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[103];
struct node
{
    int mat[103][103];
}M_tom,M_hxl;

void make_first(node *cur,int len)
{
    int i,j;
    for(i=1;i<=len;i++)
    for(j=1;j<=len;j++)
    if(i==j) cur->mat[i][j]=1;
    else cur->mat[i][j]=0;
}

void cheng2(node cur,char a[],int len)
{
    node ww;
    int i,j,k;
    memset(ww.mat,0,sizeof(ww.mat));
    for(i=1;i<=len;i++)
    for(k=1;k<=len;k++)
    if(cur.mat[i][k])
    {
        for(j=1;j<=1;j++)
        {
           ww.mat[i][j]=(ww.mat[i][j]^(cur.mat[i][k]&(a[k]-'0')))&1;
        }
    }
    for(i=1;i<=len;i++)
    printf("%d",ww.mat[i][1]);
    printf("
");
}

struct node cheng(node cur,node now,int len)
{
    node ww;
    int i,j,k;
    memset(ww.mat,0,sizeof(ww.mat));
    for(i=1;i<=len;i++)
    for(k=1;k<=len;k++)
    if(cur.mat[i][k])//对稀疏矩阵的优化
    {
        for(j=1;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]))&1;
            }
        }
    }
    return ww;
}

void make_init(int len)
{

    for(int i=1;i<=len;i++)
    M_hxl.mat[1][i]=0;

    M_hxl.mat[1][1]=1;
    M_hxl.mat[1][len]=1;
    for(int i=2;i<=len;i++)
    for(int j=1;j<=len;j++)
    if(i==j || i-1==j)
    M_hxl.mat[i][j]=1;
    else M_hxl.mat[i][j]=0;
}


void power_sum2(int n,int len,char a[])
{
    make_first(&M_tom,len);
    while(n)
    {
        if(n&1)
        {
            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>>1;
        M_hxl=cheng(M_hxl,M_hxl,len);
    }
   // cs(len);
    cheng2(M_tom,a,len);
}

int main()
{
    int n,len;
    while(scanf("%d",&n)>0)
    {
        scanf("%s",a+1);
        len=strlen(a+1);
        make_init(len);
        power_sum2(n,len,a);
    }
    return 0;
}