## 2016

 Accepted : 105 Submit : 394 Time Limit : 2000 MS Memory Limit : 65536 KB

# 2016

Given a 2×2 matrix

A=(a11a21a12a22),
find An where A1=A,An=A×An1.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 called2016 if you either:

1. find it hard to solve;
2. 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 (1n<10100000).

The second line contains 2 integers a11,a12.

The third line contains 2 integers a21,a22.

(0aij<7,(a11a22a12a21) 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


#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

#define LL long long
#define MAXN 100005
#define mod 7
char str[MAXN];
struct mat
{
int m[3][3];
}base,ans;

mat multiply(mat a,mat b)
{
mat temp;
for(int i=0;i<2;i++)
for(int j=0;j<2;j++){
temp.m[i][j]=0;
for(int k=0;k<2;k++)
temp.m[i][j]=(temp.m[i][j]+a.m[i][k]*b.m[k][j])%mod;
}
return temp;
}

mat quick_mod(LL n)
{
ans.m[0][0]=ans.m[1][1]=1;
ans.m[1][0]=ans.m[0][1]=0;

while(n)
{
if(n&1)
ans=multiply(ans,base);
base=multiply(base,base);
n>>=1;
}
return ans;
}

int main()
{
int num;
//freopen("in.txt","r",stdin);
while(scanf("%s",str)!=EOF)
{
num=0;
int len=strlen(str);
for(int i=0;i<len;i++)
num=(num*10+str[i]-'0')%2016;
scanf("%d%d",&base.m[0][0],&base.m[0][1]);
scanf("%d%d",&base.m[1][0],&base.m[1][1]);

base=quick_mod(num);
printf("%d %d\n",base.m[0][0],base.m[0][1]);
printf("%d %d\n",base.m[1][0],base.m[1][1]);
}
return 0;
}