Codeforces 450B-Jzzhu and Sequences (矩阵快速幂)

B. Jzzhu and Sequences

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Jzzhu has invented a kind of sequences, they meet the following property:

You are given x and y, please calculate fn modulo 1000000007 (109 + 7).

Input

The first line contains two integers x and y (|x|, |y| ≤ 109). The second line contains a single integer n (1 ≤ n ≤ 2·109).

Output

Output a single integer representing fn modulo 1000000007 (109 + 7).

Examples

input

2 3
3

output

1

input

0 -1
2

output

1000000006

Note

In the first sample, f2 = f1 + f3, 3 = 2 + f3, f3 = 1.

In the second sample, f2 =  - 1;  - 1 modulo (109 + 7) equals (109 + 6).

很明显 构造矩阵  

F(N) = F(N-1) +F(N+1) 则有  F(N-1) =F(N-2) + F( N)   故  F(N) =F(N-1) --F(N-2) 

代码:

#include <iostream>
#include <stdio.h>
#include <cmath>
#include <algorithm>
#include <cstring>

typedef long long ll;

const int N= 55;
const int MOD=1000000007;
const int MAXN=2;
using namespace std;
struct Matrix{
	ll arr[N][N];
	void init()
	{
		memset(arr,0,sizeof(arr));
		for(int i=0;i<MAXN;i++)
			arr[i][i]=1;//初始化
	}
	void iinit()
	{
	    memset(arr,0,sizeof(arr));
	    arr[0][0]=arr[1][0]=1;
	    arr[0][1]=-1;
	}
}A;
Matrix mul(Matrix X,Matrix Y)// 矩阵乘法
{
	Matrix ans;
	for(int i=0;i<MAXN;i++)
		for(int j=0;j<MAXN;j++){
			ans.arr[i][j]=0;
			for(int k=0;k<MAXN;k++){
				ans.arr[i][j]+=X.arr[i][k]*Y.arr[k][j];
			ans.arr[i][j]%=MOD;
			}
		}
	return ans;
}
Matrix Q_pow(Matrix B,int n)// 矩阵快速幂
{
	Matrix ans;
	ans.init();
	while(n)
	{
		if(n&1)
			ans=mul(ans,B);
		n>>=1;
		B=mul(B,B);
	}
	return ans;
}
Matrix Add(Matrix a,Matrix b)  //(a+b)%mod 矩阵加法
{
    int i,j,k;
    Matrix ans;
    for(i=0;i<MAXN;i++)
        for(j=0;j<MAXN;j++)
        {
            ans.arr[i][j]=a.arr[i][j]+b.arr[i][j];
            ans.arr[i][j]%=MOD;
        }
    return ans;
}
Matrix Sum(Matrix a,int n)// 矩阵和
{
	int m;
	Matrix ans,pre;
	if(n==1) return ans;
	m=n/2;
	pre=Sum(a,m);
	ans=Add(pre,mul(pre,Q_pow(a,m)));
	if(n&1)
		ans=Add(ans,Q_pow(a,n));
	return ans;
}

int main()
{
    int x,y,n;
    while(~scanf("%d %d",&x,&y))
    {
        scanf("%d",&n);
        Matrix ans;
        ans.iinit();
        if(n==1)
        {
            printf("%d\n",(x+MOD)%MOD);
        }
        else if(n==2)
        {
            printf("%d\n",(y+MOD)%MOD);
        }
        else
        {
            ans=Q_pow(ans,n-2);
            ll res=(ans.arr[0][0]*y+ans.arr[0][1]*x)%MOD;
             printf("%lld\n",(res+MOD)%MOD);

        }
    }
    return 0;
}