Codeforces 450B div.2 Jzzhu and Sequences 矩阵快速幂or规律

Posted 2020-07-06 ( m Lweleth)

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).

题意：给出一个递推式和前两项，求第n项模1e9+7后的值。

题解：这题其实本来是很水的..只是最近都在尝试写一些矩阵快速幂的题目，最难的在于化递推式并构造矩阵上，而这道题直接给出了递推式，心痒想使用矩阵。_(:3」∠)_

由f(i)=f(i+1)+f(i-1)可以得出f(i+1)=f(i)-f(i-1)

又由于i>=2，从f(1)开始，于是

f(3)=(1) * f(2) + (-1) * f(1)

f(2)=(1) * f(1) + (0) * f(0)

另外要注意的是，得到的值是负数还得再处理一下。（最近总WA在这上）

 1 #include <stdio.h>
 2 #include <algorithm>
 3 #include <iostream>
 4 #include <string.h>
 5 #define ll __int64
 6 using namespace std;
 7 
 8 const int mod = 1000000007;
 9 struct matrix
10 {
11     ll x[2][2];
12 };
13 matrix mul(matrix a, matrix b)
14 {
15     matrix c;
16     c.x[0][0] = c.x[0][1] = c.x[1][0] = c.x[1][1] = 0;
17     for( int i = 0; i < 2; i++)
18         for(int j = 0; j < 2; j++)
19         {
20             for(int k = 0; k < 2; k++)
21             {
22                 c.x[i][j] += a.x[i][k] * b.x[k][j];
23             }
24             c.x[i][j] %= mod;
25         }
26     return c;
27 }
28 matrix powe(matrix x, ll n)
29 {
30     matrix r;
31     r.x[1][1] = r.x[0][0] = 1;  //注意初始化
32     r.x[0][1] = r.x[1][0] = 0;
33     while(n)
34     {
35         if(n & 1)
36             r = mul(r , x);
37         x = mul(x , x);
38         n >>= 1;
39     }
40     return r;
41 }
42 int main()
43 {
44 
45     ll x, y, n, ans;
46     while(~scanf("%I64d%I64d%I64d", &x, &y, &n))
47     {
48         if(n == 2)
49             printf("%I64d\\n",(y%mod + mod)%mod);  //负数情况下的考虑
50         else if(n == 1)
51             printf("%I64d\\n",(x%mod + mod)%mod);
52         else
53         {
54             matrix d;
55             d.x[0][0] = 1;
56             d.x[0][1] = -1;
57             d.x[1][0] = 1;
58             d.x[1][1] = 0;
59 
60             d = powe(d, n - 2);
61             ans = d.x[0][0] * y +d.x[0][1]*x;
62             printf("%I64d\\n", (ans%mod+mod)%mod );
63         }
64 
65     }
66 }