2018年湘潭大学程序设计竞赛 G- 又见斐波那契

推一推矩阵直接快速幂。

#include<bits/stdc++.h>
#define LL long long
#define pii pair<int,int>
#define mk make_pair
#define fi first
#define se second
using namespace std;

const int N=1e5+7;
const int M=1e5+7;
const int inf=0x3f3f3f3f;
const LL INF=0x3f3f3f3f3f3f3f3f;
const int mod=1e9+7;

LL n;
struct Matrix {
    LL a[6][6];
    Matrix() {memset(a, 0, sizeof(a));}
    void init() {
        for(int i = 0; i < 6; i++)
            for(int j = 0; j < 6; j++)
                a[i][j] = (i == j);
    }

    Matrix operator * (const Matrix &B) const {
        Matrix C;
        for(int i = 0; i < 6; i++) {
            for(int j = 0; j < 6; j++) {
                for(int k = 0; k < 6; k++) {
                    C.a[i][j] = (C.a[i][j] + a[i][k] * B.a[k][j]) % mod;
                }
            }
        }
        return C;
    }

    Matrix operator ^ (const LL &p) const {
        Matrix A =  *this, res;
        res.init();
        LL t = p;
        while(t) {
            if(t & 1) res = res * A;
            A = A * A; t >>= 1;
        }
        return res;
    }
}A;


int main() {

    int a[6][6] = {
        {1, 1, 1, 1, 1, 1},
        {1, 0, 0, 0, 0, 0},
        {0, 0, 1, 3, 3, 1},
        {0, 0, 0, 1, 2, 1},
        {0, 0, 0, 0, 1, 1},
        {0, 0, 0, 0, 0, 1}
    };

    for(int i = 0; i < 6; i++)
        for(int j = 0; j < 6; j++)
            A.a[i][j] = a[i][j];

    int T; scanf("%d", &T);
    while(T--) {
        scanf("%lld", &n);
        if(n == 1) {
            puts("1");
            continue;
        }
        Matrix B = A ^ (n - 1);
        LL ans = B.a[0][0] + B.a[0][2] * 8 + B.a[0][3] * 4 + B.a[0][4] * 2 + B.a[0][5];
        ans %= mod;
        printf("%lld
", ans);
    }
    return 0;
}
/*
*/