Ideal Pyramid
Gym - 102411I
https://vjudge.net/problem/Gym-102411I/origin
主要突破点在45度，而且这个正方形的边是平行于坐标轴的，让每个柱子全都沿着4个方向摔倒就可以了，然后分别取最大最小值，柱子就在中间

#include <bits/stdc++.h>
#define inf 2333333333333333
#define N 1000010
#define p(a) putchar(a)
#define For(i,a,b) for(long long i=a;i<=b;++i)
//by war
//2020.9.21
using namespace std;
long long n,l,r,u,d;
struct node{
    long long x;
    long long y;
    long long h;
}a[N];
void in(long long &x){
    long long y=1;char c=getchar();x=0;
    while(c<'0'||c>'9'){if(c=='-')y=-1;c=getchar();}
    while(c<='9'&&c>='0'){ x=(x<<1)+(x<<3)+c-'0';c=getchar();}
    x*=y;
}
void o(long long x){
    if(x<0){p('-');x=-x;}
    if(x>9)o(x/10);
    p(x%10+'0');
}

signed main(){
    in(n);
    l=inf;r=-inf;d=inf;u=-inf;
    For(i,1,n){
        in(a[i].x);in(a[i].y);in(a[i].h);
        l=min(l,a[i].x-a[i].h);
        r=max(r,a[i].x+a[i].h);
        d=min(d,a[i].y-a[i].h);
        u=max(u,a[i].y+a[i].h);
    }
    o((l+r)/2);p(' ');o((u+d)/2);p(' ');
    if(r-l>=u-d) o((r-l+(r-l)%2)/2);
    else o((u-d+(u-d)%2)/2);
    return 0;
}