time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard output

There are n servers in a laboratory, each of them can perform tasks. Each server has a unique id — integer from 1 to n.

It is known that during the day q tasks will come, the i-th of them is characterized with three integers: ti — the moment in seconds in which the task will come, ki — the number of servers needed to perform it, and di — the time needed to perform this task in seconds. All ti are distinct.

To perform the i-th task you need ki servers which are unoccupied in the second ti. After the servers begin to perform the task, each of them will be busy over the next di seconds. Thus, they will be busy in seconds ti, ti + 1, …, ti + di - 1. For performing the task, ki servers with the smallest ids will be chosen from all the unoccupied servers. If in the second ti there are not enough unoccupied servers, the task is ignored.

Write the program that determines which tasks will be performed and which will be ignored.

Input

The first line contains two positive integers n and q (1 ≤ n ≤ 100, 1 ≤ q ≤ 105) — the number of servers and the number of tasks.

Next q lines contains three integers each, the i-th line contains integers ti, ki and di (1 ≤ ti ≤ 106, 1 ≤ ki ≤ n, 1 ≤ di ≤ 1000) — the moment in seconds in which the i-th task will come, the number of servers needed to perform it, and the time needed to perform this task in seconds. The tasks are given in a chronological order and they will come in distinct seconds.

Output

Print q lines. If the i-th task will be performed by the servers, print in the i-th line the sum of servers’ ids on which this task will be performed. Otherwise, print -1.

Examples

input

4 3

1 3 2

2 2 1

3 4 3

output

6

-1

10

input

3 2

3 2 3

5 1 2

output

3

3

input

8 6

1 3 20

4 2 1

6 5 5

10 1 1

15 3 6

21 8 8

output

6

9

30

-1

15

36

Note

In the first example in the second 1 the first task will come, it will be performed on the servers with ids 1, 2 and 3 (the sum of the ids equals 6) during two seconds. In the second 2 the second task will come, it will be ignored, because only the server 4 will be unoccupied at that second. In the second 3 the third task will come. By this time, servers with the ids 1, 2 and 3 will be unoccupied again, so the third task will be done on all the servers with the ids 1, 2, 3 and 4 (the sum of the ids is 10).

In the second example in the second 3 the first task will come, it will be performed on the servers with ids 1 and 2 (the sum of the ids is 3) during three seconds. In the second 5 the second task will come, it will be performed on the server 3, because the first two servers will be busy performing the first task.

【题目链接】:http://codeforces.com/contest/747/problem/C

【题解】



一个任务一个任务地枚举

因为时间是升序的、所以不用管。

在枚举的时候看看这个任务开始的时候有哪些人是空闲的?

->如何确定某个人是否空闲???

->这个人完成任务的时间.

每让一个人接任务过后,就记录每个人任务完成的时间是什么时候.

时间复杂度O(n*q);



【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x) typedef pair<int,int> pii;
typedef pair<LL,LL> pll; const int MAXP = 1e2+10;
const int MAXQ = 1e5+10;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0); struct abc
{
int t,k,d;
}; int last[MAXP];
abc a[MAXQ];
int n,q;
int b[MAXP];
int ans[MAXQ];
int sum[MAXP]; int main()
{
//freopen("F:\\rush.txt","r",stdin);
rei(n);rei(q);
rep1(i,1,q)
{
rei(a[i].t);rei(a[i].k);rei(a[i].d);
}
rep1(i,1,q)
{
b[0] = 0;
sum[0] = 0;
rep1(j,1,n)
if (last[j]<a[i].t)
{
b[0]++;
b[b[0]] = j;
sum[b[0]] = sum[b[0]-1]+b[b[0]];
if (b[0]==a[i].k)
break;
}
if (b[0]==a[i].k)
{
ans[i] = sum[b[0]];
rep1(j,1,b[0])
last[b[j]] = a[i].t + a[i].d-1;
}
else
ans[i] = -1;
}
rep1(i,1,q)
printf("%d\n",ans[i]);
return 0;
}

【50.00%】【codeforces 747C】Servers的更多相关文章

  1. 【50.00%】【codeforces 602C】The Two Routes

    time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...

