Problem:
Score : 100 points

Problem Statement

Takahashi is meeting up with Aoki.

They have planned to meet at a place that is D meters away from Takahashi's house in T minutes from now.

Takahashi will leave his house now and go straight to the place at a speed of S meters per minute.

Will he arrive in time?

Constraints

1 \leq D \leq 10000

1 \leq T \leq 10000

1 \leq S \leq 10000

All values in input are integers.

Input

Input is given from Standard Input in the following format:

D T S

Output

If Takahashi will reach the place in time, print Yes; otherwise, print No.

Sample Input 1

1000 15 80

Sample Output 1

Yes

It takes 12.5 minutes to go 1000 meters to the place at a speed of 80 meters per minute. They have planned to meet in 15 minutes so he will arrive in time.

Sample Input 2

2000 20 100

Sample Output 2

Yes

It takes 20 minutes to go 2000 meters to the place at a speed of 100 meters per minute. They have planned to meet in 20 minutes so he will arrive just on time.

Sample Input 3

10000 1 1

Sample Output 3

No

He will be late.

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define sz(a) (a).size()
#define all(c) (c).begin(), (c).end()
#define rep(i, n) for (int i = 0; i < n; i++)
#define forn(i, n) for (int i = 1; i <= n; i++)
#define ford(i, n) for (int i = n; i >= 0; i--)
#define REP(i, a, n) for (int i = a; i < n; i++)
#define foreach(itr, c) for (__typeof((c).begin(), (c).end()) itr = c.begin(); itr != c.end(); itr++)
#define fill(a, b) memset(a, b, sizeof(a))
#define re return
#define INF 1e9
#define LINF 1e18
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pi;
typedef vector<int> vi;
typedef vector<char> vs;
typedef vector<long long> vl;
typedef vector<pi> vpi;

using namespace std;

int main()
{
    int n;
    int temp = 0;
    ll ans = 0;
    int S = pow(10, 9) + 7;

    cin >> n;

    vl a(n);

    rep(i, n)
    {
        cin >> a[i];
    }

    rep(i, n)
    {
        for (int j = i + 1; j < n; j++)
        {
            temp += (a[i] % S) * (a[j] % S);
        }
    }

    ans = temp % S;

    cout << ans;
}