uva839--Balance __ACM

Description:

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

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's lever with the fulcrum on the point where the string ties the wire. From the lever principle we know this to balance a simple mobile the product of the weight of the objects by their Distanc E to the fulcrum must is equal. This is wlxdl = wRxdr where Dwas the left distance, DR are the right Distance, WL is the "left weight and wR" right weight.

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

Analysis: in short, enter a balance system, input from top to bottom, if the position has a sub balance, it is entered in order from left to right to determine whether the system is balanced (that is, left weight * left arm = right weight * Right arm) The problem is entered in a recursive definition, so it is also convenient to write the problems recursively. , where the code can be written concisely by using a value reference. P.S This problem submitted when encountered a little problem, to now do not understand, doubt has been written in the comments, ask the expert enlighten. Code:

#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #include < cmath> #include <string> #include <vector> #include <queue> #include <stack> #include <set > #include <map>//#define file #define PI 3.1415926 #define MAX (a,b) ((A) > (B)? (

A):(B)) #define LL long #define Fordo (A,B,C) for (int (A) = (B);(a) <= (C);(a) + +) using namespace Std;
    BOOL Solve (int& W) {int w1,d1,w2,d2;
    BOOL B1=true;
    BOOL B2=true;
    cin>>w1>>d1>>w2>>d2; if (! W1)//Here is very strange, when I use w1==0 VJ will report WA, only.
    W1 This notation can be passed, it seems that the two are no different ... b1=solve (W1);//When there are sub balances, solve. if (!
    W2) B2=solve (W2);
    w=w1+w2;//the weight of the balance is calculated by weight return b1&&b2&& (W1*D1==W2*D2);
Recursive solution: When the sub balance of the pallet is balanced and the balance is balanced, the balance of the balance is returned.
    int main () {#ifdef file Freopen ("Test.in", "R", stdin);
    Freopen ("Test.out", "w", stdout);
    #endif//file int t,w;
    cin>>t; while (t--) {if Solve (W) cout<< "YES" <<endl;
        else cout<< "NO" <<endl;
    if (T) cout<<endl;
return 0;
 }