1e5+10; 4 int n,tot,rt[maxn*24]; 5 6 struct Node{ 7 int l,r,sum; 8 } tree[maxn*24]; 9 10 inline void build(int l,int r,int &pos) 11 { 12 pos = ++tot; 13 tree[pos].sum = 0; 14 if(l==r) return; 15 int m = (l+r) >> 1; 16 build(l,m,tree[pos].l); 17 build(m+1,r,tree[pos].r); 18 } 19 20 inline void update(int p,int c,int pre,int l,int r,int &pos) 21 { 22 pos=++tot; tree[pos]=tree[pre]; tree[pos].sum+=c; 23 if(l==r)return; 24 int mid=(l+r)>>1; 25 if(p<=mid) update(p,c,tree[pre].l,l,mid,tree[pos].l); 26 else update(p,c,tree[pre].r,mid+1,r,tree[pos].r); 27 } 28 29 inline int Rank(int s,int e,int L,int R,int l,int r) 30 { 31 if(L<=l&&r<=R) return tree[e].sum-tree[s].sum; 32 int mid=(l+r)>>1; 33 int ans=0; 34 if(L<=mid) ans+=Rank(tree[s].l,tree[e].l,L,R,l,mid); 35 if(R>mid) ans+=Rank(tree[s].r,tree[e].r,L,R,mid+1,r); 36 return ans; 37 } 38 39 inline int query(int L,int R,int l,int r,int k) 40 { 41 if(l==r)return l; 42 int mid=(l+r)>>1; 43 int x=tree[tree[R].l].sum-tree[tree[L].l].sum; 44 if(k<=x) return query(tree[L].l,tree[R].l,l,mid,k); 45 else return query(tree[L].r,tree[R].r,mid+1,r,k-x); 46 } 47 48 int main() 49 { 50 int right=maxn; 51 while(~scanf("%d",&n)) 52 { 53 tot=0; 54 build(1,right,rt[0]); 55 for(int i=1;i<=n;i++) 56 { 57 int x; 58 scanf("%d",&x); 59 if(x==0) 60 { 61 int z; 62 scanf("%d",&z); 63 update(z,1,rt[i-1],1,right,rt[i]); 64 } 65 else if(x==1) 66 { 67 int z; 68 scanf("%d",&z); 69 if(Rank(rt[0],rt[i-1],z,z,1,right)) update(z,-1,rt[i-1],1,right,rt[i]); 70 else 71 { 72 update(z,0,rt[i-1],1,right,rt[i]); 73 printf("No Elment! "); 74 } 75 } 76 else 77 { 78 update(1,0,rt[i-1],1,right,rt[i]); 79 int a,k; 80 scanf("%d%d",&a,&k); 81 int v=Rank(rt[0],rt[i],1,a,1,right); 82 k+=v; 83 int q=Rank(rt[0],rt[i],1,right,1,right); 84 if(k>q) printf("Not Find! "); 85 else printf("%d ",query(rt[0],rt[i],1,right,k)); 86 } 87 } 88 } 89 return 0; 90 }