Niuke Xiaobai monthly race 52 E group logarithmic sum (inclusion exclusion theorem + dichotomy)

Sign in — major IT Written interview preparation platform _ Cattle from Niuke.com is the magic weapon of Internet Job Hunting ,C++、Java、 front end 、 product 、 Operational skills learning / Test preparation / Job search question bank , Online Baidu Alibaba Tencent Netease and other Internet famous enterprises written interview simulation test practice , Discuss the classic test questions with Niu Ren , Improve your technical ability in an all-round way https://ac.nowcoder.com/acm/contest/11229/E You can put all the numbers directly into a sequence , Then enumerate each number , Through inclusion and exclusion, we can know the number that satisfies the condition in the total sequence - The number in the child , You can get the combination of the numbers , And then divide by 2 that will do

#include <cstdio>
#include <cstring>
#include <iostream>
#include <string>
#include <cmath>
#include <algorithm>
#include <vector>
#include <map>
#include <queue>
#define lowbit(x) ((x)&(-x))
using namespace std;
using ll = long long;
using ull = unsigned long long;
using P = pair<int,int>;
const int MAXN=1e6+5;
const int INF=0x3f3f3f3f;
const ll NNF=0x3f3f3f3f3f3f3f3f;


const int N=1e5;
int n,k;
const int mod=998244353;
vector<int> s;
vector<int> rec[MAXN];
void solve(){
	scanf("%d %d",&n,&k);
	int t,tmp;
	for(int i=1;i<=n;i++){
		scanf("%d",&t);
		for(int j=1;j<=t;j++){
			scanf("%d",&tmp);
			rec[i].push_back(tmp);
			s.push_back(tmp);
		}
		sort(rec[i].begin(),rec[i].end());
	}
	ll ans=0;
	sort(s.begin(),s.end());
	for(int i=1;i<=n;i++){
		int res=rec[i].size();
		for(int j=0;j<res;j++){
			int tmp1=rec[i].end()-lower_bound(rec[i].begin(),rec[i].end(),k-rec[i][j]);
			int tmp2=s.end()-lower_bound(s.begin(),s.end(),k-rec[i][j]);
			ans+=tmp2-tmp1;
		}
	}
	printf("%lld",ans/2%mod);
}


int main(){
	solve();
	return 0;
}