hdu 2276（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): 2301    Accepted Submission(s): 1176

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

 

Source

HDU 8th Programming Contest Site（1）

 

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/*  能够把转换关系找出来的话，这题就是个比较裸的快速幂了  用a[i]*代表上一层的a[i],通过推样例会发现如下规律：  a[i]=(a[i]*+a[i-1]*)%2,其中a[1]=(a[1]*a[n]*)%2    第一次敲快速幂，存做模板*/#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#include <string>#include <queue>#include <map>#include <stack>#include <set>#include <cmath>#include <vector>#include<cstdlib>#define INF 1<<30#pragma comment (linker,"/STACK:102400000,102400000")using namespace std;#define maxn 150#define moo 2char s[maxn];int n,m;struct Mat{    int a[maxn][maxn];};Mat res1;Mat muti(Mat a,Mat b){    Mat res;    memset(res.a,0,sizeof(res.a));    for(int i=0;i<n;i++)    for(int j=0;j<n;j++)    {       for(int k=0;k<n;k++)            res.a[i][j]=(res.a[i][j] + a.a[i][k]*b.a[k][j])%moo;    }    return res;}Mat pow(Mat res,int m){    Mat ss;    memset(ss.a,0,sizeof(ss.a));    for(int i=0;i<n;i++)        ss.a[i][i]=1;    while(m)    {        if(m%2)        ss=muti(ss,res);        res=muti(res,res);        m/=2;    }    return ss;}int main(){    int ans[maxn];    Mat oo;    while(~scanf("%d",&m))    {        scanf("%s",s);        n=strlen(s);        for(int i=0;i<n;i++)        ans[i]=s[i]-'0';        int ant[maxn];        memset(ant,0,sizeof(ant));        memset(res1.a,0,sizeof(res1.a));        res1.a[0][0]=res1.a[n-1][0]=res1.a[n-1][n-1]=1;        for(int i=0;i<n-1;i++)            res1.a[i][i]=res1.a[i][i+1]=1;        oo=pow(res1,m);        for(int j=0;j<n;j++)            for(int i=0;i<n;i++)            ant[j]=(ant[j]+ans[i]*oo.a[i][j])%moo;            for(int i=0;i<n;i++)                cout<<ant[i];            cout<<endl;    }    return 0;}