luogu P5644 [PKUWC2018]猎人杀

https://www.luogu.com.cn/problem/P5644

好题啊好题！！
首先考虑鞭尸，即可以打死去的人，但是还要继续开枪，直到打到活人

设$tot=\sum w_i，则p_i=\frac{w_i}{tot}$

设$F(S)表示恰好集合S中的人在1之后被打死的概率$
设$G(S)表示钦定集合S中的人在1之后被打死的概率$
然后考虑子集反演可得
$$ANS=F(\varnothing )=\sum (-1{)}^{|S|}G(S)$$
问题变为如何计算G
由于可以鞭尸，所以变成无限开枪，每次可以打到集合外的人，直到打到1为止
$$G(S)=\sum _{i=0}^{\infty }(\frac{tot-w_1-sum(S)}{tot}{)}^i\frac{w_1}{tot}$$
前面那个显然是收敛的，用等比数列求和可得
$$G(S)=\frac{w_1}{w_1+sum(S)}$$
带入回去可得
$$ANS=\sum (-1{)}^{|S|}\frac{w_1}{w_1+sum(S)}$$
枚举子集显然起飞，不太行
考虑枚举$sum(S)$，把前面的容斥系数求个和即可
设$C(i)=\sum _{sum(S)=i}(-1{)}^{|S|}$
$$ANS=\sum _{i=0}^{tot}C(i)\frac{w_1}{w_1+i}$$

$C显然就是=\prod _{i=2}^n(1-x^{w_i})$

分治FFT即可

code:

#include<bits/stdc++.h>
#define V vector<int>
#define N 800050
#define mod 998244353
using namespace std;
int add(int x, int y) { x += y;
    if(x >= mod) x -= mod;
    return x;
}
int sub(int x, int y) { x -= y;
    if(x < 0) x += mod;
    return x;
}
int mul(int x, int y) {
    return 1ll * x * y % mod;
}
int qpow(int x, int y) {
    int ret = 1;
    for(; y; y >>= 1, x = mul(x, x)) if(y & 1) ret = mul(ret, x);
    return ret;
}
const int G = 3;
const int G_inv = qpow(3, mod - 2);
int rev[N], w[N];
void ntt(V &a, int n, int o) {
    while(a.size() < n) a.push_back(0);
    for(int i = 1; i < n; i ++) if(i > rev[i]) swap(a[i], a[rev[i]]);
    for(int len = 2; len <= n; len <<= 1) {
        int w0 = qpow(o == 1? G : G_inv, (mod - 1) / len);
        for(int j = 0; j < n; j += len) {
            int wn = 1;
            for(int k = j; k < j + (len >> 1); k ++, wn = mul(wn, w0)) {
                int X = a[k], Y = mul(wn, a[k + (len >> 1)]);
                a[k] = add(X, Y), a[k + (len >> 1)] = sub(X, Y);
            }
        }
    }
    int ninv = qpow(n, mod - 2);
    if(o == -1)
        for(int i = 0; i < n; i ++) a[i] = mul(a[i], ninv);
}
void prt(V C) {
    printf("%d\n", (int)C.size());
    for(int i = 0; i < C.size(); i ++) printf("%d ", C[i]); printf("\n");
}
V cdq(int l, int r) {
    if(l == r) {
        V a; a.clear();
        a.push_back(1);
        for(int i = 1; i < w[l]; i ++) a.push_back(0);
        a.push_back(mod - 1);
        return a;
    }
    int mid = (l + r) >> 1;
    V a = cdq(l, mid), b = cdq(mid + 1, r);
//    prt(a), prt(b);
    int n = a.size() - 1, m = b.size() - 1, len = 1;
    for(; len <= n + m; ) len <<= 1;
    for(int i = 1; i < len; i ++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) * (len >> 1));
    ntt(a, len, 1), ntt(b, len, 1);
  //  prt(a), prt(b);
    for(int i = 0; i < len; i ++) a[i] = mul(a[i], b[i]);
    ntt(a, len, - 1);
    while(a.size() > n + m + 1) a.pop_back();
//    prt(a);
 //   printf("----%d %d--\n", l, r);
    return a;
}
int n;
int main() {
    scanf("%d", &n); int tot = 0;
    for(int i = 1; i <= n; i ++) scanf("%d", &w[i]), tot += w[i];
    tot -= w[1];
    V C = cdq(2, n); int ans = 0;
  //  prt(C);
    //printf("** %d\n", tot);
    for(int i = 0; i <= tot; i ++) ans = add(ans, mul(C[i], mul(w[1], qpow(add(w[1], i), mod - 2))));
    printf("%d", ans);
    return 0;
}