Codeforces 1279D

Description

思路

做法一：组合数学

操作分三步，前两步选孩子再选礼物，最后一步再选孩子。
因此求出两步内选到礼物i的概率，再乘上想要礼物i的孩子的占比，求和所有礼物的结果就是答案。

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
#define maxn 2000000
#define M 998244353

set<int> gi;
ll sum[maxn];
int cnt[maxn];


ll qpow(ll a, ll b) {
    ll res = 1;
    while(b) {
        if(b & 1) res = res * a % M;
        a = (a * a) % M;
        b = b >> 1;
    }
    return res;
}

ll ddiv(ll a, ll b) { // a/b
    return a * qpow(b, M - 2) % M;
}

 
int main() {
    ios::sync_with_stdio(false);
    int n;
    cin >> n;
    ll tmp = ddiv(1, n);
    for(int i = 1; i <= n; i++) {
        int k;
        cin >> k;
        ll ttmp = ddiv(1, k);
        for(int i = 0; i < k; i++) {
            int g;
            cin >> g;
            gi.insert(g);
            sum[g] = (sum[g] + (tmp * ttmp % M)) % M;
            cnt[g]++;
        }
    }
    ll ans = 0;
    for(auto g : gi) {
        ans = (ans + (sum[g] * ddiv(cnt[g], n)) % M) % M;
    }
    cout << ans ;
}

做法二：dp
dp[i]代表选了第i个孩子的概率。然后有四种情况分类讨论，详情见代码。
注意，对于这个解法，用了太多次的快速幂会超时，因此要保存一些快速幂的结果减少计算量。
显然，做法一更好

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
#define maxn 2000000
#define M 998244353
 
ll dp[maxn];
vector<int> ch[maxn];
vector<int> gi[maxn];
ll sum[maxn];
 
 
ll qpow(ll a, ll b) {
    ll res = 1;
    while(b) {
        if(b & 1) res = res * a % M;
        a = (a * a) % M;
        b = b >> 1;
    }
    return res;
}
 
ll ddiv(ll a, ll b) { // a/b
    return a * qpow(b, M - 2) % M;
}
 
inline ll pow2(ll a) {
    return (a * a) % M;
}
 
int main() {
    ios::sync_with_stdio(false);
    int n;
    cin >> n;
    for(int i = 1; i <= n; i++) {
        int k;
        cin >> k;
        while(k--) {
            int g;
            cin >> g;
            ch[i].push_back(g);
        }
    }
    dp[0] = 1;
    for(int i = 1; i <= n; i++) {
        ll a = pow2(ddiv(i - 1, i)) * dp[i - 1] % M ;
        ll b = pow2(ddiv(1, i));
        ll c = 0;
        ll d = 0;
        for(int g : ch[i]) {
            c = (c + ddiv(1, ch[i].size()) * ddiv(gi[g].size(), i - 1) % M) % M;
            d = (d + (ddiv(1, i - 1) * sum[g]) % M) % M;
        }
        for(int g : ch[i]) {
            gi[g].push_back(i);
            if(sum[g] == 0) sum[g] = ddiv(1, ch[i].size());
            else sum[g] = (sum[g] + ddiv(1, ch[i].size())) % M;
        }
        ll tmp =  (ddiv(1, i) * ddiv(i - 1, i) % M);
        d = tmp * d % M;
        c = tmp * c % M;
        dp[i] = (a + b + c + d) % M;
    }
    cout << dp[n];
}