CodeForces 450B 矩阵

A - Jzzhu and Sequences

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 450B

Appoint description:  System Crawler  (2016-04-23)

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 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).

 

观察他的公式f[1] = x,f[2] = y,f[i] = f[i] - f[i-1];

 

可以得到     |    -1     1   |  ^((n%2 ? n:(--n))/2)       ×  |    f[1]   |     =      |  f[n]     | 

                  |    -1    0   |                                             |    f[2]  |             |  f[n+1]  |

这样只要处理矩阵的次方后就能得出答案。

#include<map>
#include<set>
#include<string>
#include<queue>
#include<stack>
#include<cmath>
#include<vector>
#include<cstdio>
#include<time.h>
#include<cstring>
#include<iostream>
#include<algorithm>
#define INF 1000000001
#define ll long long
#define MOD 1000000007
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
const int MAXN = 100010;
int x,y;
ll n;
struct Mat
{
    ll a[4][4];
};
Mat operator *(Mat a,Mat b)
{
    Mat c;
    memset(c.a,0,sizeof(c.a));
    for(int i = 0; i < 2; i++){
        for(int j = 0; j < 2; j++){
            for(int k = 0; k < 2;k++){
                c.a[i][j] += ((a.a[i][k] * b.a[k][j]) % MOD + MOD) % MOD;
                c.a[i][j] = ((c.a[i][j])% MOD + MOD) % MOD;
            }
        }
    }
    return c;
}
Mat mod_pow(Mat b,int n)
{
    Mat c;
    c.a[0][0] = c.a[1][1] = 1;
    c.a[0][1] = c.a[1][0] = 0;
    while(n){
        if(n & 1){
            c = c * b;
        }
        b = b * b;
        n >>= 1;
    }
    return c;
}
int main()
{
    while(~scanf("%d%d%lld",&x,&y,&n)){
        if(n == 1){
            x = (x%MOD + MOD)%MOD;
            cout<<x<<endl;
            continue;
        }
        else if(n == 2){
            y = (y%MOD + MOD)%MOD;
            cout<<y<<endl;
            continue;
        }
        else {
            Mat b;
            b.a[0][0] = -1;
            b.a[0][1] = 1;
            b.a[1][0] = -1;
            b.a[1][1] = 0;
            int t = n;
            if(t % 2 == 0)t --;
            b = mod_pow(b,t/2);
            ll ans;
            if(n % 2){
                ans = ((b.a[0][0] * x)%MOD + (b.a[0][1] * y)%MOD + MOD)%MOD;
            }
            else {
                ans = ((b.a[1][0] * x)%MOD + (b.a[1][1] * y)%MOD + MOD)%MOD;
            }
            cout<<(ans + MOD) % MOD<<endl;
        }
    }
    return 0;
}