codeforces 672D Robin Hood (二分)

[align=center]D. Robin Hood[/align]

time limit per test：1 second

memory limit per test：256 megabytes

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.

题目链接：http://codeforces.com/problemset/problem/672/D

题目大意：n个数字，每次把最大的拿出1后给最小的，进行k次后问当前最大值和最小值的差最小是多少

题目分析：典型二分，分别二分k次后最大值的最小值和最小值的最大值
#include <cstdio>
#include <cstring>
#include <algorithm>
#define ll long long
using namespace std;
int const INF = 2e9;
int const MAX = 5e5 + 5;
int n, k, ma, mi;
ll c[MAX];

bool ok1(ll x)
{
ll res = 0;
for(int i = n - 1; i >= 0; i--)
{
if(c[i] > x)
res += c[i] - x;
else
break;
}
return res <= k;
}

bool ok2(ll x)
{
ll res = 0;
for(int i = 0; i < n; i++)
{
if(x > c[i])
res += x - c[i];
else
break;
}
return res <= k;
}

int main()
{
ll sum = 0;
ll MA = 0, MI = INF;
scanf("%d %d", &n, &k);
for(int i = 0; i < n; i++)
{
scanf("%I64d", &c[i]);
sum += c[i];
MA = max(MA, c[i]);
MI = min(MI, c[i]);
}
if(sum % n == 0)
ma = mi = sum / n;
else
{
ma = sum / n + 1;
mi = sum / n;
}
int rma = 0, lma = 0;
sort(c, c + n);

int l = ma, r = MA;
while(l <= r)
{
ll mid = (l + r) >> 1;
if(ok1(mid))
{
r = mid - 1;
rma = mid;
}
else
l = mid + 1;
}

l = MI, r = mi;
while(l <= r)
{
ll mid = (l + r) >> 1;
if(ok2(mid))
{
l = mid + 1;
lma = mid;
}
else
r = mid - 1;
}
printf("%d\n", rma - lma);
}