  2. 【 BowWow and the Timetable CodeForces - 1204A 】【思维】

    题目链接 可以发现 十进制4 对应 二进制100 十进制16 对应 二进制10000 十进制64 对应 二进制1000000 可以发现每多两个零,4的次幂就增加1. 用string读入题目给定的二进制 ...

  3. Codeforces 747C:Servers(模拟)

    http://codeforces.com/problemset/problem/747/C 题意:有n台机器,q个操作.每次操作从ti时间开始,需要ki台机器,花费di的时间.每次选择机器从小到大开 ...

  4. 【25.00%】【codeforces 584E】Anton and Ira

    time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard ou ...

  5. 【50.88%】【Codeforces round 382B】Urbanization

    time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...

  6. 【74.00%】【codeforces 747A】Display Size

    time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard ou ...

  7. 【codeforces 750A】New Year and Hurry

    time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard ou ...

  8. [CodeForces - 1225E]Rock Is Push 【dp】【前缀和】

    [CodeForces - 1225E]Rock Is Push [dp][前缀和] 标签:题解 codeforces题解 dp 前缀和 题目描述 Time limit 2000 ms Memory ...

  9. 【codeforces 709D】Recover the String

    [题目链接]:http://codeforces.com/problemset/problem/709/D [题意] 给你一个序列; 给出01子列和10子列和00子列以及11子列的个数; 然后让你输出 ...

随机推荐

  1. dialog的进度条

    import android.app.Activity; import android.app.ProgressDialog; import android.os.Bundle; import and ...

  2. Emacs用JDEE编写Android程序

    版权声明:本文为博主原创文章.未经博主同意不得转载. https://blog.csdn.net/sheismylife/article/details/24842669 前文介绍了怎样用Maven构 ...

  3. 重磅!容器集群监控利器 阿里云Prometheus 正式免费公测

    Prometheus 作为容器生态下集群监控的首选方案,是一套开源的系统监控报警框架.它启发于 Google 的 borgmon 监控系统,并于 2015 年正式发布.2016 年,Prometheu ...

  4. 草地排水 改了又改(DCOJ6013)

    题目描述 在农夫约翰的农场上,每逢下雨,贝茜最喜欢的三叶草地就积聚了一潭水.这意味着草地被水淹没了,并且小草要继续生长还要花相当长一段时间.因此,农夫约翰修建了一套排水系统来使贝茜的草地免除被大水淹没 ...

  5. Ubuntu里node命令出错,找不到

    ubuntu里用sudo apt-get install nodejs安装Node.js后, 发现terminals里运行node命令(比如node –-version)时候会有No such fil ...

  6. JavaScript--轮播图_带计时器

    轮播图效果: 实现的功能: 1.鼠标移入,左右按钮显示 2.右下叫小圆点鼠标移入,进入下一张图 3.左右按钮点击,右下小圆点页跟随变更 4.自动开启计时器,鼠标移入右下叫小圆点区,计时器停止,鼠标移出 ...

  7. 【JZOJ4819】【NOIP2016提高A组模拟10.15】算循环

    题目描述 输入 输出 样例输入 167 198 样例输出 906462341 数据范围 解法 令f(n)=∑ni=1i,g(n)=∑ni=1i2 易得ans=∑ni=1∑mj=1f(n−i+1)∗f( ...

  8. hdu2897 巴什博奕

    n%(q+p)==0,也就是说先手必胜; n%(q+p)<=p,先手必输; n%(q+p)==k if(k>p&&k<=q)先手必胜; if(k>p&& ...

  9. vuex之仓库数据的设置与获取

    如果你之前使用过vue.js,你一定知道在vue中各个组件之间传值的痛苦,在vue中我们可以使用vuex来保存我们需要管理的状态值,值一旦被修改,所有引用该值的地方就会自动更新,那么接下来我们就来学习 ...

  10. c++:三

    C++:三   对于共享数据的保护,我们可以使用常量,在使用数据的同时也防止了数据被修改,即可有效的保护数据. 常对象   常对象必须在定义对象时就使用"const"关键字将指定对 ...