Hdoj HDU 2082 find the word ACM 2082 in HDU

Miyu original, post Please note: Reprinted from __________ White House

Question address:
Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 2082
Description: Problem description
Assume there are x1 letters A and X2 letters B ,.. assume that the value of letter A is 1, and the value of letter B is 2 ,.. the value of Z is 26. How much value can be found for a given letter? <= What about 50 words? The value of a word is the sum of the values of all the letters of a word. For example, the value of word ACM is 1. + 3 + 14 = 18 The value of HDU is 8. + 4 + 21 = 33 . (The words are irrelevant to the order. For example, ACM and CMA think they are the same word ).
 

Input
The input is an integer N, indicating the number of test instances.
Then there are n rows of data, and each row contains 26<=Integer x1, x2,... x26.
 

Output
For each test instance, output the total value that can be found<=The number of words in 50, and the output of each instance occupies one line.
 

Sample Input
2
1   1   1   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0
9   2   6   2   10   2   2   5   6   1   0   2   7   0   2   2   7   5   10   6   10   2   10   6   1   9
 

Sample output
7
379297

Question Analysis:
There is a very standard primary function question (for more information about the primary function, see <primary function details> ). however, this question can be properly optimized. Because the data is input in 26 data orders, rather than sparse polynomials, you can simply input one variable.

CodeAs follows:Miyu original, post Please note: Reprinted from __________ White House

# Include < Iostream >
Using   Namespace STD;
Int Num1 [ 51 ];
Int Num2 [ 51 ];
Int Main ()
{
Int T;
While (CIN > T)
{
While (T -- )
{
Memset (num1, 0 , Sizeof (Num1 ));
Memset (num2, 0 , Sizeof (Num2 ));
Num1 [ 0 ] =   1 ;
For ( Int I =   1 ; I <=   26 ; ++ I)
{
Int CNT;
CIN > CNT;
If (CNT =   0 )
{
Continue ;
}
For ( Int J =   0 ; J <=   50 ; ++ J)
{
For ( Int K =   0 ; K <= CNT && K * I + J <=   50 ; K ++ )
{
Num2 [K * I + J] + = Num1 [J];
}
}
For ( Int J =   0 ; J <=   50 ; ++ J)
{
Num1 [J] = Num2 [J];
Num2 [J] =   0 ;
}
}
Int Total =   0 ;
For ( Int I =   1 ; I <=   50 ; ++ I)
{
Total + = Num1 [I];
}
Cout < Total < Endl;
}
}
Return   0 ;
}