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通俗的说，就是矩阵的乘方。
封装类
核心代码
class CMat { public: // 矩阵乘法 static vector<vector<long long>> multiply(const vector<vector<long long>>& a, const vector<vector<long long>>& b) { const int r = a.size(), c = b.front().size(),iK = a.front().size(); assert(iK == b.size()); vector<vector<long long>> ret(r, vector<long long>(c)); for (int i = 0; i < r; i++) { for (int j = 0; j < c ; j++) { for (int k = 0; k < iK; k++) { ret[i][j] = (ret[i][j] + a[i][k] * b[k][j] ) % s_llMod; } } } return ret; } // 矩阵快速幂 static vector<vector<long long>> pow( const vector<vector<long long>>& a, vector<vector<long long>> b, long long n) { vector<vector<long long>> res = a; for (; n; n /= 2) { if (n % 2) { res = multiply(res, b); } b = multiply(b, b); } return res; } static vector<vector<long long>> TotalRow(const vector<vector<long long>>& a) { vector<vector<long long>> b(a.front().size(), vector<long long>(1, 1)); return multiply(a, b); } protected: const static long long s_llMod = 1e9 + 7; };
测试用例
template<class T> void Assert(const T& t1, const T& t2) { assert(t1 == t2); } template<class T> void Assert(const vector<T>& v1, const vector<T>& v2) { if (v1.size() != v2.size()) { assert(false); return; } for (int i = 0; i < v1.size(); i++) { Assert(v1[i], v2[i]); } } int main() { vector<vector<long long>> pre = { {1,2} }; vector<vector<long long>> mat = { {2,3},{1,10} }; { auto res = CMat::pow(pre, mat, 0); Assert(pre, res); } { auto res = CMat::multiply(pre, mat); Assert(vector<vector<long long>>{ {4, 23}}, res); auto res2 = CMat::pow(pre, mat,1); Assert(res2, res); } { auto res = CMat::pow(pre, mat, 2); auto res1 = CMat::multiply(pre, mat); auto res2 = CMat::multiply(res1, mat); Assert(res2, res); Assert(vector<vector<long long>>{ {31, 242}}, res); }; for (int i = 3; i < 100; i++) { auto res = pre; for (int j = 0; j < i; j++) { res = CMat::multiply(res, mat); } auto res2 = CMat::pow(pre, mat, i); Assert(res2, res); } }
具体例子
题目、分析和原理见：
【动态规划】【矩阵快速幂】【滚动向量】C++算法552. 学生出勤记录 II
原解法用二维表示状态，改成一维。 i是缺勤数量,j是连续迟到数，新的状态为：3*i+j
6种状态，故转移矩阵为6行6列，故结果矩阵为6列，6个数据1行就足够了。
令旧结果矩阵为mat1，转移矩阵为mat2，新矩阵为mat3，K mat1的列数，mat2的行数。则：
mat3[r][c] = Sum[ 0 , k ) i ^{i}_{[0,k)}[0,k)i(mat1[r][i]*mat2[i][c])
i在mat1中列号，在mat2中是行号。 也就是旧状态在第几列，mat2就在第几行。
新状态就是mat2的行号。
class Solution { public: int checkRecord(int n) { vector<vector<long long>> pre(1, vector<long long>(6));//1行6列 pre[0][0] = 1; vector<vector<long long>> mat(6, vector<long long>(6)); { //之前的状态在pre是第几列，矩阵中就是第几行。新状态的列号就矩阵的列号 //处理一次缺勤 ，缺勤两次排除 for (int i = 0; i < 3; i++) { mat[i][3]++; } //处理请假 for (int i = 0; i < 2; i++) { for (int j = 0; j < 2; j++) { const int pre = 3 * i + j; mat[pre][pre + 1]++; } } //处理正常 for (int i = 0; i < 2; i++) { for (int j = 0; j < 3; j++) { const int pre = 3 * i + j; const int cur = 3 * i; mat[pre][cur]++; } } } auto res = CMat::pow(pre, mat, n); res = CMat::TotalRow(res); return res[0][0]; } };
测试用例
template
void Assert(const T& t1, const T& t2)
{
assert(t1 == t2);
}
template
void Assert(const vector& v1, const vector& v2)
{
if (v1.size() != v2.size())
{
assert(false);
return;
}
for (int i = 0; i < v1.size(); i++)
{
Assert(v1[i], v2[i]);
}
}
int main()
{
int n;
{
Solution sln;
n = 0;
auto res = sln.checkRecord(n);
Assert(1, res);
}
{
Solution sln;
n = 1;
auto res = sln.checkRecord(n);
Assert(3, res);
}
{
Solution sln;
n = 2;
auto res = sln.checkRecord(n);
Assert(8, res);
}
{
Solution sln;
n = 3;
auto res = sln.checkRecord(n);
Assert(19, res);
}
{
Solution sln;
n = 4;
auto res = sln.checkRecord(n);
Assert(43, res);
}
{
Solution sln;
n = 5;
auto res = sln.checkRecord(n);
Assert(94, res);
}
{
Solution sln;
n = 6;
auto res = sln.checkRecord(n);
Assert(200, res);
}
{
Solution sln;
n = 7;
auto res = sln.checkRecord(n);
Assert(418, res);
}
{
Solution sln;
n = 10101;
auto res = sln.checkRecord(n);
Assert(183236316, res);
}
}
