bzoj 4332 FFT型的快速幂（需要强有力的推导公式能力）

 有n个小朋友，m颗糖，你要把所有糖果分给这些小朋友。

规则第 i 个小朋友没有糖果，那么他之后的小朋友都没有糖果。、
如果一个小朋友分到了 xx 个糖果，那么的他的权值是 f(x) = ox^2 + sx + u

没有分到糖果的小朋友的权值是 1 

每种方案的权值是各个小朋友权值的乘积

求出所有方案的权值和 

 

设g(i,j)表示前i个小朋友分j个糖果的权值乘积和

很容易得到一个式子

这个显然是一个卷积用FFT就可以处理

但是问题来了 我们如何得到ans呢

　　

n<=1e8  朴素的算法不太行

要想办法优化一下

然后我就被卡死了  去看了网上的各种题解

顺便拔下来了一个封装性很好的FFT板子

看推导过程的点这个   想要迅速理解的看这个

然后你就看的懂下面这个式子了

然后就可以套快速幂了

#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <algorithm>
#include <set>
#include <iostream>
#include <map>
#include <stack>
#include <string>
#include <vector>
#define  eps 1e-9
#define  fi first
#define  se second
#define  rtl   rt<<1
#define  rtr   rt<<1|1
#define  bug         printf("******
")
#define  mem(a,b)    memset(a,b,sizeof(a))
#define  name2str(x) #x
#define  fuck(x)     cout<<#x" = "<<x<<endl
#define  f(a)        a*a
#define  sf(n)       scanf("%d", &n)
#define  sff(a,b)    scanf("%d %d", &a, &b)
#define  sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define  sffff(a,b,c,d) scanf("%d %d %d %d", &a, &b, &c, &d)
#define  pf          printf
#define  FRE(i,a,b)  for(i = a; i <= b; i++)
#define  FREE(i,a,b) for(i = a; i >= b; i--)
#define  FRL(i,a,b)  for(i = a; i < b; i++)+
#define  FRLL(i,a,b) for(i = a; i > b; i--)
#define  FIN         freopen("data.txt","r",stdin)
#define  gcd(a,b)    __gcd(a,b)
#define  lowbit(x)   x&-x
#define rep(i,a,b) for(int i=a;i<b;++i)
#define per(i,a,b) for(int i=a-1;i>=b;--i)
using namespace std;
typedef long long  LL;
typedef unsigned long long ULL;
int modu;
namespace FFT {
const double pi = acos ( -1.0 );
struct cpx {
    double a, b;
    cpx ( double a = 0, double b = 0 ) : a ( a ), b ( b ) {}
    inline void init() {
        a = 0, b = 0;
    }
    inline cpx operator + ( const cpx &ano ) const {
        return cpx ( a + ano.a, b + ano.b );
    }
    inline cpx operator - ( const cpx &ano ) const {
        return cpx ( a - ano.a, b - ano.b );
    }
    inline cpx operator * ( const cpx &ano ) const {
        return cpx ( a * ano.a - b * ano.b, b * ano.a + a * ano.b );
    }
};
typedef cpx C;
typedef vector<C> vc;
typedef vector<int> vi;

vc a, b;

void DFT ( vc &a, int oper = 1 ) {
    int n = a.size();
    for ( int i = 0, j = 0; i < n; ++i ) {
        if ( i > j ) swap ( a[i], a[j] );
        for ( int l = n >> 1; ( j ^= l ) < l; l >>= 1 );
    }
    for ( int l = 1, ll = 2; l < n; l <<= 1, ll <<= 1 ) {
        double x = oper * pi / l;
        C omega = 1, omegan ( cos ( x ), sin ( x ) );
        for ( int k = 0; k < l; ++k, omega = omega * omegan ) {
            for ( int st = k; st < n; st += ll ) {
                C tmp = omega * a[st + l];
                a[st + l] = a[st] - tmp;
                a[st] = a[st] + tmp;
            }
        }
    }
    if ( oper == -1 ) for ( int i = 0; i < n; ++i ) a[i].a /= n;
}

vi& operator * ( const vi &v1, const vi &v2 ) {
    int s = 1, ss = ( int ) v1.size() + ( int ) v2.size();
    while ( s < ss ) s <<= 1;
    a.resize ( s ), b.resize ( s );
    for ( int i = 0; i < s; ++i ) a[i].init(), b[i].init();
    for (  int i = 0; i < v1.size(); ++i ) a[i] = v1[i];
    for (  int i = 0; i < v2.size(); ++i ) b[i] = v2[i];
    DFT ( a ), DFT ( b );
    for ( int i = 0; i < s; ++i ) a[i] = a[i] * b[i];
    DFT ( a, -1 );
    static vi res;
    res.resize ( v1.size() );
    for (  int i = 0; i < v1.size(); ++i ) res[i] = ( a[i].a + 0.5 ), res[i] %= modu ;
    return res;
}

void operator *= ( vi &v1, const vi &v2 ) {
    int s = 1, ss = ( int ) v1.size() + ( int ) v2.size();
    while ( s < ss ) s <<= 1;
    a.resize ( s ), b.resize ( s );
    for ( int i = 0; i < s; ++i ) a[i].init(), b[i].init();
    for (  int i = 0; i < v1.size(); ++i ) a[i] = v1[i];
    for (  int i = 0; i < v2.size(); ++i ) b[i] = v2[i];
    DFT ( a ), DFT ( b );
    for ( int i = 0; i < s; ++i ) a[i] = a[i] * b[i];
    DFT ( a, -1 );
    for (  int i = 0; i < v1.size(); ++i ) v1[i] =  ( a[i].a + 0.5 ), v1[i] %= modu ;
}

void operator += ( vi &v1, const vi &v2 ) {
    for (  int i = 0; i < v1.size(); ++i ) v1[i] = ( v1[i] + v2[i] + modu ) % modu;
}
}

using namespace FFT;
int m, p, n, o, s, u;
vi f;
vi expmod ( const vi&v, int b ) {
    vi res ( v.size(), 0 ), tmp = v;
    res[0] = 1;
    while ( b ) {
        if ( b & 1 ) res *= tmp;
        tmp *= tmp;
        b = b >> 1;
    }
    return res;
}
vi& solve ( int n ) {
    static vi res, ghalf;
    if ( n == 1 ) return res = ghalf = f;
    solve ( n / 2 );
    res += res * ghalf;
    ghalf *= ghalf;
    if ( n & 1 ) res += expmod ( f, n ), ghalf *= f;
    return res;
}
int main() {
    //FIN;
    sff ( m, modu );
    sffff ( n, o, s, u );
    f = vi ( m + 1, 0 );
    for ( int i = 1 ; i < m + 1 ; i++ ) f[i] = ( 1LL * o * i * i + s * i + u ) % modu;
    vi &res = solve ( min ( n, m ) );
    printf ( "%d
", res[m] );
    return 0;
}