A - Jzzhu and Sequences
Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
Appoint description:
System Crawler (2016-04-23)
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 modulo 1000000007(10 + 7).
Input
The first line contains two integers x and y(|x|, |y| ≤ 10). The second line contains a single integer n(1 ≤ n ≤ 2·10).
Output
Output a single integer representing f modulo 1000000007(10 + 7).
Sample Input
Input
2 3
3
Output
1
Input
0 -1
2
Output
1000000006
Hint
In the first sample, f = f + f, 3 = 2 + f, f = 1.
In the second sample, f = - 1; - 1 modulo (10 + 7) equals (10 + 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 ;
}