Poj 3233 矩阵快速幂，暑假训练专题中的某一道题目，矩阵快速幂的模板

题目链接 请猛戳~

Description

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

结题思路：

当k为偶数时，s(k)=(1+A^(k/2))*(A+A^2+…+A^(k/2))

当k为奇数时，s(k)=A+(A+A^(k/2+1))*(A+A^2+…+A^(k/2))加以递归求解

#include <iostream>
#include <string.h>
#include <cstdio>
using namespace std;
const int MAX = 32;
struct Matrix{
int v[MAX][MAX];
};
Matrix E;//定义单位矩阵 E
int n,k,M;
Matrix add(Matrix a,Matrix b){//矩阵加法运算
Matrix c;
for(int i = 1;i <= n;i++)
for(int j = 1;j <= n;j++){
c.v[i][j] = (a.v[i][j] + b.v[i][j]) % M;
}
return c;
}
Matrix mul(Matrix a,Matrix b){//矩阵乘法运算
Matrix c;
for(int i = 1;i <= n;i++)
for(int j = 1;j <= n;j++){
c.v[i][j] = 0;
for(int k = 1;k <= n;k++)
c.v[i][j] = (a.v[i][k] * b.v[k][j] + c.v[i][j]) % M;
}
return c;
}
Matrix pow(Matrix a,int k){//矩阵幂运算
if(k == 0){
memset(a.v,0,sizeof(a.v));
for(int i = 1;i <= n;i++)
a.v[i][i] = 1;
return a;
}else if(k == 1) return a;
Matrix c = pow(a,k / 2);
if(k & 1){
return mul(a,mul(c,c));
}else
return mul(c,c);
}
Matrix sum(Matrix a,int k){ //连续升幂求和
if(k == 1)return a;
Matrix b = pow(a,(k + 1) / 2);
Matrix c = sum(a,k / 2);
if(k % 2){
return add(a,mul(add(a,b),c));
}else
return mul(add(E,b),c);
}
void initE(){ //初始化单位矩阵E
memset(E.v,0,sizeof(E.v));
for(int i = 1;i <= n;i++) E.v[i][i] = 1;
}
void print(Matrix a){
for(int i = 1;i <= n;i++){
for(int j = 1;j <= n;j++)
printf("%d ",a.v[i][j]);
puts("");
}
}
Matrix get(){
Matrix a;
for(int i = 1;i <= n;i++)
for(int j = 1;j <= n;j++)scanf(" %d",&a.v[i][j]);
return a;
}
int main()
{

while(~scanf("%d%d%d",&n,&k,&M)){
initE();
Matrix a = get();
Matrix b = sum(a,k);
print(b);
}
return 0;
}