Codeforces #257B Jzzhu and Sequences(矩阵快速幂）

Description

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

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

Sample Input

Input

2 33

Output

1

Input

0 -12

Output

1000000006

题比较简单，和求斐波那契数列一样，唯一要注意的就是前两项的输出。

值为负的时候要先转为正的再取模。

#include<cstdio>#include<cstring>#include<algorithm>#include<vector>#include<string>#include<iostream>#include<queue>#include<cmath>#include<map>#include<stack>#include<bitset>using namespace std;#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )#define CLEAR( a , x ) memset ( a , x , sizeof a )typedef long long LL;typedef pair<int,int>pil;const int MOD=1e9+7;const int INF = 0x3f3f3f3f;struct Matrix{   LL mat[2][2];   void Clear()   {       CLEAR(mat,0);   }};Matrix mult(Matrix m1,Matrix m2){    Matrix ans;    for(int i=0;i<2;i++)    {        for(int j=0;j<2;j++)        {            ans.mat[i][j]=0;            for(int k=0;k<2;k++)            {                ans.mat[i][j]=(ans.mat[i][j]+m1.mat[i][k]*m2.mat[k][j]);                while(ans.mat[i][j]<0)  ans.mat[i][j]+=MOD;                ans.mat[i][j]%=MOD;            }        }    }    return ans;}Matrix Pow(Matrix m1,LL b){    Matrix ans;ans.Clear();    for(int i=0;i<2;i++)        ans.mat[i][i]=1;    while(b)    {        if(b&1)            ans=mult(ans,m1);        b>>=1;        m1=mult(m1,m1);    }    return ans;}LL x,y,n;int main(){    while(~scanf("%lld%lld",&x,&y))    {        while(x<0)  x+=MOD;while(y<0) y+=MOD;        x%=MOD;y%=MOD;        Matrix A;scanf("%lld",&n);        if(n==1)        {            printf("%lld\n",x);            continue;        }        if(n==2)        {            printf("%d\n",y);            continue;        }        A.mat[0][0]=1;A.mat[0][1]=-1;        A.mat[1][0]=1;A.mat[1][1]=0;        A=Pow(A,n-2);        LL ans=0;        ans=A.mat[0][0]*y;while(ans<0)  ans+=MOD;        ans%=MOD;ans+=A.mat[0][1]*x;while(ans<0)  ans+=MOD;        ans%=MOD;        printf("%lld\n",ans);    }    return 0;}