Problem Description
You might have noticed that there is the new fashion among rich people to have their yards tiled with black and white tiles, forming a pattern. The company Broken Tiles is well known as the best tiling company in our region. It provides the widest choices of nice patterns to tile your yard with. The pattern is nice if there is no square of size 2 × 2, such that all tiles in it have the same color. So patterns on the figure 1 are nice, while patterns on the figure 2 are not.
The president of the company wonders whether the variety of nice patterns he can provide to the clients is large enough. Thus he asks you to find out the number of nice patterns that can be used to tile the yard of size N × M . Now he is interested in the long term estimation, so he suggests N ≤ 10100. However, he does not like big numbers, so he asks you to find the answer modulo P .
Input
The input file contains three integer numbers: N (1 ≤ N ≤ 10100), M (1 ≤ M ≤ 5) and P (1 ≤ P ≤10000).
Output
Write the number of nice patterns of size N × M modulo P to the output file.
Sample Input
2 2 5
3 3 23
Sample Output
4
0
Source
Andrew Stankevich Contest 1
Manager
题意：给出一个n*m的矩阵(n <= 10^100, m <= 5)，对于2*2的子方格若全是黑色或全是白色的是非法的，用黑白两色去染n*m的方格，问共有多少种合法的染色方案。
思路：构造出转移矩阵，上一行向下一行的转移矩阵，因为m<=5，每行最多有32个状态，可以进行状态压缩构造出一个32*32的转移矩阵A，A[i][j] = 1表示上一行i状态可以向下一行的j状态转移，否则不能转移。要求最后的合法方案数，就再构造一个B矩阵，是一个32*1的矩阵，表示了到达第一行每一个状态的方案数。那么A*B就表示到达第二行每一个状态的方案数，以此类推，A^n-1 * B表示到达第n行每一个状态的合法方案数，那么所有状态对应方案数的和就是总的方案数。
代码如下：
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
using namespace std; #define DIGIT 4 //四位隔开,即万进制
#define DEPTH 10000 //万进制
#define MAX 100+5 //题目最大位数/4，要不大直接设为最大位数也行
typedef int bignum_t[MAX+]; /************************************************************************/
/* 读取操作数，对操作数进行处理存储在数组里 */
/************************************************************************/
int read(bignum_t a,istream&is=cin)
{
char buf[MAX*DIGIT+],ch ;
int i,j ;
memset((void*)a,,sizeof(bignum_t));
if(!(is>>buf))return ;
for(a[]=strlen(buf),i=a[]/-;i>=;i--)
ch=buf[i],buf[i]=buf[a[]--i],buf[a[]--i]=ch ;
for(a[]=(a[]+DIGIT-)/DIGIT,j=strlen(buf);j<a[]*DIGIT;buf[j++]='');
for(i=;i<=a[];i++)
for(a[i]=,j=;j<DIGIT;j++)
a[i]=a[i]*+buf[i*DIGIT--j]-'' ;
for(;!a[a[]]&&a[]>;a[]--);
return ;
} void write(const bignum_t a,ostream&os=cout)
{
int i,j ;
for(os<<a[i=a[]],i--;i;i--)
for(j=DEPTH/;j;j/=)
os<<a[i]/j% ;
} int comp(const bignum_t a,const bignum_t b)
{
int i ;
if(a[]!=b[])
return a[]-b[];
for(i=a[];i;i--)
if(a[i]!=b[i])
return a[i]-b[i];
return ;
} int comp(const bignum_t a,const int b)
{
int c[]=
{ }
;
for(c[]=b;c[c[]]>=DEPTH;c[c[]+]=c[c[]]/DEPTH,c[c[]]%=DEPTH,c[]++);
return comp(a,c);
} int comp(const bignum_t a,const int c,const int d,const bignum_t b)
{
int i,t=,O=-DEPTH* ;
if(b[]-a[]<d&&c)
return ;
for(i=b[];i>d;i--)
{
t=t*DEPTH+a[i-d]*c-b[i];
if(t>)return ;
if(t<O)return ;
}
for(i=d;i;i--)
{
t=t*DEPTH-b[i];
if(t>)return ;
if(t<O)return ;
