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C2. Exam in BerSU (hard version)

The only difference between easy and hard versions is constraints.

If you write a solution in Python, then prefer to send it in PyPy to speed up execution time.

A session has begun at Beland State University. Many students are taking exams.

Polygraph Poligrafovich is going to examine a group of n students. Students will take the exam one-by-one in order from 1 -th to n -th. Rules of the exam are following:

• The i -th student randomly chooses a ticket.
• if this ticket is too hard to the student, he doesn't answer and goes home immediately (this process is so fast that it's considered no time elapses). This student fails the exam.
• if the student finds the ticket easy, he spends exactly ti minutes to pass the exam. After it, he immediately gets a mark and goes home.

Students take the exam in the fixed order, one-by-one, without any interruption. At any moment of time, Polygraph Poligrafovich takes the answer from one student.

The duration of the whole exam for all students is M minutes (maxti≤M ), so students at the end of the list have a greater possibility to run out of time to pass the exam.

For each student i , you should count the minimum possible number of students who need to fail the exam so the i -th student has enough time to pass the exam.

For each student i , find the answer independently. That is, if when finding the answer for the student i1 some student j should leave, then while finding the answer for i2 (i2>i1 ) the student j student does not have to go home.

Input

The first line of the input contains two integers n and M (1≤n≤2⋅105 , 1≤M≤2⋅107 ) — the number of students and the total duration of the exam in minutes, respectively.

The second line of the input contains n integers ti (1≤ti≤100 ) — time in minutes that i -th student spends to answer to a ticket.

It's guaranteed that all values of ti are not greater than M .

Output

Print n numbers: the i -th number must be equal to the minimum number of students who have to leave the exam in order to i -th student has enough time to pass the exam.

题意:这题困难版就是一个可能会让你超时的问题，我就被卡了第14组(实在太弱)，然后在这上面找题解，发现一位大佬写的较为简便的方法，因为每个人的时间都被限制在100以内，所以我们可以用一个数组来记录所有的数出现的次数，这样我们每次判断这个同学能通过考试所需最少不及格同学人数时，我们只需要尽可能地从小到大取掉最大足够小的时间的同学，最后得到的答案必然是最小。下面附上代码。

#include
   
    using namespace std;int number[105];int main(){    int n,m;    int t,ans,sum,x;    scanf("%d%d",&n,&m);    memset(number,0,sizeof(number));    for(int i=1; i<=n; i++)    {        scanf("%d",&t);        ans=0;        sum=m-t;        for(int j=1; j<=100; j++)        {            x=min(sum/j,number[j]);            ans+=x;            sum-=x*j;        }        number[t]++;        printf("%d ",i-1-ans);    }}