In the capital city of Berland, Bertown, demonstrations are against the recent election of the King of Berland. Berland opposition, led by Mr. Ovalny, believes that the elections were not fair enough and wants to organize a demonstration at one of the squares.

Bertown has n squares, numbered from 1 to n, they are numbered in the order of increasing distance between them and the city center. That is, square number 1 is central, and square number n is the farthest from the center. Naturally, the opposition wants to hold a meeting as close to the city center as possible (that is, they want an square with the minimum number).

There are exactly k (k < n) days left before the demonstration. Now all squares are free. But the Bertown city administration never sleeps, and the approval of an application for the demonstration threatens to become a very complex process. The process of approval lasts several days, but every day the following procedure takes place:

  • The opposition shall apply to hold a demonstration at a free square (the one which isn't used by the administration).
  • The administration tries to move the demonstration to the worst free square left. To do this, the administration organizes some long-term activities on the square, which is specified in the application of opposition. In other words, the administration starts using the square and it is no longer free. Then the administration proposes to move the opposition demonstration to the worst free square. If the opposition has applied for the worst free square then request is accepted and administration doesn't spend money. If the administration does not have enough money to organize an event on the square in question, the opposition's application is accepted. If administration doesn't have enough money to organize activity, then rest of administration's money spends and application is accepted
  • If the application is not accepted, then the opposition can agree to the administration's proposal (that is, take the worst free square), or withdraw the current application and submit another one the next day. If there are no more days left before the meeting, the opposition has no choice but to agree to the proposal of City Hall. If application is accepted opposition can reject it. It means than opposition still can submit more applications later, but square remains free.

In order to organize an event on the square i, the administration needs to spend ai bourles. Because of the crisis the administration has only b bourles to confront the opposition. What is the best square that the opposition can take, if the administration will keep trying to occupy the square in question each time? Note that the administration's actions always depend only on the actions of the opposition.

Input

The first line contains two integers n and k — the number of squares and days left before the meeting, correspondingly (1 ≤ k < n ≤ 105).

The second line contains a single integer b — the number of bourles the administration has (1 ≤ b ≤ 1018).

The third line contains n space-separated integers ai — the sum of money, needed to organise an event on square i (1 ≤ ai ≤ 109).

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Output

Print a single number — the minimum number of the square where the opposition can organize the demonstration.

Examples

Input
5 2
8
2 4 5 3 1
Output
2
Input
5 2
8
3 2 4 1 5
Output
5
Input
5 4
1000000000000000
5 4 3 2 1
Output
5

Note

In the first sample the opposition can act like this. On day one it applies for square 3. The administration has to organize an event there and end up with 3 bourles. If on the second day the opposition applies for square 2, the administration won't have the money to intervene.

In the second sample the opposition has only the chance for the last square. If its first move occupies one of the first four squares, the administration is left with at least 4 bourles, which means that next day it can use its next move to move the opposition from any square to the last one.

In the third sample administration has a lot of money, so opposition can occupy only last square.

OJ-ID:
CodeForce 192D

author:
Caution_X

date of submission:
20191017

tags:
贪心

description modelling:
n个广场,每个广场都有相应的使用费,现在反对派有m天时间,政府有k块钱,反对派想要尽可能使用点数小的广场,现在反对派每天可以申请一个广场,政府则会用手上的钱租用这个广场(除非政府没钱或者这个广场是最后一个广场),问反对派可以拿到的最好的广场。

major steps to solve it:
(1).为了结果最优,假定反对派在最后一天到来之前都挑选最贵的广场
(2).到了最后一天:从广场1开始遍历,如果我们选择的这个广场在之前还没有被使用,那么这个广场就是最优广场,如果这个广场在之前已经被使用过了,那么我们就找找有没有还没有使用过的广场(序号偏后)可以替代这个广场。

AC code:

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll a[],b[];
bool cmp(ll a,ll b)
{
return a>b;
}
int main()
{
ll n,m,tot;
scanf("%lld%lld%lld",&n,&m,&tot);
for(int i=;i<=n;i++) {
scanf("%lld",&a[i]);
}
ll sum=;
for(int i=;i<n;i++) {
b[i]=a[i];
}
sort(b+,b+n,cmp);
for(int i=;i<m;i++) {
sum+=b[i];
}
for(int i=;i<=n;i++) {
if(a[i]>=b[m]) {
if(sum+b[m]>tot) {
printf("%d\n",i);
return ;
}
}
else if(sum+a[i]>tot){
printf("%d\n",i);
return ;
}
}
printf("%d\n",n);
return ;
}

CodeForce 192D Demonstration的更多相关文章

  1. Codeforce - Street Lamps

    Bahosain is walking in a street of N blocks. Each block is either empty or has one lamp. If there is ...

