Codeforce 1118 D2 Coffee and Coursework (Hard Version)

D2. Coffee and Coursework (Hard Version)

time limit per test

2.5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

The only difference between easy and hard versions is the constraints.

Polycarp has to write a coursework. The coursework consists of mm pages.

Polycarp also has nn cups of coffee. The coffee in the ii -th cup Polycarp has aiai caffeine in it. Polycarp can drink some cups of coffee (each one no more than once). He can drink cups in any order. Polycarp drinks each cup instantly and completely (i.e. he cannot split any cup into several days).

Surely, courseworks are not being written in a single day (in a perfect world of Berland, at least).

Let's consider some day of Polycarp's work. Consider Polycarp drinks kk cups of coffee during this day and caffeine dosages of cups Polycarp drink during this day are ai1,ai2,…,aikai1,ai2,…,aik . Then the first cup he drinks gives him energy to write ai1ai1 pages of coursework, the second cup gives him energy to write max(0,ai2−1)max(0,ai2−1) pages, the third cup gives him energy to write max(0,ai3−2)max(0,ai3−2) pages, ..., the kk -th cup gives him energy to write max(0,aik−k+1)max(0,aik−k+1) pages.

If Polycarp doesn't drink coffee during some day, he cannot write coursework at all that day.

Polycarp has to finish his coursework as soon as possible (spend the minimum number of days to do it). Your task is to find out this number of days or say that it is impossible.

Input

The first line of the input contains two integers nn and mm (1≤n≤2⋅1051≤n≤2⋅105 , 1≤m≤1091≤m≤109 ) — the number of cups of coffee and the number of pages in the coursework.

The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109 ), where aiai is the caffeine dosage of coffee in the ii -th cup.

Output

If it is impossible to write the coursework, print -1. Otherwise print the minimum number of days Polycarp needs to do it.

Examples

Input

Copy

5 8
2 3 1 1 2

Output

Copy

4

Input

Copy

7 10
1 3 4 2 1 4 2

Output

Copy

2

Input

Copy

5 15
5 5 5 5 5

Output

Copy

1

Input

Copy

5 16
5 5 5 5 5

Output

Copy

2

Input

Copy

5 26
5 5 5 5 5

Output

Copy

-1

Note

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.

 

题意:n杯咖啡,m页论文,每杯咖啡有自己的咖啡因含量~,对应着喝完这杯咖啡能写页多少论文。在一天内喝第一杯咖啡,效果等于原值(ai的咖啡因量能肝ai页论文),每多喝一杯,这杯咖啡能产生的效果就会减一。问最少多少天完成论文?

二分

贪心时优先天数,每天的第i杯咖啡

#include<bits/stdc++.h>
using namespace std;
int a[200005];
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    long long p=0;
    for(int i=0; i<n; i++)
        scanf("%d",&a[i]),p+=a[i];
    sort(a,a+n,greater<int>());
    if(p<m)
    {
        printf("-1");
        return 0;
    }

    int l=1,r=n;
    while(r-l>1)
    {
        int day=(l+r)/2;
        long long sum=0;
        int u=0;
        int cnt=0;
        for(int i=0; i<n; i++)
        {
            if(a[i]>u)
            {
                sum+=a[i]-u;
                cnt++;
            }
            else
                break;
            if(cnt==day)
                cnt=0,u++;
        }
        if(sum>=m)
        {
            r=day;
        }
        else
        {
            l=day;
        }
    }
    int day=l;
    long long sum=0;
    int u=0;
    int cnt=0;
    for(int i=0; i<n; i++)
    {
        if(a[i]>u)
        {
            sum+=a[i]-u;
            cnt++;
        }
        else
            break;
        if(cnt==day)
            cnt=0,u++;
    }
    if(sum>=m)
    {
        r=day;
    }
    printf("%d\n",r);
    return 0;
}

 

 

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风中翠竹:真的真的真的没有kpi。。。面试官是没有任何kpi的,捞是真的想试试看这个行不行,碰碰运气,或者是面试官比较闲现在,没事捞个人看看。kpi算HR那边,但是只有你入职了,kpi才作数,面试是没有的。
双非有机会进大厂吗
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07-11 22:27
中南大学 Java
程序员牛肉:学历的话没问题。但是没问题的也就只有学历了。 其实你的整体架构是正确的,博客接着干。但是项目有点过于简单了。从后端的角度上讲,你这也就是刚入门的水平,所以肯定约面试够呛。 如果你要应聘后端岗位,那你第一个项目竟然是仿写操作系统。这个你要面试官咋问你。你一定要记住一点,你简历上写的所有的东西,都是为了证明你有能力胜任当前的岗位,而不是为了证明你自己会什么。 如果你只是浅浅的做几个项目,描述也都是烂大街。技术点也都是各种混水类的配置类需求,那你就不要幻想自己能走多远。一定要保持思考,保持学习。
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