Uva 839 Not so Mobile-白红宇

Uva 839 Not so Mobile

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发布时间：2019-06-18

本文共 2836 字，大约阅读时间需要 9 分钟。

Not so Mobile

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.

$\epsfbox{p839a.eps}$

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.

$\epsfbox{p839b.eps}$

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 本题极为重要！！！

1 #include
      
          2 #include
       
           3 #include
        
            4 #include
         
             5 #include
          
              6 #include
           
             7 #include
            
              8 #include
             
               9 #include
              
                10 #include
               
                 11 #include
                 12 #include
                 
                   13 #include
                  
                    14 using namespace std; 15 bool solve(int& W)16 {17 int W1,D1,W2,D2;18 bool b1=true,b2=true;19 cin>>W1>>D1>>W2>>D2;20 if(!W1)b1=solve(W1);21 if(!W2)b2=solve(W2);22 W=W1+W2;23 return b1 && b2 && (W1*D1==W2*D2); 24 } 25 int main()26 {27 int T,W;28 cin>>T;29 while(T--)30 {31 if(solve(W))32 cout<<"YES\n";33 else34 cout<<"NO\n";35 if(T)36 printf("\N");37 }38 return 0;39 }

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