UVa 839 Not so Mobile(树的递归输入)
判断一个树状天平是否平衡，每个测试样例每行4个数 wl,dl,wr,dr，当wl*dl=wr*dr时，视为这个天平平衡，当wl或wr等于0是，下一行将是一个子天平，如果子天平平衡，wl为子天平的wl+wr ，否则整个天平不平衡
1 #include<iostream> 2 using namespace std; 3 bool solve(int &w) 4 { 5 int wl,dl,wr,dr; 6 cin>>wl>>dl>>wr>>dr; 7 bool b1 = true,b2=true; 8 if(wl == 0) b1 = solve(wl);///进入左子天平 9 if(wr == 0) b2 = solve(wr);///进入右子天平 10 w = wl+wr; 11 return b1 && b2 &&(wl*dl == wr*dr); 12 } 13 14 int main() 15 { 16 int n; 17 cin>>n; 18 while(n--) 19 { 20 int weight; 21 bool s = solve(weight); 22 if(s) cout<<"YES"<<endl; 23 else cout<<"NO"<<endl; 24 if(n) cout<<endl; 25 } 26 return 0; 27 }
