poj 1450 Gridland

//简单题，经过分析只要将m*n的积分奇数和偶数讨论就可以了！ #include <iostream>#include <cstdio>#include <cmath>using namespace std;int main(){    int tc, m, n, c = 0;    double sum;    cin >> tc;    while (tc--)    {          c++;          cin >> m >> n;          sum = m * n;          cout << "Scenario #" << c << ":" << endl;          if ((int)sum % 2 == 0)              printf("%.2lf\n", sum);          else           {               sum = sum-1+ sqrt(double (2));               printf("%.2lf\n", sum);          }          cout << endl;    }        system("pause");}