C. Anya and Ghosts
time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Anya loves to watch horror movies. In the best traditions of horror, she will be visited by m ghosts tonight. Anya has lots of candles prepared for the visits, each candle can produce light for exactly t seconds. It takes the girl one second to light one candle. More formally, Anya can spend one second to light one candle, then this candle burns for exactly t seconds and then goes out and can no longer be used.

For each of the m ghosts Anya knows the time at which it comes: the i-th visit will happen wi seconds after midnight, all wi's are distinct. Each visit lasts exactly one second.

What is the minimum number of candles Anya should use so that during each visit, at least r candles are burning? Anya can start to light a candle at any time that is integer number of seconds from midnight, possibly, at the time before midnight. That means, she can start to light a candle integer number of seconds before midnight or integer number of seconds after a midnight, or in other words in any integer moment of time.

Input

The first line contains three integers mtr (1 ≤ m, t, r ≤ 300), representing the number of ghosts to visit Anya, the duration of a candle's burning and the minimum number of candles that should burn during each visit.

The next line contains m space-separated numbers wi (1 ≤ i ≤ m, 1 ≤ wi ≤ 300), the i-th of them repesents at what second after the midnight the i-th ghost will come. All wi's are distinct, they follow in the strictly increasing order.

Output

If it is possible to make at least r candles burn during each visit, then print the minimum number of candles that Anya needs to light for that.

If that is impossible, print  - 1.

Examples
input
1 8 3
10
output
3
input
2 10 1
5 8
output
1
input
1 1 3
10
output
-1
Note

Anya can start lighting a candle in the same second with ghost visit. But this candle isn't counted as burning at this visit.

It takes exactly one second to light up a candle and only after that second this candle is considered burning; it means that if Anya starts lighting candle at moment x, candle is buring from second x + 1 to second x + t inclusively.

In the first sample test three candles are enough. For example, Anya can start lighting them at the 3-rd, 5-th and 7-th seconds after the midnight.

In the second sample test one candle is enough. For example, Anya can start lighting it one second before the midnight.

In the third sample test the answer is  - 1, since during each second at most one candle can burn but Anya needs three candles to light up the room at the moment when the ghost comes.

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
const int N=1e5+,M=4e6+,inf=1e9+,mod=1e9+;
const ll INF=1e18+;
int a[N];
int c[N];
int main()
{
int m,t,r;
scanf("%d%d%d",&m,&t,&r);
for(int i=;i<=m;i++)
scanf("%d",&a[i]);
if(t<r)return puts("-1");
int flag=;
for(int i=r;i>=;i--)
c[flag++]=a[]-i;
int ans=r;
for(int i=;i<=m;i++)
{
int j;
for(j=;j<=r;j++)
if(c[j]+t+>a[i])break;
flag=;
for(int k=j;k<=r;k++)
c[flag++]=c[k];
int p=r-j+;
p=r-p;
if(p>)
{
ans+=p;
j=p;
for(int k=;k<j;k++)
c[flag++]=a[i]-p,p--;
}
}
printf("%d\n",ans);
return ;
}

Codeforces Round #288 (Div. 2) C. Anya and Ghosts 模拟的更多相关文章

  1. Codeforces Round #288 (Div. 2) C. Anya and Ghosts 模拟 贪心

    C. Anya and Ghosts time limit per test 2 seconds memory limit per test 256 megabytes input standard ...

  2. 贪心+模拟 Codeforces Round #288 (Div. 2) C. Anya and Ghosts

    题目传送门 /* 贪心 + 模拟:首先,如果蜡烛的燃烧时间小于最少需要点燃的蜡烛数一定是-1(蜡烛是1秒点一支), num[g[i]]记录每个鬼访问时已点燃的蜡烛数,若不够,tmp为还需要的蜡烛数, ...

