矩阵快速幂 例题+模板
用矩阵表示递推公式是一个非常方便的方法，可以在O（log（n））的时间复杂度里面求解f（n），一般来说，递推项的系数要是常数或者能转化为常数。
情形1：F(n) = a1F(n - 1) + a2F(n - 2) + a3F(n - 3) + … + akF(n - k)
观察上式可以得出下面等价方程组
很明显可以构造一个矩阵
例题部分：
1. http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1752
构造矩阵为
将左边的矩阵乘n-1次
#include
using namespace std;
typedef long long llt;
const int Cube_SIZE = 3;///矩阵大小
struct Cube{
llt mat[Cube_SIZE][Cube_SIZE];
};
//单位矩阵
Cube _UnitCube = {
1, 0, 0,
0, 1, 0,
0, 0, 1
};
///矩阵乘机
Cube _Multiply(Cube A,Cube B,llt mod){
Cube _tmpCube;
for (int i = 0;i < Cube_SIZE;++i)
for (int j = 0;j < Cube_SIZE;++j){
_tmpCube.mat[i][j] = 0;
for (int k = 0;k < Cube_SIZE;++k){
_tmpCube.mat[i][j] += A.mat[i][k]*B.mat[k][j];
_tmpCube.mat[i][j] %= mod;
}
}
return _tmpCube;
}
///矩阵快速幂
Cube power_Cube(Cube A,llt n,llt mod) //矩阵快速幂
{
Cube _tmpCube = _UnitCube;
while(n){
if(n & 1)
_tmpCube = _Multiply(_tmpCube, A ,mod);
A = _Multiply(A, A, mod);
n >>= 1;
}
return _tmpCube;
}
int main(){
int t;cin>>t;
Cube tmp,arr;
llt n,x1,a,b,MOD;
while (t--){
cin>>x1>>a>>b>>MOD>>n;
arr.mat[0][0] = a;arr.mat[0][1] = b;arr.mat[0][2] = 1;
arr.mat[1][0] = 0;arr.mat[1][1] = 1;arr.mat[1][2] = 0;
arr.mat[2][0] = 0;arr.mat[2][1] = 0;arr.mat[2][2] = 1;
tmp.mat[0][0] = x1;tmp.mat[0][1] = 0;tmp.mat[0][2] = 0;
tmp.mat[1][0] = 1; tmp.mat[1][1] = 0;tmp.mat[1][2] = 0;
tmp.mat[2][0] = 1; tmp.mat[2][1] = 0;tmp.mat[2][2] = 0;
arr = power_Cube(arr,n-1,MOD);
arr = _Multiply(arr,tmp,MOD);
llt ans = arr.mat[0][0]%MOD;
cout < 2. http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1769
题目是计算非波拉契数列，n非常大，所以用矩阵快速幂
构造相当简单
#include
using namespace std;
typedef long long llt;
const int MOD = 1000000007;
const int Cube_SIZE = 2;///矩阵大小
struct Cube{
llt mat[Cube_SIZE][Cube_SIZE];
};
//单位矩阵
Cube _UnitCube = {
1, 0,
0, 1,
};
///矩阵乘机
Cube _Multiply(Cube A,Cube B,llt mod){
Cube _tmpCube;
for (int i = 0;i < Cube_SIZE;++i)
for (int j = 0;j < Cube_SIZE;++j){
_tmpCube.mat[i][j] = 0;
for (int k = 0;k < Cube_SIZE;++k){
_tmpCube.mat[i][j] += A.mat[i][k]*B.mat[k][j];
_tmpCube.mat[i][j] %= mod;
}
}
return _tmpCube;
}
///矩阵快速幂
Cube power_Cube(Cube A,llt n,llt mod) //矩阵快速幂
{
Cube _tmpCube = _UnitCube;
while(n){
if(n & 1)
_tmpCube = _Multiply(_tmpCube, A ,mod);
A = _Multiply(A, A, mod);
n >>= 1;
}
return _tmpCube;
}
int main(){
Cube arr;
llt n,hp1,hp2,hp3;
while (cin>>hp1>>hp2>>hp3>>n){
if (n == 0){
cout < 3.http://acm.hunnu.edu.cn/online/?action=problem&type=show&id=10188
| A Simple Math Problem |
| Time Limit: 1000ms, Special Time Limit:2500ms, Memory Limit:32768KB |
| Total submit users: 4, Accepted users: 3 |
| Problem 10188 : No special judgement |
