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stringlengths 52
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#include <bits/stdc++.h>
using namespace std;
pair<int, int> p[21];
int n, m;
int a[7];
bool d[6][6];
int ans(int ind) {
if (ind == n) {
int res = 0;
for (int i = 0; i < m; ++i) {
int f = min(a[p[i].first], a[p[i].second]),
s = max(a[p[i].first], a[p[i].second]);
if (f != 6 && s != 6 && !d[f][s]) {
++res;
d[f][s] = true;
}
}
for (int i = 0; i < 6; ++i) {
for (int j = 0; j < 6; ++j) {
d[i][j] = false;
}
}
return res;
}
int maxi = 0;
for (int i = 0; i < 7; ++i) {
a[ind] = i;
maxi = max(maxi, ans(ind + 1));
}
return maxi;
}
int main() {
cin >> n >> m;
int ai, bi;
for (int i = 0; i < m; ++i) {
cin >> ai >> bi;
p[i].first = ai - 1;
p[i].second = bi - 1;
}
for (int i = 0; i < 6; ++i) {
for (int j = 0; j < 6; ++j) {
d[i][j] = false;
}
}
cout << ans(0);
return 0;
}
| ### Prompt
Generate a Cpp solution to the following problem:
Anadi has a set of dominoes. Every domino has two parts, and each part contains some dots. For every a and b such that 1 ≤ a ≤ b ≤ 6, there is exactly one domino with a dots on one half and b dots on the other half. The set contains exactly 21 dominoes. Here is an exact illustration of his set:
<image>
Also, Anadi has an undirected graph without self-loops and multiple edges. He wants to choose some dominoes and place them on the edges of this graph. He can use at most one domino of each type. Each edge can fit at most one domino. It's not necessary to place a domino on each edge of the graph.
When placing a domino on an edge, he also chooses its direction. In other words, one half of any placed domino must be directed toward one of the endpoints of the edge and the other half must be directed toward the other endpoint. There's a catch: if there are multiple halves of dominoes directed toward the same vertex, each of these halves must contain the same number of dots.
How many dominoes at most can Anadi place on the edges of his graph?
Input
The first line contains two integers n and m (1 ≤ n ≤ 7, 0 ≤ m ≤ (n⋅(n-1))/(2)) — the number of vertices and the number of edges in the graph.
The next m lines contain two integers each. Integers in the i-th line are a_i and b_i (1 ≤ a, b ≤ n, a ≠ b) and denote that there is an edge which connects vertices a_i and b_i.
The graph might be disconnected. It's however guaranteed that the graph doesn't contain any self-loops, and that there is at most one edge between any pair of vertices.
Output
Output one integer which denotes the maximum number of dominoes which Anadi can place on the edges of the graph.
Examples
Input
4 4
1 2
2 3
3 4
4 1
Output
4
Input
7 0
Output
0
Input
3 1
1 3
Output
1
Input
7 21
1 2
1 3
1 4
1 5
1 6
1 7
2 3
2 4
2 5
2 6
2 7
3 4
3 5
3 6
3 7
4 5
4 6
4 7
5 6
5 7
6 7
Output
16
Note
Here is an illustration of Anadi's graph from the first sample test:
<image>
And here is one of the ways to place a domino on each of its edges:
<image>
Note that each vertex is faced by the halves of dominoes with the same number of dots. For instance, all halves directed toward vertex 1 have three dots.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
pair<int, int> p[21];
int n, m;
int a[7];
bool d[6][6];
int ans(int ind) {
if (ind == n) {
int res = 0;
for (int i = 0; i < m; ++i) {
int f = min(a[p[i].first], a[p[i].second]),
s = max(a[p[i].first], a[p[i].second]);
if (f != 6 && s != 6 && !d[f][s]) {
++res;
d[f][s] = true;
}
}
for (int i = 0; i < 6; ++i) {
for (int j = 0; j < 6; ++j) {
d[i][j] = false;
}
}
return res;
}
int maxi = 0;
for (int i = 0; i < 7; ++i) {
a[ind] = i;
maxi = max(maxi, ans(ind + 1));
}
return maxi;
}
int main() {
cin >> n >> m;
int ai, bi;
for (int i = 0; i < m; ++i) {
cin >> ai >> bi;
p[i].first = ai - 1;
p[i].second = bi - 1;
}
for (int i = 0; i < 6; ++i) {
for (int j = 0; j < 6; ++j) {
d[i][j] = false;
}
}
cout << ans(0);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
string s, t;
int dp[200 + 10][200 + 10][400 + 10];
tuple<int, int, int> fukugen[200 + 10][200 + 10][400 + 10];
const int INF = 1e9;
int main() {
cin >> s >> t;
for (int i = 0; i <= s.size(); i++)
for (int j = 0; j <= t.size(); j++)
for (int k = 0; k <= s.size() + t.size(); k++) dp[i][j][k] = INF;
dp[0][0][0] = 0;
for (int i = 0; i <= s.size(); i++) {
for (int j = 0; j <= t.size(); j++) {
for (int k = 0; k <= s.size() + t.size(); k++) {
if (dp[i][j][k] == INF) continue;
int ni = i + (i < s.size() && s[i] == '(');
int nj = j + (j < t.size() && t[j] == '(');
if (k + 1 <= s.size() + t.size() &&
dp[ni][nj][k + 1] > dp[i][j][k] + 1) {
dp[ni][nj][k + 1] = dp[i][j][k] + 1;
fukugen[ni][nj][k + 1] = make_tuple(i, j, k);
}
ni = i + (i < s.size() && s[i] == ')');
nj = j + (j < t.size() && t[j] == ')');
if (k - 1 >= 0 && dp[ni][nj][k - 1] > dp[i][j][k] + 1) {
dp[ni][nj][k - 1] = dp[i][j][k] + 1;
fukugen[ni][nj][k - 1] = make_tuple(i, j, k);
}
}
}
}
string ansstr, strplus;
int mans = INF, K = -1;
for (int i = 0; i <= s.size() + t.size(); i++) {
if (mans > dp[s.size()][t.size()][i] + i) {
mans = dp[s.size()][t.size()][i] + i;
K = i;
}
}
for (int i = 0; i < K; i++) strplus += ')';
int i = s.size(), j = t.size();
while (true) {
if (i == j && i == 0 && K == 0) {
break;
}
int ni = get<0>(fukugen[i][j][K]), nj = get<1>(fukugen[i][j][K]),
nk = get<2>(fukugen[i][j][K]);
ansstr += (K == nk + 1 ? '(' : ')');
i = ni, j = nj, K = nk;
}
reverse(ansstr.begin(), ansstr.end());
cout << ansstr << strplus << endl;
return 0;
}
| ### Prompt
Your challenge is to write a CPP solution to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
string s, t;
int dp[200 + 10][200 + 10][400 + 10];
tuple<int, int, int> fukugen[200 + 10][200 + 10][400 + 10];
const int INF = 1e9;
int main() {
cin >> s >> t;
for (int i = 0; i <= s.size(); i++)
for (int j = 0; j <= t.size(); j++)
for (int k = 0; k <= s.size() + t.size(); k++) dp[i][j][k] = INF;
dp[0][0][0] = 0;
for (int i = 0; i <= s.size(); i++) {
for (int j = 0; j <= t.size(); j++) {
for (int k = 0; k <= s.size() + t.size(); k++) {
if (dp[i][j][k] == INF) continue;
int ni = i + (i < s.size() && s[i] == '(');
int nj = j + (j < t.size() && t[j] == '(');
if (k + 1 <= s.size() + t.size() &&
dp[ni][nj][k + 1] > dp[i][j][k] + 1) {
dp[ni][nj][k + 1] = dp[i][j][k] + 1;
fukugen[ni][nj][k + 1] = make_tuple(i, j, k);
}
ni = i + (i < s.size() && s[i] == ')');
nj = j + (j < t.size() && t[j] == ')');
if (k - 1 >= 0 && dp[ni][nj][k - 1] > dp[i][j][k] + 1) {
dp[ni][nj][k - 1] = dp[i][j][k] + 1;
fukugen[ni][nj][k - 1] = make_tuple(i, j, k);
}
}
}
}
string ansstr, strplus;
int mans = INF, K = -1;
for (int i = 0; i <= s.size() + t.size(); i++) {
if (mans > dp[s.size()][t.size()][i] + i) {
mans = dp[s.size()][t.size()][i] + i;
K = i;
}
}
for (int i = 0; i < K; i++) strplus += ')';
int i = s.size(), j = t.size();
while (true) {
if (i == j && i == 0 && K == 0) {
break;
}
int ni = get<0>(fukugen[i][j][K]), nj = get<1>(fukugen[i][j][K]),
nk = get<2>(fukugen[i][j][K]);
ansstr += (K == nk + 1 ? '(' : ')');
i = ni, j = nj, K = nk;
}
reverse(ansstr.begin(), ansstr.end());
cout << ansstr << strplus << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 2e2 + 5, M = 5e2 + 5;
int n, m;
char s[N], t[N];
int mem[N][N][M];
int solve(int i, int j, int o) {
if (o == M) return 1e9;
if (i == n && j == m) return o;
int& ret = mem[i][j][o];
if (~ret) return ret;
ret = solve(i, j, o + 1) + 1;
if (o) ret = min(ret, solve(i, j, o - 1) + 1);
if (i < n && j < m && s[i] == '(' && t[j] == '(')
ret = min(ret, solve(i + 1, j + 1, o + 1) + 1);
if (i < n && s[i] == '(') ret = min(ret, solve(i + 1, j, o + 1) + 1);
if (j < m && t[j] == '(') ret = min(ret, solve(i, j + 1, o + 1) + 1);
if (o) {
if (i < n && j < m && s[i] == ')' && t[j] == ')')
ret = min(ret, solve(i + 1, j + 1, o - 1) + 1);
if (i < n && s[i] == ')') ret = min(ret, solve(i + 1, j, o - 1) + 1);
if (j < m && t[j] == ')') ret = min(ret, solve(i, j + 1, o - 1) + 1);
}
return ret;
}
void build(int i, int j, int o) {
if (i == n && j == m) {
int x = o;
while (x--) printf(")");
return;
}
char ch = '(';
int best = solve(i, j, o + 1) + 1;
int ii = i, jj = j, oo = o + 1;
if (o) {
int x = solve(i, j, o - 1) + 1;
if (x < best) best = x, ch = ')', ii = i, jj = j, oo = o - 1;
}
if (i < n && j < m && s[i] == '(' && t[j] == '(') {
int x = solve(i + 1, j + 1, o + 1) + 1;
if (x < best) best = x, ch = '(', ii = i + 1, jj = j + 1, oo = o + 1;
}
if (i < n && s[i] == '(') {
int x = solve(i + 1, j, o + 1) + 1;
if (x < best) best = x, ch = '(', ii = i + 1, jj = j, oo = o + 1;
}
if (j < m && t[j] == '(') {
int x = solve(i, j + 1, o + 1) + 1;
if (x < best) best = x, ch = '(', ii = i, jj = j + 1, oo = o + 1;
}
if (o) {
if (i < n && j < m && s[i] == ')' && t[j] == ')') {
int x = solve(i + 1, j + 1, o - 1) + 1;
if (x < best) best = x, ch = ')', ii = i + 1, jj = j + 1, oo = o - 1;
}
if (i < n && s[i] == ')') {
int x = solve(i + 1, j, o - 1) + 1;
if (x < best) best = x, ch = ')', ii = i + 1, jj = j, oo = o - 1;
}
if (j < m && t[j] == ')') {
int x = solve(i, j + 1, o - 1) + 1;
if (x < best) best = x, ch = ')', ii = i, jj = j + 1, oo = o - 1;
}
}
printf("%c", ch);
build(ii, jj, oo);
}
int main() {
scanf("%s%s", s, t);
n = strlen(s);
m = strlen(t);
memset(mem, -1, sizeof mem);
build(0, 0, 0);
return 0;
}
| ### Prompt
Please provide a Cpp coded solution to the problem described below:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 2e2 + 5, M = 5e2 + 5;
int n, m;
char s[N], t[N];
int mem[N][N][M];
int solve(int i, int j, int o) {
if (o == M) return 1e9;
if (i == n && j == m) return o;
int& ret = mem[i][j][o];
if (~ret) return ret;
ret = solve(i, j, o + 1) + 1;
if (o) ret = min(ret, solve(i, j, o - 1) + 1);
if (i < n && j < m && s[i] == '(' && t[j] == '(')
ret = min(ret, solve(i + 1, j + 1, o + 1) + 1);
if (i < n && s[i] == '(') ret = min(ret, solve(i + 1, j, o + 1) + 1);
if (j < m && t[j] == '(') ret = min(ret, solve(i, j + 1, o + 1) + 1);
if (o) {
if (i < n && j < m && s[i] == ')' && t[j] == ')')
ret = min(ret, solve(i + 1, j + 1, o - 1) + 1);
if (i < n && s[i] == ')') ret = min(ret, solve(i + 1, j, o - 1) + 1);
if (j < m && t[j] == ')') ret = min(ret, solve(i, j + 1, o - 1) + 1);
}
return ret;
}
void build(int i, int j, int o) {
if (i == n && j == m) {
int x = o;
while (x--) printf(")");
return;
}
char ch = '(';
int best = solve(i, j, o + 1) + 1;
int ii = i, jj = j, oo = o + 1;
if (o) {
int x = solve(i, j, o - 1) + 1;
if (x < best) best = x, ch = ')', ii = i, jj = j, oo = o - 1;
}
if (i < n && j < m && s[i] == '(' && t[j] == '(') {
int x = solve(i + 1, j + 1, o + 1) + 1;
if (x < best) best = x, ch = '(', ii = i + 1, jj = j + 1, oo = o + 1;
}
if (i < n && s[i] == '(') {
int x = solve(i + 1, j, o + 1) + 1;
if (x < best) best = x, ch = '(', ii = i + 1, jj = j, oo = o + 1;
}
if (j < m && t[j] == '(') {
int x = solve(i, j + 1, o + 1) + 1;
if (x < best) best = x, ch = '(', ii = i, jj = j + 1, oo = o + 1;
}
if (o) {
if (i < n && j < m && s[i] == ')' && t[j] == ')') {
int x = solve(i + 1, j + 1, o - 1) + 1;
if (x < best) best = x, ch = ')', ii = i + 1, jj = j + 1, oo = o - 1;
}
if (i < n && s[i] == ')') {
int x = solve(i + 1, j, o - 1) + 1;
if (x < best) best = x, ch = ')', ii = i + 1, jj = j, oo = o - 1;
}
if (j < m && t[j] == ')') {
int x = solve(i, j + 1, o - 1) + 1;
if (x < best) best = x, ch = ')', ii = i, jj = j + 1, oo = o - 1;
}
}
printf("%c", ch);
build(ii, jj, oo);
}
int main() {
scanf("%s%s", s, t);
n = strlen(s);
m = strlen(t);
memset(mem, -1, sizeof mem);
build(0, 0, 0);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int dp[205][205][205], zagrada[205][205][205];
char s[205], t[205];
int n, m;
int resi(int l, int d, int zbir) {
if (dp[l][d][zbir] + 1) return dp[l][d][zbir];
if (zbir == 0) {
dp[l][d][zbir] = 1 + resi((s[l] == ')' ? l - 1 : l),
(t[d] == ')' ? d - 1 : d), zbir + 1);
zagrada[l][d][zbir] = -1;
} else if (zbir >= 205) {
return 505;
} else {
dp[l][d][zbir] = 1 + resi((s[l] == ')' ? l - 1 : l),
(t[d] == ')' ? d - 1 : d), zbir + 1);
zagrada[l][d][zbir] = -1;
int tmp = 1 + resi((s[l] == '(' ? l - 1 : l), (t[d] == '(' ? d - 1 : d),
zbir - 1);
if (tmp < dp[l][d][zbir]) {
zagrada[l][d][zbir] = 1;
dp[l][d][zbir] = tmp;
}
}
return dp[l][d][zbir];
}
vector<char> res;
void nadji(int l, int d, int zbir) {
if (!l && !d && !zbir) return;
if (zagrada[l][d][zbir] == 1) {
res.push_back('(');
nadji((s[l] == '(' ? l - 1 : l), (t[d] == '(' ? d - 1 : d), zbir - 1);
} else {
res.push_back(')');
nadji((s[l] == ')' ? l - 1 : l), (t[d] == ')' ? d - 1 : d), zbir + 1);
}
}
int main() {
scanf("%s", s + 1);
scanf("%s", t + 1);
for (int i = 0; i < 205; i++) {
for (int j = 0; j < 205; j++) {
for (int l = 0; l < 205; l++) {
dp[i][j][l] = -1;
}
}
}
dp[0][0][0] = 0;
s[0] = t[0] = 'A';
n = strlen(s) - 1;
m = strlen(t) - 1;
resi(n, m, 0);
nadji(n, m, 0);
reverse(res.begin(), res.end());
for (auto x : res) putchar(x);
return 0;
}
| ### Prompt
Construct a CPP code solution to the problem outlined:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int dp[205][205][205], zagrada[205][205][205];
char s[205], t[205];
int n, m;
int resi(int l, int d, int zbir) {
if (dp[l][d][zbir] + 1) return dp[l][d][zbir];
if (zbir == 0) {
dp[l][d][zbir] = 1 + resi((s[l] == ')' ? l - 1 : l),
(t[d] == ')' ? d - 1 : d), zbir + 1);
zagrada[l][d][zbir] = -1;
} else if (zbir >= 205) {
return 505;
} else {
dp[l][d][zbir] = 1 + resi((s[l] == ')' ? l - 1 : l),
(t[d] == ')' ? d - 1 : d), zbir + 1);
zagrada[l][d][zbir] = -1;
int tmp = 1 + resi((s[l] == '(' ? l - 1 : l), (t[d] == '(' ? d - 1 : d),
zbir - 1);
if (tmp < dp[l][d][zbir]) {
zagrada[l][d][zbir] = 1;
dp[l][d][zbir] = tmp;
}
}
return dp[l][d][zbir];
}
vector<char> res;
void nadji(int l, int d, int zbir) {
if (!l && !d && !zbir) return;
if (zagrada[l][d][zbir] == 1) {
res.push_back('(');
nadji((s[l] == '(' ? l - 1 : l), (t[d] == '(' ? d - 1 : d), zbir - 1);
} else {
res.push_back(')');
nadji((s[l] == ')' ? l - 1 : l), (t[d] == ')' ? d - 1 : d), zbir + 1);
}
}
int main() {
scanf("%s", s + 1);
scanf("%s", t + 1);
for (int i = 0; i < 205; i++) {
for (int j = 0; j < 205; j++) {
for (int l = 0; l < 205; l++) {
dp[i][j][l] = -1;
}
}
}
dp[0][0][0] = 0;
s[0] = t[0] = 'A';
n = strlen(s) - 1;
m = strlen(t) - 1;
resi(n, m, 0);
nadji(n, m, 0);
reverse(res.begin(), res.end());
for (auto x : res) putchar(x);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 200 + 5;
const int INF = 0X3F3F3F3F;
char a[N], b[N];
int n, m;
int dp[N][N][N];
vector<char> ans;
struct node {
int x, y, z;
char c;
node(int _x = 0, int _y = 0, int _z = 0, char _c = 0) {
x = _x, y = _y, z = _z, c = _c;
}
} p[N][N][N];
void bfs() {
memset(dp, INF, sizeof(dp));
queue<node> que;
que.push(node(0, 0, 0));
dp[0][0][0] = 0;
while (!que.empty()) {
node top = que.front();
que.pop();
int nx = top.x + (top.x < n && a[top.x] == '(');
int ny = top.y + (top.y < m && b[top.y] == '(');
int nz = top.z + 1;
if (nz <= 200 && nz >= 0 && dp[nx][ny][nz] == INF) {
dp[nx][ny][nz] = dp[top.x][top.y][top.z] + 1;
p[nx][ny][nz] = node(top.x, top.y, top.z, '(');
que.push(node(nx, ny, nz));
}
nx = top.x + (top.x < n && a[top.x] == ')');
ny = top.y + (top.y < m && b[top.y] == ')');
nz = top.z - 1;
if (nz <= 200 && nz >= 0 && dp[nx][ny][nz] == INF) {
dp[nx][ny][nz] = dp[top.x][top.y][top.z] + 1;
p[nx][ny][nz] = node(top.x, top.y, top.z, ')');
que.push(node(nx, ny, nz));
}
}
}
int main() {
scanf("%s%s", a, b);
n = strlen(a);
m = strlen(b);
bfs();
node k = p[n][m][0];
while (k.x > 0 || k.y > 0 || k.z > 0) {
ans.push_back(k.c);
k = p[k.x][k.y][k.z];
}
ans.push_back(k.c);
reverse(ans.begin(), ans.end());
for (int i = 0; i < ans.size(); i++) printf("%c", ans[i]);
puts("");
}
| ### Prompt
In CPP, your task is to solve the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 200 + 5;
const int INF = 0X3F3F3F3F;
char a[N], b[N];
int n, m;
int dp[N][N][N];
vector<char> ans;
struct node {
int x, y, z;
char c;
node(int _x = 0, int _y = 0, int _z = 0, char _c = 0) {
x = _x, y = _y, z = _z, c = _c;
}
} p[N][N][N];
void bfs() {
memset(dp, INF, sizeof(dp));
queue<node> que;
que.push(node(0, 0, 0));
dp[0][0][0] = 0;
while (!que.empty()) {
node top = que.front();
que.pop();
int nx = top.x + (top.x < n && a[top.x] == '(');
int ny = top.y + (top.y < m && b[top.y] == '(');
int nz = top.z + 1;
if (nz <= 200 && nz >= 0 && dp[nx][ny][nz] == INF) {
dp[nx][ny][nz] = dp[top.x][top.y][top.z] + 1;
p[nx][ny][nz] = node(top.x, top.y, top.z, '(');
que.push(node(nx, ny, nz));
}
nx = top.x + (top.x < n && a[top.x] == ')');
ny = top.y + (top.y < m && b[top.y] == ')');
nz = top.z - 1;
if (nz <= 200 && nz >= 0 && dp[nx][ny][nz] == INF) {
dp[nx][ny][nz] = dp[top.x][top.y][top.z] + 1;
p[nx][ny][nz] = node(top.x, top.y, top.z, ')');
que.push(node(nx, ny, nz));
}
}
}
int main() {
scanf("%s%s", a, b);
n = strlen(a);
m = strlen(b);
bfs();
node k = p[n][m][0];
while (k.x > 0 || k.y > 0 || k.z > 0) {
ans.push_back(k.c);
k = p[k.x][k.y][k.z];
}
ans.push_back(k.c);
reverse(ans.begin(), ans.end());
for (int i = 0; i < ans.size(); i++) printf("%c", ans[i]);
puts("");
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long inf = (1ll << 61);
const int MX = 205;
string s, t;
int n, m;
int dp[MX][MX][1205];
int DP(int x, int y, int sum) {
if (x == n && y == m) {
return sum;
}
if (sum == 1200) return 100000;
int &ret = dp[x][y][sum];
if (ret != -1) return ret;
ret = 100000;
int d = (s[x] == '(');
int d1 = (t[y] == '(');
ret = min(ret, DP(x, y, sum + 1) + 1);
ret = min(ret, DP(x + d, y, sum + 1) + 1);
ret = min(ret, DP(x, y + d1, sum + 1) + 1);
ret = min(ret, DP(x + d, y + d1, sum + 1) + 1);
if (sum != 0) {
int d = (s[x] == ')');
int d1 = (t[y] == ')');
ret = min(ret, DP(x, y, sum - 1) + 1);
ret = min(ret, DP(x + d, y, sum - 1) + 1);
ret = min(ret, DP(x, y + d1, sum - 1) + 1);
ret = min(ret, DP(x + d, y + d1, sum - 1) + 1);
}
return ret;
}
void FDP(int x, int y, int sum) {
if (x == n && y == m) {
while (sum) {
sum--;
cout << ')';
}
return;
}
if (sum == 1200) return;
int ret = DP(x, y, sum);
int d = (s[x] == '(');
int d1 = (t[y] == '(');
if (ret == DP(x, y, sum + 1) + 1) {
cout << '(';
FDP(x, y, sum + 1);
return;
}
if (ret == DP(x + d, y, sum + 1) + 1) {
cout << '(';
FDP(x + d, y, sum + 1);
return;
}
if (ret == DP(x, y + d1, sum + 1) + 1) {
cout << '(';
FDP(x, y + d1, sum + 1);
return;
}
if (ret == DP(x + d, y + d1, sum + 1) + 1) {
cout << '(';
FDP(x + d, y + d1, sum + 1);
return;
}
if (sum != 0) {
int d = (s[x] == ')');
int d1 = (t[y] == ')');
if (ret == DP(x, y, sum - 1) + 1) {
cout << ')';
FDP(x, y, sum - 1);
return;
}
if (ret == DP(x + d, y, sum - 1) + 1) {
cout << ')';
FDP(x + d, y, sum - 1);
return;
}
if (ret == DP(x, y + d1, sum - 1) + 1) {
cout << ')';
FDP(x, y + d1, sum - 1);
return;
}
if (ret == DP(x + d, y + d1, sum - 1) + 1) {
cout << ')';
FDP(x + d, y + d1, sum - 1);
return;
}
}
return;
}
int main() {
cin >> s >> t;
n = s.size();
m = t.size();
s += "####";
t += "####";
memset(dp, -1, sizeof(dp));
FDP(0, 0, 0);
}
| ### Prompt
Create a solution in CPP for the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long inf = (1ll << 61);
const int MX = 205;
string s, t;
int n, m;
int dp[MX][MX][1205];
int DP(int x, int y, int sum) {
if (x == n && y == m) {
return sum;
}
if (sum == 1200) return 100000;
int &ret = dp[x][y][sum];
if (ret != -1) return ret;
ret = 100000;
int d = (s[x] == '(');
int d1 = (t[y] == '(');
ret = min(ret, DP(x, y, sum + 1) + 1);
ret = min(ret, DP(x + d, y, sum + 1) + 1);
ret = min(ret, DP(x, y + d1, sum + 1) + 1);
ret = min(ret, DP(x + d, y + d1, sum + 1) + 1);
if (sum != 0) {
int d = (s[x] == ')');
int d1 = (t[y] == ')');
ret = min(ret, DP(x, y, sum - 1) + 1);
ret = min(ret, DP(x + d, y, sum - 1) + 1);
ret = min(ret, DP(x, y + d1, sum - 1) + 1);
ret = min(ret, DP(x + d, y + d1, sum - 1) + 1);
}
return ret;
}
void FDP(int x, int y, int sum) {
if (x == n && y == m) {
while (sum) {
sum--;
cout << ')';
}
return;
}
if (sum == 1200) return;
int ret = DP(x, y, sum);
int d = (s[x] == '(');
int d1 = (t[y] == '(');
if (ret == DP(x, y, sum + 1) + 1) {
cout << '(';
FDP(x, y, sum + 1);
return;
}
if (ret == DP(x + d, y, sum + 1) + 1) {
cout << '(';
FDP(x + d, y, sum + 1);
return;
}
if (ret == DP(x, y + d1, sum + 1) + 1) {
cout << '(';
FDP(x, y + d1, sum + 1);
return;
}
if (ret == DP(x + d, y + d1, sum + 1) + 1) {
cout << '(';
FDP(x + d, y + d1, sum + 1);
return;
}
if (sum != 0) {
int d = (s[x] == ')');
int d1 = (t[y] == ')');
if (ret == DP(x, y, sum - 1) + 1) {
cout << ')';
FDP(x, y, sum - 1);
return;
}
if (ret == DP(x + d, y, sum - 1) + 1) {
cout << ')';
FDP(x + d, y, sum - 1);
return;
}
if (ret == DP(x, y + d1, sum - 1) + 1) {
cout << ')';
FDP(x, y + d1, sum - 1);
return;
}
if (ret == DP(x + d, y + d1, sum - 1) + 1) {
cout << ')';
FDP(x + d, y + d1, sum - 1);
return;
}
}
return;
}
int main() {
cin >> s >> t;
n = s.size();
m = t.size();
s += "####";
t += "####";
memset(dp, -1, sizeof(dp));
FDP(0, 0, 0);
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long int logx = 30;
const long long int N = 1e5 + 5;
const long long int M = 3e5 + 5;
const long long int mod = 1e9 + 7;
const long long int INF = 1e18 + 5;
const double PI = 3.14159265358979323846;
inline long long int mul(long long int a, long long int b) {
return (a * 1ll * b) % mod;
}
inline long long int sub(long long int a, long long int b) {
long long int c = a - b;
if (c < 0) c += mod;
return c;
}
inline long long int add(long long int a, long long int b) {
long long int c = a + b;
if (c > mod) c -= mod;
return c;
}
template <typename T>
T power(T x, T y, long long int m) {
T ans = 1;
while (y > 0) {
if (y & 1LL) ans = (ans * x) % m;
y >>= 1ll;
x = (x * x) % m;
}
return ans % m;
}
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
cerr << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
const char* comma = strchr(names + 1, ',');
cerr.write(names, comma - names) << " : " << arg1 << " | ";
__f(comma + 1, args...);
}
long long int n, m;
string second, t;
long long int dp[205][205][410];
long long int F(long long int i, long long int j, long long int open) {
long long int res = INF;
if (open < 0) {
return INF;
}
if (open >= 410) {
return INF;
}
if (open == 0) {
if (i == n && j == m) {
return 0;
}
}
if (dp[i][j][open] != -1) {
return dp[i][j][open];
}
if (i == n && j == m) {
return open;
}
if (i == n && j != m) {
res = min(res, 1 + F(i, j + (t[j] == '('), open + 1));
res = min(res, 1 + F(i, j + (t[j] == ')'), open - 1));
}
if (i != n && j == m) {
res = min(res, 1 + F(i + (second[i] == '('), j, open + 1));
res = min(res, 1 + F(i + (second[i] == ')'), j, open - 1));
}
if (i != n && j != m) {
res = min(res, 1 + F(i + (second[i] == '('), j + (t[j] == '('), open + 1));
res = min(res, 1 + F(i + (second[i] == ')'), j + (t[j] == ')'), open - 1));
}
return dp[i][j][open] = res;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
;
memset(dp, -1, sizeof(dp));
cin >> second >> t;
n = second.length();
m = t.length();
string out = "\0";
long long int ans = F(0, 0, 0), prevans = 0, prei = 0, prej = 0, preopen = 0;
for (long long int i = 0; i < ans; i++) {
long long int tempans =
1 + prevans +
F(prei + (second[prei] == '('), prej + (t[prej] == '('), preopen + 1);
if (tempans == ans) {
out += "(";
prevans++;
preopen++;
prei += (second[prei] == '(');
prej += (t[prej] == '(');
continue;
}
tempans =
1 + prevans +
F(prei + (second[prei] == ')'), prej + (t[prej] == ')'), preopen - 1);
if (tempans == ans) {
out += ")";
prevans++;
preopen--;
prei += (second[prei] == ')');
prej += (t[prej] == ')');
continue;
}
}
cout << out << endl;
}
| ### Prompt
Construct a Cpp code solution to the problem outlined:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long int logx = 30;
const long long int N = 1e5 + 5;
const long long int M = 3e5 + 5;
const long long int mod = 1e9 + 7;
const long long int INF = 1e18 + 5;
const double PI = 3.14159265358979323846;
inline long long int mul(long long int a, long long int b) {
return (a * 1ll * b) % mod;
}
inline long long int sub(long long int a, long long int b) {
long long int c = a - b;
if (c < 0) c += mod;
return c;
}
inline long long int add(long long int a, long long int b) {
long long int c = a + b;
if (c > mod) c -= mod;
return c;
}
template <typename T>
T power(T x, T y, long long int m) {
T ans = 1;
while (y > 0) {
if (y & 1LL) ans = (ans * x) % m;
y >>= 1ll;
x = (x * x) % m;
}
return ans % m;
}
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
cerr << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
const char* comma = strchr(names + 1, ',');
cerr.write(names, comma - names) << " : " << arg1 << " | ";
__f(comma + 1, args...);
}
long long int n, m;
string second, t;
long long int dp[205][205][410];
long long int F(long long int i, long long int j, long long int open) {
long long int res = INF;
if (open < 0) {
return INF;
}
if (open >= 410) {
return INF;
}
if (open == 0) {
if (i == n && j == m) {
return 0;
}
}
if (dp[i][j][open] != -1) {
return dp[i][j][open];
}
if (i == n && j == m) {
return open;
}
if (i == n && j != m) {
res = min(res, 1 + F(i, j + (t[j] == '('), open + 1));
res = min(res, 1 + F(i, j + (t[j] == ')'), open - 1));
}
if (i != n && j == m) {
res = min(res, 1 + F(i + (second[i] == '('), j, open + 1));
res = min(res, 1 + F(i + (second[i] == ')'), j, open - 1));
}
if (i != n && j != m) {
res = min(res, 1 + F(i + (second[i] == '('), j + (t[j] == '('), open + 1));
res = min(res, 1 + F(i + (second[i] == ')'), j + (t[j] == ')'), open - 1));
}
return dp[i][j][open] = res;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
;
memset(dp, -1, sizeof(dp));
cin >> second >> t;
n = second.length();
m = t.length();
string out = "\0";
long long int ans = F(0, 0, 0), prevans = 0, prei = 0, prej = 0, preopen = 0;
for (long long int i = 0; i < ans; i++) {
long long int tempans =
1 + prevans +
F(prei + (second[prei] == '('), prej + (t[prej] == '('), preopen + 1);
if (tempans == ans) {
out += "(";
prevans++;
preopen++;
prei += (second[prei] == '(');
prej += (t[prej] == '(');
continue;
}
tempans =
1 + prevans +
F(prei + (second[prei] == ')'), prej + (t[prej] == ')'), preopen - 1);
if (tempans == ans) {
out += ")";
prevans++;
preopen--;
prei += (second[prei] == ')');
prej += (t[prej] == ')');
continue;
}
}
cout << out << endl;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int dp[205][205][205];
pair<pair<int, int>, int> pv[205][205][205];
char s[205], t[205];
queue<int> q;
int main() {
memset(dp, -1, sizeof(dp));
scanf("%s%s", s + 1, t + 1);
int n = strlen(s + 1), m = strlen(t + 1);
dp[0][0][0] = 0;
queue<pair<pair<int, int>, int> > q;
q.push(pair<pair<int, int>, int>(pair<int, int>(0, 0), 0));
while (!q.empty()) {
int a = q.front().first.first, b = q.front().first.second,
k = q.front().second;
q.pop();
int na = a, nb = b, nk = k + 1;
if (s[a + 1] == '(') na++;
if (t[b + 1] == '(') nb++;
if (dp[na][nb][nk] == -1 && nk <= 200) {
dp[na][nb][nk] = dp[a][b][k] + 1;
pv[na][nb][nk] = pair<pair<int, int>, int>(pair<int, int>(a, b), k);
q.push(pair<pair<int, int>, int>(pair<int, int>(na, nb), nk));
}
na = a, nb = b, nk = k - 1;
if (s[a + 1] == ')') na++;
if (t[b + 1] == ')') nb++;
if (dp[na][nb][nk] == -1 && nk >= 0) {
dp[na][nb][nk] = dp[a][b][k] + 1;
pv[na][nb][nk] = pair<pair<int, int>, int>(pair<int, int>(a, b), k);
q.push(pair<pair<int, int>, int>(pair<int, int>(na, nb), nk));
}
}
string ans = "";
int a = n, b = m, k = 0;
while (a != 0 || b != 0 || k != 0) {
int na = pv[a][b][k].first.first, nb = pv[a][b][k].first.second,
nk = pv[a][b][k].second;
if (k + 1 == nk)
ans += ')';
else
ans += '(';
a = na;
b = nb;
k = nk;
}
reverse(ans.begin(), ans.end());
cout << ans << "\n";
}
| ### Prompt
Generate a cpp solution to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int dp[205][205][205];
pair<pair<int, int>, int> pv[205][205][205];
char s[205], t[205];
queue<int> q;
int main() {
memset(dp, -1, sizeof(dp));
scanf("%s%s", s + 1, t + 1);
int n = strlen(s + 1), m = strlen(t + 1);
dp[0][0][0] = 0;
queue<pair<pair<int, int>, int> > q;
q.push(pair<pair<int, int>, int>(pair<int, int>(0, 0), 0));
while (!q.empty()) {
int a = q.front().first.first, b = q.front().first.second,
k = q.front().second;
q.pop();
int na = a, nb = b, nk = k + 1;
if (s[a + 1] == '(') na++;
if (t[b + 1] == '(') nb++;
if (dp[na][nb][nk] == -1 && nk <= 200) {
dp[na][nb][nk] = dp[a][b][k] + 1;
pv[na][nb][nk] = pair<pair<int, int>, int>(pair<int, int>(a, b), k);
q.push(pair<pair<int, int>, int>(pair<int, int>(na, nb), nk));
}
na = a, nb = b, nk = k - 1;
if (s[a + 1] == ')') na++;
if (t[b + 1] == ')') nb++;
if (dp[na][nb][nk] == -1 && nk >= 0) {
dp[na][nb][nk] = dp[a][b][k] + 1;
pv[na][nb][nk] = pair<pair<int, int>, int>(pair<int, int>(a, b), k);
q.push(pair<pair<int, int>, int>(pair<int, int>(na, nb), nk));
}
}
string ans = "";
int a = n, b = m, k = 0;
while (a != 0 || b != 0 || k != 0) {
int na = pv[a][b][k].first.first, nb = pv[a][b][k].first.second,
nk = pv[a][b][k].second;
if (k + 1 == nk)
ans += ')';
