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 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.

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_l D_l W_r D_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

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;

}