【poj3233】Matrix Power Series 矩阵+快速幂

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

题意：给你一个n*n的矩阵A，求

∑k=1nAｋ

其中每个元素mod m。

二分，可以像快速幂一样递归处理。

先把A想象成数字，则：

A1+A2+A3+A4+A5+A6

=(A1+A2+A3)+A3∗(A1+A2+A3)

=(1+A3)∗(A1+A2+A3)

对于矩阵，重载了+和*，1为单位矩阵，就一样做了。

代码（不知道为何，很慢）：

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

int n,mod;

struct matrix{
int num[32][32];
void init()
{
memset(num,0,sizeof(num));
for(int i = 1;i <= n;i ++) num[i][i] = 1;
}
};

matrix operator *(const matrix &a,const matrix &b)
{
matrix ans;
for(int i = 1;i <= n;i ++)
{
for(int j = 1;j <= n;j ++)
{
ans.num[i][j] = 0;
for(int k = 1;k <= n;k ++)
{
ans.num[i][j] += a.num[i][k] * b.num[k][j];
ans.num[i][j] %= mod;
}
}
}
return ans;
}

matrix operator +(const matrix &a,const matrix &b)
{
matrix ans;
for(int i = 1;i <= n;i ++)
{
for(int j = 1;j <= n;j ++)
{
ans.num[i][j] = (a.num[i][j] + b.num[i][j]) % mod;
}
}
return ans;
}

matrix ksm(matrix a,int b)
{
matrix ans;
ans.init();
while(b)
{
if(b & 1) ans = ans * a;
a = a * a;
b >>= 1;
}
return ans;
}

void print(const matrix &ans)
{
for(int i = 1;i <= n;i ++)
{
for(int j = 1;j <= n;j ++)
{
printf("%d ",ans.num[i][j]);
}
puts("");
}
}

matrix ask(const matrix &a,int b)
{
if(b == 1) return a;
matrix ans;
ans.init();
ans = (ans + ksm(a,b >> 1)) * ask(a,b >> 1);
//  print(ans);
if(b & 1)
return ans + ksm(a,b);
else
return ans;
}

int main()
{
matrix ans;
int k;
while(~scanf("%d%d%d",&n,&k,&mod))
{
for(int i = 1;i <= n;i ++)
{
for(int j = 1;j <= n;j ++)
{
scanf("%d",&ans.num[i][j]);
}
}
ans = ask(ans,k);
print(ans);
}
return 0;
}