"JSOI2016" bit computing

Title Description

JYY recent operations in the study site. He found the most interesting bit computing is the exclusive OR * (xor) * operation. For an exclusive OR operation of two numbers, JYY found a conclusion: two different numbers or value of $ 0 $ if and only if they are equal. So JYY began to think, what is the nature or value will be for the exclusive $ N $ the number of it?

JYY wondered if $ 0 $ to the range of $ R-1 $ to elect $ N $ different integers, and so that $ N $ integers exclusive OR value of $ 0 $, then a number of species selected Total the way to do that? (Do not select a different order for double counting, see sample)

JYY is a computer scientist, so his mind is very, very large $ R $. In order to facilitate the expression, if R & lt we $ $ $ $ 01 written as a string, then R & lt $ $ $ is a short string of $ 01 S $ $ $ repeated K times to obtain $. For example, if $ S = 101, K = 2 $, then R & lt $ $ $ compared to the binary representation of $ 101,101. As a result of the calculation will be very large, JYY results can only tell you the total number of his choice of $ 10 ^ 9 + 7 $ modulo.

data range

$ 3 \ N \ the 7.1 \ k \ 10 ^ 5,1 \ the | S | \ le $ 50

answer

Considering the number of the selected different $ x_i $, $ and $ n-small, so consider like pressure, $ f_ {i, s} $ I $ $ represents the front position, the first state represents bit $ J $ $ $ less than or x_j equal to $ x_ {j + 1} $, special, we let $ x_ {n + 1} = R $.

String can be found for each repetition state transition paths is the same, then to take the moment.

Code

#include <bits/stdc++.h>
using namespace std;
const int N=130,P=1e9+7;
int n,k,m,a[N],al,f[N][N];
char ch[N];bool c[N];
struct Mat{int p[N][N];}F,G,V;
int X(int x){if (x>=P) x-=P;return x;}
int work(int l,int r,int x){
    int v=0;
    for (int y,z,j=1;j<n;j++)
        if (l&(1<<j)){
            y=(r>>j)&1;z=(r>>(j-1))&1;
            if (y>z) return -1;
            if (y==z) v|=(1<<j);
        }
    if (l&1){
        if ((r&1)>a[x]) return -1;
        if ((r&1)==a[x]) v|=1;
    }
    return v;
}
Mat C(Mat A,Mat B){
    for (int i=0;i<al;i++)
        for (int j=0;j<al;j++){
            V.p[i][j]=0;
            for (int k=0;k<al;k++)
                V.p[i][j]=X(V.p[i][j]+1ll*A.p[i][k]*B.p[k][j]%P);
        }
    return V;
}
int main(){
    scanf("%d%d%s",&n,&k,ch+1);m=strlen(ch+1);
    for (int i=1;i<=m;i++) a[i]=ch[i]^48;al=1<<n;c[0]=1;
    for (int s=1;s<al;s++) c[s]=(!c[s-(s&-s)]);
    for (int s=0;s<al;s++){
        for (int i=0;i<=m;i++)
            for (int j=0;j<al;j++)
                f[i][j]=0;
        f[0][s]=1;
        for (int i=1,j;i<=m;i++)
            for (int l=0;l<al;l++) if (f[i-1][l])
                for (int r=0;r<al;r++) if (c[r])
                    if (~(j=work(l,r,i)))
                        f[i][j]=X(f[i][j]+f[i-1][l]);
        for (int i=0;i<al;i++)
            F.p[i][s]=f[m][i];
    }
    for (int i=0;i<al;i++) G.p[i][i]=1;
    for (;k;k>>=1,F=C(F,F)) if (k&1) G=C(G,F);
    printf("%d\n",G.p[0][al-1]);
    return 0;
}