HDU 1575 Tr A【矩阵幂】

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题目点这

求对角线和 [ (A^k)%9973 ]

矩阵幂即可，轻松

#include <cstdio>#include <cmath>#include <cstring>#include <set>#include <stack>#include <queue>#include <vector>#include <iostream>using namespace std;#include <algorithm>#define ll long long#define INF 0x7FFFFFFF#define INT_MIN -(1<<31)#define eps 10^(-6)#define Q_CIN ios::sync_with_stdio(false)#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )#define REV( i , n ) for ( int i = n - 1 ; i >= 0 ; -- i )#define FOR( i , a , b ) for ( int i = a ; i <= b ; ++ i )#define FOV( i , a , b ) for ( int i = a ; i >= b ; -- i )#define CLR( a , x ) memset ( a , x , sizeof (a) )#define RE freopen("1.in","r",stdin)#define WE freopen("1.out","w",stdout)#define MOD 10009#define NMAX 10002#define min(a,b) ((a)>(b)?(b):(a))int n,k;struct Matrix{    int v[15][15];}m0;Matrix mul(Matrix A,Matrix B){    Matrix C;    REP(i,n)    {        REP(j,n)        {            C.v[i][j] = 0;            REP(k,n)            {                C.v[i][j] += A.v[i][k] * B.v[k][j];            }            C.v[i][j] %= 9973;        }    }    return C;}Matrix mtPow(Matrix A, int k){    if(k == 1)        return A;    A = mtPow(A, k / 2);    if(k % 2 == 0)        return mul(A, A);    else        return mul(mul(A, A), m0);}int main(){    int t;    //RE;    cin>>t;    while(t--)    {        cin>>n>>k;        REP(i,n)            REP(j,n)                cin>>m0.v[i][j];        Matrix m1 = mtPow(m0, k);    /*    REP(i,n){            REP(j,n)            cout<<m1.v[i][j]<<" ";            cout<<endl;}*/        int ans = 0;        REP(i,n)            ans += m1.v[i][i];        cout<<ans%9973<<endl;    }return 0;}

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