else
ans += '(';
a = na;
b = nb;
k = nk;
}
reverse(ans.begin(), ans.end());
cout << ans << "\n";
}
``` |
#include <bits/stdc++.h>
using namespace std;
constexpr int inf = 1001001001;
int dp[205][205][805];
int main() {
string s, t;
cin >> s >> t;
for (int i = 0; i < 205; i++) {
for (int j = 0; j < 205; j++) {
for (int k = 0; k < 805; k++) {
dp[i][j][k] = inf;
}
}
}
dp[0][0][0] = 0;
for (int i = 0; i <= s.size(); i++) {
for (int j = 0; j <= t.size(); j++) {
for (int k = 0; k <= 799; k++) {
dp[i][j][k + 1] = min(dp[i][j][k + 1], dp[i][j][k] + 1);
if (i < s.size() && s[i] == '(') {
dp[i + 1][j][k + 1] = min(dp[i + 1][j][k + 1], dp[i][j][k] + 1);
}
if (j < t.size() && t[j] == '(') {
dp[i][j + 1][k + 1] = min(dp[i][j + 1][k + 1], dp[i][j][k] + 1);
}
if (i < s.size() && s[i] == '(' && j < t.size() && t[j] == '(') {
dp[i + 1][j + 1][k + 1] =
min(dp[i + 1][j + 1][k + 1], dp[i][j][k] + 1);
}
}
for (int k = 800; k >= 1; k--) {
dp[i][j][k - 1] = min(dp[i][j][k - 1], dp[i][j][k] + 1);
if (i < s.size() && s[i] == ')') {
dp[i + 1][j][k - 1] = min(dp[i + 1][j][k - 1], dp[i][j][k] + 1);
}
if (j < t.size() && t[j] == ')') {
dp[i][j + 1][k - 1] = min(dp[i][j + 1][k - 1], dp[i][j][k] + 1);
}
if (i < s.size() && s[i] == ')' && j < t.size() && t[j] == ')') {
dp[i + 1][j + 1][k - 1] =
min(dp[i + 1][j + 1][k - 1], dp[i][j][k] + 1);
}
}
}
}
int now = dp[s.size()][t.size()][0];
int l = (int)s.size(), r = (int)t.size(), sum = 0;
string ans = "";
for (int i = now - 1; i >= 0; i--) {
bool flag = false;
for (int j = -1; j <= 0; j++) {
for (int k = -1; k <= 0; k++) {
for (int m = -1; m <= 1; m += 2) {
if (0 <= l + j && 0 <= r + k && 0 <= sum + m) {
bool ok = true;
if (j == -1 && m == -1 && s[l + j] == ')') {
ok = false;
}
if (j == -1 && m == 1 && s[l + j] == '(') {
ok = false;
}
if (k == -1 && m == -1 && t[r + k] == ')') {
ok = false;
}
if (k == -1 && m == 1 && t[r + k] == '(') {
ok = false;
}
if (ok && dp[l + j][r + k][sum + m] == i) {
if (m == -1) {
ans += '(';
} else {
ans += ')';
}
l += j;
r += k;
sum += m;
flag = true;
}
}
if (flag) {
break;
}
}
if (flag) {
break;
}
}
if (flag) {
break;
}
}
}
reverse(ans.begin(), ans.end());
cout << ans << endl;
}
| ### Prompt
Your challenge is to write a CPP solution to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
constexpr int inf = 1001001001;
int dp[205][205][805];
int main() {
string s, t;
cin >> s >> t;
for (int i = 0; i < 205; i++) {
for (int j = 0; j < 205; j++) {
for (int k = 0; k < 805; k++) {
dp[i][j][k] = inf;
}
}
}
dp[0][0][0] = 0;
for (int i = 0; i <= s.size(); i++) {
for (int j = 0; j <= t.size(); j++) {
for (int k = 0; k <= 799; k++) {
dp[i][j][k + 1] = min(dp[i][j][k + 1], dp[i][j][k] + 1);
if (i < s.size() && s[i] == '(') {
dp[i + 1][j][k + 1] = min(dp[i + 1][j][k + 1], dp[i][j][k] + 1);
}
if (j < t.size() && t[j] == '(') {
dp[i][j + 1][k + 1] = min(dp[i][j + 1][k + 1], dp[i][j][k] + 1);
}
if (i < s.size() && s[i] == '(' && j < t.size() && t[j] == '(') {
dp[i + 1][j + 1][k + 1] =
min(dp[i + 1][j + 1][k + 1], dp[i][j][k] + 1);
}
}
for (int k = 800; k >= 1; k--) {
dp[i][j][k - 1] = min(dp[i][j][k - 1], dp[i][j][k] + 1);
if (i < s.size() && s[i] == ')') {
dp[i + 1][j][k - 1] = min(dp[i + 1][j][k - 1], dp[i][j][k] + 1);
}
if (j < t.size() && t[j] == ')') {
dp[i][j + 1][k - 1] = min(dp[i][j + 1][k - 1], dp[i][j][k] + 1);
}
if (i < s.size() && s[i] == ')' && j < t.size() && t[j] == ')') {
dp[i + 1][j + 1][k - 1] =
min(dp[i + 1][j + 1][k - 1], dp[i][j][k] + 1);
}
}
}
}
int now = dp[s.size()][t.size()][0];
int l = (int)s.size(), r = (int)t.size(), sum = 0;
string ans = "";
for (int i = now - 1; i >= 0; i--) {
bool flag = false;
for (int j = -1; j <= 0; j++) {
for (int k = -1; k <= 0; k++) {
for (int m = -1; m <= 1; m += 2) {
if (0 <= l + j && 0 <= r + k && 0 <= sum + m) {
bool ok = true;
if (j == -1 && m == -1 && s[l + j] == ')') {
ok = false;
}
if (j == -1 && m == 1 && s[l + j] == '(') {
ok = false;
}
if (k == -1 && m == -1 && t[r + k] == ')') {
ok = false;
}
if (k == -1 && m == 1 && t[r + k] == '(') {
ok = false;
}
if (ok && dp[l + j][r + k][sum + m] == i) {
if (m == -1) {
ans += '(';
} else {
ans += ')';
}
l += j;
r += k;
sum += m;
flag = true;
}
}
if (flag) {
break;
}
}
if (flag) {
break;
}
}
if (flag) {
break;
}
}
}
reverse(ans.begin(), ans.end());
cout << ans << endl;
}
``` |
#include <bits/stdc++.h>
using namespace std;
string s, t;
int dp[202][202][2 * 202];
bool to[202][202][2 * 202];
int solve(int i, int j, int o) {
if (i == s.size() && j == t.size()) return o;
if (o > s.size() + t.size()) return 1e9;
if (~dp[i][j][o]) return dp[i][j][o];
int op1 = 1 + solve(i + (i < s.size() && s[i] == '('),
j + (j < t.size() && t[j] == '('), o + 1),
op2 = 1e9;
if (o)
op2 = 1 + solve(i + (i < s.size() && s[i] == ')'),
j + (j < t.size() && t[j] == ')'), o - 1);
to[i][j][o] = op1 < op2;
return dp[i][j][o] = min(op1, op2);
}
int main() {
cin >> s >> t;
memset(dp, -1, sizeof dp);
solve(0, 0, 0);
int i = 0, j = 0, o = 0;
while (i < s.size() || j < t.size()) {
printf(to[i][j][o] ? "(" : ")");
if (to[i][j][o]) {
i += (i < s.size() && s[i] == '(');
j += (j < t.size() && t[j] == '(');
o++;
} else {
i += (i < s.size() && s[i] == ')');
j += (j < t.size() && t[j] == ')');
o--;
}
}
while (o--) printf(")");
cout << endl;
i = 0, j = 0, o = 0;
while (i < s.size() || j < t.size()) {
if (to[i][j][o]) {
i += (i < s.size() && s[i] == '(');
j += (j < t.size() && t[j] == '(');
o++;
} else {
i += (i < s.size() && s[i] == ')');
j += (j < t.size() && t[j] == ')');
o--;
}
}
printf("\n");
}
| ### Prompt
Generate a cpp solution to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
string s, t;
int dp[202][202][2 * 202];
bool to[202][202][2 * 202];
int solve(int i, int j, int o) {
if (i == s.size() && j == t.size()) return o;
if (o > s.size() + t.size()) return 1e9;
if (~dp[i][j][o]) return dp[i][j][o];
int op1 = 1 + solve(i + (i < s.size() && s[i] == '('),
j + (j < t.size() && t[j] == '('), o + 1),
op2 = 1e9;
if (o)
op2 = 1 + solve(i + (i < s.size() && s[i] == ')'),
j + (j < t.size() && t[j] == ')'), o - 1);
to[i][j][o] = op1 < op2;
return dp[i][j][o] = min(op1, op2);
}
int main() {
cin >> s >> t;
memset(dp, -1, sizeof dp);
solve(0, 0, 0);
int i = 0, j = 0, o = 0;
while (i < s.size() || j < t.size()) {
printf(to[i][j][o] ? "(" : ")");
if (to[i][j][o]) {
i += (i < s.size() && s[i] == '(');
j += (j < t.size() && t[j] == '(');
o++;
} else {
i += (i < s.size() && s[i] == ')');
j += (j < t.size() && t[j] == ')');
o--;
}
}
while (o--) printf(")");
cout << endl;
i = 0, j = 0, o = 0;
while (i < s.size() || j < t.size()) {
if (to[i][j][o]) {
i += (i < s.size() && s[i] == '(');
j += (j < t.size() && t[j] == '(');
o++;
} else {
i += (i < s.size() && s[i] == ')');
j += (j < t.size() && t[j] == ')');
o--;
}
}
printf("\n");
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int maxn = 205;
const int INF = 2e7 + 5;
int n, m, len;
char S[maxn];
int A[maxn], B[maxn], ans[maxn << 1];
int F[maxn][maxn][maxn << 1], G[maxn][maxn][maxn << 1],
H[maxn][maxn][maxn << 1], P[maxn][maxn][maxn << 1];
int main() {
gets(S + 1);
n = strlen(S + 1);
for (int i = 1; i <= n; ++i)
if (S[i] == '(')
A[i] = 1;
else
A[i] = -1;
gets(S + 1);
m = strlen(S + 1);
for (int i = 1; i <= m; ++i)
if (S[i] == '(')
B[i] = 1;
else
B[i] = -1;
len = n + m;
for (int i = 0; i <= n; ++i)
for (int j = 0; j <= m; ++j)
for (int k = 0; k <= len; ++k) F[i][j][k] = INF;
F[0][0][0] = 0;
for (int i = 0; i <= n; ++i)
for (int j = 0; j <= m; ++j)
for (int k = 0; k <= len; ++k) {
for (int tp = -1; tp <= 1; ++tp) {
if (!tp || F[i][j][k] >= INF) continue;
int u = i, v = j;
if (A[u + 1] == tp) ++u;
if (B[v + 1] == tp) ++v;
if (k + tp < 0 || k + tp > len) continue;
if (F[u][v][k + tp] > F[i][j][k] + 1) {
F[u][v][k + tp] = F[i][j][k] + 1;
G[u][v][k + tp] = tp, H[u][v][k + tp] = i, P[u][v][k + tp] = j;
}
}
}
int sans = INF, sp = 0, cnt = 0;
for (int k = 0; k <= len; ++k)
if (F[n][m][k] + k < sans) {
sans = F[n][m][k] + k;
sp = k;
}
for (int i = 1; i <= sp; ++i) ans[++cnt] = ')';
int u = n, v = m;
while (u || v || sp) {
int s = G[u][v][sp], s1 = u, s2 = v;
if (s == 1)
ans[++cnt] = '(';
else
ans[++cnt] = ')';
u = H[s1][s2][sp], v = P[s1][s2][sp];
sp -= s;
}
for (int i = cnt; i >= 1; --i) putchar(ans[i]);
return 0;
}
| ### Prompt
In Cpp, your task is to solve the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = 205;
const int INF = 2e7 + 5;
int n, m, len;
char S[maxn];
int A[maxn], B[maxn], ans[maxn << 1];
int F[maxn][maxn][maxn << 1], G[maxn][maxn][maxn << 1],
H[maxn][maxn][maxn << 1], P[maxn][maxn][maxn << 1];
int main() {
gets(S + 1);
n = strlen(S + 1);
for (int i = 1; i <= n; ++i)
if (S[i] == '(')
A[i] = 1;
else
A[i] = -1;
gets(S + 1);
m = strlen(S + 1);
for (int i = 1; i <= m; ++i)
if (S[i] == '(')
B[i] = 1;
else
B[i] = -1;
len = n + m;
for (int i = 0; i <= n; ++i)
for (int j = 0; j <= m; ++j)
for (int k = 0; k <= len; ++k) F[i][j][k] = INF;
F[0][0][0] = 0;
for (int i = 0; i <= n; ++i)
for (int j = 0; j <= m; ++j)
for (int k = 0; k <= len; ++k) {
for (int tp = -1; tp <= 1; ++tp) {
if (!tp || F[i][j][k] >= INF) continue;
int u = i, v = j;
if (A[u + 1] == tp) ++u;
if (B[v + 1] == tp) ++v;
if (k + tp < 0 || k + tp > len) continue;
if (F[u][v][k + tp] > F[i][j][k] + 1) {
F[u][v][k + tp] = F[i][j][k] + 1;
G[u][v][k + tp] = tp, H[u][v][k + tp] = i, P[u][v][k + tp] = j;
}
}
}
int sans = INF, sp = 0, cnt = 0;
for (int k = 0; k <= len; ++k)
if (F[n][m][k] + k < sans) {
sans = F[n][m][k] + k;
sp = k;
}
for (int i = 1; i <= sp; ++i) ans[++cnt] = ')';
int u = n, v = m;
while (u || v || sp) {
int s = G[u][v][sp], s1 = u, s2 = v;
if (s == 1)
ans[++cnt] = '(';
else
ans[++cnt] = ')';
u = H[s1][s2][sp], v = P[s1][s2][sp];
sp -= s;
}
for (int i = cnt; i >= 1; --i) putchar(ans[i]);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int MAX = 205;
const int INF = 0x3f3f3f3f;
string a, b;
int dp[MAX][MAX][405];
int call(int x, int y, int s) {
if (x >= a.size() and y >= b.size()) return s;
if (dp[x][y][s] != -1) return dp[x][y][s];
int ret = INF;
if (x < a.size() and y < b.size() and a[x] == b[y]) {
if (a[x] == '(') {
ret = min(ret, 1 + call(x + 1, y + 1, s + 1));
} else {
if (s) {
ret = min(ret, 1 + call(x + 1, y + 1, s - 1));
} else {
ret = min(ret, 2 + call(x + 1, y + 1, s));
}
}
} else {
if (x < a.size()) {
if (a[x] == '(') {
ret = min(ret, 1 + call(x + 1, y, s + 1));
} else {
if (s) {
ret = min(ret, 1 + call(x + 1, y, s - 1));
} else {
ret = min(ret, 2 + call(x + 1, y, s));
}
}
}
if (y < b.size()) {
if (b[y] == '(') {
ret = min(ret, 1 + call(x, y + 1, s + 1));
} else {
if (s) {
ret = min(ret, 1 + call(x, y + 1, s - 1));
} else {
ret = min(ret, 2 + call(x, y + 1, s));
}
}
}
}
return dp[x][y][s] = ret;
}
void go(int x, int y, int s) {
if (x >= a.size() and y >= b.size()) {
for (int i = 0; i < s; i++) cout << ')';
cout << endl;
return;
}
int ret = call(x, y, s);
if (x < a.size() and y < b.size() and a[x] == b[y]) {
if (a[x] == '(') {
if (ret == 1 + call(x + 1, y + 1, s + 1)) {
cout << '(';
go(x + 1, y + 1, s + 1);
return;
}
} else {
if (s) {
if (ret == 1 + call(x + 1, y + 1, s - 1)) {
cout << ')';
go(x + 1, y + 1, s - 1);
return;
}
} else {
if (ret == 2 + call(x + 1, y + 1, s)) {
cout << "()";
go(x + 1, y + 1, s);
return;
}
}
}
} else {
if (x < a.size()) {
if (a[x] == '(') {
if (ret == 1 + call(x + 1, y, s + 1)) {
cout << '(';
go(x + 1, y, s + 1);
return;
}
} else {
if (s) {
if (ret == 1 + call(x + 1, y, s - 1)) {
cout << ')';
go(x + 1, y, s - 1);
return;
}
} else {
if (ret == 2 + call(x + 1, y, s)) {
cout << "()";
go(x + 1, y, s);
return;
}
}
}
}
if (y < b.size()) {
if (b[y] == '(') {
if (ret == 1 + call(x, y + 1, s + 1)) {
cout << '(';
go(x, y + 1, s + 1);
return;
}
} else {
if (s) {
if (ret == 1 + call(x, y + 1, s - 1)) {
cout << ')';
go(x, y + 1, s - 1);
return;
}
} else {
if (ret == 2 + call(x, y + 1, s)) {
cout << "()";
go(x, y + 1, s);
return;
}
}
}
}
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cin >> a >> b;
memset(dp, -1, sizeof(dp));
go(0, 0, 0);
return 0;
}
| ### Prompt
Generate a Cpp solution to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int MAX = 205;
const int INF = 0x3f3f3f3f;
string a, b;
int dp[MAX][MAX][405];
int call(int x, int y, int s) {
if (x >= a.size() and y >= b.size()) return s;
if (dp[x][y][s] != -1) return dp[x][y][s];
int ret = INF;
if (x < a.size() and y < b.size() and a[x] == b[y]) {
if (a[x] == '(') {
ret = min(ret, 1 + call(x + 1, y + 1, s + 1));
} else {
if (s) {
ret = min(ret, 1 + call(x + 1, y + 1, s - 1));
} else {
ret = min(ret, 2 + call(x + 1, y + 1, s));
}
}
} else {
if (x < a.size()) {
if (a[x] == '(') {
ret = min(ret, 1 + call(x + 1, y, s + 1));
} else {
if (s) {
ret = min(ret, 1 + call(x + 1, y, s - 1));
} else {
ret = min(ret, 2 + call(x + 1, y, s));
}
}
}
if (y < b.size()) {
if (b[y] == '(') {
ret = min(ret, 1 + call(x, y + 1, s + 1));
} else {
if (s) {
ret = min(ret, 1 + call(x, y + 1, s - 1));
} else {
ret = min(ret, 2 + call(x, y + 1, s));
}
}
}
}
return dp[x][y][s] = ret;
}
void go(int x, int y, int s) {
if (x >= a.size() and y >= b.size()) {
for (int i = 0; i < s; i++) cout << ')';
cout << endl;
return;
}
int ret = call(x, y, s);
if (x < a.size() and y < b.size() and a[x] == b[y]) {
if (a[x] == '(') {
if (ret == 1 + call(x + 1, y + 1, s + 1)) {
cout << '(';
go(x + 1, y + 1, s + 1);
return;
}
} else {
if (s) {
if (ret == 1 + call(x + 1, y + 1, s - 1)) {
cout << ')';
go(x + 1, y + 1, s - 1);
return;
}
} else {
if (ret == 2 + call(x + 1, y + 1, s)) {
cout << "()";
go(x + 1, y + 1, s);
return;
}
}
}
} else {
if (x < a.size()) {
if (a[x] == '(') {
if (ret == 1 + call(x + 1, y, s + 1)) {
cout << '(';
go(x + 1, y, s + 1);
return;
}
} else {
if (s) {
if (ret == 1 + call(x + 1, y, s - 1)) {
cout << ')';
go(x + 1, y, s - 1);
return;
}
} else {
if (ret == 2 + call(x + 1, y, s)) {
cout << "()";
go(x + 1, y, s);
return;
}
}
}
}
if (y < b.size()) {
if (b[y] == '(') {
if (ret == 1 + call(x, y + 1, s + 1)) {
cout << '(';
go(x, y + 1, s + 1);
return;
}
} else {
if (s) {
if (ret == 1 + call(x, y + 1, s - 1)) {
cout << ')';
go(x, y + 1, s - 1);
return;
}
} else {
if (ret == 2 + call(x, y + 1, s)) {
cout << "()";
go(x, y + 1, s);
return;
}
}
}
}
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cin >> a >> b;
memset(dp, -1, sizeof(dp));
go(0, 0, 0);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
string a, b;
cin >> a >> b;
int n1 = a.length(), n2 = b.length();
int n3 = max(n1, n2);
int dp[n1 + 1][n2 + 1][n3 + 1];
array<int, 4> prv[n1 + 1][n2 + 1][n3 + 1];
for (int i = 0; i <= n1; i++)
for (int j = 0; j <= n2; j++)
for (int k = 0; k <= n3; k++) dp[i][j][k] = 1000000;
dp[0][0][0] = 0;
for (int i = 0; i <= n1; i++) {
for (int j = 0; j <= n2; j++) {
for (int k = 0; k <= n3; k++) {
if (dp[i][j][k] == 1000000) continue;
int nx = i, ny = j, nk = k + 1;
if (i < n1 and a[i] == '(') nx++;
if (j < n2 and b[j] == '(') ny++;
if (k + 1 <= n3 and dp[i][j][k] + 1 < dp[nx][ny][nk]) {
dp[nx][ny][nk] = min(dp[nx][ny][nk], dp[i][j][k] + 1);
prv[nx][ny][nk] = {i, j, k, 0};
}
nx = i, ny = j, nk = k - 1;
if (i < n1 and a[i] == ')') nx++;
if (j < n2 and b[j] == ')') ny++;
if (nk >= 0 and dp[i][j][k] + 1 < dp[nx][ny][nk]) {
dp[nx][ny][nk] = min(dp[nx][ny][nk], dp[i][j][k] + 1);
prv[nx][ny][nk] = {i, j, k, 1};
}
}
}
}
int bal = 1000000, x, y, z;
for (int k = 0; k <= n3; k++) {
if (dp[n1][n2][k] + k < bal) {
bal = dp[n1][n2][k] + k, x = n1, y = n2, z = k;
}
}
string ans = "";
int ad = z;
while (x or y or z) {
if (prv[x][y][z][3])
ans += ')';
else
ans += '(';
int nx, ny, nz;
nx = prv[x][y][z][0];
ny = prv[x][y][z][1];
nz = prv[x][y][z][2];
x = nx, y = ny, z = nz;
}
reverse(ans.begin(), ans.end());
cout << ans;
for (int i = 0; i < ad; i++) cout << ')';
return 0;
}
| ### Prompt
In CPP, your task is to solve the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
string a, b;
cin >> a >> b;
int n1 = a.length(), n2 = b.length();
int n3 = max(n1, n2);
int dp[n1 + 1][n2 + 1][n3 + 1];
array<int, 4> prv[n1 + 1][n2 + 1][n3 + 1];
for (int i = 0; i <= n1; i++)
for (int j = 0; j <= n2; j++)
for (int k = 0; k <= n3; k++) dp[i][j][k] = 1000000;
dp[0][0][0] = 0;
for (int i = 0; i <= n1; i++) {
for (int j = 0; j <= n2; j++) {
for (int k = 0; k <= n3; k++) {
if (dp[i][j][k] == 1000000) continue;
int nx = i, ny = j, nk = k + 1;
if (i < n1 and a[i] == '(') nx++;
if (j < n2 and b[j] == '(') ny++;
if (k + 1 <= n3 and dp[i][j][k] + 1 < dp[nx][ny][nk]) {
dp[nx][ny][nk] = min(dp[nx][ny][nk], dp[i][j][k] + 1);
prv[nx][ny][nk] = {i, j, k, 0};
}
nx = i, ny = j, nk = k - 1;
if (i < n1 and a[i] == ')') nx++;
if (j < n2 and b[j] == ')') ny++;
if (nk >= 0 and dp[i][j][k] + 1 < dp[nx][ny][nk]) {
dp[nx][ny][nk] = min(dp[nx][ny][nk], dp[i][j][k] + 1);
prv[nx][ny][nk] = {i, j, k, 1};
}
}
}
}
int bal = 1000000, x, y, z;
for (int k = 0; k <= n3; k++) {
if (dp[n1][n2][k] + k < bal) {
bal = dp[n1][n2][k] + k, x = n1, y = n2, z = k;
}
}
string ans = "";
int ad = z;
while (x or y or z) {
if (prv[x][y][z][3])
ans += ')';
else
ans += '(';
int nx, ny, nz;
nx = prv[x][y][z][0];
ny = prv[x][y][z][1];
nz = prv[x][y][z][2];
x = nx, y = ny, z = nz;
}
reverse(ans.begin(), ans.end());
cout << ans;
for (int i = 0; i < ad; i++) cout << ')';
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 210, MX = 2 * N;
int cost[MX][2 * N][N];
int n, m, cur, i, j;
char s[N], t[N];
queue<pair<int, pair<int, int> > > q;
inline void add(int cur, int i, int j, int C) {
if (cur < 0 || cur >= MX || cost[cur][i][j] != -1) return;
cost[cur][i][j] = C;
q.push(make_pair(cur, make_pair(i, j)));
}
inline bool check(int cur1, int i, int j, int cur2, int i2, int j2, char x) {
if (x == '(')
cur1++;
else
cur1--;
if (i < n && s[i] == x) i++;
if (j < m && t[j] == x) j++;
if (cur1 == cur2 && i == i2 && j == j2) return true;
return false;
}
string ans;
void build_path(int cur, int i, int j) {
if (cur == 0 && i == 0 && j == 0) return;
for (int cur1 = max(0, cur - 1); cur1 <= cur + 1; cur1++) {
if (cur1 == cur) continue;
for (int i2 = i - 1; i2 <= i; i2++) {
for (int j2 = j - 1; j2 <= j; j2++) {
if (check(cur1, i2, j2, cur, i, j, '(') &&
cost[cur1][i2][j2] == cost[cur][i][j] - 1) {
ans += '(';
build_path(cur1, i2, j2);
return;
}
if (check(cur1, i2, j2, cur, i, j, ')') &&
cost[cur1][i2][j2] == cost[cur][i][j] - 1) {
ans += ')';
build_path(cur1, i2, j2);
return;
}
}
}
}
assert(false);
}
int main() {
memset(cost, -1, sizeof(cost));
scanf("%s", s);
scanf("%s", t);
n = strlen(s);
m = strlen(t);
q.push(make_pair(0, make_pair(0, 0)));
cost[0][0][0] = 0;
while (!q.empty()) {
cur = q.front().first;
i = q.front().second.first;
j = q.front().second.second;
q.pop();
if (cur == 0 && i == n && j == m) {
break;
}
add(cur + 1, i + (i < n && s[i] == '('), j + (j < m && t[j] == '('),
cost[cur][i][j] + 1);
add(cur - 1, i + (i < n && s[i] == ')'), j + (j < m && t[j] == ')'),
cost[cur][i][j] + 1);
}
build_path(0, n, m);
reverse(ans.begin(), ans.end());
puts(ans.c_str());
return 0;
}
| ### Prompt
Please create a solution in CPP to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 210, MX = 2 * N;
int cost[MX][2 * N][N];
int n, m, cur, i, j;
char s[N], t[N];
queue<pair<int, pair<int, int> > > q;
inline void add(int cur, int i, int j, int C) {
if (cur < 0 || cur >= MX || cost[cur][i][j] != -1) return;
cost[cur][i][j] = C;
q.push(make_pair(cur, make_pair(i, j)));
}
inline bool check(int cur1, int i, int j, int cur2, int i2, int j2, char x) {
if (x == '(')
cur1++;
else
cur1--;
if (i < n && s[i] == x) i++;
if (j < m && t[j] == x) j++;
if (cur1 == cur2 && i == i2 && j == j2) return true;
return false;
}
string ans;
void build_path(int cur, int i, int j) {
if (cur == 0 && i == 0 && j == 0) return;
for (int cur1 = max(0, cur - 1); cur1 <= cur + 1; cur1++) {
if (cur1 == cur) continue;
for (int i2 = i - 1; i2 <= i; i2++) {
for (int j2 = j - 1; j2 <= j; j2++) {
if (check(cur1, i2, j2, cur, i, j, '(') &&
cost[cur1][i2][j2] == cost[cur][i][j] - 1) {
ans += '(';
build_path(cur1, i2, j2);
return;
}
if (check(cur1, i2, j2, cur, i, j, ')') &&
cost[cur1][i2][j2] == cost[cur][i][j] - 1) {
ans += ')';
build_path(cur1, i2, j2);
return;
}
}
}
}
assert(false);
}
int main() {
memset(cost, -1, sizeof(cost));
scanf("%s", s);
scanf("%s", t);
n = strlen(s);
m = strlen(t);
q.push(make_pair(0, make_pair(0, 0)));
cost[0][0][0] = 0;
while (!q.empty()) {
cur = q.front().first;
i = q.front().second.first;
j = q.front().second.second;
q.pop();
if (cur == 0 && i == n && j == m) {
break;
}
add(cur + 1, i + (i < n && s[i] == '('), j + (j < m && t[j] == '('),
cost[cur][i][j] + 1);
add(cur - 1, i + (i < n && s[i] == ')'), j + (j < m && t[j] == ')'),
cost[cur][i][j] + 1);
}
build_path(0, n, m);
reverse(ans.begin(), ans.end());
puts(ans.c_str());
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
string a, b, ans;
int dp[205 << 1][205][205], n, m;
int solve(int e, int i, int j) {
if (e < 0 || e > n + m) return 1000000009;
if (i == 0 && j == 0) return e;
if (dp[e][i][j] == -1) {
int op1 = solve(e + 1, i - (i > 0 && a[i - 1] == ')'),
j - (j > 0 && b[j - 1] == ')'));
int op2 = solve(e - 1, i - (i > 0 && a[i - 1] == '('),
j - (j > 0 && b[j - 1] == '('));
dp[e][i][j] = 1 + min(op1, op2);
}
return dp[e][i][j];
}
void print(int e, int i, int j) {
if (i == 0 && j == 0) {
while (e > 0) {
ans.push_back('(');
e--;
}
return;
}
int op1 = solve(e + 1, i - (i > 0 && a[i - 1] == ')'),
j - (j > 0 && b[j - 1] == ')'));
int op2 = solve(e - 1, i - (i > 0 && a[i - 1] == '('),
j - (j > 0 && b[j - 1] == '('));
if (op1 < op2) {
ans.push_back(')');
print(e + 1, i - (i > 0 && a[i - 1] == ')'),
j - (j > 0 && b[j - 1] == ')'));
} else {
ans.push_back('(');
print(e - 1, i - (i > 0 && a[i - 1] == '('),
j - (j > 0 && b[j - 1] == '('));
}
}
int main() {
memset(dp, -1, sizeof(dp));
cin >> a >> b;
n = a.size(), m = b.size();
print(0, n, m);
reverse(ans.begin(), ans.end());
cout << ans << endl;
return 0;
}
| ### Prompt
In CPP, your task is to solve the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
string a, b, ans;
int dp[205 << 1][205][205], n, m;
int solve(int e, int i, int j) {
if (e < 0 || e > n + m) return 1000000009;
if (i == 0 && j == 0) return e;
if (dp[e][i][j] == -1) {
int op1 = solve(e + 1, i - (i > 0 && a[i - 1] == ')'),
j - (j > 0 && b[j - 1] == ')'));
int op2 = solve(e - 1, i - (i > 0 && a[i - 1] == '('),
j - (j > 0 && b[j - 1] == '('));
dp[e][i][j] = 1 + min(op1, op2);
}
return dp[e][i][j];
}
void print(int e, int i, int j) {
if (i == 0 && j == 0) {
while (e > 0) {
ans.push_back('(');
e--;
}
return;
}
int op1 = solve(e + 1, i - (i > 0 && a[i - 1] == ')'),
j - (j > 0 && b[j - 1] == ')'));
int op2 = solve(e - 1, i - (i > 0 && a[i - 1] == '('),
j - (j > 0 && b[j - 1] == '('));
if (op1 < op2) {
ans.push_back(')');
print(e + 1, i - (i > 0 && a[i - 1] == ')'),
j - (j > 0 && b[j - 1] == ')'));
} else {
ans.push_back('(');
print(e - 1, i - (i > 0 && a[i - 1] == '('),
j - (j > 0 && b[j - 1] == '('));
}
}
int main() {
memset(dp, -1, sizeof(dp));
cin >> a >> b;
n = a.size(), m = b.size();
print(0, n, m);
reverse(ans.begin(), ans.end());
cout << ans << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 200 + 11;
int dp[N][N][N];
pair<pair<int, int>, pair<int, int> > pr[N][N][N];
vector<pair<int, pair<int, int> > > vv;
int n, m;
void up(int l, int r, int k, int ans, int c1, int c2, int c3, int c4) {
if (k < 0 || k > max(n, m)) return;
if (dp[l][r][k] != -1) return;
dp[l][r][k] = ans;
pr[l][r][k] = make_pair(make_pair(c1, c2), make_pair(c3, c4));
vv.push_back(make_pair(l, make_pair(r, k)));
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
string s, t;
cin >> s >> t;
n = s.size();
m = t.size();
s = "." + s;
t = "." + t;
for (int i = 0; i <= n; i++)
for (int j = 0; j <= m; j++)
for (int bal = 0; bal <= max(n, m); bal++) dp[i][j][bal] = -1;
dp[0][0][0] = 0;
vv.push_back(make_pair(0, make_pair(0, 0)));
for (int j = 0; j < vv.size(); j++) {
int l = vv[j].first;
int r = vv[j].second.first;
int k = vv[j].second.second;
int ans = dp[l][r][k] + 1;
up(l, r, k + 1, ans, 1, l, r, k);
up(l, r, k - 1, ans, -1, l, r, k);
if (l < n && r < m && s[l + 1] == '(' && t[r + 1] == '(')
up(l + 1, r + 1, k + 1, ans, 1, l, r, k);
if (l < n && r < m && s[l + 1] == ')' && t[r + 1] == ')')
up(l + 1, r + 1, k - 1, ans, -1, l, r, k);
if (l < n && s[l + 1] == '(') up(l + 1, r, k + 1, ans, 1, l, r, k);
if (l < n && s[l + 1] == ')') up(l + 1, r, k - 1, ans, -1, l, r, k);
if (r < m && t[r + 1] == '(') up(l, r + 1, k + 1, ans, 1, l, r, k);
if (r < m && t[r + 1] == ')') up(l, r + 1, k - 1, ans, -1, l, r, k);
}
string ans = "";
int k = 0;
while (n != 0 || m != 0 || k != 0) {
pair<pair<int, int>, pair<int, int> > p = pr[n][m][k];
if (p.first.first == 1)
ans += "(";
else
ans += ")";
n = p.first.second;
m = p.second.first;
k = p.second.second;
}
reverse(ans.begin(), ans.end());
cout << ans;
}
| ### Prompt
Your challenge is to write a cpp solution to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 200 + 11;
int dp[N][N][N];
pair<pair<int, int>, pair<int, int> > pr[N][N][N];
vector<pair<int, pair<int, int> > > vv;
int n, m;
void up(int l, int r, int k, int ans, int c1, int c2, int c3, int c4) {
if (k < 0 || k > max(n, m)) return;
if (dp[l][r][k] != -1) return;
dp[l][r][k] = ans;
pr[l][r][k] = make_pair(make_pair(c1, c2), make_pair(c3, c4));
vv.push_back(make_pair(l, make_pair(r, k)));
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
string s, t;
cin >> s >> t;
n = s.size();
m = t.size();
s = "." + s;
t = "." + t;
for (int i = 0; i <= n; i++)
for (int j = 0; j <= m; j++)
for (int bal = 0; bal <= max(n, m); bal++) dp[i][j][bal] = -1;
dp[0][0][0] = 0;
vv.push_back(make_pair(0, make_pair(0, 0)));
for (int j = 0; j < vv.size(); j++) {
int l = vv[j].first;
int r = vv[j].second.first;
int k = vv[j].second.second;
int ans = dp[l][r][k] + 1;
up(l, r, k + 1, ans, 1, l, r, k);
up(l, r, k - 1, ans, -1, l, r, k);
if (l < n && r < m && s[l + 1] == '(' && t[r + 1] == '(')
up(l + 1, r + 1, k + 1, ans, 1, l, r, k);
if (l < n && r < m && s[l + 1] == ')' && t[r + 1] == ')')
up(l + 1, r + 1, k - 1, ans, -1, l, r, k);
if (l < n && s[l + 1] == '(') up(l + 1, r, k + 1, ans, 1, l, r, k);
if (l < n && s[l + 1] == ')') up(l + 1, r, k - 1, ans, -1, l, r, k);
if (r < m && t[r + 1] == '(') up(l, r + 1, k + 1, ans, 1, l, r, k);
if (r < m && t[r + 1] == ')') up(l, r + 1, k - 1, ans, -1, l, r, k);
}
string ans = "";
int k = 0;
while (n != 0 || m != 0 || k != 0) {
pair<pair<int, int>, pair<int, int> > p = pr[n][m][k];
if (p.first.first == 1)
ans += "(";
else
ans += ")";
n = p.first.second;
m = p.second.first;
k = p.second.second;
}
reverse(ans.begin(), ans.end());
cout << ans;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
string s, t;
cin >> s >> t;
int n = s.size(), m = t.size();
vector<vector<vector<int>>> dp0(
n + 1, vector<vector<int>>(m + 1, vector<int>(n + m + 5, 1 << 29)));
vector<vector<vector<pair<pair<int, int>, int>>>> dp1(
n + 1, vector<vector<pair<pair<int, int>, int>>>(
m + 1, vector<pair<pair<int, int>, int>>(
n + m + 5, make_pair(make_pair(-1, -1), -1))));
dp0[0][0][0] = 0;
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= m; j++) {
for (int k = 0; k <= n + m; k++) {
if (dp0[i][j][k] == (1 << 29)) continue;
int x = i, y = j;
if (i < n && s[i] == '(') x++;
if (j < m && t[j] == '(') y++;
if (dp0[x][y][k + 1] > (dp0[i][j][k] + 1)) {
dp0[x][y][k + 1] = dp0[i][j][k] + 1;
dp1[x][y][k + 1] = make_pair(make_pair(i, j), k);
}
x = i, y = j;
if (i < n && s[i] == ')') x++;
if (j < m && t[j] == ')') y++;
if (k > 0 && dp0[x][y][k - 1] > (dp0[i][j][k] + 1)) {
dp0[x][y][k - 1] = dp0[i][j][k] + 1;
dp1[x][y][k - 1] = make_pair(make_pair(i, j), k);
}
}
}
}
int res = 1 << 29;
int x = n, y = m, z = 0;
for (int k = 0; k <= n + m; k++) {
if (res > (dp0[n][m][k] + k)) {
res = dp0[n][m][k] + k;
z = k;
}
}
string str;
for (int i = 0; i < z; i++) str.push_back(')');
while (x != 0 || y != 0 || z != 0) {
if (dp1[x][y][z].second == z) {
str.push_back(')');
str.push_back('(');
} else if (dp1[x][y][z].second < z) {
str.push_back('(');
} else {
str.push_back(')');
}
int nx = dp1[x][y][z].first.first;
int ny = dp1[x][y][z].first.second;
int nz = dp1[x][y][z].second;
x = nx, y = ny, z = nz;
}
reverse(str.begin(), str.end());
cout << str << "\n";
return 0;
}
| ### Prompt
Please formulate a Cpp solution to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
string s, t;
cin >> s >> t;
int n = s.size(), m = t.size();
vector<vector<vector<int>>> dp0(
n + 1, vector<vector<int>>(m + 1, vector<int>(n + m + 5, 1 << 29)));
vector<vector<vector<pair<pair<int, int>, int>>>> dp1(
n + 1, vector<vector<pair<pair<int, int>, int>>>(
m + 1, vector<pair<pair<int, int>, int>>(
n + m + 5, make_pair(make_pair(-1, -1), -1))));
dp0[0][0][0] = 0;
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= m; j++) {
for (int k = 0; k <= n + m; k++) {
if (dp0[i][j][k] == (1 << 29)) continue;
int x = i, y = j;
if (i < n && s[i] == '(') x++;
if (j < m && t[j] == '(') y++;
if (dp0[x][y][k + 1] > (dp0[i][j][k] + 1)) {
