提出 #69793927

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ソースコード 拡げる

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define ll long long
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
using namespace std;
const int N = 1e5+5;
int n;
ll mx[N],a[N],x;
struct node{ll l,r;}b[N];
char buf[1<<21],*p1,*p2;
inline int rd()
{
    char c;int f = 1;
    while(!isdigit(c = getchar()))if(c=='-')f = -1;
    int x = c-'0';
    while(isdigit(c = getchar()))x = x*10+(c^48);
    return x*f;
}
int main()
{
    // freopen(".in","r",stdin);
    // freopen(".out","w",stdout);
    n = rd();x = rd();
    for(int i = 1;i <= n;i++)a[i] = rd();
    sort(a+1,a+n+1);
    for(int i = 1;i <= n;i++)
    {
        ll l = a[i],r = a[i];
        while(r <= a[n])
        {
            if(r*2+x > a[n])b[i].l = r;
            l = l*2;r = r*2+x;
        }
        b[i].r = max(a[n],l);
    }
    sort(b+1,b+n+1,[](node x,node y){return x.l > y.l;});
    for(int i = n;i;i--)mx[i] = max(mx[i+1],b[i].r);
    ll mi = min(mx[1]-a[n],a[n]-b[n].l);
    for(int i = 1;i < n;i++)
        mi = min(mi,mx[i+1]-b[i].l);
    cout << (mi<x?0:mi) << endl;
    return 0;
}

提出情報

ジャッジ結果