UVaOJ839---Not so Mobile
来源:互联网 发布:房地产行业 黑心 知乎 编辑:程序博客网 时间:2024/05/28 15:38
839 - Not so Mobile
Time limit: 3.000 seconds
839 - Not so Mobile
Time limit: 3.000 secondsBefore 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.
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, Dris 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
10 2 0 40 3 0 11 1 1 12 4 4 21 6 3 2
Sample Output
YES
#include<stdio.h>#include<string.h>#define MAXD 1000010int size, wl[MAXD], wr[MAXD], dl[MAXD], dr[MAXD], d[MAXD], ok;int dfs(){ int k; ++ size; k = size; scanf("%d%d%d%d", &wl[k], &dl[k], &wr[k], &dr[k]); if(wl[k] == 0) wl[k] = dfs(); if(wr[k] == 0) wr[k] = dfs(); if(wl[k] * dl[k] != wr[k] * dr[k]) ok = 0; return wl[k] + wr[k];}void solve(){ size = 0; ok = 1; dfs(); if(ok) printf("YES\n"); else printf("NO\n");}int main(){ int t, tt; scanf("%d", &t); for(tt = 0; tt < t; tt ++) { if(tt) printf("\n"); solve(); } return 0;}
Jose Paulo Leal, ACM-UP'2001
- UVaOJ839---Not so Mobile
- 839 - Not so Mobile
- 839 - Not so Mobile
- UVA839 - Not so Mobile
- 839 - Not so Mobile
- UVA839 - Not so Mobile
- 839 - Not so Mobile
- Not so Mobile uva
- UVa839 Not so Mobile
- 839 - Not so Mobile
- Not so Mobile
- Not so Mobile
- uva839 not so mobile
- 839 - Not so Mobile
- 天平(Not so Mobile)
- Not so Mobile
- UVA839 - Not so Mobile
- Not so Mobile
- 调用约定
- 图片跳动显示对应图片
- 百度地图无法正常画线的问题
- java.util.concurrent.ArrayBlockingQueue
- <数据结构>线性表
- UVaOJ839---Not so Mobile
- [Point Cloud Library] 基本结构 PointCloud
- P11.Image separation
- 在共享DLL中使用MFC 和在静态库中使用MFC的区别
- android--BroadCastReceiver 简介
- svn中导出来的项目运行出错
- php遍历目录下的文件夹
- android基础学习(7)-------spinner的使用
- UVaOJ10562---Undraw the Trees