【AtCoder】AGC005F - Many Easy Problems

题解

我们把一个点的贡献转化为一条边的贡献，因为边的数量是点的数量-1，最后再加上选点方案数\(\binom{n}{k}\)即可

一条边的贡献是\(\binom{n}{k} - \binom{a}{k} - \binom{n - a}{k}\)就是在n个点里选k个点，去掉不合法的情况也就是k个点都在去掉这条边的两个子树里

然后我们要统计的就是\(\binom{a}{k} + \binom{n - a}{k}\)

这个可以转化成\(ans_{k} = \sum_{i = 1}^{n} b_{i} \binom{i}{k}\)

\(ans_{k} = \frac{1}{k!} \sum_{i = 1}^{n} b_{i} i! \frac{1}{(i - k)!}\)
这个数组是可以卷积的，只要把一个倒过来就行
设\(f(i) = \frac{1}{(n - i)!}\)
\(g(i) = b_{i}i!\)
\(h = g * f\)
\(ans_{k} = h(N + k)\)

代码

#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <map>
//#define ivorysi
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define mo 974711
#define RG register
#define MAXN 200005
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
    while(c < '0' || c > '9') {
    if(c == '-') f = -1;
    c = getchar();
    }
    while(c >= '0' && c <= '9') {
    res = res * 10 + c - '0';
    c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {putchar('-');x = -x;}
    if(x >= 10) {
    out(x / 10);
    }
    putchar('0' + x % 10);
}
struct node {
    int to,next;
}E[MAXN * 2];
int head[MAXN],sumE;
void add(int u,int v) {
    E[++sumE].to = v;
    E[sumE].next = head[u];
    head[u] = sumE;
}
const int MOD = 924844033,L = (1 << 19),G = 5;
int N,f[L + 5],fac[MAXN],invfac[MAXN],b[L + 5],W[L + 5];
int mul(int a,int b) {return 1LL * a * b % MOD;}
int inc(int a,int b) {a = a + b;if(a >= MOD) a -= MOD;return a;}
int fpow(int x,int c) {
    int t = x,res = 1;
    while(c) {
    if(c & 1) res = mul(res,t);
    t = mul(t,t);
    c >>= 1;
    }
    return res;
}
int C(int n,int m) {
    if(n < m) return 0;
    return mul(mul(fac[n],invfac[m]),invfac[n - m]);
}
int dfs(int u,int fa) {
    int siz = 1;
    for(int i = head[u] ; i ; i = E[i].next) {
    int v = E[i].to;
    if(v != fa) {
        siz += dfs(v,u);
    }
    }
    if(fa != 0) {
    b[siz]++;
    b[N - siz]++;
    }
    return siz;
}
void NTT(int *a,int LEN,int on) {
    for(int i = 1 , j = LEN / 2 ; i < LEN - 1 ; ++i) {
    if(i < j) swap(a[i],a[j]);
    int k = LEN / 2;
    while(j >= k) {
        j -= k;
        k >>= 1;
    }
    j += k;
    }
    for(int h = 2 ; h <= LEN ; h <<= 1) {
    int wn = W[(L + on * L / h) % L];
    for(int k = 0 ; k < LEN ; k += h) {
        int w = 1;
        for(int j = k ; j < k + h / 2 ; ++j) {
        int64 u = a[j],t = 1LL * a[j + h / 2] * w;
        a[j] = (u + t) % MOD;
        a[j + h / 2] = (u - t + 1LL * MOD * MOD) % MOD;
        w = mul(w,wn);
        }
    }
    }
    if(on == -1) {
    int invL = fpow(LEN,MOD - 2);
    for(int i = 0 ; i < LEN ; ++i) a[i] = mul(a[i],invL);
    }
}
void Solve() {
    read(N);
    int u,v;
    for(int i = 1 ; i < N ; ++i) {
    read(u);read(v);add(u,v);add(v,u);
    }
    dfs(1,0);
    fac[0] = 1;
    for(int i = 1 ; i <= N ; ++i) fac[i] = mul(fac[i - 1],i);
    invfac[N] = fpow(fac[N],MOD - 2);
    for(int i = N - 1 ; i >= 0 ; --i) invfac[i] = mul(invfac[i + 1],i + 1);
    for(int i = 0 ; i <= N ; ++i) f[N - i] = invfac[i];
    for(int i = 1 ; i <= N ; ++i) b[i] = mul(b[i],fac[i]);
    W[0] = 1;W[1] = fpow(G,(MOD - 1) / L);
    for(int i = 2 ; i < L ; ++i) W[i] = mul(W[i - 1],W[1]);
    int t = 1;
    while(t <= 2 * N) t <<= 1;
    NTT(b,t,1);NTT(f,t,1);
    for(int i = 0 ; i < t ; ++i) f[i] = mul(f[i],b[i]);
    NTT(f,t,-1);
    for(int i = 1 ; i <= N ; ++i) {
    int ans = mul(f[i + N],invfac[i]);
    ans = inc(mul(N,C(N,i)),MOD - ans);
    out(ans);enter;
    }
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

巴特西

【AtCoder】AGC005F - Many Easy Problems

题解

代码

最新文章

热门文章