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solve: In fact, the replacement operation can finally, we can merge first m matrix operations (multiply, multiply note the order) and then finally calculated k matrix% m remaining to be done on the line

#pragma GCC optimize("O3")
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define re register
#define pb push_back
#define fi first
#define se second
const int N=1e6+10;
const int mod=1e9+7;
void read(int &a)
{
    a=0;int d=1;char ch;
    while(ch=getchar(),ch>'9'||ch<'0')
        if(ch=='-')
            d=-1;
    a=ch^48;
    while(ch=getchar(),ch>='0'&&ch<='9')
        a=(a<<3)+(a<<1)+(ch^48);
    a*=d;
}
struct note
{
    int a[105][105];
}ans,a,b[15];
int n,m,k;
note mul(note x,note y)
{
    note c;
    for(re int i=1;i<=n;i++)
    {
        for(re int j=1;j<=n;j++)
        {
            c.a[i][j]=0;
            for(re int k=1;k<=n;k++) c.a[i][j]=(c.a[i][j]+x.a[i][k]*y.a[k][j]);
        }
    }
    return c;
}
note quickmod(note x,int y)
{
    note res,base=x;
    for(re int i=1;i<=n;i++)
        for(re int j=1;j<=n;j++)
            res.a[i][j]=i==j?1:0;
    while(y)
    {
        if(y&1) res=mul(res,base);
        base=mul(base,base);
        y>>=1;
    }
    return res;
}
int main()
{
    read(n),read(m),read(k);
    for(re int i=1;i<=n;i++)
        for(re int j=1;j<=n;j++)
            a.a[i][j]=i==j?1:0;
    for(re int i=1;i<=n;i++) ans.a[i][1]=i;
    for(re int i=1;i<=m;i++)
    {
        note c;
        for(re int j=1;j<=n;j++)
            for(re int k=1;k<=n;k++)
                c.a[j][k]=0;
        for(re int j=1,x;j<=n;j++) read(x),c.a[j][x]=1;
        b[i]=c;
        a=mul(c,a);
    }
    int t=k/m;
    ans=mul(quickmod(a,t),ans);
    t=k%m;
    for(re int i=1;i<=t;i++) ans=mul(b[i],ans);
    for(re int i=1;i<=n;i++) printf("%d ",ans.a[i][1]);
    return 0;
}