牛客测试 - 序列和

思路参考：

C++代码实现

#include<iostream>

using namespace std;

int main() {
    long n,l;
    cin >> n >> l;
    double a = 0;
    long i = l;
    for(; i<=100; i++) {
        double b = (double)(2*n - i*(i-1)) / (2 * i);
        if(b - long(b) == 0) {
            a = b;
            break;
        }
    }

    if(i>100) puts("No");
    else {
        for(int j = 0; j<i; j++) {
            if(j==i-1)
                cout << (long)(a + j);
            else
                cout << (long)(a + j) << " ";
        }
    }

    return 0;
}