hdu6470 Count 矩阵快速幂

hdu6470 Count

 

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 6, mod = 123456789;

struct MAT {
    ll a[maxn][maxn];
    MAT(){ memset(a,0,sizeof(a)); }
    MAT operator*(MAT p) {
        MAT res;
        for (int i = 0; i < maxn; i++)
            for (int j = 0; j < maxn; j++)
                for(int k = 0; k < maxn; k++)
                    res.a[i][j] = (res.a[i][j]+a[i][k]*p.a[k][j])%mod;
        return res;
    }
};
MAT mat_qpow(MAT A, ll b) {
    MAT res;
    res.a[0][0] = 1;
    for (int i = 0; i < maxn; i++) {
        res.a[i][i] = 1;
    }
    while(b) {
        if(b&1) res = res*A;
        A = A*A;
        b >>= 1;
    }
    return res;
}
void init(MAT& A, MAT& B) {
    A.a[0][0]=1, A.a[0][1]=2, A.a[0][2]=1, A.a[0][3]=0, A.a[0][4]=0, A.a[0][5]=0;
    A.a[1][0]=1, A.a[1][1]=0, A.a[1][2]=0, A.a[1][3]=0, A.a[1][4]=0, A.a[1][5]=0;
    A.a[2][0]=0, A.a[2][1]=0, A.a[2][2]=1, A.a[2][3]=3, A.a[2][4]=3, A.a[2][5]=1;
    A.a[3][0]=0, A.a[3][1]=0, A.a[3][2]=0, A.a[3][3]=1, A.a[3][4]=2, A.a[3][5]=1;
    A.a[4][0]=0, A.a[4][1]=0, A.a[4][2]=0, A.a[4][3]=0, A.a[4][4]=1, A.a[4][5]=1;
    A.a[5][0]=0, A.a[5][1]=0, A.a[5][2]=0, A.a[5][3]=0, A.a[5][4]=0, A.a[5][5]=1;

    B.a[0][0]=2, B.a[1][0]=1, B.a[2][0]=27, B.a[3][0]=9, B.a[4][0]=3, B.a[5][0]=1;
}

int main() {
    int t; scanf("%d",&t);
    while(t--) {
        ll n; scanf("%lld",&n);
        MAT A; MAT ans;
        init(A,ans);
        ans = mat_qpow(A,n-2)*ans;
        printf("%lld\n",ans.a[0][0]);
    }
    return 0;
}