CF 352 D 罗宾汉发钱 模拟题+贪心

D. Robin Hood

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

We all know the impressive story of Robin Hood. Robin Hood uses his archery skills and his wits to steal the money from rich, and return it to the poor.

There are n citizens in Kekoland, each person has ci coins. Each day, Robin Hood will take exactly 1 coin from the richest person in the city and he will give it to the poorest person (poorest person right after taking richest's 1 coin). In case the choice is not unique, he will select one among them at random. Sadly, Robin Hood is old and want to retire in k days. He decided to spend these last days with helping poor people.

After taking his money are taken by Robin Hood richest person may become poorest person as well, and it might even happen that Robin Hood will give his money back. For example if all people have same number of coins, then next day they will have same number of coins too.

Your task is to find the difference between richest and poorest persons wealth after k days. Note that the choosing at random among richest and poorest doesn't affect the answer.

Input

The first line of the input contains two integers n and k (1 ≤ n ≤ 500 000, 0 ≤ k ≤ 109) — the number of citizens in Kekoland and the number of days left till Robin Hood's retirement.

The second line contains n integers, the i-th of them is ci (1 ≤ ci ≤ 109) — initial wealth of thei-th person.

Output

Print a single line containing the difference between richest and poorest peoples wealth.

Examples

input

4 1
1 1 4 2

output

2

input

3 1
2 2 2

output

0

Note

Lets look at how wealth changes through day in the first sample.

1. [1, 1, 4, 2]
2. [2, 1, 3, 2] or [1, 2, 3, 2]

So the answer is 3 - 1 = 2

In second sample wealth will remain the same for each person.

 题意：给出n个人,每个人有一个初始的价值ai,给出k个操作,每个操作必须从当前最富有的那个人拿一个单位的价值给当前最穷的人,问在经历k次操作之后,贫富差距多大

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <algorithm>
#include <set>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define MM(a,b) memset(a,b,sizeof(a));
const double eps = 1e-10;
const int  inf =0x7f7f7f7f;
const double pi=acos(-1);
const int maxn=100+500000;

ll a[maxn];
int main()
{
    int n,k;
    while(~scanf("%d %d",&n,&k))
    {
        ll sum=0;
        for(int i=0;i<n;i++)
            {
                scanf("%lld",&a[i]);
                sum+=a[i];
            }
        sort(a,a+n);

        int maxn=a[n-1],minn=a[0],tp=k;
        for(int i=n-1;i>=0;i--)
          if(maxn>a[i])
           {
                if((n-1-i)*(maxn-a[i])<=tp)//乘法int会爆
                {
                    tp-=(n-1-i)*(maxn-a[i]);
                    maxn=a[i];
                }//能全部减下去
                else
                {
                    maxn-=tp/(n-1-i);
                    break;
                }//无法进一步减下去
           }

        tp=k;minn=a[0];
        for(int i=0;i<n;i++)
           if(minn<a[i])
           {
               if(i*(a[i]-minn)<=tp)
               {
                   tp-=i*(a[i]-minn);
                   minn=a[i];
               }
               else
               {
                   minn+=tp/i;
                   break;
               }
           }

        if(minn>=maxn)  printf("%d\n",sum%n==0?0:1);
        else printf("%d\n",maxn-minn);
    }
    return 0;
}

分析：一开始被吓到了，，后来稍微看了下题解，再自己仔细做了下，，还是觉得很水的.....

建立模型：将所有人的财富按从小到大排序，然后首先从最右边开始切，最多切k个格子，算出

切了后的最大的值maxn，然后再从左往右补，算出补了后的最小的值minn,如果maxn>=minn，说明穷人与富人之间出现了反转，则要特判一下所有才丰富能否均分。