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.
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.
Print a single line containing the difference between richest and poorest peoples wealth.
4 1 1 1 4 2
2
3 1 2 2 2
0
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个数，k次操作， 每次操作把最大的减一，最小的加一， 如果最大的减一变成最小的，就吧这减掉的1再加回去
直接暴力900多ms躺过，，，，
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <stack>
#include <queue>
#include <set>
#include <map>
using namespace std;
const int inf = 2147483647;
const double PI = acos(-1.0);
const double e = 2.718281828459;
const int mod = 1000000007;
typedef long long LL;
#pragma comment(linker, "/STACK:102400000,102400000")
//freopen("in.txt","r",stdin); //输入重定向,输入数据将从in.txt文件中读取
//freopen("out.txt","w",stdout); //输出重定向,输出数据将保存在out.txt文件中cin
int a[500005];
map<LL, LL>mp;
set<int>st;
set<int>::iterator it, tt;
set<int>::reverse_iterator rt, rtt;
int main()
{
int n, k, i, j;
while (cin >> n >> k)
{
LL s = 0;
mp.clear();
st.clear();
for (i = 1; i <= n; ++i)
{
scanf("%d", &a[i]);
st.insert(a[i]);
mp[a[i]]++;
s += a[i] * 1LL;
}
sort(a + 1, a + 1 + n);
LL x = s / n;
LL y = s % n;
LL num = 0;
for (i = 1; i <= n; ++i)
{
LL xx = x;
if (i >= n - y + 1)
xx++;
if (a[i] <= xx)
{
num += xx - a[i];
}
}
if (k >= num)//如果k比num大,最后肯定是一种平衡状态，怎么操作都不会变了，要么是0，要么是1
{
if (y == 0)
printf("0\n");
else
printf("1\n");
}
else
{
int k1 = k, k2 = k;
int mmin, mmax;
for (it = ++st.begin(); it != st.end(); ++it)//正着跑一边，看k次操作后的最小值是多少
{
tt = --it;
++it;
if (mp[*tt] * (*it - *tt) * 1LL > k1 )//注意两个相乘可能爆int，所以map直接longlong即可
{
mmin = (*tt + k1 / mp[*tt]);
break;
}
else if (mp[*tt] * (*it - *tt) * 1LL == k1)
{
mmin = *it;
break;
}
else
{
k1 -= mp[*tt] * (*it - *tt);
mp[*it] += mp[*tt];
}
}
for (rt = st.rbegin(); rt != --st.rend(); ++ rt)//反着跑一边求出最大值
{
rtt = ++rt;
--rt;
if (mp[*rt] * (*rt - *rtt) * 1LL > k2)
{
mmax = *rt - (k2 / mp[*rt]);
break;
}
else if (mp[*rt] * (*rt - *rtt) * 1LL == k2)
{
mmax = *rtt;
break;
}
else
{
k2 -= mp[*rt] * (*rt - *rtt);
mp[*rtt] += mp[*rt];
}
}
printf("%d\n", mmax - mmin);
}
}
}