题解 | #序列和#

#include <iostream>
using namespace std;

int main() {
    int N, L;
    cin >> N >> L;
    bool is_find = false;
    for (int i = L; i <= 100; i++) {
        //当N的大小小于从0开始的连续L个数字之和时，必不可能成功
        if (N < (i - 1)*i / 2) break;
        // 对当前连续数字是奇数个还是偶数个进行区分
        if (i % 2 == 0) {
            // 偶数逻辑
            // 如果当前无法均分，则继续
            if (N % (i / 2) != 0 || (N / (i / 2)) % 2 == 0) continue;
            int center = N / (i / 2);
            int left = center / 2, right = center / 2 + 1;
            if (left - i / 2 + 1 < 0) continue;
            for (int j = left - i / 2 + 1; j < right + i / 2; j++) {
                cout << j << " ";
            }
            cout << endl;
            is_find = true;
            break;
        } else {
            if (N % i != 0) continue;
            int center = N / i;
            if (center - i / 2 < 0) continue;
            for (int j = center - i / 2; j <= center + i / 2; j++) {
                cout << j << " ";
            }
            cout << endl;
            is_find = true;
            break;
        }
    }
    if (!is_find) cout << "No" << endl;
    return 0;
}