dp0[x][y][k + 1] = dp0[i][j][k] + 1;
dp1[x][y][k + 1] = make_pair(make_pair(i, j), k);
}
x = i, y = j;
if (i < n && s[i] == ')') x++;
if (j < m && t[j] == ')') y++;
if (k > 0 && dp0[x][y][k - 1] > (dp0[i][j][k] + 1)) {
dp0[x][y][k - 1] = dp0[i][j][k] + 1;
dp1[x][y][k - 1] = make_pair(make_pair(i, j), k);
}
}
}
}
int res = 1 << 29;
int x = n, y = m, z = 0;
for (int k = 0; k <= n + m; k++) {
if (res > (dp0[n][m][k] + k)) {
res = dp0[n][m][k] + k;
z = k;
}
}
string str;
for (int i = 0; i < z; i++) str.push_back(')');
while (x != 0 || y != 0 || z != 0) {
if (dp1[x][y][z].second == z) {
str.push_back(')');
str.push_back('(');
} else if (dp1[x][y][z].second < z) {
str.push_back('(');
} else {
str.push_back(')');
}
int nx = dp1[x][y][z].first.first;
int ny = dp1[x][y][z].first.second;
int nz = dp1[x][y][z].second;
x = nx, y = ny, z = nz;
}
reverse(str.begin(), str.end());
cout << str << "\n";
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1000000007;
int opt[201][201][201];
int par[201][201][201];
void print(int n, int m, int K) {
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
for (int k = 0; k < K; k++) {
cout << opt[i][j][k] << " ";
}
cout << endl;
}
cout << endl;
}
}
void findstring(int n, int m, string &s, string &t) {
int i = 0, j = 0, k = 0, p = par[i][j][k];
string ans;
int ni, nj, c;
while (p != -1) {
switch (p) {
case 1:
ans += ')';
k--;
break;
case 2:
ni = i < n ? i + 1 : i;
nj = j < m ? j + 1 : j;
c = i == n ? t[j] : s[i];
if (c == '(') {
ans += '(';
k++;
} else if (k > 0) {
ans += ')';
k--;
} else {
ans += "()";
}
i = ni;
j = nj;
break;
case 3:
ans += '(';
i++;
k++;
break;
case 4:
ans += ')';
j++;
k--;
break;
case 5:
ans += "()";
j++;
break;
case 6:
ans += '(';
j++;
k++;
break;
case 7:
ans += ')';
i++;
k--;
break;
case 8:
ans += "()";
i++;
break;
default:
assert(false);
}
p = par[i][j][k];
}
cout << ans << endl;
}
int dp(int i, int j, int k, string &s, string &t) {
if (opt[i][j][k] != -1) return opt[i][j][k];
int n = s.length(), m = t.length();
int ans = INT_MAX;
if (i == n && j == m && k > 0) {
ans = 1 + dp(i, j, k - 1, s, t);
par[i][j][k] = 1;
} else if (i == n || j == m || s[i] == t[j]) {
int ni = i < n ? i + 1 : i;
int nj = j < m ? j + 1 : j;
char c = i == n ? t[j] : s[i];
if (c == '(') {
ans = 1 + dp(ni, nj, k + 1, s, t);
} else if (k > 0) {
ans = 1 + dp(ni, nj, k - 1, s, t);
} else {
ans = 2 + dp(ni, nj, k, s, t);
}
par[i][j][k] = 2;
} else if (s[i] == '(') {
ans = 1 + dp(i + 1, j, k + 1, s, t);
par[i][j][k] = 3;
if (k > 0) {
int x = 1 + dp(i, j + 1, k - 1, s, t);
if (x < ans) {
ans = x;
par[i][j][k] = 4;
}
} else {
int x = 2 + dp(i, j + 1, k, s, t);
if (x < ans) {
ans = x;
par[i][j][k] = 5;
}
}
} else {
ans = 1 + dp(i, j + 1, k + 1, s, t);
par[i][j][k] = 6;
if (k > 0) {
int x = 1 + dp(i + 1, j, k - 1, s, t);
if (x < ans) {
ans = x;
par[i][j][k] = 7;
}
} else {
int x = 2 + dp(i + 1, j, k, s, t);
if (x < ans) {
ans = x;
par[i][j][k] = 8;
}
}
}
opt[i][j][k] = ans;
return ans;
}
void sherlock(int x) {
string s, t;
cin >> s >> t;
int n = s.length(), m = t.length(), K = max(n, m);
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= m; j++) {
for (int k = 0; k <= K; k++) {
opt[i][j][k] = -1;
par[i][j][k] = -1;
}
}
}
opt[n][m][0] = 0;
dp(0, 0, 0, s, t);
findstring(n, m, s, t);
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int tt;
tt = 1;
for (int t = 1; t <= tt; t++) sherlock(t);
return 0;
}
| ### Prompt
Develop a solution in CPP to the problem described below:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1000000007;
int opt[201][201][201];
int par[201][201][201];
void print(int n, int m, int K) {
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
for (int k = 0; k < K; k++) {
cout << opt[i][j][k] << " ";
}
cout << endl;
}
cout << endl;
}
}
void findstring(int n, int m, string &s, string &t) {
int i = 0, j = 0, k = 0, p = par[i][j][k];
string ans;
int ni, nj, c;
while (p != -1) {
switch (p) {
case 1:
ans += ')';
k--;
break;
case 2:
ni = i < n ? i + 1 : i;
nj = j < m ? j + 1 : j;
c = i == n ? t[j] : s[i];
if (c == '(') {
ans += '(';
k++;
} else if (k > 0) {
ans += ')';
k--;
} else {
ans += "()";
}
i = ni;
j = nj;
break;
case 3:
ans += '(';
i++;
k++;
break;
case 4:
ans += ')';
j++;
k--;
break;
case 5:
ans += "()";
j++;
break;
case 6:
ans += '(';
j++;
k++;
break;
case 7:
ans += ')';
i++;
k--;
break;
case 8:
ans += "()";
i++;
break;
default:
assert(false);
}
p = par[i][j][k];
}
cout << ans << endl;
}
int dp(int i, int j, int k, string &s, string &t) {
if (opt[i][j][k] != -1) return opt[i][j][k];
int n = s.length(), m = t.length();
int ans = INT_MAX;
if (i == n && j == m && k > 0) {
ans = 1 + dp(i, j, k - 1, s, t);
par[i][j][k] = 1;
} else if (i == n || j == m || s[i] == t[j]) {
int ni = i < n ? i + 1 : i;
int nj = j < m ? j + 1 : j;
char c = i == n ? t[j] : s[i];
if (c == '(') {
ans = 1 + dp(ni, nj, k + 1, s, t);
} else if (k > 0) {
ans = 1 + dp(ni, nj, k - 1, s, t);
} else {
ans = 2 + dp(ni, nj, k, s, t);
}
par[i][j][k] = 2;
} else if (s[i] == '(') {
ans = 1 + dp(i + 1, j, k + 1, s, t);
par[i][j][k] = 3;
if (k > 0) {
int x = 1 + dp(i, j + 1, k - 1, s, t);
if (x < ans) {
ans = x;
par[i][j][k] = 4;
}
} else {
int x = 2 + dp(i, j + 1, k, s, t);
if (x < ans) {
ans = x;
par[i][j][k] = 5;
}
}
} else {
ans = 1 + dp(i, j + 1, k + 1, s, t);
par[i][j][k] = 6;
if (k > 0) {
int x = 1 + dp(i + 1, j, k - 1, s, t);
if (x < ans) {
ans = x;
par[i][j][k] = 7;
}
} else {
int x = 2 + dp(i + 1, j, k, s, t);
if (x < ans) {
ans = x;
par[i][j][k] = 8;
}
}
}
opt[i][j][k] = ans;
return ans;
}
void sherlock(int x) {
string s, t;
cin >> s >> t;
int n = s.length(), m = t.length(), K = max(n, m);
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= m; j++) {
for (int k = 0; k <= K; k++) {
opt[i][j][k] = -1;
par[i][j][k] = -1;
}
}
}
opt[n][m][0] = 0;
dp(0, 0, 0, s, t);
findstring(n, m, s, t);
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int tt;
tt = 1;
for (int t = 1; t <= tt; t++) sherlock(t);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
string S, T;
int A, B;
int hoge[202][202][204];
vector<int> from[202][202][204];
void solve() {
int i, j, k, l, r, x, y;
string s;
cin >> S >> T;
A = S.size();
B = T.size();
memset(hoge, 0x11, sizeof(hoge));
hoge[0][0][0] = 0;
for (x = 0; x < (A + 1); x++)
for (y = 0; y < (B + 1); y++) {
for (i = 0; i < (102); i++) {
int nx = x;
int ny = y;
if (x < A && S[x] == '(') nx++;
if (y < B && T[y] == '(') ny++;
if (hoge[x][y][i] + 1 < hoge[nx][ny][i + 1]) {
hoge[nx][ny][i + 1] = hoge[x][y][i] + 1;
from[nx][ny][i + 1] = {x, y, i, '('};
}
nx = x, ny = y;
if (x < A && S[x] == ')') nx++;
if (y < B && T[y] == ')') ny++;
if (i && hoge[x][y][i] + 1 < hoge[nx][ny][i - 1]) {
hoge[nx][ny][i - 1] = hoge[x][y][i] + 1;
from[nx][ny][i - 1] = {x, y, i, ')'};
}
}
for (i = 100; i >= 0; i--) {
int nx = x;
int ny = y;
if (x < A && S[x] == '(') nx++;
if (y < B && T[y] == '(') ny++;
if (hoge[x][y][i] + 1 < hoge[nx][ny][i + 1]) {
hoge[nx][ny][i + 1] = hoge[x][y][i] + 1;
from[nx][ny][i + 1] = {x, y, i, '('};
}
nx = x, ny = y;
if (x < A && S[x] == ')') nx++;
if (y < B && T[y] == ')') ny++;
if (i && hoge[x][y][i] + 1 < hoge[nx][ny][i - 1]) {
hoge[nx][ny][i - 1] = hoge[x][y][i] + 1;
from[nx][ny][i - 1] = {x, y, i, ')'};
}
}
}
x = A, y = B, i = 0;
while (hoge[x][y][i]) {
s.push_back(from[x][y][i][3]);
int nx = from[x][y][i][0];
int ny = from[x][y][i][1];
int ni = from[x][y][i][2];
x = nx, y = ny, i = ni;
}
reverse((s.begin()), (s.end()));
cout << s << endl;
}
int main(int argc, char** argv) {
string s;
int i;
if (argc == 1) ios::sync_with_stdio(false), cin.tie(0);
for (i = 0; i < (argc - 1); i++) s += argv[i + 1], s += '\n';
for (i = 0; i < (s.size()); i++) ungetc(s[s.size() - 1 - i], stdin);
cout.tie(0);
solve();
return 0;
}
| ### Prompt
Please provide a cpp coded solution to the problem described below:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
string S, T;
int A, B;
int hoge[202][202][204];
vector<int> from[202][202][204];
void solve() {
int i, j, k, l, r, x, y;
string s;
cin >> S >> T;
A = S.size();
B = T.size();
memset(hoge, 0x11, sizeof(hoge));
hoge[0][0][0] = 0;
for (x = 0; x < (A + 1); x++)
for (y = 0; y < (B + 1); y++) {
for (i = 0; i < (102); i++) {
int nx = x;
int ny = y;
if (x < A && S[x] == '(') nx++;
if (y < B && T[y] == '(') ny++;
if (hoge[x][y][i] + 1 < hoge[nx][ny][i + 1]) {
hoge[nx][ny][i + 1] = hoge[x][y][i] + 1;
from[nx][ny][i + 1] = {x, y, i, '('};
}
nx = x, ny = y;
if (x < A && S[x] == ')') nx++;
if (y < B && T[y] == ')') ny++;
if (i && hoge[x][y][i] + 1 < hoge[nx][ny][i - 1]) {
hoge[nx][ny][i - 1] = hoge[x][y][i] + 1;
from[nx][ny][i - 1] = {x, y, i, ')'};
}
}
for (i = 100; i >= 0; i--) {
int nx = x;
int ny = y;
if (x < A && S[x] == '(') nx++;
if (y < B && T[y] == '(') ny++;
if (hoge[x][y][i] + 1 < hoge[nx][ny][i + 1]) {
hoge[nx][ny][i + 1] = hoge[x][y][i] + 1;
from[nx][ny][i + 1] = {x, y, i, '('};
}
nx = x, ny = y;
if (x < A && S[x] == ')') nx++;
if (y < B && T[y] == ')') ny++;
if (i && hoge[x][y][i] + 1 < hoge[nx][ny][i - 1]) {
hoge[nx][ny][i - 1] = hoge[x][y][i] + 1;
from[nx][ny][i - 1] = {x, y, i, ')'};
}
}
}
x = A, y = B, i = 0;
while (hoge[x][y][i]) {
s.push_back(from[x][y][i][3]);
int nx = from[x][y][i][0];
int ny = from[x][y][i][1];
int ni = from[x][y][i][2];
x = nx, y = ny, i = ni;
}
reverse((s.begin()), (s.end()));
cout << s << endl;
}
int main(int argc, char** argv) {
string s;
int i;
if (argc == 1) ios::sync_with_stdio(false), cin.tie(0);
for (i = 0; i < (argc - 1); i++) s += argv[i + 1], s += '\n';
for (i = 0; i < (s.size()); i++) ungetc(s[s.size() - 1 - i], stdin);
cout.tie(0);
solve();
return 0;
}
``` |
#include <bits/stdc++.h>
const int Max = 210;
int f[Max][Max][Max * 2];
bool is[Max][Max][Max * 2];
int main() {
std::ios::sync_with_stdio(false);
std::cin.tie(0);
std::cout.tie(0);
std::string s1, s2;
std::cin >> s1 >> s2;
memset(f, 0x3f, sizeof f);
const int inf = f[0][0][0];
f[0][0][0] = 0;
int N = s1.size(), M = s2.size();
for (int i = 0; i <= N; ++i) {
for (int j = 0; j <= M; ++j) {
int a = i + (i < N && s1[i] == '(');
int b = j + (j < M && s2[j] == '(');
for (int k = 0; k <= N + M; ++k) {
if (f[i][j][k] == inf) {
continue;
}
if (f[a][b][k + 1] > f[i][j][k] + 1) {
f[a][b][k + 1] = f[i][j][k] + 1;
is[a][b][k + 1] = true;
}
}
a = i + (i < N && s1[i] == ')');
b = j + (j < M && s2[j] == ')');
for (int k = N + M; k >= 1; --k) {
if (f[i][j][k] == inf) {
continue;
}
if (f[a][b][k - 1] > f[i][j][k] + 1) {
f[a][b][k - 1] = f[i][j][k] + 1;
is[a][b][k - 1] = false;
}
}
}
}
int mn = inf, p = 0;
for (int i = 0; i <= N + M; ++i) {
if (mn > f[N][M][i] + i) {
mn = f[N][M][i] + i;
p = i;
}
}
std::string ans;
for (int i = 0; i < p; ++i) {
ans.push_back(')');
}
for (int i = N, j = M, k = p; i > 0 || j > 0 || k > 0;) {
int x = f[i][j][k] - 1;
char c;
if (is[i][j][k] == true) {
c = '(';
--k;
} else {
c = ')';
++k;
}
ans.push_back(c);
int a = (i > 0 && s1[i - 1] == c);
int b = (j > 0 && s2[j - 1] == c);
if (f[i][j][k] == x) {
continue;
} else if (a > 0 && f[i - 1][j][k] == x) {
--i;
} else if (b > 0 && f[i][j - 1][k] == x) {
--j;
} else {
--i;
--j;
}
}
std::reverse(ans.begin(), ans.end());
std::cout << ans;
return 0;
}
| ### Prompt
Create a solution in cpp for the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
const int Max = 210;
int f[Max][Max][Max * 2];
bool is[Max][Max][Max * 2];
int main() {
std::ios::sync_with_stdio(false);
std::cin.tie(0);
std::cout.tie(0);
std::string s1, s2;
std::cin >> s1 >> s2;
memset(f, 0x3f, sizeof f);
const int inf = f[0][0][0];
f[0][0][0] = 0;
int N = s1.size(), M = s2.size();
for (int i = 0; i <= N; ++i) {
for (int j = 0; j <= M; ++j) {
int a = i + (i < N && s1[i] == '(');
int b = j + (j < M && s2[j] == '(');
for (int k = 0; k <= N + M; ++k) {
if (f[i][j][k] == inf) {
continue;
}
if (f[a][b][k + 1] > f[i][j][k] + 1) {
f[a][b][k + 1] = f[i][j][k] + 1;
is[a][b][k + 1] = true;
}
}
a = i + (i < N && s1[i] == ')');
b = j + (j < M && s2[j] == ')');
for (int k = N + M; k >= 1; --k) {
if (f[i][j][k] == inf) {
continue;
}
if (f[a][b][k - 1] > f[i][j][k] + 1) {
f[a][b][k - 1] = f[i][j][k] + 1;
is[a][b][k - 1] = false;
}
}
}
}
int mn = inf, p = 0;
for (int i = 0; i <= N + M; ++i) {
if (mn > f[N][M][i] + i) {
mn = f[N][M][i] + i;
p = i;
}
}
std::string ans;
for (int i = 0; i < p; ++i) {
ans.push_back(')');
}
for (int i = N, j = M, k = p; i > 0 || j > 0 || k > 0;) {
int x = f[i][j][k] - 1;
char c;
if (is[i][j][k] == true) {
c = '(';
--k;
} else {
c = ')';
++k;
}
ans.push_back(c);
int a = (i > 0 && s1[i - 1] == c);
int b = (j > 0 && s2[j - 1] == c);
if (f[i][j][k] == x) {
continue;
} else if (a > 0 && f[i - 1][j][k] == x) {
--i;
} else if (b > 0 && f[i][j - 1][k] == x) {
--j;
} else {
--i;
--j;
}
}
std::reverse(ans.begin(), ans.end());
std::cout << ans;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e18;
struct par {
long long i, j, bal;
};
int32_t main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
map<char, long long> key = {{'(', 1}, {')', -1}};
string s, t;
cin >> s >> t;
long long n = (long long)s.size(), m = (long long)t.size();
vector<vector<vector<long long>>> dp(
n + 1, vector<vector<long long>>(m + 1, vector<long long>(205, INF)));
vector<vector<vector<par>>> p(n + 1,
vector<vector<par>>(m + 1, vector<par>(205)));
dp[0][0][0] = 0;
for (long long i = 0; i <= n; ++i)
for (long long j = 0; j <= m; ++j)
for (long long bal = 0; bal <= 200; ++bal) {
if (dp[i][j][bal] == INF) continue;
if (bal != 0) {
long long ni = i, nj = j, nbal = bal - 1;
if (i < n && s[i] == ')') ni++;
if (j < m && t[j] == ')') nj++;
if (dp[ni][nj][nbal] > dp[i][j][bal] + 1) {
dp[ni][nj][nbal] = dp[i][j][bal] + 1;
p[ni][nj][nbal] = {i, j, bal};
}
}
if (bal != 200) {
long long ni = i, nj = j, nbal = bal + 1;
if (i < n && s[i] == '(') ni++;
if (j < m && t[j] == '(') nj++;
if (dp[ni][nj][nbal] > dp[i][j][bal] + 1) {
dp[ni][nj][nbal] = dp[i][j][bal] + 1;
p[ni][nj][nbal] = {i, j, bal};
}
}
}
for (long long bal = 1; bal <= 200; ++bal) {
if (dp[n][m][0] > dp[n][m][bal] + bal) {
dp[n][m][0] = dp[n][m][bal] + bal;
p[n][m][0] = {n, m, bal};
}
}
long long ti = n, tj = m, tbal = 0;
string ans;
while (ti != 0 || tj != 0 || tbal != 0) {
auto tp = p[ti][tj][tbal];
ti = tp.i;
tj = tp.j;
for (long long i = 0; i < (tbal - tp.bal); ++i) ans += '(';
for (long long i = 0; i < (tp.bal - tbal); ++i) ans += ')';
tbal = tp.bal;
}
reverse(ans.begin(), ans.end());
cout << ans;
return 0;
}
| ### Prompt
Please provide a cpp coded solution to the problem described below:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e18;
struct par {
long long i, j, bal;
};
int32_t main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
map<char, long long> key = {{'(', 1}, {')', -1}};
string s, t;
cin >> s >> t;
long long n = (long long)s.size(), m = (long long)t.size();
vector<vector<vector<long long>>> dp(
n + 1, vector<vector<long long>>(m + 1, vector<long long>(205, INF)));
vector<vector<vector<par>>> p(n + 1,
vector<vector<par>>(m + 1, vector<par>(205)));
dp[0][0][0] = 0;
for (long long i = 0; i <= n; ++i)
for (long long j = 0; j <= m; ++j)
for (long long bal = 0; bal <= 200; ++bal) {
if (dp[i][j][bal] == INF) continue;
if (bal != 0) {
long long ni = i, nj = j, nbal = bal - 1;
if (i < n && s[i] == ')') ni++;
if (j < m && t[j] == ')') nj++;
if (dp[ni][nj][nbal] > dp[i][j][bal] + 1) {
dp[ni][nj][nbal] = dp[i][j][bal] + 1;
p[ni][nj][nbal] = {i, j, bal};
}
}
if (bal != 200) {
long long ni = i, nj = j, nbal = bal + 1;
if (i < n && s[i] == '(') ni++;
if (j < m && t[j] == '(') nj++;
if (dp[ni][nj][nbal] > dp[i][j][bal] + 1) {
dp[ni][nj][nbal] = dp[i][j][bal] + 1;
p[ni][nj][nbal] = {i, j, bal};
}
}
}
for (long long bal = 1; bal <= 200; ++bal) {
if (dp[n][m][0] > dp[n][m][bal] + bal) {
dp[n][m][0] = dp[n][m][bal] + bal;
p[n][m][0] = {n, m, bal};
}
}
long long ti = n, tj = m, tbal = 0;
string ans;
while (ti != 0 || tj != 0 || tbal != 0) {
auto tp = p[ti][tj][tbal];
ti = tp.i;
tj = tp.j;
for (long long i = 0; i < (tbal - tp.bal); ++i) ans += '(';
for (long long i = 0; i < (tp.bal - tbal); ++i) ans += ')';
tbal = tp.bal;
}
reverse(ans.begin(), ans.end());
cout << ans;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9;
int dp[209][209][409];
string s, s1, s2;
int bt(int id1, int id2, int curr) {
if (curr > 400) return inf;
if (id1 == s1.size() && id2 == s2.size() && !curr) return 0;
int &ret = dp[id1][id2][curr];
if (ret != -1) return ret;
int Nid1, Nid2;
ret = inf;
Nid1 = id1 + (id1 < s1.size() && s1[id1] == '(');
Nid2 = id2 + (id2 < s2.size() && s2[id2] == '(');
ret = min(ret, bt(Nid1, Nid2, curr + 1) + 1);
if (!curr) return ret;
Nid1 = id1 + (id1 < s1.size() && s1[id1] == ')');
Nid2 = id2 + (id2 < s2.size() && s2[id2] == ')');
ret = min(ret, bt(Nid1, Nid2, curr - 1) + 1);
return ret;
}
void path(int id1, int id2, int curr) {
if (id1 == s1.size() && id2 == s2.size() && !curr) return;
int ret = dp[id1][id2][curr], Nid1, Nid2;
Nid1 = id1 + (id1 < s1.size() && s1[id1] == '(');
Nid2 = id2 + (id2 < s2.size() && s2[id2] == '(');
if (ret == dp[Nid1][Nid2][curr + 1] + 1) {
s += '(', path(Nid1, Nid2, curr + 1);
return;
}
Nid1 = id1 + (id1 < s1.size() && s1[id1] == ')');
Nid2 = id2 + (id2 < s2.size() && s2[id2] == ')');
if (ret == dp[Nid1][Nid2][curr - 1] + 1) {
s += ')', path(Nid1, Nid2, curr - 1);
return;
}
}
int main() {
ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0), cin >> s1 >> s2;
memset(dp, -1, sizeof(dp));
dp[s1.size()][s2.size()][0] = 0;
int x = bt(0, 0, 0);
path(0, 0, 0);
cout << s << endl;
}
| ### Prompt
Your challenge is to write a Cpp solution to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9;
int dp[209][209][409];
string s, s1, s2;
int bt(int id1, int id2, int curr) {
if (curr > 400) return inf;
if (id1 == s1.size() && id2 == s2.size() && !curr) return 0;
int &ret = dp[id1][id2][curr];
if (ret != -1) return ret;
int Nid1, Nid2;
ret = inf;
Nid1 = id1 + (id1 < s1.size() && s1[id1] == '(');
Nid2 = id2 + (id2 < s2.size() && s2[id2] == '(');
ret = min(ret, bt(Nid1, Nid2, curr + 1) + 1);
if (!curr) return ret;
Nid1 = id1 + (id1 < s1.size() && s1[id1] == ')');
Nid2 = id2 + (id2 < s2.size() && s2[id2] == ')');
ret = min(ret, bt(Nid1, Nid2, curr - 1) + 1);
return ret;
}
void path(int id1, int id2, int curr) {
if (id1 == s1.size() && id2 == s2.size() && !curr) return;
int ret = dp[id1][id2][curr], Nid1, Nid2;
Nid1 = id1 + (id1 < s1.size() && s1[id1] == '(');
Nid2 = id2 + (id2 < s2.size() && s2[id2] == '(');
if (ret == dp[Nid1][Nid2][curr + 1] + 1) {
s += '(', path(Nid1, Nid2, curr + 1);
return;
}
Nid1 = id1 + (id1 < s1.size() && s1[id1] == ')');
Nid2 = id2 + (id2 < s2.size() && s2[id2] == ')');
if (ret == dp[Nid1][Nid2][curr - 1] + 1) {
s += ')', path(Nid1, Nid2, curr - 1);
return;
}
}
int main() {
ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0), cin >> s1 >> s2;
memset(dp, -1, sizeof(dp));
dp[s1.size()][s2.size()][0] = 0;
int x = bt(0, 0, 0);
path(0, 0, 0);
cout << s << endl;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 210;
const int INF = 0x3f3f3f3f;
int dp[N][N][N];
pair<pair<int, int>, pair<int, char>> p[N][N][N];
int n, m;
int main() {
string s, t;
cin >> s >> t;
n = s.size(), m = t.size();
for (int i = 0; i <= n; ++i) {
for (int j = 0; j <= m; ++j) {
for (int bal = 0; bal < N; ++bal) {
dp[i][j][bal] = INF;
}
}
}
dp[0][0][0] = 0;
for (int i = 0; i <= n; ++i) {
for (int j = 0; j <= m; ++j) {
for (int bal = 0; bal < N; ++bal) {
if (dp[i][j][bal] == INF) continue;
int nxti = i + (i < n && s[i] == '(');
int nxtj = j + (j < m && t[j] == '(');
if (bal + 1 < N && dp[nxti][nxtj][bal + 1] > dp[i][j][bal] + 1) {
dp[nxti][nxtj][bal + 1] = dp[i][j][bal] + 1;
p[nxti][nxtj][bal + 1] =
make_pair(make_pair(i, j), make_pair(bal, '('));
}
nxti = i + (i < n && s[i] == ')');
nxtj = j + (j < m && t[j] == ')');
if (bal > 0 && dp[nxti][nxtj][bal - 1] > dp[i][j][bal] + 1) {
dp[nxti][nxtj][bal - 1] = dp[i][j][bal] + 1;
p[nxti][nxtj][bal - 1] =
make_pair(make_pair(i, j), make_pair(bal, ')'));
}
}
}
}
int ci = n, cj = m, cbal = 0;
for (int bal = 0; bal < N; ++bal) {
if (dp[n][m][bal] + bal < dp[n][m][cbal] + cbal) {
cbal = bal;
}
}
string res = string(cbal, ')');
while (ci > 0 || cj > 0 || cbal != 0) {
int nci = p[ci][cj][cbal].first.first;
int ncj = p[ci][cj][cbal].first.second;
int ncbal = p[ci][cj][cbal].second.first;
res += p[ci][cj][cbal].second.second;
ci = nci;
cj = ncj;
cbal = ncbal;
}
reverse(res.begin(), res.end());
cout << res << endl;
return 0;
}
| ### Prompt
Construct a Cpp code solution to the problem outlined:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 210;
const int INF = 0x3f3f3f3f;
int dp[N][N][N];
pair<pair<int, int>, pair<int, char>> p[N][N][N];
int n, m;
int main() {
string s, t;
cin >> s >> t;
n = s.size(), m = t.size();
for (int i = 0; i <= n; ++i) {
for (int j = 0; j <= m; ++j) {
for (int bal = 0; bal < N; ++bal) {
dp[i][j][bal] = INF;
}
}
}
dp[0][0][0] = 0;
for (int i = 0; i <= n; ++i) {
for (int j = 0; j <= m; ++j) {
for (int bal = 0; bal < N; ++bal) {
if (dp[i][j][bal] == INF) continue;
int nxti = i + (i < n && s[i] == '(');
int nxtj = j + (j < m && t[j] == '(');
if (bal + 1 < N && dp[nxti][nxtj][bal + 1] > dp[i][j][bal] + 1) {
dp[nxti][nxtj][bal + 1] = dp[i][j][bal] + 1;
p[nxti][nxtj][bal + 1] =
make_pair(make_pair(i, j), make_pair(bal, '('));
}
nxti = i + (i < n && s[i] == ')');
nxtj = j + (j < m && t[j] == ')');
if (bal > 0 && dp[nxti][nxtj][bal - 1] > dp[i][j][bal] + 1) {
dp[nxti][nxtj][bal - 1] = dp[i][j][bal] + 1;
p[nxti][nxtj][bal - 1] =
make_pair(make_pair(i, j), make_pair(bal, ')'));
}
}
}
}
int ci = n, cj = m, cbal = 0;
for (int bal = 0; bal < N; ++bal) {
if (dp[n][m][bal] + bal < dp[n][m][cbal] + cbal) {
cbal = bal;
}
}
string res = string(cbal, ')');
while (ci > 0 || cj > 0 || cbal != 0) {
int nci = p[ci][cj][cbal].first.first;
int ncj = p[ci][cj][cbal].first.second;
int ncbal = p[ci][cj][cbal].second.first;
res += p[ci][cj][cbal].second.second;
ci = nci;
cj = ncj;
cbal = ncbal;
}
reverse(res.begin(), res.end());
cout << res << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 205, INF = 0x3f3f3f3f;
char a[N], b[N];
int n, m, f[N][N][N * 2];
int dp(int i, int j, int d) {
if (d < 0 || d > 400) return INF;
if (i == n && j == m && d == 0) return 0;
if (f[i][j][d] != -1) return f[i][j][d];
int minv = INF;
if (i != n || j != m)
minv = dp(i + (a[i] == '('), j + (b[j] == '('), d + 1) + 1;
if (d > 0)
minv = (minv > dp(i + (a[i] == ')'), j + (b[j] == ')'), d - 1) + 1
? dp(i + (a[i] == ')'), j + (b[j] == ')'), d - 1) + 1
: minv);
return f[i][j][d] = minv;
}
int main() {
scanf("%s%s", a, b);
memset(f, -1, sizeof(f));
n = strlen(a);
m = strlen(b);
a[n] = '.';
b[m] = '.';
int minv = dp(0, 0, 0);
int i = 0, j = 0, d = 0;
while (minv--) {
if (f[i + (a[i] == '(')][j + (b[j] == '(')][d + 1] == minv) {
printf("(");
i += (a[i] == '(');
j += (b[j] == '(');
d++;
} else {
printf(")");
i += (a[i] == ')');
j += (b[j] == ')');
d--;
}
}
return 0;
}
| ### Prompt
Please create a solution in cpp to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 205, INF = 0x3f3f3f3f;
char a[N], b[N];
int n, m, f[N][N][N * 2];
int dp(int i, int j, int d) {
if (d < 0 || d > 400) return INF;
if (i == n && j == m && d == 0) return 0;
if (f[i][j][d] != -1) return f[i][j][d];
int minv = INF;
if (i != n || j != m)
minv = dp(i + (a[i] == '('), j + (b[j] == '('), d + 1) + 1;
if (d > 0)
minv = (minv > dp(i + (a[i] == ')'), j + (b[j] == ')'), d - 1) + 1
? dp(i + (a[i] == ')'), j + (b[j] == ')'), d - 1) + 1
: minv);
return f[i][j][d] = minv;
}
int main() {
scanf("%s%s", a, b);
memset(f, -1, sizeof(f));
n = strlen(a);
m = strlen(b);
a[n] = '.';
b[m] = '.';
int minv = dp(0, 0, 0);
int i = 0, j = 0, d = 0;
while (minv--) {
if (f[i + (a[i] == '(')][j + (b[j] == '(')][d + 1] == minv) {
printf("(");
i += (a[i] == '(');
j += (b[j] == '(');
d++;
} else {
printf(")");
i += (a[i] == ')');
j += (b[j] == ')');
d--;
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
string s, t;
cin >> s >> t;
int dp[s.size() + 1][t.size() + 1][201];
for (int i = 0; i < s.size() + 1; i++)
for (int j = 0; j < t.size() + 1; j++)
for (int k = 0; k < 201; k++) dp[i][j][k] = 10000;
dp[0][0][0] = 0;
tuple<int, int, int> trace[s.size() + 1][t.size() + 1][201];
for (int i = 0; i < s.size() + 1; i++)
for (int j = 0; j < t.size() + 1; j++)
for (int k = 0; k < 201; k++) {
if (k + 1 <= 200) {
int ni;
if (i < s.size() && s[i] == '(')
ni = i + 1;
else
ni = i;
int nj;
if (j < t.size() && t[j] == '(')
nj = j + 1;
else
nj = j;
if (dp[ni][nj][k + 1] > dp[i][j][k] + 1) {
dp[ni][nj][k + 1] = dp[i][j][k] + 1;
trace[ni][nj][k + 1] = make_tuple(i, j, k);
}
}
if (k - 1 >= 0) {
int ni;
if (i < s.size() && s[i] == ')')
ni = i + 1;
else
ni = i;
int nj;
if (j < t.size() && t[j] == ')')
nj = j + 1;
else
nj = j;
if (dp[ni][nj][k - 1] > dp[i][j][k] + 1) {
dp[ni][nj][k - 1] = dp[i][j][k] + 1;
trace[ni][nj][k - 1] = make_tuple(i, j, k);
}
}
}
string ans = "";
int now_i = s.size();
int now_j = t.size();
int now_k;
int m = 10000;
for (int k = 0; k < 201; k++) m = min(m, dp[s.size()][t.size()][k] + k);
for (int k = 0; k < 201; k++)
if (dp[s.size()][t.size()][k] + k == m) now_k = k;
for (int k = 0; k < now_k; k++) ans += ')';
while (true) {
int prev_i = get<0>(trace[now_i][now_j][now_k]);
int prev_j = get<1>(trace[now_i][now_j][now_k]);
int prev_k = get<2>(trace[now_i][now_j][now_k]);
if (now_k - prev_k > 0)
ans += '(';
else
ans += ')';
now_i = prev_i;
now_j = prev_j;
now_k = prev_k;
if (now_i == 0 && now_j == 0 && now_k == 0) break;
}
reverse(ans.begin(), ans.end());
cout << ans << endl;
}
| ### Prompt
Please formulate a Cpp solution to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
string s, t;
cin >> s >> t;
int dp[s.size() + 1][t.size() + 1][201];
for (int i = 0; i < s.size() + 1; i++)
for (int j = 0; j < t.size() + 1; j++)
for (int k = 0; k < 201; k++) dp[i][j][k] = 10000;
dp[0][0][0] = 0;
tuple<int, int, int> trace[s.size() + 1][t.size() + 1][201];
for (int i = 0; i < s.size() + 1; i++)
for (int j = 0; j < t.size() + 1; j++)
for (int k = 0; k < 201; k++) {
if (k + 1 <= 200) {
int ni;
if (i < s.size() && s[i] == '(')
ni = i + 1;
else
ni = i;
int nj;
if (j < t.size() && t[j] == '(')
nj = j + 1;
else
nj = j;
if (dp[ni][nj][k + 1] > dp[i][j][k] + 1) {
dp[ni][nj][k + 1] = dp[i][j][k] + 1;
trace[ni][nj][k + 1] = make_tuple(i, j, k);
}
}
if (k - 1 >= 0) {
int ni;
if (i < s.size() && s[i] == ')')
ni = i + 1;
else
ni = i;
int nj;
if (j < t.size() && t[j] == ')')
nj = j + 1;
else
nj = j;
if (dp[ni][nj][k - 1] > dp[i][j][k] + 1) {
dp[ni][nj][k - 1] = dp[i][j][k] + 1;
trace[ni][nj][k - 1] = make_tuple(i, j, k);
}
}
}
string ans = "";
int now_i = s.size();
int now_j = t.size();
int now_k;
int m = 10000;
for (int k = 0; k < 201; k++) m = min(m, dp[s.size()][t.size()][k] + k);
for (int k = 0; k < 201; k++)
if (dp[s.size()][t.size()][k] + k == m) now_k = k;
for (int k = 0; k < now_k; k++) ans += ')';
while (true) {
int prev_i = get<0>(trace[now_i][now_j][now_k]);
int prev_j = get<1>(trace[now_i][now_j][now_k]);
int prev_k = get<2>(trace[now_i][now_j][now_k]);
if (now_k - prev_k > 0)
ans += '(';
else
ans += ')';
now_i = prev_i;
now_j = prev_j;
now_k = prev_k;
if (now_i == 0 && now_j == 0 && now_k == 0) break;
}
reverse(ans.begin(), ans.end());
cout << ans << endl;
}
``` |
#include <bits/stdc++.h>
using namespace std;
char s[205], t[205];
int n, m, f[205][205][405], nxt[205][205][405];
inline int dfs(int a, int b, int c) {
if (f[a][b][c] != -1) return f[a][b][c];
if (a == n + 1 && b == m + 1 && c == 0) return 0;
if (a == n + 1 && b == m + 1) {
nxt[a][b][c] = 1;
return f[a][b][c] = 1 + dfs(a, b, c - 1);
}
if (c > 400) return 1000000;
f[a][b][c] = 1000000;
if (c == 0) {
nxt[a][b][c] = 0;
return f[a][b][c] = 1 + dfs(a + (s[a] == '('), b + (t[b] == '('), c + 1);
}
int x = 1 + dfs(a + (s[a] == '('), b + (t[b] == '('), c + 1);
int y = 1 + dfs(a + (s[a] == ')'), b + (t[b] == ')'), c - 1);
if (x < y) {
nxt[a][b][c] = 0;
return f[a][b][c] = x;
} else {
nxt[a][b][c] = 1;
return f[a][b][c] = y;
}
return 0;
}
inline void out(int a, int b, int c) {
if (a == n + 1 && b == m + 1 && c == 0) return;
if (nxt[a][b][c] == 0) {
putchar('(');
out(a + (s[a] == '('), b + (t[b] == '('), c + 1);
} else {
putchar(')');
out(a + (s[a] == ')'), b + (t[b] == ')'), c - 1);
}
return;
}
int main() {
scanf("%s%s", s + 1, t + 1);
n = strlen(s + 1), m = strlen(t + 1);
s[n + 1] = t[m + 1] = '$';
memset(f, -1, sizeof(f));
dfs(1, 1, 0);
out(1, 1, 0);
cout << endl;
return 0;
}
| ### Prompt
In Cpp, your task is to solve the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
char s[205], t[205];
int n, m, f[205][205][405], nxt[205][205][405];
inline int dfs(int a, int b, int c) {
if (f[a][b][c] != -1) return f[a][b][c];
if (a == n + 1 && b == m + 1 && c == 0) return 0;
if (a == n + 1 && b == m + 1) {
nxt[a][b][c] = 1;
return f[a][b][c] = 1 + dfs(a, b, c - 1);
}
if (c > 400) return 1000000;
f[a][b][c] = 1000000;
if (c == 0) {
nxt[a][b][c] = 0;
return f[a][b][c] = 1 + dfs(a + (s[a] == '('), b + (t[b] == '('), c + 1);
