2019牛客暑期多校训练营（第九场）A——The power of Fibonacci（循环节+中国剩余定理（互质）||广义BM）

链接：https://ac.nowcoder.com/acm/contest/889/A
来源：牛客网

时间限制：C/C++ 2秒，其他语言4秒
空间限制：C/C++ 262144K，其他语言524288K
64bit IO Format: %lld

题目描述

Amy asks Mr. B problem A. Please help Mr. B to solve the following problem.

Let Fi be fibonacci number.

F0 = 0, F1 = 1, Fi = Fi-1 + Fi-2

Given n and m, please calculate

∑i=0nFim\sum_{i = 0}^n F_i^m∑i=0nFim

As the answer might be very large, output it module 1000000000.

输入描述:

The first and only line contains two integers n, m(1 <= n <= 1000000000, 1 <= m <= 1000).

输出描述:

Output a single integer, the answer module 1000000000.

示例1

输入

复制

5 5

输出

复制

说明

05 + 15 + 15 + 25 + 35 + 55 = 3402

示例2

输入

复制

10 10

输出

复制

696237975

说明

Don't forget to mod 1000000000

示例3

输入

复制

1000000000 1000

输出

复制

641796875

说明

Time is precious.

题意：求1-n的斐波那契数列的m次方的和。

题解：考虑循环节，斐波那契数列在模意义下是有循环节的，即使它取幂了（记住！），求循环节见https://blog.csdn.net/lgz0921/article/details/99679155，1e9的循环节是1.5e9显然不行，我们考虑把1e9拆开求循环节，然后利用中国剩余定理求解，拆成1e9=2^9*5^9,这样就可以求了，解释一下中国剩余定理，它是解决x%a1=b1,x%a2=b2.......求x的，这样我们把两个循环节的求出来，利用中国剩余定理的性质再合起来，就ok了~~最后说一下，因为2^9与5^9互质，所以用一般的中国剩余定理就ok！！

上代码：

#include <iostream>
#include <cstdio>
using namespace std;
typedef long long ll;
const int MAX = 8e6+10;
const int mod = 1000000000;
const int mod1=512;//2^9
const int mod2=1953125;//5^9
const int p1=768;//2^9的循环节
const int p2=7812500;//5^9的循环节
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}
}
ll inv(ll t,ll p){ll d,x,y;ex_gcd(t,p,d,x,y);return d == 1?(x%p+p)%p:-1;
}
ll China(int n,ll *a,ll *m){//中国剩余定理互质 x%m[i]=a[i];ll M = 1,ret = 0;for(int i=0;i<n;i++) M*=m[i];for(int i=0;i<n;i++){ll w = M/m[i];ret = (ret+w*inv(w,m[i])*a[i])%M;}return ret;
}
ll Chinaa(int len,ll *a,ll *r){//中国剩余定理非互质 x%r[i]=a[i];ll M=a[0],R=r[0],x,y,d;for(int i=1;i<len;i++){ex_gcd(M,a[i],d,x,y);if((R-r[i])%d!=0) return -1;x=(R-r[i])/d*x%a[i];R-=x*M;M=M/d*a[i];R%=M;}return (R%M+M)%M;
}
ll f[MAX];
void init(){f[0]=0;f[1]=1;for (int i = 2; i < MAX;i++){f[i]=(f[i-1]+f[i-2])%mod;}
}
ll quick(ll a,ll b,ll c){ll ans=1;while(b){if(b&1) ans=(ans*a)%c;b>>=1;a=(a*a)%c;}return ans;
}
ll solve(ll n,ll m,ll mod,ll p){//f[i]^m%mod int k=n%p;ll ans=0;for (int i = 1; i <= p;i++){if(i<=k) ans+=quick(f[i],m,mod)*(n/p+1);else ans+=quick(f[i],m,mod)*(n/p);ans%=mod;}return ans;
}
ll a[3],b[3];//x%a[i]=b[i];
int main(){init();ll n,m;scanf("%lld%lld",&n,&m);a[0]=mod1;a[1]=mod2;b[0]=solve(n,m,mod1,p1);b[1]=solve(n,m,mod2,p2);ll ans=China(2,b,a);printf("%lld\n",(ans%mod+mod)%mod);return 0;
}

