Not so Mobile

Description

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 both W_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

这道题我一开始理解有偏差，认为天平平衡只需要最底层平衡就可以了，但实际上是必须所有的天平都得平衡。这道题的巧妙之处是输入就用了递归的思想。本题还使用了“传引用”的方式来改变W的值。代码如下：

#include<iostream>using namespace std;int w;bool check(int &w){bool b1=true,b2=true;int w1,d1,w2,d2;cin>>w1>>d1>>w2>>d2;if(!w1) b1=check(w1);if(!w2) b2=check(w2);w=w1+w2;return b1&&b2&&(w1*d1==w2*d2);}int main(){int T;cin>>T;while(T--){if(check(w)) cout<<"YES\n";else cout<<"NO\n";if(T) cout<<endl;}return 0;}

这道题我犯了一个错误，就是把w1，d1，w2，d2成全局变量了，这样显然是错的。因为每一次都要输入这四个数，如果是全局变量，那么它们的值只会保存为最后一次输入的结果。