# Need help in problem (ICM2008)

Problem: ICM2008 Problem - CodeChef
My solution: kishita94251 | CodeChef User Profile for Muskan | CodeChef
Showing that my answer is not correct.

1 Like

If you need test case for which your code fails is : 20 10 3 5

Actually in this question what you need to do, is if a and b are already equal then answer is obviously yes, but if they aren’t equal but c and d are equal then answer can not be Yes, because they can’t meet.

Otherwise, let k1 number of times Mr. Pr chooses c unit to move and k2 number of times he chosses d units.
In, that case if they meet then the following condition must be true:

a + k1c + k2d = b + k2c + k1d
which simplifies to:
a - b = (k2-k1)(c-d)
Since, k2 and k1 are constant, we conclude that a-b is divisble by c-d.

Implementation is simple:

``````    /*author - Ashutosh Chitranshi*/
#include<bits/stdc++.h>
using namespace std;
#pragma GCC push_options
#pragma GCC optimize ("unroll-loops")
#define watch(x) cout << (#x) << " is " << (x) << "\n"
#define watch2(x,y) cout <<(#x)<<" is "<<(x)<<" and "<<(#y)<<" is "<<(y)<<"\n"
#define pow2(x) ((x)*(x))
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
#define ll long long
#define ld long double
#define eb emplace_back
#define pb push_back
#define pf push_front
#define mod 1000000007
#define clock (clock() * 1000.0 / CLOCKS_PER_SEC)
#define mp make_pair
#define ff first
#define ss second
#define null NULL
#define all(c) (c).begin(),(c).end()
#define nl "\n"
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector< vi > vvi;
typedef vector< vl > vvl;
typedef pair< int,int > ii;
typedef pair< ll,ll> pll;
typedef map< ll,ll> mll;
typedef map< int,int> mii;
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
ll t;
cin >> t;
while(t--)
{
ll a,b,c,d;
cin >> a >> b >> c >> d;
if(a == b)
cout << "YES\n";
else if(c!=d)
{
if(abs(a-b)%abs(c-d)==0)
cout << "YES\n";
else
cout << "NO\n";
}
else
cout << "NO\n";
}
return 0;
}``````