Coffee and Coursework (Hard Version)

In the first example Polycarp can drink fourth cup during first day (and write 11 page), first and second cups during second day (and write 2+(3−1)=42+(3−1)=4 pages), fifth cup during the third day (and write 22 pages) and third cup during the fourth day (and write 11 page) so the answer is 44 . It is obvious that there is no way to write the coursework in three or less days.

In the second example Polycarp can drink third, fourth and second cups during first day (and write 4+(2−1)+(3−2)=64+(2−1)+(3−2)=6 pages) and sixth cup during second day (and write 44 pages) so the answer is 22 . It is obvious that Polycarp cannot write the whole coursework in one day in this test.

In the third example Polycarp can drink all cups of coffee during first day and write 5+(5−1)+(5−2)+(5−3)+(5−4)=155+(5−1)+(5−2)+(5−3)+(5−4)=15 pages of coursework.

In the fourth example Polycarp cannot drink all cups during first day and should drink one of them during the second day. So during first day he will write 5+(5−1)+(5−2)+(5−3)=145+(5−1)+(5−2)+(5−3)=14 pages of coursework and during second day he will write 55 pages of coursework. This is enough to complete it.

In the fifth example Polycarp cannot write the whole coursework at all, even if he will drink one cup of coffee during each day, so the answer is -1.

分析：大数据，既然已经使用sort排序了，直接二分就可以了鸭

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm>
 4 using namespace std;
 5 #define ll long long
 6 int n,m;
 7 vector<int> a;
 8 
 9 bool can(int i){
10     ll sum=0;
11     for( int j=0; j<n; j++ ){
12         sum+=max(a[j]-j/i,0);
13     }
14     if(sum>=m)return true;
15     else return false;
16 }
17 
18 int main(int argc, char const *argv[])
19 {
20     cin>>n>>m;
21     a=vector<int>(n);
22 
23     for( int i=0; i<n; i++ ){
24         cin>>a[i];
25     }
26 
27     sort(a.rbegin(),a.rend());
28 
29     int l=1,r=n;
30     while(r-1>l){
31         ll mid=(l+r) >> 1;
32         if(can(mid)) r=mid;
33         else l=mid;
34     }
35 
36     if(can(l)) cout<<l<<endl;
37     else if(can(r)) cout<<r<<endl;
38     else{
39         cout<<-1<<endl;
40     }
41 
42     return 0;
43 }