zoukankan      html  css  js  c++  java
  • uva 839 Not so Mobile-S.B.S.

    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.
    The gure 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 gure.
    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 de ne
    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 de ne two sub-mobiles: rst 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

    ------------------------我是分割线----------------------------------------------------------------------------------------

    这道题从题目就可以看出是递归关系定义的,所以使用递归进行输入;

    并且可以在输入过程中进行判断;

    使用引用传值而不用全局变量,极大简化代码,增加可读性。

     1 #include<iostream>
     2 #include<cstdio>
     3 #include<cstring>
     4 #include<queue>
     5 #include<cmath>
     6 #include<algorithm>
     7 #include<cstdlib>
     8 using namespace std;
     9 int read(){
    10     int x=0,f=1;char ch=getchar();
    11     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    12     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    13     return x*f;
    14 }
    15 bool build(int &w)
    16 {
    17     int w1,w2,d1,d2;
    18     bool b1=true,b2=true;
    19     w1=read();d1=read();w2=read();d2=read();
    20     if(!w1) b1=build(w1);
    21     if(!w2) b2=build(w2);
    22     w=w1+w2;
    23     return b1&&b2&&(w1*d1==w2*d2);
    24 }
    25 int main()
    26 {
    27     int t,w;
    28     t=read();
    29     while(t--)
    30     {
    31         if(build(w)) cout<<"YES"<<endl;
    32         else cout<<"NO"<<endl;
    33         if(t) cout<<"
    ";
    34     }
    35     return 0;
    36 }
  • 相关阅读:
    fetch的优点
    gitignore不起作用
    css动画和js动画区别
    工业家居气象空气环境质量监测仪记录数据甲醛PM2.5二氧化碳大气压温湿度
    摆脱淘宝、京东、拼多多内部引流消费规则,自建网站利用其完成支付
    语音朗读模块TTS文本变量实时转语音朗读科大讯飞XFS5152CE芯片AI
    PCB altium designer AD10 AD20 导出DWG CAD文件 过孔问题
    【Creator3】如何在3D场景中实现炫酷传送门,和简单的小地图功能,RenderTexture技术应用
    B站视频:【Creator3】好玩的编队代码 魔性排列停不下来 附源码及出处
    B站视频:《四图猜词》 Part3 | CocosCreator游戏开发教程
  • 原文地址:https://www.cnblogs.com/SBSOI/p/5575013.html
Copyright © 2011-2022 走看看