Matrix Power Series

Description

Time Limit: 3000MS Memory Limit: 131072K

Given a n∗n matrix A and a positive integer k, find the sum S=A+A2+A3+…+Ak.

Input

The input contains exactly one test case. The first line of input contains three positive integers n(n≤30), k(k≤109) and m(m<104). Then follow n lines each containing n nonnegative integers below 32,768, giving A’s elements in row-major order.

Output

Output the elements of S modulo m in the same way as A is given.

Sample Input

2 2 4

0 1

1 1

Sample Output

1 2

2 3

Solution & Code

法一：k的范围是109，考虑分治。

Si=A+A2+...+Ai

S2i=Ai+1+Ai+2+...+A2i

S2i=Si∗Ai

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

typedef long long ll;

const int maxr = 35;

int n, k;
ll m;

struct matrix{ll val[maxr][maxr];};
matrix U, V, Z, I;

matrix multiply(matrix A, matrix B){

matrix C = Z;
for(int i = 1; i <= n; ++i)
for(int j = 1; j <= n; ++j)
for(int k = 1; k <= n; ++k){
C.val[i][j] += A.val[i][k] * B.val[k][j];
C.val[i][j] %= m;
}
return C;
}

matrix add(matrix A, matrix B){

matrix C = Z;
for(int i = 1; i <= n; ++i)
for(int j = 1; j <= n; ++j){
C.val[i][j] = A.val[i][j] + B.val[i][j];
C.val[i][j] %= m;
}
return C;
}

matrix powmod(matrix A, int x){

matrix B = I;
while(x){
if(x & 1) B = multiply(B, A);
x = x >> 1;
A = multiply(A, A);
}
return B;
}

matrix calc(int x){

if(x == 1) return U;
int y = x >> 1;
matrix L = calc(y);
matrix R = multiply(L, powmod(U, y));
if(y * 2 != x) return add(add(L, R), powmod(U, x));
else return add(L, R);
}

int main(){

cin >> n >> k >> m;
for(int i = 1; i <= n; ++i){
for(int j = 1; j <= n; ++j) cin >> U.val[i][j];
I.val[i][i] = 1;
}
V = calc(k);
for(int i = 1; i <= n; ++i){
for(int j = 1; j <= n; ++j) printf("%lld ", V.val[i][j]);
puts("");
}

return 0;
}

法二：如果求s=a+a2+...ak我们可以考虑用矩阵快速幂对其进行加速，求S=A+A2+A3+…+Ak时则可以构造一个元素为矩阵的矩阵，然后用矩阵快速幂加速。