hdu 2276 矩阵 有点小发现，矩阵mod的使用太多易造成TLE

Kiki & Little Kiki 2

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1232    Accepted Submission(s): 645

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）

 

Recommend

lcy

题意：

题目大意：有一圈的灯，其中0和1分别表示灯的暗亮，灯每秒会以一种规律改变状态，规则是

当前一盏灯亮时，就改变状态，否则就不改变状态。

思路：  很容易看出

可以用矩阵去运算，先要构造一个矩阵！

| 1 0 0 0 ....1 |

| 1 1 0 0.....0 |     

| 0 1 1 0 ....0|  

     ......

我们只要把这个矩阵^M再乘以初始状态的就可以了

注意：

在代码中注意mod 的应用  如果没有用的mod用多了  就会造成超时  所以除了必须的那个mod  能省的mod还是要省去的  不然很超时

#include<stdio.h>
#include<string.h>
#define ll int
struct Mat
{
      int martix[102][102];
};
int N,n,d,mod=2;
Mat temp,res,init,q;
Mat Martix_Add(Mat a,Mat b)  
{  
    int i,j;  
    Mat c;  
    for (i=0;i<N;i++)  
    {  
        for (j=0;j<N;j++)  
        {  
            c.martix[i][j]=(a.martix[i][j]+b.martix[i][j])%mod;  
        }  
    }  
    return c;  
}  
Mat Martix_Mul(Mat a,Mat b)  
{  
    int i,j,l;  
    Mat c;  
    for (i=0;i<N;i++)  
    {  
        for (j=0;j<N;j++)  
        {  
            c.martix[i][j]=0;  
            for (l=0;l<N;l++)  
            {  
                c.martix[i][j]+=a.martix[i][l]*b.martix[l][j];  
                
            } 
			c.martix[i][j]%=mod;  
        }  
    }  
    return c;  
}  
  
Mat er_fun(Mat e,ll x)  //求矩阵e^x  
{  
    Mat tp; 
    tp=e;
    e=res;  //res是单位矩阵
    while(x)  
    {  
        if(x&1)  
            e=Martix_Mul(e,tp);  
        tp=Martix_Mul(tp,tp);  
        x>>=1;  
    }  
    return e;  
}  
int main()
{
    int n,g[102],i,j,k;
    char s[102];
    memset(res.martix,0,sizeof(res.martix));
    for(i=0;i<=100;i++)
        res.martix[i][i]=1;
    while(scanf("%d",&n)!=EOF)
    {
        scanf("%s",s);
        d=strlen(s);
            for(i=0;i<d;i++)
                  g[i]=s[i]-'0';
            N=d;
            memset(init.martix,0,sizeof(init.martix));
            init.martix[0][0]=init.martix[0][d-1]=1;
            k=0;
            for(i=1;i<d;i++)
            {
                init.martix[i][k]=init.martix[i][k+1]=1;
                k++;
            }
            q=er_fun(init,n);
            for(i=0;i<d;i++)
            {
                int ans=0;
                for(j=0;j<d;j++)
                    ans+=q.martix[i][j]*g[j]%mod;
                printf("%d",ans%2);
            }
            printf("\n");

    }
    return 0;
}