Codeforces 671B Robin Hood 【思维】

题目链接：Codeforces 671B Robin Hood

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：给定n个人的钱数，每天从最富裕的人那里拿走1元钱，给最穷的人。问k天后最大的贫富差距。

思路：仔细想想会发现是一道很水的题目。

我们先升序排列记录元素个数并去重，然后从左到右预处理前缀元素个数以及处理到第i个元素的代价和（将所有比它小的元素提上来）。算出在k天的限制下将最低标准提到多少，同理求出k天限制将最高标准降到多少，差就是结果。但是会有一些特殊情况——k很大，导致最低标准过大，最高标准过小。其实对于这样的情况，我们发现无论如何总是可以将它们的差距最小化的。这样我们求出总和，看综合能否被n整除，可以的话最小差就是0，反之是1。

AC代码：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <map>
#include <vector>
#define CLR(a, b) memset(a, (b), sizeof(a))
using namespace std;
typedef long long LL;
const int MAXN = 5 * 1e5 + 10;
int a[MAXN], numl[MAXN], numr[MAXN];
map<int, int> cnt;
LL suml[MAXN], sumr[MAXN];
int main()
{
int n, k;
while(scanf("%d%d", &n, &k) != EOF) {
cnt.clear(); LL total = 0;
for(int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
cnt[a[i]]++;
total += a[i];
}
sort(a + 1, a + n + 1);
int N = unique(a + 1, a + n + 1) - a - 1;
int L = -1, R = -1;

numl[0] = 0; suml[0] = 0; a[0] = a[1];
for(int i = 1; i <= N; i++) {
numl[i] = numl[i-1] + cnt[a[i]];
suml[i] = suml[i-1] + 1LL * numl[i-1] * (a[i] - a[i-1]);
if(suml[i] > k && L == -1) L = i - 1;
}

numr[N+1] = 0; sumr[N+1] = 0; a[N+1] = a
;
for(int i = N; i >= 1; i--) {
numr[i] = numr[i+1] + cnt[a[i]];
sumr[i] = sumr[i+1] + 1LL * numr[i+1] * (a[i+1] - a[i]);
if(sumr[i] > k && R == -1) R = i + 1;
}
//cout << L << ' ' << R << endl;
if(L >= R) {
printf(total % n ? "1\n" : "0\n");
}
else {
int Min = a[L] + (k - suml[L]) / numl[L];
int Max = a[R] - (k - sumr[R]) / numr[R];
if(Min >= Max) {
printf(total % n ? "1\n" : "0\n");
}
else {
printf("%d\n", Max - Min);
}
}
}
return 0;
}