XTU 1243 2016

最新推荐文章于 2020-07-12 18:02:04 发布

ACM__dongsheng

最新推荐文章于 2020-07-12 18:02:04 发布

阅读量824

点赞数

CC 4.0 BY-SA版权

分类专栏：矩阵快速幂

本文链接：https://blog.youkuaiyun.com/ACM__dongsheng/article/details/51698746

矩阵快速幂专栏收录该内容

3 篇文章

订阅专栏

本文介绍了一个关于2×2矩阵的快速幂算法题目，通过大数取余的方法解决了指数可能非常大的问题。该算法适用于求解特定形式的矩阵乘方，并提供了完整的C++实现代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

2016

Given 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.

Input

The 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)
Output

For 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
分析：不要被它的n吓到了，用大数取余就好了，其他的就是快速幂了！

#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<cmath>
#include<set>
#include<map>
#include<queue>
#include<cstdio>
using namespace std;
char s[100005]; 
#define N 7
struct r
{
    long long int m[2][2];
}ans, tmp;

r multi(r a,r b)
{
    r cnt;
    cnt.m[0][0]=(a.m[0][0]*b.m[0][0]+a.m[0][1]*b.m[1][0])%N;
    cnt.m[0][1]=(a.m[0][0]*b.m[0][1]+a.m[0][1]*b.m[1][1])%N;
    cnt.m[1][0]=(a.m[1][0]*b.m[0][0]+a.m[1][1]*b.m[1][0])%N;
    cnt.m[1][1]=(a.m[1][0]*b.m[0][1]+a.m[1][1]*b.m[1][1])%N;
    return cnt;
}

 void  Matrix_power(int n)
{
    tmp.m[0][0]=tmp.m[1][1] =1;
    tmp.m[0][1]=tmp.m[1][0]=0;
    while(n)
    {
        if(n&1)
        tmp=multi(tmp,ans);
        ans=multi(ans,ans);
        n>>=1;
    }
    cout<<tmp.m[0][0]<<" "<<tmp.m[0][1]<<endl<<tmp.m[1][0]<<" "<<tmp.m[1][1]<<endl;
}

int main()
{
    while(cin>>s)
    {
         scanf("%lld%lld%lld%lld",&ans.m[0][0],&ans.m[0][1],&ans.m[1][0],&ans.m[1][1]);
         int mod=0,len=strlen(s);
         for(int i=0;i<len;i++)
         {
            mod=mod*10+s[i]-'0';
            mod%=2016;  
         }
         if(mod==0)
         {
            cout<<"1 0"<<endl<<"0 1"<<endl;
         }
         else
         Matrix_power(mod);
    }
    return 0;
}