CF 672D Robin Hood（二分答案）

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 c_i 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 c_i (1 ≤ c_i ≤ 10⁹) — initial wealth of the i-th person.

Output

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

Examples
input
Copy
4 1
1 1 4 2
output
Copy
2
input
Copy
3 1
2 2 2
output
Copy
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】

给你N个人的财富值，每天最有钱的人会给最没钱的人一块钱，问K天之后，最穷的人和最富有的人之间差多少钱。

【分析】

初始时硬币数比平均数少的人最终会增加k个硬币，比平均数多的会减少k个硬币
因为时间越长（天数经过的越多），最大值就会越小，同理，最小值就会越大，那么我们二分一个最小值，接下来二分一个最大值。

对于枚举最小值的时候，如果需要的最少天数小于等于k，那么就加大最小值，否则减小最小值即可。那么二分枚举最大值的时候同理即可。

【代码】

Select Code

#include<cstdio>
#include<iostream>
#define debug(x) cerr<<#x<<" "<<x<<"
";
using namespace std;
typedef long long ll;
inline int read(){
	register char ch=getchar();register int x=0;
	for(;ch<'0'||ch>'9';ch=getchar());
	for(;ch>='0'&&ch<='9';ch=getchar()) x=(x<<3)+(x<<1)+ch-'0';
	return x;
}
const int N=5e5+5;
int n,a[N];ll k,sum;
inline bool check1(ll m){
	ll res=0;
	for(int i=1;i<=n;i++){
		if(a[i]<m){
			res+=m-a[i];
		}
	}
	return res<=k;
}
inline bool check2(ll m){
	ll res=0;
	for(int i=1;i<=n;i++){
		if(a[i]>m){
			res+=a[i]-m;
		}
	}
	return res<=k;
}
int main(){
	n=read();k=read();
	for(int i=1;i<=n;i++) a[i]=read(),sum+=a[i];
	ll l=1,r=sum/n,mid,ans1=0,ans2=0;
	while(l<=r){
		mid=l+r>>1;
		if(check1(mid)){
			ans1=mid;
			l=mid+1;
		}
		else{
			r=mid-1;
		}
	}
	l=sum/n,r=sum;
	if(sum%n) l++;
	while(l<=r){
		mid=l+r>>1;
		if(check2(mid)){
			ans2=mid;
			r=mid-1;
		}
		else{
			l=mid+1;
		}
	}
	printf("%d
",ans2-ans1);
	return 0;
}