hdu 2276 Kiki & Little Kiki 2矩阵快速幂

Kiki & Little Kiki 2

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2300    Accepted Submission(s): 1175

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

1010111110100000001

 

Sample Output

1111000001000010

 

题意：告诉一个01序列表示灯的开关情况，0就是关，1就是开，现在要操作次，如果一个数左边是1就改变这个数的状态，第一个数左边的数是最右边那个数。

分析:最开始第一反应是找循环节，但是直接存起来会MLE，找到后再算会TLE，赛后才知道是举证快速幂。对于每个数，操作时只与当前以及它左边的数有关，跟其他数没有任何关系，于是构造矩阵时，就能想到只有当前位置和它左边位置有关，只有这两个位置的数是1，其他都是0。

#include <iostream>#include <stdio.h>#include <string>#include <cstring>#include <cmath>typedef __int64 ll;using namespace std;struct matrix{    int f[102][102];}s,t;char str[109];int mp[109][109];int ss[109];matrix mul(matrix a,matrix b,int len){    matrix c;    memset(c.f,0,sizeof c.f);    for(int i=0;i<len;i++)    {        for(int j=0;j<len;j++)        {            for(int k=0;k<len;k++)            {                c.f[i][j]=(c.f[i][j]+a.f[i][k]*b.f[k][j])%2;            }        }    }    return c;}matrix fun(matrix a,ll n,int len){    matrix b;    for(int i=0;i<len;i++)        for(int j=0;j<len;j++)        if(i==j) b.f[i][j]=1;         else b.f[i][j]=0;         while(n)         {             if(n&1)                b=mul(b,a,len);             a=mul(a,a,len);             n>>=1;         }         return b;}int main(){    ll n;    while(~scanf("%I64d",&n))    {        scanf("%s",str);        int len=strlen(str);        memset(s.f,0,sizeof s.f);        for(int i=0;i<len;i++)//构造矩阵            if(i==0 || i==len-1) s.f[i][0]=1;        for(int j=1;j<len;j++)            for(int i=0;i<len;i++)                if(i==j || i==j-1) s.f[i][j]=1;        for(int i=0;i<len;i++)            ss[i]=str[i]-'0';            t=fun(s, n ,len);            for(int i=0;i<len;i++)            {                int ans=0;                for(int j=0;j<len;j++)                {                    ans=(ans+ss[j]*t.f[j][i])%2;                }                printf("%d",ans);            }            puts("");    }    return 0;}