HDU - 2276 Kiki & Little Kiki 2 (矩阵快速幂)

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

 

思路:

想到矩阵快速幂是关键.

其实这个公式我推的时候也没有想到,但实在是不应该.

s[i]=(s[i]+s[i-1])%2;

代码:

#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<ctime>

#define fuck(x) cout<<#x<<" = "<<x<<endl;
#define debug(a, x) cout<<#a<<"["<<x<<"] = "<<a[x]<<endl;
#define ls (t<<1)
#define rs ((t<<1)|1)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 108;
const int maxm = 100086;
const int inf = 0x3f3f3f3f;
const ll Inf = 999999999999999999;
const int mod = 2;
const double eps = 1e-6;
const double pi = acos(-1);

struct Matrix{
    int mp[108][108];
};
Matrix mul(Matrix a,Matrix b,int n){
    Matrix ans;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            ans.mp[i][j]=0;
            for(int k=1;k<=n;k++){
                ans.mp[i][j]+=a.mp[i][k]*b.mp[k][j];
            }
            ans.mp[i][j]%=mod;
        }
    }
    return ans;
}
Matrix q_pow(Matrix a,int b,int n){
    Matrix ans;
    memset(ans.mp,0,sizeof(ans.mp));
    for(int i=1;i<=n;i++){
        ans.mp[i][i]=1;
    }
    while (b){
        if(b&1){
            ans=mul(ans,a,n);
        }
        b>>=1;
        a=mul(a,a,n);
    }
    return ans;
}

int num[maxn];
char s[maxn];
int main() {
    int n,m;
    while (scanf("%d",&m)!=EOF){
        scanf("%s",s+1);
        n=strlen(s+1);
        for(int i=1;i<=n;i++){
            num[i]=s[i]-'0';
        }

        Matrix tmp;
        memset(tmp.mp,0,sizeof(tmp.mp));
        tmp.mp[1][1]=tmp.mp[1][n]=1;
        for(int i=2;i<=n;i++){
            tmp.mp[i][i]=tmp.mp[i][i-1]=1;
        }
        tmp=q_pow(tmp,m,n);
        for(int i=1;i<=n;i++){
            int ans=0;
            for(int j=1;j<=n;j++){
                ans+=tmp.mp[i][j]*num[j];
                ans%=mod;
            }
            printf("%d",ans);
        }
        printf("\n");
    }
    return 0;
}