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

2
3 2
1 1
10 3
3 9 1

 
Sample Output

Case 1: 3
Case 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
#include
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        {
            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);
    }
}