967B - Watering System - CodeForces Solution

/*      Jai Ganesh Ji
        Jai Mata Dii
        Ritesh Sharma      */
#include<bits/stdc++.h>
using namespace std;

void Lets_Gooo() {
// fast_io;
#ifndef ONLINE_JUDGE

freopen("input1.txt", "r", stdin);
freopen("output.txt", "w", stdout);

#endif // ONLINE_JUDGE 
}

/* ascii value
A=65,Z=90,a=97,z=122
*/
 
// Techniques :
// divide into cases, brute force, pattern finding
// sort, greedy, binary search, two pointer
// transform into graph



#define RITESH              ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define ll                  long long int
// #define mp                  make_pair
#define nt                  _int128
#define ld                  long double
#define vi                  vector<int>
#define vll                 vector< long long int>
#define vb                  vector<bool>
#define pii                 pair<int, int>
#define pll                 pair<long long int, long long int>
#define ull                 unsigned long long
long double                 PI =3.14159265358979323846;
#define vvi                 vector<vector<int> >
// #define vll                 vector<long long>
#define pb                  push_back
#define eb                  emplace_back
// #define l                   long long 
#define MOD                 1000000007 // 998244353
#define vi                  vector<int>
#define mp(a, b)            make_pair(a, b)
#define vpii                vector<pair<int, int> >
// #define mk(arr, n, type)    type *arr = new type[n];
#define For(i, a, b)        for (int(i) = (a); (i) < (b); ++(i))
#define rfor(i, a, b)       for (int(i) = (a)-1; (i) >= (b); --(i))
#define FORALL(i, a)        for (auto& (i) : (a))
#define printall(a)         for (auto& (i) : (a)) cout << i << ' '
#define print(a)            cout << a << '\n'
#define sortarr(a)          sort((a).begin(), (a).end())
#define rsort(a)            sort((a).rbegin(), (a).rend())
#define w(x)                int x; cin >> x; while(x--)
#define all(c)              (c).begin(), (c).end()
#define sz(c)               (int)(c).size()
#define pqmx                priority_queue<int>
#define pqmn                priority_queue<int, vector<int>, greater<int> >
#define max3(a, b, c)       max((a), max((b), (c)))
#define min3(a, b, c)       min((a), min((b), (c)))
#define mx_all(c)           *max_element((c).begin(), (c).end())
#define mn_all(c)           *min_element((c).begin(), (c).end())
#define lwr_b(c, a)         lower_bound((c).begin(), (c).end(), (a)) - ((c).begin())
#define upr_b(c, a)         upper_bound((c).begin(), (c).end(), (a)) - ((c).begin())
#define llmax               9223372036854775807
#define endl                '\n'
#define debug(n) cout<<(n)<<endl;
//..........................................................................

// ll gcd(ll a, ll b){if (b == 0)return a;return gcd(b, a % b);}


long long int PowR(ll a , ll b){if( b == 0 ) return 1;long rec = PowR(a,b/2);if(b&1){    return a * rec * rec;}else return rec * rec;}

const ll mod = 1e9 + 7;

ll Pow(ll a , ll b){ll ans = 1;while(b){if(b&1){ans  = (ans * 1ll * a );}a *= a;b >>= 1;}return ans;}

ll inv(ll i) {if (i == 1) return 1; return (mod - ((mod / i) * inv(mod % i)) % mod) % mod;}

ll gcd(ll a, ll b) { if (b == 0) return a; return gcd(b, a % b);}

ll lcm(ll a, ll b){return (a / gcd(a, b)) * b;}

ll ceil_div(ll a, ll b) {return a % b == 0 ? a / b : a / b + 1;}

ll maxx(ll a , ll b) {if(a > b) return a;else return b;}
 
ll pwr(ll a, ll b) {a %= mod; ll res = 1; while (b > 0) {if (b & 1) res = res * a % mod; a = a * a % mod; b >>= 1;} return res;}

ll mod_add(ll a, ll b, ll m) {a = a % m; b = b % m; return (((a + b) % m) + m) % m;}

ll mod_mul(ll a, ll b, ll m) {a = a % m; b = b % m; return (((a * b) % m) + m) % m;}

ll mod_sub(ll a, ll b, ll m) {a = a % m; b = b % m; return (((a - b) % m) + m) % m;}

// ll mod_div(ll a, ll b, ll m) {a = a % m; b = b % m; return (mod_mul(a, mminvprime(b, m), m) + m) % m;}  //only for prime m

ll phin(ll n) {ll number = n; if (n % 2 == 0) {number /= 2; while (n % 2 == 0) n /= 2;} for (ll i = 3; i <= sqrt(n); i += 2) {if (n % i == 0) {while (n % i == 0)n /= i; number = (number / i * (i - 1));}} if (n > 1)number = (number / n * (n - 1)) ; return number;} //O(sqrt(N))
 
// ll mod_div(ll a, ll b, ll m) {a = a % m; b = b % m; return (mod_mul(a, mminvprime(b, m), m) + m) % m;}  //only for prime m

//****************************Template Ends*******************************//

const int N = 1e5 + 7; 
// const ll N = 1e7;
// const ll N = 1e9+7;
// ll dp[N];
// const ll M = 1e9+7;

void solve(){
    int n , a , b , total = 0 , count = 0;
    cin >> n >> a >> b;
    vi arr(n);

    for(auto &i : arr) {
        cin >>i;
        total += i;
    }
    ll ans = arr[0];
    sort(arr.begin() + 1 , arr.end());
    ll sum = 0;
    for(int i = n-1 ; i > 0 ; i--){
        if((a * ans)/total >= b){
            break;
        } 
        count++;
        total -= arr[i];
    }
    print(count);

}
int main() {
    Lets_Gooo();
    RITESH;      
    // long long testCase;
    // cin>>testCase;
    // while(testCase--){
    // if (solve())cout<<"YES"<<endl;
        // else cout<<"NO"<<endl;
        solve();
        // cout<<abs(-6);
        // }   
    // auto sum = [](int a , int b){return a + b;} ;
    // cout<<sum(3,5);

    return 0; 

}