我的方法是先將陣列 num[] 由小到大排序:

For MAX=(陣列大小-1) To 2

  若: num[m]+num[n]>num[MAX] 則: num[m]+num[n+1]>num[MAX] 且 num[m+1]+num[n]>num[MAX] 算出對應  (m,n)的個數

  否則: 還是算出對應(m,n)的個數~.~

End For

看了半天看不出錯在哪~

與正確輸出不相符(line:35)
您的答案為: 15
正確答案為: 16

int sumfromitoj(int i, int j) //i<=j
{
  int sum=0;
  if (i>j) return -9999999;
  else {
      for (int k=i; k<=j; k++)
      {
         sum += k;  
      }
      return sum;  
  }     
}

int count_remains_of_the_row(int num[], int original_sum_m_n, int n, int MAX)
{
    int cnt=0;
    for (int i=n+1; i<MAX; i++)
    {
      if (original_sum_m_n + num[i] > num[MAX]) cnt++;
    }
    if (n>=MAX) return 0;
    else return cnt+count_remains_of_the_row(num, original_sum_m_n+num[n+1], n+1, MAX);
}


/*
   Given: sorted_num[size-1] the last(largest) number of sorted array sorted_num[]
   For MAX=size-1 To MAX=2 Do the Following thing
     For (m=0; m<MAX; m++)
         For (n=m+1; n<MAX; n++)
             If sorted_num[m] + sorted_num[n] > num[MAX] Then
                 Count += Count_On_MAP(m,n,MAX) to Count_On_MAP(m,MAX-1,MAX)
             Else
                 Count += RemainCount_On_MAP(m,n,MAX)
         End For
     End for
   End For
*/
int num_of_polygons(int sorted_num[], int size)
{
     int count=0, count_the_pass=0, count_the_row=0;
     int MAX;
        for (MAX=size-1; MAX>=2; MAX--)
        {     
            int m=0,n=0;
            for (m=0; m<MAX; m++)
            {
                for (n=m+1; n<MAX; n++)
                {
                    if (sorted_num[m]+sorted_num[n] > sorted_num[MAX]) {
                        count_the_row += sumfromitoj(1,MAX-n); break;
                    }
                    else {
                        count_the_row += count_remains_of_the_row(sorted_num, sorted_num[m]+sorted_num[n], n, MAX );
                    }
                }
                count_the_pass += count_the_row;
                count_the_row = 0;    
            }
            count += count_the_pass;
            count_the_pass = 0;
        }
        return count;
}

int main()
{
    int num[105],i;
    stringstream ss;
    string s;
    
    while ( getline(cin, s) )
    {
        ss.str(s);
        i=0;
        while (ss >> num[i]) {
            i++;
            if (num[i-1]==0) i--;
            if ( i>105 ) { cout << "Too many numbers\n"; break; }  
        }
        if (i==0) break;
        else {
            qsort(num,i,sizeof(int),compare);
            cout << num_of_polygons(num, i) << endl;
            ss.clear();
        }
    }
    

    return 0;
}