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

 

  1 /*

  2 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

  3       1变成0 0变成1 。第一个要以最后一个为参考。

  4       

  5       矩阵的变化，

  6 

  7       | 1 0 0 0 0 1|  a1      | （a1+a6）%2 |

  8       | 1 1 0 0 0 0|  a2      | （a1+a2）%2 |

  9       | 0 1 1 0 0 0|  a3 ===  | （a2+a3）%2 |

 10       | 0 0 1 1 0 0|  a4      | （a3+a4）%2 |

 11       | 0 0 0 1 1 0|  a5      | （a4+a5）%2 |

 12       | 0 0 0 0 1 1|  a6      | （a5+a6）%2 |

 13       

 14       利用矩阵相乘就可以快速到求取出来.

 15       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！！~

 16       优化2.用位操作。怎么用位操作实现？

 17       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

 18             就会有提高。代码中有...

 19 */

 20 

 21 #include<iostream>

 22 #include<cstdio>

 23 #include<cstring>

 24 #include<cstdlib>

 25 using namespace std;

 26 

 27 

 28 char a[103];

 29 struct node

 30 {

 31     int mat[103][103];

 32 }M_tom,M_hxl;

 33 

 34 void make_first(node *cur,int len)

 35 {

 36     int i,j;

 37     for(i=1;i<=len;i++)

 38     for(j=1;j<=len;j++)

 39     if(i==j) cur->mat[i][j]=1;

 40     else cur->mat[i][j]=0;

 41 }

 42 

 43 void cheng2(node cur,char a[],int len)

 44 {

 45     node ww;

 46     int i,j,k;

 47     memset(ww.mat,0,sizeof(ww.mat));

 48     for(i=1;i<=len;i++)

 49     for(k=1;k<=len;k++)

 50     if(cur.mat[i][k])

 51     {

 52         for(j=1;j<=1;j++)

 53         {

 54            ww.mat[i][j]=(ww.mat[i][j]^(cur.mat[i][k]&(a[k]-'0')))&1;

 55         }

 56     }

 57     for(i=1;i<=len;i++)

 58     printf("%d",ww.mat[i][1]);

 59     printf("\n");

 60 }

 61 

 62 struct node cheng(node cur,node now,int len)

 63 {

 64     node ww;

 65     int i,j,k;

 66     memset(ww.mat,0,sizeof(ww.mat));

 67     for(i=1;i<=len;i++)

 68     for(k=1;k<=len;k++)

 69     if(cur.mat[i][k])//对稀疏矩阵的优化

 70     {

 71         for(j=1;j<=len;j++)

 72         {

 73             if(now.mat[k][j])//同理

 74             {

 75                ww.mat[i][j]=(ww.mat[i][j]^(cur.mat[i][k]&now.mat[k][j]))&1;

 76             }

 77         }

 78     }

 79     return ww;

 80 }

 81 

 82 void make_init(int len)

 83 {

 84 

 85     for(int i=1;i<=len;i++)

 86     M_hxl.mat[1][i]=0;

 87 

 88     M_hxl.mat[1][1]=1;

 89     M_hxl.mat[1][len]=1;

 90     for(int i=2;i<=len;i++)

 91     for(int j=1;j<=len;j++)

 92     if(i==j || i-1==j)

 93     M_hxl.mat[i][j]=1;

 94     else M_hxl.mat[i][j]=0;

 95 }

 96 

 97 

 98 void power_sum2(int n,int len,char a[])

 99 {

100     make_first(&M_tom,len);

101     while(n)

102     {

103         if(n&1)

104         {

105             M_tom=cheng(M_hxl,M_tom,len);//这也是优化。顺序改变结果就不同了

106         }

107         n=n>>1;

108         M_hxl=cheng(M_hxl,M_hxl,len);

109     }

110    // cs(len);

111     cheng2(M_tom,a,len);

112 }

113 

114 int main()

115 {

116     int n,len;

117     while(scanf("%d",&n)>0)

118     {

119         scanf("%s",a+1);

120         len=strlen(a+1);

121         make_init(len);

122         power_sum2(n,len,a);

123     }

124     return 0;

125 }