Codeforces 1244E Minimizing Difference （二分答案）

链接：https://codeforces.com/problemset/problem/1244/E

题意：长度为n的序列，k次操作，每次可将任意一个数加一或减一，问k次操作后 最小的 最大值和最小值的差是多少

题解：很明显，序列在向中间收缩， 答案满足单调性，可二分答案，分别暴力枚举最大值和最小值，查询上下界，判断是否能在k次内完成。复杂度nlogn*logn

#include <bits/stdc++.h>
#define IO_read ios::sync_with_stdio(false);cin.tie(0)
#define fre freopen("in.txt", "r", stdin)
#define _for(i,a,b) for(int i=a; i< b; i++)
#define _rep(i,a,b) for(int i=a; i<=b; i++)
#define lowbit(a) ((a)&-(a))
using namespace std;
typedef long long ll;
template <class T>
void read(T &x)
{
    char c; bool op=0;
    while(c=getchar(), c<'0'||c>'9') if(c=='-') op=1;
    x=c-'0';
    while(c=getchar(), c>='0'&&c<='9') x=x*10+c-'0';
    if(op) x=-x;
}
template <class T>
void write(T x)
{
    if(x<0) putchar('-'), x=-x;
    if(x>=10) write(x/10);
    putchar('0'+x%10);
}

const int maxn=1e5+5;

int n;
int a[maxn];
ll k, sum[maxn];

bool check(int val)
{
    bool ok=false;
    for(int i=1; i<=n; i++)
    {
        int low=i;
        int high=lower_bound(a+1, a+1+n, a[i]+val)-a;
        ll area=1ll*low*a[low]-sum[low]+sum[n]-sum[high-1]-1ll*(n-high+1)*(a[low]+val);
        //printf("! val=%d, low=%d, high=%d, area=%I64d\n", val, low, high, area);
        if(area<=k) ok=true;
    }
    if(ok) return ok;
    for(int i=1; i<=n; i++)
    {
        int low=lower_bound(a+1, a+1+n, a[i]-val)-a-1;  //注意这里要减去1
        int high=i;
        ll area=1ll*low*(a[high]-val)-sum[low]+sum[n]-sum[high-1]-1ll*(n-high+1)*a[high];
        //printf("! val=%d, low=%d, high=%d, area=%I64d\n", val, low, high, area);
        if(area<=k) ok=true;
    }
    return ok;
}

int main()
{
    IO_read;
    //fre;
    cin>>n>>k;
    for(int i=1; i<=n; i++) cin>>a[i];
    sort(a+1, a+1+n);
    for(int i=1; i<=n; i++) sum[i]=sum[i-1]+a[i];
    int l=0, r=a[n]-a[1];
    while(l<r)
    {
        int mid=l+r>>1;
        if(check(mid)) r=mid;
        else l=mid+1;
        //printf("! l=%d, r=%d, mid=%d\n", l, r, mid);
    }
    cout<<l<<"\n";
}