Data Structure?

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

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
 思路:taobanzi;
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
const int N=3e5+,M=4e6+,inf=1e9+,mod=1e9+;
const ll INF=1e18+;
int tree[N],n,k;
int lowbit(int x)
{
return x&-x;
}
void update(int x,int change)
{
while(x<=n)
{
tree[x]+=change;
x+=lowbit(x);
}
}
int k_thfind(int K)//树状数组求第K小
{
int sum=;
for(int i=;i>=;i--)
{
if(sum+(<<i)<=n&&tree[sum+(<<i)]<K)
{
K-=tree[sum+(<<i)];
sum+=<<i;
}
}
return sum+;
}
int main(){
int T,cas=;
scanf("%d",&T);
while(T--)
{
memset(tree,,sizeof(tree));
scanf("%d%d",&n,&k);
for(int i=;i<=n;i++)
update(i,);
ll ans=;
for(int i=;i<k;i++)
{
int z;
scanf("%d",&z);
int v=k_thfind(z);
ans+=v;
update(v,-);
}
printf("Case %d: %lld\n",cas++,ans);
}
return ;
}

hdu 4217 Data Structure? 树状数组求第K小的更多相关文章

  1. 树状数组求第k小的元素

    int find_kth(int k) { int ans = 0,cnt = 0; for (int i = 20;i >= 0;i--) //这里的20适当的取值,与MAX_VAL有关,一般 ...

  2. 树状数组求第K小值 (spoj227 Ordering the Soldiers &amp;&amp; hdu2852 KiKi&#39;s K-Number)

    题目:http://www.spoj.com/problems/ORDERS/ and pid=2852">http://acm.hdu.edu.cn/showproblem.php? ...

  3. UVA11525 Permutation[康托展开 树状数组求第k小值]

    UVA - 11525 Permutation 题意:输出1~n的所有排列,字典序大小第∑k1Si∗(K−i)!个 学了好多知识 1.康托展开 X=a[n]*(n-1)!+a[n-1]*(n-2)!+ ...

  4. *HDU2852 树状数组(求第K小的数)

    KiKi's K-Number Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)T ...

  5. poj 2985 The k-th Largest Group 树状数组求第K大

    The k-th Largest Group Time Limit: 2000MS   Memory Limit: 131072K Total Submissions: 8353   Accepted ...

  6. HDU 5249 离线树状数组求第k大+离散化

    KPI Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submiss ...

  7. hdu 5147 Sequence II (树状数组 求逆序数)

    题目链接 Sequence II Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) ...

  8. POJ2985 The k-th Largest Group[树状数组求第k大值+并查集||treap+并查集]

    The k-th Largest Group Time Limit: 2000MS   Memory Limit: 131072K Total Submissions: 8807   Accepted ...

  9. hdu 2838 Cow Sorting 树状数组求所有比x小的数的个数

    Cow Sorting Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total ...

随机推荐

  1. Redis常用命令速查 02_转

    一.Key Key命令速查: 命令 说明 DEL 删除给定的一个或多个 key,不存在的 key 会被忽略,返回值:被删除 key 的数量 DUMP 序列化给定 key,返回被序列化的值,使用 RES ...

  2. css改变背景透明度【转】

    兼容主流浏览器的CSS透明代码: .transparent_class { filter:alpha(opacity=50); -moz-opacity:0.5; -khtml-opacity: 0. ...

  3. 手机端上传未知图片大小,js设置宽高比例

    <style rel="stylesheet" type="text/css"> .lunboimg{ width: 100%; height: a ...

  4. jquery 当前链接激活传递参数|div的切换显示

    一.链接激活时传递参数 $("a").click(function(){ var obj=$(this).attr("field"); //获取当前field ...

  5. ACM题目————STL练习之字符串替换

    描述 编写一个程序实现将字符串中的所有"you"替换成"we" 输入 输入包含多行数据 每行数据是一个字符串,长度不超过1000 数据以EOF结束 输出 对于输 ...

  6. 使用Window Live Writer写博客

    1.打开“日志账户”—>“日志选项”. 2.点击“更新账户信息”. 3.输入博客地址,用户名和密码,点击“下一步”. 4.耐心等待片刻... 5.设置“日志昵称”,点击“完成”. 这样就大功告成 ...

  7. 使用repeater实现gridview的功能

    <asp:Repeater ID="rptfindData" runat="server"> <HeaderTemplate> < ...

  8. oracle中的自动增长

    create table test( id int not null primary key, name varchar2(20), sex int) ; create sequence t -> ...

  9. HashMap, HashTable, CurrentHashMap的区别

    转载:http://www.jianshu.com/p/c00308c32de4 HashMap vs ConcurrentHashMap 引入ConcurrentHashMap是为了在同步集合Has ...

  10. 基于@AspectJ和schema的aop(一)

    在前面我们使用Pointcut和Advice描述切点和增强, 并使用Advisor整合两者描述切面.@AspectJ使用注解来描述切点和增强.两者使用的方式不同, 但是在本质上都是一样的. 我们还是用 ...