}
return t> ;
}
/************************************************************************/
/* 大数与大数相加 */
/************************************************************************/
void add(bignum_t a,const bignum_t b)
{
int i ;
for(i=;i<=b[];i++)
if((a[i]+=b[i])>=DEPTH)
a[i]-=DEPTH,a[i+]++;
if(b[]>=a[])
a[]=b[];
else
for(;a[i]>=DEPTH&&i<a[];a[i]-=DEPTH,i++,a[i]++);
a[]+=(a[a[]+]>);
}
/************************************************************************/
/* 大数与小数相加 */
/************************************************************************/
void add(bignum_t a,const int b)
{
int i= ;
for(a[]+=b;a[i]>=DEPTH&&i<a[];a[i+]+=a[i]/DEPTH,a[i]%=DEPTH,i++);
for(;a[a[]]>=DEPTH;a[a[]+]=a[a[]]/DEPTH,a[a[]]%=DEPTH,a[]++);
}
/************************************************************************/
/* 大数相减(被减数>=减数) */
/************************************************************************/
void sub(bignum_t a,const bignum_t b)
{
int i ;
for(i=;i<=b[];i++)
if((a[i]-=b[i])<)
a[i+]--,a[i]+=DEPTH ;
for(;a[i]<;a[i]+=DEPTH,i++,a[i]--);
for(;!a[a[]]&&a[]>;a[]--);
}
/************************************************************************/
/* 大数减去小数(被减数>=减数) */
/************************************************************************/
void sub(bignum_t a,const int b)
{
int i= ;
for(a[]-=b;a[i]<;a[i+]+=(a[i]-DEPTH+)/DEPTH,a[i]-=(a[i]-DEPTH+)/DEPTH*DEPTH,i++);
for(;!a[a[]]&&a[]>;a[]--);
} void sub(bignum_t a,const bignum_t b,const int c,const int d)
{
int i,O=b[]+d ;
for(i=+d;i<=O;i++)
if((a[i]-=b[i-d]*c)<)
a[i+]+=(a[i]-DEPTH+)/DEPTH,a[i]-=(a[i]-DEPTH+)/DEPTH*DEPTH ;
for(;a[i]<;a[i+]+=(a[i]-DEPTH+)/DEPTH,a[i]-=(a[i]-DEPTH+)/DEPTH*DEPTH,i++);
for(;!a[a[]]&&a[]>;a[]--);
}
/************************************************************************/
/* 大数相乘，读入被乘数a，乘数b，结果保存在c[] */
/************************************************************************/
void mul(bignum_t c,const bignum_t a,const bignum_t b)
{
int i,j ;
memset((void*)c,,sizeof(bignum_t));
for(c[]=a[]+b[]-,i=;i<=a[];i++)
for(j=;j<=b[];j++)
if((c[i+j-]+=a[i]*b[j])>=DEPTH)
c[i+j]+=c[i+j-]/DEPTH,c[i+j-]%=DEPTH ;
for(c[]+=(c[c[]+]>);!c[c[]]&&c[]>;c[]--);
}
/************************************************************************/
/* 大数乘以小数，读入被乘数a，乘数b，结果保存在被乘数 */
/************************************************************************/
void mul(bignum_t a,const int b)
{
int i ;
for(a[]*=b,i=;i<=a[];i++)
{
a[i]*=b ;
if(a[i-]>=DEPTH)
a[i]+=a[i-]/DEPTH,a[i-]%=DEPTH ;
}
for(;a[a[]]>=DEPTH;a[a[]+]=a[a[]]/DEPTH,a[a[]]%=DEPTH,a[]++);
for(;!a[a[]]&&a[]>;a[]--);
} void mul(bignum_t b,const bignum_t a,const int c,const int d)
{
int i ;
memset((void*)b,,sizeof(bignum_t));
for(b[]=a[]+d,i=d+;i<=b[];i++)
if((b[i]+=a[i-d]*c)>=DEPTH)
b[i+]+=b[i]/DEPTH,b[i]%=DEPTH ;
for(;b[b[]+];b[]++,b[b[]+]=b[b[]]/DEPTH,b[b[]]%=DEPTH);
for(;!b[b[]]&&b[]>;b[]--);
}
/**************************************************************************/
/* 大数相除,读入被除数a，除数b，结果保存在c[]数组 */
