题解 | #序列和#

#include<iostream>

using namespace std;
int main(){
    
    int N=0,L=0;
    bool have_out=0;
    while(cin>>N>>L){
        have_out=0;
        while(L<=100&&N>=(L*(L-1)/2)){      
            if((2*N-L*L+L)%(2*L)==0){
                int n=(2*N-L*L+L)/(2*L);
                for(int i=0;i<L;i++){
                    cout<<n+i;
                    if(i!=L-1){
                        cout<<' ';
                    }
                }
                cout<<endl;
                have_out=1;
                break;
            }
            L++;
        }
        
        if(have_out==0){
            cout<<"No"<<endl;
        }
    }
    return 0;
    
    
}