1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cassert>
#include <iostream>
#include <algorithm>
#include <iterator>
#include <string>
#include <vector>
#include <queue>
#include <bitset>
#include <utility>
#include <stack>

using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
typedef vector<int> VI;
#define MP make_pair
#define FOR(v,p,k) for(int v=(p);v<=(k);++v)
#define FORD(v,p,k) for(int v=(p);v>=(k);--v)
#define REP(i,n) for(int i=0;i<(n);++i)
#define VAR(v,i) __typeof(i) v=(i)
#define FOREACH(i,c) for(VAR(i,(c).begin());i!=(c).end();++i)
#define PB push_back
#define ST first
#define ND second
#define SIZE(x) (int)x.size()
#define ALL(c) c.begin(),c.end()

#define ODD(x) ((x)%2)
#define EVEN(x) (!(ODD(x)))

constexpr int mod = 1000000007;

// based on: https://ideone.com/fkHhL8 and https://hdqdgqnpvfesazkq.quora.com/Solution-for-Brackets-Subsequences


vector<vector<int>> f;
vector<vector<int>> g;
int solution(const std::string &s) {
    f.clear();
    g.clear();
    f.push_back(vector<int>(s.size()+1,0));
    f.push_back(vector<int>(s.size()+1,0));
    g.push_back(vector<int>(s.size()+1,0));
    g.push_back(vector<int>(s.size()+1,0));
    // f[0][x] denotes the number of distinct balanced
    // subsequences ending with '(' and having extra 
    // (unmatched) open parenthesis equal to 'x'.
    for(int i=0;i<s.size();++i){
        int x=(s[i]=='L')?1:-1;
        int y=(s[i]=='L')?0:1;
        for(int j=s.size()-1;j>=0;--j){
            if (x==-1 and j==0) continue;
            f[y][j+x]=(g[0][j]+g[1][j]);
        }
        if (x==1) f[y][1]+=1;
        f[1-y]=g[1-y];
        g = f;
        f[0].assign(s.size()+1,0);
        f[1].assign(s.size()+1,0);
    }
    return g[1][0];
}

int main() {
    f.reserve(2000);
    g.reserve(2000);
    std::ios::sync_with_stdio(false);
    int t;
    cin >> t;
    vector<string> seqs(t);
    REP(i, t) {
        cin >> seqs[i];
    }

    REP(i, t) {
        REP(j, t) {
            f.clear();
            g.clear();
	    cout << solution(seqs[i]+seqs[j]);
            cout << ((j < t-1)? " " : "\n");
        }
    }

}