hdu 2276 Kiki & Little Kiki 2

Kiki & Little Kiki 2

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2915    Accepted Submission(s): 1546

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

 

有n盏灯，每秒后如果左边的等是开了，那么当前的灯就变成反状态，否则不变。

左边   原来   现在

1          0         1

1         1          0

0          1         1

0          0         0

 

现在=原来^左边    由于用加法公式才符合用矩阵快速幂

所以可以变成 现在 = （左边+原来）% 2

所以得到公式：

a_n                   1   1   .   .   .   0  0           a_n

a_n-1                0   1   1  .  .   .    0           a_n-1

 .                                      .                       .

 .             =                       .                       .

 .                                      .                       .

 a₂                  0    0  .   .   .    1  1            a₂

 a₁                  1    0   .  .   .    0   1            a₁

 

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int mod = 200907;
const int N = 110;
char str[N];
int len, a[N];
struct mat{
    ll m[N][N];
    mat(){
        memset(m, 0, sizeof(m));
    }
};

mat mul(mat &A, mat &B) {
    mat C;
    for(int i = 0; i < len; i ++) {
        for(int j = 0; j < len; j ++) {
            for(int k = 0; k < len; k ++) {
                C.m[i][j] = (C.m[i][j] + A.m[i][k]*B.m[k][j]) % 2;
            }
        }
    }
    return C;
}

mat pow(mat A, int n) {
    mat B;
    for(int i = 0; i < len; i ++) B.m[i][i] = 1;
    while(n) {
        if(n&1) B = mul(B, A);
        A = mul(A, A);
        n >>= 1;
    }
    return B;
}
int main() {
    int m;
    while(scanf("%d%s",&m,str) != EOF){
        memset(a, 0, sizeof(a));
        len = strlen(str);
        mat A;
        for(int i = 0; i < len; i ++) {
            A.m[i][i] = A.m[i][(i+1)%len] = 1;
        }
        A = pow(A, m);
        for(int i = 0; i < len; i ++) {
            for(int j = 0; j < len; j ++) {
                a[i] += A.m[len-i-1][j]*(str[(len-j-1)%len]-'0');
            }
        }
        for(int i = 0; i < len; i ++) printf("%d",a[i]%2);
        printf("\n");
    }
    return 0;
}