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

原创于 2018-09-08 08:32:09 发布 · 225 阅读

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本文解析了CodeForces-450B题目，介绍了一种特殊的序列，通过递推公式f1=x,f2=y,对所有i(i≥2)，fi=fi−1+fi+1进行分析。利用矩阵快速幂方法，解决了求fn模10^9+7的问题。

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Jzzhu and Sequences CodeForces - 450B

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

这里写图片描述
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).

题意：

给定递推式

f 1 = x, f 2 = y, \forall i (i \geq 2) f i = f i - 1 + f i + 1

$f_1 = x,f_2 = y,\forall i(i\ge 2)f_i = f_{i-1} + f_{i+1}$

分析：

我们可以转化递推式

f i + 1 = f i - f i - 1

$f_{i+1} = f_i - f_{i-1}$

即：

f i = f i - 1 - f i - 2

$f_{i} = f_{i-1} - f_{i-2}$

因此可以得到关系矩阵

对于 $n \ge 2$

(f n f n - 1) = (1 - 1 10) n - 2 (y x) (41)

$\begin{equation} \left( \begin{matrix} f_n\\ f_{n-1} \end{matrix} \right) = \left( \begin{matrix} 1\ \ \ \ -1\\ 1\ \ \ \ \ \ \ \ 0\\ \end{matrix} \right)^{n-2} \left( \begin{matrix} y\\ x \end{matrix} \right) \end{equation}$
注意步步取模
code:

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long ll;
const int maxn = 1e5+10;
const ll mod = 1e9+7;
ll x,y,n;
struct Matrix{
    ll mat[3][3];
    Matrix operator * (const Matrix &b)const{
        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] + mat[i][k] * b.mat[k][j] % mod + mod) % mod;
                }
            }
        }
        return ans;
    }
};
Matrix q_pow(Matrix a,ll b){
    Matrix ans;
    memset(ans.mat,0,sizeof(ans.mat));
    for(int i = 0; i < 2; i++){
        ans.mat[i][i] = 1;
    }
    while(b){
        if(b & 1)
            ans = ans * a;
        b >>= 1;
        a = a * a;
    }
    return ans;
}
int main(){
    scanf("%lld%lld",&x,&y);
    scanf("%lld",&n);
    if(n == 1){
        printf("%lld\n",(x % mod + mod) % mod);
        return 0;
    }
    if(n == 2){
        printf("%lld\n",(y % mod + mod) % mod);
        return 0;
    }
    Matrix ans;
    ans.mat[0][0] = 1;
    ans.mat[0][1] = -1;
    ans.mat[1][0] = 1;
    ans.mat[1][1] = 0;
    ans = q_pow(ans,n-2);
    ll ret = ((ans.mat[0][0] * y % mod + ans.mat[0][1] * x % mod) % mod + mod) % mod;
    printf("%lld\n",ret);
    return 0;
}