| Problem description |
| Lele now is thinking about a simple function f(x). If x < 10 f(x) = x. If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10); And ai(0<=i<=9) can only be 0 or 1 . Now, I will give a0 ~ a9 and two positive integers k and m ,and could you help Lele to caculate f(k)%m. |
| Input |
| The problem contains mutiple test cases.Please process to the end of file. In each case, there will be two lines. In the first line , there are two positive integers k and m. ( k<2*109 , m < 104 ) In the second line , there are ten integers represent a0 ~ a9. |
| Output |
| For each case, output f(k) % m in one line. |
| Sample Input |
10 9999 1 1 1 1 1 1 1 1 1 1 20 500 1 0 1 0 1 0 1 0 1 0 |
| Sample Output |
45 104 |
情形一直接套公式
#include
using namespace std;
typedef long long llt;
const int Cube_SIZE = 10;///矩阵大小
struct Cube{
llt mat[Cube_SIZE][Cube_SIZE];
};
//单位矩阵
Cube _UnitCube = {
1, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 1, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 1, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 1, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 1, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 1, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 1, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 1, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 1,
};
Cube tmp = {
1, 1, 0, 0, 0, 0, 0, 0, 0, 0,
1, 0, 1, 0, 0, 0, 0, 0, 0, 0,
1, 0, 0, 1, 0, 0, 0, 0, 0, 0,
1, 0, 0, 0, 1, 0, 0, 0, 0, 0,
1, 0, 0, 0, 0, 1, 0, 0, 0, 0,
1, 0, 0, 0, 0, 0, 1, 0, 0, 0,
1, 0, 0, 0, 0, 0, 0, 1, 0, 0,
1, 0, 0, 0, 0, 0, 0, 0, 1, 0,
1, 0, 0, 0, 0, 0, 0, 0, 0, 1,
1, 0, 0, 0, 0, 0, 0, 0, 0, 0,
};
///矩阵乘机
Cube _Multiply(Cube A,Cube B,llt mod){
Cube _tmpCube;
for (int i = 0;i < Cube_SIZE;++i)
for (int j = 0;j < Cube_SIZE;++j){
_tmpCube.mat[i][j] = 0;
for (int k = 0;k < Cube_SIZE;++k){
_tmpCube.mat[i][j] += A.mat[i][k]*B.mat[k][j];
_tmpCube.mat[i][j] %= mod;
}
}
return _tmpCube;
}
///矩阵快速幂
Cube power_Cube(Cube A,llt n,llt mod) //矩阵快速幂
{
Cube _tmpCube = _UnitCube;
while(n){
if(n & 1)
_tmpCube = _Multiply(_tmpCube, A ,mod);
A = _Multiply(A, A, mod);
n >>= 1;
}
return _tmpCube;
}
int main(){
Cube arr;
llt k,MOD;
while (cin>>k>>MOD){
for (int i = 0;i < 10;++i){
cin>>tmp.mat[i][0];
}
arr = power_Cube(tmp,k - 9LL,MOD);
llt ans = 0;
for (int i = 0;i < 10;++i){
ans += (10-i-1) * arr.mat[i][0];
ans %= MOD;
}
cout < 部分参考自：
http://wenku.baidu.com/link?url=b7uA6oQcOQmGmxIgHJq3vvE7NMWo1dzV9oTUCuEcqq1A5lO9L-76US3W8S7W4mfA_s5bMfKa5qDDSTTjq6n2lFPZDhMhba_iES8i_nbnxpe
http://luowei828.blog.163.com/blog/static/31031204201110133125750/