Codeforces Round #325 (Div. 2) C. Gennady the Dentist 暴力

C. Gennady the Dentist

Time Limit: 1 Sec  

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/586/problem/C

Description

Gennady is one of the best child dentists in Berland. Today n children got an appointment with him, they lined up in front of his office.

All children love to cry loudly at the reception at the dentist. We enumerate the children with integers from 1 to n in the order they go in the line. Every child is associated with the value of his cofidence p_i. The children take turns one after another to come into the office; each time the child that is the first in the line goes to the doctor.

While Gennady treats the teeth of the i-th child, the child is crying with the volume of v_i. At that the confidence of the first child in the line is reduced by the amount of v_i, the second one — by value v_i - 1, and so on. The children in the queue after the v_i-th child almost do not hear the crying, so their confidence remains unchanged.

If at any point in time the confidence of the j-th child is less than zero, he begins to cry with the volume of d_j and leaves the line, running towards the exit, without going to the doctor's office. At this the confidence of all the children after the j-th one in the line is reduced by the amount of d_j.

All these events occur immediately one after the other in some order. Some cries may lead to other cries, causing a chain reaction. Once in the hallway it is quiet, the child, who is first in the line, goes into the doctor's office.

Help Gennady the Dentist to determine the numbers of kids, whose teeth he will cure. Print their numbers in the chronological order.

Input

The first line of the input contains a positive integer n (1 ≤ n ≤ 4000) — the number of kids in the line.

Next n lines contain three integers each v_i, d_i, p_i (1 ≤ v_i, d_i, p_i ≤ 10⁶) — the volume of the cry in the doctor's office, the volume of the cry in the hall and the confidence of the i-th child.

i​​,C​i​​，即此题的初始分值、每分钟减少的分值、dxy做这道题需要花费的时间。

Output

In the first line print number k — the number of children whose teeth Gennady will cure.

In the second line print k integers — the numbers of the children who will make it to the end of the line in the increasing order.

Sample Input

5
4 2 2
4 1 2
5 2 4
3 3 5
5 1 2

Sample Output

2
1 3

HINT

 

题意

一堆小孩要去看牙医，小孩进入牙医之后，就会发出叫声，使得接下来的v[i]个孩子的信心分别下降v[i],v[i]-1......1这么多

如果小孩被吓跑了，他们又会叫，使得接下来的孩子发出d[i]的叫声

然后问你一共有多少人能够看病，并且是哪些人

题解：

数据范围只有4000，那就n^2暴力就好了

有两个坑点：

1.爆int

2.得v[i]减完之后，大家再一起叫d[i]的，不是边v[i]边d[i]

代码：

#include<stdio.h>
#include<iostream>
#include<math.h>
#include<iostream>
#include<cstring>
#include<vector>
using namespace std;

long long v[4005],d[4005],p[4005];
long long ans[4005];
long long flag[4005];
int n;
int main()
{
    memset(flag,0,sizeof(flag));
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%lld%lld%lld",&v[i],&d[i],&p[i]);
    int tot = 0;
    for(int i=1;i<=n;i++)
    {
        if(p[i]<0)continue;
        ans[tot++]=i;
        long long sum=0;
        long long time = v[i];
        for(int j=i+1;j<=n;j++)
        {
            flag[j]=0;
            if(p[j]>=0)
            {
                flag[j]=1;
                if(time>0)
                    p[j]-=time;
                time--;
            }
        }
        for(int j=i+1;j<=n;j++)
        {
            if(p[j]>=0)
                p[j]-=sum;
            if(p[j]<0&&flag[j])
                sum+=d[j];
        }
    }
    printf("%d\n",tot);
    for(int i=0;i<tot;i++)
        printf("%lld ",ans[i]);
    printf("\n");
}