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大连网站代运营的公司有哪些,seo课程培训课程,专业做网站较好的公司,网站宣传有文化事业建设费吗本文涉及知识点【矩阵快速幂】封装类及测试用例及样例 P10502 Matrix Power Series 题目描述给定一个 n n nn nn 矩阵 A A A 和一个正整数 k k k，找出和 S A A 2 A 3 . . . A k SAA^2 A^3 ...A^k SAA2A3...Ak。输入格式输入包含一个测试用例。输…

本文涉及知识点

【矩阵快速幂】封装类及测试用例及样例

P10502 Matrix Power Series

题目描述

给定一个 $n \times n$ 矩阵 $A$ 和一个正整数 $k$ ，找出和 $S=A+A^2 +A^3 +...+A^k$ 。

输入格式

输入包含一个测试用例。输入的第一行包含三个正整数 $n$ （ $\le 30$ ）、 $k$ （ $\le 10^9$ ）和 $m$ （ $m < 10^4$ ）。接下来的 $n$ 行每行包含 $n$ 个小于 32,768 的非负整数，按行主序给出 $A$ 的元素。

输出格式

以与给定 $A$ 相同的方式输出 $S$ 的元素对 $m$ 取模。

翻译来自于：ChatGPT

输入输出样例 #1

输入 #1

2 2 4 
0 1 
1 1

输出 #1

1 2
2 3

矩阵快速幂

扩容为2n2n矩阵matn，分成4个矩阵：左上角 $A_k$ ,右上 $S_k$ ，左下角全0，右下单位矩阵。
matmat^k-1的右上角就是答案。

核心代码

#include <iostream>
#include <sstream>
#include <vector>
#include<map>
#include<unordered_map>
#include<set>
#include<unordered_set>
#include<string>
#include<algorithm>
#include<functional>
#include<queue>
#include <stack>
#include<iomanip>
#include<numeric>
#include <math.h>
#include <climits>
#include<assert.h>
#include<cstring>
#include<list>
#include <array>#include <bitset>
using namespace std;template<class T1, class T2>
std::istream& operator >> (std::istream& in, pair<T1, T2>& pr) {in >> pr.first >> pr.second;return in;
}template<class T1, class T2, class T3 >
std::istream& operator >> (std::istream& in, tuple<T1, T2, T3>& t) {in >> get<0>(t) >> get<1>(t) >> get<2>(t);return in;
}template<class T1, class T2, class T3, class T4 >
std::istream& operator >> (std::istream& in, tuple<T1, T2, T3, T4>& t) {in >> get<0>(t) >> get<1>(t) >> get<2>(t) >> get<3>(t);return in;
}template<class T = int>
vector<T> Read() {int n;scanf("%d", &n);vector<T> ret(n);for (int i = 0; i < n; i++) {cin >> ret[i];}return ret;
}template<class T = int>
vector<T> Read(int n) {vector<T> ret(n);for (int i = 0; i < n; i++) {cin >> ret[i];}return ret;
}template<int N = 1'000'000>
class COutBuff
{
public:COutBuff() {m_p = puffer;}template<class T>void write(T x) {int num[28], sp = 0;if (x < 0)*m_p++ = '-', x = -x;if (!x)*m_p++ = 48;while (x)num[++sp] = x % 10, x /= 10;while (sp)*m_p++ = num[sp--] + 48;AuotToFile();}void writestr(const char* sz) {strcpy(m_p, sz);m_p += strlen(sz);AuotToFile();}inline void write(char ch){*m_p++ = ch;AuotToFile();}inline void ToFile() {fwrite(puffer, 1, m_p - puffer, stdout);m_p = puffer;}~COutBuff() {ToFile();}
private:inline void AuotToFile() {if (m_p - puffer > N - 100) {ToFile();}}char  puffer[N], * m_p;
};template<int N = 1'000'000>
class CInBuff
{
public:inline CInBuff() {}inline CInBuff<N>& operator>>(char& ch) {FileToBuf();ch = *S++;return *this;}inline CInBuff<N>& operator>>(int& val) {FileToBuf();int x(0), f(0);while (!isdigit(*S))f |= (*S++ == '-');while (isdigit(*S))x = (x << 1) + (x << 3) + (*S++ ^ 48);val = f ? -x : x; S++;//忽略空格换行		return *this;}inline CInBuff& operator>>(long long& val) {FileToBuf();long long x(0); int f(0);while (!isdigit(*S))f |= (*S++ == '-');while (isdigit(*S))x = (x << 1) + (x << 3) + (*S++ ^ 48);val = f ? -x : x; S++;//忽略空格换行return *this;}template<class T1, class T2>inline CInBuff& operator>>(pair<T1, T2>& val) {*this >> val.first >> val.second;return *this;}template<class T1, class T2, class T3>inline CInBuff& operator>>(tuple<T1, T2, T3>& val) {*this >> get<0>(val) >> get<1>(val) >> get<2>(val);return *this;}template<class T1, class T2, class T3, class T4>inline CInBuff& operator>>(tuple<T1, T2, T3, T4>& val) {*this >> get<0>(val) >> get<1>(val) >> get<2>(val) >> get<3>(val);return *this;}template<class T = int>inline CInBuff& operator>>(vector<T>& val) {int n;*this >> n;val.resize(n);for (int i = 0; i < n; i++) {*this >> val[i];}return *this;}template<class T = int>vector<T> Read(int n) {vector<T> ret(n);for (int i = 0; i < n; i++) {*this >> ret[i];}return ret;}template<class T = int>vector<T> Read() {vector<T> ret;*this >> ret;return ret;}
