HDU 2276 矩阵快速幂

Posted 2020-09-27 zzq

Kiki & Little Kiki 2

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2650    Accepted Submission(s): 1393

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

 

 

Source

HDU 8th Programming Contest Site（1）

 

一开始没想到公式。。。看了一眼别人的公示后推开了矩阵。。脑子秀逗了总把乘号写成加号

公式a[i]=(a[i]+a[i-1])%2;,特别的对于a[1]=(a[i]+a[N])%2

M最大10亿显然朴素法不可取，由于是看专题进来的所以直接想的就是矩阵= =

假设第零次的数组为原始01串(a,b,c,d)

则第一次 ((a+d)%2,(b+a)%2,(c+b)%2,(d+c)%2)

   第二次 (((a+d)%2+(d+c)%2)%2,......)

不难构造出一个N*N的矩阵，第i列的第i和i-1个元素置为1其余元素置0即可。

然后计算出这个转移矩阵的N次幂后再与原始行矩阵相乘得到答案。

此处还用到了同余定理 (A+B)%M=(A%M+B%M)%M;

 

 

#include<bits/stdc++.h>
using namespace std;
int N,M;
struct Matrix
{
    int a[105][105];
    Matrix operator*(Matrix tmp){
        Matrix ans;
        memset(ans.a,0,sizeof(ans.a));
        for(int i=1;i<=N;++i){
            for(int k=1;k<=N;++k){
                for(int j=1;j<=N;++j){
                    ans.a[i][j]+=a[i][k]*tmp.a[k][j];
                    ans.a[i][j]%=2;
                }
            }
        }
    return ans;
    }
};
void show(Matrix a)
{int i,j,k;
    for(i=1;i<=N;++i){
        for(j=1;j<=N;++j){
            cout<<a.a[i][j]<<" ";
        }cout<<endl;
    }cout<<endl;
}
Matrix qpow(Matrix A,int n)
{
    Matrix ans;
    memset(ans.a,0,sizeof(ans.a));
    for(int i=0;i<=N;++i) ans.a[i][i]=1;
    while(n){
        if(n&1) ans=ans*A;
        A=A*A;
        n>>=1;
    }
    return ans;
}

void solve(string s)
{
    Matrix A;
    int i,j,k,u[105];
    char ans[105];
    for(i=0;i<s.size();++i) u[i+1]=s[i]-‘0‘;
    memset(A.a,0,sizeof(A.a));
    A.a[1][1]=A.a[N][1]=1;
    for(i=2;i<=N;++i){
            A.a[i-1][i]=A.a[i][i]=1;
    }
    A=qpow(A,M);

    for(i=1;i<=N;++i){int d=0;
        for(j=1;j<=N;++j){
            d+=u[j]*A.a[j][i];
            d%=2;
        }
       cout<<d%2;
    }cout<<endl;
}
int main()
{
    string s;
    while(cin>>M>>s){
            N=s.size();
              solve(s);
    }
    return 0;
}