  3. 贪心 Codeforces Round #288 (Div. 2) B. Anton and currency you all know

    题目传送门 /* 题意:从前面找一个数字和末尾数字调换使得变成偶数且为最大 贪心:考虑两种情况:1. 有偶数且比末尾数字大(flag标记):2. 有偶数但都比末尾数字小(x位置标记) 仿照别人写的,再 ...

  4. Codeforces Round #297 (Div. 2)E. Anya and Cubes 折半搜索

    Codeforces Round #297 (Div. 2)E. Anya and Cubes Time Limit: 2 Sec  Memory Limit: 512 MBSubmit: xxx  ...

  5. Codeforces Round #367 (Div. 2) B. Interesting drink (模拟)

    Interesting drink 题目链接: http://codeforces.com/contest/706/problem/B Description Vasiliy likes to res ...

  6. Codeforces Round #288 (Div. 2)

    A. Pasha and Pixels     题意就是给一个n*m的矩阵,k次操作,一开始矩阵全白,一次操作可以染黑一个格子,问第几次操作可以使得矩阵中存在一个2*2的黑色矩阵.直接模拟即可 代码: ...

  7. Codeforces Round #288 (Div. 2) E. Arthur and Brackets

    题目链接:http://codeforces.com/contest/508/problem/E 输入一个n,表示有这么多对括号,然后有n行,每行输入一个区间,第i行的区间表示从前往后第i对括号的左括 ...

  8. Codeforces Round #288 (Div. 2)D. Tanya and Password 欧拉通路

    D. Tanya and Password Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/508 ...

  9. Codeforces Round #293 (Div. 2) C. Anya and Smartphone 数学题

    C. Anya and Smartphone time limit per test 1 second memory limit per test 256 megabytes input standa ...

随机推荐

  1. source insight资源

    http://www.cnblogs.com/Red_angelX/p/3713935.html https://github.com/redxu/sihook

  2. 为什么drop table的时候要在checking permissions花很长时间?

    昨天,我drop一个表的时候在checking permissions花了20s+,这个时间花在哪里了呢?经常查找发现我的配置文件innodb_file_per_table=1的,innodb需要遍历 ...

  3. linux配置java环境变量(详细)【转】

    转自:http://www.cnblogs.com/samcn/archive/2011/03/16/1986248.html 一. 解压安装jdk 在shell终端下进入jdk-6u14-linux ...

  4. java多线程中的生产者与消费者之等待唤醒机制@Version1.0

    一.生产者消费者模式的学生类成员变量生产与消费demo,第一版1.等待唤醒:    Object类中提供了三个方法:    wait():等待    notify():唤醒单个线程    notify ...

  5. CSS修改方法

    1.在IE中,大部分情况下默认margin = 0px padding = 0px,但在Chrome中需要写明 在css.css文件开头加上(要加在最上面) html,body,ul,ol,li,ta ...

  6. spring.xml命名空间

    <beans xmlns="http://www.springframework.org/schema/beans" xmlns:xsi="http://www.w ...

  7. oracle日期函数2!

    1.日期时间间隔操作  当前时间减去7分钟的时间 select sysdate,sysdate - interval '7' MINUTE from dual 当前时间减去7小时的时间 ...

  8. JavaEE基础(十五)/集合

    1.集合框架(对象数组的概述和使用) A:案例演示 需求:我有5个学生,请把这个5个学生的信息存储到数组中,并遍历数组,获取得到每一个学生信息. Student[] arr = new Student ...

  9. RAC例子

    我个人非常推崇ReactiveCocoa,它就像中国的太极,太极生两仪,两仪生四象,四象生八卦,八卦生万物.ReactiveCocoa是一个高度抽象的编程框架,它真的很抽象,初看你不知道它是要干嘛的, ...

  10. node-webkit 新建实例窗口间通信问题解决办法

    终于弄明白这问题了,只要在js文件里加上段代码,就可解决两窗口间通信问题. var str = { username: User.name, userrole: User.role }; var ne ...