Counting (KMP + Thinking)

topic link

https://ac.nowcoder.com/acm/contest/27589/B

topic

ideas

We think we can know that for a $f(s,t)$ only$The\; length\; of\; t$ is less than or equal to$s$ can at least match successfully, then we will find that we only need to process the string with the smallest length, because other strings will definitely have an$f(s,t)$ is 0, because a long string cannot be called a substring of a short string, but if it is only a string with the smallest length,nnFind kmp kmp$for\; n$ strings$If\; kmp$ matches, it will also time out. In case all strings have the same length, we further observe and find that if there is a pair of differences between any two strings with the shortest length, it will contribute to the cumulative multiplication a$0$ , otherwise all short strings have theanssame value, we only need to loop it oncekmp, and for other strings, we directly output$0$ is fine

code

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define mod 998244353
#define endl "\n"
#define PII pair<int,int>
#define INF 0x3f3f3f3f

const int N = 2e6+10;

int n;

struct KMP{
    
    
	int nextt[N];
	void get_next(string &S){
    
    
		int n = S.size(),i = 0,j = -1;
		nextt[0] = -1;
		while(i < n) {
    
    
			if(j == -1 || S[i] == S[j])
				nextt[++i] = ++j;
			else 
				j = nextt[j];
		}
	}
	//S是匹配串、T是模板串
	int get_times(string &S,string &T){
    
    //在S中查找T出现了多少次
		int cnt = 0,i = 0,j = 0;
		int len1 = S.size();
		int len2 = T.size();
		while(i < len1) {
    
    
			if(j == -1 || S[i] == T[j]) 
				i++,j++;
			else 
				j = nextt[j];
			if(j == len2) 
				j = nextt[j],cnt++;
		}
		return cnt;
	}
}kmp;

string ch[N];
int lench[N];

int main()
{
    
    
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin>>n;
	int min_len = INF;
	for(int i = 1;i <= n; ++i) {
    
    
		cin>>ch[i];
		lench[i] = ch[i].size();
		min_len = min(lench[i],min_len);
	}
	vector<string> Vec;
	for(int i = 1;i <= n; ++i) {
    
    
		if(lench[i] == min_len) 
			Vec.push_back(ch[i]);
	}
	bool fg = true;
	int l = Vec.size();
	for(int i = 0;i < l - 1; ++i) {
    
    
		if(Vec[i] != Vec[i + 1]) {
    
    
			fg = false;
			break;
		}
	}
	if(fg) {
    
    
		kmp.get_next(Vec[0]);
		ll ans = 1LL;
		for(int i = 1;i <= n; ++i) {
    
    
			ans = ans * (ll)kmp.get_times(ch[i],Vec[0]) % mod;
		}
		for(int i = 1;i <= n; ++i) {
    
    
			if(lench[i] == min_len) cout<<ans<<endl;
			else cout<<0<<endl;
		}
	} else {
    
    
		for(int i = 1;i <= n; ++i)
			cout<<0<<endl;
	}
	
	return 0;
}

/*
2
AA
AAA

2
0
*/