codeforces 450B B. Jzzhu and Sequences(矩阵快速幂)

题目链接：

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 (10^9 + 7) equals (10^9 + 6).

题意：

水题,不行说;

思路：

矩阵快速幂的水题;

AC代码：

#include <bits/stdc++.h>
using namespace std;
const int N=1e4+6;
typedef long long ll;
const ll mod=1e9+7;
ll n,x,y;
struct matrix
{
ll a[2][2];
};
matrix mul(matrix A,matrix B)
{
matrix s;
s.a[0][0]=s.a[1][1]=0;
s.a[0][1]=s.a[1][0]=0;
for(int i=0;i<2;i++)
{
for(int j=0;j<2;j++)
{
s.a[i][j]=0;
for(int k=0;k<2;k++)
{
s.a[i][j]+=A.a[i][k]*B.a[k][j];
s.a[i][j]%=mod;
}
}
}
return s;
}
ll fast_pow(matrix A,ll num)
{
matrix s,base;
for(int i=0;i<2;i++)
{
for(int j=0;j<2;j++)
{
s.a[i][j]=(i==j);
base.a[i][j]=A.a[i][j];
}
}

while(num)
{
if(num&1)
{
s=mul(s,base);
}
base=mul(base,base);
num=(num>>1);
}
return (s.a[0][0]*y%mod+s.a[0][1]*x%mod)%mod;

}

int main()
{
cin>>x>>y;
cin>>n;
matrix ma;
ma.a[0][0]=ma.a[1][0]=1;
ma.a[0][1]=-1;
ma.a[1][1]=0;
if(n>2)cout<<(fast_pow(ma,n-2)%mod+mod)%mod<<"\n";
else if(n==2)cout<<(y%mod+mod)%mod<<"\n";
else cout<<(x%mod+mod)%mod<<"\n";

}