题目链接：
http://acm.hust.edu.cn/vjudge/contest/122094#problem/G
Power of Matrix
Time Limit:3000MSMemory Limit:0KB
#### 问题描述
> 给你一个矩阵A，求A+A^2+A^3+...+A^k
#### 输入
> Input consists of no more than 20 test cases. The first line for each case contains two positive integers n
> (≤ 40) and k (≤ 1000000). This is followed by n lines, each containing n non-negative integers, giving
> the matrix A.
> Input is terminated by a case where n = 0. This case need NOT be processed.
输出
For each case, your program should compute the matrix A + A2 + A3 + . . . + Ak
. Since the values may
be very large, you only need to print their last digit. Print a blank line after each case.
样例
sample input
3 2
0 2 0
0 0 2
0 0 0
0 0sample output
0 2 4
0 0 2
0 0 0
题解
A+A2+...Ak=(I+A(n/2))(A+...+A(n/2))=...
一直递推下去，深度只有logn，只要logn次快速幂求A^x。
代码
WA四次的代码：
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
const int maxn = 55;
const int mod = 10;
typedef int LL;
struct Matrix {
LL mat[maxn][maxn];
Matrix() { memset(mat, 0, sizeof(mat)); }
friend Matrix operator *(const Matrix& A, const Matrix& B);
friend Matrix operator +(const Matrix &A,const Matrix &B);
friend Matrix operator ^(Matrix A, int n);
};
Matrix I;
Matrix operator +(const Matrix& A, const Matrix& B) {
Matrix ret;
for (int i = 0; i < maxn; i++) {
for (int j = 0; j < maxn; j++) {
ret.mat[i][j] = (A.mat[i][j] + B.mat[i][j])%mod;
}
}
return ret;
}
Matrix operator *(const Matrix& A, const Matrix& B) {
Matrix ret;
for (int i = 0; i < maxn; i++) {
for (int j = 0; j < maxn; j++) {
for (int k = 0; k < maxn; k++) {
ret.mat[i][j] = (ret.mat[i][j]+A.mat[i][k] * B.mat[k][j]) % mod;
}
}
}
return ret;
}
Matrix operator ^(Matrix A, int n) {
Matrix ret=I;
while (n) {
if (n & 1) ret = ret*A;
A = A*A;
n /= 2;
}
return ret;
}
Matrix solve(Matrix A, int n) {
if (!n) return I;
if (n == 1) return A;
//这里要加括号！！！ ret=I+(A^(n/2))!!!
Matrix ret = I + A ^ (n / 2);
ret = ret*solve(A, n / 2);
//这里也要加！！！！ ret=ret+(A^n)!!!
if (n % 2) ret = ret + A^n;
return ret;
}
int n, k;
int main() {
for (int i = 0; i < maxn; i++) I.mat[i][i] = 1;
while (scanf("%d%d", &n, &k) == 2 && n) {
Matrix A;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
scanf("%d", &A.mat[i][j]);
A.mat[i][j] %= mod;
}
}
Matrix ans=solve(A, k);
for (int i = 0; i < n; i++) {
for (int j = 0; j < n-1; j++) {
printf("%d ",ans.mat[i][j]);
}
printf("%d\n", ans.mat[i][n - 1]);
}
printf("\n");
}
return 0;
}
保险一些的写法：
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
const int maxn = 55;
const int mod = 10;
typedef int LL;
struct Matrix {
LL mat[maxn][maxn];
Matrix() { memset(mat, 0, sizeof(mat)); }
friend Matrix operator *(const Matrix& A, const Matrix& B);
friend Matrix operator +(const Matrix &A,const Matrix &B);
friend Matrix pow(Matrix A, int n);
};
Matrix I;
Matrix operator +(const Matrix& A, const Matrix& B) {
Matrix ret;
for (int i = 0; i < maxn; i++) {
for (int j = 0; j < maxn; j++) {
ret.mat[i][j] = (A.mat[i][j] + B.mat[i][j])%mod;
}
}
return ret;
}
Matrix operator *(const Matrix& A, const Matrix& B) {
Matrix ret;
for (int i = 0; i < maxn; i++) {
for (int j = 0; j < maxn; j++) {
for (int k = 0; k < maxn; k++) {
ret.mat[i][j] = (ret.mat[i][j]+A.mat[i][k] * B.mat[k][j]) % mod;
}
}
}
return ret;
}
Matrix pow(Matrix A, int n) {
Matrix ret=I;
while (n) {
if (n & 1) ret = ret*A;
A = A*A;
n /= 2;
}
return ret;
}
Matrix solve(Matrix A, int n) {
if (!n) return I;
if (n == 1) return A;
Matrix ret = I + pow(A,n/2);
ret = ret*solve(A, n / 2);
if (n % 2) ret = ret + pow(A,n);
return ret;
}
int n, k;
int main() {
for (int i = 0; i < maxn; i++) I.mat[i][i] = 1;
while (scanf("%d%d", &n, &k) == 2 && n) {
Matrix A;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
scanf("%d", &A.mat[i][j]);
A.mat[i][j] %= mod;
}
}
Matrix ans=solve(A, k);
for (int i = 0; i < n; i++) {
for (int j = 0; j < n-1; j++) {
printf("%d ",ans.mat[i][j]);
}
printf("%d\n", ans.mat[i][n - 1]);
}
printf("\n");
}
return 0;
}