矩阵优化可以经常利用在递推式中。
首先了解一下矩阵乘法的法则。
(egin{bmatrix}a&b\c&dend{bmatrix}) ( imes) (egin{bmatrix}e&f\g&hend{bmatrix}) (=) (egin{bmatrix} a imes e + b imes g & a imes f + b imes h \ c imes e + d imes g & c imes f + d imes h end{bmatrix})
这就貌似非常简单了。
看一个例题,斐波那契数列不同于一般的斐波那契数列,(n)在(long) (long)之内,所以(O(n))绝对会超时,这是需要矩阵快速幂,复杂度是(O(3^3logn)) (=) (O(27log n)),忽略常数,那么复杂度就是(O(logn))
我们定义一个矩阵等式,然后去求问号矩阵
(egin{bmatrix} f[i+1] & f[i]end{bmatrix}) ( imes) (egin{bmatrix}? end{bmatrix}) (=) (egin{bmatrix}f[i+2] & f[i+1] end{bmatrix})
(f[i+2] = f[i] + f[i+1])
构造的(?)号矩阵就是
(egin{bmatrix} 1 & 1 \ 1 & 0end{bmatrix})
带回检验
(egin{bmatrix}f[i+1] & f[i] \ 0 & 0end{bmatrix} imes egin{bmatrix} 1 & 1 \ 1 & 0end{bmatrix} = egin{bmatrix}f[i+1] imes 1 + f[i] imes 1 & f[i+1] imes 1 + f[i] imes 0 \ 0 imes 1 + 0 imes 1 & 0 imes 1 + 0 imes 0 end{bmatrix})
上式化简为
(egin{bmatrix}f[i+2] & f[i+1] \ 0 & 0end{bmatrix} = egin{bmatrix}f[i+2] & f[i+1]end{bmatrix})
所以成立.
矩阵快速幂模板代码就是
struct Mat {
int a[3][3];
Mat() {memset(a,0,sizeof a);}
inline void build() {
memset(a,0,sizeof a);
for(re int i = 1 ; i <= 2 ; ++ i) a[i][i]=1;
}
};
Mat operator*(Mat &a,Mat &b)
{
Mat c;
for(re int k = 1 ; k <= 2 ; ++ k)
for(re int i = 1 ; i <= 2 ; ++ i)
for(re int j = 1 ; j <= 2 ; ++ j)
c.a[i][j]=(c.a[i][j]+a.a[i][k]*b.a[k][j]%mod)%mod;
return c;
}
Mat quick_Mat(int x)
{
Mat ans;ans.build();
while(x) {
if((x&1)==1) ans = ans * a;
a = a * a;
x >>= 1;
}
return ans;
}
例题代码是
#include <cstdio>
#include <iostream>
#include <cmath>
#include <cstring>
#include <queue>
#include <stack>
#define re register
#define Max 200000012
#define int long long
int n;
const int mod=1000000007;
struct Mat {
int a[3][3];
Mat() {memset(a,0,sizeof a);}
inline void build() {
memset(a,0,sizeof a);
for(re int i = 1 ; i <= 2 ; ++ i) a[i][i]=1;
}
};
Mat operator*(Mat &a,Mat &b)
{
Mat c;
for(re int k = 1 ; k <= 2 ; ++ k)
for(re int i = 1 ; i <= 2 ; ++ i)
for(re int j = 1 ; j <= 2 ; ++ j)
c.a[i][j]=(c.a[i][j]+a.a[i][k]*b.a[k][j]%mod)%mod;
return c;
}
Mat a;
Mat quick_Mat(int x)
{
Mat ans;ans.build();
while(x) {
if((x&1)==1) ans = ans * a;
a = a * a;
x >>= 1;
}
return ans;
}
signed main()
{
scanf("%lld",&n);
a.a[1][1]=1;a.a[1][2]=1;
a.a[2][1]=1;Mat b;
b.a[1][1]=1;b.a[2][1]=1;
if(n>=1 && n<=2) {
printf("1");return 0;
}
Mat ans=quick_Mat(n-2);
ans=ans*b;
printf("%lld",ans.a[1][1]);
return 0;
}