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CODECHEF TV Discount Problem Solution (TVDISC)

In this Codechef Tv discount problem solution Chef is looking to buy a TV and has shortlisted two models. The first one costs AA rupees, while the second one costs BB rupees. Since there is a huge sale coming up on Chefzon, Chef can get a flat discount of CC rupees on the first TV, and a flat discount of DD rupees on the second one. Help Chef determine which of the two TVs would be cheaper to buy during the sale.

CODECHEF TV Discount Problem Solution (TVDISC)

CODECHEF TV Discount Problem Solution (TVDISC)

//Divyansh Vinayak
using namespace std;
#define gc getchar_unlocked
#define fo(i,n) for(i=0;i<n;i++)
#define Fo(i,k,n) for(i=k;k<n?i<n:i>n;k<n?i+=1:i-=1)
#define ll long long
#define si(x)	scanf("%d",&x)
#define sl(x)	scanf("%lld",&x)
#define ss(s)	scanf("%s",s)
#define pi(x)	printf("%d",x)
#define pl(x)	printf("%lld",x)
#define ps(s)	printf("%s",s)
#define pnl()   printf("\n")
#define deb(x) cout << #x << "=" << x << endl
#define deb2(x, y) cout << #x << "=" << x << "," << #y << "=" << y << endl
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define all(x) x.begin(), x.end()
#define clr(x) memset(x, 0, sizeof(x))
#define sortall(x) sort(all(x))
#define tr(it, a) for(auto it = a.begin(); it != a.end(); it++)
#define PI 3.1415926535897932384626
typedef pair<int, int>	pii;
typedef pair<ll, ll>	pl;
typedef vector<int>		vi;
typedef vector<ll>		vl;
typedef vector<pii>		vpii;
typedef vector<pl>		vpl;
typedef vector<vi>		vvi;
typedef vector<vl>		vvl;
int mpow(int base, int exp); 
void ipgraph(int m);
void dfs(int u, int par);
const int mod = 1000000007;
const int N = 18 * 103, M = N;
#define top(a,b) (ll)((a+b-1)/b)
vi g[N];
void matrix(ll n, ll m){
  char s[101][101];
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){

int fact(ll n);

// Returns factorial of n
int fact(ll n)
    ll res = 1;
    for (int i = 2; i <= n; i++)
        res = res * i;
    return res;

int gcd(int a, int b)
    if (a == 0)
        return b;
    return gcd(b % a, a);

int main() {
  ll t;
  cin >> t;
    ll a , b, c, d;
    cin >> a >> b >> c >> d;
    a = a - c;
    b = b - d;
    if(a > b)
        cout << "SECOND" << endl;
    else if(a < b)
        cout << "FIRST" << endl;
    else if(a == b)
        cout << "ANY" << endl;
  return 0;

int mpow(int base, int exp) {
  base %= mod;
  int result = 1;
  while (exp > 0) {
    if (exp & 1) result = ((ll)result * base) % mod;
    base = ((ll)base * base) % mod;
    exp >>= 1;
  return result;

void ipgraph(int n, int m){
	int i, u, v;

void dfs(int u, int par){
	for(int v:g[u]){
		if (v == par) continue;
		dfs(v, u);

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