根据需计算的数据量打表，分析各数作用，可有效降低运行时间。

方法一：直接开数组存储

#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
const int N = 5e6;
int a[N],b[N];
bool c[N];
int main(){
    int k;
    int n;
    cin>>n;
    for(int i=0;i*i<=n/2;i++){
        for(int j=i;i*i+j*j<=n;j++){
            k = i*i + j*j;
            if(k<5e6 && !c[k]){
                a[k] = i;
                b[k] = j;
                c[k] = true;
            }
        }
    }
    for(int i=0;i*i<=n/2;i++){
        for(int j=i;i*i+j*j<n/2;j++){
            if(c[n-(i*i+j*j)]){
                cout<<i<<" "<<j<<" "<<a[n-(i*i+j*j)]<<" "<<b[n-(i*i+j*j)];
                return 0;
            }
        }
    }
}

方法二：用unorder_map记录(开启O2/O3优化可过)

#pragma GCC optimize(2)
#include <cstring>
#include <iostream>
#include <algorithm>
#include <unordered_map>
#define x first
#define y second
using namespace std;
typedef pair<int, int> PII;
int n, m;
unordered_map<int, PII> S;

int main(){
    cin>>n;
    for (int c = 0; c * c <= n / 2; c ++ )
        for (int d = c; c * c + d * d <= n; d ++ ){
            int t = c * c + d * d;
            if (S.count(t) == 0) S[t] = {c, d};
        }
    for (int a = 0; a * a <= n; a ++ )
        for (int b = a; a * a + b * b <= n / 2; b ++ ){
            int t = n - a * a - b * b;
            if (S.count(t)){
                printf("%d %d %d %d\n", a, b, S[t].x, S[t].y);
                return 0;
            }
        }
    return 0;
}

此时发现超时，弃用unorder_map，直接采用pair数组+状态数组，速度与数组实现一致，远快于采用unorder_map。

方法三：用pair记录

#include <cstring>
#include <iostream>
#include <algorithm>
#define x first
#define y second
using namespace std;
typedef pair<int, int> PII;
int n, m;
const int N = 5e6;
PII S[N];
bool H[N];

int main(){
    cin >> n;
    for (int a = 0; a * a <= n/2; a ++ )
        for (int b = a; a * a + b * b <= n; b ++ ){
            int t = a * a + b * b;
            if (!H[t]){
                S[t] = {a, b};
                H[t] = true;
            } 
        }
    for (int a = 0; a * a <= n/2; a ++ )
        for (int b = 0; a * a + b * b <= n/2; b ++ ){
            int t = n - a * a - b * b;
            if (H[t]){
                printf("%d %d %d %d\n", a, b, S[t].x, S[t].y);
                return 0;
            }
        }
    return 0;
}

方法四：用结构体记录，并进行符号重载。sort排序，用二分的方式查找对应值所在的结构体。(开启O2优化可发现优化效果明显)
注意：二分查找时，应采用查找第一个值的二分查找法。

//查找不小于x的第一个位置
int bsearch1(int l, int r){
    while(l<r){
        int mid = (l+r)>>1;
        if(a[mid]<x) l = mid+1;
        else r = mid;
    }
    return l;
}

#pragma GCC optimize(2)
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
const int N = 2500010;
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 = 0; 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;
}