}
int x = 1 + dfs(a + (s[a] == '('), b + (t[b] == '('), c + 1);
int y = 1 + dfs(a + (s[a] == ')'), b + (t[b] == ')'), c - 1);
if (x < y) {
nxt[a][b][c] = 0;
return f[a][b][c] = x;
} else {
nxt[a][b][c] = 1;
return f[a][b][c] = y;
}
return 0;
}
inline void out(int a, int b, int c) {
if (a == n + 1 && b == m + 1 && c == 0) return;
if (nxt[a][b][c] == 0) {
putchar('(');
out(a + (s[a] == '('), b + (t[b] == '('), c + 1);
} else {
putchar(')');
out(a + (s[a] == ')'), b + (t[b] == ')'), c - 1);
}
return;
}
int main() {
scanf("%s%s", s + 1, t + 1);
n = strlen(s + 1), m = strlen(t + 1);
s[n + 1] = t[m + 1] = '$';
memset(f, -1, sizeof(f));
dfs(1, 1, 0);
out(1, 1, 0);
cout << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 205;
int dp[N][N][N * 2];
char s[N], t[N];
int solveDp(int x, int y, int diff, int n, int m) {
if (diff == N * 2 || diff < 0) {
return 1e9;
}
if (diff == 0 && x == n && y == m) {
return 0;
}
int &ret = dp[x][y][diff];
if (ret != -1) {
return ret;
}
ret = 1e9;
char c[2] = {'(', ')'};
for (int i = 0; i < 2; ++i) {
int nx = x, ny = y;
if (s[x] == c[i]) ++nx;
if (t[y] == c[i]) ++ny;
ret = min(ret, 1 + solveDp(nx, ny, diff + (c[i] == '(' ? 1 : -1), n, m));
}
return ret;
}
void construct(int x, int y, int diff, int n, int m) {
int ret = solveDp(x, y, diff, n, m);
if (diff == 0 && x == n && y == m) {
putchar('\n');
return;
}
char c[2] = {'(', ')'};
for (int i = 0; i < 2; ++i) {
int nx = x, ny = y;
if (s[x] == c[i]) ++nx;
if (t[y] == c[i]) ++ny;
if (1 + solveDp(nx, ny, diff + (c[i] == '(' ? 1 : -1), n, m) == ret) {
putchar(c[i]);
construct(nx, ny, diff + (c[i] == '(' ? 1 : -1), n, m);
break;
}
}
}
void solve() {
scanf("%s %s ", s, t);
int n = strlen(s), m = strlen(t);
memset(dp, -1, sizeof(dp));
int len = solveDp(0, 0, 0, n, m);
construct(0, 0, 0, n, m);
}
int main() {
solve();
return 0;
}
| ### Prompt
In cpp, your task is to solve the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 205;
int dp[N][N][N * 2];
char s[N], t[N];
int solveDp(int x, int y, int diff, int n, int m) {
if (diff == N * 2 || diff < 0) {
return 1e9;
}
if (diff == 0 && x == n && y == m) {
return 0;
}
int &ret = dp[x][y][diff];
if (ret != -1) {
return ret;
}
ret = 1e9;
char c[2] = {'(', ')'};
for (int i = 0; i < 2; ++i) {
int nx = x, ny = y;
if (s[x] == c[i]) ++nx;
if (t[y] == c[i]) ++ny;
ret = min(ret, 1 + solveDp(nx, ny, diff + (c[i] == '(' ? 1 : -1), n, m));
}
return ret;
}
void construct(int x, int y, int diff, int n, int m) {
int ret = solveDp(x, y, diff, n, m);
if (diff == 0 && x == n && y == m) {
putchar('\n');
return;
}
char c[2] = {'(', ')'};
for (int i = 0; i < 2; ++i) {
int nx = x, ny = y;
if (s[x] == c[i]) ++nx;
if (t[y] == c[i]) ++ny;
if (1 + solveDp(nx, ny, diff + (c[i] == '(' ? 1 : -1), n, m) == ret) {
putchar(c[i]);
construct(nx, ny, diff + (c[i] == '(' ? 1 : -1), n, m);
break;
}
}
}
void solve() {
scanf("%s %s ", s, t);
int n = strlen(s), m = strlen(t);
memset(dp, -1, sizeof(dp));
int len = solveDp(0, 0, 0, n, m);
construct(0, 0, 0, n, m);
}
int main() {
solve();
return 0;
}
``` |
#include <bits/stdc++.h>
#pragma GCC target( \
"sse,sse2,sse3,ssse3,sse4,sse4.2,popcnt,abm,mmx,avx2,tune=native")
#pragma GCC optimize("-ffast-math")
#pragma GCC optimize("-funroll-loops")
#pragma GCC optimize("-funroll-all-loops,-fpeel-loops,-funswitch-loops")
using namespace std;
long long power(long long x, long long y, long long m) {
long long temp;
if (y == 0) return 1;
temp = (power(x, y / 2, m)) % m;
if (y % 2 == 0)
return ((temp % m) * temp);
else
return ((x * temp % m) * temp % m) % m;
}
long long inv(long long x, long long m = 1000000007) {
return (power(x, m - 2, m)) % m;
}
string s, t;
long long DP[201][201][201], BEST = 0;
long long REC(long long idx, long long idx2, long long C) {
if (C >= 201) return INT_MAX;
if (idx == s.size() && idx2 == t.size()) return C;
if (DP[idx][idx2][C] + 1) return DP[idx][idx2][C];
bool a = 0, b = 0;
if (idx < s.size())
if (s[idx] == '(') a = 1;
if (idx2 < t.size())
if (t[idx2] == '(') b = 1;
long long miner = INT_MAX;
if (C) {
miner = REC(min(idx + (!a), (long long)s.size()),
min(idx2 + (!b), (long long)t.size()), C - 1) +
1;
}
miner = min(miner, REC(min(idx + a, (long long)s.size()),
min(idx2 + b, (long long)t.size()), C + 1) +
1);
return DP[idx][idx2][C] = miner;
}
int32_t main() {
cin.tie(0), iostream::sync_with_stdio(0);
cin >> s >> t;
for (long long i = 0; i < 201; i++)
for (long long l = 0; l < 201; l++)
for (long long x = 0; x < 201; x++) DP[i][l][x] = -1;
BEST = REC(0, 0, 0);
long long A = 0, B = 0, C = 0;
BEST--;
while (BEST >= 0) {
bool F = 0, F2 = 0;
if (A < s.size())
if (s[A] == '(') F = 1;
if (B < t.size())
if (t[B] == '(') F2 = 1;
if (C) {
if (REC(A + (!F), B + (!F2), C - 1) == BEST) {
cout << ')', BEST--, C--, A = min(A + (!F), (long long)s.size()),
B = min(B + (!F2), (long long)t.size());
goto a;
}
}
if (REC(A + F, B + F2, C + 1) == BEST)
cout << '(', BEST--, C++, A = min(A + (F), (long long)s.size()),
B = min(B + (F2), (long long)t.size());
else {
while (C--) {
cout << ')';
}
return 0;
}
a:;
}
return 0;
}
| ### Prompt
Please create a solution in Cpp to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
#pragma GCC target( \
"sse,sse2,sse3,ssse3,sse4,sse4.2,popcnt,abm,mmx,avx2,tune=native")
#pragma GCC optimize("-ffast-math")
#pragma GCC optimize("-funroll-loops")
#pragma GCC optimize("-funroll-all-loops,-fpeel-loops,-funswitch-loops")
using namespace std;
long long power(long long x, long long y, long long m) {
long long temp;
if (y == 0) return 1;
temp = (power(x, y / 2, m)) % m;
if (y % 2 == 0)
return ((temp % m) * temp);
else
return ((x * temp % m) * temp % m) % m;
}
long long inv(long long x, long long m = 1000000007) {
return (power(x, m - 2, m)) % m;
}
string s, t;
long long DP[201][201][201], BEST = 0;
long long REC(long long idx, long long idx2, long long C) {
if (C >= 201) return INT_MAX;
if (idx == s.size() && idx2 == t.size()) return C;
if (DP[idx][idx2][C] + 1) return DP[idx][idx2][C];
bool a = 0, b = 0;
if (idx < s.size())
if (s[idx] == '(') a = 1;
if (idx2 < t.size())
if (t[idx2] == '(') b = 1;
long long miner = INT_MAX;
if (C) {
miner = REC(min(idx + (!a), (long long)s.size()),
min(idx2 + (!b), (long long)t.size()), C - 1) +
1;
}
miner = min(miner, REC(min(idx + a, (long long)s.size()),
min(idx2 + b, (long long)t.size()), C + 1) +
1);
return DP[idx][idx2][C] = miner;
}
int32_t main() {
cin.tie(0), iostream::sync_with_stdio(0);
cin >> s >> t;
for (long long i = 0; i < 201; i++)
for (long long l = 0; l < 201; l++)
for (long long x = 0; x < 201; x++) DP[i][l][x] = -1;
BEST = REC(0, 0, 0);
long long A = 0, B = 0, C = 0;
BEST--;
while (BEST >= 0) {
bool F = 0, F2 = 0;
if (A < s.size())
if (s[A] == '(') F = 1;
if (B < t.size())
if (t[B] == '(') F2 = 1;
if (C) {
if (REC(A + (!F), B + (!F2), C - 1) == BEST) {
cout << ')', BEST--, C--, A = min(A + (!F), (long long)s.size()),
B = min(B + (!F2), (long long)t.size());
goto a;
}
}
if (REC(A + F, B + F2, C + 1) == BEST)
cout << '(', BEST--, C++, A = min(A + (F), (long long)s.size()),
B = min(B + (F2), (long long)t.size());
else {
while (C--) {
cout << ')';
}
return 0;
}
a:;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
template <typename T>
istream& operator>>(istream& is, vector<T>& a) {
for (T& ai : a) is >> ai;
return is;
}
template <typename T>
ostream& operator<<(ostream& os, vector<T> const& a) {
os << "[ ";
for (const T& ai : a) os << ai << " ";
return os << "]";
}
template <typename T1, typename T2>
istream& operator>>(istream& is, pair<T1, T2>& a) {
return is >> a.first >> a.second;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& os, pair<T1, T2> const& a) {
return os << "(" << a.first << ", " << a.second << ")";
}
string f() {
string s, t;
cin >> s >> t;
int n = s.size(), m = t.size();
int l = max(n, m);
vector<int> prev((n + 1) * (m + 1) * (l + 1), -1);
queue<int> q;
q.push(0);
int d = ((n + 1) * (m + 1) - 1) * (l + 1);
while (!q.empty() && prev[d] < 0) {
int u = q.front();
q.pop();
int i = u;
int k = i % (l + 1);
i /= l + 1;
int j = i % (m + 1);
i /= m + 1;
if (k < l) {
int v = u + 1;
if (i < n && s[i] == '(') v += (l + 1) * (m + 1);
if (j < m && t[j] == '(') v += l + 1;
if (prev[v] < 0) {
prev[v] = u;
q.push(v);
}
}
if (k > 0) {
int v = u - 1;
if (i < n && s[i] == ')') v += (l + 1) * (m + 1);
if (j < m && t[j] == ')') v += l + 1;
if (prev[v] < 0) {
prev[v] = u;
q.push(v);
}
}
}
string r;
int u = d;
while (prev[u] >= 0) {
r.push_back((u % (l + 1) - prev[u] % (l + 1) > 0) ? '(' : ')');
u = prev[u];
}
reverse((r).begin(), (r).end());
return r;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout << f() << "\n";
return 0;
}
| ### Prompt
Construct a CPP code solution to the problem outlined:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
template <typename T>
istream& operator>>(istream& is, vector<T>& a) {
for (T& ai : a) is >> ai;
return is;
}
template <typename T>
ostream& operator<<(ostream& os, vector<T> const& a) {
os << "[ ";
for (const T& ai : a) os << ai << " ";
return os << "]";
}
template <typename T1, typename T2>
istream& operator>>(istream& is, pair<T1, T2>& a) {
return is >> a.first >> a.second;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& os, pair<T1, T2> const& a) {
return os << "(" << a.first << ", " << a.second << ")";
}
string f() {
string s, t;
cin >> s >> t;
int n = s.size(), m = t.size();
int l = max(n, m);
vector<int> prev((n + 1) * (m + 1) * (l + 1), -1);
queue<int> q;
q.push(0);
int d = ((n + 1) * (m + 1) - 1) * (l + 1);
while (!q.empty() && prev[d] < 0) {
int u = q.front();
q.pop();
int i = u;
int k = i % (l + 1);
i /= l + 1;
int j = i % (m + 1);
i /= m + 1;
if (k < l) {
int v = u + 1;
if (i < n && s[i] == '(') v += (l + 1) * (m + 1);
if (j < m && t[j] == '(') v += l + 1;
if (prev[v] < 0) {
prev[v] = u;
q.push(v);
}
}
if (k > 0) {
int v = u - 1;
if (i < n && s[i] == ')') v += (l + 1) * (m + 1);
if (j < m && t[j] == ')') v += l + 1;
if (prev[v] < 0) {
prev[v] = u;
q.push(v);
}
}
}
string r;
int u = d;
while (prev[u] >= 0) {
r.push_back((u % (l + 1) - prev[u] % (l + 1) > 0) ? '(' : ')');
u = prev[u];
}
reverse((r).begin(), (r).end());
return r;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout << f() << "\n";
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int Get() {
char c;
bool o = false;
while (!isdigit(c = getchar()))
if (c == '-') o = true;
int x = c - '0';
while (isdigit(c = getchar())) x = x * 10 + c - '0';
return o ? -x : x;
}
const int maxn = 2e2 + 7;
int f[maxn][maxn][maxn << 1];
int prei[maxn][maxn][maxn << 1];
int prej[maxn][maxn][maxn << 1];
int prek[maxn][maxn][maxn << 1];
int ansn;
char ansseq[maxn << 2];
char s[maxn], t[maxn];
int main() {
scanf("%s\n%s", s + 1, t + 1);
int n = strlen(s + 1), m = strlen(t + 1);
memset(f, 127 / 3, sizeof(f));
f[0][0][0] = 0;
for (int i = 0; i <= n; ++i)
for (int j = 0; j <= m; ++j)
for (int k = 0; k <= n + m; ++k) {
if (i == n && j == m) continue;
int u = i, v = j, w = k + 1;
if (w <= n + m) {
if (i < n && s[i + 1] == '(') u = i + 1;
if (j < m && t[j + 1] == '(') v = j + 1;
if (f[u][v][w] > f[i][j][k] + 1) {
f[u][v][w] = f[i][j][k] + 1;
prei[u][v][w] = i;
prej[u][v][w] = j;
prek[u][v][w] = k;
}
}
u = i, v = j, w = k - 1;
if (w >= 0) {
if (i < n && s[i + 1] == ')') u = i + 1;
if (j < m && t[j + 1] == ')') v = j + 1;
if (f[u][v][w] > f[i][j][k] + 1) {
f[u][v][w] = f[i][j][k] + 1;
prei[u][v][w] = i;
prej[u][v][w] = j;
prek[u][v][w] = k;
}
}
}
int ans = 2147483647, ansi = 0;
for (int i = 0; i <= n + m; ++i)
if (ans > f[n][m][i] + i) {
ans = f[n][m][i] + i;
ansi = i;
}
for (int i = 1; i <= ansi; ++i) ansseq[++ansn] = ')';
int u = n, v = m, w = ansi;
while (u || v || w) {
int i = prei[u][v][w], j = prej[u][v][w], k = prek[u][v][w];
if (w == k + 1) ansseq[++ansn] = '(';
if (w == k - 1) ansseq[++ansn] = ')';
u = i, v = j, w = k;
}
for (int i = ansn; i >= 1; --i) putchar(ansseq[i]);
}
| ### Prompt
Please provide a CPP coded solution to the problem described below:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int Get() {
char c;
bool o = false;
while (!isdigit(c = getchar()))
if (c == '-') o = true;
int x = c - '0';
while (isdigit(c = getchar())) x = x * 10 + c - '0';
return o ? -x : x;
}
const int maxn = 2e2 + 7;
int f[maxn][maxn][maxn << 1];
int prei[maxn][maxn][maxn << 1];
int prej[maxn][maxn][maxn << 1];
int prek[maxn][maxn][maxn << 1];
int ansn;
char ansseq[maxn << 2];
char s[maxn], t[maxn];
int main() {
scanf("%s\n%s", s + 1, t + 1);
int n = strlen(s + 1), m = strlen(t + 1);
memset(f, 127 / 3, sizeof(f));
f[0][0][0] = 0;
for (int i = 0; i <= n; ++i)
for (int j = 0; j <= m; ++j)
for (int k = 0; k <= n + m; ++k) {
if (i == n && j == m) continue;
int u = i, v = j, w = k + 1;
if (w <= n + m) {
if (i < n && s[i + 1] == '(') u = i + 1;
if (j < m && t[j + 1] == '(') v = j + 1;
if (f[u][v][w] > f[i][j][k] + 1) {
f[u][v][w] = f[i][j][k] + 1;
prei[u][v][w] = i;
prej[u][v][w] = j;
prek[u][v][w] = k;
}
}
u = i, v = j, w = k - 1;
if (w >= 0) {
if (i < n && s[i + 1] == ')') u = i + 1;
if (j < m && t[j + 1] == ')') v = j + 1;
if (f[u][v][w] > f[i][j][k] + 1) {
f[u][v][w] = f[i][j][k] + 1;
prei[u][v][w] = i;
prej[u][v][w] = j;
prek[u][v][w] = k;
}
}
}
int ans = 2147483647, ansi = 0;
for (int i = 0; i <= n + m; ++i)
if (ans > f[n][m][i] + i) {
ans = f[n][m][i] + i;
ansi = i;
}
for (int i = 1; i <= ansi; ++i) ansseq[++ansn] = ')';
int u = n, v = m, w = ansi;
while (u || v || w) {
int i = prei[u][v][w], j = prej[u][v][w], k = prek[u][v][w];
if (w == k + 1) ansseq[++ansn] = '(';
if (w == k - 1) ansseq[++ansn] = ')';
u = i, v = j, w = k;
}
for (int i = ansn; i >= 1; --i) putchar(ansseq[i]);
}
``` |
#include <bits/stdc++.h>
#pragma GCC optimize("O3", "unroll-loops")
#pragma GCC target("avx2")
using namespace std;
template <class T, class U>
inline void checkmin(T &x, U y) {
if (y < x) x = y;
}
template <class T, class U>
inline void checkmax(T &x, U y) {
if (y > x) x = y;
}
template <class T, class U>
inline bool ifmax(T &x, U y) {
if (y > x) return x = y, true;
return false;
}
template <class T, class U>
inline bool ifmin(T &x, U y) {
if (y < x) return x = y, true;
return false;
}
template <class T>
inline void sort(T &a) {
sort(a.begin(), a.end());
}
template <class T>
inline void rsort(T &a) {
sort(a.rbegin(), a.rend());
}
template <class T>
inline void reverse(T &a) {
reverse(a.begin(), a.end());
}
template <class T, class U>
inline istream &operator>>(istream &str, pair<T, U> &p) {
return str >> p.first >> p.second;
}
template <class T>
inline istream &operator>>(istream &str, vector<T> &a) {
for (auto &i : a) str >> i;
return str;
}
template <class T>
inline T sorted(T a) {
sort(a);
return a;
}
const int N = 205;
int dp[N][N][N];
struct pos {
int p1 = -1, p2 = -1, balance;
pos() {}
pos(int p1, int p2, int balance) {
this->p1 = p1;
this->p2 = p2;
this->balance = balance;
}
};
pos p[N][N][N];
signed main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cout << fixed << setprecision(12);
srand(94385);
string second, t;
cin >> second >> t;
second = '0' + second;
t = '0' + t;
for (int i = 0; i < N; ++i)
for (int j = 0; j < N; ++j)
for (int k = 0; k < N; ++k) dp[i][j][k] = 1000000007;
for (int balance = 0; balance < N; ++balance) dp[0][0][balance] = balance;
for (int p1 = 0; p1 < signed(second.size()); ++p1)
for (int p2 = 0; p2 < signed(t.size()); ++p2)
if (p1 > 0 || p2 > 0) {
for (int balance = 0; balance < N; ++balance) {
pos &past = p[p1][p2][balance];
if (balance > 0) {
if (second[p1] == '(') {
if (ifmin(dp[p1][p2][balance], dp[p1 - 1][p2][balance - 1] + 1))
past = {p1 - 1, p2, balance - 1};
}
if (t[p2] == '(') {
if (ifmin(dp[p1][p2][balance], dp[p1][p2 - 1][balance - 1] + 1))
past = {p1, p2 - 1, balance - 1};
}
if (second[p1] == '(' && t[p2] == '(') {
if (ifmin(dp[p1][p2][balance],
dp[p1 - 1][p2 - 1][balance - 1] + 1))
past = {p1 - 1, p2 - 1, balance - 1};
}
}
if (second[p1] == ')') {
if (ifmin(dp[p1][p2][balance], dp[p1 - 1][p2][balance + 1] + 1))
past = {p1 - 1, p2, balance + 1};
}
if (t[p2] == ')') {
if (ifmin(dp[p1][p2][balance], dp[p1][p2 - 1][balance + 1] + 1))
past = {p1, p2 - 1, balance + 1};
}
if (second[p1] == ')' && t[p2] == ')') {
if (ifmin(dp[p1][p2][balance], dp[p1 - 1][p2 - 1][balance + 1] + 1))
past = {p1 - 1, p2 - 1, balance + 1};
}
}
}
pair<int, pos> ans;
ans.first = 1000000007;
for (int balance = 0; balance < N; ++balance)
if (ifmin(ans.first,
dp[signed(second.size()) - 1][signed(t.size()) - 1][balance] +
balance))
ans.second =
pos(signed(second.size()) - 1, signed(t.size()) - 1, balance);
pos curr = ans.second;
string answer(curr.balance, ')');
pos check = p[1][2][0];
while (curr.p1 > 0 || curr.p2 > 0) {
if (curr.balance > p[curr.p1][curr.p2][curr.balance].balance)
answer = '(' + answer;
else
answer = ')' + answer;
curr = p[curr.p1][curr.p2][curr.balance];
}
while (curr.balance) answer = '(' + answer, --curr.balance;
cout << answer;
return 0;
}
| ### Prompt
Generate a cpp solution to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("O3", "unroll-loops")
#pragma GCC target("avx2")
using namespace std;
template <class T, class U>
inline void checkmin(T &x, U y) {
if (y < x) x = y;
}
template <class T, class U>
inline void checkmax(T &x, U y) {
if (y > x) x = y;
}
template <class T, class U>
inline bool ifmax(T &x, U y) {
if (y > x) return x = y, true;
return false;
}
template <class T, class U>
inline bool ifmin(T &x, U y) {
if (y < x) return x = y, true;
return false;
}
template <class T>
inline void sort(T &a) {
sort(a.begin(), a.end());
}
template <class T>
inline void rsort(T &a) {
sort(a.rbegin(), a.rend());
}
template <class T>
inline void reverse(T &a) {
reverse(a.begin(), a.end());
}
template <class T, class U>
inline istream &operator>>(istream &str, pair<T, U> &p) {
return str >> p.first >> p.second;
}
template <class T>
inline istream &operator>>(istream &str, vector<T> &a) {
for (auto &i : a) str >> i;
return str;
}
template <class T>
inline T sorted(T a) {
sort(a);
return a;
}
const int N = 205;
int dp[N][N][N];
struct pos {
int p1 = -1, p2 = -1, balance;
pos() {}
pos(int p1, int p2, int balance) {
this->p1 = p1;
this->p2 = p2;
this->balance = balance;
}
};
pos p[N][N][N];
signed main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cout << fixed << setprecision(12);
srand(94385);
string second, t;
cin >> second >> t;
second = '0' + second;
t = '0' + t;
for (int i = 0; i < N; ++i)
for (int j = 0; j < N; ++j)
for (int k = 0; k < N; ++k) dp[i][j][k] = 1000000007;
for (int balance = 0; balance < N; ++balance) dp[0][0][balance] = balance;
for (int p1 = 0; p1 < signed(second.size()); ++p1)
for (int p2 = 0; p2 < signed(t.size()); ++p2)
if (p1 > 0 || p2 > 0) {
for (int balance = 0; balance < N; ++balance) {
pos &past = p[p1][p2][balance];
if (balance > 0) {
if (second[p1] == '(') {
if (ifmin(dp[p1][p2][balance], dp[p1 - 1][p2][balance - 1] + 1))
past = {p1 - 1, p2, balance - 1};
}
if (t[p2] == '(') {
if (ifmin(dp[p1][p2][balance], dp[p1][p2 - 1][balance - 1] + 1))
past = {p1, p2 - 1, balance - 1};
}
if (second[p1] == '(' && t[p2] == '(') {
if (ifmin(dp[p1][p2][balance],
dp[p1 - 1][p2 - 1][balance - 1] + 1))
past = {p1 - 1, p2 - 1, balance - 1};
}
}
if (second[p1] == ')') {
if (ifmin(dp[p1][p2][balance], dp[p1 - 1][p2][balance + 1] + 1))
past = {p1 - 1, p2, balance + 1};
}
if (t[p2] == ')') {
if (ifmin(dp[p1][p2][balance], dp[p1][p2 - 1][balance + 1] + 1))
past = {p1, p2 - 1, balance + 1};
}
if (second[p1] == ')' && t[p2] == ')') {
if (ifmin(dp[p1][p2][balance], dp[p1 - 1][p2 - 1][balance + 1] + 1))
past = {p1 - 1, p2 - 1, balance + 1};
}
}
}
pair<int, pos> ans;
ans.first = 1000000007;
for (int balance = 0; balance < N; ++balance)
if (ifmin(ans.first,
dp[signed(second.size()) - 1][signed(t.size()) - 1][balance] +
balance))
ans.second =
pos(signed(second.size()) - 1, signed(t.size()) - 1, balance);
pos curr = ans.second;
string answer(curr.balance, ')');
pos check = p[1][2][0];
while (curr.p1 > 0 || curr.p2 > 0) {
if (curr.balance > p[curr.p1][curr.p2][curr.balance].balance)
answer = '(' + answer;
else
answer = ')' + answer;
curr = p[curr.p1][curr.p2][curr.balance];
}
while (curr.balance) answer = '(' + answer, --curr.balance;
cout << answer;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int maxn = 202;
string to, ans;
string s, t;
int ls, lt;
int f[2 * maxn][maxn][maxn];
int solve(int sum, int ps, int pt) {
if (sum < 0) return 1 << 30;
if (sum == 0 && ps == ls && pt == lt) return 0;
if (sum == 2 * maxn) return 1 << 30;
if (f[sum][ps][pt] != -1) return f[sum][ps][pt];
int ans = 1 << 30;
f[sum][ps][pt] = ans;
ans = min(ans, solve(sum + 1, ps + (s[ps] == '('), pt + (t[pt] == '(')) + 1);
ans = min(ans, solve(sum - 1, ps + (s[ps] == ')'), pt + (t[pt] == ')')) + 1);
return f[sum][ps][pt] = ans;
}
void roll(int sum, int ps, int pt) {
if (sum == 0 && ps == ls && pt == lt) return;
if (f[sum][ps][pt] ==
solve(sum + 1, ps + (s[ps] == '('), pt + (t[pt] == '(')) + 1) {
ans.push_back('(');
roll(sum + 1, ps + (s[ps] == '('), pt + (t[pt] == '('));
return;
}
ans.push_back(')');
roll(sum - 1, ps + (s[ps] == ')'), pt + (t[pt] == ')'));
}
int main() {
cin >> s >> t;
ls = s.size();
lt = t.size();
s.push_back('#');
t.push_back('#');
memset(f, -1, sizeof(f));
solve(0, 0, 0);
roll(0, 0, 0);
cout << ans << endl;
return 0;
}
| ### Prompt
Your task is to create a CPP solution to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = 202;
string to, ans;
string s, t;
int ls, lt;
int f[2 * maxn][maxn][maxn];
int solve(int sum, int ps, int pt) {
if (sum < 0) return 1 << 30;
if (sum == 0 && ps == ls && pt == lt) return 0;
if (sum == 2 * maxn) return 1 << 30;
if (f[sum][ps][pt] != -1) return f[sum][ps][pt];
int ans = 1 << 30;
f[sum][ps][pt] = ans;
ans = min(ans, solve(sum + 1, ps + (s[ps] == '('), pt + (t[pt] == '(')) + 1);
ans = min(ans, solve(sum - 1, ps + (s[ps] == ')'), pt + (t[pt] == ')')) + 1);
return f[sum][ps][pt] = ans;
}
void roll(int sum, int ps, int pt) {
if (sum == 0 && ps == ls && pt == lt) return;
if (f[sum][ps][pt] ==
solve(sum + 1, ps + (s[ps] == '('), pt + (t[pt] == '(')) + 1) {
ans.push_back('(');
roll(sum + 1, ps + (s[ps] == '('), pt + (t[pt] == '('));
return;
}
ans.push_back(')');
roll(sum - 1, ps + (s[ps] == ')'), pt + (t[pt] == ')'));
}
int main() {
cin >> s >> t;
ls = s.size();
lt = t.size();
s.push_back('#');
t.push_back('#');
memset(f, -1, sizeof(f));
solve(0, 0, 0);
roll(0, 0, 0);
cout << ans << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9;
int sn, tn;
string s, t;
int memo[201][201][201];
char mc[201][201][201];
int dp(int sp, int tp, int p) {
if (sp == sn && tp == tn) return p;
int &ret = memo[sp][tp][p];
if (ret != -1) return ret;
ret = inf;
mc[sp][tp][p] = '(';
if (p < 200) {
int ct = 0, cs = 0;
if (sp < sn) cs = s[sp] == '(';
if (tp < tn) ct = t[tp] == '(';
ret = min(ret, dp(sp + cs, tp + ct, p + 1));
}
if (p > 0) {
int ct = 0, cs = 0;
if (sp < sn) cs = s[sp] == ')';
if (tp < tn) ct = t[tp] == ')';
if (ret > dp(sp + cs, tp + ct, p - 1)) {
ret = dp(sp + cs, tp + ct, p - 1);
mc[sp][tp][p] = ')';
}
}
return ++ret;
}
int main() {
cin >> s >> t;
sn = s.size(), tn = t.size();
memset(memo, -1, 4 * 201 * 201 * 201);
dp(0, 0, 0);
string ans;
int sp = 0, tp = 0, p = 0;
while (sp < sn || tp < tn) {
char c = mc[sp][tp][p];
ans += c;
if (sp < sn && s[sp] == c) sp++;
if (tp < tn && t[tp] == c) tp++;
p += (c == '(') ? 1 : -1;
}
while (p--) ans += ')';
cout << ans;
cin >> s;
}
| ### Prompt
Construct a Cpp code solution to the problem outlined:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9;
int sn, tn;
string s, t;
int memo[201][201][201];
char mc[201][201][201];
int dp(int sp, int tp, int p) {
if (sp == sn && tp == tn) return p;
int &ret = memo[sp][tp][p];
if (ret != -1) return ret;
ret = inf;
mc[sp][tp][p] = '(';
if (p < 200) {
int ct = 0, cs = 0;
if (sp < sn) cs = s[sp] == '(';
if (tp < tn) ct = t[tp] == '(';
ret = min(ret, dp(sp + cs, tp + ct, p + 1));
}
if (p > 0) {
int ct = 0, cs = 0;
if (sp < sn) cs = s[sp] == ')';
if (tp < tn) ct = t[tp] == ')';
if (ret > dp(sp + cs, tp + ct, p - 1)) {
ret = dp(sp + cs, tp + ct, p - 1);
mc[sp][tp][p] = ')';
}
}
return ++ret;
}
int main() {
cin >> s >> t;
sn = s.size(), tn = t.size();
memset(memo, -1, 4 * 201 * 201 * 201);
dp(0, 0, 0);
string ans;
int sp = 0, tp = 0, p = 0;
while (sp < sn || tp < tn) {
char c = mc[sp][tp][p];
ans += c;
if (sp < sn && s[sp] == c) sp++;
if (tp < tn && t[tp] == c) tp++;
p += (c == '(') ? 1 : -1;
}
while (p--) ans += ')';
cout << ans;
cin >> s;
}
``` |
#include <bits/stdc++.h>
using namespace std;
void fast() {
ios::sync_with_stdio(0);
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
const int MAN = 210;
int dp[MAN][MAN][MAN];
tuple<int, int, int> pred[MAN][MAN][MAN];
int INF = 1e9;
signed main() {
fast();
int n1, n2, i, j, n2i, n2j, a, b, c, d, bal, ni, nj;
string s1, s2;
cin >> s1 >> s2;
char sk;
n1 = ((int)s1.size());
n2 = ((int)s2.size());
s1 = '#' + s1;
s2 = '#' + s2;
for (i = 0; i <= n1; i++)
for (j = 0; j <= n2; j++)
for (bal = 0; bal < MAN; bal++) dp[i][j][bal] = INF;
dp[0][0][0] = 0;
for (i = 0; i <= n1; i++) {
for (j = 0; j <= n2; j++) {
for (bal = 0; bal < MAN; bal++) {
if (dp[i][j][bal] == INF) continue;
ni = i + (i + 1 <= n1 && s1[i + 1] == '(');
nj = j + (j + 1 <= n2 && s2[j + 1] == '(');
if (bal + 1 < MAN && dp[ni][nj][bal + 1] > dp[i][j][bal] + 1) {
dp[ni][nj][bal + 1] = dp[i][j][bal] + 1;
pred[ni][nj][bal + 1] = make_tuple(i, j, bal);
}
ni = i + (i + 1 <= n1 && s1[i + 1] == ')');
nj = j + (j + 1 <= n2 && s2[j + 1] == ')');
if (bal - 1 >= 0 && dp[ni][nj][bal - 1] > dp[i][j][bal] + 1) {
dp[ni][nj][bal - 1] = dp[i][j][bal] + 1;
pred[ni][nj][bal - 1] = make_tuple(i, j, bal);
}
}
}
}
int balmi = INF;
for (i = 0; i < MAN; i++) {
if (dp[n1][n2][i] + i < balmi) balmi = i;
}
string ans = "";
for (i = 0; i < balmi; i++) {
ans += ')';
}
i = n1;
j = n2;
bal = balmi;
int i2, j2, bal2;
while (pred[i][j][bal] != make_tuple(0, 0, 0)) {
tie(i2, j2, bal2) = pred[i][j][bal];
if (bal2 < bal)
ans += '(';
else
ans += ')';
i = i2;
j = j2;
bal = bal2;
}
ans += '(';
for (i = ((int)ans.size()) - 1; i >= 0; i--) cout << ans[i];
return 0;
}
| ### Prompt
In CPP, your task is to solve the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void fast() {
ios::sync_with_stdio(0);
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
const int MAN = 210;
int dp[MAN][MAN][MAN];
tuple<int, int, int> pred[MAN][MAN][MAN];
int INF = 1e9;
signed main() {
fast();
int n1, n2, i, j, n2i, n2j, a, b, c, d, bal, ni, nj;
string s1, s2;
cin >> s1 >> s2;
char sk;
n1 = ((int)s1.size());
n2 = ((int)s2.size());
s1 = '#' + s1;
s2 = '#' + s2;
for (i = 0; i <= n1; i++)
for (j = 0; j <= n2; j++)
for (bal = 0; bal < MAN; bal++) dp[i][j][bal] = INF;
dp[0][0][0] = 0;
for (i = 0; i <= n1; i++) {
for (j = 0; j <= n2; j++) {
for (bal = 0; bal < MAN; bal++) {
if (dp[i][j][bal] == INF) continue;
ni = i + (i + 1 <= n1 && s1[i + 1] == '(');
nj = j + (j + 1 <= n2 && s2[j + 1] == '(');
if (bal + 1 < MAN && dp[ni][nj][bal + 1] > dp[i][j][bal] + 1) {
dp[ni][nj][bal + 1] = dp[i][j][bal] + 1;
pred[ni][nj][bal + 1] = make_tuple(i, j, bal);
}
ni = i + (i + 1 <= n1 && s1[i + 1] == ')');
nj = j + (j + 1 <= n2 && s2[j + 1] == ')');
if (bal - 1 >= 0 && dp[ni][nj][bal - 1] > dp[i][j][bal] + 1) {
dp[ni][nj][bal - 1] = dp[i][j][bal] + 1;
pred[ni][nj][bal - 1] = make_tuple(i, j, bal);
}
}
}
}
int balmi = INF;
for (i = 0; i < MAN; i++) {
if (dp[n1][n2][i] + i < balmi) balmi = i;
}
string ans = "";
for (i = 0; i < balmi; i++) {
ans += ')';
}
i = n1;
j = n2;
bal = balmi;
int i2, j2, bal2;
while (pred[i][j][bal] != make_tuple(0, 0, 0)) {
tie(i2, j2, bal2) = pred[i][j][bal];
if (bal2 < bal)
ans += '(';
else
ans += ')';
i = i2;
j = j2;
bal = bal2;
}
ans += '(';
for (i = ((int)ans.size()) - 1; i >= 0; i--) cout << ans[i];
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3fll;
void sc(int& x) { scanf("%d", &x); }
void sc(long long& x) { scanf("%lld", &x); }
template <class T, class... Ts>
void sc(T& t, Ts&... ts) {
sc(t), sc(ts...);
}
const int MAX = 210;
int dp[MAX][MAX][MAX];
char s1[MAX], s2[MAX];
int n1, n2;
pair<pair<int, int>, pair<int, int> > get(int i1, int i2) {
bool a = (i1 != n1), b = (i2 != n2);
pair<pair<int, int>, pair<int, int> > ret;
if (s1[i1] == '(')
ret.first = {a, 0};
else
ret.first = {0, a};
if (s2[i2] == '(')
ret.second = {b, 0};
else
ret.second = {0, b};
return ret;
}
int solve(int i1, int i2, int c, bool p = 0) {
if (c < 0) return INF;
if (c >= MAX) return INF;
if (i1 == n1 and i2 == n2) {
if (p)
for (int i = 0; i < c; i++) printf(")");
return c;
}
auto& x = dp[i1][i2][c];
if (x + 1 and !p) return x;
auto t = get(i1, i2);
int a = 1 + solve(i1 + t.first.first, i2 + t.second.first, c + 1);
int b = 1 + solve(i1 + t.first.second, i2 + t.second.second, c - 1);
if (p) {
if (a <= b)
printf("("), solve(i1 + t.first.first, i2 + t.second.first, c + 1, 1);
else
printf(")"), solve(i1 + t.first.second, i2 + t.second.second, c - 1, 1);
