链接：
https://codeforces.com/contest/1185/problem/C2
题意：
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.
思路：
c1和c2范围不同，就只写一个c2了
每次循环从100-1的数找有几个，优先用掉最大的就行，原本以为过不了hard版本。
代码：
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MAXN = 2e7 + 10;
const int MOD = 1e9 + 7;
int n, m, k, t;
int Num[MAXN];
int main()
{
cin >> n >> m;
int sum = 0, v, cnt;
for (int i = 1;i <= n;i++)
{
cnt = 0;
int tmp = sum;
cin >> v;
if (m - sum < v)
{
for (int j = 100;j >= 1;j--)
{
if (m - (tmp-j*Num[j]) >= v)
{
cnt += (tmp-m+v+j-1)/j;
break;
}
tmp -= j*Num[j];
cnt += Num[j];
}
}
cout << cnt << ' ' ;
Num[v]++;
sum += v;
}
cout << endl;
return 0;
}