bzoj5118 number of columns Fib + 2 matrix quadratic residue Fast Power

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https://lydsy.com/JudgeOnline/problem.php?id=5118

answer

This title is a look do not look.

Seeking Fibonacci number of \ (n-\) key, \ (n-\ Leq ^ 2 ^ {15} {10} \) ? ? ?

Such people how fast matrix power ah.

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Etc. This module amazing ah.

\ (1125899839733759 \) seems to be a prime number, but also to \ (9 \) at the end.

Then \ (5 \) to \ (1125899839733759 \) must have a quadratic residue slightly.

Then The term formulas, Fib
\ [f (n) = (
\ frac {\ sqrt 5 + 1} {2}) ^ n + (\ frac {\ sqrt 5 - 1} 2) ^ n \] then the \ (2 ^ n \) can be can be transformed according to Fermat's little theorem \ (n \ bmod P-1 \) a.

Then the \ (2 ^ n \) in the range of a lot of small moment.

So you can directly power the matrix quickly, pay attention to use multiplication multiply quickly realized.

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UPD: discovery seems silly out, there is a general term formula, but also what to write fast power matrix.

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Time complexity \ (O (T \ log ^ n-2) \) .

#include<bits/stdc++.h>

#define fec(i, x, y) (int i = head[x], y = g[i].to; i; i = g[i].ne, y = g[i].to)
#define dbg(...) fprintf(stderr, __VA_ARGS__)
#define File(x) freopen(#x".in", "r", stdin), freopen(#x".out", "w", stdout)
#define fi first
#define se second
#define pb push_back

template<typename A, typename B> inline char smax(A &a, const B &b) {return a < b ? a = b, 1 : 0;}
template<typename A, typename B> inline char smin(A &a, const B &b) {return b < a ? a = b, 1 : 0;}

typedef long long ll; typedef unsigned long long ull; typedef std::pair<int, int> pii;

template<typename I> inline void read(I &x) {
    int f = 0, c;
    while (!isdigit(c = getchar())) c == '-' ? f = 1 : 0;
    x = c & 15;
    while (isdigit(c = getchar())) x = (x << 1) + (x << 3) + (c & 15);
    f ? x = -x : 0;
}

const ll P = 1125899839733759;

ll n;

inline void sadd(ll &x, ll y, const ll &P = ::P) { x += y; x >= P ? x -= P : x; }
inline ll smod(const ll &x, const ll &P = ::P) { return x >= P ? x - P : x; }
inline ll fmul(ll x, ll y, const ll &P = ::P) {
    ll ans = 0;
    for (; y; y >>= 1, sadd(x, x, P)) if (y & 1) sadd(ans, x, P);
    return ans;
}
inline ll fpow(ll x, ll y, const ll &P = ::P) {
    ll ans = 1;
    for (; y; y >>= 1, x = fmul(x, x, P)) if (y & 1) ans = fmul(ans, x, P);
    return ans;
}

struct Matrix {
    ll a[2][2];
    
    inline Matrix() { memset(a, 0, sizeof(a)); }
    inline Matrix(const ull &x) {
        memset(a, 0, sizeof(a));
        a[0][0] = a[1][1] = x;
    }
    
    inline Matrix operator * (const Matrix &b) {
        Matrix c;
        c.a[0][0] = smod(fmul(a[0][0], b.a[0][0]) + fmul(a[0][1], b.a[1][0]));
        c.a[0][1] = smod(fmul(a[0][0], b.a[0][1]) + fmul(a[0][1], b.a[1][1]));
        c.a[1][0] = smod(fmul(a[1][0], b.a[0][0]) + fmul(a[1][1], b.a[1][0]));
        c.a[1][1] = smod(fmul(a[1][0], b.a[0][1]) + fmul(a[1][1], b.a[1][1]));
        return c;
    }
} A, B;

inline Matrix fpow(Matrix x, ll y) {
    Matrix ans(1);
    for (; y; y >>= 1, x = x * x) if (y & 1) ans = ans * x;
    return ans;
}

inline void work() {
    A.a[0][0] = 1, A.a[0][1] = 1;
    A.a[1][0] = 1, A.a[1][1] = 0;
    B.a[0][0] = 0, B.a[1][0] = 1;
    A = fpow(A, n) * B;
    printf("%lld\n", A.a[0][0]);
}

inline void init() {
    read(n);
    n = fpow(2, n, P - 1);
}

int main() {
#ifdef hzhkk
    freopen("hkk.in", "r", stdin);
#endif
    int T;
    read(T);
    while (T--) {
        init();
        work();
    }
    fclose(stdin), fclose(stdout);
    return 0;
}