/* 需要comp()函数 */
/**************************************************************************/
void div(bignum_t c,bignum_t a,const bignum_t b)
{
int h,l,m,i ;
memset((void*)c,,sizeof(bignum_t));
c[]=(b[]<a[]+)?(a[]-b[]+): ;
for(i=c[];i;sub(a,b,c[i]=m,i-),i--)
for(h=DEPTH-,l=,m=(h+l+)>>;h>l;m=(h+l+)>>)
if(comp(b,m,i-,a))h=m- ;
else l=m ;
for(;!c[c[]]&&c[]>;c[]--);
c[]=c[]>?c[]: ;
} void div(bignum_t a,const int b,int&c)
{
int i ;
for(c=,i=a[];i;c=c*DEPTH+a[i],a[i]=c/b,c%=b,i--);
for(;!a[a[]]&&a[]>;a[]--);
}
/************************************************************************/
/* 大数平方根，读入大数a，结果保存在b[]数组里 */
/* 需要comp()函数 */
/************************************************************************/
void sqrt(bignum_t b,bignum_t a)
{
int h,l,m,i ;
memset((void*)b,,sizeof(bignum_t));
for(i=b[]=(a[]+)>>;i;sub(a,b,m,i-),b[i]+=m,i--)
for(h=DEPTH-,l=,b[i]=m=(h+l+)>>;h>l;b[i]=m=(h+l+)>>)
if(comp(b,m,i-,a))h=m- ;
else l=m ;
for(;!b[b[]]&&b[]>;b[]--);
for(i=;i<=b[];b[i++]>>=);
}
/************************************************************************/
/* 返回大数的长度 */
/************************************************************************/
int length(const bignum_t a)
{
int t,ret ;
for(ret=(a[]-)*DIGIT,t=a[a[]];t;t/=,ret++);
return ret>?ret: ;
}
/************************************************************************/
/* 返回指定位置的数字，从低位开始数到第b位，返回b位上的数 */
/************************************************************************/
int digit(const bignum_t a,const int b)
{
int i,ret ;
for(ret=a[(b-)/DIGIT+],i=(b-)%DIGIT;i;ret/=,i--);
return ret% ;
}
/************************************************************************/
/* 返回大数末尾0的个数 */
/************************************************************************/
int zeronum(const bignum_t a)
{
int ret,t ;
for(ret=;!a[ret+];ret++);
for(t=a[ret+],ret*=DIGIT;!(t%);t/=,ret++);
return ret ;
} void comp(int*a,const int l,const int h,const int d)
{
int i,j,t ;
for(i=l;i<=h;i++)
for(t=i,j=;t>;j++)
while(!(t%j))
a[j]+=d,t/=j ;
} void convert(int*a,const int h,bignum_t b)
{
int i,j,t= ;
memset(b,,sizeof(bignum_t));
for(b[]=b[]=,i=;i<=h;i++)
if(a[i])
for(j=a[i];j;t*=i,j--)
if(t*i>DEPTH)
mul(b,t),t= ;
mul(b,t);
}
/************************************************************************/
/* 组合数 */
/************************************************************************/
void combination(bignum_t a,int m,int n)
{
int*t=new int[m+];
memset((void*)t,,sizeof(int)*(m+));
comp(t,n+,m,);
comp(t,,m-n,-);
convert(t,m,a);
delete[]t ;
}
/************************************************************************/
/* 排列数 */
/************************************************************************/
void permutation(bignum_t a,int m,int n)
{
int i,t= ;
memset(a,,sizeof(bignum_t));
a[]=a[]= ;
for(i=m-n+;i<=m;t*=i++)
if(t*i>DEPTH)
mul(a,t),t= ;
mul(a,t);
} #define SGN(x) ((x)>0?1:((x)<0?-1:0))
#define ABS(x) ((x)>0?(x):-(x)) int read(bignum_t a,int&sgn,istream&is=cin)
