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Not so Mobile

来源:互联网 作者:佚名 时间:2013-03-28 13:39
Not so Mobile Before being an ubiquous communications gadget, amobilewas 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 illustr

  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.

 

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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 Wl×Dl = Wr×Dr where Dl is the left distance, Dr is the right distance, Wl is the left weight and Wr 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.

 

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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: Wl Dl Wr Dr

If Wl or Wr 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 Wl and Wr 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  1 0 2 0 4 0 3 0 1 1 1 1 1 2 4 4 2 1 6 3 2

 

Sample Output  YES

 

 

Jose Paulo Leal, ACM-UP'2001 

#include<stdio.h> #include<malloc.h> #include<string.h> /* typedef struct BiTree{ int weight, value; struct BiTree *lch, *rch; }BiTree, BiNTree; */ int Traverse(int& balance) { int wl, dl, wr, dr; scanf(, &wl, &dl, &wr, &dr); /* BiNTree left, right; Tree->lch = (BiNTree)malloc(sizeof(BiTree)); Tree->lch->weight = dl; Tree->lch->lch = Tree->lch->rch = NULL; if(wl == 0) Tree->lch->value = Traverse(Tree->lch, balance); else Tree->lch->value = wl; Tree->rch = (BiNTree)malloc(sizeof(BiTree)); Tree->rch->weight = dr; Tree->rch->lch = Tree->rch->rch = NULL; if(wr == 0) Tree->rch->value = Traverse(Tree->rch, balance); else Tree->rch->value = wr; */ if(wl == 0) wl = Traverse(balance); if(wr == 0) wr = Traverse(balance); if(wl*dl != wr*dr) balance = 0; return wl+wr; } int main() { int T, n, i, j, balance; scanf(, &T); while(T--) { /* Tree = (BiNTree)malloc(sizeof(BiTree)); Tree->weight = Tree->value = NULL; Tree->lch = Tree->rch = NULL; */ balance = 1; Traverse(balance); ); ); ); } return 0; }

解题思路:

题目的意思是:类似于杠杆原理你要保持平衡那么左边的重量乘以距离 == 右边的重量乘以右边的距离,只不过这里的意思是说如果mobile的左边或右边的重量变为零了,服务器空间,那么它下面坑定还有另外一辆mobile,香港空间,此时的重量替代为下面这辆mobile的左边和右边重量的相加,

赤裸裸的水题啊,害的我要看题目两边,而且代码中可以看到,被坑加忽悠了!!!

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