C. Maximum Median
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
You are given an array aa of nn integers, where nn is odd. You can make the following operation with it:
Choose one of the elements of the array (for example aiai) and increase it by 11 (that is, replace it with ai+1ai+1).
You want to make the median of the array the largest possible using at most kk operations.
The median of the odd-sized array is the middle element after the array is sorted in non-decreasing order. For example, the median of the array [1,5,2,3,5][1,5,2,3,5] is 33.
Input
The first line contains two integers nn and kk (1≤n≤2⋅1051≤n≤2⋅105, nn is odd, 1≤k≤1091≤k≤109) — the number of elements in the array and the largest number of operations you can make.
The second line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109).
Output
Print a single integer — the maximum possible median after the operations.
Examples
input
Copy
3 2
1 3 5
output
Copy
5
input
Copy
5 5
1 2 1 1 1
output
Copy
3
input
Copy
7 7
4 1 2 4 3 4 4
output
Copy
5
Note
In the first example, you can increase the second element twice. Than array will be [1,5,5][1,5,5] and it's median is 55.
In the second example, it is optimal to increase the second number and than increase third and fifth. This way the answer is 33.
In the third example, you can make four operations: increase first, fourth, sixth, seventh element. This way the array will be [5,1,2,5,3,5,5][5,1,2,5,3,5,5] and the median will be 55.
题意主要是说：给一个数组，每次选择给一个数加一，执行b此操作之后，问中位数最大是多少；
显然，中位数是数组排序之后的中位数，先给数组排序，前一半不做任何处理，对后一半进行操作。之后就类似于填坑，只要当前和后边齐平就可以。
二分去找那个数。
要注意开ll；
1 #include<stdio.h>
2 #include<algorithm>
3 #include<iostream>
4 #include<string.h>
5 #include<queue>
6 #include<vector>
7 #include<map>
8 #include<math.h>
9 using namespace std;
10 typedef long long ll;
11 const int mod=1e9+7;
12 const int PI=acos(-1.0);
13 const int maxn=200010;
14 const int inf=1e7+5;
15 ll n,k;
16 ll a[maxn];
17 ll check(ll m)
18 {
19 ll sum=0;
20 int t=n/2+1;//找中位数
21 for(int i=n;i>=t;i--){
22 if(m>a[i]){
23 sum+=m-a[i];//填到mid所需要的代价
24 }
25 }
26 if(sum<=k){//填坑操作次数如果<=k说明操作可以实现
27 return 1;
28 }
29 return 0;
30 }
31 int main()
32 {
33 cin>>n>>k;
34 ll ans=0;
35 for(int i=1;i<=n;i++)
36 cin>>a[i];
37 sort(a+1,a+1+n);
38 ll l=1,r=2e9;
39 while(l<=r)
40 {
41 int mid=(l+r)/2;
42 if(check(mid))
43 {
44 ans=mid;
45 l=mid+1;
46 }
47 else
48 r=mid-1;
49 }
50 cout<<ans;
51 }