XTOJ 1243 2016【矩阵快速幂取模】

最新推荐文章于 2019-11-27 18:51:17 发布

AC_Dreameng

最新推荐文章于 2019-11-27 18:51:17 发布

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分类专栏：（矩阵）快速幂取模 Other OJ ACM_HDU刷题录

本文链接：https://blog.youkuaiyun.com/hurmishine/article/details/53000691

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   2016 
Accepted : 100 Submit : 366
Time Limit : 2000 MS Memory Limit : 65536 KB
 

 
  2016Given a  
          
          2×2 
          matrix
 
         
          
          A=(a11a21a12a22), 
          
       
 find  
        
         
         An 
         where  
        
         
         A1=A,An=A×An−1 
        . As the result may be large, you are going to find only the remainder after division by  
        
         
         7 
        . 
       
Special Note: The problem is intended to be easy. Feel free to think why the problem is called 2016 if you either:
find it hard to solve;
or, solved all the other problems easily.
 InputThe input contains at most  
          
          40 
          sets. For each set:
The first line contains an integer  
          
          n 
          ( 
          
          1≤n<10100000 
         ).
The second line contains  
          
          2 
          integers  
          
          a11,a12 
         .
The third line contains  
          
          2 
          integers  
          
          a21,a22 
         .
( 
          
          0≤aij<7 
         ,  
          
          (a11a22−a12a21) 
          is not a multiple of  
          
          7 
         )
 OutputFor each set, a  
          
          2×2 
          matrix denotes the remainder of  
          
          An 
          after division by  
          
          7 
         .
 Sample Input 
        2
1 1
1 2
2016
1 1
1 2
 
       
 Sample Output 
        2 3
3 5
1 0
0 1
 
       

 Source XTU OnlineJudge
 

原题链接：http://202.197.224.59/OnlineJudge2/index.php/Problem/read/id/1243

当时做的时候没注意n的范围，矩阵快速幂取模，n对2016取模，具体为什么也还没有想清楚。

剩下的就是模板的事了。

AC代码：

/**
  * 博客地址:http://blog.youkuaiyun.com/hurmishine
*/

#include <iostream>
#include <cstdio>
using namespace std;
const int maxn=100000+5;
struct Matrix
{
    int m[2][2];
};
Matrix I=
{
    1,0,
    0,1
};
Matrix p;

Matrix mul(Matrix a,Matrix b)
{
    Matrix ans;
    for(int i=0; i<2; i++)
    {
        for(int j=0; j<2; j++)
        {
            ans.m[i][j]=0;
            for(int k=0; k<2; k++)
            {
                ans.m[i][j]+=(a.m[i][k]*b.m[k][j])%7;
                ans.m[i][j]%=7;
            }
        }
    }
    return ans;
}

Matrix quick_mod(Matrix a,int p)
{
    Matrix ans=I;
    Matrix tmp=a;
    while(p)
    {
        if(p&1)
            ans=mul(ans,tmp);
        tmp=mul(tmp,tmp);
        p>>=1;
    }
    return ans;
}

int main()
{
    char a[maxn];
    while(cin>>a)
    {
        cin>>p.m[0][0]>>p.m[0][1]>>p.m[1][0]>>p.m[1][1];
        int n=0;
        for(int i=0; a[i]; i++)
        {
            n=(n*10+a[i]-'0')%2016;
        }
        Matrix ans=quick_mod(p,n);
        cout<<ans.m[0][0]<<" "<<ans.m[0][1]<<endl;
        cout<<ans.m[1][0]<<" "<<ans.m[1][1]<<endl;
    }
    return 0;
}

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