题解：显然斐波那契数列的m次方前缀和依然是线性递推，考虑exBM解决，套模板就可以啦，这个模板在我的电脑上编译错误，交上能过~~~我的编译器太喽了~~

上代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int mod=1e9;
struct LinearRecurrence {using vec = std::vector<LL>;static void extand(vec &a, size_t d, LL value = 0) {if (d <= a.size()) return;a.resize(d, value);}static void exgcd(LL a, LL b, LL &g, LL &x, LL &y) {if (!b) x = 1, y = 0, g = a;else {exgcd(b, a % b, g, y, x);y -= x * (a / b);}}static LL crt(const vec &c, const vec &m) {int n = c.size();LL M = 1, ans = 0;for (int i = 0; i < n; ++i) M *= m[i];for (int i = 0; i < n; ++i) {LL x, y, g, tm = M / m[i];exgcd(tm, m[i], g, x, y);ans = (ans + tm * x * c[i] % M) % M;}return (ans + M) % M;}static vec ReedsSloane(const vec &s, LL mod) {auto inverse = [](LL a, LL m) {LL d, x, y;exgcd(a, m, d, x, y);return d == 1 ? (x % m + m) % m : -1;};auto L = [](const vec &a, const vec &b) {int da = (a.size() > 1 || (a.size() == 1 && a[0])) ? a.size() - 1 : -1000;int db = (b.size() > 1 || (b.size() == 1 && b[0])) ? b.size() - 1 : -1000;return std::max(da, db + 1);};auto prime_power = [&](const vec &s, LL mod, LL p, LL e) {// linear feedback shift register mod p^e, p is primestd::vector<vec> a(e), b(e), an(e), bn(e), ao(e), bo(e);vec t(e), u(e), r(e), to(e, 1), uo(e), pw(e + 1);pw[0] = 1;for (int i = pw[0] = 1; i <= e; ++i) pw[i] = pw[i - 1] * p;for (LL i = 0; i < e; ++i) {a[i] = {pw[i]}, an[i] = {pw[i]};b[i] = {0}, bn[i] = {s[0] * pw[i] % mod};t[i] = s[0] * pw[i] % mod;if (t[i] == 0)t[i] = 1, u[i] = e;elsefor (u[i] = 0; t[i] % p == 0; t[i] /= p, ++u[i]);}for (size_t k = 1; k < s.size(); ++k) {for (int g = 0; g < e; ++g)if (L(an[g], bn[g]) > L(a[g], b[g])) {ao[g] = a[e - 1 - u[g]];bo[g] = b[e - 1 - u[g]];to[g] = t[e - 1 - u[g]];uo[g] = u[e - 1 - u[g]];r[g] = k - 1;}a = an, b = bn;for (int o = 0; o < e; ++o) {LL d = 0;for (size_t i = 0; i < a[o].size() && i <= k; ++i) {d = (d + a[o][i] * s[k - i]) % mod;}if (d == 0) {t[o] = 1, u[o] = e;} else {for (u[o] = 0, t[o] = d; t[o] % p == 0; t[o] /= p, ++u[o]);int g = e - 1 - u[o];if (L(a[g], b[g]) == 0) {extand(bn[o], k + 1);bn[o][k] = (bn[o][k] + d) % mod;} else {LL coef =t[o] * inverse(to[g], mod) % mod * pw[u[o] - uo[g]] % mod;int m = k - r[g];extand(an[o], ao[g].size() + m);extand(bn[o], bo[g].size() + m);for (size_t i = 0; i < ao[g].size(); ++i) {an[o][i + m] -= coef * ao[g][i] % mod;if (an[o][i + m] < 0) an[o][i + m] += mod;}while (an[o].size() && an[o].back() == 0) an[o].pop_back();for (size_t i = 0; i < bo[g].size(); ++i) {bn[o][i + m] -= coef * bo[g][i] % mod;if (bn[o][i + m] < 0) bn[o][i + m] -= mod;}while (bn[o].size() && bn[o].back() == 0) bn[o].pop_back();}}}}return std::make_pair(an[0], bn[0]);};std::vector<std::tuple<LL, LL, int>> fac;for (LL i = 2; i * i <= mod; ++i)if (mod % i == 0) {LL cnt = 0, pw = 1;while (mod % i == 0) mod /= i, ++cnt, pw *= i;fac.emplace_back(pw, i, cnt);}if (mod > 1) fac.emplace_back(mod, mod, 1);std::vector<vec> as;size_t n = 0;for (auto &&x: fac) {LL mod, p, e;vec a, b;std::tie(mod, p, e) = x;auto ss = s;for (auto &&x: ss) x %= mod;std::tie(a, b) = prime_power(ss, mod, p, e);as.emplace_back(a);n = std::max(n, a.size());}vec a(n), c(as.size()), m(as.size());for (size_t i = 0; i < n; ++i) {for (size_t j = 0; j < as.size(); ++j) {m[j] = std::get<0>(fac[j]);c[j] = i < as[j].size() ? as[j][i] : 0;}a[i] = crt(c, m);}return a;}LinearRecurrence(const vec &s, const vec &c, LL mod) :init(s), trans(c), mod(mod), m(s.size()) {}LinearRecurrence(const vec &s, LL mod, bool is_prime = true) : mod(mod) {vec A;A = ReedsSloane(s, mod);if (A.empty()) A = {0};m = A.size() - 1;trans.resize(m);for (int i = 0; i < m; ++i) trans[i] = (mod - A[i + 1]) % mod;std::reverse(trans.begin(), trans.end());init = {s.begin(), s.begin() + m};}LL calc(LL n) {if (mod == 1) return 0;if (n < m) return init[n];vec v(m), u(m << 1);int msk = !!n;for (LL m = n; m > 1; m >>= 1) msk <<= 1;v[0] = 1 % mod;for (int x = 0; msk; msk >>= 1, x <<= 1) {std::fill_n(u.begin(), m * 2, 0);x |= !!(n & msk);if (x < m) u[x] = 1 % mod;else {// can be optimized by fft/nttfor (int i = 0; i < m; ++i)for (int j = 0, t = i + (x & 1); j < m; ++j, ++t)u[t] = (u[t] + v[i] * v[j]) % mod;for (int i = m * 2 - 1; i >= m; --i)for (int j = 0, t = i - m; j < m; ++j, ++t)u[t] = (u[t] + trans[j] * u[i]) % mod;}v = {u.begin(), u.begin() + m};}LL ret = 0;for (int i = 0; i < m; ++i) ret = (ret + v[i] * init[i]) % mod;return ret;}vec init, trans;LL mod;int m;
};
int quick_pow(int a,int b)
{int ans=1;while(b){if(b&1) ans=1LL*a*ans%mod;a=1LL*a*a%mod;b>>=1;}return ans;
}
int main()
{int n,m;scanf("%d%d",&n,&m);vector<LL> f;f.push_back(0),f.push_back(1);for(int i=2;i<=2000;i++){f.push_back((f[i-1]+f[i-2])%mod);}for(int i=1;i<=2000;i++) f[i]=quick_pow(f[i],m);for(int i=1;i<=2000;i++) f[i]=(f[i]+f[i-1])%mod;LinearRecurrence qw(f,mod,false);printf("%lld\n",qw.calc(n));return 0;
}