poj 3233（矩阵快速幂+二分）

Matrix Power Series

Time Limit: 3000MS		Memory Limit: 131072K
Total Submissions: 19323		Accepted: 8152

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

Source
POJ Monthly--2007.06.03, Huang, Jinsong
a^1+a^2+a^3+a^4+a^5=(a^1+a^2+a^3+a^4)+a^5
(a^1+a^2+a^3+a^4)=(a^1+a^2+a^2(a^1+a^2))=(a^1+a^2)*(1+a^2)

(a^1+a^2)=a^1(1+a^1)

#include <algorithm>
#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
int n,k,m;

struct Matrix
{
int ma[30][30];
}p,q;

Matrix Multi(Matrix a,Matrix b)
{
Matrix res;
memset(res.ma,0,sizeof(res.ma));
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
for(int k=0;k<n;k++)
res.ma[i][j]=(res.ma[i][j]+a.ma[i][k]*b.ma[k][j])%m;
return res;
}

Matrix quick_pow(Matrix m,int k)
{
Matrix res=q;
while(k)
{
if(k&1)
res=Multi(res,m);
k>>=1;
m=Multi(m,m);
}
return res;
}

Matrix sum(Matrix a,Matrix b)
{
for(int i=0;i<n;i++)
for(int k=0;k<n;k++)
a.ma[i][k]=(a.ma[i][k]+b.ma[i][k])%m;
return a;
}

Matrix recursion(Matrix m,int k)
{
if(k==1)
return m;
else if(k&1)
return sum(recursion(m,k-1),quick_pow(m,k));
else
return Multi(recursion(m,k/2),sum(q,quick_pow(m,k/2)));
}

int main()
{
while(~scanf("%d%d%d",&n,&k,&m))
{
memset(q.ma,0,sizeof(q.ma));
for(int i=0;i<n;i++)
q.ma[i][i]=1;
for(int i=0;i<n;i++)
for(int k=0;k<n;k++)
scanf("%d",&p.ma[i][k]);
Matrix res;
res=recursion(p,k);
for(int i=0;i<n;i++)
{
for(int k=0;k<n-1;k++)
printf("%d ",res.ma[i][k]);
printf("%d\n",res.ma[i][n-1]);
}
}
return 0;
}