【POJ 3420】Quad Tiling（dp|递推 +矩阵快速幂）

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Quad Tiling

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Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 4122 Accepted: 1885

Description
Tired of the Tri Tiling game finally, Michael turns to a more challengeable game, Quad Tiling:
In how many ways can you tile a 4 × N (1 ≤ N ≤ 109) rectangle with 2 × 1 dominoes? For the answer would be very big, output the answer modulo M (0 < M ≤ 105).

Input
Input consists of several test cases followed by a line containing double 0. Each test case consists of two integers, N and M, respectively.

Output
For each test case, output the answer modules M.

Sample Input
1 10000
3 10000
5 10000
0 0

Sample Output
1
11
95

Source
POJ Monthly–2007.10.06, Dagger

【题解】【递推+矩阵快速幂】
【刚开始，推式子ing….一段漫长而依赖std的时光】
【result : f[i]=f[i-1]+5*f[i-2]+f[i-3]-f[i-4]】
【具体这个式子是如何得来的？呵呵，我可不可以说我是凑的？！（我刚开始用f[1]、f[2]凑f[3]，然后发现f[4]不满足，failed! 然后，用f[1]、f[2]、f[3]凑f[4]，然后，f[5]不满足，failed！然后，我用前四个凑f[5],往后试了几个，发现accept!）】
【好了，朴素的算法已经得出，那么，构造矩阵即可】
初始矩阵为： {f[4],f[3],f[2],f[1]} => {36,11,5,1}
转移矩阵为：

⎡⎣⎢⎢⎢1，1，0，05，0，1，01，0，0，1−1，0，0，0⎤⎦⎥⎥⎥

【注意：转移矩阵在进行快速幂时，需要%m,但由于有负数需要+m%m】

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct node{
    long long k[5][5];
}d;
int f[5],n,m,ans;
inline node jc(node a,node b)
{
    node c;
    memset(c.k,0,sizeof(c.k));
    for(int i=1;i<=4;++i)
     for(int j=1;j<=4;++j)
      for(int l=1;l<=4;++l)
       {
          long long sm=a.k[i][l]*b.k[l][j];
          sm=((sm+m)%m+m)%m;
          c.k[i][j]=(c.k[i][j]+sm)%m;
       }
    return c;
}
node poww(node x,int p)
{
    if(p==1) return d;
    if(p%2==0)
     {
        node a=poww(x,p/2);
        a=jc(a,a);
        return a;
     }
    else
     {
        node a=poww(x,p/2);
        a=jc(a,a);
        a=jc(a,x);
        return a;
     }
}
inline int solve(int p)
{
    node as=poww(d,p);
    int sum[5];
    memset(sum,0,sizeof(sum));
    for(int i=1;i<=4;++i)
     for(int j=1;j<=4;++j)
      sum[j]=(sum[j]+as.k[i][j]*f[i]%m)%m;
    return sum[1];
}
int main()
{
    //freopen("int.txt","r",stdin);
    //freopen("my.txt","w",stdout);
    f[1]=36; f[2]=11; f[3]=5; f[4]=1;
    d.k[1][1]=d.k[1][2]=d.k[2][3]=d.k[3][1]=d.k[3][4]=1;
    d.k[2][1]=5; d.k[4][1]=-1;
    while((scanf("%d%d",&n,&m)==2)&&!(n==0&&m==0))
     {
        if(n==1||n==2||n==3||n==4) {printf("%d\n",f[4-n+1]%m); continue;}
        ans=solve(n-4);
        printf("%d\n",ans);
     }
}