洛谷P3390 【模板】矩阵快速幂

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从今天开始学习矩阵快速幂.jpg

//minamoto
#include<iostream>
#include<cstdio>
#include<cstring>
#define ll long long
using namespace std;
#define getc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
char buf[1<<21],*p1=buf,*p2=buf;
inline ll read(){
    #define num ch-'0'
    char ch;bool flag=0;ll res;
    while(!isdigit(ch=getc()))
    (ch=='-')&&(flag=true);
    for(res=num;isdigit(ch=getc());res=res*10+num);
    (flag)&&(res=-res);
    #undef num
    return res;
}
const int N=105,mod=1e9+7;
ll n,k;
struct Matrix{
    int n;ll g[N][N];
    Matrix(){memset(g,0,sizeof(g));}
    inline Matrix operator *(Matrix b){
        Matrix ans;
        for(int i=1;i<=n;++i)
        for(int j=1;j<=n;++j)
        for(int k=1;k<=n;++k)
        (ans.g[i][j]+=g[i][k]*b.g[k][j])%=mod;
        ans.n=n;
        return ans;
    }
}x,res;
inline Matrix qpow(Matrix a,ll b){
    while(b){
        if(b&1) res=res*a;
        a=a*a,b>>=1;
    }
    return res;
}
inline void print(Matrix x){
    int n=x.n;
    for(int i=1;i<=n;++i){
        for(int j=1;j<=n;++j) printf("%d ",x.g[i][j]);
        putchar(10);
    }
}
int main(){
    //freopen("testdata.in","r",stdin);
    n=read(),k=read();
    x.n=n;
    for(int i=1;i<=n;++i)
    for(int j=1;j<=n;++j)
    x.g[i][j]=read();
    for(int i=1;i<=n;++i) res.g[i][i]=1;res.n=n;
    print(qpow(x,k));
    return 0;
}