题解 | #序列和#

#include<bits/stdc++.h>
using namespace std;
const double EPS = 1e-6;
int main(){
    int n, l;
    cin >> n >> l;
    int curLen = l;
    while(curLen <= 100) {
        double a = (double)(2 * n - curLen * (curLen - 1)) / (2 * curLen);
        if(a >= 0 && fabs(a - (int)a) < EPS){
            int b = a;
            for(int j = 0; j < curLen; j++){
                printf("%.0lf ", a++);
            }
            break;
        }
        curLen++;
    }
    if(curLen > 100){
        cout << "No" << endl;
    }
    return 0;
}