CodeForces 450B 矩阵

A - Jzzhu and Sequences

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

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

Description

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

CodeForces 450B 矩阵

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 + f3f3 = 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 = ;
int x,y;
ll n;
struct Mat
{
ll a[][];
};
Mat operator *(Mat a,Mat b)
{
Mat c;
memset(c.a,,sizeof(c.a));
for(int i = ; i < ; i++){
for(int j = ; j < ; j++){
for(int k = ; k < ;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[][] = c.a[][] = ;
c.a[][] = c.a[][] = ;
while(n){
if(n & ){
c = c * b;
}
b = b * b;
n >>= ;
}
return c;
}
int main()
{
while(~scanf("%d%d%lld",&x,&y,&n)){
if(n == ){
x = (x%MOD + MOD)%MOD;
cout<<x<<endl;
continue;
}
else if(n == ){
y = (y%MOD + MOD)%MOD;
cout<<y<<endl;
continue;
}
else {
Mat b;
b.a[][] = -;
b.a[][] = ;
b.a[][] = -;
b.a[][] = ;
int t = n;
if(t % == )t --;
b = mod_pow(b,t/);
ll ans;
if(n % ){
ans = ((b.a[][] * x)%MOD + (b.a[][] * y)%MOD + MOD)%MOD;
}
else {
ans = ((b.a[][] * x)%MOD + (b.a[][] * y)%MOD + MOD)%MOD;
}
cout<<(ans + MOD) % MOD<<endl;
}
}
return ;
}
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