【HDU - 4217 】Data Structure? （线段树求第k小数）

题干：

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

解题报告：

      这题是暑假集训的时候的线段树的题，用线段树维护在某一个区间还剩下几个数。跟这题对比【HDU - 4006】The kth great number

AC代码：

//又忘了开longlong了吧。。。
#include<bits/stdc++.h>

using namespace std;
const int MAX =262144 + 100000;
int n,m;

struct TREE {
    int l,r;
    int val;
} tree[MAX*4];//数组要不要再大一点？

void pushup(int cur) {
    tree[cur].val = tree[cur*2].val+ tree[cur*2+1].val;
}
void build(int l,int r,int cur) {
    tree[cur].l = l;
    tree[cur].r = r;
    if(tree[cur].l == tree[cur].r) {
        tree[cur].val = 1;return ;
    }
    int m = (l + r)/ 2;
    build(l,m,cur*2);
    build(m+1,r,cur*2+1);
    pushup(cur);

}
int query(int k,int cur) {
    if(tree[cur].l == tree[cur].r){
 //       tree[cur].val = 0;
        return tree[cur].l;
    }
    if(k > tree[cur*2].val) return query(k-tree[cur*2].val,cur*2+1);
    else return query(k,cur*2);
}
void update(int tar,int cur) {
    if(tree[cur].l == tree[cur].r) {
        tree[cur].val = 0;
        return ;
    }
    if(tar <= tree[cur*2].r) update(tar,2*cur);
    else update(tar,2*cur+1);
    pushup(cur);
}

int main()
{
    int t,tmp;
    int k,iCase = 0;;
    long long ans = 0;
    cin>>t;
    while(t--) {
        //初始化
        ans = 0;

        scanf("%d%d",&n,&m);
        build(1,n,1);
        while(m--) {
            scanf("%d",&k);
            tmp = query(k,1);
            update(tmp,1);
            ans += tmp;

        }
        printf("Case %d: %lld\n",++iCase,ans);
    }

    return 0 ;
}