題目

簡述：

有一段長度為n的貝殼，將其劃分為若幹段，給出劃分為每種長度的方案數，問有多少種劃分方案

題解

設\(f[i]\)表示長度為\(i\)時的方案數
不難得dp方程：
\[f[i] = \sum\limits_{j=0}^{i} a[j] * f[i - j]\]

考慮轉移
直接轉移是\(O(n^2)\)的
如何優化？
容易發現這個轉移方程非常特別，是一個卷積的形式
考慮fft

分治fft

分治fft解決的就是這樣一個轉移方程的快速計算的問題
\[f[i] = \sum\limits_{j=0}^{i} a[j] * f[i - j]\]

考慮cdq分治的模式：
我們先處理左半區間，然後用左半區間的\(f[i]\)

我們便解決了這道題

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
 

#define LL long long int
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define BUG(s,n) for (int i = 1; i <= (n); i++) cout<<s[i]<<‘ ‘; puts("");
using namespace std;
const int maxn = 400005,maxm = 100005,INF = 1000000000,P = 313;
inline int
 
 read(){
    int out = 0,flag = 1; char c = getchar();
    while (c < 48 || c > 57){if (c == ‘-‘) flag = -1; c = getchar();}
    while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
    return out * flag;
}
const double pi = acos(-1);
struct E{
    double r,i;
    E(){}
    E(double a,double b):r(a),i(b){}
    E operator =(const int& b){
        r = b; i = 0;
        return *this;
    }
};
inline E operator +(const E& a,const E& b){
    return E(a.r + b.r,a.i + b.i);
}
inline E operator -(const E& a,const E& b){
    return E(a.r - b.r,a.i - b.i);
}
inline E operator *(const E& a,const E& b){
    return E(a.r * b.r - a.i * b.i,a.r * b.i + b.r * a.i);
}
inline E operator *=(E& a,const E& b){
    return (a = a * b);
}
inline E operator /(E& a,const double& b){
    return E(a.r / b,a.i / b);
}
inline E operator /=(E& a,const double& b){
    return (a = a / b);
}
int n,m,L,R[maxn];
void fft(E* a,int f){
    for (int i = 0; i < n; i++) if (i < R[i]) swap(a[i],a[R[i]]);
    for (int i = 1; i < n; i <<= 1){
        E wn(cos(pi / i),f * sin(pi / i));
        for (int j = 0; j < n; j += (i << 1)){
            E w(1,0);
            for (int k = 0; k < i; k++,w *= wn){
                E x = a[j + k],y = w * a[j + k + i];
                a[j + k] = x + y; a[j + k + i] = x - y;
            }
        }
    }
    if (f == -1) for (int i = 0; i < n; i++) a[i] /= n;
}
E A[maxn],B[maxn];
int N,a[maxn],f[maxn];
void solve(int l,int r){
    if (l == r){
        f[l] = (f[l] + a[l]) % P;
        return;
    }
    int mid = l + r >> 1;
    solve(l,mid);
    n = mid - l + 1;
    for (int i = 0; i < n; i++) A[i] = f[l + i];
    m = r - l + 1;
    for (int i = 0; i < m; i++) B[i] = a[i + 1];
    m = n + m; L = 0;
    for (n = 1; n <= m; n <<= 1) L++;
    for (int i = 0; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
    fft(A,1); fft(B,1);
    for (int i = 0; i < n; i++) A[i] *= B[i];
    fft(A,-1);
    for (int i = mid + 1; i <= r; i++){
        f[i] = (f[i] + (int)floor(A[i - l - 1].r + 0.5) % P) % P;
    }
    for (int i = 0; i < n; i++) A[i] = B[i] = 0;
    solve(mid + 1,r);
}
int main(){
    while ((~scanf("%d",&N)) && N){
        for (int i = 1; i <= N; i++){
            a[i] = read() % P;
            f[i] = 0;
        }
        solve(1,N);
        printf("%d\n",f[N]);
    }
    return 0;
}

hdu5730 Shell Necklace 【分治fft】