Hdu 4217 Data Structure?【二分+树状数组】

Data Structure?

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3359    Accepted Submission(s): 1077

Problem Description

Data structure is one of the basic skills for Computer Science students, which is a particular way of storing and organizing data in a computer so that it can be used efficiently. Today let me introduce a data-structure-like problem for you.
Original, there are N numbers, namely 1, 2, 3...N. Each round, iSea find out the Ki-th smallest number and take it away, your task is reporting him the total sum of the numbers he has taken away.

 

Input

The first line contains a single integer T, indicating the number of test cases.
Each test case includes two integers N, K, K indicates the round numbers. Then a line with K numbers following, indicating in i (1-based) round, iSea take away the Ki-th smallest away.

Technical Specification
1. 1 <= T <= 128
2. 1 <= K <= N <= 262 144
3. 1 <= Ki <= N - i + 1

 

Output

For each test case, output the case number first, then the sum.

 

Sample Input

23 21 110 33 9 1

 

Sample Output

Case 1: 3Case 2: 14

 

Author

iSea@WHU

 

Source

首届华中区程序设计邀请赛暨第十届武汉大学程序设计大赛

题目大意：

给你N个数，从1~N，每次从中拿第ki小的数，然后去掉这个数，一共拿K次，问累加的和是多少。

思路：

一开始我们让数组都设定为1，表示各个数都没有被拿过，然后我们每一次拿的时候，二分一下树上剩余的第k个数的位子就行了。

过程维护一下即可。

Ac代码：

#include<stdio.h>#include<string.h>using namespace std;int tree[400005];int n,k;int lowbit(int x){    return x&(-x);}int sum(int x){    int sum=0;    while(x>0)    {        sum+=tree[x];        x-=lowbit(x);    }    return sum;}void add(int x,int c){    while(x<=n)    {        tree[x]+=c;        x+=lowbit(x);    }}int main(){    int t;    int kase=0;    scanf("%d",&t);    while(t--)    {        memset(tree,0,sizeof(tree));        scanf("%d%d",&n,&k);        for(int i=1;i<=n;i++)add(i,1);        long long int output=0;        for(int i=0;i<k;i++)        {            int x;scanf("%d",&x);            int l=1;            int r=n;            int pos=-1;            while(r-l>=0)            {                int mid=(l+r)/2;                if(sum(mid)>=x)                {                    if(sum(mid)==x)pos=mid;                    r=mid-1;                }                else l=mid+1;            }            if(pos==-1)continue;            else output+=pos,add(pos,-1);        }        printf("Case %d: %lld\n",++kase,output);    }}