hdu2276 Kiki & Little Kiki 2（矩阵的幂运算）

 

Kiki & Little Kiki 2

时间限制：3000 ms  |  内存限制：65535 KB

难度：4

 

描述

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)

 

输入

The input contains no more than 1000 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.

输出

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

样例输入

1
0101111
10
100000001

样例输出

1111000
001000010

分析：这个题目的意思是给你一个字符串其中字符串中只含 0 和 1 在 t 秒时如果s[i-1]==1  则是s[i]^1。

这个题目我是想了好久都没做出来了，后来看了解题报告，才知道通过矩阵运算可以解的。

还未解决：1、不知道为什么这么做就行？？？

              2、在杭电AC了，可是到南阳就TLE？？？

              3、为什么矩阵运算时使用递归会错？？？

 

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #define N 105
 
 using namespace std;
 
 struct matrix{ int m[N][N];}a,b,unit;
 char s[N];
 
 void init(int n)
 {
     int i,j;
     for(i=0;i<n;i++)
     {
         for(j=0;j<n;j++)
         {
             unit.m[i][j]=a.m[i][j]=(i==j);
             b.m[i][j]=0;
         }
     }
     a.m[0][0]=a.m[0][n-1]=1;
     for(i=1;i<n;i++)
     {
         a.m[i][i-1]=a.m[i][i];
     }
     for(i=0;i<n;i++)
     {
         b.m[i][0]=s[i]-'0';
     }
 }
 
 matrix matr_mult(matrix x,matrix y,int n)
 {
     int i,j,k;
     matrix z;
     for(i=0;i<n;i++)
     {
         for(j=0;j<n;j++)
         {
             z.m[i][j]=0;
             for(k=0;k<n;k++)
             {
                 z.m[i][j]+=(x.m[i][k]*y.m[k][j]);
             }
             z.m[i][j]=z.m[i][j]%2;
         }
     }
     return z;
 }
 
 matrix matr_pow_mod(matrix A,int n,int m)
 {
     /*
     递归为什么不行呀？？？
     matrix t;
     if(m==1) return A;
     t=matr_pow_mod(A,n,m/2);
     t=matr_mult(t,t,n);
     if(m%2==1) t=matr_mult(A,t,n);
     return t;
     */
     matrix t,tt;
     t=A;tt=unit;
     while(m)
     {
         if(m%2==1) tt=matr_mult(tt,t,n);
         m=m/2;
         t=matr_mult(t,t,n);
     }
     return tt;    
 }
 
 int main()
 {
     int len,i,t;
     while(scanf("%d",&t)!=EOF)
     {
         scanf("%s",s);
         matrix ans;
         len=strlen(s);
         init(len);
         ans=matr_pow_mod(a,len,t);
         ans=matr_mult(ans,b,len);
         for(i=0;i<len;i++)
             printf("%d",ans.m[i][0]);
         printf("\n");
     }
     return 0;
 }