  2. Codeforce Round #216 Div2

    e,还是写一下这次的codeforce吧...庆祝这个月的开始,看自己有能,b到什么样! cf的第二题,脑抽的交了错两次后过了pretest然后system的挂了..脑子里还有自己要挂的感觉,果然回头 ...

  3. Codeforce 水题报告(2)

    又水了一发Codeforce ,这次继续发发题解顺便给自己PKUSC攒攒人品吧 CodeForces 438C:The Child and Polygon: 描述:给出一个多边形,求三角剖分的方案数( ...

  4. codeforce 375_2_b_c

    codeforce 375_2 标签: 水题 好久没有打代码,竟然一场比赛两次卡在边界条件上....跪 b.题意很简单...纯模拟就可以了,开始忘记了当字符串结束的时候也要更新两个值,所以就错了 #i ...

  5. codeforce 367dev2_c dp

    codeforce 367dev2_c dp 标签: dp 题意: 你可以通过反转任意字符串,使得所给的所有字符串排列顺序为字典序,每次反转都有一定的代价,问你最小的代价 题解:水水的dp...仔细想 ...

  6. 三维dp&codeforce 369_2_C

    三维dp&codeforce 369_2_C 标签: dp codeforce 369_2_C 题意: 一排树,初始的时候有的有颜色,有的没有颜色,现在给没有颜色的树染色,给出n课树,用m种燃 ...

  7. 强连通分量&hdu_1269&Codeforce 369D

    强连通分量 标签: 图论 算法介绍 还记得割点割边算法吗.回顾一下,tarjan算法,dfs过程中记录当前点的时间戳,并通过它的子节点的low值更新它的low,low值是这个点不通过它的父亲节点最远可 ...

  8. Demonstration of DB Query Analyzer 6.03 Installation and Running on Microsoft Windows 8

    Demonstration of DB Query Analyzer 6.03 Installation and Running on Microsoft Windows 8 Ma Genfeng ( ...

  9. 【树状数组】区间出现偶数次数的异或和(区间不同数的异或和)@ codeforce 703 D

    [树状数组]区间出现偶数次数的异或和(区间不同数的异或和)@ codeforce 703 D PROBLEM 题目描述 初始给定n个卡片拍成一排,其中第i个卡片上的数为x[i]. 有q个询问,每次询问 ...

随机推荐

  1. php laravel请求处理管道(装饰者模式)

    laravel的中间件使用了装饰者模式.比如,验证维护模式,cookie加密,开启会话等等.这些处理有些在响应前,有些在响应之后,使用装饰者模式动态减少或增加功能,使得框架可扩展性大大增强. 接下来简 ...

  2. 关于C#异常的处理

    关于异常,从我们一开始写代码的时候就开始伴随着我们,只不过那时还没入门,脑子里并没有产生意识这就是异常.异常:程序运行期间发生错误,异常对象: 将程序发生的各种错误封装成对象曾记得第一次面试的时候,面 ...

  3. 使用ADO.NET实体数据模型

    前景:要操作的数据表必须添加主键(方式:进入数据库-->数据表名-->设计-->列名右键-->设置主键) 可在服务器资源管理器中查看是否设置了主键(主键会有一把钥匙的图样) 1 ...

  4. go实现整型的二进制转化

    go中已经实现了int->bin的转化函数,我这里只是化过程逻辑的实现,至于原理我就假设大家都知道了 本案例只考虑 int->bin  的转化 包含了正整数,负整数,0 的转化 packa ...

  5. arcgis api 3.x for js 解决 textSymbol 文本换行显示(附源码下载)

    前言 关于本篇功能实现用到的 api 涉及类看不懂的,请参照 esri 官网的 arcgis api 3.x for js:esri 官网 api,里面详细的介绍 arcgis api 3.x 各个类 ...

  6. [b0029] python 归纳 (十四)_队列Queue实现生产者消费者

    # -*- coding: UTF-8 -*- """ 多线程的生产者,消费者 使用队列Queue """ import Queue imp ...

  7. 我写了个项目,帮你学习HTTP接口测试!

    端午节我写了一个项目,帮助你学习HTTP接口测试. GitHub地址: https://github.com/defnngj/learning-API-test 整个项目基于Flask和 Reques ...

  8. Python创建virtualenv虚拟环境方法

    一.通过virtualenv软件创建 安装:        -pip3 install virtualenv    创建虚拟环境:        -(1)virtualenv wannings-ms- ...

  9. 如何将MagicaVoxel模型导入UE4中(1)

    前言 当初在选择自己项目的美术风格时,由于自己的美术基础实在是太差,所以选择了体素风格来构建(其实还是MagicaVoxel的建模操作很容易上手),但是将自己千辛万苦做好的模型导入至项目中时,出现了这 ...

  10. Jmeter设置默认中文启动

    安装好jmeter,启动的页面是全英文,对于不熟悉jmeter的人来说,理解起来还是很困难的.如何设置成中文页面呢? 方法一:打开jmeter后,设置Options—choose language—C ...