private:inline void FileToBuf() {const int canRead = m_iWritePos - (S - buffer);if (canRead >= 100) { return; }if (m_bFinish) { return; }for (int i = 0; i < canRead; i++){buffer[i] = S[i];//memcpy出错			}m_iWritePos = canRead;buffer[m_iWritePos] = 0;S = buffer;int readCnt = fread(buffer + m_iWritePos, 1, N - m_iWritePos, stdin);if (readCnt <= 0) { m_bFinish = true; return; }m_iWritePos += readCnt;buffer[m_iWritePos] = 0;S = buffer;}int m_iWritePos = 0; bool m_bFinish = false;char buffer[N + 10], * S = buffer;
};template<class T = long long>
class CMatMul
{
public:CMatMul(T llMod = 1e9 + 7) :m_llMod(llMod) {}// 矩阵乘法vector<vector<T>> multiply(const vector<vector<T>>& a, const vector<vector<T>>& b) {const int r = a.size(), c = b.front().size(), iK = a.front().size();assert(iK == b.size());vector<vector<T>> ret(r, vector<T>(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]) % m_llMod;}}}return ret;}// 矩阵快速幂vector<vector<T>> pow(const vector<vector<T>>& a, vector<vector<T>> b, T n) {vector<vector<T>> res = a;for (; n; n /= 2) {if (n % 2) {res = multiply(res, b);}b = multiply(b, b);}return res;}vector<vector<T>> pow(vector<vector<T>> pre, vector<vector<T>> mat, const string& str){for (int i = str.length() - 1; i >= 0; i--) {const int t = str[i] - '0';pre = pow(pre, mat, t);mat = pow(mat, mat, 9);}return pre;}vector<vector<T>> TotalRow(const vector<vector<T>>& a){vector<vector<T>> b(a.front().size(), vector<T>(1, 1));return multiply(a, b);}vector<vector<T>> CreateRow(int C) {return vector<vector<T>>(1, vector<T>(C));}vector<vector<T>> CreateUint(int RC) {vector<vector<T>> ret(RC, vector<T>(RC));for (int i = 0; i < RC; i++) { ret[i][i] = 1; }return ret;}
protected:const  T m_llMod;
};class KMPEx
{
public:static vector<int> ZFunction(string s) {int n = (int)s.length();vector<int> z(n);z[0] = n;for (int i = 1, left = 0, r = 0; i < n; ++i) {if (i <= r) {//如果此if,r-i+1可能为负数z[i] = min(z[i - left], r - i + 1);}while ((i + z[i] < n) && (s[z[i]] == s[i + z[i]])) {z[i]++;}if (i + z[i] - 1 > r) left = i, r = i + z[i] - 1;}return z;//z[i] 表示S与其后缀S[i,n]的最长公共前缀(LCP)的长度}
};class Solution {
public:vector<vector<int>> Ans(const vector<vector<int>>& A, int k, int MOD) {const int N = A.size();CMatMul<> matMul(MOD);vector<vector<long long>> pre(N, vector<long long>(N * 2));vector<vector<long long>> mat(N * 2, vector<long long>(N * 2));for (int r = 0; r < N; r++) {mat[r + N][r + N] = 1;for (int c = 0; c < N; c++) {pre[r][c] = pre[r][c + N] = A[r][c];mat[r][c] = mat[r][c + N] = A[r][c];}}auto matAns = matMul.pow(pre, mat, k - 1);vector<vector<int>> ans(N, vector<int>(N));for (int r = 0; r < N; r++) {for (int c = 0; c < N; c++) {ans[r][c] = matAns[r][c + N];}}return ans;}
};int main() {
#ifdef _DEBUGfreopen("a.in", "r", stdin);
#endif // DEBUG	ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);int n,k,m;cin >> n >> k >>m ;vector<vector<int>> A(n);for (int i = 0; i < n; i++) {A[i] = Read<int>(n);}auto res = Solution().Ans(A,k,m);for (const auto& v : res) {for (const auto& i : v) {cout << i << " ";}cout <<  "\n";}#ifdef _DEBUG		//printf("start=%d,end=%d,T=%d", start,end,T);//Out(edge, "edge=");//Out(fish, ",fish=");/*Out(edge, "edge=");Out(que, "que=");*/
#endif // DEBUG		return 0;
}

单元测试

TEST_METHOD(TestMethod1){auto res = Solution().Ans({ {0,1},{1,1} },1, 4);AssertV(vector<vector<int>>{ {0,1},{1,1} }, res);}TEST_METHOD(TestMethod2){auto res = Solution().Ans({ {0,1},{1,1} }, 2, 4);AssertV(vector<vector<int>>{ {1, 2}, { 2,3 } }, res);}

扩展阅读

我想对大家说的话
工作中遇到的问题，可以按类别查阅鄙人的算法文章，请点击《算法与数据汇总》。
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闻缺陷则喜(喜缺)是一个美好的愿望，早发现问题，早修改问题，给老板节约钱。
子墨子言之：事无终始，无务多业。也就是我们常说的专业的人做专业的事。
如果程序是一条龙，那算法就是他的是睛
失败+反思=成功成功+反思=成功