{
char str[MAX*DIGIT+],ch,*buf ;
int i,j ;
memset((void*)a,,sizeof(bignum_t));
if(!(is>>str))return ;
buf=str,sgn= ;
if(*buf=='-')sgn=-,buf++;
for(a[]=strlen(buf),i=a[]/-;i>=;i--)
ch=buf[i],buf[i]=buf[a[]--i],buf[a[]--i]=ch ;
for(a[]=(a[]+DIGIT-)/DIGIT,j=strlen(buf);j<a[]*DIGIT;buf[j++]='');
for(i=;i<=a[];i++)
for(a[i]=,j=;j<DIGIT;j++)
a[i]=a[i]*+buf[i*DIGIT--j]-'' ;
for(;!a[a[]]&&a[]>;a[]--);
if(a[]==&&!a[])sgn= ;
return ;
}
struct bignum
{
bignum_t num ;
int sgn ;
public :
inline bignum()
{
memset(num,,sizeof(bignum_t));
num[]= ;
sgn= ;
}
inline int operator!()
{
return num[]==&&!num[];
}
inline bignum&operator=(const bignum&a)
{
memcpy(num,a.num,sizeof(bignum_t));
sgn=a.sgn ;
return*this ;
}
inline bignum&operator=(const int a)
{
memset(num,,sizeof(bignum_t));
num[]= ;
sgn=SGN (a);
add(num,sgn*a);
return*this ;
}
;
inline bignum&operator+=(const bignum&a)
{
if(sgn==a.sgn)add(num,a.num);
else if
(sgn&&a.sgn)
{
int ret=comp(num,a.num);
if(ret>)sub(num,a.num);
else if(ret<)
{
bignum_t t ;
memcpy(t,num,sizeof(bignum_t));
memcpy(num,a.num,sizeof(bignum_t));
sub (num,t);
sgn=a.sgn ;
}
else memset(num,,sizeof(bignum_t)),num[]=,sgn= ;
}
else if(!sgn)
memcpy(num,a.num,sizeof(bignum_t)),sgn=a.sgn ;
return*this ;
}
inline bignum&operator+=(const int a)
{
if(sgn*a>)add(num,ABS(a));
else if(sgn&&a)
{
int ret=comp(num,ABS(a));
if(ret>)sub(num,ABS(a));
else if(ret<)
{
bignum_t t ;
memcpy(t,num,sizeof(bignum_t));
memset(num,,sizeof(bignum_t));
num[]= ;
add(num,ABS (a));
sgn=-sgn ;
sub(num,t);
}
else memset(num,,sizeof(bignum_t)),num[]=,sgn= ;
}
else if
(!sgn)sgn=SGN(a),add(num,ABS(a));
return*this ;
}
inline bignum operator+(const bignum&a)
{
bignum ret ;
memcpy(ret.num,num,sizeof (bignum_t));
ret.sgn=sgn ;
ret+=a ;
return ret ;
}
inline bignum operator+(const int a)
{
bignum ret ;
memcpy(ret.num,num,sizeof (bignum_t));
ret.sgn=sgn ;
ret+=a ;
return ret ;
}
inline bignum&operator-=(const bignum&a)
{
if(sgn*a.sgn<)add(num,a.num);
else if
(sgn&&a.sgn)
{
int ret=comp(num,a.num);
if(ret>)sub(num,a.num);
else if(ret<)
{
bignum_t t ;
memcpy(t,num,sizeof(bignum_t));
memcpy(num,a.num,sizeof(bignum_t));
sub(num,t);
sgn=-sgn ;
}
else memset(num,,sizeof(bignum_t)),num[]=,sgn= ;
}
else if(!sgn)add (num,a.num),sgn=-a.sgn ;
return*this ;
}
inline bignum&operator-=(const int a)
{
if(sgn*a<)add(num,ABS(a));
else if(sgn&&a)
{
int ret=comp(num,ABS(a));
if(ret>)sub(num,ABS(a));
else if(ret<)
{
bignum_t t ;
memcpy(t,num,sizeof(bignum_t));
memset(num,,sizeof(bignum_t));
num[]= ;
add(num,ABS(a));
sub(num,t);
sgn=-sgn ;
}
else memset(num,,sizeof(bignum_t)),num[]=,sgn= ;
}
else if
(!sgn)sgn=-SGN(a),add(num,ABS(a));
return*this ;
}
inline bignum operator-(const bignum&a)
{
bignum ret ;
memcpy(ret.num,num,sizeof(bignum_t));
ret.sgn=sgn ;
ret-=a ;
return ret ;
}
inline bignum operator-(const int a)
{
bignum ret ;
memcpy(ret.num,num,sizeof(bignum_t));
ret.sgn=sgn ;
ret-=a ;
return ret ;
}
inline bignum&operator*=(const bignum&a)
