Idea: It is to construct an arithmetic sequence with n elements in it, and ensure that there must be x and y in it, and construct an arithmetic sequence with the smallest sum.
Because the data range of this topic is very small, we should try our best to be in x and y Put some more elements in between, then directly enumerate between x and y to find the smallest tolerance, find the first element after determining the tolerance, and then output it.
I have been thinking about the 1200 question for a long time. It's too bad! !
#include <bits/stdc++.h>
using namespace std;
int main()
{
int t; cin>>t;
while(t--){
int x,y,n; cin>>n>>x>>y;
int s = y-x;
int ans = 0;
for(int i=0;i<=n-2;i++){
if(s%(i+1)==0){
ans = i;
}
}
//cout<<ans<<endl;
int step = s/(ans+1);
//cout<<step<<endl;
int begin1 = x;
for(int i=n-ans-2;i>0;i--){
if(begin1-step>0){
begin1-=step;
}
else break;
}
for(int i=0;i<n;i++){
cout<<begin1+step*i<<" ";
}
cout<<endl;
}
return 0;
}