joj 2510 Product简单的深度优先搜索

joj 2510 Product简单的深度优先搜索

#include < iostream >
using   namespace  std;
int  num[ 100 ];
int  a,s;
int  MAX;
void  search( int  i, int  sum)
{
      int  j;
      if (i == a) // 到达终点的处理
     {
      if (sum * (s - sum) > MAX)
        MAX = sum * (s - sum);
     }
      else
      for (j = 0 ;j < 2 ;j ++ ) // 各个支路
     {
           // 理论上应当添加条件判断
          search(i + 1 ,sum + j * num[i - 1 ]);
     }
}
int  main()
{
    int  b,c,sum;
    while (cin >> a)
   {       
         sum = 0 ;
         s = 0 ;
         MAX = 0 ;       
          for (b = 0 ;b < a;b ++ )
         {
             cin >> num[b];
             s += num[b];            
         }
         search( 0 , sum); // 从这里开始
         cout << MAX << endl;
                
   } 
}

还有一点启发，max不能用，可以用MAX，无需记录

There are multiple test cases, the first line of each test case is a integer n. (2<=n<=20) The next line have n integers (every integer in the range [1,1000]). We can divide these n integers into two parts and get the sum of the integers in these two parts, define them as sumA and sumB. Please calc the max number of sumA*sumB and output it in a single line.

Sample Input

3
5 3 7
2
4 9

Sample Output

56
36

Hint

(5,3),(7)  sumA=8  sumB=7  sumA*sumB=56
(5),(3,7)  sumA=5  sumB=10 sumA*sumB=50
(3),(5,7)  sumA=3  sumB=12 sumA*sumB=36
(),(3,5,7) sumA=0  sumB=15 sumA*sumB=0
The max is 56.