解题报告

题目传送门

题意：

意思很好理解。

思路：

每次操作是100000次，数据大小100000，又是多组输入。普通模拟肯定不行。

线段树结点记录区间里存在数字的个数，加点删点操作就让该点个数+1，判断x存在就查询[1,x]区间的个数和[1,x-1]的个数。

求x之后第k大的数就先确定小于x的个数t，第t+k小的数就是要求的。

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;
int sum[1000000],ans;
void push_up(int rt) {
    sum[rt]=sum[rt*2]+sum[rt*2+1];
}
void update(int rt,int l,int r,int p,int v) {
    if(l==r) {
        sum[rt]+=v;
        return;
    }
    int mid=(l+r)/2;
    if(p<=mid)
        update(rt*2,l,mid,p,v);
    else update(rt*2+1,mid+1,r,p,v);
    push_up(rt);
}
int q_sum(int rt,int l,int r,int ql,int qr) {
    if(ql>r||qr<l)return 0;
    if(ql<=l&&r<=qr)
        return sum[rt];
    int mid=(l+r)/2;
    return q_sum(rt*2,l,mid,ql,qr)+q_sum(rt*2+1,mid+1,r,ql,qr);
}
void find_rt(int rt,int l,int r,int t) {
    int mid=(l+r)/2;
    if(t>sum[rt*2]) {
        if(l==r) {
            ans=r;
            return ;
        }
        find_rt(rt*2+1,mid+1,r,t-sum[rt*2]);
    } else find_rt(rt*2,l,mid,t);
}
int main() {
    int n,m,i,j,p,e,k;
    while(~scanf("%d",&m)) {
        n=100000;
        memset(sum,0,sizeof(sum));
        for(i=0; i<m; i++) {
            scanf("%d",&p);
            if(p==0) {
                scanf("%d",&e);
                update(1,1,n,e,1);
            } else if(p==1) {
                scanf("%d",&e);
                if(q_sum(1,1,n,1,e)==q_sum(1,1,n,1,e-1))
                    printf("No Elment!\n");
                else update(1,1,n,e,-1);
            } else if(p==2) {
                scanf("%d%d",&e,&k);
                int t=q_sum(1,1,n,1,e)+k;
                find_rt(1,1,n,t);
                if(ans==n)
                    printf("Not Find!\n");
                else
                    printf("%d\n",ans);
            }
        }
    }
    return 0;
}

KiKi‘s K-Number

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem 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 4

Sample Output

No Elment!
6
Not Find!
2
2
4
Not Find!

Source

2009 Multi-University Training
Contest 4 - Host by HDU

HDU2852_KiKi's K-Number(线段树/单点更新),布布扣,bubuko.com

HDU2852_KiKi's K-Number(线段树/单点更新)