The Clocks
Time Limit:1000 MS 

Memory Limit:65536 K 
Total Submissions:12403 

Accepted:4908 
Description
  
      
O  O   O 
     
  
A B C
  
     
 O   O   O 
        
  
D E F
  
     
 O   O  O 
       
  
G H I
(Figure 1)
There are nine clocks in a 3*3 array (figure 1 ). the goal is to return all the dials to 12 o'clock with as few moves as possible. there are nine different allowed ways to turn the dials on the clocks. each such way is called a move. select for each move a number 1 to 9. that number will turn the dials 90' (degrees) clockwise on those clocks which are affected according to figure 2 below.
Move Affected clocks
1 ABDE
2 ABC
3 BCEF
4 ADG
5 BDEFH
6 CFI
7 DEGH
8 GHI
9 EFHI
(Figure 2)
Input
Your program is to read from standard input. nine numbers give the start positions of the dials. 0 = 12 o 'clock, 1 = 3 o 'clock, 2 = 6 o 'clock, 3 = 9 o' clock.
Output
Your program is to write to standard output. Output a shortest sorted sequence of moves (numbers), which returns all the dials to 12 o 'clock. You are convinced that the answer is unique.
Sample Input
3 3 02 2 22 1 2
Sample Output
4 5 8 9
Source
IOI 1994 is boring... The brute force search is over .. I wanted to practice Gaussian elimination, but I found that 4 is not a prime number and there will be problems .. It seems that we can do this using an inverse matrix... Still thinking
#include <string.h>#include <stdio.h>int main(){ int i,a[10],b[10],c[10]; for(i=1;i<=9;i++) scanf("%d",&a[i]); for(b[1]=0;b[1]<=3;b[1]++) for(b[2]=0;b[2]<=3;b[2]++) for(b[3]=0;b[3]<=3;b[3]++) for(b[4]=0;b[4]<=3;b[4]++) for(b[5]=0;b[5]<=3;b[5]++) for(b[6]=0;b[6]<=3;b[6]++) for(b[7]=0;b[7]<=3;b[7]++) for(b[8]=0;b[8]<=3;b[8]++) for(b[9]=0;b[9]<=3;b[9]++) { c[1]=(a[1]+b[1]+b[2]+b[4])%4; c[2]=(a[2]+b[1]+b[2]+b[3]+b[5])%4; c[3]=(a[3]+b[2]+b[3]+b[6])%4; c[4]=(a[4]+b[1]+b[4]+b[5]+b[7])%4; c[5]=(a[5]+b[1]+b[3]+b[5]+b[7]+b[9])%4; c[6]=(a[6]+b[3]+b[5]+b[6]+b[9])%4; c[7]=(a[7]+b[4]+b[7]+b[8])%4; c[8]=(a[8]+b[5]+b[7]+b[8]+b[9])%4; c[9]=(a[9]+b[6]+b[8]+b[9])%4; if(c[1]+c[2]+c[3]+c[4]+c[5]+c[6]+c[7]+c[8]+c[9]==0) { for(i=0;i<b[1];i++) printf("1 "); for(i=0;i<b[2];i++) printf("2 "); for(i=0;i<b[3];i++) printf("3 "); for(i=0;i<b[4];i++) printf("4 "); for(i=0;i<b[5];i++) printf("5 "); for(i=0;i<b[6];i++) printf("6 "); for(i=0;i<b[7];i++) printf("7 "); for(i=0;i<b[8];i++) printf("8 "); for(i=0;i<b[9];i++) printf("9 "); printf("\n"); return(0); } }}