#include <bits/stdc++.h>
#include "gift.h"
using namespace std;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef long long ll;
typedef pair<ll,ll> pll;
typedef tuple<int,int,int> ti;
typedef unsigned long long ull;
typedef long double ld;
typedef vector<ll> vll;
typedef pair<ld,ld> pld;
#define pb push_back
#define popb pop_back()
#define pf push_front
#define popf pop_front
#define ff first
#define ss second
#define MOD (int)(1e8)
#define INF (ll) (1e18)
#define all(v) (v).begin(),(v).end()
ll gcd(ll a , ll b) {return b ? gcd(b , a % b) : a ;}
ll lcm(ll a , ll b) {return (a * b) / gcd(a , b);}
ld pointdist(ld x, ld y, ld point) { return ((x-point)*(y-point))/abs(x-y); }
ld dist(ld x, ld y, ld a, ld b){ return sqrt((x-a)*(x-a) + (y-b)*(y-b)); }
const int nx[4] = {0, 0, 1, -1}, ny[4] = {1, -1, 0, 0}; //East, West, South, North+
////////////******SOLUTION******\\\\\\\\\\\
int sz;
vi A, B, X;
int construct(int n, int r, vi a, vi b, vi x)
{
sz = n;
A = a;
B = B;
X = x;
string s(n, 'R');
bool all_one = true;
bool all_two = true;
for(int i = 0; i <n; i ++)
{
if(x[i] != 1)
all_one = false;
if(x[i] != 2)
all_two = false;
}
if(all_one)
{
craft(s);
return 1;
}
else if(all_two)
{
for(int i = 0; i <n; i ++)
{
if(a[i] == b[i] || n == 1)
return 0;
}
for(int j = 1; j < n; j += 2)
s[j] = 'B';
craft(s);
return 1;
}
return 0;
}
/*
Identify problem diagram: Brute force, Greedy, Dynamic Programming, Divide and Conquer
Reformulate the problem into something more theoretical
!!!!! IMPLICIT GRAPH ??????
!!!!! PAY ATTENTION TO THE CONSTRAINTS: DP nD ? BF ? BITMASKING ?
!!!!! SOLVE THE SUBTASKS FIRST: IT'S TOTALLY OK NOT TO SOLVE THE PROBLEM ENTIRELY
Search for multiple approaches: select the seemingly optimal one
Remember that you're the king of CP
Change your approach
Imagine Corner cases before submitting
Don't spend too much time on the problem: move on !
*/
