【NOIP2016提高A组模拟9.17】数格子

题目

分析

设表示每一行的状态，用一个4位的二进制来表示，当前这一行中的每一个位数对下一位有没有影响。

设fi,s表示，做完了的i行，其状态为s，的方案数。

两个状态之间是否可以转移就留给读者自己思考了。

答案就是fn,0因为最后一行对下一行不能造成影响。

然而，这样只有60分。

100分是个矩阵快速幂，

B矩阵构造很简单，当两个状态s、s′可以转移，那么，B矩阵gs,s′=1。

当i等于零时， A矩阵为{1, 0 <repeats 15 times>}

#include <cmath>
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <queue>
const long long maxlongint=2147483647;
const long long N=500005;
using namespace std;
long long m,n,e[16][16]=
{
{1,0,0,1,0,0,0,0,0,1,0,0,1,0,0,1},
{0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0},
{0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0},
{1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0},
{0,0,0,0,0,0,0,0,1,0,0,1,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0},
{0,1,0,0,1,0,0,1,0,0,0,0,0,0,0,0},
{1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0},
{1,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0}
};
long long r[16]={1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},f[16],g[16][16];
long long time()
{
long long f1[16];
for(int i=0;i<=15;i++)
f1[i]=f[i];
for(int i=0;i<=15;i++)
{
f[i]=0;
for(int j=0;j<=15;j++)
f[i]=(f[i]+f1[j]*g[j][i]%m)%m;
}
}
long long time1()
{
long long f1[16][16];
for(int i=0;i<=15;i++)
for(int j=0;j<=15;j++)
f1[i][j]=g[i][j];
for(int i=0;i<=15;i++)
for(int j=0;j<=15;j++)
{
g[i][j]=0;
for(int k=0;k<=15;k++)
{
g[i][j]=(g[i][j]+f1[i][k]*f1[k][j]%m)%m;
}
}
}
long long mi(long long x)
{
while(x)
{
if(x&1) time();
time1();
x/=2;
}
}
int main()
{
while(1)
{
for(int i=0;i<=15;i++)
f[i]=r[i];
for(int i=0;i<=15;i++)
for(int j=0;j<=15;j++)
g[i][j]=e[i][j];
scanf("%lld%lld",&n,&m);
if(n==0 && m==0) break;
mi(n);
cout<<f[0]<<endl;
}
}