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：
#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：
#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;
}