Hdu 2276 Kiki & Little Kiki 2

Kiki & Little Kiki 2

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1376    Accepted Submission(s): 726

Problem Description

There are n lights in a circle numbered from 1 to n. The left of light 1 is light n, and the left of light k (1< k<= n) is the light k-1.At time of 0, some of them turn on, and others turn off.
Change the state of light i (if it's on, turn off it; if it is not on, turn on it) at t+1 second (t >= 0), if the left of light i is on !!! Given the initiation state, please find all lights’ state after M second. (2<= n <= 100, 1<= M<= 10^8)

 

Input

The input contains one or more data sets. The first line of each data set is an integer m indicate the time, the second line will be a string T, only contains '0' and '1' , and its length n will not exceed 100. It means all lights in the circle from 1 to n.
If the ith character of T is '1', it means the light i is on, otherwise the light is off.

 

Output

For each data set, output all lights' state at m seconds in one line. It only contains character '0' and '1.

 

Sample Input

1 0101111 10 100000001

 

Sample Output

1111000 001000010

 

package acm.hdu;
import java.util.Scanner;
public class Hdu2276_20130606_2119 {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		long m;
		String str ;
		while(in.hasNextLong()){
			m = in.nextLong();
			str = in.next();
			int [][] result = solve(str,m);
			showA(result);
		}
	}
	private static int [][]  solve(String str,long m){
		int n = str.length();
		int [][] init = new int[1][n];
		int [][] unit = new int[n][n];
		int [][] temp = getArray(n);
		for(int i = 0; i < n; ++i){
			init[0][i] = str.charAt(i) - '0';
			unit[i][i] = 1;
		}
		while(m > 0){
			if((m&1)==1){
				unit = multipyMatrix(temp,unit);
			}
			temp = multipyMatrix(temp,temp);
			m >>= 1;
		}
		return multipyMatrix(init,unit);
	}
	private static int [][] multipyMatrix(int [][] a, int [][] b){
		int [][] c = new int[a.length][b[0].length];
		for(int i = 0; i < a.length; ++i){
			for(int j = 0;j < b[0].length; ++j){
				for(int k = 0; k < b.length; ++k){
					c[i][j] = c[i][j]^(a[i][k]&b[k][j]);
				}
			}
		}
		return c;
	}
	private static int [][] getArray(int n){
		int [][] array = new int[n][n];
		for(int i = 0; i < n-1; ++i){
			array[i][i] = 1;
			array[i][i+1] = 1;
		}
		array[n-1][n-1] = 1;
		array[n-2][n-1] = 1;
		array[n-1][0] = 1;
		return array;
	}
	private static void showA(int [][] array){
		StringBuilder sb = new StringBuilder();
		for(int [] as:array){
			for(int a: as){
				sb.append(a);
			}
		}
		System.out.println(sb);
	}
}