D. Robin Hood
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 11 1 4 2
output
2
input
3 12 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：给你n个数，每次能从最大的数一个移给最小的数，问k次操作后，最大和最小的数的差。
思路 : 题目可以简化成找最终状态下的最大值和最小值，那么分别进行二分最大值和最小值，此外注意二分最大、最小时的不同之处。
1 #include <stdio.h>
2 #include <iostream>
3 #include <algorithm>
4 #include <string.h>
5 using namespace std;
6
7 const int MAX = 500010;
8 const int INF = 0x3f3f3f3f;
9 int a[MAX], n;
10
11 int findmax(int x, int k)
12 {
13 for(int i = 0; i < n; i++)
14 {
15 if(a[i] > x)
16 k-=a[i] - x;
17 if(k < 0)
18 return 0;
19 }
20 return 1;
21 }
22 int findmin(int x, int k)
23 {
24 for(int i = 0; i < n; i++)
25 {
26 if(a[i] < x)
27 k-=x - a[i];
28 if(k < 0)
29 return 0;
30 }
31 return 1;
32 }
33
34 int main()
35 {
36 int k;
37 __int64 sum = 0;
38 int t1, t2;
39 cin >> n >> k;
40 t1 = 0;
41 t2 = INF;
42 for(int i = 0; i < n; i++)
43 {
44 scanf("%d", a + i);
45 sum += a[i];
46 t1 = max(t1, a[i]);
47 t2 = min(t2, a[i]);
48 }
49 int mx = sum / n + (sum%n>0 ?1 : 0);
50 int mi = sum / n;
51
52 int l , r, maxx , minn;
53 /////mi
54 l = t2, r = mi;
55 while(l <= r) //注意等于
56 {
57 int mid = (l + r) >> 1;
58 if(findmin(mid, k))
59 {
60 l = mid + 1;
61 minn = mid;
62 }
63 else r = mid - 1;
64 }
65 ////ma
66 l = mx, r = t1;
67 while(l < r)
68 {
69 int mid = (l + r) >> 1;
70 if(findmax(mid, k))
71 {
72 r = mid;
73 maxx = max(mid, maxx);
74 }
75 else l = mid + 1;
76 }
77 maxx = (l + r) >>1; //结束状态时取
78 printf("%d\n", maxx - minn);
79 }