UVa 11149 Power of Matrix (矩阵快速幂，倍增法或构造矩阵)

目录

作者：@dwtfukgv
本文为作者原创，转载请注明出处：https://www.cnblogs.com/dwtfukgv/p/7244041.html

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题意：求A + A^2 + A^3 + ... + A^m。

析：主要是两种方式，第一种是倍增法，把A + A^2 + A^3 + ... + A^m，拆成两部分，一部分是（E + A^(m/2)）（A + A^2 + A^3 + ... + A^(m/2)），然后依次计算下去，就可以分解，logn的复杂度分解，注意要分奇偶。

另一种是直接构造矩阵，，然后就可以用辞阵快速幂计算了，注意要用分块矩阵的乘法。

代码如下：

倍增法：

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#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <sstream> #include <list> #define debug() puts("++++"); #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std;   typedef long long LL; typedef unsigned long long ULL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1e3 + 10; const int mod = 10; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline bool is_in(int r, int c) {     return r > 0 && r <= n && c > 0 && c <= m; }   struct Matrix{   int a[40][40];   int n;     friend Matrix operator + (const Matrix &lhs, const Matrix &rhs){     Matrix res;     res.n = lhs.n;     for(int i = 0; i < lhs.n; ++i)       for(int j = 0; j < lhs.n; ++j)         res.a[i][j] = (lhs.a[i][j] + rhs.a[i][j]) % mod;     return res;   }     friend Matrix operator * (const Matrix &lhs, const Matrix &rhs){     Matrix res;     res.n = lhs.n;     for(int i = 0; i < lhs.n; ++i)       for(int j = 0; j < lhs.n; ++j){         res.a[i][j] = 0;         for(int k = 0; k < lhs.n; ++k)           res.a[i][j] += lhs.a[i][k] * rhs.a[k][j];         res.a[i][j] %= mod;       }     return res;   } };   Matrix E;   Matrix fast_pow(Matrix a, int m){   Matrix res;   res.n = n;   memset(res.a, 0, sizeof res.a);   for(int i = 0; i < res.n; ++i)     res.a[i][i] = 1;   while(m){     if(m & 1)  res = res * a;     m >>= 1;     a = a * a;   }   return res; }   Matrix dfs(int m, Matrix x){   if(m == 1)  return x;   if(m == 0)  return E;   Matrix ans = (E + fast_pow(x, m/2)) * dfs(m/2, x);   if(m & 1)  ans = ans + fast_pow(x, m);   return ans; }   int main(){   while(scanf("%d %d", &n, &m) == 2 && n){     Matrix x;  x.n = n;     E.n = n;     memset(E.a, 0, sizeof E.a);     for(int i = 0; i < n; ++i)       E.a[i][i] = 1;     for(int i = 0; i < n; ++i)       for(int j = 0; j < n; ++j){         scanf("%d", &x.a[i][j]);         x.a[i][j] %= mod;       }       Matrix ans = dfs(m, x);     for(int i = 0; i < n; ++i)       for(int j = 0; j < n; ++j)         if(j + 1 == n)  printf("%d\n", ans.a[i][j]);         else printf("%d ", ans.a[i][j]);     printf("\n");   }   return 0; }

　　

构造法：

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#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <sstream> #include <list> #define debug() puts("++++"); #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std;   typedef long long LL; typedef unsigned long long ULL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1e3 + 10; const int mod = 10; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline bool is_in(int r, int c) {     return r > 0 && r <= n && c > 0 && c <= m; }   struct Node{   int a[80][80];     friend void add(const Node &lhs, const Node &rhs, Node &res, int x, int y, int l, int r){     for(int i = x; i < y; ++i)       for(int j = l; j < r; ++j)         res.a[i][j] = (lhs.a[i-x][j-l] + rhs.a[i-x][j-l]) % mod;   }     friend void solve(int x, int y, int l, int r, int p, int q, const Node &lhs, const Node &rhs, Node &res){     for(int i = x; i < y; ++i)       for(int j = l; j < r; ++j){         res.a[i-x][j-l] = 0;         for(int k = p; k < q; ++k)           res.a[i-x][j-l] += lhs.a[i][k] * rhs.a[k][j];       }   }     friend Node operator * (const Node &lhs, const Node &rhs){     Node res, x, y;     solve(0, n, 0, n, 0, n, lhs, rhs, x);     solve(0, n, 0, n, n, n+n, lhs, rhs, y);     add(x, y, res, 0, n, 0, n);       solve(0, n, n, n+n, 0, n, lhs, rhs, x);     solve(0, n, n, n+n, n, n+n, lhs, rhs, y);     add(x, y, res, 0, n, n, n+n);       solve(n, n+n, 0, n, 0, n, lhs, rhs, x);     solve(n, n+n, 0, n, n, n+n, lhs, rhs, y);     add(x, y, res, n, n+n, 0, n);       solve(n, n+n, n, n+n, 0, n, lhs, rhs, x);     solve(n, n+n, n, n+n, n, n+n, lhs, rhs, y);     add(x, y, res, n, n+n, n, n+n);       return res;   } };   Node fast_pow(Node a, int m){   Node res;   memset(res.a, 0, sizeof res.a);   for(int i = 0; i < n; ++i)     res.a[i][i] = res.a[i+n][i] = 1;     while(m){     if(m & 1)  res = res * a;     m >>= 1;     a = a * a;   }   return res; }   int main(){   while(scanf("%d %d", &n, &m) == 2 && n){     Node x, y;     memset(y.a, 0, sizeof y.a);     for(int i = 0; i < n; ++i)       for(int j = n; j < n+n; ++j){         scanf("%d", &y.a[i][j]);         y.a[i][j] %= mod;       }     for(int i = 0; i < n; ++i)       y.a[i][i] = 1;     for(int i = n; i < n + n; ++i)       for(int j = n; j < n + n; ++j)         y.a[i][j] = y.a[i-n][j];     memset(x.a, 0, sizeof x.a);     for(int i = n; i < n+n; ++i)       x.a[i][i-n] = 1;       Node ans = fast_pow(y, m);       if(m)  ans = ans * x;     for(int i = 0; i < n; ++i)       for(int j = 0; j < n; ++j)         if(j == n-1)  printf("%d\n", ans.a[i][j]);         else  printf("%d ", ans.a[i][j]);     printf("\n");   }   return 0; }