A - Jzzhu and Sequences
Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u
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Status
 
Practice
 
CodeForces 450B
Appoint description: 
Description

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).

Sample Input

Input
2 3
3
Output
1
Input
0 -1
2
Output
1000000006
Hint

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).
 

题意：

 f1 = x, f2 = y; fi = f(i-1) + f(i+1)，求f[n]
       

分析：

因为

fi = f(i-1) + f(i+1)

所以

f(i+1)=f(i)-f(i-1)

fi = f(i-1)-f(i-2)

所以，

f(i)=1*f(i-1)-1*f(i-2)

f(i-1)=1*f(i-1)

构造矩阵A

1 -1

1 0

f（n）                        f(2)

f（n-1）= A^（k-2）*f(1)

注意取mod

 

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<iomanip>
#include<cmath>
#include<cstring>
#include<vector>
#define N 10005  
using namespace std;
const int MAXN= 100;
const int p=1000000007;
typedef long long ll;
ll len;
struct Matrix
{
    long long M[MAXN][MAXN];
    Matrix(const bool I = 0)  ///初始化对角矩阵
    {
        memset(M, 0, sizeof(M));
        if (I)
        for(int i = 0; i < len; i++) 
              M[i][i] = 1;
    }
    Matrix operator *(const Matrix &y) ///矩阵乘法，对乘法重新定义
    {
        Matrix z;
        for (int i = 0; i < len; i++)
            for (int j = 0; j < len; j++)
                for (int k = 0; k < len; k++)
                    z.M[i][j] = (z.M[i][j]+M[i][k]*y.M[k][j]%p)%p;
        return z;
    }
    void out()
    {
        for (int i = 0; i <len; i++)
        {
            for (int j = 0; j < len; j++)
                 cout << M[i][j] << " ";
            cout << "\n";
        }
    }
};
Matrix pow(Matrix A, long long b)///矩阵的快速幂
    {
        Matrix ans = Matrix(1);
        for (; b; b>>=1)
        {
          if (b&1) ans = ans*A;
          A = A*A;
        }
       return ans; 
    }
int main()  
{  
    int T;
    len=2;  ///构造矩阵的长度
    ll x,y,k;
    while(scanf("%lld%lld%lld",&x,&y,&k)!=-1)
    {
        if(k==2) printf("%lld\n",(p+y%p)%p);
        else if(k==1) printf("%lld\n",(x%p+p)%p) ;
        else
        {
        Matrix ans;
        
        ans.M[0][0] = 1;
        ans.M[0][1] = -1;
        ans.M[1][0] = 1;
        ans.M[1][1] = 0;
        
        ans = pow(ans,k-2);
        //ans.out();
        ll ans1=(ans.M[0][0]*y%p+ans.M[0][1]*x%p+p)%p;
        while(ans1<0) ans1+=p;
        printf("%lld\n",ans1);  
        
        }
        
    }
     

    
    
    return 0;  
}