题目：
C. Servers
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard 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: t**i — the moment in seconds in which the task will come, k**i — the number of servers needed to perform it, and d**i — the time needed to perform this task in seconds. Allt**i are distinct.
To perform the i-th task you need k**i servers which are unoccupied in the second t**i. After the servers begin to perform the task, each of them will be busy over the next d**i seconds. Thus, they will be busy in seconds t**i, t**i + 1, ..., t**i + d**i - 1. For performing the task, k**iservers with the smallest ids will be chosen from all the unoccupied servers. If in the second t**i 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 t**i, k**i and d**i (1 ≤ t**i ≤ 106, 1 ≤ k**i ≤ n, 1 ≤ d**i ≤ 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.
代码：
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <math.h>
#include <stack>
#include <queue>
#include <vector>
#include <algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
struct nnd
{
int start;//开始时间
int num;//所需服务器数量
int need;//需要用服务器的时间
} a[100005];
int vis[101];//用来标记当前时间这个服务器有没有被使用（使用服务器的最后时间）
int main()
{
int n,q;
scanf("%d%d",&n,&q);//输入服务器数量和询问次数
mem(vis,0);
for(int i=0; i<q; i++)
scanf("%d%d%d",&a[i].start,&a[i].num,&a[i].need);//读入数据
for(int i=0; i<q; i++)//遍历询问次数
{
int knum=0;//标记服务器数量
for(int j=1; j<=n&&knum<a[i].num; j++)//分配服务器
if(vis[j]<a[i].start)//判断当前时间服务器有没有被使用
knum++;
if(knum==a[i].num)//如果把服务器的数量足够
{
long long sum=0;//计算编号和
for(int j=1;j<=n&&knum>0;j++)
{
if(vis[j]<a[i].start)//当前服务器可以使用时
{
vis[j]=a[i].start+a[i].need-1;//开始的时间加上需要的时间
knum--;
sum+=j;//累加
}
}
printf("%lld\n",sum);//输出编号和
}
else
printf("-1\n");
}
return 0;
}
本来看不懂，看了学长的代码才会，还是要有思路。
