题目链接：https://codeforc.es/contest/1279/problem/D
Santa Claus has received letters from n different kids throughout this year. Of course, each kid wants to get some presents from Santa: in particular, the i-th kid asked Santa to give them one of ki different items as a present. Some items could have been asked by multiple kids.
Santa is really busy, so he wants the New Year Bot to choose the presents for all children. Unfortunately, the Bot’s algorithm of choosing presents is bugged. To choose a present for some kid, the Bot does the following:
choose one kid x equiprobably among all n kids;
choose some item y equiprobably among all kx items kid x wants;
choose a kid z who will receive the present equipropably among all n kids (this choice is independent of choosing x and y); the resulting triple (x,y,z) is called the decision of the Bot.
If kid z listed item y as an item they want to receive, then the decision valid. Otherwise, the Bot’s choice is invalid.
Santa is aware of the bug, but he can’t estimate if this bug is really severe. To do so, he wants to know the probability that one decision generated according to the aforementioned algorithm is valid. Can you help him?
Input
The first line contains one integer n (1≤n≤106) — the number of kids who wrote their letters to Santa.
Then n lines follow, the i-th of them contains a list of items wanted by the i-th kid in the following format: ki ai,1 ai,2 … ai,ki (1≤ki,ai,j≤106), where ki is the number of items wanted by the i-th kid, and ai,j are the items themselves. No item is contained in the same list more than once.
It is guaranteed that ∑i=1nki≤106.
Output
Print the probatility that the Bot produces a valid decision as follows:
Let this probability be represented as an irreducible fraction xy. You have to print x?y?1mod998244353, where y?1 is the inverse element of y modulo 998244353 (such integer that y?y?1 has remainder 1 modulo 998244353).
Examples
input
2
2 2 1
1 1
output
124780545
input
5
2 1 2
2 3 1
3 2 4 3
2 1 4
3 4 3 2
output
798595483
题意
有 n 个人，第 i 个人有 ki 个想要的礼物。
随机选一个人 x ，再随机选其中一个他喜欢的礼物 y ，再随机选一个人 z ，问 y 这个礼物也是 z 喜欢的礼物的概率。
分析
随机选一个人： 1 / n
随机选一个他喜欢的礼物： 1 / ki
随机选一个人他喜欢的礼物包含 y 的概率： (喜欢 y 的人数）/ n
每个礼物的这个积的和即答案。
代码
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;ll n;
ll k;
const ll mod = 998244353;
map<ll,ll> sum;struct node{
ll id;ll p;
};
vector<node> s;ll qm(ll x, ll y)
{
ll res = 1;while(y){
if(y & 1) res = (res * x) % mod;x = (x * x) % mod;y >>= 1;}return res;
}int main()
{
sum.clear();scanf("%lld",&n);ll tmp = qm(n, mod - 2);for(int i=1;i<=n;i++){
scanf("%lld",&k);ll temp = qm(k, mod - 2);ll t = tmp * temp % mod;for(int j=1;j<=k;j++){
ll x;scanf("%lld",&x);sum[x]++;s.push_back({
x, t});}}int len = s.size();ll ans = 0;for(int i=0;i<len;i++){
ans = (ans + sum[s[i].id] * qm(n, mod - 2) % mod * s[i].p % mod) % mod;}printf("%lld\n",ans);return 0;
}