{
bignum_t t ;
mul(t,num,a.num);
memcpy(num,t,sizeof(bignum_t));
sgn*=a.sgn ;
return*this ;
}
inline bignum&operator*=(const int a)
{
mul(num,ABS(a));
sgn*=SGN(a);
return*this ;
}
inline bignum operator*(const bignum&a)
{
bignum ret ;
mul(ret.num,num,a.num);
ret.sgn=sgn*a.sgn ;
return ret ;
}
inline bignum operator*(const int a)
{
bignum ret ;
memcpy(ret.num,num,sizeof (bignum_t));
mul(ret.num,ABS(a));
ret.sgn=sgn*SGN(a);
return ret ;
}
inline bignum&operator/=(const bignum&a)
{
bignum_t t ;
div(t,num,a.num);
memcpy (num,t,sizeof(bignum_t));
sgn=(num[]==&&!num[])?:sgn*a.sgn ;
return*this ;
}
inline bignum&operator/=(const int a)
{
int t ;
div(num,ABS(a),t);
sgn=(num[]==&&!num [])?:sgn*SGN(a);
return*this ;
}
inline bignum operator/(const bignum&a)
{
bignum ret ;
bignum_t t ;
memcpy(t,num,sizeof(bignum_t));
div(ret.num,t,a.num);
ret.sgn=(ret.num[]==&&!ret.num[])?:sgn*a.sgn ;
return ret ;
}
inline bignum operator/(const int a)
{
bignum ret ;
int t ;
memcpy(ret.num,num,sizeof(bignum_t));
div(ret.num,ABS(a),t);
ret.sgn=(ret.num[]==&&!ret.num[])?:sgn*SGN(a);
return ret ;
}
inline bignum&operator%=(const bignum&a)
{
bignum_t t ;
div(t,num,a.num);
if(num[]==&&!num[])sgn= ;
return*this ;
}
inline int operator%=(const int a)
{
int t ;
div(num,ABS(a),t);
memset(num,,sizeof (bignum_t));
num[]= ;
add(num,t);
return t ;
}
inline bignum operator%(const bignum&a)
{
bignum ret ;
bignum_t t ;
memcpy(ret.num,num,sizeof(bignum_t));
div(t,ret.num,a.num);
ret.sgn=(ret.num[]==&&!ret.num [])?:sgn ;
return ret ;
}
inline int operator%(const int a)
{
bignum ret ;
int t ;
memcpy(ret.num,num,sizeof(bignum_t));
div(ret.num,ABS(a),t);
memset(ret.num,,sizeof(bignum_t));
ret.num[]= ;
add(ret.num,t);
return t ;
}
inline bignum&operator++()
{
*this+= ;
return*this ;
}
inline bignum&operator--()
{
*this-= ;
return*this ;
}
;
inline int operator>(const bignum&a)
{
return sgn>?(a.sgn>?comp(num,a.num)>:):(sgn<?(a.sgn<?comp(num,a.num)<:):a.sgn<);
}
inline int operator>(const int a)
{
return sgn>?(a>?comp(num,a)>:):(sgn<?(a<?comp(num,-a)<:):a<);
}
inline int operator>=(const bignum&a)
{
return sgn>?(a.sgn>?comp(num,a.num)>=:):(sgn<?(a.sgn<?comp(num,a.num)<=:):a.sgn<=);
}
inline int operator>=(const int a)
{
return sgn>?(a>?comp(num,a)>=:):(sgn<?(a<?comp(num,-a)<=:):a<=);
}
inline int operator<(const bignum&a)
{
return sgn<?(a.sgn<?comp(num,a.num)>:):(sgn>?(a.sgn>?comp(num,a.num)<:):a.sgn>);
}
inline int operator<(const int a)
{
return sgn<?(a<?comp(num,-a)>:):(sgn>?(a>?comp(num,a)<:):a>);
}
inline int operator<=(const bignum&a)
{
return sgn<?(a.sgn<?comp(num,a.num)>=:):(sgn>?(a.sgn>?comp(num,a.num)<=:):a.sgn>=);
}
inline int operator<=(const int a)
{
return sgn<?(a<?comp(num,-a)>=:):
(sgn>?(a>?comp(num,a)<=:):a>=);
}
inline int operator==(const bignum&a)
{
return(sgn==a.sgn)?!comp(num,a.num): ;
}
inline int operator==(const int a)
{
return(sgn*a>=)?!comp(num,ABS(a)): ;
}
inline int operator!=(const bignum&a)
