Zju2759 perfect weighing skill-balance weighing Problem

Description:

The weight quality of a balance is the Npower of 3, that is, 1, 3, 9, 27 ,...... Evaluate a measurement scheme to name an item with a mass of M (M <= 500000000 ). The weight can be placed on both sides of the balance.

Analysis:

If the data volume is small, you can use the search method to calculate all the measurement methods in advance and then create a table.

This item is of great quality. How can we determine which side of the weight is placed on the balance?

We can consider using a triple representation of item quality. See the following table:

M tridecimal weight 9 3 1(-1 indicates the same side as the item, 1 indicates the other side of the item, 0 does not need to be used)
1 0 0 1 0 1
2 0 0 2 0 1-1
3 0 1 0 0 1 0
4 0 1 1 0 1 1
5 0 1 2 1-1-1
6 0 2 0 1-1 0
7 0 2 1 1-1 1
8 0 2 2 1 0-1
9 1 0 0 1 0 0

We can see that there is a rule that the method of placing an item corresponds to the three-digit representation of the item quality.

Terminate the rule as follows:

Convert M to a triplicate system. In the conversion process,

If one digit is 0, no matter;

If one digit is 1, place a weight on the opposite side of the item. The weight quality is the base of the three-digit mechanism.

If one digit is 2, place a weight on the same side of the item and carry forward (m + = 1 ).

Do not have spaces at the end of the number sequence during output. It seems that it is not PE but wa.

Code:

/*
Zju2759
*/

# Include <stdio. h>
# Define N 20

Int s [N];

Void printa (int A [], int N ){
Int I;
For (I = 0; I <n; I ++ ){
If (I) printf ("");
Printf ("% d", a [I]);
}
Printf ("/N ");
}

Int main ()
{
Int I, J, K, X, M, N, R, P;
Int A [n], B [N];

// Ready
K = 1; s [0] = 1;
For (I = 1; I <20; I ++ ){
K * = 3;
S [I] = K;
}
N = 20;

While (scanf ("% d", & M )! = EOF)
{
R = 0; I = J = 0;
N = 0; P = 0;

While (m ){
R = m % 3;
M/= 3;
If (r = 2 ){
N ++;
B [j ++] = s [p];
M + = 1;
} Else if (r = 1 ){
A [I ++] = s [p];

}
P ++;
}

// Output
If (! J) printf ("-1/n ");
Else printa (B, j );
If (! I) printf ("-1/n ");
Else printa (a, I );
Puts ("");
}

Return 0;
}