Before being an ubiquous communications gadget, a mobile was just a structure made of strings and wires suspending colourfull things. This kind of mobile is usually found hanging over cradles of small babies.

The figure illustrates a simple mobile. It is just a wire, suspended by a string, with an object on each side. It can also be seen as a kind of lever with the fulcrum on the point where the string ties the wire. From the lever principle we know that to balance a simple mobile the product of the weight of the objects by their distance to the fulcrum must be equal. That is W_l×D_l = W_r×D_r where D_l is the left distance, D_r is the right distance, W_l is the left weight and W_r is the right weight.

In a more complex mobile the object may be replaced by a sub-mobile, as shown in the next figure. In this case it is not so straightforward to check if the mobile is balanced so we need you to write a program that, given a description of a mobile as input, checks whether the mobile is in equilibrium or not.

Input 

The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.

The input is composed of several lines, each containing 4 integers separated by a single space. The 4 integers represent the distances of each object to the fulcrum and their weights, in the format: W_l D_l W_r D_r

If W_l or W_r is zero then there is a sub-mobile hanging from that end and the following lines define the the sub-mobile. In this case we compute the weight of the sub-mobile as the sum of weights of all its objects, disregarding the weight of the wires and strings. If bothW_l and W_r are zero then the following lines define two sub-mobiles: first the left then the right one.

Output 

For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.

Write `YES‘ if the mobile is in equilibrium, write `NO‘ otherwise.

Sample Input 

10 2 0 40 3 0 11 1 1 12 4 4 21 6 3 2

Sample Output 

YES

 1 #include <iostream> 2 #include <cstdio> 3 using namespace std; 4  5 int ok; 6  7 int dfs() { 8     int wl,dl,wr,dr; 9     cin >> wl >> dl >> wr >> dr;10     if (wl == 0) wl = dfs();11     if (wr == 0) wr = dfs();12     if (wl * dl != wr * dr) ok = 0;13     else return wl + wr; 14 }15 16 int main () {17     int T;18     freopen("1.in","r",stdin);19     cin >> T;20     while (T--) {21         ok = 1;22         dfs();23         cout << (ok == 0 ? "NO" : "YES") << endl;24         if (T) cout << endl;25     }26 }

View Code

UVA - 839