8.3.6 Kiki & Little Kiki 2

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

大水题啊，，

自己多想想，应该想的出来的，提醒一句就是尝试用矩阵表示灯的状态，再转移即可

/*
Author:wuhuajun
*/
#include <cmath>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <cstdlib>
using namespace std;

typedef long long ll;
typedef double dd;
const int maxn=110;
struct matrix
{
    int m[maxn][maxn];
} ans,base,c,a;
int n,m,l;
char s[maxn];

void close()
{
exit(0);
}

void print(matrix a)
{
    for (int i=1;i<=n;i++)
    {
        for (int j=1;j<=n;j++)
            printf("%d ",a.m[i][j]);
        puts("");
    }
}

matrix mul(matrix a,matrix b)
{
    memset(c.m,0,sizeof(c.m));
    for (int i=1;i<=n;i++)
        for (int j=1;j<=n;j++)
            for (int k=1;k<=n;k++)
            {
                c.m[i][j]+=a.m[i][k]*b.m[k][j];
                c.m[i][j] %= 2;
            }
    return c;
}

void work()
{
    memset(ans.m,0,sizeof(ans.m));
    memset(base.m,0,sizeof(base.m));
    for (int i=1;i<=n;i++)
        ans.m[i][i]=1;
    base.m[1][n]=1;
    for (int i=1;i<=n;i++)
        base.m[i][i-1]=base.m[i][i]=1;
    while (m!=0)
    {
        if (m & 1)
            ans=mul(ans,base);
        m/=2;
        base=mul(base,base);
    }
    a=mul(ans,a);
    for (int i=1;i<=n;i++)
        printf("%d",a.m[i][1]);
    puts("");
}


void init()
{
    while (scanf("%d",&m)!=EOF)
    {
        scanf("%s",s);
        l=strlen(s);
        n=l;
        for (int i=0;i<l;i++)
            a.m[i+1][1]=s[i]-'0';
        work();
    }
}

int main ()
{
    init();
    close();
    return 0;
}