}
return x = min(a, b);
}
int main() {
scanf(" %s %s", s1, s2);
n1 = strlen(s1), n2 = strlen(s2);
memset(dp, -1, sizeof(dp));
solve(0, 0, 0, 1);
printf("\n");
exit(0);
}
| ### Prompt
Develop a solution in Cpp to the problem described below:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3fll;
void sc(int& x) { scanf("%d", &x); }
void sc(long long& x) { scanf("%lld", &x); }
template <class T, class... Ts>
void sc(T& t, Ts&... ts) {
sc(t), sc(ts...);
}
const int MAX = 210;
int dp[MAX][MAX][MAX];
char s1[MAX], s2[MAX];
int n1, n2;
pair<pair<int, int>, pair<int, int> > get(int i1, int i2) {
bool a = (i1 != n1), b = (i2 != n2);
pair<pair<int, int>, pair<int, int> > ret;
if (s1[i1] == '(')
ret.first = {a, 0};
else
ret.first = {0, a};
if (s2[i2] == '(')
ret.second = {b, 0};
else
ret.second = {0, b};
return ret;
}
int solve(int i1, int i2, int c, bool p = 0) {
if (c < 0) return INF;
if (c >= MAX) return INF;
if (i1 == n1 and i2 == n2) {
if (p)
for (int i = 0; i < c; i++) printf(")");
return c;
}
auto& x = dp[i1][i2][c];
if (x + 1 and !p) return x;
auto t = get(i1, i2);
int a = 1 + solve(i1 + t.first.first, i2 + t.second.first, c + 1);
int b = 1 + solve(i1 + t.first.second, i2 + t.second.second, c - 1);
if (p) {
if (a <= b)
printf("("), solve(i1 + t.first.first, i2 + t.second.first, c + 1, 1);
else
printf(")"), solve(i1 + t.first.second, i2 + t.second.second, c - 1, 1);
}
return x = min(a, b);
}
int main() {
scanf(" %s %s", s1, s2);
n1 = strlen(s1), n2 = strlen(s2);
memset(dp, -1, sizeof(dp));
solve(0, 0, 0, 1);
printf("\n");
exit(0);
}
``` |
#include <bits/stdc++.h>
using namespace std;
string a, b;
int n, m;
int dp[205][205][405];
pair<pair<int, int>, int> asdasfsadj[205][205][405];
pair<pair<int, int>, int> ansPos;
int ansLen = -1;
string ans;
int main() {
getline(cin, a);
getline(cin, b);
n = (int)a.size();
m = (int)b.size();
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= m; j++) {
for (int k = 0; k <= 400; k++) {
dp[i][j][k] = -1;
}
}
}
dp[0][0][0] = 0;
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= m; j++) {
for (int k = 0; k < 400; k++) {
if (dp[i][j][k] == -1) {
continue;
}
if (i == n && j == m && k == 0) {
if (ansLen == -1 || dp[i][j][k] < ansLen) {
ansPos = make_pair(make_pair(i, j), k);
ansLen = dp[i][j][k];
}
continue;
}
int addI = (i < n && a[i] == '(') ? 1 : 0;
int addJ = (j < m && b[j] == '(') ? 1 : 0;
if (dp[i + addI][j + addJ][k + 1] == -1 ||
dp[i + addI][j + addJ][k + 1] > dp[i][j][k] + 1) {
dp[i + addI][j + addJ][k + 1] = dp[i][j][k] + 1;
asdasfsadj[i + addI][j + addJ][k + 1] = make_pair(make_pair(i, j), k);
}
}
for (int k = 400; k >= 0; k--) {
if (dp[i][j][k] == -1) {
continue;
}
if (i == n && j == m && k == 0) {
if (ansLen == -1 || dp[i][j][k] < ansLen) {
ansPos = make_pair(make_pair(i, j), k);
ansLen = dp[i][j][k];
}
continue;
}
int addI = (i < n && a[i] == ')') ? 1 : 0;
int addJ = (j < m && b[j] == ')') ? 1 : 0;
if (dp[i + addI][j + addJ][k - 1] == -1 ||
dp[i + addI][j + addJ][k - 1] > dp[i][j][k] + 1) {
dp[i + addI][j + addJ][k - 1] = dp[i][j][k] + 1;
asdasfsadj[i + addI][j + addJ][k - 1] = make_pair(make_pair(i, j), k);
}
}
}
}
cerr << ansLen << endl;
while (true) {
int i = ansPos.first.first;
int j = ansPos.first.second;
int k = ansPos.second;
if (dp[i][j][k] == 0) {
break;
}
ansPos = asdasfsadj[i][j][k];
int newK = ansPos.second;
if (newK < k) {
ans.append(1, '(');
} else {
ans.append(1, ')');
}
}
reverse(ans.begin(), ans.end());
cout << ans << endl;
return 0;
}
| ### Prompt
Generate a cpp solution to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
string a, b;
int n, m;
int dp[205][205][405];
pair<pair<int, int>, int> asdasfsadj[205][205][405];
pair<pair<int, int>, int> ansPos;
int ansLen = -1;
string ans;
int main() {
getline(cin, a);
getline(cin, b);
n = (int)a.size();
m = (int)b.size();
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= m; j++) {
for (int k = 0; k <= 400; k++) {
dp[i][j][k] = -1;
}
}
}
dp[0][0][0] = 0;
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= m; j++) {
for (int k = 0; k < 400; k++) {
if (dp[i][j][k] == -1) {
continue;
}
if (i == n && j == m && k == 0) {
if (ansLen == -1 || dp[i][j][k] < ansLen) {
ansPos = make_pair(make_pair(i, j), k);
ansLen = dp[i][j][k];
}
continue;
}
int addI = (i < n && a[i] == '(') ? 1 : 0;
int addJ = (j < m && b[j] == '(') ? 1 : 0;
if (dp[i + addI][j + addJ][k + 1] == -1 ||
dp[i + addI][j + addJ][k + 1] > dp[i][j][k] + 1) {
dp[i + addI][j + addJ][k + 1] = dp[i][j][k] + 1;
asdasfsadj[i + addI][j + addJ][k + 1] = make_pair(make_pair(i, j), k);
}
}
for (int k = 400; k >= 0; k--) {
if (dp[i][j][k] == -1) {
continue;
}
if (i == n && j == m && k == 0) {
if (ansLen == -1 || dp[i][j][k] < ansLen) {
ansPos = make_pair(make_pair(i, j), k);
ansLen = dp[i][j][k];
}
continue;
}
int addI = (i < n && a[i] == ')') ? 1 : 0;
int addJ = (j < m && b[j] == ')') ? 1 : 0;
if (dp[i + addI][j + addJ][k - 1] == -1 ||
dp[i + addI][j + addJ][k - 1] > dp[i][j][k] + 1) {
dp[i + addI][j + addJ][k - 1] = dp[i][j][k] + 1;
asdasfsadj[i + addI][j + addJ][k - 1] = make_pair(make_pair(i, j), k);
}
}
}
}
cerr << ansLen << endl;
while (true) {
int i = ansPos.first.first;
int j = ansPos.first.second;
int k = ansPos.second;
if (dp[i][j][k] == 0) {
break;
}
ansPos = asdasfsadj[i][j][k];
int newK = ansPos.second;
if (newK < k) {
ans.append(1, '(');
} else {
ans.append(1, ')');
}
}
reverse(ans.begin(), ans.end());
cout << ans << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int i, i0, n, m;
int dp[205][205][405];
int main() {
string s, t;
cin >> s >> t;
for (int i = 0; i <= s.length(); i++) {
for (int i0 = 0; i0 <= t.length(); i0++) {
for (int p = 0; p <= 400; p++) {
dp[i][i0][p] = 10000;
}
}
}
dp[0][0][0] = 0;
for (int i = 0; i <= s.length(); i++) {
for (int i0 = 0; i0 <= t.length(); i0++) {
if (i == 0 && i0 == 0) continue;
for (int p = 0; p <= 400; p++) {
if (i && i0 && s[i - 1] == t[i0 - 1]) {
if (s[i - 1] == '(') {
if (p) {
dp[i][i0][p] = min(dp[i][i0][p], dp[i - 1][i0 - 1][p - 1] + 1);
}
} else {
dp[i][i0][p] = min(dp[i][i0][p], dp[i - 1][i0 - 1][p + 1] + 1);
dp[i][i0][p] = min(dp[i][i0][p], dp[i - 1][i0 - 1][p] + 2);
}
}
if (i && s[i - 1] == '(') {
if (p) {
dp[i][i0][p] = min(dp[i][i0][p], dp[i - 1][i0][p - 1] + 1);
}
} else if (i) {
dp[i][i0][p] = min(dp[i][i0][p], dp[i - 1][i0][p + 1] + 1);
dp[i][i0][p] = min(dp[i][i0][p], dp[i - 1][i0][p] + 2);
}
if (i0 && t[i0 - 1] == '(') {
if (p) {
dp[i][i0][p] = min(dp[i][i0][p], dp[i][i0 - 1][p - 1] + 1);
}
} else if (i0) {
dp[i][i0][p] = min(dp[i][i0][p], dp[i][i0 - 1][p + 1] + 1);
dp[i][i0][p] = min(dp[i][i0][p], dp[i][i0 - 1][p] + 2);
}
}
}
}
int len = INT_MAX;
for (int i = 0; i <= 400; i++) {
len = min(len, i + dp[s.length()][t.length()][i]);
}
for (int x = 0; x <= 400; x++) {
if (len == x + dp[s.length()][t.length()][x]) {
string ans;
int i = s.length(), i0 = t.length(), p = x;
while (i != 0 || i0 != 0 || p != 0) {
if (i && i0 && s[i - 1] == t[i0 - 1]) {
if (s[i - 1] == '(') {
if (p) {
if (dp[i][i0][p] == dp[i - 1][i0 - 1][p - 1] + 1) {
ans += s[i - 1];
i--, i0--, p--;
continue;
}
}
} else {
if (dp[i][i0][p] == dp[i - 1][i0 - 1][p + 1] + 1) {
ans += s[i - 1];
i--, i0--, p++;
continue;
}
if (dp[i][i0][p] == dp[i - 1][i0 - 1][p] + 2) {
i--, i0--;
ans += ")(";
continue;
}
}
}
if (i && s[i - 1] == '(') {
if (p) {
if (dp[i][i0][p] == dp[i - 1][i0][p - 1] + 1) {
ans += s[i - 1];
i--, p--;
continue;
}
}
} else if (i) {
if (dp[i][i0][p] == dp[i - 1][i0][p + 1] + 1) {
ans += s[i - 1];
i--, p++;
continue;
}
if (dp[i][i0][p] == dp[i - 1][i0][p] + 2) {
i--;
ans += ")(";
continue;
}
}
if (i0 && t[i0 - 1] == '(') {
if (p) {
if (dp[i][i0][p] == dp[i][i0 - 1][p - 1] + 1) {
ans += t[i0 - 1];
i0--, p--;
continue;
}
}
} else if (i0) {
if (dp[i][i0][p] == dp[i][i0 - 1][p + 1] + 1) {
ans += t[i0 - 1];
i0--, p++;
continue;
}
if (dp[i][i0][p] == dp[i][i0 - 1][p] + 2) {
i0--;
ans += ")(";
continue;
}
}
}
reverse(ans.begin(), ans.end());
while (x--) ans += ')';
cout << ans << endl;
break;
}
}
return 0;
}
| ### Prompt
Your challenge is to write a CPP solution to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int i, i0, n, m;
int dp[205][205][405];
int main() {
string s, t;
cin >> s >> t;
for (int i = 0; i <= s.length(); i++) {
for (int i0 = 0; i0 <= t.length(); i0++) {
for (int p = 0; p <= 400; p++) {
dp[i][i0][p] = 10000;
}
}
}
dp[0][0][0] = 0;
for (int i = 0; i <= s.length(); i++) {
for (int i0 = 0; i0 <= t.length(); i0++) {
if (i == 0 && i0 == 0) continue;
for (int p = 0; p <= 400; p++) {
if (i && i0 && s[i - 1] == t[i0 - 1]) {
if (s[i - 1] == '(') {
if (p) {
dp[i][i0][p] = min(dp[i][i0][p], dp[i - 1][i0 - 1][p - 1] + 1);
}
} else {
dp[i][i0][p] = min(dp[i][i0][p], dp[i - 1][i0 - 1][p + 1] + 1);
dp[i][i0][p] = min(dp[i][i0][p], dp[i - 1][i0 - 1][p] + 2);
}
}
if (i && s[i - 1] == '(') {
if (p) {
dp[i][i0][p] = min(dp[i][i0][p], dp[i - 1][i0][p - 1] + 1);
}
} else if (i) {
dp[i][i0][p] = min(dp[i][i0][p], dp[i - 1][i0][p + 1] + 1);
dp[i][i0][p] = min(dp[i][i0][p], dp[i - 1][i0][p] + 2);
}
if (i0 && t[i0 - 1] == '(') {
if (p) {
dp[i][i0][p] = min(dp[i][i0][p], dp[i][i0 - 1][p - 1] + 1);
}
} else if (i0) {
dp[i][i0][p] = min(dp[i][i0][p], dp[i][i0 - 1][p + 1] + 1);
dp[i][i0][p] = min(dp[i][i0][p], dp[i][i0 - 1][p] + 2);
}
}
}
}
int len = INT_MAX;
for (int i = 0; i <= 400; i++) {
len = min(len, i + dp[s.length()][t.length()][i]);
}
for (int x = 0; x <= 400; x++) {
if (len == x + dp[s.length()][t.length()][x]) {
string ans;
int i = s.length(), i0 = t.length(), p = x;
while (i != 0 || i0 != 0 || p != 0) {
if (i && i0 && s[i - 1] == t[i0 - 1]) {
if (s[i - 1] == '(') {
if (p) {
if (dp[i][i0][p] == dp[i - 1][i0 - 1][p - 1] + 1) {
ans += s[i - 1];
i--, i0--, p--;
continue;
}
}
} else {
if (dp[i][i0][p] == dp[i - 1][i0 - 1][p + 1] + 1) {
ans += s[i - 1];
i--, i0--, p++;
continue;
}
if (dp[i][i0][p] == dp[i - 1][i0 - 1][p] + 2) {
i--, i0--;
ans += ")(";
continue;
}
}
}
if (i && s[i - 1] == '(') {
if (p) {
if (dp[i][i0][p] == dp[i - 1][i0][p - 1] + 1) {
ans += s[i - 1];
i--, p--;
continue;
}
}
} else if (i) {
if (dp[i][i0][p] == dp[i - 1][i0][p + 1] + 1) {
ans += s[i - 1];
i--, p++;
continue;
}
if (dp[i][i0][p] == dp[i - 1][i0][p] + 2) {
i--;
ans += ")(";
continue;
}
}
if (i0 && t[i0 - 1] == '(') {
if (p) {
if (dp[i][i0][p] == dp[i][i0 - 1][p - 1] + 1) {
ans += t[i0 - 1];
i0--, p--;
continue;
}
}
} else if (i0) {
if (dp[i][i0][p] == dp[i][i0 - 1][p + 1] + 1) {
ans += t[i0 - 1];
i0--, p++;
continue;
}
if (dp[i][i0][p] == dp[i][i0 - 1][p] + 2) {
i0--;
ans += ")(";
continue;
}
}
}
reverse(ans.begin(), ans.end());
while (x--) ans += ')';
cout << ans << endl;
break;
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
int inp() {
char c = getchar();
while (c < '0' || c > '9') c = getchar();
int sum = 0;
while (c >= '0' && c <= '9') {
sum = sum * 10 + c - '0';
c = getchar();
}
return sum;
}
int f[210][210][410];
char s[210], t[210];
int fi[210][210][410], fj[210][210][410], fu[210][210][410];
char br[3] = {'(', ' ', ')'}, ans[10010];
void ckmin(int i, int j, int u, int b, int _i, int _j, int _u) {
f[i][j][u] = std::min(f[i][j][u], b);
if (f[i][j][u] == b) {
fi[i][j][u] = _i;
fj[i][j][u] = _j;
fu[i][j][u] = _u;
}
}
int main() {
scanf("%s", s + 1);
scanf("%s", t + 1);
int n = strlen(s + 1);
int m = strlen(t + 1);
memset(f, 0x3f, sizeof(f));
f[0][0][0] = 0;
for (int i = 0; i <= n; i++)
for (int j = 0; j <= m; j++) {
for (int u = 0; u <= 400; u++)
ckmin(i + (s[i + 1] == '('), j + (t[j + 1] == '('), u + 1,
f[i][j][u] + 1, i, j, u);
for (int u = 400; u > 0; u--)
ckmin(i + (s[i + 1] == ')'), j + (t[j + 1] == ')'), u - 1,
f[i][j][u] + 1, i, j, u);
}
int i = n;
int j = m;
int u = 0;
while (i != 0 || j != 0 || u != 0) {
ans[f[i][j][u]] = br[fu[i][j][u] - u + 1];
int _i = fi[i][j][u];
int _j = fj[i][j][u];
int _u = fu[i][j][u];
i = _i;
j = _j;
u = _u;
}
for (int i = 1; i <= f[n][m][0]; i++) putchar(ans[i]);
putchar('\n');
}
| ### Prompt
Develop a solution in cpp to the problem described below:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
int inp() {
char c = getchar();
while (c < '0' || c > '9') c = getchar();
int sum = 0;
while (c >= '0' && c <= '9') {
sum = sum * 10 + c - '0';
c = getchar();
}
return sum;
}
int f[210][210][410];
char s[210], t[210];
int fi[210][210][410], fj[210][210][410], fu[210][210][410];
char br[3] = {'(', ' ', ')'}, ans[10010];
void ckmin(int i, int j, int u, int b, int _i, int _j, int _u) {
f[i][j][u] = std::min(f[i][j][u], b);
if (f[i][j][u] == b) {
fi[i][j][u] = _i;
fj[i][j][u] = _j;
fu[i][j][u] = _u;
}
}
int main() {
scanf("%s", s + 1);
scanf("%s", t + 1);
int n = strlen(s + 1);
int m = strlen(t + 1);
memset(f, 0x3f, sizeof(f));
f[0][0][0] = 0;
for (int i = 0; i <= n; i++)
for (int j = 0; j <= m; j++) {
for (int u = 0; u <= 400; u++)
ckmin(i + (s[i + 1] == '('), j + (t[j + 1] == '('), u + 1,
f[i][j][u] + 1, i, j, u);
for (int u = 400; u > 0; u--)
ckmin(i + (s[i + 1] == ')'), j + (t[j + 1] == ')'), u - 1,
f[i][j][u] + 1, i, j, u);
}
int i = n;
int j = m;
int u = 0;
while (i != 0 || j != 0 || u != 0) {
ans[f[i][j][u]] = br[fu[i][j][u] - u + 1];
int _i = fi[i][j][u];
int _j = fj[i][j][u];
int _u = fu[i][j][u];
i = _i;
j = _j;
u = _u;
}
for (int i = 1; i <= f[n][m][0]; i++) putchar(ans[i]);
putchar('\n');
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 205;
int dp[N][N][N << 1];
pair<pair<int, int>, int> p[N][N][N << 1];
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
string s, t;
cin >> s >> t;
int n = int((s).size()), m = int((t).size());
for (int i = int(0); i < int(n + 1); ++i)
for (int j = int(0); j < int(m + 1); ++j)
for (int k = int(0); k < int(N << 1); ++k) dp[i][j][k] = 1e9;
dp[0][0][0] = 0;
queue<pair<pair<int, int>, int> > Q;
Q.push(make_pair(make_pair(0, 0), 0));
for (; !Q.empty(); Q.pop()) {
int x = Q.front().first.first, y = Q.front().first.second,
d = Q.front().second;
if (x == n && y == m && !d) break;
if (d + 1 < (N << 1)) {
int nxtX = x + (x < n && s[x] == '(');
int nxtY = y + (y < m && t[y] == '(');
if (dp[x][y][d] + 1 < dp[nxtX][nxtY][d + 1]) {
dp[nxtX][nxtY][d + 1] = dp[x][y][d] + 1;
p[nxtX][nxtY][d + 1] = make_pair(make_pair(x, y), d);
Q.push(make_pair(make_pair(nxtX, nxtY), d + 1));
}
}
if (d) {
int nxtX = x + (x < n && s[x] == ')');
int nxtY = y + (y < m && t[y] == ')');
if (dp[x][y][d] + 1 < dp[nxtX][nxtY][d - 1]) {
dp[nxtX][nxtY][d - 1] = dp[x][y][d] + 1;
p[nxtX][nxtY][d - 1] = make_pair(make_pair(x, y), d);
Q.push(make_pair(make_pair(nxtX, nxtY), d - 1));
}
}
}
int i = n, j = m, k = 0;
string ans = "";
while (i || j || k) {
int nxtI = p[i][j][k].first.first;
int nxtJ = p[i][j][k].first.second;
int nxtK = p[i][j][k].second;
if (nxtK < k)
ans.push_back('(');
else
ans.push_back(')');
i = nxtI, j = nxtJ, k = nxtK;
}
reverse((ans).begin(), (ans).end());
cout << ans << endl;
return 0;
}
| ### Prompt
Develop a solution in Cpp to the problem described below:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 205;
int dp[N][N][N << 1];
pair<pair<int, int>, int> p[N][N][N << 1];
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
string s, t;
cin >> s >> t;
int n = int((s).size()), m = int((t).size());
for (int i = int(0); i < int(n + 1); ++i)
for (int j = int(0); j < int(m + 1); ++j)
for (int k = int(0); k < int(N << 1); ++k) dp[i][j][k] = 1e9;
dp[0][0][0] = 0;
queue<pair<pair<int, int>, int> > Q;
Q.push(make_pair(make_pair(0, 0), 0));
for (; !Q.empty(); Q.pop()) {
int x = Q.front().first.first, y = Q.front().first.second,
d = Q.front().second;
if (x == n && y == m && !d) break;
if (d + 1 < (N << 1)) {
int nxtX = x + (x < n && s[x] == '(');
int nxtY = y + (y < m && t[y] == '(');
if (dp[x][y][d] + 1 < dp[nxtX][nxtY][d + 1]) {
dp[nxtX][nxtY][d + 1] = dp[x][y][d] + 1;
p[nxtX][nxtY][d + 1] = make_pair(make_pair(x, y), d);
Q.push(make_pair(make_pair(nxtX, nxtY), d + 1));
}
}
if (d) {
int nxtX = x + (x < n && s[x] == ')');
int nxtY = y + (y < m && t[y] == ')');
if (dp[x][y][d] + 1 < dp[nxtX][nxtY][d - 1]) {
dp[nxtX][nxtY][d - 1] = dp[x][y][d] + 1;
p[nxtX][nxtY][d - 1] = make_pair(make_pair(x, y), d);
Q.push(make_pair(make_pair(nxtX, nxtY), d - 1));
}
}
}
int i = n, j = m, k = 0;
string ans = "";
while (i || j || k) {
int nxtI = p[i][j][k].first.first;
int nxtJ = p[i][j][k].first.second;
int nxtK = p[i][j][k].second;
if (nxtK < k)
ans.push_back('(');
else
ans.push_back(')');
i = nxtI, j = nxtJ, k = nxtK;
}
reverse((ans).begin(), (ans).end());
cout << ans << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int maxn = 205;
const int oo = 1000000009;
char s[maxn], t[maxn];
int n, m, f[maxn][maxn][2 * maxn];
pair<int, pair<int, int> > trace[maxn][maxn][2 * maxn];
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> s + 1 >> t + 1;
n = strlen(s + 1);
m = strlen(t + 1);
s[n + 1] = 'K';
t[m + 1] = 'K';
for (int i = 0; i <= n; ++i)
for (int j = 0; j <= m; ++j)
for (int k = 0; k <= n + m; ++k) f[i][j][k] = oo;
f[0][0][0] = 0;
for (int i = 0; i <= n; ++i)
for (int j = 0; j <= m; ++j) {
for (int k = 0; k < n + m; ++k)
if (f[i][j][k] < oo) {
int ii = i + (s[i + 1] == '('), jj = j + (t[j + 1] == '(');
if (f[ii][jj][k + 1] > f[i][j][k] + 1) {
f[ii][jj][k + 1] = f[i][j][k] + 1;
trace[ii][jj][k + 1] =
pair<int, pair<int, int> >(1, pair<int, int>(i, j));
}
}
for (int k = n + m; k > 0; --k)
if (f[i][j][k] < oo) {
int ii = i + (s[i + 1] == ')'), jj = j + (t[j + 1] == ')');
if (f[ii][jj][k - 1] > f[i][j][k] + 1) {
f[ii][jj][k - 1] = f[i][j][k] + 1;
trace[ii][jj][k - 1] =
pair<int, pair<int, int> >(-1, pair<int, int>(i, j));
}
}
}
vector<char> res;
int i = n, j = m, k = 0;
while (i > 0 || j > 0 || k > 0) {
if (trace[i][j][k].first == 1) {
res.push_back('(');
int ti = i;
i = trace[i][j][k].second.first;
j = trace[ti][j][k].second.second;
--k;
} else {
res.push_back(')');
int ti = i;
i = trace[i][j][k].second.first;
j = trace[ti][j][k].second.second;
++k;
}
}
for (auto i = res.rbegin(); i != res.rend(); ++i) cout << *i;
}
| ### Prompt
Please create a solution in Cpp to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = 205;
const int oo = 1000000009;
char s[maxn], t[maxn];
int n, m, f[maxn][maxn][2 * maxn];
pair<int, pair<int, int> > trace[maxn][maxn][2 * maxn];
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> s + 1 >> t + 1;
n = strlen(s + 1);
m = strlen(t + 1);
s[n + 1] = 'K';
t[m + 1] = 'K';
for (int i = 0; i <= n; ++i)
for (int j = 0; j <= m; ++j)
for (int k = 0; k <= n + m; ++k) f[i][j][k] = oo;
f[0][0][0] = 0;
for (int i = 0; i <= n; ++i)
for (int j = 0; j <= m; ++j) {
for (int k = 0; k < n + m; ++k)
if (f[i][j][k] < oo) {
int ii = i + (s[i + 1] == '('), jj = j + (t[j + 1] == '(');
if (f[ii][jj][k + 1] > f[i][j][k] + 1) {
f[ii][jj][k + 1] = f[i][j][k] + 1;
trace[ii][jj][k + 1] =
pair<int, pair<int, int> >(1, pair<int, int>(i, j));
}
}
for (int k = n + m; k > 0; --k)
if (f[i][j][k] < oo) {
int ii = i + (s[i + 1] == ')'), jj = j + (t[j + 1] == ')');
if (f[ii][jj][k - 1] > f[i][j][k] + 1) {
f[ii][jj][k - 1] = f[i][j][k] + 1;
trace[ii][jj][k - 1] =
pair<int, pair<int, int> >(-1, pair<int, int>(i, j));
}
}
}
vector<char> res;
int i = n, j = m, k = 0;
while (i > 0 || j > 0 || k > 0) {
if (trace[i][j][k].first == 1) {
res.push_back('(');
int ti = i;
i = trace[i][j][k].second.first;
j = trace[ti][j][k].second.second;
--k;
} else {
res.push_back(')');
int ti = i;
i = trace[i][j][k].second.first;
j = trace[ti][j][k].second.second;
++k;
}
}
for (auto i = res.rbegin(); i != res.rend(); ++i) cout << *i;
}
``` |
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = 210;
const int inf = 1.01e9;
int dp[2 * N][N][N];
string res = "";
string foo, bar;
int n, m;
int rec(int p, int i, int j) {
if (p < 0 || p >= 2 * N) return inf;
if (i == n && j == m) return p;
if (dp[p][i][j] != -1) return dp[p][i][j];
return dp[p][i][j] =
1 + min(rec(p + 1, i + (foo[i] == '('), j + (bar[j] == '(')),
rec(p - 1, i + (foo[i] == ')'), j + (bar[j] == ')')));
}
void trace(int p, int i, int j) {
if (p < 0 || p >= 2 * N) assert(false);
if (i == n && j == m) {
res += string(p, ')');
return;
}
if (rec(p, i, j) ==
1 + rec(p + 1, i + (foo[i] == '('), j + (bar[j] == '('))) {
res += "(";
trace(p + 1, i + (foo[i] == '('), j + (bar[j] == '('));
} else {
res += ")";
trace(p - 1, i + (foo[i] == ')'), j + (bar[j] == ')'));
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
memset(dp, -1, sizeof dp);
cin >> foo >> bar;
n = (int)foo.size();
m = (int)bar.size();
if (n > m) swap(n, m), swap(foo, bar);
foo += '+';
bar += '+';
res = "";
trace(0, 0, 0);
cout << res << '\n';
return 0;
}
| ### Prompt
Please formulate a Cpp solution to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = 210;
const int inf = 1.01e9;
int dp[2 * N][N][N];
string res = "";
string foo, bar;
int n, m;
int rec(int p, int i, int j) {
if (p < 0 || p >= 2 * N) return inf;
if (i == n && j == m) return p;
if (dp[p][i][j] != -1) return dp[p][i][j];
return dp[p][i][j] =
1 + min(rec(p + 1, i + (foo[i] == '('), j + (bar[j] == '(')),
rec(p - 1, i + (foo[i] == ')'), j + (bar[j] == ')')));
}
void trace(int p, int i, int j) {
if (p < 0 || p >= 2 * N) assert(false);
if (i == n && j == m) {
res += string(p, ')');
return;
}
if (rec(p, i, j) ==
1 + rec(p + 1, i + (foo[i] == '('), j + (bar[j] == '('))) {
res += "(";
trace(p + 1, i + (foo[i] == '('), j + (bar[j] == '('));
} else {
res += ")";
trace(p - 1, i + (foo[i] == ')'), j + (bar[j] == ')'));
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
memset(dp, -1, sizeof dp);
cin >> foo >> bar;
n = (int)foo.size();
m = (int)bar.size();
if (n > m) swap(n, m), swap(foo, bar);
foo += '+';
bar += '+';
res = "";
trace(0, 0, 0);
cout << res << '\n';
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
inline int add(int a, int b) {
a += b;
return a >= 1000000007 ? a - 1000000007 : a;
}
inline int sub(int a, int b) {
a -= b;
return a < 0 ? a + 1000000007 : a;
}
inline int mul(int a, int b) { return (long long int)a * b % 1000000007; }
int Set(int N, int pos) { return N = N | (1 << pos); }
int Reset(int N, int pos) { return N = N & ~(1 << pos); }
bool Cheek(int N, int pos) { return (bool)(N & (1 << pos)); }
string s1, s2, ans = "";
int len1, len2, dp[205][205][205];
int FuN(int pos1, int pos2, int open) {
if (pos1 >= len1 && pos2 >= len2) return dp[pos1][pos2][open] = open;
if (dp[pos1][pos2][open] != -1) return dp[pos1][pos2][open];
int ret1 = (1 << 28), ret2 = (1 << 28);
if (pos1 < len1 && pos2 < len2 && s1[pos1] == s2[pos2]) {
if (s1[pos1] == '(')
ret1 = 1 + FuN(pos1 + 1, pos2 + 1, open + 1);
else {
if (open > 0)
ret2 = 1 + FuN(pos1 + 1, pos2 + 1, open - 1);
else
ret2 = 2 + FuN(pos1 + 1, pos2 + 1, open);
}
} else {
if ((pos1 < len1 && s1[pos1] == '(') || (pos2 < len2 && s2[pos2] == '('))
ret1 = 1 + FuN(pos1 + (pos1 < len1 && s1[pos1] == '('),
pos2 + (pos2 < len2 && s2[pos2] == '('), open + 1);
if ((pos1 < len1 && s1[pos1] == ')') || (pos2 < len2 && s2[pos2] == ')')) {
if (open > 0)
ret2 = 1 + FuN(pos1 + (pos1 < len1 && s1[pos1] == ')'),
pos2 + (pos2 < len2 && s2[pos2] == ')'), open - 1);
else
ret2 = 2 + FuN(pos1 + (pos1 < len1 && s1[pos1] == ')'),
pos2 + (pos2 < len2 && s2[pos2] == ')'), open);
}
}
return dp[pos1][pos2][open] = min(ret1, ret2);
}
void Print(int pos1, int pos2, int open) {
if (pos1 >= len1 && pos2 >= len2) {
for (int i = 1; i <= open; i++) printf(")");
printf("\n");
return;
}
int ret1 = (1 << 28), ret2 = (1 << 28), pr = 0;
if (pos1 < len1 && pos2 < len2 && s1[pos1] == s2[pos2]) {
if (s1[pos1] == '(')
ret1 = 1 + dp[pos1 + 1][pos2 + 1][open + 1];
else {
if (open > 0)
ret2 = 1 + dp[pos1 + 1][pos2 + 1][open - 1];
else
ret2 = 2 + dp[pos1 + 1][pos2 + 1][open], pr = 1;
}
if (ret1 <= ret2) {
printf("(");
Print(pos1 + 1, pos2 + 1, open + 1);
} else {
if (!pr) {
printf(")");
Print(pos1 + 1, pos2 + 1, open - 1);
} else {
printf("()");
Print(pos1 + 1, pos2 + 1, open);
}
}
} else {
if ((pos1 < len1 && s1[pos1] == '(') || (pos2 < len2 && s2[pos2] == '('))
ret1 = 1 + dp[pos1 + (pos1 < len1 && s1[pos1] == '(')]
[pos2 + (pos2 < len2 && s2[pos2] == '(')][open + 1];
if ((pos1 < len1 && s1[pos1] == ')') || (pos2 < len2 && s2[pos2] == ')')) {
if (open > 0)
ret2 = 1 + dp[pos1 + (pos1 < len1 && s1[pos1] == ')')]
[pos2 + (pos2 < len2 && s2[pos2] == ')')][open - 1];
else
ret2 = 2 + dp[pos1 + (pos1 < len1 && s1[pos1] == ')')]
[pos2 + (pos2 < len2 && s2[pos2] == ')')][open],
pr = 1;
}
if (ret1 <= ret2) {
printf("(");
Print(pos1 + (pos1 < len1 && s1[pos1] == '('),
pos2 + (pos2 < len2 && s2[pos2] == '('), open + 1);
} else {
if (!pr) {
printf(")");
Print(pos1 + (pos1 < len1 && s1[pos1] == ')'),
pos2 + (pos2 < len2 && s2[pos2] == ')'), open - 1);
} else {
printf("()");
Print(pos1 + (pos1 < len1 && s1[pos1] == ')'),
pos2 + (pos2 < len2 && s2[pos2] == ')'), open);
}
}
}
}
void Solve(int t) {
int i, j, k, ans;
cin >> s1 >> s2;
len1 = (int)s1.size();
len2 = (int)s2.size();
memset(dp, -1, sizeof(dp));
ans = FuN(0, 0, 0);
Print(0, 0, 0);
}
int main() {
int t, T;
T = 1;
for (__typeof(t) t = 1; t <= T; t++) {
Solve(t);
}
return 0;
}
| ### Prompt
Construct a CPP code solution to the problem outlined:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
inline int add(int a, int b) {
a += b;
return a >= 1000000007 ? a - 1000000007 : a;
}
inline int sub(int a, int b) {
a -= b;
return a < 0 ? a + 1000000007 : a;
}
inline int mul(int a, int b) { return (long long int)a * b % 1000000007; }
int Set(int N, int pos) { return N = N | (1 << pos); }
int Reset(int N, int pos) { return N = N & ~(1 << pos); }
bool Cheek(int N, int pos) { return (bool)(N & (1 << pos)); }
string s1, s2, ans = "";
int len1, len2, dp[205][205][205];
int FuN(int pos1, int pos2, int open) {
if (pos1 >= len1 && pos2 >= len2) return dp[pos1][pos2][open] = open;
if (dp[pos1][pos2][open] != -1) return dp[pos1][pos2][open];
int ret1 = (1 << 28), ret2 = (1 << 28);
if (pos1 < len1 && pos2 < len2 && s1[pos1] == s2[pos2]) {
if (s1[pos1] == '(')
ret1 = 1 + FuN(pos1 + 1, pos2 + 1, open + 1);
else {
if (open > 0)
ret2 = 1 + FuN(pos1 + 1, pos2 + 1, open - 1);
else
ret2 = 2 + FuN(pos1 + 1, pos2 + 1, open);
}
} else {
if ((pos1 < len1 && s1[pos1] == '(') || (pos2 < len2 && s2[pos2] == '('))
ret1 = 1 + FuN(pos1 + (pos1 < len1 && s1[pos1] == '('),
pos2 + (pos2 < len2 && s2[pos2] == '('), open + 1);
if ((pos1 < len1 && s1[pos1] == ')') || (pos2 < len2 && s2[pos2] == ')')) {
if (open > 0)
ret2 = 1 + FuN(pos1 + (pos1 < len1 && s1[pos1] == ')'),
pos2 + (pos2 < len2 && s2[pos2] == ')'), open - 1);
else
ret2 = 2 + FuN(pos1 + (pos1 < len1 && s1[pos1] == ')'),
pos2 + (pos2 < len2 && s2[pos2] == ')'), open);
}
}
return dp[pos1][pos2][open] = min(ret1, ret2);
}
void Print(int pos1, int pos2, int open) {
if (pos1 >= len1 && pos2 >= len2) {
for (int i = 1; i <= open; i++) printf(")");
printf("\n");
return;
}
int ret1 = (1 << 28), ret2 = (1 << 28), pr = 0;
if (pos1 < len1 && pos2 < len2 && s1[pos1] == s2[pos2]) {
if (s1[pos1] == '(')
ret1 = 1 + dp[pos1 + 1][pos2 + 1][open + 1];
else {
if (open > 0)
ret2 = 1 + dp[pos1 + 1][pos2 + 1][open - 1];
else
ret2 = 2 + dp[pos1 + 1][pos2 + 1][open], pr = 1;
}
if (ret1 <= ret2) {
printf("(");
Print(pos1 + 1, pos2 + 1, open + 1);
} else {
if (!pr) {
printf(")");
Print(pos1 + 1, pos2 + 1, open - 1);
} else {
printf("()");
Print(pos1 + 1, pos2 + 1, open);
}
}
} else {
if ((pos1 < len1 && s1[pos1] == '(') || (pos2 < len2 && s2[pos2] == '('))
ret1 = 1 + dp[pos1 + (pos1 < len1 && s1[pos1] == '(')]
[pos2 + (pos2 < len2 && s2[pos2] == '(')][open + 1];
if ((pos1 < len1 && s1[pos1] == ')') || (pos2 < len2 && s2[pos2] == ')')) {
if (open > 0)
ret2 = 1 + dp[pos1 + (pos1 < len1 && s1[pos1] == ')')]
[pos2 + (pos2 < len2 && s2[pos2] == ')')][open - 1];
else
ret2 = 2 + dp[pos1 + (pos1 < len1 && s1[pos1] == ')')]
[pos2 + (pos2 < len2 && s2[pos2] == ')')][open],
pr = 1;
}
if (ret1 <= ret2) {
printf("(");
Print(pos1 + (pos1 < len1 && s1[pos1] == '('),
pos2 + (pos2 < len2 && s2[pos2] == '('), open + 1);
} else {
if (!pr) {
printf(")");
Print(pos1 + (pos1 < len1 && s1[pos1] == ')'),
pos2 + (pos2 < len2 && s2[pos2] == ')'), open - 1);
} else {
printf("()");
Print(pos1 + (pos1 < len1 && s1[pos1] == ')'),
pos2 + (pos2 < len2 && s2[pos2] == ')'), open);
}
}
}
}
void Solve(int t) {
int i, j, k, ans;
cin >> s1 >> s2;
len1 = (int)s1.size();
len2 = (int)s2.size();
memset(dp, -1, sizeof(dp));
ans = FuN(0, 0, 0);
Print(0, 0, 0);
}
int main() {
int t, T;
T = 1;
for (__typeof(t) t = 1; t <= T; t++) {
Solve(t);
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
using ll = long long int;
using ii = pair<int, int>;
using vi = vector<int>;
using vl = vector<ll>;
using vii = vector<ii>;
using graph = vector<vi>;
using matrix = graph;
const int MAXN = 2e5 + 5;
const int INF = 0x3f3f3f3f;
const ll INFL = 1e18 + 4;
const ll mod = 1e9 + 7;
int n, m, dp[405][405][405];
string s, f;
int sol(int a, int b, int open) {
if (open == 400) return INF;
if (a >= n && b >= m) return dp[a][b][open] = open;
if (dp[a][b][open] > 0) return dp[a][b][open];
int ans = min(INF, 1 + sol(a + (a < n && s[a] == '('),
