1928E - Modular Sequence - CodeForces Solution

#include <iostream>
#include <algorithm>
#include <vector>
#include <utility>
#include <cstdio>
#include <stack>
#include <fstream>
#include <deque>
#include <bitset>
#include <queue>
#include <set>
#include <map>
#include <cmath>
#include <climits>
#include <cstdint>
#include <string>
#include <cassert>
 
using namespace std;
 
using ll = long long;
using ld = long double;
 
#define f first
#define s second
#define mp make_pair
 
const ll MOD = 1e9+7; //998244353
const int dx[4] = {0,-1,0,1}, dy[4] = {-1,0,1,0};
//const int MXN = 5e5;
 
int main(){
	pair<int,int> dp[400001];
	dp[0] = mp(0,0);
	for(int x=1; x<=4e5; x++){
		dp[x] = mp(2e5+1,2e5+1);
		for(int y=2; y*(y-1)<=x; y++){
			int last = x-y*(y-1);
			if(dp[last].f+y<dp[x].f){
				dp[x] = mp(dp[last].f+y,y);
			}
		}
	}
	int tt; 
	bool READ = 1;
	if(READ) cin>>tt;
	else tt = 1;
	while(tt--){
		int n;
		ll a,b,s; cin>>n>>a>>b>>s;
		vector<int> ans;
		for(int x=1; x<n && ans.size()==0; x++){
			s-=a+b*(x-1);
			ll val = (a%b)*(n-x);
			if(s<val) continue;
			if((s-val)%b!=0) continue;
			val = 2*(s-val)/b;
			if(val>4e5 || val<0) continue;
			if(dp[(int)val].f<=n-x){
				for(int y=0; y<x; y++) ans.push_back(a+b*y);
				int cur = val;
				while(cur>0){
					for(int y=0; y<dp[cur].s; y++){
						ans.push_back((a%b)+b*y);
					}
					cur-=dp[cur].s*(dp[cur].s-1);
				}
				while(ans.size()<n) ans.push_back((a%b));
			}
		}
		if(ans.size()==0 && s==a+b*(n-1)){
			for(int x=0; x<n; x++) ans.push_back(a+b*x);
		}
		cout<<(ans.size()==0?"No":"Yes")<<endl;
		if(ans.size()){
			for(int x: ans) cout<<x<<" ";
			cout<<endl;
		}
	}
}