[数论] 51nod 1365 Fib(N) mod Fib(K)

Description

Fib(N)表示斐波那契数列的第N项（F(0)=0,F(1)=1），给出N和K，求Fib(N)modFib(K)。由于结果太大，输出Mod 1000000007的结果。

Solution

先贴几个公式。

F−n=(−1)n−1Fn

Fn=FkFn−k+1+Fk−1Fn−k(1)

F2k−1+Fk−1Fk−F2k=(−1)k(2)

(2)式可以考虑对

(1110)n=(Fn+1FnFnFn−1)

这个等式两边同时计算行列式。

Fn≡F⌊nk⌋k−1Fnmodk(modFk)

这就是考虑按(1)式展开就好了。设

i=⌊nk⌋,j=nmodk

所以

Fn≡Fik−1Fj(modFk)

这里需要分类讨论i的奇偶性：

若i为偶数，考虑将(2)式代入

Fik−1Fj==(F2k−1+Fk−1Fk−F2k)i2Fj(−1)i2kFj

若i为奇数，考虑将(2)式代入

Fik−1Fj====Fik−1(Fk−1Fj−k+FkFj−k+1)Fi+1k−1Fj−kFi+1k−1(−1)k−j−1Fk−j(−1)i+12k+k−j−1Fk−j

到这里就可以通过矩阵乘法快速幂得到答案啦！

#include <bits/stdc++.h>using namespace std;const int MOD = 1000000007;typedef long long ll;inline void Add(ll &x, ll b) {    x += b; while (x >= MOD) x -= MOD;}inline char get(void) {    static char buf[100000], *S = buf, *T = buf;    if (S == T) {        T = (S = buf) + fread(buf, 1, 100000, stdin);        if (S == T) return EOF;    }    return *S++;}template<typename T>inline void read(T &x) {    static char c; x = 0;    for (c = get(); c < '0' || c > '9'; c = get());    for (; c >= '0' && c <= '9'; c = get()) x = x * 10 + c - '0';}ll n, i, j, k, f, ans, fk;int test;struct Matrix {    ll a[3][3];    inline ll* operator [](ll x) {        return a[x];    }    inline Matrix(void) {        memset(a, 0, sizeof a);    }    inline friend Matrix operator *(Matrix a, Matrix b) {        Matrix x;        for (ll i = 1; i <= 2; i++)            for (ll j = 1; j <= 2; j++)                for (ll k = 1; k <= 2; k++)                    Add(x[i][j], a[i][k] * b[k][j] % MOD);        return x;    }};Matrix Fib, I;inline Matrix Pow(Matrix a, ll b) {    Matrix c = I;    while (b) {        if (b & 1) c = c * a;        b >>= 1; a = a * a;    }    return c;}inline ll F(ll x, ll y, ll z = 0) {    if ((x & 1) && (y & 1)) return z & 1 ? 1 : -1;    return z & 1 ? -1 : 1;}int main(void) {    freopen("1.in", "r", stdin);    I[1][1] = I[2][2] = 1;    Fib[1][1] = Fib[1][2] = Fib[2][1] = 1;    read(test);    while (test--) {        read(n); read(k);        i = n / k; j = n % k;        if (i & 1) {            if (k - j == k) f = 0;            else f = Pow(Fib, k - j)[1][2];            fk = Pow(Fib, k)[1][2];            ans = F((i + 1) / 2, k, k - j - 1) * f;            if (ans < 0) ans += fk;            Add(ans, MOD);            printf("%d\n", ans);        } else {            if (j == k) f = 0;            else f = Pow(Fib, j)[1][2];            ans = F(i / 2, k) * f;            fk = Pow(Fib, k)[1][2];            if (ans < 0) ans += fk;            Add(ans, MOD);            printf("%d\n", ans);        }    }    return 0;}