uva 839 - Not so Mobile
2012-07-20 11:13
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Not so Mobile |
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.
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
题目背景是杠杆原理,读入的数据的顺序类似二叉树的先序遍历顺序,遇到wl==0读入一组数据作为左儿子,wr==0读入一组数据作为右儿子;父亲节点weight=0则w=w左+w右;很明显是递归的
#include <stdio.h>
int flag;
int sort(int wl,int dl,int wr,int dr)
{int Wl,Dl,Wr,Dr,k1=wl,k2=wr;
if (wl==0) {scanf("%d%d%d%d",&Wl,&Dl,&Wr,&Dr);
k1=sort(Wl,Dl,Wr,Dr);
}
if (wr==0) {scanf("%d%d%d%d",&Wl,&Dl,&Wr,&Dr);
k2=sort(Wl,Dl,Wr,Dr);
}
if (k1*dl!=k2*dr) flag=0;
return k1+k2;
};
int main()
{int wl,dl,wr,dr,t;
scanf("%d",&t);
while (t--)
{
flag=1;
scanf("%d%d%d%d",&wl,&dl,&wr,&dr);
sort(wl,dl,wr,dr);
if (flag) printf("YES\n");
else printf("NO\n");
if (t) printf("\n");
}
return 0;
}
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