AcWing 四平方和 二分

 

 考虑下数据范围：N = 5e10

所以a，b，c，d每个数都一定<= sqrt(N) < 2300

最简单的想法是三层for循环枚举a，b，c，然后计算出d，但是在AcWing上这一题的数据加强后会超时

//超时 
#include <bits/stdc++.h>
using namespace std;
int main() { 
    int n;
    cin >> n;
    for (int a = 0; a * a <= n; a++) {
        for (int b = a; a * a + b * b <= n; b++) {
            for (int c = b; a * a + b * b + c * c <= n; c++) {
                int t = n - a * a - b * b - c * c;
                int d = sqrt(t);
                if (d * d == t) {
                    cout << a << " " << b << " " << c << " " << d << endl;
                    return 0;
                }
            }
        }
    }
    return 0;
}

按照2300的数据范围看时间复杂度，最多只能两层for循环枚举两个数

考虑用空间换时间

二分做法AC代码，时间复杂度n ^ 2 log n

#include <bits/stdc++.h>
using namespace std;
const int N = 2500010;
struct Sum { //用结构体存c ^ 2 + d ^ 2 和 c 和 d这三个数 
    int s, c, d;
} sum[N];
bool cmp(Sum s1, Sum s2) {
    if (s1.s != s2.s) {
        return s1.s < s2.s;
    }
    if (s1.c != s2.c) {
        return s1.c < s2.c;
    }
    return s1.d < s2.d;
}
int main() {
    int n, m; //m表示所有组合的个数 
    cin >> n;
    for (int c = 0; c * c <= n; c++) { //存下c和d的组合 
        for (int d = c; c * c + d * d <= n; d++) {
            Sum t;
            t.s = c * c + d * d;
            t.c = c;
            t.d = d;
            sum[m++] = t;
            //sum[m++] = {c * c + d * d, c, d}; //c++11的新特性 
        }
    } 
    sort (sum, sum + m, cmp);
    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; //差值
            //找到 >= 这个差值的最小的一个数 
            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) {
                cout << a << " " << b << " " << sum[l].c << " " << sum[l].d;
                return 0;
            }
        }
    } 
    return 0;
}

然后是unordered_map哈希做法，时间复杂度n ^ 2，结果正确但是超时了，不能AC，不过可以熟悉下unordered_map的用法

#include <bits/stdc++.h>
using namespace std;
typedef pair<int, int> PII;
unordered_map<int, PII> S;
int main() {
    int n;
    cin >> n;
    for (int c = 0; c * c <= n; c++) {
        for (int d = c; c * c + d * d <= n; d++) {
            int t = c * c + d * d;
            if (S.count(t) == 0) { //如果没有t的话 
                //S[t] = {c, d};
                S[t] = make_pair(c, d);
            }
        }
    }
    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;
            if (S.count(t)) {
                cout << a << " " << b << " " << S[t].first << " " << S[t].second << endl;
                return 0;
            }
        }
    }
    return 0;
}