Problem Description

Input

Output

Sample Input

1
3 2
0 2 0
0 0 2
0 0 0

Sample Output

0 2 4
0 0 2
0 0 0

解法：矩陣的等比求和，知道這個就用模版寫

#include<bits/stdc++.h>
#define clr(x) memset(x,0,sizeof(x))
#define LL long long
using namespace std;
#define INF 0x3f3f3f3f
typedef long long ll;
const int N= 30 +9;
struct Matrix
{
    int m[N][N];
};
Matrix I;
int n,k,M;

Matrix add(Matrix a,Matrix b)
{
    Matrix c;
    for
 
(int i=0;i<n;i++)
        for(int j=0;j<n;j++)
            c.m[i][j]=(a.m[i][j]+b.m[i][j])%M;
    return c;
}

Matrix multi(Matrix a,Matrix b)
{
    Matrix c;
    for(int i=0;i<n;i++){
        for(int j=0;j<n;j++){
            c.m[i][j]=0;
            for(int k=0;k<n;k++)
                c.m[i][j]
 
=(c.m[i][j]+a.m[i][k]*b.m[k][j])%M;
        }
    }
    return c;
}
Matrix power(Matrix A,ll n)
{
    Matrix ans=I;
    while(n){
        if(n&1)
            ans=multi(ans,A);
        A=multi(A,A);
        n>>=1;
    }
    return ans;
}

Matrix sum(Matrix A,ll k)
{
    if(k==1) return A;
    Matrix t=sum(A,k/2);
    Matrix cur=power(A,k/2+(k&1));
    t=add(t,multi(t,cur));
    if(k&1) t=add(t,cur);
    return t;
}

int main()
{
    int T;
    ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    while(cin>>T){
        while(T--){
            cin>>n>>k;
            M=10;
            Matrix A;
            for(int i=0;i<n;i++){
                for(int j=0;j<n;j++){
                    cin>>A.m[i][j];
                    A.m[i][j]%=M;
                }
                I.m[i][i]=1;
            }
            Matrix ans=sum(A,k);
            for(int i=0;i<n;i++){
                for(int j=0;j<n;j++){
                    if(j+1!=n) cout<<ans.m[i][j]<<" ";
                    else cout<<ans.m[i][j]<<endl;
                }
            }
        }
    }
    return 0;
}

華東交通大學2017年ACM“雙基”程序設計競賽 1005