UVA 839 (13.08.20)

于 2013-08-20 18:08:00 发布 · 60 阅读

本文探讨了如何通过递归算法解决平衡复杂天平的问题，包括处理嵌套天平的平衡计算，涉及数学原理和编程实现。

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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.

$\epsfbox{p839a.eps}$

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 isW_l×D_l = W_r×D_rwhere D_l is the left distance, D_r is theright distance, W_l is the left weight and W_ris 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.

$\epsfbox{p839b.eps}$

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: W_lD_lW_rD_r

If W_l or W_r 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 W_l and W_r 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

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;
}