题目链接:https://codeforc.es/blog/entry/68911
题意
对与n个数,有m次+1操作,可以选择任意数字进行+1,要求m次操作后,n个数排序后的中位数最大,n满足一定是奇数。
The first line contains two integers n and k (1≤n≤2⋅10^5, n is odd, 1≤k≤10^9) — the number of elements in the array and the largest number of operations you can make.
思路
二分答案
首先排序,再二分答案,对n/2+1到n之间的数计算当前的最小操作,设x为中位数,则最少次数为
,当前中位数判断是否满足操作次数<=m,不满足取较小区间,满足取较大区间。
#include<bits/stdc++.h>
using namespace std;
#define MAXN 200005
#define ll long long
int n,m;
void read(int &x)
{
x=0;
bool flag=0;
char ch=getchar();
if(ch=='-') flag=1;
while(ch<'0'||ch>'9')ch=getchar();
while(ch>='0'&&ch<='9')x*=10,x+=ch-'0',ch=getchar();
if(flag) x=-x;
}
int ff[MAXN];
bool check(int x)
{
ll sum=0;
for(int i=n/2+1;i<=n;i++)
sum+=max(x-ff[i],0);
if(sum<=m)
return true;
else
return false;
}
int main()
{
read(n),read(m);
for(int i=1;i<=n;i++)
read(ff[i]);
sort(ff+1,ff+n+1);
ll l = 0, r = 2000000009;
while (l<r)
{
ll mid = (l + r + 1) / 2;
if (check(mid))l = mid;
else
r = mid - 1;
}
printf("%lld",l);
return 0;
}