{
return(sgn==a.sgn)?comp(num,a.num): ;
}
inline int operator!=(const int a)
{
return(sgn*a>=)?comp(num,ABS(a)): ;
}
inline int operator[](const int a)
{
return digit(num,a);
}
friend inline istream&operator>>(istream&is,bignum&a)
{
read(a.num,a.sgn,is);
return is ;
}
friend inline ostream&operator<<(ostream&os,const bignum&a)
{
if(a.sgn<)
os<<'-' ;
write(a.num,os);
return os ;
}
friend inline bignum sqrt(const bignum&a)
{
bignum ret ;
bignum_t t ;
memcpy(t,a.num,sizeof(bignum_t));
sqrt(ret.num,t);
ret.sgn=ret.num[]!=||ret.num[];
return ret ;
}
friend inline bignum sqrt(const bignum&a,bignum&b)
{
bignum ret ;
memcpy(b.num,a.num,sizeof(bignum_t));
sqrt(ret.num,b.num);
ret.sgn=ret.num[]!=||ret.num[];
b.sgn=b.num[]!=||ret.num[];
return ret ;
}
inline int length()
{
return :: length(num);
}
inline int zeronum()
{
return :: zeronum(num);
}
inline bignum C(const int m,const int n)
{
combination(num,m,n);
sgn= ;
return*this ;
}
inline bignum P(const int m,const int n)
{
permutation(num,m,n);
sgn= ;
return*this ;
}
}; ///=======================================================================================
const int MAXN = ;
const int MAXM = ;
int MOD;
typedef struct
{
int mat[MAXN][MAXM];
}Matrix;
int kk;
Matrix Init(Matrix I)///单位矩阵
{
for(int i=; i<(<<kk); i++)
{
for(int j=; j<(<<kk); j++)
{
if(i == j)
I.mat[i][j] = ;
else
I.mat[i][j] = ;
}
}
return I;
}
Matrix Multi_Matrix(Matrix a, Matrix b)///矩阵乘法
{
Matrix c;
for(int i=; i<(<<kk); i++)
{
for(int j=; j<(<<kk); j++)
{
c.mat[i][j] = ;
for(int k=; k<(<<kk); k++)
{
c.mat[i][j] += a.mat[i][k]*b.mat[k][j];
c.mat[i][j] %= MOD;
}
}
}
return c;
}
int a[];
bool Judge_get_Binary(int i, int j)
{
int cnt1=, cnt2=;
for(int k=; k<kk; k++)
{
cnt1 = i%;
cnt2 = j%;
if(cnt1== && cnt2==)
a[k] = ;
else if(!cnt1 && !cnt2)
a[k] = ;
else
a[k] = -;
i/=, j/=;
}
for(int k=; k<kk-; k++)
{
if(a[k]== && a[k+]==)
return ;
if(a[k]== && a[k+]==)
return ;
}
return ;
}
Matrix BuildA()
{
Matrix A;
for(int i=; i<(<<kk); i++)
{
for(int j=; j<(<<kk); j++)
{
if(Judge_get_Binary(i, j))
A.mat[i][j] = ;
else
A.mat[i][j] = ;
}
}
return A;
}
int Eular(int m) ///欧拉函数，本题没用上；
{
int ans = m;
for(int i=; i*i<=m; i++)
{
if(m%i==)
ans = ans-ans/i;
while(m%i==)
m /= i;
}
if(m > )
ans = ans-ans/m;
return ans;
}
Matrix quick_MOD_Matrix(bignum n)
{
n = n-;
Matrix ans ;
ans= Init(ans);
Matrix A = BuildA();
while(n>)
{
if(n%==)
ans = Multi_Matrix(ans, A);
n=n/;
A = Multi_Matrix(A, A);
}
return ans;
} int main()
{
bignum n;
while(cin>>n>>kk>>MOD)
{
Matrix tmp = quick_MOD_Matrix(n);
Matrix ans;
Matrix tmp1;
for(int i=; i<(<<kk); i++)
for(int j=; j<; j++)
tmp1.mat[i][j] = ; for(int i=; i<(<<kk); i++)
{
for(int j=; j<; j++)
{
ans.mat[i][j] = ;
for(int k=; k<(<<kk); k++)
{
ans.mat[i][j] += tmp.mat[i][k]*tmp1.mat[k][j];
ans.mat[i][j] %= MOD;
}
}
}
int sum = ;
for(int i=; i<(<<kk); i++)
{
sum = sum + ans.mat[i][];
sum %= MOD;
}
sum = (sum%MOD+MOD)%MOD;
cout<<sum<<endl;
}
return ;
}