b + (b < m && f[b] == '('), open + 1));
if (open > 0) {
ans = min(ans, 1 + sol(a + (a < n && s[a] == ')'),
b + (b < m && f[b] == ')'), open - 1));
}
return dp[a][b][open] = ans;
}
string ans = "";
void retrieve(int a, int b, int open) {
if (open == dp[a][b][open]) {
while (open--) ans += ')';
return;
}
if (dp[a][b][open] ==
dp[a + (a < n && s[a] == '(')][b + (b < m && f[b] == '(')][open + 1] +
1) {
ans += '(';
retrieve(a + (a < n && s[a] == '('), b + (b < m && f[b] == '('), open + 1);
} else {
ans += ')';
retrieve(a + (a < n && s[a] == ')'), b + (b < m && f[b] == ')'), open - 1);
}
}
int main() {
cin >> s >> f;
n = s.size();
m = f.size();
memset(dp, -1, sizeof dp);
sol(0, 0, 0);
retrieve(0, 0, 0);
cout << ans << endl;
}
| ### Prompt
Your task is to create a Cpp solution to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long int;
using ii = pair<int, int>;
using vi = vector<int>;
using vl = vector<ll>;
using vii = vector<ii>;
using graph = vector<vi>;
using matrix = graph;
const int MAXN = 2e5 + 5;
const int INF = 0x3f3f3f3f;
const ll INFL = 1e18 + 4;
const ll mod = 1e9 + 7;
int n, m, dp[405][405][405];
string s, f;
int sol(int a, int b, int open) {
if (open == 400) return INF;
if (a >= n && b >= m) return dp[a][b][open] = open;
if (dp[a][b][open] > 0) return dp[a][b][open];
int ans = min(INF, 1 + sol(a + (a < n && s[a] == '('),
b + (b < m && f[b] == '('), open + 1));
if (open > 0) {
ans = min(ans, 1 + sol(a + (a < n && s[a] == ')'),
b + (b < m && f[b] == ')'), open - 1));
}
return dp[a][b][open] = ans;
}
string ans = "";
void retrieve(int a, int b, int open) {
if (open == dp[a][b][open]) {
while (open--) ans += ')';
return;
}
if (dp[a][b][open] ==
dp[a + (a < n && s[a] == '(')][b + (b < m && f[b] == '(')][open + 1] +
1) {
ans += '(';
retrieve(a + (a < n && s[a] == '('), b + (b < m && f[b] == '('), open + 1);
} else {
ans += ')';
retrieve(a + (a < n && s[a] == ')'), b + (b < m && f[b] == ')'), open - 1);
}
}
int main() {
cin >> s >> f;
n = s.size();
m = f.size();
memset(dp, -1, sizeof dp);
sol(0, 0, 0);
retrieve(0, 0, 0);
cout << ans << endl;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int maxn = 205;
string s, t;
struct Node {
int i, j, k;
Node(int i = 0, int j = 0, int k = 0) : i(i), j(j), k(k) {}
};
Node f[maxn][maxn][maxn << 1];
void solve() {
int n = s.length(), m = t.length();
memset(f, -1, sizeof(f));
queue<Node> q;
q.push(Node(0, 0, 0));
f[0][0][0] = Node(0, 0, 0);
while (!q.empty()) {
Node u = q.front();
q.pop();
int i = u.i, j = u.j, k = u.k;
int vi, vj, vk;
vi = i, vj = j, vk = k + 1;
if (s[i] == '(') vi++;
if (t[j] == '(') vj++;
if (vk <= n + m && f[vi][vj][vk].i == -1) {
f[vi][vj][vk] = u;
q.push(Node(vi, vj, vk));
}
vi = i, vj = j, vk = k - 1;
if (s[i] == ')') vi++;
if (t[j] == ')') vj++;
if (vk >= 0 && f[vi][vj][vk].i == -1) {
f[vi][vj][vk] = u;
q.push(Node(vi, vj, vk));
}
}
Node t;
int i = n, j = m, k = 0;
string ans = "";
while (i || j || k) {
t = f[i][j][k];
if (t.k > k)
ans = ")" + ans;
else
ans = "(" + ans;
i = t.i;
j = t.j;
k = t.k;
}
cout << ans << endl;
}
int main() {
cin >> s >> t;
solve();
return 0;
}
| ### Prompt
Please create a solution in CPP to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = 205;
string s, t;
struct Node {
int i, j, k;
Node(int i = 0, int j = 0, int k = 0) : i(i), j(j), k(k) {}
};
Node f[maxn][maxn][maxn << 1];
void solve() {
int n = s.length(), m = t.length();
memset(f, -1, sizeof(f));
queue<Node> q;
q.push(Node(0, 0, 0));
f[0][0][0] = Node(0, 0, 0);
while (!q.empty()) {
Node u = q.front();
q.pop();
int i = u.i, j = u.j, k = u.k;
int vi, vj, vk;
vi = i, vj = j, vk = k + 1;
if (s[i] == '(') vi++;
if (t[j] == '(') vj++;
if (vk <= n + m && f[vi][vj][vk].i == -1) {
f[vi][vj][vk] = u;
q.push(Node(vi, vj, vk));
}
vi = i, vj = j, vk = k - 1;
if (s[i] == ')') vi++;
if (t[j] == ')') vj++;
if (vk >= 0 && f[vi][vj][vk].i == -1) {
f[vi][vj][vk] = u;
q.push(Node(vi, vj, vk));
}
}
Node t;
int i = n, j = m, k = 0;
string ans = "";
while (i || j || k) {
t = f[i][j][k];
if (t.k > k)
ans = ")" + ans;
else
ans = "(" + ans;
i = t.i;
j = t.j;
k = t.k;
}
cout << ans << endl;
}
int main() {
cin >> s >> t;
solve();
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int INF = 1 << 30;
int MOD = 1000000007;
int dp[210][210][210];
pair<pair<int, int>, int> par[210][210][210];
void solve() {
string S, T;
cin >> S >> T;
for (int i = 0; i < (210); i++)
for (int j = 0; j < (210); j++)
for (int k = 0; k < (210); k++) dp[i][j][k] = INF;
dp[0][0][0] = 0;
for (int i = 0; i < (S.size() + 1); i++) {
for (int j = 0; j < (T.size() + 1); j++) {
for (int k = 0; k < (210); k++) {
if (i == S.size() && j == T.size()) break;
if (i == S.size()) {
if (T[j] == '(') {
if (dp[i][j + 1][k + 1] > dp[i][j][k] + 1) {
dp[i][j + 1][k + 1] = dp[i][j][k] + 1;
par[i][j + 1][k + 1] = {{i, j}, k};
}
} else {
if (k == 0) {
if (dp[i][j + 1][k] > dp[i][j][k] + 2) {
dp[i][j + 1][k] = dp[i][j][k] + 2;
par[i][j + 1][k] = {{i, j}, k};
}
} else {
if (dp[i][j + 1][k - 1] > dp[i][j][k] + 1) {
dp[i][j + 1][k - 1] = dp[i][j][k] + 1;
par[i][j + 1][k - 1] = {{i, j}, k};
}
}
}
} else if (j == T.size()) {
if (S[i] == '(') {
if (dp[i + 1][j][k + 1] > dp[i][j][k] + 1) {
dp[i + 1][j][k + 1] = dp[i][j][k] + 1;
par[i + 1][j][k + 1] = {{i, j}, k};
}
} else {
if (k == 0) {
if (dp[i + 1][j][k] > dp[i][j][k] + 2) {
dp[i + 1][j][k] = dp[i][j][k] + 2;
par[i + 1][j][k] = {{i, j}, k};
}
} else {
if (dp[i + 1][j][k - 1] > dp[i][j][k] + 1) {
dp[i + 1][j][k - 1] = dp[i][j][k] + 1;
par[i + 1][j][k - 1] = {{i, j}, k};
}
}
}
} else if (S[i] == T[j]) {
if (S[i] == '(') {
if (dp[i + 1][j + 1][k + 1] > dp[i][j][k] + 1) {
dp[i + 1][j + 1][k + 1] = dp[i][j][k] + 1;
par[i + 1][j + 1][k + 1] = {{i, j}, k};
}
} else {
if (k == 0) {
if (dp[i + 1][j + 1][k] > dp[i][j][k] + 2) {
dp[i + 1][j + 1][k] = dp[i][j][k] + 2;
par[i + 1][j + 1][k] = {{i, j}, k};
}
} else {
if (dp[i + 1][j + 1][k - 1] > dp[i][j][k] + 1) {
dp[i + 1][j + 1][k - 1] = dp[i][j][k] + 1;
par[i + 1][j + 1][k - 1] = {{i, j}, k};
}
}
}
} else {
if (S[i] == '(') {
if (dp[i + 1][j][k + 1] > dp[i][j][k] + 1) {
dp[i + 1][j][k + 1] = dp[i][j][k] + 1;
par[i + 1][j][k + 1] = {{i, j}, k};
}
if (k != 0) {
if (dp[i][j + 1][k - 1] > dp[i][j][k] + 1) {
dp[i][j + 1][k - 1] = dp[i][j][k] + 1;
par[i][j + 1][k - 1] = {{i, j}, k};
}
}
} else {
if (dp[i][j + 1][k + 1] > dp[i][j][k] + 1) {
dp[i][j + 1][k + 1] = dp[i][j][k] + 1;
par[i][j + 1][k + 1] = {{i, j}, k};
}
if (k != 0) {
if (dp[i + 1][j][k - 1] > dp[i][j][k] + 1) {
dp[i + 1][j][k - 1] = dp[i][j][k] + 1;
par[i + 1][j][k - 1] = {{i, j}, k};
}
}
}
}
}
}
}
int second = S.size(), t = T.size(), d;
int m = INF;
for (int i = 0; i < (210); i++) {
if (m > dp[second][t][i] + i) {
m = dp[second][t][i] + i;
d = i;
}
}
string ans = "";
for (int i = 0; i < (d); i++) ans += ')';
int cnt = 0;
while (second != 0 || t != 0) {
cnt++;
int a, b, c;
a = par[second][t][d].first.first;
b = par[second][t][d].first.second;
c = par[second][t][d].second;
if (a == S.size()) {
if (T[b] == '(') {
ans += '(';
} else {
if (c == 0) {
ans += ')';
ans += '(';
} else {
ans += ')';
}
}
} else if (b == T.size()) {
if (S[a] == '(') {
ans += '(';
} else {
if (c == 0) {
ans += ')';
ans += '(';
} else {
ans += ')';
}
}
} else if (S[a] == T[b]) {
if (S[a] == '(') {
ans += '(';
} else {
if (c == 0) {
ans += ')';
ans += '(';
} else {
ans += ')';
}
}
} else {
if (S[a] == '(') {
if (a == second) {
ans += ')';
} else {
ans += '(';
}
} else {
if (a == second) {
ans += '(';
} else {
ans += ')';
}
}
}
second = a;
t = b;
d = c;
}
reverse(ans.begin(), ans.end());
cout << ans << endl;
}
int main() { solve(); }
| ### Prompt
Please create a solution in Cpp to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int INF = 1 << 30;
int MOD = 1000000007;
int dp[210][210][210];
pair<pair<int, int>, int> par[210][210][210];
void solve() {
string S, T;
cin >> S >> T;
for (int i = 0; i < (210); i++)
for (int j = 0; j < (210); j++)
for (int k = 0; k < (210); k++) dp[i][j][k] = INF;
dp[0][0][0] = 0;
for (int i = 0; i < (S.size() + 1); i++) {
for (int j = 0; j < (T.size() + 1); j++) {
for (int k = 0; k < (210); k++) {
if (i == S.size() && j == T.size()) break;
if (i == S.size()) {
if (T[j] == '(') {
if (dp[i][j + 1][k + 1] > dp[i][j][k] + 1) {
dp[i][j + 1][k + 1] = dp[i][j][k] + 1;
par[i][j + 1][k + 1] = {{i, j}, k};
}
} else {
if (k == 0) {
if (dp[i][j + 1][k] > dp[i][j][k] + 2) {
dp[i][j + 1][k] = dp[i][j][k] + 2;
par[i][j + 1][k] = {{i, j}, k};
}
} else {
if (dp[i][j + 1][k - 1] > dp[i][j][k] + 1) {
dp[i][j + 1][k - 1] = dp[i][j][k] + 1;
par[i][j + 1][k - 1] = {{i, j}, k};
}
}
}
} else if (j == T.size()) {
if (S[i] == '(') {
if (dp[i + 1][j][k + 1] > dp[i][j][k] + 1) {
dp[i + 1][j][k + 1] = dp[i][j][k] + 1;
par[i + 1][j][k + 1] = {{i, j}, k};
}
} else {
if (k == 0) {
if (dp[i + 1][j][k] > dp[i][j][k] + 2) {
dp[i + 1][j][k] = dp[i][j][k] + 2;
par[i + 1][j][k] = {{i, j}, k};
}
} else {
if (dp[i + 1][j][k - 1] > dp[i][j][k] + 1) {
dp[i + 1][j][k - 1] = dp[i][j][k] + 1;
par[i + 1][j][k - 1] = {{i, j}, k};
}
}
}
} else if (S[i] == T[j]) {
if (S[i] == '(') {
if (dp[i + 1][j + 1][k + 1] > dp[i][j][k] + 1) {
dp[i + 1][j + 1][k + 1] = dp[i][j][k] + 1;
par[i + 1][j + 1][k + 1] = {{i, j}, k};
}
} else {
if (k == 0) {
if (dp[i + 1][j + 1][k] > dp[i][j][k] + 2) {
dp[i + 1][j + 1][k] = dp[i][j][k] + 2;
par[i + 1][j + 1][k] = {{i, j}, k};
}
} else {
if (dp[i + 1][j + 1][k - 1] > dp[i][j][k] + 1) {
dp[i + 1][j + 1][k - 1] = dp[i][j][k] + 1;
par[i + 1][j + 1][k - 1] = {{i, j}, k};
}
}
}
} else {
if (S[i] == '(') {
if (dp[i + 1][j][k + 1] > dp[i][j][k] + 1) {
dp[i + 1][j][k + 1] = dp[i][j][k] + 1;
par[i + 1][j][k + 1] = {{i, j}, k};
}
if (k != 0) {
if (dp[i][j + 1][k - 1] > dp[i][j][k] + 1) {
dp[i][j + 1][k - 1] = dp[i][j][k] + 1;
par[i][j + 1][k - 1] = {{i, j}, k};
}
}
} else {
if (dp[i][j + 1][k + 1] > dp[i][j][k] + 1) {
dp[i][j + 1][k + 1] = dp[i][j][k] + 1;
par[i][j + 1][k + 1] = {{i, j}, k};
}
if (k != 0) {
if (dp[i + 1][j][k - 1] > dp[i][j][k] + 1) {
dp[i + 1][j][k - 1] = dp[i][j][k] + 1;
par[i + 1][j][k - 1] = {{i, j}, k};
}
}
}
}
}
}
}
int second = S.size(), t = T.size(), d;
int m = INF;
for (int i = 0; i < (210); i++) {
if (m > dp[second][t][i] + i) {
m = dp[second][t][i] + i;
d = i;
}
}
string ans = "";
for (int i = 0; i < (d); i++) ans += ')';
int cnt = 0;
while (second != 0 || t != 0) {
cnt++;
int a, b, c;
a = par[second][t][d].first.first;
b = par[second][t][d].first.second;
c = par[second][t][d].second;
if (a == S.size()) {
if (T[b] == '(') {
ans += '(';
} else {
if (c == 0) {
ans += ')';
ans += '(';
} else {
ans += ')';
}
}
} else if (b == T.size()) {
if (S[a] == '(') {
ans += '(';
} else {
if (c == 0) {
ans += ')';
ans += '(';
} else {
ans += ')';
}
}
} else if (S[a] == T[b]) {
if (S[a] == '(') {
ans += '(';
} else {
if (c == 0) {
ans += ')';
ans += '(';
} else {
ans += ')';
}
}
} else {
if (S[a] == '(') {
if (a == second) {
ans += ')';
} else {
ans += '(';
}
} else {
if (a == second) {
ans += '(';
} else {
ans += ')';
}
}
}
second = a;
t = b;
d = c;
}
reverse(ans.begin(), ans.end());
cout << ans << endl;
}
int main() { solve(); }
``` |
#include <bits/stdc++.h>
using namespace std;
int memo[201][201][201];
int nxt[201][201][201];
class FTwoBracketSequences {
string a;
string b;
bool is_open(string &x, int p) { return p < x.length() && x[p] == '('; }
bool is_close(string &x, int p) { return p < x.length() && x[p] == ')'; }
int dp(int x, int y, int o) {
if (x == a.length() && y == b.length() && o == 0) return 0;
if (memo[x][y][o] != -1) return memo[x][y][o];
auto &ans = memo[x][y][o];
ans = 800;
bool a_open = is_open(a, x);
bool b_open = is_open(b, y);
if (a_open || b_open || o == 0) {
int xp = x + a_open;
int yp = y + b_open;
int cur = 1 + dp(xp, yp, o + 1);
if (cur < ans) {
ans = cur;
nxt[x][y][o] = +1;
}
}
if (o) {
int xp = x + is_close(a, x);
int yp = y + is_close(b, y);
int cur = 1 + dp(xp, yp, o - 1);
if (cur < ans) {
ans = cur;
nxt[x][y][o] = -1;
}
}
return ans;
}
public:
void solve(std::istream &in, std::ostream &out) {
memset(memo, -1, sizeof memo);
in >> a >> b;
dp(0, 0, 0);
int x = 0, y = 0, o = 0;
while (x < a.length() || y < b.length() || o > 0) {
int d = nxt[x][y][o];
o += d;
assert(o >= 0);
if (d == +1) {
out << "(";
if (is_open(a, x)) ++x;
if (is_open(b, y)) ++y;
} else {
out << ")";
if (is_close(a, x)) ++x;
if (is_close(b, y)) ++y;
}
}
out << endl;
}
};
int main() {
FTwoBracketSequences solver;
std::istream &in(std::cin);
std::ostream &out(std::cout);
solver.solve(in, out);
return 0;
}
| ### Prompt
Develop a solution in Cpp to the problem described below:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int memo[201][201][201];
int nxt[201][201][201];
class FTwoBracketSequences {
string a;
string b;
bool is_open(string &x, int p) { return p < x.length() && x[p] == '('; }
bool is_close(string &x, int p) { return p < x.length() && x[p] == ')'; }
int dp(int x, int y, int o) {
if (x == a.length() && y == b.length() && o == 0) return 0;
if (memo[x][y][o] != -1) return memo[x][y][o];
auto &ans = memo[x][y][o];
ans = 800;
bool a_open = is_open(a, x);
bool b_open = is_open(b, y);
if (a_open || b_open || o == 0) {
int xp = x + a_open;
int yp = y + b_open;
int cur = 1 + dp(xp, yp, o + 1);
if (cur < ans) {
ans = cur;
nxt[x][y][o] = +1;
}
}
if (o) {
int xp = x + is_close(a, x);
int yp = y + is_close(b, y);
int cur = 1 + dp(xp, yp, o - 1);
if (cur < ans) {
ans = cur;
nxt[x][y][o] = -1;
}
}
return ans;
}
public:
void solve(std::istream &in, std::ostream &out) {
memset(memo, -1, sizeof memo);
in >> a >> b;
dp(0, 0, 0);
int x = 0, y = 0, o = 0;
while (x < a.length() || y < b.length() || o > 0) {
int d = nxt[x][y][o];
o += d;
assert(o >= 0);
if (d == +1) {
out << "(";
if (is_open(a, x)) ++x;
if (is_open(b, y)) ++y;
} else {
out << ")";
if (is_close(a, x)) ++x;
if (is_close(b, y)) ++y;
}
}
out << endl;
}
};
int main() {
FTwoBracketSequences solver;
std::istream &in(std::cin);
std::ostream &out(std::cout);
solver.solve(in, out);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9;
int dp[209][209][409];
string s, s1, s2;
int bt(int id1, int id2, int curr) {
if (curr > 400) return inf;
if (id1 == s1.size() && id2 == s2.size() && !curr) return 0;
int &ret = dp[id1][id2][curr];
if (ret != -1) return ret;
int Nid1, Nid2;
ret = inf;
Nid1 = id1 + (id1 < s1.size() && s1[id1] == '(');
Nid2 = id2 + (id2 < s2.size() && s2[id2] == '(');
ret = min(ret, bt(Nid1, Nid2, curr + 1) + 1);
if (!curr) return ret;
Nid1 = id1 + (id1 < s1.size() && s1[id1] == ')');
Nid2 = id2 + (id2 < s2.size() && s2[id2] == ')');
ret = min(ret, bt(Nid1, Nid2, curr - 1) + 1);
return ret;
}
int q;
void path(int id1, int id2, int curr) {
if (id1 == s1.size() && id2 == s2.size() && !curr) return;
int ret = dp[id1][id2][curr], Nid1, Nid2;
Nid1 = id1 + (id1 < s1.size() && s1[id1] == '(');
Nid2 = id2 + (id2 < s2.size() && s2[id2] == '(');
if (ret == dp[Nid1][Nid2][curr + 1] + 1) {
s += '(', path(Nid1, Nid2, curr + 1);
return;
}
Nid1 = id1 + (id1 < s1.size() && s1[id1] == ')');
Nid2 = id2 + (id2 < s2.size() && s2[id2] == ')');
if (ret == dp[Nid1][Nid2][curr - 1] + 1) {
s += ')', path(Nid1, Nid2, curr - 1);
return;
}
}
int main() {
ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0), cin >> s1 >> s2;
memset(dp, -1, sizeof(dp));
dp[s1.size()][s2.size()][0] = 0;
int x = bt(0, 0, 0);
path(0, 0, 0);
cout << s << endl;
}
| ### Prompt
Your challenge is to write a CPP solution to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9;
int dp[209][209][409];
string s, s1, s2;
int bt(int id1, int id2, int curr) {
if (curr > 400) return inf;
if (id1 == s1.size() && id2 == s2.size() && !curr) return 0;
int &ret = dp[id1][id2][curr];
if (ret != -1) return ret;
int Nid1, Nid2;
ret = inf;
Nid1 = id1 + (id1 < s1.size() && s1[id1] == '(');
Nid2 = id2 + (id2 < s2.size() && s2[id2] == '(');
ret = min(ret, bt(Nid1, Nid2, curr + 1) + 1);
if (!curr) return ret;
Nid1 = id1 + (id1 < s1.size() && s1[id1] == ')');
Nid2 = id2 + (id2 < s2.size() && s2[id2] == ')');
ret = min(ret, bt(Nid1, Nid2, curr - 1) + 1);
return ret;
}
int q;
void path(int id1, int id2, int curr) {
if (id1 == s1.size() && id2 == s2.size() && !curr) return;
int ret = dp[id1][id2][curr], Nid1, Nid2;
Nid1 = id1 + (id1 < s1.size() && s1[id1] == '(');
Nid2 = id2 + (id2 < s2.size() && s2[id2] == '(');
if (ret == dp[Nid1][Nid2][curr + 1] + 1) {
s += '(', path(Nid1, Nid2, curr + 1);
return;
}
Nid1 = id1 + (id1 < s1.size() && s1[id1] == ')');
Nid2 = id2 + (id2 < s2.size() && s2[id2] == ')');
if (ret == dp[Nid1][Nid2][curr - 1] + 1) {
s += ')', path(Nid1, Nid2, curr - 1);
return;
}
}
int main() {
ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0), cin >> s1 >> s2;
memset(dp, -1, sizeof(dp));
dp[s1.size()][s2.size()][0] = 0;
int x = bt(0, 0, 0);
path(0, 0, 0);
cout << s << endl;
}
``` |
#include <bits/stdc++.h>
using namespace std;
inline int read() {
char c = getchar();
int x = 0, f = 1;
for (; !isdigit(c); c = getchar())
if (c == '-') f = -1;
for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
return x * f;
}
const int MAXN = 210;
int f[MAXN][MAXN][MAXN], M[3], Orzxyy[MAXN][MAXN][MAXN], ans[MAXN];
int N1, N2;
char s[MAXN], t[MAXN];
pair<int, pair<int, int> > xyyAKIOI[MAXN][MAXN][MAXN];
inline void Work() {
int A = 0, B = N1, C = N2, Len = 0;
while (A + B + C != 0) {
ans[++Len] = Orzxyy[A][B][C];
pair<int, pair<int, int> > p = xyyAKIOI[A][B][C];
A = p.first, B = p.second.first, C = p.second.second;
}
for (int i = Len; i; --i) putchar((ans[i] == 1) ? '(' : ')');
puts("");
}
inline void Debug() { assert(1); }
inline void Isco() {
int cnt1 = 0;
for (int i = 1; i <= 20; ++i) ++cnt1;
}
int main() {
scanf("%s", s + 1);
scanf("%s", t + 1);
N1 = strlen(s + 1);
N2 = strlen(t + 1);
s[N1 + 1] = t[N2 + 1] = 'a';
memset(f, 0x3f, sizeof(f));
f[0][0][0] = 0;
for (int j = 0; j <= N1; ++j) {
for (int k = 0; k <= N2; ++k) {
for (int i = 0; i <= 200; ++i) {
int J = (s[j + 1] == '(') ? (j + 1) : j,
K = (t[k + 1] == '(') ? (k + 1) : k;
if (f[i + 1][J][K] > f[i][j][k] + 1) {
f[i + 1][J][K] = f[i][j][k] + 1;
pair<int, pair<int, int> > p;
p.first = i, p.second.first = j, p.second.second = k;
Debug();
int cnt = 10;
while (cnt > 5) --cnt;
xyyAKIOI[i + 1][J][K] = p, Orzxyy[i + 1][J][K] = 1;
}
J = (s[j + 1] == ')') ? (j + 1) : j;
K = (t[k + 1] == ')') ? (k + 1) : k;
if (i) {
if (f[i - 1][J][K] > f[i][j][k] + 1) {
f[i - 1][J][K] = f[i][j][k] + 1;
pair<int, pair<int, int> > p;
Debug();
int cnt = 10;
while (cnt > 5) --cnt;
p.first = i, p.second.first = j, p.second.second = k;
xyyAKIOI[i - 1][J][K] = p;
Orzxyy[i - 1][J][K] = -1;
while (true) break;
}
}
}
Isco();
for (int i = 0; i <= 200; ++i) {
int J = (s[j + 1] == '(') ? (j + 1) : j,
K = (t[k + 1] == '(') ? (k + 1) : k;
if (f[i + 1][J][K] > f[i][j][k] + 1) {
f[i + 1][J][K] = f[i][j][k] + 1;
pair<int, pair<int, int> > p;
Debug();
int cnt = 10;
while (cnt > 5) --cnt;
p.first = i, p.second.first = j, p.second.second = k;
xyyAKIOI[i + 1][J][K] = p;
Orzxyy[i + 1][J][K] = 1;
}
J = (s[j + 1] == ')') ? (j + 1) : j,
K = (t[k + 1] == ')') ? (k + 1) : k;
if (i) {
if (f[i - 1][J][K] > f[i][j][k] + 1) {
f[i - 1][J][K] = f[i][j][k] + 1;
pair<int, pair<int, int> > p;
Debug();
int cnt = 10;
while (cnt > 5) --cnt;
p.first = i, p.second.first = j, p.second.second = k;
while (true) break;
xyyAKIOI[i - 1][J][K] = p;
Orzxyy[i - 1][J][K] = -1;
while (true) break;
}
}
}
}
}
Work();
return 0;
}
| ### Prompt
Please provide a cpp coded solution to the problem described below:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
inline int read() {
char c = getchar();
int x = 0, f = 1;
for (; !isdigit(c); c = getchar())
if (c == '-') f = -1;
for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
return x * f;
}
const int MAXN = 210;
int f[MAXN][MAXN][MAXN], M[3], Orzxyy[MAXN][MAXN][MAXN], ans[MAXN];
int N1, N2;
char s[MAXN], t[MAXN];
pair<int, pair<int, int> > xyyAKIOI[MAXN][MAXN][MAXN];
inline void Work() {
int A = 0, B = N1, C = N2, Len = 0;
while (A + B + C != 0) {
ans[++Len] = Orzxyy[A][B][C];
pair<int, pair<int, int> > p = xyyAKIOI[A][B][C];
A = p.first, B = p.second.first, C = p.second.second;
}
for (int i = Len; i; --i) putchar((ans[i] == 1) ? '(' : ')');
puts("");
}
inline void Debug() { assert(1); }
inline void Isco() {
int cnt1 = 0;
for (int i = 1; i <= 20; ++i) ++cnt1;
}
int main() {
scanf("%s", s + 1);
scanf("%s", t + 1);
N1 = strlen(s + 1);
N2 = strlen(t + 1);
s[N1 + 1] = t[N2 + 1] = 'a';
memset(f, 0x3f, sizeof(f));
f[0][0][0] = 0;
for (int j = 0; j <= N1; ++j) {
for (int k = 0; k <= N2; ++k) {
for (int i = 0; i <= 200; ++i) {
int J = (s[j + 1] == '(') ? (j + 1) : j,
K = (t[k + 1] == '(') ? (k + 1) : k;
if (f[i + 1][J][K] > f[i][j][k] + 1) {
f[i + 1][J][K] = f[i][j][k] + 1;
pair<int, pair<int, int> > p;
p.first = i, p.second.first = j, p.second.second = k;
Debug();
int cnt = 10;
while (cnt > 5) --cnt;
xyyAKIOI[i + 1][J][K] = p, Orzxyy[i + 1][J][K] = 1;
}
J = (s[j + 1] == ')') ? (j + 1) : j;
K = (t[k + 1] == ')') ? (k + 1) : k;
if (i) {
if (f[i - 1][J][K] > f[i][j][k] + 1) {
f[i - 1][J][K] = f[i][j][k] + 1;
pair<int, pair<int, int> > p;
Debug();
int cnt = 10;
while (cnt > 5) --cnt;
p.first = i, p.second.first = j, p.second.second = k;
xyyAKIOI[i - 1][J][K] = p;
Orzxyy[i - 1][J][K] = -1;
while (true) break;
}
}
}
Isco();
for (int i = 0; i <= 200; ++i) {
int J = (s[j + 1] == '(') ? (j + 1) : j,
K = (t[k + 1] == '(') ? (k + 1) : k;
if (f[i + 1][J][K] > f[i][j][k] + 1) {
f[i + 1][J][K] = f[i][j][k] + 1;
pair<int, pair<int, int> > p;
Debug();
int cnt = 10;
while (cnt > 5) --cnt;
p.first = i, p.second.first = j, p.second.second = k;
xyyAKIOI[i + 1][J][K] = p;
Orzxyy[i + 1][J][K] = 1;
}
J = (s[j + 1] == ')') ? (j + 1) : j,
K = (t[k + 1] == ')') ? (k + 1) : k;
if (i) {
if (f[i - 1][J][K] > f[i][j][k] + 1) {
f[i - 1][J][K] = f[i][j][k] + 1;
pair<int, pair<int, int> > p;
Debug();
int cnt = 10;
while (cnt > 5) --cnt;
p.first = i, p.second.first = j, p.second.second = k;
while (true) break;
xyyAKIOI[i - 1][J][K] = p;
Orzxyy[i - 1][J][K] = -1;
while (true) break;
}
}
}
}
}
Work();
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
string a, b;
int dp[202][202][402];
int aux[3][202][202][402];
int main() {
int ia, ib, na, nb, ct, i, j, k, bal, ii, jj, rez;
cin >> a;
cin >> b;
na = a.size();
nb = b.size();
ct = 0;
for (i = 0; i <= na; i++)
for (j = 0; j <= nb; j++)
for (k = 0; k <= na + nb; k++) dp[i][j][k] = 1e6;
dp[0][0][0] = 0;
for (i = 0; i <= na; i++)
for (j = 0; j <= nb; j++)
for (k = 0; k <= na + nb; k++) {
if (dp[i][j][k] == 1e6) continue;
ii = i + (i < na && a[i] == '(');
jj = j + (j < nb && b[j] == '(');
if (dp[ii][jj][k + 1] > dp[i][j][k] + 1) {
dp[ii][jj][k + 1] = dp[i][j][k] + 1;
aux[0][ii][jj][k + 1] = i;
aux[1][ii][jj][k + 1] = j;
aux[2][ii][jj][k + 1] = 0;
}
ii = i + (i < na && a[i] == ')');
jj = j + (j < nb && b[j] == ')');
if (k > 0 && dp[ii][jj][k - 1] > dp[i][j][k] + 1) {
dp[ii][jj][k - 1] = dp[i][j][k] + 1;
aux[0][ii][jj][k - 1] = i;
aux[1][ii][jj][k - 1] = j;
aux[2][ii][jj][k - 1] = 1;
}
}
bal = 0;
rez = 1e6;
for (k = 0; k <= na + nb; k++)
if (rez > dp[na][nb][k] + k) {
rez = dp[na][nb][k] + k;
bal = k;
}
i = na;
j = nb;
k = bal;
string sir = string(bal, ')');
;
while (i > 0 || j > 0 || k > 0) {
if (aux[2][i][j][k] == 0) {
sir.push_back('(');
bal--;
} else {
sir.push_back(')');
bal++;
}
ii = aux[0][i][j][k];
jj = aux[1][i][j][k];
i = ii;
j = jj;
k = bal;
}
reverse(sir.begin(), sir.end());
cout << sir;
return 0;
}
| ### Prompt
Please formulate a CPP solution to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
string a, b;
int dp[202][202][402];
int aux[3][202][202][402];
int main() {
int ia, ib, na, nb, ct, i, j, k, bal, ii, jj, rez;
cin >> a;
cin >> b;
na = a.size();
nb = b.size();
ct = 0;
for (i = 0; i <= na; i++)
for (j = 0; j <= nb; j++)
for (k = 0; k <= na + nb; k++) dp[i][j][k] = 1e6;
dp[0][0][0] = 0;
for (i = 0; i <= na; i++)
for (j = 0; j <= nb; j++)
for (k = 0; k <= na + nb; k++) {
if (dp[i][j][k] == 1e6) continue;
ii = i + (i < na && a[i] == '(');
jj = j + (j < nb && b[j] == '(');
if (dp[ii][jj][k + 1] > dp[i][j][k] + 1) {
dp[ii][jj][k + 1] = dp[i][j][k] + 1;
aux[0][ii][jj][k + 1] = i;
aux[1][ii][jj][k + 1] = j;
aux[2][ii][jj][k + 1] = 0;
}
ii = i + (i < na && a[i] == ')');
jj = j + (j < nb && b[j] == ')');
if (k > 0 && dp[ii][jj][k - 1] > dp[i][j][k] + 1) {
dp[ii][jj][k - 1] = dp[i][j][k] + 1;
aux[0][ii][jj][k - 1] = i;
aux[1][ii][jj][k - 1] = j;
aux[2][ii][jj][k - 1] = 1;
}
}
bal = 0;
rez = 1e6;
for (k = 0; k <= na + nb; k++)
if (rez > dp[na][nb][k] + k) {
rez = dp[na][nb][k] + k;
bal = k;
}
i = na;
j = nb;
k = bal;
string sir = string(bal, ')');
;
while (i > 0 || j > 0 || k > 0) {
if (aux[2][i][j][k] == 0) {
sir.push_back('(');
bal--;
} else {
sir.push_back(')');
bal++;
}
ii = aux[0][i][j][k];
jj = aux[1][i][j][k];
i = ii;
j = jj;
k = bal;
}
reverse(sir.begin(), sir.end());
cout << sir;
return 0;
}
``` |
#include <bits/stdc++.h>
int faster_in() {
int r = 0, c;
for (c = getchar(); c <= 32; c = getchar())
;
if (c == '-') return -faster_in();
for (; c > 32; r = (r << 1) + (r << 3) + c - '0', c = getchar())
;
return r;
}
using namespace std;
const int INF = int(1e9 + 7);
const int tam = 210;
int dp[tam][tam][2 * tam];
int n, m;
string a, b;
int f(int i, int j, int diff) {
if (diff < 0 || diff > n + m) return (INF >> 1);
if (i == n && j == m) return diff;
int& res = dp[i][j][diff];
if (res != -1) return res;
res = INF;
res = min(res, f(i + (a[i] == '('), j + (b[j] == '('), diff + 1) + 1);
res = min(res, f(i + (a[i] == ')'), j + (b[j] == ')'), diff - 1) + 1);
return res;
}
string ans = "";
void build(int i, int j, int diff) {
if (i == n && j == m) {
while (diff--) {
ans += ')';
}
return;
}
int& res = dp[i][j][diff];
if (res == f(i + (a[i] == '('), j + (b[j] == '('), diff + 1) + 1) {
ans += '(';
build(i + (a[i] == '('), j + (b[j] == '('), diff + 1);
} else {
ans += ')';
build(i + (a[i] == ')'), j + (b[j] == ')'), diff - 1);
}
}
int main() {
std::ios::sync_with_stdio(false);
cin.tie(0);
memset(dp, (-1), sizeof(dp));
cin >> a >> b;
n = a.size();
m = b.size();
a += '$';
b += '$';
f(0, 0, 0);
build(0, 0, 0);
cout << ans << '\n';
return 0;
}
| ### Prompt
Generate a Cpp solution to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
int faster_in() {
int r = 0, c;
for (c = getchar(); c <= 32; c = getchar())
;
if (c == '-') return -faster_in();
for (; c > 32; r = (r << 1) + (r << 3) + c - '0', c = getchar())
;
return r;
}
using namespace std;
const int INF = int(1e9 + 7);
const int tam = 210;
int dp[tam][tam][2 * tam];
int n, m;
string a, b;
int f(int i, int j, int diff) {
if (diff < 0 || diff > n + m) return (INF >> 1);
if (i == n && j == m) return diff;
int& res = dp[i][j][diff];
if (res != -1) return res;
res = INF;
res = min(res, f(i + (a[i] == '('), j + (b[j] == '('), diff + 1) + 1);
res = min(res, f(i + (a[i] == ')'), j + (b[j] == ')'), diff - 1) + 1);
return res;
}
string ans = "";
void build(int i, int j, int diff) {
if (i == n && j == m) {
while (diff--) {
ans += ')';
}
return;
}
int& res = dp[i][j][diff];
if (res == f(i + (a[i] == '('), j + (b[j] == '('), diff + 1) + 1) {
ans += '(';
build(i + (a[i] == '('), j + (b[j] == '('), diff + 1);
} else {
ans += ')';
build(i + (a[i] == ')'), j + (b[j] == ')'), diff - 1);
}
}
int main() {
std::ios::sync_with_stdio(false);
cin.tie(0);
memset(dp, (-1), sizeof(dp));
cin >> a >> b;
n = a.size();
m = b.size();
a += '$';
b += '$';
f(0, 0, 0);
build(0, 0, 0);
cout << ans << '\n';
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int dcmp(double x, double y) {
return fabs(x - y) <= 1e-9 ? 0 : x < y ? -1 : 1;
}
const int MAX = 220;
char s[MAX], t[MAX];
int n, m;
int memo[MAX][MAX][1500];
int dp(int i, int j, int cnt) {
if (cnt > 1000 || cnt < 0) return (int)1e9 + 9;
if (i == n && j == m) {
return cnt;
}
int &ret = memo[i][j][cnt];
if (ret != -1) return ret;
int path1 = (int)1e9 + 9;
int path2 = (int)1e9 + 9;
int ni = i + (s[i] == ')');
int nj = j + (t[j] == ')');
path1 = 1 + dp(ni, nj, cnt - 1);
ni = i + (s[i] == '(');
nj = j + (t[j] == '(');
path2 = 1 + dp(ni, nj, cnt + 1);
return ret = min(path1, path2);
}
void get_path(int i, int j, int cnt) {
if (i == n && j == m) {
if (cnt > 0) {
for (int k = 0; k < cnt; k++) {
cout << ')';
}
}
return;
}
int path1 = (int)1e9 + 9;
int path2 = (int)1e9 + 9;
int &ret = memo[i][j][cnt];
int ni = i + (s[i] == ')');
int nj = j + (t[j] == ')');
path1 = 1 + dp(ni, nj, cnt - 1);
int ni1 = i + (s[i] == '(');
int nj1 = j + (t[j] == '(');
path2 = 1 + dp(ni1, nj1, cnt + 1);
if (path1 < path2) {
cout << ')';
get_path(ni, nj, cnt - 1);
return;
} else {
cout << '(';
get_path(ni1, nj1, cnt + 1);
return;
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
;
memset(memo, -1, sizeof(memo));
cin >> s >> t;
n = strlen(s);
m = strlen(t);
int ans = dp(0, 0, 0);
get_path(0, 0, 0);
}
| ### Prompt
Your task is to create a CPP solution to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int dcmp(double x, double y) {
return fabs(x - y) <= 1e-9 ? 0 : x < y ? -1 : 1;
