Sgu139help needed! Determine whether there is a solution for the 15th digital, and determine whether there is a deduction for the n digital.

This is the case. You need to determine whether there is a solution for the 15th digital.

I won't. I found this method.

Store 16 numbers in one-dimensional arrays in order of appearance,

Then accumulate the number of reverse pairs of each number,

Add the distance from 0 to the final state of Manhattan to get a number X.

Due to the final state, the number of Reverse Order pairs for the entire graph is 15,

Is an odd number, so if X is also an odd number,

Then there will be solutions for this 15 digital,

Otherwise, no solution is available.

All other digital products follow this method,

My code is as follows:

# Include <iostream> using namespace STD; int main () {int map [16], ANS = 0; For (INT I = 0; I <16; I ++) {scanf ("% d", & map [I]); If (! Map [I]) ans + = 6-i % 4-I/4; for (Int J = 0; j <I; j ++) if (Map [J]> map [I]) ans ++;} If (ANS & 1) printf ("yes"); elseprintf ("no ");}