暴力穷举跟哈希map都会超时，故用二分方法来写
用结构体存放s = c * c + d * d,c,d确立区间[l,r]为s的取值范围
判断条件：s按从小到大的顺序排序，具有二段性
字典序排序，重载小于号
循环枚举a和b,二分是找到最小的大于等于n−a×a−b×b的数对,更新区间
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>

using namespace std;
const int N = 25 * 1e6 + 10;
struct Sum{
    int s,c,d;
    bool operator< (const Sum &t)const
    {
        if(s != t.s)    return s < t.s;
        if(c != t.c)    return c < t.c;
        return d < t.d;
    }
}sum[N];
int n,m;

int main(){
    cin >> n;
    for(int c = 0;c * c <= n;++c){
        for(int d = c;c * c + d * d <= n;++d){
            sum[m++] = {c * c + d * d,c,d};
        }
    }
    sort(sum,sum + m);
    for(int a = 0;a * a <= n;++a){
        for(int b = a;a * a + b * b <= n;++b){
            int t = n - a * a - b * b;
            int l = 0,r = m - 1;
            while(l < r){
                int mid = l + r >> 1;
                if(sum[mid].s >= t) r = mid;
                else    l = mid + 1;
            }
            if(sum[l].s == t){
                printf("%d %d %d %d\n",a,b,sum[l].c,sum[l].d);
                return 0;
            }
        }
    }
    return 0;
}