UVA839 - Not so Mobile

题目较为简单，不做多余解释，递归函数的简单应用

代码如下：

#include <iostream>using namespace std;int cacult(int &qual){    int w1, d1, w2, d2, temp;    if(cin>>w1>>d1>>w2>>d2 == 0)return -1;    if(!w1)        w1 = cacult(temp);    if(!w2)        w2 = cacult(temp);    if(w1*d1 == w2*d2&&temp)qual = 1;    else qual = 0;    return w1+w2;}int main (){    int temp, n;    cin>>n;    while(n--)    {        if(cacult(temp) == -1)break;        if(temp)cout<<"YES"<<endl;        else cout<<"NO"<<endl;        if(n)cout<<endl;    }    return 0;}