UVA 839 (13.08.20)
| Not so Mobile |
Before being an ubiquous communications gadget, a mobile wasjust a structure made of strings and wires suspending colourfullthings. This kind of mobile is usually found hanging over cradlesof 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 beseen as a kind of lever with the fulcrum on the point where thestring ties the wire. From the lever principle we know that tobalance a simple mobile the product of the weight of the objectsby their distance to the fulcrum must be equal. That isWl×Dl = Wr×Drwhere Dl is the left distance, Dr is theright distance, Wl is the left weight and Wris the right weight.
In a more complex mobile the object may be replaced by asub-mobile, as shown in the next figure. In this case it is not sostraightforward to check if the mobile is balanced so we need youto 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 indicatingthe 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 betweentwo consecutive inputs.
The input is composed of several lines, each containing 4 integersseparated by a single space. The 4 integers represent thedistances of each object to the fulcrum and their weights, in theformat: Wl Dl Wr Dr
If Wl or Wr is zero then there is a sub-mobile hanging fromthat end and the following lines define the the sub-mobile. Inthis case we compute the weight of the sub-mobile as the sum ofweights of all its objects, disregarding the weight of the wiresand strings. If both Wl and Wr are zero then the followinglines 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
题意：
输入数据，描绘了一个天平，要求天平要平衡，其中有嵌套进去的天平，也要求不能倾斜～
然后平衡的公式应该都知道吧：w1 * d1 = w2 * d2；
思路：
递归不解释，数据出现0的时候就是递归的入口～
AC代码：
#include<stdio.h>
int flag;
int getW(int w) {
int tw1, td1, tw2, td2;
int p1, p2;
if(w == 0) {
scanf("%d %d %d %d", &tw1, &td1, &tw2, &td2);
p1 = getW(tw1) * td1;
p2 = getW(tw2) * td2;
if(p1 == p2)
return (p1/td1 + p2/td2);
else {
flag = 0;
return -1;
}
}
else
return w;
}
int main() {
int T;
scanf("%d", &T);
while(T--) {
int w1, d1, w2, d2;
int p1, p2;
flag = 1;
scanf("%d %d %d %d", &w1, &d1, &w2, &d2);
p1 = getW(w1) * d1;
p2 = getW(w2) * d2;
if(p1 == p2 && flag == 1) {
printf("YES\n");
if(T)
printf("\n");
}
else {
printf("NO\n");
if(T)
printf("\n");
}
}
return 0;
}