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TOJ1003

 

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 1 #include<iostream>
 2 using namespace std;
 3 
 4 int main()
 5 {
 6     int n, m;
 7     while (cin >> n >> m,n!=0,m!=0)
 8     {
 9         int n_dove, n_rabbit;
10         if (m % 2 == 0&&(m>2*n)&&(4*n>m))
11         {
12             n_dove = (4 * n - m) / 2;
13             n_rabbit = n-n_dove;
14             cout << n_dove << " " << n_rabbit << endl;
15         }
16         else
17             cout << "Error" << endl;
18     }
19     return 0;
20 }
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要考虑结果的正负性(m-2*n>0&&4*n-m>0)及int情况下的整数性(m%2==0)

TOJ1003