hdu 2276 Kiki & Little Kiki 2 矩阵的应用

http://acm.hdu.edu.cn/showproblem.php?pid=2276

题目描述：

 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)

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.

由题意很容易得出：

light[0] = (light[n-1] + light[0])%2;

light[1] = (light[0] + light[1])%2;

light[2] = (light[1] + light[[2])%2;

所以，可以模拟。担考虑到M很大，所要优化。

???怎么优化呢。我也不会，看了人家的Code才Understand。所以说要多做题目，找experience.

对于2^M次方，要想快速求出可以分治:

.==>   if M is even

                       2^M=2^(M/2) * 2^(M/2);

             else

                      2^M=2*(2^(M/2)*2^(M/2);

所以可以利用矩阵的模仿上面的方法。

用；matrix^M * strmatrix = result;

 

先要构造一个矩阵！

| 1 0 0 0 ....1 |

| 1 1 0 0.....0 |     

| 0 1 1 0 ....0|  

另外还要用到单位矩阵。

Code:

 

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
using namespace std;
#define MAXV 105

struct Node{
    int Graph[MAXV][MAXV];
};
Node unit,init,Right;
int t,len;
//string str;
char str[MAXV];
void Init(){
    int i,j;
    for(i=0;i<len;++i)
        for(j=0;j<len;++j){
            unit.Graph[i][j]=(i==j);
            init.Graph[i][j]=Right.Graph[i][j]=0;
        }
    init.Graph[0][0]=init.Graph[0][len-1]=1;

    for(i=1;i<len;++i)
        init.Graph[i][i-1]=init.Graph[i][i]=1;
    for(i=0;i<len;++i)
        Right.Graph[i][0]=str[i]-'0';
}
Node Multy(Node &a,Node &b){
    Node c;
    int i,j,k;
    for(i=0;i<len;++i)
        for(j=0;j<len;++j){
            c.Graph[i][j]=0;
            for(k=0;k<len;++k)
                c.Graph[i][j]+=a.Graph[i][k]*b.Graph[k][j];
            c.Graph[i][j]%=2;
        }
    return c;
}
Node Cal(){
    Node p,q;
    p=unit;q=init;
    while(t){
        if(t&1)
            p=Multy(p,q);
        t>>=1;
        q=Multy(q,q);
    }
    p=Multy(p,Right);
    return p;
}
int main() {
    int i;
//    while(cin>>t){
    while(scanf("%d",&t)!=EOF){
//        cin>>str;
        scanf("%s",str);
//        len=str.length();
        len=strlen(str);
        Init();
        Node res=Cal();
        for(i=0;i<len;++i)
            printf("%d",res.Graph[i][0]);
        printf("\n");
//            cout<<res.Graph[i][0];
//        cout<<endl;
    }
    return 0;
}