Gym - 102035D:Coach Ayoub

思路：找到兩邊邊界值輸出即可。有坑點。 

#include <bits/stdc++.h>
using namespace std;
int main(){
    int n,a,b,s;
    cin >>n>>a>>b>>s;
    int l,r;
    l = (n-1)*a+b,r=(n-1)*b+a;
    if(s>=l&&s<=r)cout <<"YES"<<endl;
    else cout <<"NO"<<endl;    
    return 0;
}