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. scala泛型

    package com.ming.test /** * scala泛型 * 类型参数测试 */ object TypeParamsTest { //泛型函数 def getMiddle[T](a:Ar ...

  2. linux内核参数优化

    net.ipv4.ip_forward = 0net.ipv4.conf.default.rp_filter = 1net.ipv4.conf.default.accept_source_route ...

  3. android 开发中的常见问题

    Android studio 使用极光推送, 显示获取sdk版本失败 在 build.gradle(Module.app) 添加 android {    sourceSets.main {      ...

  4. Java 基本数据类型 sizeof 功能【转】

    转自:http://blog.csdn.net/sunboy_2050/article/details/7310008 版权声明:本文为博主原创文章,未经博主允许不得转载. Java基本数据类型int ...

  5. 坑爹的VS2012

    2.2.2.如果卸载 Visual Studio 2010 Service Pack 1,则必须先重新安装 Visual Studio 2010,然后才能再次安装 SP1 如果卸载 Visual St ...

  6. Pro ASP.NET MVC 5 Framework.学习笔记.6.3.MVC的必备工具

    每个MVC程序员的军火库中,都有这三个工具:一个依赖注入(DI)容器,一个单元测试框架,一个模拟工具. 1.准备一个示例项目 创建一个ASP.NET MVC Web Application的Empty ...

  7. export

    export export PATH=$PATH:/ROOT

  8. php const define 区别有那些呢?

    (1) 编译器处理方式不同 define宏是在预处理阶段展开. const常量是编译运行阶段使用. (2) 类型和安全检查不同 define宏没有类型,不做任何类型检查,仅仅是展开. const常量有 ...

  9. [c++][语言语法]函数模板和模板函数 及参数类型的运行时判断

    参考:http://blog.csdn.net/beyondhaven/article/details/4204345 参考:http://blog.csdn.net/joeblackzqq/arti ...

  10. css杂记

    1,font-variant: 设置文本是否为小型的大写字母,值可以为normal,small-caps; 2,a:link: 未访问过的 a:visited: 访问过的 a:active: 活动的链 ...