Matrix Power Series
Time Limit: 3000MS Memory Limit: 131072K
Total Submissions: 23165 Accepted: 9651
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

• 錯誤總結

今天才知道取模操作多了就會超時以及陣列開大了會RE。。。可怕，乘法裡先是取了兩次模就超時了，取一次 1800MS，嗯，又多了一種TLE和RE的姿勢，還需要努力啊。

• 其中等比數列的求和其實是二分求解的操作，當K為偶數時，F(A,k)=F(A,k/2)(A^k/2+E)，當k為奇數時，F(A,k)=F(A,k/2)(A^(k+1/2)+E)+A^（k+1/2）

#include <iostream>
#include <cmath>
#include <cstring>
#include <cstdio>
#include <algorithm>
#define maxn 1000
#define kuma 35

using namespace std;
typedef long long ll;
ll mod;
ll n;
struct matrix
{
    ll a[kuma][kuma];//陣列開大了居然會RE.....
};
matrix E;
matrix mul(matrix
 
 x,matrix y)
{
    matrix ans=E;
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<n;j++)
        {
            ans.a[i][j]=0;
            for(int k=0;k<n;k++)
            {
                ans.a[i][j]=(ans.a[i][j]+x.a[i][k]*y.a[k][j])%mod;//取模取多了居然會TLE....
            }
        }
    }
    return ans;
}
matrix add(matrix x,matrix y)
{
    matrix ans=E;
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<n;j++)
        {
            ans.a[i][j]=(x.a[i][j]+y.a[i][j])%mod;
        }
    }
    return ans;
}

matrix quickpow(matrix a,ll p)
{

    matrix ans;
    ans=E;
    matrix t;
    t=a;
    while(p!=0)
    {
        if(p&1)ans=mul(ans,t);
        p>>=1;
        t=mul(t,t);
    }
    return ans;
}
matrix dos(matrix A,int k)
{
    if(k==1)return A;
    if(k%2==0)
    {
        matrix t=dos(A,k/2);
        return mul(add(quickpow(A,k/2),E),t);
    }
    else
    {
        matrix t=dos(A,k/2);
        matrix q=quickpow(A,(k+1)/2);
        return add(q,mul(t,add(q,E)));
    }
}
void init()
{
    for(int i=0;i<n;i++)
        E.a[i][i]=1;
}
int main()
{

    matrix nico;
    int k;
    scanf("%I64d%I64d%I64d",&n,&k,&mod);
       init();
       for(int i=0;i<n;i++)
       {
           for(int j=0;j<n;j++)
           {
               scanf("%I64d",&nico.a[i][j]);
           }
       }
       matrix ans=dos(nico,k);
       for(int i=0;i<n;i++)
       {
           for(int j=0;j<n;j++)
           {
               printf("%I64d ",ans.a[i][j]);
           }
           printf("\n");
       }
    return 0;
}