E. Minimizing Difference
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
You are given a sequence a1,a2,…,ana1,a2,…,an consisting of nn integers.
You may perform the following operation on this sequence: choose any element and either increase or decrease it by one.
Calculate the minimum possible difference between the maximum element and the minimum element in the sequence, if you can perform the aforementioned operation no more than kk times.
Input
The first line contains two integers nn and kk (2≤n≤105,1≤k≤1014)(2≤n≤105,1≤k≤1014) — the number of elements in the sequence and the maximum number of times you can perform the operation, respectively.
The second line contains a sequence of integers a1,a2,…,ana1,a2,…,an (1≤ai≤109)(1≤ai≤109).
Output
Print the minimum possible difference between the maximum element and the minimum element in the sequence, if you can perform the aforementioned operation no more than kk times.
Examples
input
Copy
4 5 3 1 7 5
output
Copy
2
input
Copy
3 10 100 100 100
output
Copy
0
input
Copy
10 9 4 5 5 7 5 4 5 2 4 3
output
Copy
1
Note
In the first example you can increase the first element twice and decrease the third element twice, so the sequence becomes [3,3,5,5][3,3,5,5], and the difference between maximum and minimum is 22. You still can perform one operation after that, but it's useless since you can't make the answer less than 22.
In the second example all elements are already equal, so you may get 00 as the answer even without applying any operations.
刚看见这道题的时候满脑子都是二分策略
其实在纸上随便写写就能发现排好顺序周要么升高头
要么就降低尾部 对于这两种方法我们贪心的选择消耗少的即可
#include<bits/stdc++.h>
using namespace std;
int a[100005];
int main()
{
int n;
long long k;
scanf("%d%lld",&n,&k);
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
}
sort(a+1,a+1+n);
int pos1=1;
int pos2=n;
long long ans=a[n]-a[1];
while(pos1<pos2)
{
long long tmp=0;
if(pos1-1<=n-pos2)
{
pos1++;
tmp=min(1LL*a[pos1]-a[pos1-1],k/(pos1-1));
ans-=tmp;
k-=1LL*tmp*(pos1-1);
}
else
{
pos2--;
tmp=min(1LL*a[pos2+1]-a[pos2],k/(n-pos2));
ans-=tmp;
k-=1LL*tmp*(n-pos2);
}
// cout<<tmp<<endl;
//if(tmp==0) break;
}
printf("%lld\n",ans);
}
