HDU 2276 Kiki & Little Kiki 2 矩阵构造

Kiki & Little Kiki 2

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1659    Accepted Submission(s): 837

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

 

Source

HDU 8th Programming Contest Site（1）

 

Recommend

lcy

 

 /*
 题意：一个灯环，输出进行N此变化的结果，变化的规则是：如果左边是1，那么自己变成相反的
       1变成0 0变成1 。第一个要以最后一个为参考。

       矩阵的变化，

       | 1 0 0 0 0 1|  a1      | （a1+a6）%2 |
       | 1 1 0 0 0 0|  a2      | （a1+a2）%2 |
       | 0 1 1 0 0 0|  a3 ===  | （a2+a3）%2 |
       | 0 0 1 1 0 0|  a4      | （a3+a4）%2 |
       | 0 0 0 1 1 0|  a5      | （a4+a5）%2 |
       | 0 0 0 0 1 1|  a6      | （a5+a6）%2 |

       利用矩阵相乘就可以快速到求取出来.
       优化1.由于左边是稀疏矩阵，那么在矩阵相乘到时候，可以优化。代码中有！！~
       优化2.用位操作。怎么用位操作实现？
       优化3.在快速幂的时候，顺序的变化。因为稀疏矩阵的存在，所以改变一下顺序
             就会有提高。代码中有...
 */

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cstdlib>
 using namespace std;

 char a[];
 struct node
 {
     int mat[][];
 }M_tom,M_hxl;

 void make_first(node *cur,int len)
 {
     int i,j;
     for(i=;i<=len;i++)
     for(j=;j<=len;j++)
     if(i==j) cur->mat[i][j]=;
     else cur->mat[i][j]=;
 }

 void cheng2(node cur,char a[],int len)
 {
     node ww;
     int i,j,k;
     memset(ww.mat,,sizeof(ww.mat));
     for(i=;i<=len;i++)
     for(k=;k<=len;k++)
     if(cur.mat[i][k])
     {
         for(j=;j<=;j++)
         {
            ww.mat[i][j]=(ww.mat[i][j]^(cur.mat[i][k]&(a[k]-'')))&;
         }
     }
     for(i=;i<=len;i++)
     printf("%d",ww.mat[i][]);
     printf("\n");
 }

 struct node cheng(node cur,node now,int len)
 {
     node ww;
     int i,j,k;
     memset(ww.mat,,sizeof(ww.mat));
     for(i=;i<=len;i++)
     for(k=;k<=len;k++)
     if(cur.mat[i][k])//对稀疏矩阵的优化
     {
         for(j=;j<=len;j++)
         {
             if(now.mat[k][j])//同理
             {
                ww.mat[i][j]=(ww.mat[i][j]^(cur.mat[i][k]&now.mat[k][j]))&;
             }
         }
     }
     return ww;
 }

 void make_init(int len)
 {

     for(int i=;i<=len;i++)
     M_hxl.mat[][i]=;

     M_hxl.mat[][]=;
     M_hxl.mat[][len]=;
     for(int i=;i<=len;i++)
     for(int j=;j<=len;j++)
     if(i==j || i-==j)
     M_hxl.mat[i][j]=;
     else M_hxl.mat[i][j]=;
 }

 void power_sum2(int n,int len,char a[])
 {
     make_first(&M_tom,len);
     while(n)
     {
         if(n&)
         {
             M_tom=cheng(M_hxl,M_tom,len);//这也是优化。顺序改变结果就不同了
         }
         n=n>>;
         M_hxl=cheng(M_hxl,M_hxl,len);
     }
    // cs(len);
     cheng2(M_tom,a,len);
 }

 int main()
 {
     int n,len;
     while(scanf("%d",&n)>)
     {
         scanf("%s",a+);
         len=strlen(a+);
         make_init(len);
         power_sum2(n,len,a);
     }
     return ;
 }