}
const int MAX = 220;
char s[MAX], t[MAX];
int n, m;
int memo[MAX][MAX][1500];
int dp(int i, int j, int cnt) {
if (cnt > 1000 || cnt < 0) return (int)1e9 + 9;
if (i == n && j == m) {
return cnt;
}
int &ret = memo[i][j][cnt];
if (ret != -1) return ret;
int path1 = (int)1e9 + 9;
int path2 = (int)1e9 + 9;
int ni = i + (s[i] == ')');
int nj = j + (t[j] == ')');
path1 = 1 + dp(ni, nj, cnt - 1);
ni = i + (s[i] == '(');
nj = j + (t[j] == '(');
path2 = 1 + dp(ni, nj, cnt + 1);
return ret = min(path1, path2);
}
void get_path(int i, int j, int cnt) {
if (i == n && j == m) {
if (cnt > 0) {
for (int k = 0; k < cnt; k++) {
cout << ')';
}
}
return;
}
int path1 = (int)1e9 + 9;
int path2 = (int)1e9 + 9;
int &ret = memo[i][j][cnt];
int ni = i + (s[i] == ')');
int nj = j + (t[j] == ')');
path1 = 1 + dp(ni, nj, cnt - 1);
int ni1 = i + (s[i] == '(');
int nj1 = j + (t[j] == '(');
path2 = 1 + dp(ni1, nj1, cnt + 1);
if (path1 < path2) {
cout << ')';
get_path(ni, nj, cnt - 1);
return;
} else {
cout << '(';
get_path(ni1, nj1, cnt + 1);
return;
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
;
memset(memo, -1, sizeof(memo));
cin >> s >> t;
n = strlen(s);
m = strlen(t);
int ans = dp(0, 0, 0);
get_path(0, 0, 0);
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 205;
struct three {
int i, j, k;
three() {}
three(int _i, int _j, int _k) {
i = _i;
j = _j;
k = _k;
}
};
int dp[2 * N][N][N];
three trace[2 * N][N][N];
string s, t;
int n, m;
bool umax(int &a, int b) {
if (a < b) {
a = b;
return true;
}
return false;
}
void gogo() {
cin >> s >> t;
n = ((int)(s).size()), m = ((int)(t).size());
s = ' ' + s, t = ' ' + t;
for (int i = 0; i <= 400; ++i)
for (int j = 0; j <= n; ++j)
for (int k = 0; k <= 200; ++k) {
dp[i][j][k] = -1;
}
dp[0][0][0] = 0;
for (int i = 0; i <= 400; ++i) {
for (int j = 0; j <= n; ++j) {
for (int k = 0; k <= 200; ++k) {
int cur = dp[i][j][k];
if (cur == -1) continue;
if (s[j + 1] == '(') {
if (umax(dp[i + 1][j + 1][k + 1],
dp[i][j][k] + (t[cur + 1] == '('))) {
trace[i + 1][j + 1][k + 1] = three(i, j, k);
}
} else {
if (umax(dp[i + 1][j][k + 1], dp[i][j][k] + (t[cur + 1] == '('))) {
trace[i + 1][j][k + 1] = three(i, j, k);
}
}
if (k == 0) continue;
if (s[j + 1] == ')') {
if (umax(dp[i + 1][j + 1][k - 1],
dp[i][j][k] + (t[cur + 1] == ')'))) {
trace[i + 1][j + 1][k - 1] = three(i, j, k);
}
} else {
if (umax(dp[i + 1][j][k - 1], dp[i][j][k] + (t[cur + 1] == ')'))) {
trace[i + 1][j][k - 1] = three(i, j, k);
}
}
}
}
}
int res = 1e9;
for (int i = 0; i <= 2 * (n + m); ++i) {
if (dp[i][n][0] >= m) {
res = i;
break;
}
}
string ans = "";
int i = res, j = n, k = 0;
while (1) {
three tmp = trace[i][j][k];
if (k - tmp.k == 1)
ans += '(';
else
ans += ')';
i = tmp.i;
j = tmp.j;
k = tmp.k;
if (i == 0) break;
}
reverse((ans).begin(), (ans).end());
cout << ans;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
if (fopen("sol"
".inp",
"r")) {
freopen(
"sol"
".inp",
"r", stdin);
freopen(
"sol"
".out",
"w", stdout);
}
gogo();
return 0;
}
| ### Prompt
In CPP, your task is to solve the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 205;
struct three {
int i, j, k;
three() {}
three(int _i, int _j, int _k) {
i = _i;
j = _j;
k = _k;
}
};
int dp[2 * N][N][N];
three trace[2 * N][N][N];
string s, t;
int n, m;
bool umax(int &a, int b) {
if (a < b) {
a = b;
return true;
}
return false;
}
void gogo() {
cin >> s >> t;
n = ((int)(s).size()), m = ((int)(t).size());
s = ' ' + s, t = ' ' + t;
for (int i = 0; i <= 400; ++i)
for (int j = 0; j <= n; ++j)
for (int k = 0; k <= 200; ++k) {
dp[i][j][k] = -1;
}
dp[0][0][0] = 0;
for (int i = 0; i <= 400; ++i) {
for (int j = 0; j <= n; ++j) {
for (int k = 0; k <= 200; ++k) {
int cur = dp[i][j][k];
if (cur == -1) continue;
if (s[j + 1] == '(') {
if (umax(dp[i + 1][j + 1][k + 1],
dp[i][j][k] + (t[cur + 1] == '('))) {
trace[i + 1][j + 1][k + 1] = three(i, j, k);
}
} else {
if (umax(dp[i + 1][j][k + 1], dp[i][j][k] + (t[cur + 1] == '('))) {
trace[i + 1][j][k + 1] = three(i, j, k);
}
}
if (k == 0) continue;
if (s[j + 1] == ')') {
if (umax(dp[i + 1][j + 1][k - 1],
dp[i][j][k] + (t[cur + 1] == ')'))) {
trace[i + 1][j + 1][k - 1] = three(i, j, k);
}
} else {
if (umax(dp[i + 1][j][k - 1], dp[i][j][k] + (t[cur + 1] == ')'))) {
trace[i + 1][j][k - 1] = three(i, j, k);
}
}
}
}
}
int res = 1e9;
for (int i = 0; i <= 2 * (n + m); ++i) {
if (dp[i][n][0] >= m) {
res = i;
break;
}
}
string ans = "";
int i = res, j = n, k = 0;
while (1) {
three tmp = trace[i][j][k];
if (k - tmp.k == 1)
ans += '(';
else
ans += ')';
i = tmp.i;
j = tmp.j;
k = tmp.k;
if (i == 0) break;
}
reverse((ans).begin(), (ans).end());
cout << ans;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
if (fopen("sol"
".inp",
"r")) {
freopen(
"sol"
".inp",
"r", stdin);
freopen(
"sol"
".out",
"w", stdout);
}
gogo();
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3fll;
void sc(int& x) { scanf("%d", &x); }
void sc(long long& x) { scanf("%lld", &x); }
template <class T, class... Ts>
void sc(T& t, Ts&... ts) {
sc(t), sc(ts...);
}
const int MAX = 210;
int dp[MAX][MAX][MAX];
char s1[MAX], s2[MAX];
int n1, n2;
pair<pair<int, int>, pair<int, int> > get(int i1, int i2) {
bool a = (i1 != n1), b = (i2 != n2);
pair<pair<int, int>, pair<int, int> > ret;
if (s1[i1] == '(')
ret.first = {a, 0};
else
ret.first = {0, a};
if (s2[i2] == '(')
ret.second = {b, 0};
else
ret.second = {0, b};
return ret;
}
int solve(int i1, int i2, int c, bool p = 0) {
if (c < 0) return INF;
if (c >= MAX) return INF;
if (i1 == n1 and i2 == n2) {
if (p)
for (int i = 0; i < c; i++) printf(")");
return c;
}
auto& x = dp[i1][i2][c];
if (x + 1 and !p) return x;
auto t = get(i1, i2);
int a = 1 + solve(i1 + t.first.first, i2 + t.second.first, c + 1);
int b = 1 + solve(i1 + t.first.second, i2 + t.second.second, c - 1);
if (p) {
if (a < b)
printf("("), solve(i1 + t.first.first, i2 + t.second.first, c + 1, 1);
else
printf(")"), solve(i1 + t.first.second, i2 + t.second.second, c - 1, 1);
}
return x = min(a, b);
}
int main() {
scanf(" %s %s", s1, s2);
n1 = strlen(s1), n2 = strlen(s2);
memset(dp, -1, sizeof(dp));
solve(0, 0, 0, 1);
printf("\n");
exit(0);
}
| ### Prompt
Develop a solution in CPP to the problem described below:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3fll;
void sc(int& x) { scanf("%d", &x); }
void sc(long long& x) { scanf("%lld", &x); }
template <class T, class... Ts>
void sc(T& t, Ts&... ts) {
sc(t), sc(ts...);
}
const int MAX = 210;
int dp[MAX][MAX][MAX];
char s1[MAX], s2[MAX];
int n1, n2;
pair<pair<int, int>, pair<int, int> > get(int i1, int i2) {
bool a = (i1 != n1), b = (i2 != n2);
pair<pair<int, int>, pair<int, int> > ret;
if (s1[i1] == '(')
ret.first = {a, 0};
else
ret.first = {0, a};
if (s2[i2] == '(')
ret.second = {b, 0};
else
ret.second = {0, b};
return ret;
}
int solve(int i1, int i2, int c, bool p = 0) {
if (c < 0) return INF;
if (c >= MAX) return INF;
if (i1 == n1 and i2 == n2) {
if (p)
for (int i = 0; i < c; i++) printf(")");
return c;
}
auto& x = dp[i1][i2][c];
if (x + 1 and !p) return x;
auto t = get(i1, i2);
int a = 1 + solve(i1 + t.first.first, i2 + t.second.first, c + 1);
int b = 1 + solve(i1 + t.first.second, i2 + t.second.second, c - 1);
if (p) {
if (a < b)
printf("("), solve(i1 + t.first.first, i2 + t.second.first, c + 1, 1);
else
printf(")"), solve(i1 + t.first.second, i2 + t.second.second, c - 1, 1);
}
return x = min(a, b);
}
int main() {
scanf(" %s %s", s1, s2);
n1 = strlen(s1), n2 = strlen(s2);
memset(dp, -1, sizeof(dp));
solve(0, 0, 0, 1);
printf("\n");
exit(0);
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9;
string li, ne;
int dp[201][201][201];
int rp[201][201][201];
int f(int i, int j, int k) {
if (k < 0 || k > 200) return inf;
if (i == li.size() - 1 && j == ne.size() - 1) return k;
if (dp[i][j][k] > -1) return dp[i][j][k];
rp[i][j][k] = 1;
int mx = f(i + (li[i] == '('), j + (ne[j] == '('), k + 1);
if (mx > f(i + (li[i] == ')'), j + (ne[j] == ')'), k - 1)) {
mx = f(i + (li[i] == ')'), j + (ne[j] == ')'), k - 1);
rp[i][j][k] = 0;
}
return dp[i][j][k] = mx + 1;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(NULL);
cin >> li >> ne;
li += '$';
ne += '$';
memset(dp, -1, sizeof dp);
int rx = f(0, 0, 0);
int k = 0;
for (int i = 0, j = 0; rx; --rx) {
if (i == li.size() - 1 && j == ne.size() - 1) break;
if (rp[i][j][k]) {
k += 1;
i += (li[i] == '(');
j += (ne[j] == '(');
cout << '(';
} else {
k -= 1;
i += (li[i] == ')');
j += (ne[j] == ')');
cout << ')';
}
}
while (k-- > 0) cout << ')';
return 0;
}
| ### Prompt
Create a solution in Cpp for the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9;
string li, ne;
int dp[201][201][201];
int rp[201][201][201];
int f(int i, int j, int k) {
if (k < 0 || k > 200) return inf;
if (i == li.size() - 1 && j == ne.size() - 1) return k;
if (dp[i][j][k] > -1) return dp[i][j][k];
rp[i][j][k] = 1;
int mx = f(i + (li[i] == '('), j + (ne[j] == '('), k + 1);
if (mx > f(i + (li[i] == ')'), j + (ne[j] == ')'), k - 1)) {
mx = f(i + (li[i] == ')'), j + (ne[j] == ')'), k - 1);
rp[i][j][k] = 0;
}
return dp[i][j][k] = mx + 1;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(NULL);
cin >> li >> ne;
li += '$';
ne += '$';
memset(dp, -1, sizeof dp);
int rx = f(0, 0, 0);
int k = 0;
for (int i = 0, j = 0; rx; --rx) {
if (i == li.size() - 1 && j == ne.size() - 1) break;
if (rp[i][j][k]) {
k += 1;
i += (li[i] == '(');
j += (ne[j] == '(');
cout << '(';
} else {
k -= 1;
i += (li[i] == ')');
j += (ne[j] == ')');
cout << ')';
}
}
while (k-- > 0) cout << ')';
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
char s[201], t[201];
int m, n;
int f[203][203][403];
int tr[203][203][403][3];
void init() {
for (int i = 0; i <= 202; ++i)
for (int j = 0; j <= 202; ++j)
for (int k = 0; k <= 401; ++k) {
f[i][j][k] = 999999;
}
}
void atr(const int &a, const int &b, const int &c, const int &d, const int &e,
const int &g) {
tr[d][e][g][0] = a;
tr[d][e][g][1] = b;
tr[d][e][g][2] = c;
}
int dp(int i, int j, int k);
void mini(int &res, const int &i, const int &j, const int &k, const int &i0,
const int &j0, const int &k0) {
if (dp(i, j, k) + 1 < res) {
res = dp(i, j, k) + 1;
atr(i, j, k, i0, j0, k0);
}
}
int MM = 200;
int dp(int i, int j, int k) {
if (k < 0 || k > MM) return 999999 - 1;
if (i > m && j > n) {
if (k == 0) return 0;
}
int &r = f[i][j][k];
if (r < 999999) return r;
r = 999999 - 1;
int di, dj;
di = (i <= m && s[i] == '(');
dj = (j <= n && t[j] == '(');
mini(r, i + di, j + dj, k + 1, i, j, k);
dj = (j <= n && t[j] == ')');
di = (i <= m && s[i] == ')');
mini(r, i + di, j + dj, k - 1, i, j, k);
return r;
}
int main() {
scanf(" %s %s", s + 1, t + 1);
m = strlen(s + 1);
n = strlen(t + 1);
MM = min(MM, n + m);
init();
int rr;
cerr << (rr = dp(1, 1, 0));
string s;
int x = 1, y = 1, z = 0, u, v, w;
for (; s.size() < rr;) {
u = tr[x][y][z][0];
v = tr[x][y][z][1];
w = tr[x][y][z][2];
if (z < w)
s.push_back('(');
else
s.push_back(')');
x = u;
y = v;
z = w;
if (x > m && y > n) {
if (z > 0) {
string tc(z, ')');
s += tc;
} else {
string to(-z, '(');
s = to + s;
}
break;
}
}
cout << s;
return 0;
}
| ### Prompt
In cpp, your task is to solve the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
char s[201], t[201];
int m, n;
int f[203][203][403];
int tr[203][203][403][3];
void init() {
for (int i = 0; i <= 202; ++i)
for (int j = 0; j <= 202; ++j)
for (int k = 0; k <= 401; ++k) {
f[i][j][k] = 999999;
}
}
void atr(const int &a, const int &b, const int &c, const int &d, const int &e,
const int &g) {
tr[d][e][g][0] = a;
tr[d][e][g][1] = b;
tr[d][e][g][2] = c;
}
int dp(int i, int j, int k);
void mini(int &res, const int &i, const int &j, const int &k, const int &i0,
const int &j0, const int &k0) {
if (dp(i, j, k) + 1 < res) {
res = dp(i, j, k) + 1;
atr(i, j, k, i0, j0, k0);
}
}
int MM = 200;
int dp(int i, int j, int k) {
if (k < 0 || k > MM) return 999999 - 1;
if (i > m && j > n) {
if (k == 0) return 0;
}
int &r = f[i][j][k];
if (r < 999999) return r;
r = 999999 - 1;
int di, dj;
di = (i <= m && s[i] == '(');
dj = (j <= n && t[j] == '(');
mini(r, i + di, j + dj, k + 1, i, j, k);
dj = (j <= n && t[j] == ')');
di = (i <= m && s[i] == ')');
mini(r, i + di, j + dj, k - 1, i, j, k);
return r;
}
int main() {
scanf(" %s %s", s + 1, t + 1);
m = strlen(s + 1);
n = strlen(t + 1);
MM = min(MM, n + m);
init();
int rr;
cerr << (rr = dp(1, 1, 0));
string s;
int x = 1, y = 1, z = 0, u, v, w;
for (; s.size() < rr;) {
u = tr[x][y][z][0];
v = tr[x][y][z][1];
w = tr[x][y][z][2];
if (z < w)
s.push_back('(');
else
s.push_back(')');
x = u;
y = v;
z = w;
if (x > m && y > n) {
if (z > 0) {
string tc(z, ')');
s += tc;
} else {
string to(-z, '(');
s = to + s;
}
break;
}
}
cout << s;
return 0;
}
``` |
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
string a, b;
long long memo[205][205][205];
long long calc(char c) { return (c == '(') ? 1 : -1; }
long long solve(long long i, long long j, long long bal) {
if (bal < 0 || bal > 200) return 1e9;
if (i == a.size() && j == b.size()) return bal;
long long &ans = memo[i][j][bal];
if (~ans) return ans;
if (i < a.size() && j < b.size() && b[j] == a[i])
ans = solve(i + 1, j + 1, bal + calc(a[i])) + 1;
else if (i < a.size() && j < b.size())
ans = solve(i + 1, j, bal + calc(a[i])) + 1,
ans = min(ans, solve(i, j + 1, bal + calc(b[j])) + 1);
else if (i < a.size())
ans = solve(i + 1, j, bal + calc(a[i])) + 1;
else if (j < b.size())
ans = solve(i, j + 1, bal + calc(b[j])) + 1;
ans = min(ans, solve(i, j, bal + 1) + 1);
ans = min(ans, solve(i, j, bal - 1) + 1);
return ans;
}
void print(long long i, long long j, long long bal) {
if (i == a.size() && j == b.size()) {
for (int k = 0; k < bal; k++) cout << ')';
exit(0);
}
if (memo[i][j][bal] == solve(i, j, bal + 1) + 1) {
cout << '(';
print(i, j, bal + 1);
} else if (memo[i][j][bal] == solve(i, j, bal - 1) + 1) {
cout << ')';
print(i, j, bal - 1);
} else if (i < a.size() && j < b.size() && b[j] == a[i]) {
cout << a[i];
print(i + 1, j + 1, bal + calc(a[i]));
} else if (i < a.size() && j < b.size()) {
if (memo[i][j][bal] == solve(i + 1, j, bal + calc(a[i])) + 1) {
cout << a[i];
print(i + 1, j, bal + calc(a[i]));
} else {
cout << b[j];
print(i, j + 1, bal + calc(b[j]));
}
} else if (i < a.size()) {
cout << a[i];
print(i + 1, j, bal + calc(a[i]));
} else {
cout << b[j];
print(i, j + 1, bal + calc(b[j]));
}
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
memset(memo, -1, sizeof memo);
cin >> a >> b;
solve(0, 0, 0);
print(0, 0, 0);
return 0;
}
| ### Prompt
Create a solution in Cpp for the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
string a, b;
long long memo[205][205][205];
long long calc(char c) { return (c == '(') ? 1 : -1; }
long long solve(long long i, long long j, long long bal) {
if (bal < 0 || bal > 200) return 1e9;
if (i == a.size() && j == b.size()) return bal;
long long &ans = memo[i][j][bal];
if (~ans) return ans;
if (i < a.size() && j < b.size() && b[j] == a[i])
ans = solve(i + 1, j + 1, bal + calc(a[i])) + 1;
else if (i < a.size() && j < b.size())
ans = solve(i + 1, j, bal + calc(a[i])) + 1,
ans = min(ans, solve(i, j + 1, bal + calc(b[j])) + 1);
else if (i < a.size())
ans = solve(i + 1, j, bal + calc(a[i])) + 1;
else if (j < b.size())
ans = solve(i, j + 1, bal + calc(b[j])) + 1;
ans = min(ans, solve(i, j, bal + 1) + 1);
ans = min(ans, solve(i, j, bal - 1) + 1);
return ans;
}
void print(long long i, long long j, long long bal) {
if (i == a.size() && j == b.size()) {
for (int k = 0; k < bal; k++) cout << ')';
exit(0);
}
if (memo[i][j][bal] == solve(i, j, bal + 1) + 1) {
cout << '(';
print(i, j, bal + 1);
} else if (memo[i][j][bal] == solve(i, j, bal - 1) + 1) {
cout << ')';
print(i, j, bal - 1);
} else if (i < a.size() && j < b.size() && b[j] == a[i]) {
cout << a[i];
print(i + 1, j + 1, bal + calc(a[i]));
} else if (i < a.size() && j < b.size()) {
if (memo[i][j][bal] == solve(i + 1, j, bal + calc(a[i])) + 1) {
cout << a[i];
print(i + 1, j, bal + calc(a[i]));
} else {
cout << b[j];
print(i, j + 1, bal + calc(b[j]));
}
} else if (i < a.size()) {
cout << a[i];
print(i + 1, j, bal + calc(a[i]));
} else {
cout << b[j];
print(i, j + 1, bal + calc(b[j]));
}
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
memset(memo, -1, sizeof memo);
cin >> a >> b;
solve(0, 0, 0);
print(0, 0, 0);
return 0;
}
``` |
#include <bits/stdc++.h>
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
string s1, s2;
bool visited[201][201][201 + 20];
char moves[201][201][201 + 20];
tuple<long long int, long long int, long long int> par[201][201][201 + 20];
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cin >> s1 >> s2;
queue<tuple<long long int, long long int, long long int> > fringe;
memset(moves, -1, sizeof moves);
fringe.push({0, 0, 0});
visited[0][0][0] = true;
while (1) {
tuple<long long int, long long int, long long int> u = fringe.front();
long long int i, j, b;
tie(i, j, b) = u;
fringe.pop();
if (u == tuple<long long int, long long int, long long int>{s1.size(),
s2.size(), 0}) {
string ans = "";
while (moves[get<0>(u)][get<1>(u)][get<2>(u)] != -1) {
ans += moves[get<0>(u)][get<1>(u)][get<2>(u)];
u = par[get<0>(u)][get<1>(u)][get<2>(u)];
}
reverse(ans.begin(), ans.end());
cout << ans << endl;
return 0;
}
for (auto c : "()") {
tuple<long long int, long long int, long long int> v = {
((i < s1.size() && s1[i] == c) ? i + 1 : i),
((j < s2.size() && s2[j] == c) ? j + 1 : j),
(c == '(' ? b + 1 : b - 1)};
if (get<2>(v) < 0 || get<2>(v) > 210) continue;
if (visited[get<0>(v)][get<1>(v)][get<2>(v)]) continue;
fringe.push(v);
visited[get<0>(v)][get<1>(v)][get<2>(v)] = 1;
moves[get<0>(v)][get<1>(v)][get<2>(v)] = c;
par[get<0>(v)][get<1>(v)][get<2>(v)] = u;
}
}
}
| ### Prompt
Develop a solution in CPP to the problem described below:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
string s1, s2;
bool visited[201][201][201 + 20];
char moves[201][201][201 + 20];
tuple<long long int, long long int, long long int> par[201][201][201 + 20];
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cin >> s1 >> s2;
queue<tuple<long long int, long long int, long long int> > fringe;
memset(moves, -1, sizeof moves);
fringe.push({0, 0, 0});
visited[0][0][0] = true;
while (1) {
tuple<long long int, long long int, long long int> u = fringe.front();
long long int i, j, b;
tie(i, j, b) = u;
fringe.pop();
if (u == tuple<long long int, long long int, long long int>{s1.size(),
s2.size(), 0}) {
string ans = "";
while (moves[get<0>(u)][get<1>(u)][get<2>(u)] != -1) {
ans += moves[get<0>(u)][get<1>(u)][get<2>(u)];
u = par[get<0>(u)][get<1>(u)][get<2>(u)];
}
reverse(ans.begin(), ans.end());
cout << ans << endl;
return 0;
}
for (auto c : "()") {
tuple<long long int, long long int, long long int> v = {
((i < s1.size() && s1[i] == c) ? i + 1 : i),
((j < s2.size() && s2[j] == c) ? j + 1 : j),
(c == '(' ? b + 1 : b - 1)};
if (get<2>(v) < 0 || get<2>(v) > 210) continue;
if (visited[get<0>(v)][get<1>(v)][get<2>(v)]) continue;
fringe.push(v);
visited[get<0>(v)][get<1>(v)][get<2>(v)] = 1;
moves[get<0>(v)][get<1>(v)][get<2>(v)] = c;
par[get<0>(v)][get<1>(v)][get<2>(v)] = u;
}
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
void testCase() {}
int dp[202][202][404];
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
string str1, str2;
cin >> str1 >> str2;
int n = str1.size();
int m = str2.size();
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= m; j++) {
for (int k = 400; k >= 0; k--) {
if (i == 0 && j == 0)
dp[i][j][k] = k;
else if (i == 0) {
if (str2[j - 1] == '(') {
if (k == 0)
dp[i][j][k] = 1e9;
else
dp[i][j][k] = dp[i][j - 1][k - 1] + 1;
} else {
if (k == 400)
dp[i][j][k] = 1e9;
else
dp[i][j][k] = 1 + dp[i][j - 1][k + 1];
}
} else if (j == 0) {
if (str1[i - 1] == '(') {
if (k == 0)
dp[i][j][k] = 1e9;
else
dp[i][j][k] = dp[i - 1][j][k - 1] + 1;
} else {
if (k == 400)
dp[i][j][k] = 1e9;
else
dp[i][j][k] = 1 + dp[i - 1][j][k + 1];
}
} else if (str1[i - 1] == str2[j - 1]) {
if (str1[i - 1] == '(') {
if (k == 0)
dp[i][j][k] = 1e9;
else
dp[i][j][k] = 1 + dp[i - 1][j - 1][k - 1];
} else {
if (k == 400)
dp[i][j][k] = 1e9;
else
dp[i][j][k] = 1 + dp[i - 1][j - 1][k + 1];
}
} else {
dp[i][j][k] = 1e9;
if (str1[i - 1] == '(') {
if (k > 0) dp[i][j][k] = min(dp[i][j][k], 1 + dp[i - 1][j][k - 1]);
} else {
if (k <= 400)
dp[i][j][k] = min(dp[i][j][k], 1 + dp[i - 1][j][k + 1]);
}
if (str2[j - 1] == '(') {
if (k > 0) dp[i][j][k] = min(dp[i][j][k], 1 + dp[i][j - 1][k - 1]);
} else {
if (k <= 400)
dp[i][j][k] = min(dp[i][j][k], 1 + dp[i][j - 1][k + 1]);
}
}
}
}
}
int lyhin = 1e9;
int parask = -1;
for (int i = 0; i <= 400; i++) {
if (i + dp[n][m][i] < lyhin) {
lyhin = i + dp[n][m][i];
parask = i;
}
}
string ans = "";
for (int i = 0; i < parask; i++) ans += ")";
int I = n, J = m, K = parask;
while (I > 0 && J > 0) {
if (str1[I - 1] == str2[J - 1]) {
if (str1[I - 1] == '(') {
ans += "(";
K--;
} else {
ans += ")";
K++;
}
I--;
J--;
} else {
if (str1[I - 1] == '(') {
if (K != 0 && dp[I - 1][J][K - 1] < dp[I][J - 1][K + 1]) {
ans += "(";
K--;
I--;
} else {
ans += ")";
K++;
J--;
}
} else {
if (K != 0 && dp[I][J - 1][K - 1] < dp[I - 1][J][K + 1]) {
ans += "(";
K--;
J--;
} else {
ans += ")";
K++;
I--;
}
}
}
}
if (I > 0) {
for (int i = I - 1; i >= 0; i--) {
ans += str1[i];
if (str1[i] == '(')
K--;
else
K++;
}
} else if (J > 0) {
for (int i = J - 1; i >= 0; i--) {
ans += str2[i];
if (str2[i] == '(')
K--;
else
K++;
}
}
while (K > 0) {
ans += "(";
K--;
}
reverse(ans.begin(), ans.end());
cout << ans << "\n";
}
| ### Prompt
Develop a solution in Cpp to the problem described below:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void testCase() {}
int dp[202][202][404];
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
string str1, str2;
cin >> str1 >> str2;
int n = str1.size();
int m = str2.size();
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= m; j++) {
for (int k = 400; k >= 0; k--) {
if (i == 0 && j == 0)
dp[i][j][k] = k;
else if (i == 0) {
if (str2[j - 1] == '(') {
if (k == 0)
dp[i][j][k] = 1e9;
else
dp[i][j][k] = dp[i][j - 1][k - 1] + 1;
} else {
if (k == 400)
dp[i][j][k] = 1e9;
else
dp[i][j][k] = 1 + dp[i][j - 1][k + 1];
}
} else if (j == 0) {
if (str1[i - 1] == '(') {
if (k == 0)
dp[i][j][k] = 1e9;
else
dp[i][j][k] = dp[i - 1][j][k - 1] + 1;
} else {
if (k == 400)
dp[i][j][k] = 1e9;
else
dp[i][j][k] = 1 + dp[i - 1][j][k + 1];
}
} else if (str1[i - 1] == str2[j - 1]) {
if (str1[i - 1] == '(') {
if (k == 0)
dp[i][j][k] = 1e9;
else
dp[i][j][k] = 1 + dp[i - 1][j - 1][k - 1];
} else {
if (k == 400)
dp[i][j][k] = 1e9;
else
dp[i][j][k] = 1 + dp[i - 1][j - 1][k + 1];
}
} else {
dp[i][j][k] = 1e9;
if (str1[i - 1] == '(') {
if (k > 0) dp[i][j][k] = min(dp[i][j][k], 1 + dp[i - 1][j][k - 1]);
} else {
if (k <= 400)
dp[i][j][k] = min(dp[i][j][k], 1 + dp[i - 1][j][k + 1]);
}
if (str2[j - 1] == '(') {
if (k > 0) dp[i][j][k] = min(dp[i][j][k], 1 + dp[i][j - 1][k - 1]);
} else {
if (k <= 400)
dp[i][j][k] = min(dp[i][j][k], 1 + dp[i][j - 1][k + 1]);
}
}
}
}
}
int lyhin = 1e9;
int parask = -1;
for (int i = 0; i <= 400; i++) {
if (i + dp[n][m][i] < lyhin) {
lyhin = i + dp[n][m][i];
parask = i;
}
}
string ans = "";
for (int i = 0; i < parask; i++) ans += ")";
int I = n, J = m, K = parask;
while (I > 0 && J > 0) {
if (str1[I - 1] == str2[J - 1]) {
if (str1[I - 1] == '(') {
ans += "(";
K--;
} else {
ans += ")";
K++;
}
I--;
J--;
} else {
if (str1[I - 1] == '(') {
if (K != 0 && dp[I - 1][J][K - 1] < dp[I][J - 1][K + 1]) {
ans += "(";
K--;
I--;
} else {
ans += ")";
K++;
J--;
}
} else {
if (K != 0 && dp[I][J - 1][K - 1] < dp[I - 1][J][K + 1]) {
ans += "(";
K--;
J--;
} else {
ans += ")";
K++;
I--;
}
}
}
}
if (I > 0) {
for (int i = I - 1; i >= 0; i--) {
ans += str1[i];
if (str1[i] == '(')
K--;
else
K++;
}
} else if (J > 0) {
for (int i = J - 1; i >= 0; i--) {
ans += str2[i];
if (str2[i] == '(')
K--;
else
K++;
}
}
while (K > 0) {
ans += "(";
K--;
}
reverse(ans.begin(), ans.end());
cout << ans << "\n";
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long inf = 1e9;
struct node {
int i, j, bal;
char scoba;
node() {}
node(long long _i, long long _j, long long _k, char _scoba)
: i(_i), j(_j), bal(_k), scoba(_scoba) {}
};
node p[210][210][410];
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
string s, t;
cin >> s >> t;
long long n = s.size(), m = t.size();
vector<vector<vector<long long> > > dp(
n + 1, vector<vector<long long> >(m + 1, vector<long long>(410, inf)));
dp[0][0][0] = 0;
for (long long i = 0; i <= n; ++i) {
for (long long j = 0; j <= m; ++j) {
for (long long k = 0; k <= 410; ++k) {
if (dp[i][j][k] == inf) continue;
long long nxti = i + (i < n && s[i] == '('),
nxtj = j + (j < m && t[j] == '(');
if (k + 1 < 410 && dp[nxti][nxtj][k + 1] > dp[i][j][k] + 1) {
dp[nxti][nxtj][k + 1] = dp[i][j][k] + 1;
p[nxti][nxtj][k + 1] = node(i, j, k, '(');
}
nxti = i + (i < n && s[i] == ')'), nxtj = j + (j < m && t[j] == ')');
if (k > 0 && dp[nxti][nxtj][k - 1] > dp[i][j][k] + 1) {
dp[nxti][nxtj][k - 1] = dp[i][j][k] + 1;
p[nxti][nxtj][k - 1] = node(i, j, k, ')');
}
}
}
}
long long curi = n, curj = m, cur_balance = 0;
for (long long i = 0; i < 410; ++i) {
if (dp[n][m][cur_balance] + cur_balance > dp[n][m][i] + i &&
dp[n][m][i] != inf) {
cur_balance = i;
}
}
string ans = "";
long long temp = cur_balance;
while (curi > 0 || curj > 0 || cur_balance) {
node a = p[curi][curj][cur_balance];
long long ni = a.i;
long long nj = a.j;
long long nbal = a.bal;
ans += a.scoba;
curi = a.i;
curj = a.j;
cur_balance = a.bal;
}
while (temp--) ans = ')' + ans;
reverse(ans.begin(), ans.end());
cout << ans;
return 0;
}
| ### Prompt
In Cpp, your task is to solve the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long inf = 1e9;
struct node {
int i, j, bal;
char scoba;
node() {}
node(long long _i, long long _j, long long _k, char _scoba)
: i(_i), j(_j), bal(_k), scoba(_scoba) {}
};
node p[210][210][410];
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
string s, t;
cin >> s >> t;
long long n = s.size(), m = t.size();
vector<vector<vector<long long> > > dp(
n + 1, vector<vector<long long> >(m + 1, vector<long long>(410, inf)));
dp[0][0][0] = 0;
for (long long i = 0; i <= n; ++i) {
for (long long j = 0; j <= m; ++j) {
for (long long k = 0; k <= 410; ++k) {
if (dp[i][j][k] == inf) continue;
long long nxti = i + (i < n && s[i] == '('),
nxtj = j + (j < m && t[j] == '(');
if (k + 1 < 410 && dp[nxti][nxtj][k + 1] > dp[i][j][k] + 1) {
dp[nxti][nxtj][k + 1] = dp[i][j][k] + 1;
p[nxti][nxtj][k + 1] = node(i, j, k, '(');
}
nxti = i + (i < n && s[i] == ')'), nxtj = j + (j < m && t[j] == ')');
if (k > 0 && dp[nxti][nxtj][k - 1] > dp[i][j][k] + 1) {
dp[nxti][nxtj][k - 1] = dp[i][j][k] + 1;
p[nxti][nxtj][k - 1] = node(i, j, k, ')');
}
}
}
}
long long curi = n, curj = m, cur_balance = 0;
for (long long i = 0; i < 410; ++i) {
if (dp[n][m][cur_balance] + cur_balance > dp[n][m][i] + i &&
dp[n][m][i] != inf) {
cur_balance = i;
}
}
string ans = "";
long long temp = cur_balance;
while (curi > 0 || curj > 0 || cur_balance) {
node a = p[curi][curj][cur_balance];
long long ni = a.i;
long long nj = a.j;
long long nbal = a.bal;
ans += a.scoba;
curi = a.i;
curj = a.j;
cur_balance = a.bal;
}
while (temp--) ans = ')' + ans;
reverse(ans.begin(), ans.end());
cout << ans;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 205;
int dp[N][N][2 * N];
int n, m;
string s, t;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cin >> s >> t;
n = s.length();
m = t.length();
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= m; j++) {
for (int k = 0; k <= n + m; k++) {
dp[i][j][k] = 1e9;
}
}
}
for (int k = 0; k <= n + m; k++) {
dp[0][0][k] = k;
}
int i2, j2;
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= m; j++) {
if (i == 0 && j == 0) continue;
for (int k = n + m; k >= 0; k--) {
if (k > 0) {
i2 = (i > 0 && s[i - 1] == '(' ? i - 1 : i);
j2 = (j > 0 && t[j - 1] == '(' ? j - 1 : j);
if (i2 < i || j2 < j) {
dp[i][j][k] = min(dp[i][j][k], dp[i2][j2][k - 1] + 1);
}
}
if (k < n + m) {
i2 = (i > 0 && s[i - 1] == ')' ? i - 1 : i);
j2 = (j > 0 && t[j - 1] == ')' ? j - 1 : j);
dp[i][j][k] = min(dp[i][j][k], dp[i2][j2][k + 1] + 1);
}
}
}
}
int i = n, j = m, k = 0;
string ans;
while (i > 0 || j > 0) {
if (k > 0) {
i2 = (i > 0 && s[i - 1] == '(' ? i - 1 : i);
j2 = (j > 0 && t[j - 1] == '(' ? j - 1 : j);
if ((i2 < i || j2 < j) && dp[i][j][k] == dp[i2][j2][k - 1] + 1) {
i = i2;
j = j2;
k--;
ans.push_back('(');
continue;
}
}
i = (i > 0 && s[i - 1] == ')' ? i - 1 : i);
j = (j > 0 && t[j - 1] == ')' ? j - 1 : j);
k++;
ans.push_back(')');
}
for (int i = 0; i < dp[0][0][k]; i++) {
ans.push_back('(');
}
string ans2(ans.rbegin(), ans.rend());
cout << ans2 << endl;
}
| ### Prompt
In cpp, your task is to solve the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 205;
int dp[N][N][2 * N];
int n, m;
string s, t;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cin >> s >> t;
n = s.length();
m = t.length();
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= m; j++) {
for (int k = 0; k <= n + m; k++) {
