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

1010111110100000001

 

Sample Output

1111000001000010

 

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 intstruct 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;}