2016湘潭邀请赛 XTU 1243 2016

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本文介绍了一种利用矩阵快速幂解决特定数学问题的方法。给定一个2×2矩阵A，文章探讨了如何高效计算矩阵A的n次方，并取模7的余数。通过巧妙地运用快速幂算法，可以在大规模数据集上有效地进行计算。

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2016
Accepted : 117		Submit : 438
Time Limit : 2000 MS		Memory Limit : 65536 KB

2016

Given a 2×2 matrix

A = (a 11 a 21 a 12 a 22),

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

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

Sample Output

题目的这句话Feel free to think why the problem is called 2016 if you either:可以猜到循环节是2016

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
char x[100005];
int main()
{
                int n;
                while(~scanf("%s",x))
                {
                                n=0;
                                int t=strlen(x);
                                for(int i=0;i<t;i++)
                                {
                                                n=(n*10+x[i]-'0')%2016;
                                }
                                //printf("%d\n",n);
                                int a,b,c,d;
                                cin>>a>>b>>c>>d;
                                int a1,b1,c1,d1;
                                a1=a;
                                b1=b;
                                c1=c;
                                d1=d;
                                for(int i=1;i<n;i++)
                                {
                                                int a2=a1;
                                                int b2=b1;
                                                int c2=c1;
                                                int d2=d1;
                                          a1=(a*a2+b*c2)%7;
                                          b1=(a*b2+b*d2)%7;
                                          c1=(c*a2+d*c2)%7;
                                          d1=(c*b2+d*d2)%7;

                                }
                                if(n==0)
                                {
                                                a1=1;
                                                b1=0;
                                                c1=0;
                                                d1=1;
                                }
                                cout<<a1<<" "<<b1<<endl<<c1<<" "<<d1<<endl;
                }
                return 0;
}