Codeforces Round #257(Div. 2) B. Jzzhu and Sequences（矩阵高速幂）

题目链接：http://codeforces.com/problemset/problem/450/B

B. Jzzhu and Sequences

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 test(s)

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).

代码例如以下：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
struct A
{
    int mat[2][2];
};
A d,f;
__int64 n,mod;
A mul(A a,A b)
{
    A t;
    memset(t.mat,0,sizeof(t.mat));
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < n;j++)
        {
           // if(a.mat[i][k])
                for(int k = 0; k < n; k++)
                {
                    t.mat[i][j]+=a.mat[i][k]*b.mat[k][j];
                    t.mat[i][j]%=mod;
                }
        }
    }
    return t;
}
A quickP(int k)
{
    A p = d ,m;
    memset(m.mat,0,sizeof(m.mat));
    for(int i=0;i<n;++i)//单位矩阵
    {
        m.mat[i][i]=1;
    }
    while(k)
    {
        if(k & 1)
            m=mul(m,p);
        p=mul(p,p);
        k >>= 1 ;
    }
    return m;
}
int main()
{
    n=2;
    int k,t;__int64 x,y,z;
    while(scanf("%I64d%I64d",&x,&y)!=EOF)
    {
        int s=0;
        scanf("%I64d",&z);
        mod=1000000007;
        if(z == 1)
        {
            if(x < 0)
                printf("%I64d
",x+mod);
            else
                printf("%I64d
",x);
            continue;
        }
        d.mat[0][1]=-1;d.mat[1][1] = 0;
        d.mat[0][0]=d.mat[1][0]=1;
        A ret=quickP(z-2);//z-2 乘的次数
        __int64 ans=(ret.mat[0][0]*y%mod+ret.mat[0][1]*x%mod)%mod;
        if(ans < 0)
            ans+=mod;
        printf("%I64d
",ans);
    }
    return 0;
}