HDU 6470 Count 矩阵快速幂优化线性递推

传送门：http://acm.hdu.edu.cn/showproblem.php?pid=6470

题意

$已知f(1) = 1, f(2) =2，且f(n)=f(n-1)+2f(n-2)+n^3\;\;(mod\;\;1234567789)$

思路

对于线性递推，仿照快速幂，我们可以用矩阵快速幂优化（这就是线代吗？爱了爱了）。

考虑F_n为第n个矩阵，那就需要构造新的“递推矩阵”。

因为递推中有个$f_n$中$n^3$，所以$f_{n-1}$中需要有$(n-1)^3$，结合二项式定理可得：

$$n^3=[(n-1)+1]^3=C_{3}^0(n-1)^3+C_3^1(n-1)^2+C_{3}^2(n-1)^1+C_3^3(n-1)^0$$

所以说线性递推可以用矩阵快速幂优化。

但是！最近被mod坑死了，如果在矩阵乘法中每次都用mod，大概最高会有4秒，但是先判断在mod的话只有两秒，足足差了2秒！！！模运算是真的太慢了。。。

Code （967MS）

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##include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pdd;

##define INF 0x3f3f3f3f
##define lc u << 1
##define rc u << 1  1
##define mm (l + r) / 2
##define mid (t[u].l + t[u].r) / 2
##define lowbit(x) x & (-x)
##define mem(a, b) memset(a , b , sizeof(a))
##define FOR(i, x, n) for(int i = x;i <= n; i++)

// const ll mod = 998244353;
// const ll mod = 1e9 + 7;
// const double eps = 1e-6;
// const double PI = acos(-1);
// const double R = 0.57721566490153286060651209;

ll mod = 123456789;

struct Martix{
    ll m[6][6];
};

Martix ans;

Martix E;

Martix operator * (Martix a, Martix b) {
    Martix c;
    mem(c.m, 0);
    for (int i = 0; i < 6; i++) {
        for (int j = 0; j < 6; j++) {
            for (int k = 0; k < 6; k++) {
                c.m[i][j] = (c.m[i][j] + a.m[i][k] * b.m[k][j]);
                if(c.m[i][j] > mod)
                    c.m[i][j] %= mod;
            }
        }
    }
    return c;
}

Martix quick_pow_Martix(ll n) {
    Martix base = E;
    mem(ans.m, 0);
    ans.m[0][0] = 2;
    ans.m[1][0] = 1;
    ans.m[2][0] = 8;
    ans.m[3][0] = 4;
    ans.m[4][0] = 2;
    ans.m[5][0] = 1;
    while(n) {
        if(n & 1)
            ans = base * ans;
        base = base * base;
        n >>= 1;
    }
    return ans;
}

void solve() {
    E.m[0][0] = 1; E.m[0][1] = 2; E.m[0][2] = 1; E.m[0][3] = 3; E.m[0][4] = 3; E.m[0][5] = 1;
    E.m[1][0] = 1;
    E.m[2][2] = 1; E.m[2][3] = 3; E.m[2][4] = 3; E.m[2][5] = 1;
    E.m[3][3] = 1; E.m[3][4] = 2; E.m[3][5] = 1;
    E.m[4][4] = 1; E.m[4][5] = 1;
    E.m[5][5] = 1;
    int T;
    scanf("%d",&T);
    while(T--) {
        ll n;
        scanf("%lld",&n);

        // ans = quick_pow_Martix(n - 2);
        printf("%lld\n",quick_pow_Martix(n - 2).m[0][0]);
    }
}

signed main() {
    ios_base::sync_with_stdio(false);
    //cin.tie(nullptr);
    //cout.tie(nullptr);
##ifdef FZT_ACM_LOCAL
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
    signed test_index_for_debug = 1;
    char acm_local_for_debug = 0;
    do {
        if (acm_local_for_debug == '$') exit(0);
        if (test_index_for_debug > 20)
            throw runtime_error("Check the stdin!!!");
        auto start_clock_for_debug = clock();
        solve();
        auto end_clock_for_debug = clock();
        cout << "Test " << test_index_for_debug << " successful" << endl;
        cerr << "Test " << test_index_for_debug++ << " Run Time: "
             << double(end_clock_for_debug - start_clock_for_debug) / CLOCKS_PER_SEC << "s" << endl;
        cout << "--------------------------------------------------" << endl;
    } while (cin >> acm_local_for_debug && cin.putback(acm_local_for_debug));
##else
    solve();
##endif
    return 0;
}

本文作者：jujimeizuo
本文地址： https://blog.jujimeizuo.cn/2020/10/30/hdu-6470/
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