8840: Medical Checkup

8840: Medical Checkup

http://exam.upc.edu.cn/problem.php?id=8840

时间限制: 2 Sec  内存限制: 128 MB
提交: 184  解决: 52
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题目描述

Students of the university have to go for a medical checkup, consisting of lots of checkup items,numbered 1, 2, 3, and so on.
Students are now forming a long queue, waiting for the checkup to start. Students are also numbered 1, 2, 3, and so on, from the top of the queue. They have to undergo checkup items in the order of the item numbers, not skipping any of them nor changing the order. The order of students should not be changed either.
Multiple checkup items can be carried out in parallel, but each item can be carried out for only one student at a time. Students have to wait in queues of their next checkup items until all the others before them finish.
Each of the students is associated with an integer value called health condition. For a student with the health condition h, it takes h minutes to finish each of the checkup items. You may assume that no interval is needed between two students on the same checkup item or two checkup items for a single student. 
Your task is to find the items students are being checked up or waiting for at a specified time t.

 

输入

The input consists of a single test case in the following format.
n t
h1
.
.
.
hn
n and t are integers. n is the number of the students (1 ≤ n ≤ 105). t specifies the time of our concern (0 ≤ t ≤ 109). For each i, the integer hi is the health condition of student i(1 ≤ hi ≤ 109).

 

输出

Output n lines each containing a single integer. The i-th line should contain the checkup item number of the item which the student i is being checked up or is waiting for, at (t+0.5) minutes after the checkup starts. You may assume that all the students are yet to finish some of the checkup items at that moment.

 

样例输入

3 20
5
7
3

 

样例输出

5
3
2

题意：n个人依次排队且按体检点顺序体检，一个体检的地方每次只能体检一人，一个人在这个体检点没体检完后一个必须等。

第i个人体检hi时间，问t时这n个人现在在哪个体检点体检或等待？

思路：画画图就能得到公式，但比赛的时候一直输出超限。。。愁死了。。。

代码：

#include
#include
#include
#include
#include
#include
#define inf 0x3f3f3f3f
using namespace std;

int h[100005];

int main()
{
    int n,t;
    long long summ;
    cin>>n>>t;
    for(int i=1;i<=n;i++)
        cin>>h[i];
    int maxx=h[1];
    summ=t/h[1]+1;
    cout<