洛谷P4233 射命丸文的笔记【多项式求逆】

最新推荐文章于 2020-12-30 12:23:25 发布

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最新推荐文章于 2020-12-30 12:23:25 发布

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原文链接：http://www.cnblogs.com/Mychael/p/9282436.html

本文针对洛谷P4233题目提供了一种解决方案，通过计算哈密顿回路和强联通竞赛图的数量来解决。利用生成函数和多项式求逆的方法实现了O(nlogn)的时间复杂度。

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题目链接

洛谷P4233

题解

我们只需求出总的哈密顿回路个数和总的强联通竞赛图个数

对于每条哈密顿回路，我们统计其贡献
一条哈密顿回路就是一个圆排列，有\(\frac{n!}{n}\)种，剩余边随便连
所以总的贡献为
\[(n - 1)!2^{{n \choose 2} - n}\]

我们只需求出总的强联通竞赛图的个数
设\(g[n]\)表示\(n\)个点竞赛图个数，\(f[n]\)表示强联通竞赛图个数
那么有
\[g[n] = \sum\limits_{i = 1}^{n}{n \choose i}f[i]g[n - i]\]
即
\[\frac{g[n]}{n!} = \sum\limits_{i = 1}^{n}\frac{f[i]}{i!}\frac{g[n - i]}{(n - i)!}\]
设\(G(x)\)和\(F(x)\)分别为其指数型生成函数
那么有
\[G(x) = F(x)G(x) + 1\]
即
\[F(x) = \frac{G(x) - 1}{G(x)}\]
多项式求逆即可
复杂度\(O(nlogn)\)

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<cmath>
#include<map>
#define LL long long int
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define cls(s,v) memset(s,v,sizeof(s))
#define mp(a,b) make_pair<int,int>(a,b)
#define cp pair<int,int>
using namespace std;
const int maxn = 400005,maxm = 100005,INF = 0x3f3f3f3f;
inline int read(){
    int out = 0,flag = 1; char c = getchar();
    while (c < 48 || c > 57){if (c == '-') flag = 0; c = getchar();}
    while (c >= 48 && c <= 57){out = (out << 1) + (out << 3) + c - 48; c = getchar();}
    return flag ? out : -out;
}
const int P = 998244353,G = 3;
inline int qpow(int a,LL b){
    int re = 1;
    for (; b; b >>= 1,a = 1ll * a * a % P)
        if (b & 1) re = 1ll * re * a % P;
    return re;
}
inline LL C(int n){return 1ll * n * (n - 1) / 2;}
int R[maxn];
void NTT(int* a,int n,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){
        int gn = qpow(G,(P - 1) / (i << 1));
        for (int j = 0; j < n; j += (i << 1)){
            int g = 1,x,y;
            for (int k = 0; k < i; k++,g = 1ll * g * gn % P){
                x = a[j + k],y = 1ll * g * a[j + k + i] % P;
                a[j + k] = (x + y) % P,a[j + k + i] = ((x - y) % P + P) % P;
            }
        }
    }
    if (f == 1) return;
    int nv = qpow(n,P - 2); reverse(a + 1,a + n);
    for (int i = 0; i < n; i++) a[i] = 1ll * a[i] * nv % P;
}
int A[maxn],B[maxn],c[maxn],ans,N,fac[maxn],inv[maxn],fv[maxn];
void init(){
    fac[0] = fac[1] = inv[0] = inv[1] = fv[0] = fv[1] = 1;
    for (int i = 2; i <= N; i++){
        fac[i] = 1ll * fac[i - 1] * i % P;
        inv[i] = 1ll * (P - P / i) * inv[P % i] % P;
        fv[i] = 1ll * fv[i - 1] * inv[i] % P;
    }
}
void Inv(int deg,int* a,int* b){
    if (deg == 1){b[0] = qpow(a[0],P - 2); return;}
    Inv((deg + 1) >> 1,a,b);
    int L = 0,n = 1;
    while (n < (deg << 1)) n <<= 1,L++;
    for (int i = 1; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
    for (int i = 0; i < deg; i++) c[i] = a[i];
    for (int i = deg; i < n; i++) c[i] = 0;
    NTT(c,n,1); NTT(b,n,1);
    for (int i = 0; i < n; i++)
        b[i] = 1ll * ((2ll - 1ll * c[i] * b[i] % P) + P) % P * b[i] % P;
    NTT(b,n,-1);
    for (int i = deg; i < n; i++) b[i] = 0;
}
int main(){
    N = read(); init();
    for (int i = 0; i <= N; i++) A[i] = 1ll * qpow(2,C(i)) * fv[i] % P;
    Inv(N + 1,A,B);
    A[0] = 0;
    int n = 1,L = 0;
    while (n <= (N << 1)) n <<= 1,L++;
    for (int i = 1; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
    NTT(A,n,1); NTT(B,n,1);
    for (int i = 0; i < n; i++) A[i] = 1ll * A[i] * B[i] % P;
    NTT(A,n,-1);
    REP(i,N){
        if (i == 1) puts("1");
        else if (i == 2) puts("-1");
        else printf("%lld\n",1ll * fac[i - 1] * qpow(2,C(i) - i) % P * qpow(1ll * A[i] * fac[i] % P,P - 2) % P);
    }
    return 0;
}

转载于:https://www.cnblogs.com/Mychael/p/9282436.html