Not so Mobile, notsomobile

Not so Mobile, notsomobile

Not so Mobile

Before being an ubiquous communications gadget,MobileWas 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 has strates a simple mobile. it is just a wire, suincluded 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 isWL×DL =WR×DR whereDL is the left distance,DR is the right distance,WL is the left weight andWR 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:WLDLWRDR

IfWL orWR is zero then there is a sub-mobile hanging from that end and the following lines define 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 bothWL andWR 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

Binary Tree recursion: Give the Left and Right nodes of the root node, w1 = 0, d1 = 2, w2 = 0, d2 = 4; if the left node is zero, the left and right children's values of the current left node are read. The left and right children have values until they know;

# Include <iostream> # include <cstdio> using namespace std; int flag; int read () {int w1, d1, w2, d2; cin> w1> d1> w2> d2; if (w1 & w2 & d1 & d2) // if both left and right children exist, then the question is judged; {if (w1 * d1! = W2 * d2) {flag = 1; return 0;} else return (w1 + w2); // return the value of the root node;} else {if (! W1) w1 = read (); // The value of the node is zero to continue reading data; if (! W2) w2 = read (); if (w1 * d1! = W2 * d2) {flag = 1; return 0;} else return (w1 + w2) ;}} int main () {int n; cin >> n; while (n --) {flag = 0; read (); // read data; if (! Flag) cout <"YES" <endl; else cout <"NO" <endl; if (n! = 0) cout <endl;} return 0 ;}