codeforces 671B

B. Robin Hood
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard 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 the i-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, 4, 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:有n个人,另外一个人有k天时间做一件事情:把最富的一个人的一块钱给最穷的一个人,问你在k天过后最富的人与最穷的人的最小钱差距为多少.
方法:首先我们要知道怎样才能使最富的人与最穷的人的最小钱差距为多少.图无法放上去,只有直接写了.首先我们可以先找一条线使所有re,我们可以先找一个平均值,显然如达到了最佳情况,肯定所有人的钱数都在这个点或+1.然后我们就可从平均值往下二分让最穷的人钱越多,在往上二分让最富的人钱越少.

#include <cstdio>#include <cmath>#include <iostream>#include <cstring>#include <algorithm>#define ll long longusing namespace std;int n,k,a[500005];long long prt[500005],m,sum,mini,maxi,aer;void readdata(){    scanf("%d%d",&n,&k);    for (int i=0;i<n;i++)    {        scanf("%d",&a[i]);        sum+=a[i];    }    aer=sum/n;}bool checkl(int x){    int p=lower_bound(a,a+n,x)-a-1;    if ((ll)x * (p + 1) - prt[p] > k) return false;    else return true;}bool checkr(int x){    int p=upper_bound(a,a+n,x)-a;    if (prt[n-1] - prt[p-1] - (ll)(n - p) * x > k) return false;    else return true;}void work(){    sort(a,a+n);    int l=a[0]; int r=aer+1;    for (int i=0;i<n;i++)    {        prt[i]=prt[i-1]+a[i];    }    while (l+1<r)    {        int mid=(l+r)>>1;        if (checkl(mid))        {            l=mid;        }        else r=mid;    }    mini=l;    l=aer;if (sum%n>0) l++;    r=a[n-1];    while (l<r)    {        int mid=(l+r)>>1;        if (checkr(mid))        {            r=mid;        }        else l=mid+1;    }    maxi=r;    cout<<maxi-mini;}int main(){    readdata();    work();}