HDU 4217 Data Structure?

简单的线段树

val存当前节点子树的删减后存在的个数

！！！总是忘记算一下int范围够不够  这里要用int64 ！！！ 要注意相加或相乘很容易溢出啊！！！

则每次若左子树val较小  删除的树就在右子树  具体看代码

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

#include <stdio.h>
#include <string.h>
#include <algorithm>

#define MAXN 300006
int n;
struct Node
{
    int val, ld, rd;
}t[MAXN<<2];
void build(int pos, int st, int ed)
{
    t[pos].val = ed-st+1 ;
    t[pos].ld = st, t[pos].rd = ed;
    if(st == ed)
        return ;
    int mid = (st+ed)/2 ;
    build(pos*2, st, mid);
    build(pos*2+1, mid+1, ed);
}

int update(int pos, int key)
{
    t[pos].val --;   //因为key一定在当前节点的子树里 所以删除后当前节点的val要减1
    if(t[pos].ld == t[pos].rd)
        return t[pos].ld;
    if(t[pos*2].val >= key)
        return    update(pos*2, key);
    else
        return update(pos*2+1, key-t[pos*2].val);   //key若在右子树 要减去左子树的val
}

int main()
{
    int i, k, cas, nCase = 0, a;
    scanf("%d", &cas);
    while(cas--)
    {
        __int64 ans = 0;
        scanf("%d %d", &n, &k);    
        build(1, 1, n);
        while(k--)
        {
            scanf("%d", &a);
            ans += update(1, a);
        }
        printf("Case %d: %I64d\n", ++nCase, ans);
    }
    return 0;
}