Poj 2605 SETI (Math _ Gaussian deyuan)

Given a prime number p and a string str. F [I] = Num (str [I]), Num (str [I]) indicates 0 when str [I] is, str [I] is str [I]-a + 1 when it is a-z.
Then we obtain the n (str [I] length) equations. The format of each equations is (a1 * k ^ 0 + a2 * k ^ 1 +... + an * k ^ (n-1) mod p = f [k] mod p (1 <= k <= n). ai is an unknown number. Let's solve this equations, output n solutions.

Problem-solving ideas: the bare Gaussian deyuan, I copied a template, but I still won't talk about the core idea of Gaussian deyuan, the line generation teacher died early.
Let's take a look at the general steps of the Gaussian elimination class I know: 1. Use question information to establish equations 2. Convert the equations into augmented matrices 3. Convert the augmented Matrices Into row step matrices 4. solve the problems. Step 2 is both a key and a difficulty. You can apply the template later. For detailed analysis, see this article:
Test data:
3
29 hello * earth
Out: 8 13 9 13 4 27 18 10 12 24 15

C producer code:
[Cpp]
# Include <iostream>
# Include <string>
# Include <string. h>
# Include <cmath>
# Include <stdio. h>
# Include <algorithm>
Using namespace std;
# Define MAX 71
 
 
Char str [MAX];
Int n, m, p, x [MAX];
Int mat [MAX] [MAX];
 
 
Void Initial (){

Int I, j, k = 1;
Memset (x, 0, sizeof (x ));
For (I = 0; I <n; ++ I ){

Mat [I] [m] = str [I] = '*'? 0: str [I]-'A' + 1;
 
For (k = 1, j = 0; j <m; ++ j)
Mat [I] [j] = k, k = k * (I + 1) % p;
// For (j = 0; j <m; ++ j)
// Printf ("% d % c", mat [I] [j], j = M-1? '\ N ':'');
}
}
Inline int gcd (int x, int y ){

Int r = x % y;
While (r ){

X = y, y = r;
R = x % y;
}
Return y;
}
Inline int lcm (int x, int y ){
 
Return x/gcd (x, y) * y;
}
Void Gauss (){
 
Int I, j, row, max_r, col;
Int ta, tb, temp, LCM;
Int free_x_num, free_index;
 
 
Col = row = 0;
For (; row <n & col <m; ++ row, ++ col ){
 
Max_r = row;
For (I = row + 1; I <n; ++ I)
If (abs (mat [I] [col])> abs (mat [max_r] [col])
Max_r = I;
 
 
If (max_r! = Row)
For (j = row; j <= m; ++ j)
Swap (mat [row] [j], mat [max_r] [j]);
If (mat [row] [col] = 0 ){
 
Row --;
Continue;
}
 
 
For (I = row + 1; I <n; ++ I)
If (mat [I] [col]! = 0 ){
 
LCM = lcm (abs (mat [I] [col]), abs (mat [row] [col]);
Ta = LCM/abs (mat [I] [col]);
Tb = LCM/abs (mat [row] [col]);
If (mat [I] [col] * mat [row] [col] <0)
Tb =-tb;
For (j = col; j <= m; ++ j)
Mat [I] [j] = (mat [I] [j] * ta-mat [row] [j] * tb) % p;
}
}
// Printf ("yes, you can ac \ n ");
 
 
For (I = m-1; I> = 0; -- I ){
 
Temp = mat [I] [m];
For (j = I + 1; j <m; ++ j)
If (mat [I] [j]! = 0)
Temp = (temp-mat [I] [j] * x [j]) % p + p) % p;
While (temp % mat [I] [I]) temp = temp + p;
X [I] = (temp/mat [I] [I] + p) % p;
}
}
 
 
Int main () www.2cto.com
{
Int I, j, k, t;

 
Scanf ("% d", & t );
While (t --){

Scanf ("% d % s", & p, str );
N = strlen (str );
M = n, Initial ();
 
 
Gauss ();
For (I = 0; I <m; ++ I)
Printf ("% d % c", x [I], I = M-1? '\ N ':'');
}
}

Author: woshi250hua