Submission #72180731

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Source Code Expand

#include <iostream>
#include <vector>
using namespace std;

int main (){
    int N;
    cin >> N;
    vector<int> cnt(N + 1, 0);

    // 遍历所有可能的(x, y)对，计算n = x2 + y2
    for (int x = 1; x * x <= N / 2; ++x) {
        for (int y = x + 1; y * y <= N - x * x; ++y) {
            int n = x * x + y * y;
            if (n <= N) {
                ++cnt[n];
            }
        }
    }

    // 收集所有good integers
    vector<int> good;
    for (int n = 1; n <= N; ++n) {
        if (cnt[n] == 1) {
            good.push_back(n);
        }
    }

    // 输出结果
    cout << good.size() << endl;
    for (size_t i = 0; i < good.size(); ++i) {
        if (i > 0) cout << " ";
        cout << good[i];
    }
    if (!good.empty()) cout << endl;

    return 0;
}

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