dp[i][j][k] = 1e9;
}
}
}
for (int k = 0; k <= n + m; k++) {
dp[0][0][k] = k;
}
int i2, j2;
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= m; j++) {
if (i == 0 && j == 0) continue;
for (int k = n + m; k >= 0; k--) {
if (k > 0) {
i2 = (i > 0 && s[i - 1] == '(' ? i - 1 : i);
j2 = (j > 0 && t[j - 1] == '(' ? j - 1 : j);
if (i2 < i || j2 < j) {
dp[i][j][k] = min(dp[i][j][k], dp[i2][j2][k - 1] + 1);
}
}
if (k < n + m) {
i2 = (i > 0 && s[i - 1] == ')' ? i - 1 : i);
j2 = (j > 0 && t[j - 1] == ')' ? j - 1 : j);
dp[i][j][k] = min(dp[i][j][k], dp[i2][j2][k + 1] + 1);
}
}
}
}
int i = n, j = m, k = 0;
string ans;
while (i > 0 || j > 0) {
if (k > 0) {
i2 = (i > 0 && s[i - 1] == '(' ? i - 1 : i);
j2 = (j > 0 && t[j - 1] == '(' ? j - 1 : j);
if ((i2 < i || j2 < j) && dp[i][j][k] == dp[i2][j2][k - 1] + 1) {
i = i2;
j = j2;
k--;
ans.push_back('(');
continue;
}
}
i = (i > 0 && s[i - 1] == ')' ? i - 1 : i);
j = (j > 0 && t[j - 1] == ')' ? j - 1 : j);
k++;
ans.push_back(')');
}
for (int i = 0; i < dp[0][0][k]; i++) {
ans.push_back('(');
}
string ans2(ans.rbegin(), ans.rend());
cout << ans2 << endl;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3fll;
void sc(int& x) { scanf("%d", &x); }
void sc(long long& x) { scanf("%lld", &x); }
template <class T, class... Ts>
void sc(T& t, Ts&... ts) {
sc(t), sc(ts...);
}
const int MAX = 210;
int dp[MAX][MAX][MAX];
char s1[MAX], s2[MAX];
int n1, n2;
pair<pair<int, int>, pair<int, int> > get(int i1, int i2) {
bool a = (i1 != n1), b = (i2 != n2);
pair<pair<int, int>, pair<int, int> > ret;
if (s1[i1] == '(')
ret.first = {a, 0};
else
ret.first = {0, a};
if (s2[i2] == '(')
ret.second = {b, 0};
else
ret.second = {0, b};
return ret;
}
int solve(int i1, int i2, int c) {
if (c < 0) return INF;
if (c >= MAX) return INF;
if (i1 == n1 and i2 == n2) return c;
auto& x = dp[i1][i2][c];
if (x + 1) return x;
x = INF;
auto t = get(i1, i2);
x = min(x, 1 + solve(i1 + t.first.first, i2 + t.second.first, c + 1));
x = min(x, 1 + solve(i1 + t.first.second, i2 + t.second.second, c - 1));
return x;
}
void rec(int i1, int i2, int c) {
if (i1 == n1 and i2 == n2) {
while (c--) printf(")");
return;
}
int ans = solve(i1, i2, c);
auto t = get(i1, i2);
if (ans == 1 + solve(i1 + t.first.first, i2 + t.second.first, c + 1))
printf("("), rec(i1 + t.first.first, i2 + t.second.first, c + 1);
else
printf(")"), rec(i1 + t.first.second, i2 + t.second.second, c - 1);
}
int main() {
scanf(" %s %s", s1, s2);
n1 = strlen(s1), n2 = strlen(s2);
memset(dp, -1, sizeof(dp));
rec(0, 0, 0);
printf("\n");
exit(0);
}
| ### Prompt
Construct a cpp code solution to the problem outlined:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3fll;
void sc(int& x) { scanf("%d", &x); }
void sc(long long& x) { scanf("%lld", &x); }
template <class T, class... Ts>
void sc(T& t, Ts&... ts) {
sc(t), sc(ts...);
}
const int MAX = 210;
int dp[MAX][MAX][MAX];
char s1[MAX], s2[MAX];
int n1, n2;
pair<pair<int, int>, pair<int, int> > get(int i1, int i2) {
bool a = (i1 != n1), b = (i2 != n2);
pair<pair<int, int>, pair<int, int> > ret;
if (s1[i1] == '(')
ret.first = {a, 0};
else
ret.first = {0, a};
if (s2[i2] == '(')
ret.second = {b, 0};
else
ret.second = {0, b};
return ret;
}
int solve(int i1, int i2, int c) {
if (c < 0) return INF;
if (c >= MAX) return INF;
if (i1 == n1 and i2 == n2) return c;
auto& x = dp[i1][i2][c];
if (x + 1) return x;
x = INF;
auto t = get(i1, i2);
x = min(x, 1 + solve(i1 + t.first.first, i2 + t.second.first, c + 1));
x = min(x, 1 + solve(i1 + t.first.second, i2 + t.second.second, c - 1));
return x;
}
void rec(int i1, int i2, int c) {
if (i1 == n1 and i2 == n2) {
while (c--) printf(")");
return;
}
int ans = solve(i1, i2, c);
auto t = get(i1, i2);
if (ans == 1 + solve(i1 + t.first.first, i2 + t.second.first, c + 1))
printf("("), rec(i1 + t.first.first, i2 + t.second.first, c + 1);
else
printf(")"), rec(i1 + t.first.second, i2 + t.second.second, c - 1);
}
int main() {
scanf(" %s %s", s1, s2);
n1 = strlen(s1), n2 = strlen(s2);
memset(dp, -1, sizeof(dp));
rec(0, 0, 0);
printf("\n");
exit(0);
}
``` |
#include <bits/stdc++.h>
using namespace std;
int dp[205][205][205], p[205][205][205];
int main() {
char sa[205], sb[205], ans[410];
int a, b, i, j, k, x, y, z, len = 1000, c, f;
scanf("%s %s", sa, sb);
a = strlen(sa);
b = strlen(sb);
memset(dp, 0x7f, sizeof(dp));
for (i = 0; i <= a; i++) {
for (j = 0; j <= b; j++) {
for (k = 0; k <= 201; k++) {
if (i == 0 && j == 0) {
dp[i][j][k] = k;
p[i][j][k] = 0;
} else if (i == 0) {
if (sb[j - 1] == '(') {
if (k == 0) {
dp[i][j][k] = dp[i][j - 1][k] + 2;
p[i][j][k] = 32;
} else {
dp[i][j][k] = dp[i][j - 1][k - 1] + 1;
p[i][j][k] = 33;
}
} else {
if (k == 201) {
dp[i][j][k] = dp[i][j - 1][k] + 2;
p[i][j][k] = 32;
} else {
dp[i][j][k] = dp[i][j - 1][k + 1] + 1;
p[i][j][k] = 31;
}
}
} else if (j == 0) {
if (sa[i - 1] == '(') {
if (k == 0) {
dp[i][j][k] = dp[i - 1][j][k] + 2;
p[i][j][k] = 12;
} else {
dp[i][j][k] = dp[i - 1][j][k - 1] + 1;
p[i][j][k] = 13;
}
} else {
if (k == 201) {
dp[i][j][k] = dp[i - 1][j][k] + 2;
p[i][j][k] = 12;
} else {
dp[i][j][k] = dp[i - 1][j][k + 1] + 1;
p[i][j][k] = 11;
}
}
} else {
if (sa[i - 1] == sb[j - 1]) {
if (sa[i - 1] == '(') {
if (k == 0) {
if (dp[i][j][k] > dp[i - 1][j - 1][k] + 2) {
dp[i][j][k] = dp[i - 1][j - 1][k] + 2;
p[i][j][k] = 22;
}
} else {
if (dp[i][j][k] > dp[i - 1][j - 1][k - 1] + 1) {
dp[i][j][k] = dp[i - 1][j - 1][k - 1] + 1;
p[i][j][k] = 23;
}
}
} else {
if (k == 201) {
if (dp[i][j][k] > dp[i - 1][j - 1][k] + 2) {
dp[i][j][k] = dp[i - 1][j - 1][k] + 2;
p[i][j][k] = 22;
}
} else {
if (dp[i][j][k] > dp[i - 1][j - 1][k + 1] + 1) {
dp[i][j][k] = dp[i - 1][j - 1][k + 1] + 1;
p[i][j][k] = 21;
}
}
}
}
if (sa[i - 1] == '(') {
if (k == 0) {
if (dp[i][j][k] > dp[i - 1][j][k] + 2) {
dp[i][j][k] = dp[i - 1][j][k] + 2;
p[i][j][k] = 12;
}
} else {
if (dp[i][j][k] > dp[i - 1][j][k - 1] + 1) {
dp[i][j][k] = dp[i - 1][j][k - 1] + 1;
p[i][j][k] = 13;
}
}
} else {
if (k == 201) {
if (dp[i][j][k] > dp[i - 1][j][k] + 2) {
dp[i][j][k] = dp[i - 1][j][k] + 2;
p[i][j][k] = 12;
}
} else {
if (dp[i][j][k] > dp[i - 1][j][k + 1] + 1) {
dp[i][j][k] = dp[i - 1][j][k + 1] + 1;
p[i][j][k] = 11;
}
}
}
if (sb[j - 1] == '(') {
if (k == 0) {
if (dp[i][j][k] > dp[i][j - 1][k] + 2) {
dp[i][j][k] = dp[i][j - 1][k] + 2;
p[i][j][k] = 32;
}
} else {
if (dp[i][j][k] > dp[i][j - 1][k - 1] + 1) {
dp[i][j][k] = dp[i][j - 1][k - 1] + 1;
p[i][j][k] = 33;
}
}
} else {
if (k == 201) {
if (dp[i][j][k] > dp[i][j - 1][k] + 2) {
dp[i][j][k] = dp[i][j - 1][k] + 2;
p[i][j][k] = 32;
}
} else {
if (dp[i][j][k] > dp[i][j - 1][k + 1] + 1) {
dp[i][j][k] = dp[i][j - 1][k + 1] + 1;
p[i][j][k] = 31;
}
}
}
}
if (i == a && j == b) {
if (len > dp[i][j][k] + k) {
len = dp[i][j][k] + k;
z = k;
}
}
}
}
}
x = a;
y = b;
c = 0;
for (i = 0; i < z; i++) {
ans[len - c - 1] = ')';
c++;
}
while (c < len) {
f = z;
if (x == 0 && y == 0) {
ans[len - c - 1] = '(';
c++;
continue;
}
if (p[x][y][z] % 10 == 1) {
ans[len - c - 1] = ')';
z += 1;
c++;
} else if (p[x][y][z] % 10 == 3) {
ans[len - c - 1] = '(';
z -= 1;
c++;
} else {
ans[len - c - 1] = ')';
c++;
ans[len - c - 1] = '(';
c++;
}
if (p[x][y][f] / 10 == 1) {
x -= 1;
} else if (p[x][y][f] / 10 == 3) {
y -= 1;
} else {
x -= 1;
y -= 1;
}
}
for (i = 0; i < len; i++) printf("%c", ans[i]);
printf("\n");
return 0;
}
| ### Prompt
Construct a Cpp code solution to the problem outlined:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int dp[205][205][205], p[205][205][205];
int main() {
char sa[205], sb[205], ans[410];
int a, b, i, j, k, x, y, z, len = 1000, c, f;
scanf("%s %s", sa, sb);
a = strlen(sa);
b = strlen(sb);
memset(dp, 0x7f, sizeof(dp));
for (i = 0; i <= a; i++) {
for (j = 0; j <= b; j++) {
for (k = 0; k <= 201; k++) {
if (i == 0 && j == 0) {
dp[i][j][k] = k;
p[i][j][k] = 0;
} else if (i == 0) {
if (sb[j - 1] == '(') {
if (k == 0) {
dp[i][j][k] = dp[i][j - 1][k] + 2;
p[i][j][k] = 32;
} else {
dp[i][j][k] = dp[i][j - 1][k - 1] + 1;
p[i][j][k] = 33;
}
} else {
if (k == 201) {
dp[i][j][k] = dp[i][j - 1][k] + 2;
p[i][j][k] = 32;
} else {
dp[i][j][k] = dp[i][j - 1][k + 1] + 1;
p[i][j][k] = 31;
}
}
} else if (j == 0) {
if (sa[i - 1] == '(') {
if (k == 0) {
dp[i][j][k] = dp[i - 1][j][k] + 2;
p[i][j][k] = 12;
} else {
dp[i][j][k] = dp[i - 1][j][k - 1] + 1;
p[i][j][k] = 13;
}
} else {
if (k == 201) {
dp[i][j][k] = dp[i - 1][j][k] + 2;
p[i][j][k] = 12;
} else {
dp[i][j][k] = dp[i - 1][j][k + 1] + 1;
p[i][j][k] = 11;
}
}
} else {
if (sa[i - 1] == sb[j - 1]) {
if (sa[i - 1] == '(') {
if (k == 0) {
if (dp[i][j][k] > dp[i - 1][j - 1][k] + 2) {
dp[i][j][k] = dp[i - 1][j - 1][k] + 2;
p[i][j][k] = 22;
}
} else {
if (dp[i][j][k] > dp[i - 1][j - 1][k - 1] + 1) {
dp[i][j][k] = dp[i - 1][j - 1][k - 1] + 1;
p[i][j][k] = 23;
}
}
} else {
if (k == 201) {
if (dp[i][j][k] > dp[i - 1][j - 1][k] + 2) {
dp[i][j][k] = dp[i - 1][j - 1][k] + 2;
p[i][j][k] = 22;
}
} else {
if (dp[i][j][k] > dp[i - 1][j - 1][k + 1] + 1) {
dp[i][j][k] = dp[i - 1][j - 1][k + 1] + 1;
p[i][j][k] = 21;
}
}
}
}
if (sa[i - 1] == '(') {
if (k == 0) {
if (dp[i][j][k] > dp[i - 1][j][k] + 2) {
dp[i][j][k] = dp[i - 1][j][k] + 2;
p[i][j][k] = 12;
}
} else {
if (dp[i][j][k] > dp[i - 1][j][k - 1] + 1) {
dp[i][j][k] = dp[i - 1][j][k - 1] + 1;
p[i][j][k] = 13;
}
}
} else {
if (k == 201) {
if (dp[i][j][k] > dp[i - 1][j][k] + 2) {
dp[i][j][k] = dp[i - 1][j][k] + 2;
p[i][j][k] = 12;
}
} else {
if (dp[i][j][k] > dp[i - 1][j][k + 1] + 1) {
dp[i][j][k] = dp[i - 1][j][k + 1] + 1;
p[i][j][k] = 11;
}
}
}
if (sb[j - 1] == '(') {
if (k == 0) {
if (dp[i][j][k] > dp[i][j - 1][k] + 2) {
dp[i][j][k] = dp[i][j - 1][k] + 2;
p[i][j][k] = 32;
}
} else {
if (dp[i][j][k] > dp[i][j - 1][k - 1] + 1) {
dp[i][j][k] = dp[i][j - 1][k - 1] + 1;
p[i][j][k] = 33;
}
}
} else {
if (k == 201) {
if (dp[i][j][k] > dp[i][j - 1][k] + 2) {
dp[i][j][k] = dp[i][j - 1][k] + 2;
p[i][j][k] = 32;
}
} else {
if (dp[i][j][k] > dp[i][j - 1][k + 1] + 1) {
dp[i][j][k] = dp[i][j - 1][k + 1] + 1;
p[i][j][k] = 31;
}
}
}
}
if (i == a && j == b) {
if (len > dp[i][j][k] + k) {
len = dp[i][j][k] + k;
z = k;
}
}
}
}
}
x = a;
y = b;
c = 0;
for (i = 0; i < z; i++) {
ans[len - c - 1] = ')';
c++;
}
while (c < len) {
f = z;
if (x == 0 && y == 0) {
ans[len - c - 1] = '(';
c++;
continue;
}
if (p[x][y][z] % 10 == 1) {
ans[len - c - 1] = ')';
z += 1;
c++;
} else if (p[x][y][z] % 10 == 3) {
ans[len - c - 1] = '(';
z -= 1;
c++;
} else {
ans[len - c - 1] = ')';
c++;
ans[len - c - 1] = '(';
c++;
}
if (p[x][y][f] / 10 == 1) {
x -= 1;
} else if (p[x][y][f] / 10 == 3) {
y -= 1;
} else {
x -= 1;
y -= 1;
}
}
for (i = 0; i < len; i++) printf("%c", ans[i]);
printf("\n");
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
pair<int, pair<int, int>> nextt[201][201][201];
int dp[201][201][201];
string s, t;
int ans(int curr1, int curr2, int val, int siz) {
if (siz > 2000 || curr1 > 200 || curr2 > 200 || val > 200) {
return 1000000000;
}
if (curr1 == s.size() && curr2 == t.size()) {
for (int i = val; i > 0; i -= 1) {
nextt[curr1][curr2][val] = {curr1, {curr2, val - 1}};
}
return val;
}
if (dp[curr1][curr2][val] != -1) {
return dp[curr1][curr2][val];
}
int v1 = curr1, v2 = curr2;
if (curr1 < s.size() && s[curr1] == '(') {
v1 += 1;
}
if (curr2 < t.size() && t[curr2] == '(') {
v2 += 1;
}
nextt[curr1][curr2][val] = {v1, {v2, val + 1}};
int ret = 1 + ans(v1, v2, val + 1, siz + 1);
if (val > 0) {
v1 = curr1, v2 = curr2;
if (curr1 < s.size() && s[curr1] == ')') {
v1 += 1;
}
if (curr2 < t.size() && t[curr2] == ')') {
v2 += 1;
}
if (1 + ans(v1, v2, val - 1, siz + 1) < ret) {
ret = min(ret, 1 + ans(v1, v2, val - 1, siz + 1));
nextt[curr1][curr2][val] = {v1, {v2, val - 1}};
}
}
return dp[curr1][curr2][val] = ret;
}
int main() {
memset(dp, -1, sizeof dp);
cin >> s >> t;
int y = ans(0, 0, 0, 0);
pair<int, pair<int, int>> curr = {0, {0, 0}};
for (int i = 0; i < y; i += 1) {
pair<int, pair<int, int>> nex =
nextt[curr.first][curr.second.first][curr.second.second];
if (nex.second.second == curr.second.second + 1) {
cout << '(';
} else {
cout << ')';
}
curr = nex;
}
cout << endl;
}
| ### Prompt
Construct a Cpp code solution to the problem outlined:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
pair<int, pair<int, int>> nextt[201][201][201];
int dp[201][201][201];
string s, t;
int ans(int curr1, int curr2, int val, int siz) {
if (siz > 2000 || curr1 > 200 || curr2 > 200 || val > 200) {
return 1000000000;
}
if (curr1 == s.size() && curr2 == t.size()) {
for (int i = val; i > 0; i -= 1) {
nextt[curr1][curr2][val] = {curr1, {curr2, val - 1}};
}
return val;
}
if (dp[curr1][curr2][val] != -1) {
return dp[curr1][curr2][val];
}
int v1 = curr1, v2 = curr2;
if (curr1 < s.size() && s[curr1] == '(') {
v1 += 1;
}
if (curr2 < t.size() && t[curr2] == '(') {
v2 += 1;
}
nextt[curr1][curr2][val] = {v1, {v2, val + 1}};
int ret = 1 + ans(v1, v2, val + 1, siz + 1);
if (val > 0) {
v1 = curr1, v2 = curr2;
if (curr1 < s.size() && s[curr1] == ')') {
v1 += 1;
}
if (curr2 < t.size() && t[curr2] == ')') {
v2 += 1;
}
if (1 + ans(v1, v2, val - 1, siz + 1) < ret) {
ret = min(ret, 1 + ans(v1, v2, val - 1, siz + 1));
nextt[curr1][curr2][val] = {v1, {v2, val - 1}};
}
}
return dp[curr1][curr2][val] = ret;
}
int main() {
memset(dp, -1, sizeof dp);
cin >> s >> t;
int y = ans(0, 0, 0, 0);
pair<int, pair<int, int>> curr = {0, {0, 0}};
for (int i = 0; i < y; i += 1) {
pair<int, pair<int, int>> nex =
nextt[curr.first][curr.second.first][curr.second.second];
if (nex.second.second == curr.second.second + 1) {
cout << '(';
} else {
cout << ')';
}
curr = nex;
}
cout << endl;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int Inf = 1e9 + 7;
const int N = 204;
int dp[N][N][N];
char s[N], t[N], str[2 * N];
int mem[N][N][N];
int n, m;
int dfs(int i, int j, int k) {
if (k < 0 || k > 200) return Inf;
if (i == n + 1 && j == m + 1) {
if (k == 0)
return 0;
else
return k;
}
if (dp[i][j][k] != -1) return dp[i][j][k];
int res = Inf;
int f = 0, g = 0;
if (i <= n && s[i] == '(') f = 1;
if (j <= m && t[j] == '(') g = 1;
int res1 = 1 + dfs(i + f, j + g, k + 1);
f = 0, g = 0;
if (i <= n && s[i] == ')') f = 1;
if (j <= m && t[j] == ')') g = 1;
int res2 = 1 + dfs(i + f, j + g, k - 1);
if (res1 < res2)
mem[i][j][k] = 0;
else
mem[i][j][k] = 1;
res = min(res1, res2);
dp[i][j][k] = res;
return res;
}
int idx = 1;
void solve(int i, int j, int k) {
if (i == n + 1 && j == m + 1) {
while (k--) str[idx++] = ')';
return;
}
if (mem[i][j][k] == 0) {
str[idx++] = '(';
int f = 0, g = 0;
if (i <= n && s[i] == '(') f = 1;
if (j <= m && t[j] == '(') g = 1;
solve(i + f, j + g, k + 1);
} else {
str[idx++] = ')';
int f = 0, g = 0;
if (i <= n && s[i] == ')') f = 1;
if (j <= m && t[j] == ')') g = 1;
solve(i + f, j + g, k - 1);
}
}
int main() {
scanf("%s", s + 1);
scanf("%s", t + 1);
n = strlen(s + 1);
m = strlen(t + 1);
memset(dp, -1, sizeof(dp));
int ans = dfs(1, 1, 0);
solve(1, 1, 0);
printf("%s\n", str + 1);
return 0;
}
| ### Prompt
Please provide a Cpp coded solution to the problem described below:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int Inf = 1e9 + 7;
const int N = 204;
int dp[N][N][N];
char s[N], t[N], str[2 * N];
int mem[N][N][N];
int n, m;
int dfs(int i, int j, int k) {
if (k < 0 || k > 200) return Inf;
if (i == n + 1 && j == m + 1) {
if (k == 0)
return 0;
else
return k;
}
if (dp[i][j][k] != -1) return dp[i][j][k];
int res = Inf;
int f = 0, g = 0;
if (i <= n && s[i] == '(') f = 1;
if (j <= m && t[j] == '(') g = 1;
int res1 = 1 + dfs(i + f, j + g, k + 1);
f = 0, g = 0;
if (i <= n && s[i] == ')') f = 1;
if (j <= m && t[j] == ')') g = 1;
int res2 = 1 + dfs(i + f, j + g, k - 1);
if (res1 < res2)
mem[i][j][k] = 0;
else
mem[i][j][k] = 1;
res = min(res1, res2);
dp[i][j][k] = res;
return res;
}
int idx = 1;
void solve(int i, int j, int k) {
if (i == n + 1 && j == m + 1) {
while (k--) str[idx++] = ')';
return;
}
if (mem[i][j][k] == 0) {
str[idx++] = '(';
int f = 0, g = 0;
if (i <= n && s[i] == '(') f = 1;
if (j <= m && t[j] == '(') g = 1;
solve(i + f, j + g, k + 1);
} else {
str[idx++] = ')';
int f = 0, g = 0;
if (i <= n && s[i] == ')') f = 1;
if (j <= m && t[j] == ')') g = 1;
solve(i + f, j + g, k - 1);
}
}
int main() {
scanf("%s", s + 1);
scanf("%s", t + 1);
n = strlen(s + 1);
m = strlen(t + 1);
memset(dp, -1, sizeof(dp));
int ans = dfs(1, 1, 0);
solve(1, 1, 0);
printf("%s\n", str + 1);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
long long POW(long long a, long long b, long long MMM = MOD) {
long long ret = 1;
for (; b; b >>= 1, a = (a * a) % MMM)
if (b & 1) ret = (ret * a) % MMM;
return ret;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) {
if (a == 0 || b == 0) return a + b;
return a * (b / gcd(a, b));
}
int dx[] = {0, 1, 0, -1, 1, 1, -1, -1}, dy[] = {1, 0, -1, 0, 1, -1, 1, -1};
char s[222], t[222];
int n, m;
int d[202][202][402];
int go(int N, int M, int c) {
if (N == n && M == m && c == 0) return 0;
int &ret = d[N][M][c];
if (~ret) return ret;
ret = 987654321;
if (N == n && M == m) ret = min(ret, 1 + go(N, M, c - 1));
if (c < 400) {
int nn = (s[N] == '(' ? N + 1 : N), mm = (t[M] == '(' ? M + 1 : M);
ret = min(ret, 1 + go(nn, mm, c + 1));
}
if (c > 0) {
int nn = (s[N] == ')' ? N + 1 : N), mm = (t[M] == ')' ? M + 1 : M);
ret = min(ret, 1 + go(nn, mm, c - 1));
}
return ret;
}
void track(int N, int M, int c) {
if (N == n && M == m && c == 0) return;
if (N == n && M == m) {
putchar(')');
track(N, M, c - 1);
return;
}
int t1 = 987654321, t2 = 987654321;
int n1, m1, n2, m2;
if (c < 400) {
n1 = (s[N] == '(' ? N + 1 : N), m1 = (t[M] == '(' ? M + 1 : M);
t1 = 1 + go(n1, m1, c + 1);
}
if (c > 0) {
n2 = (s[N] == ')' ? N + 1 : N), m2 = (t[M] == ')' ? M + 1 : M);
t2 = 1 + go(n2, m2, c - 1);
}
if (t1 < t2) {
putchar('(');
track(n1, m1, c + 1);
return;
} else {
putchar(')');
track(n2, m2, c - 1);
return;
}
}
int main() {
memset((d), -1, sizeof(d));
;
scanf("%s%s", s, t);
n = strlen(s), m = strlen(t);
track(0, 0, 0);
}
| ### Prompt
Please create a solution in cpp to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
long long POW(long long a, long long b, long long MMM = MOD) {
long long ret = 1;
for (; b; b >>= 1, a = (a * a) % MMM)
if (b & 1) ret = (ret * a) % MMM;
return ret;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) {
if (a == 0 || b == 0) return a + b;
return a * (b / gcd(a, b));
}
int dx[] = {0, 1, 0, -1, 1, 1, -1, -1}, dy[] = {1, 0, -1, 0, 1, -1, 1, -1};
char s[222], t[222];
int n, m;
int d[202][202][402];
int go(int N, int M, int c) {
if (N == n && M == m && c == 0) return 0;
int &ret = d[N][M][c];
if (~ret) return ret;
ret = 987654321;
if (N == n && M == m) ret = min(ret, 1 + go(N, M, c - 1));
if (c < 400) {
int nn = (s[N] == '(' ? N + 1 : N), mm = (t[M] == '(' ? M + 1 : M);
ret = min(ret, 1 + go(nn, mm, c + 1));
}
if (c > 0) {
int nn = (s[N] == ')' ? N + 1 : N), mm = (t[M] == ')' ? M + 1 : M);
ret = min(ret, 1 + go(nn, mm, c - 1));
}
return ret;
}
void track(int N, int M, int c) {
if (N == n && M == m && c == 0) return;
if (N == n && M == m) {
putchar(')');
track(N, M, c - 1);
return;
}
int t1 = 987654321, t2 = 987654321;
int n1, m1, n2, m2;
if (c < 400) {
n1 = (s[N] == '(' ? N + 1 : N), m1 = (t[M] == '(' ? M + 1 : M);
t1 = 1 + go(n1, m1, c + 1);
}
if (c > 0) {
n2 = (s[N] == ')' ? N + 1 : N), m2 = (t[M] == ')' ? M + 1 : M);
t2 = 1 + go(n2, m2, c - 1);
}
if (t1 < t2) {
putchar('(');
track(n1, m1, c + 1);
return;
} else {
putchar(')');
track(n2, m2, c - 1);
return;
}
}
int main() {
memset((d), -1, sizeof(d));
;
scanf("%s%s", s, t);
n = strlen(s), m = strlen(t);
track(0, 0, 0);
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 205;
int dp[N][N][2 * N], n, m, idxi, idxj, idx1, idx2, k, idxx, k1, val;
pair<pair<int, int>, int> dp1[N][N][2 * N];
string s1, s2;
vector<char> v;
int main() {
cin >> s1 >> s2;
n = s1.size();
m = s2.size();
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= m; j++) {
for (int k = 0; k <= 400; k++) {
dp[i][j][k] = 1e9;
}
}
}
dp[0][0][0] = 0;
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= m; j++) {
for (int k = 0; k <= 400; k++) {
if (dp[i][j][k] == 1e9) continue;
idxi = i;
idxj = j;
if (i == 1 && j == 3 && k == 2) {
}
if (i < n)
if (s1[i] == '(') idxi++;
if (j < m)
if (s2[j] == '(') idxj++;
if (dp[idxi][idxj][k + 1] >= dp[i][j][k] + 1) {
dp[idxi][idxj][k + 1] = dp[i][j][k] + 1;
dp1[idxi][idxj][k + 1] = {{i, j}, 1};
}
idxi = i;
idxj = j;
if (i < n)
if (s1[i] == ')') idxi++;
if (j < m)
if (s2[j] == ')') idxj++;
if (dp[idxi][idxj][k - 1] >= dp[i][j][k] + 1) {
dp[idxi][idxj][k - 1] = dp[i][j][k] + 1;
dp1[idxi][idxj][k - 1] = {{i, j}, -1};
}
}
}
}
idx1 = n;
idx2 = m;
k = 0;
for (int i = 0; i <= 400; i++) {
if (1e9 != dp[n][m][i]) {
val = i;
break;
}
}
for (int i = 1; i <= val; i++) {
v.push_back(')');
}
k = val;
while (true) {
if (idx1 == idx2 && idx1 == 0 && k == 0) {
break;
}
k1 = k;
if (dp1[idx1][idx2][k].second == 1LL)
v.push_back('('), k--;
else
v.push_back(')'), k++;
idxx = idx1;
idx1 = dp1[idx1][idx2][k1].first.first;
idx2 = dp1[idxx][idx2][k1].first.second;
}
reverse(v.begin(), v.end());
for (int i = 0; i < v.size(); i++) {
cout << v[i];
}
}
| ### Prompt
Please create a solution in Cpp to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 205;
int dp[N][N][2 * N], n, m, idxi, idxj, idx1, idx2, k, idxx, k1, val;
pair<pair<int, int>, int> dp1[N][N][2 * N];
string s1, s2;
vector<char> v;
int main() {
cin >> s1 >> s2;
n = s1.size();
m = s2.size();
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= m; j++) {
for (int k = 0; k <= 400; k++) {
dp[i][j][k] = 1e9;
}
}
}
dp[0][0][0] = 0;
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= m; j++) {
for (int k = 0; k <= 400; k++) {
if (dp[i][j][k] == 1e9) continue;
idxi = i;
idxj = j;
if (i == 1 && j == 3 && k == 2) {
}
if (i < n)
if (s1[i] == '(') idxi++;
if (j < m)
if (s2[j] == '(') idxj++;
if (dp[idxi][idxj][k + 1] >= dp[i][j][k] + 1) {
dp[idxi][idxj][k + 1] = dp[i][j][k] + 1;
dp1[idxi][idxj][k + 1] = {{i, j}, 1};
}
idxi = i;
idxj = j;
if (i < n)
if (s1[i] == ')') idxi++;
if (j < m)
if (s2[j] == ')') idxj++;
if (dp[idxi][idxj][k - 1] >= dp[i][j][k] + 1) {
dp[idxi][idxj][k - 1] = dp[i][j][k] + 1;
dp1[idxi][idxj][k - 1] = {{i, j}, -1};
}
}
}
}
idx1 = n;
idx2 = m;
k = 0;
for (int i = 0; i <= 400; i++) {
if (1e9 != dp[n][m][i]) {
val = i;
break;
}
}
for (int i = 1; i <= val; i++) {
v.push_back(')');
}
k = val;
while (true) {
if (idx1 == idx2 && idx1 == 0 && k == 0) {
break;
}
k1 = k;
if (dp1[idx1][idx2][k].second == 1LL)
v.push_back('('), k--;
else
v.push_back(')'), k++;
idxx = idx1;
idx1 = dp1[idx1][idx2][k1].first.first;
idx2 = dp1[idxx][idx2][k1].first.second;
}
reverse(v.begin(), v.end());
for (int i = 0; i < v.size(); i++) {
cout << v[i];
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
long long power(long long a, long long b, long long m = mod) {
long long x = 1;
while (b) {
if (b & 1) {
x = 1ll * x * a % m;
}
a = 1ll * a * a % m;
b /= 2;
}
return x;
}
const int N = 2e5 + 9;
string s, t;
int n, m;
int dp[202][202][202], from[202][202][202];
const int inf = mod;
int solve(int p1, int p2, int bal) {
if (bal < 0 || bal > 200) return inf;
if (p1 == 0 && p2 == 0 && bal == 0) {
return 0;
}
int &ret = dp[p1][p2][bal];
if (ret != -1) return ret;
int u, v;
if (s[p1] == '(' && t[p2] == '(') {
u = solve(max(0, p1 - 1), max(0, p2 - 1), bal - 1);
v = solve(p1, p2, bal + 1);
} else if (s[p1] == '(') {
u = solve(max(0, p1 - 1), p2, bal - 1);
v = solve(p1, max(0, p2 - 1), bal + 1);
} else if (t[p2] == '(') {
u = solve(p1, max(0, p2 - 1), bal - 1);
v = solve(max(0, p1 - 1), p2, bal + 1);
} else if (p1 + p2 == 0) {
u = solve(p1, p2, bal - 1);
v = inf;
} else {
u = solve(p1, p2, bal - 1);
v = solve(max(0, p1 - 1), max(0, p2 - 1), bal + 1);
}
if (u <= v) {
from[p1][p2][bal] = 1;
ret = 1 + u;
} else {
from[p1][p2][bal] = 2;
ret = 1 + v;
}
return ret;
}
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> s >> t;
n = s.size();
m = t.size();
s = '_' + s;
t = '_' + t;
memset(dp, -1, sizeof dp);
solve(n, m, 0);
string ans;
int p1 = n, p2 = m, bal = 0;
while (p1 > 0 || p2 > 0 || bal > 0) {
int q = from[p1][p2][bal];
if (q == 1) {
if (s[p1] == '(') p1--;
if (t[p2] == '(') p2--;
bal--;
ans += '(';
} else {
if (s[p1] == ')') p1--;
if (t[p2] == ')') p2--;
bal++;
ans += ')';
}
}
reverse(ans.begin(), ans.end());
cout << ans << "\n";
return 0;
}
| ### Prompt
Please formulate a CPP solution to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
long long power(long long a, long long b, long long m = mod) {
long long x = 1;
while (b) {
if (b & 1) {
x = 1ll * x * a % m;
}
a = 1ll * a * a % m;
b /= 2;
}
return x;
}
const int N = 2e5 + 9;
string s, t;
int n, m;
int dp[202][202][202], from[202][202][202];
const int inf = mod;
int solve(int p1, int p2, int bal) {
if (bal < 0 || bal > 200) return inf;
if (p1 == 0 && p2 == 0 && bal == 0) {
return 0;
}
int &ret = dp[p1][p2][bal];
if (ret != -1) return ret;
int u, v;
if (s[p1] == '(' && t[p2] == '(') {
u = solve(max(0, p1 - 1), max(0, p2 - 1), bal - 1);
v = solve(p1, p2, bal + 1);
} else if (s[p1] == '(') {
u = solve(max(0, p1 - 1), p2, bal - 1);
v = solve(p1, max(0, p2 - 1), bal + 1);
} else if (t[p2] == '(') {
u = solve(p1, max(0, p2 - 1), bal - 1);
v = solve(max(0, p1 - 1), p2, bal + 1);
} else if (p1 + p2 == 0) {
u = solve(p1, p2, bal - 1);
v = inf;
} else {
u = solve(p1, p2, bal - 1);
v = solve(max(0, p1 - 1), max(0, p2 - 1), bal + 1);
}
if (u <= v) {
from[p1][p2][bal] = 1;
ret = 1 + u;
} else {
from[p1][p2][bal] = 2;
ret = 1 + v;
}
return ret;
}
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> s >> t;
n = s.size();
m = t.size();
s = '_' + s;
t = '_' + t;
memset(dp, -1, sizeof dp);
solve(n, m, 0);
string ans;
int p1 = n, p2 = m, bal = 0;
while (p1 > 0 || p2 > 0 || bal > 0) {
int q = from[p1][p2][bal];
if (q == 1) {
if (s[p1] == '(') p1--;
if (t[p2] == '(') p2--;
bal--;
ans += '(';
} else {
if (s[p1] == ')') p1--;
if (t[p2] == ')') p2--;
bal++;
ans += ')';
}
}
reverse(ans.begin(), ans.end());
cout << ans << "\n";
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 205;
const int MAXM = 1000005;
const int HASHSIZE = 2000003;
const int INF = 0x7fffffff;
const int mod = 1e8;
inline int read() {
int x = 0, f = 1;
char ch = getchar();
while (!isdigit(ch)) {
if (ch == '-') f = -1;
ch = getchar();
}
while (isdigit(ch)) {
x = (x << 1) + (x << 3) + (ch ^ 48);
ch = getchar();
}
return x * f;
}
template <typename T>
void print(T x) {
if (x < 0) putchar('-'), x = -x;
if (x > 9) print(x / 10);
putchar('0' + x % 10);
}
struct node {
int i, j, dis;
string ans;
bool operator<(const node& n1) const {
return (ans.length() + dis) > (n1.ans.length() + n1.dis);
}
};
char a[MAXN], b[MAXN];
int dp[203][203][203];
queue<node> q;
priority_queue<node> q2;
void bfs(int la, int lb) {
q.push({0, 0, 0});
while (!q.empty()) {
node now = q.front();
q.pop();
if (now.i == la || now.j == lb) {
q2.push(now);
continue;
}
int l = 0, r = 0;
if (a[now.i] == '(') l++;
if (b[now.j] == '(') r++;
if (now.dis <= la && now.ans.length() + 1 + now.dis + 1 <
dp[now.i + l][now.j + r][now.dis + 1]) {
dp[now.i + l][now.j + r][now.dis + 1] =
now.ans.length() + 1 + now.dis + 1;
q.push({now.i + l, now.j + r, now.dis + 1, now.ans + '('});
}
l = 0, r = 0;
if (!now.dis) continue;
if (a[now.i] == ')') l++;
if (b[now.j] == ')') r++;
if (now.ans.length() + now.dis < dp[now.i + l][now.j + r][now.dis - 1]) {
dp[now.i + l][now.j + r][now.dis - 1] = now.ans.length() + now.dis;
q.push({now.i + l, now.j + r, now.dis - 1, now.ans + ')'});
}
}
while (!q2.empty()) {
node now = q2.top();
q2.pop();
if (now.i == la && now.j == lb) {
while (now.dis) {
now.ans += ')';
now.dis--;
}
cout << now.ans << "\n";
return;
}
if (now.i == la) {
if (b[now.j] == '(') {
now.dis++;
now.j++;
now.ans += '(';
q2.push(now);
} else if (b[now.j] == ')' && now.dis) {
now.dis--;
now.j++;
now.ans += ')';
q2.push(now);
} else {
now.j++;
now.ans += "()";
q2.push(now);
}
} else {
if (a[now.i] == '(') {
now.dis++;
now.i++;
now.ans += '(';
q2.push(now);
} else if (a[now.i] == ')' && now.dis) {
now.dis--;
now.i++;
now.ans += ')';
q2.push(now);
} else {
now.i++;
now.ans += "()";
q2.push(now);
}
}
}
}
signed main() {
scanf("%s%s", a, b);
int la = strlen(a), lb = strlen(b);
if (la < lb) swap(a, b), swap(la, lb);
memset(dp, 0x7f, sizeof(dp));
bfs(la, lb);
return 0;
}
| ### Prompt
Generate a Cpp solution to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 205;
const int MAXM = 1000005;
const int HASHSIZE = 2000003;
const int INF = 0x7fffffff;
const int mod = 1e8;
inline int read() {
int x = 0, f = 1;
char ch = getchar();
while (!isdigit(ch)) {
if (ch == '-') f = -1;
ch = getchar();
}
while (isdigit(ch)) {
x = (x << 1) + (x << 3) + (ch ^ 48);
ch = getchar();
}
return x * f;
}
template <typename T>
void print(T x) {
if (x < 0) putchar('-'), x = -x;
if (x > 9) print(x / 10);
putchar('0' + x % 10);
}
struct node {
int i, j, dis;
string ans;
bool operator<(const node& n1) const {
return (ans.length() + dis) > (n1.ans.length() + n1.dis);
}
};
char a[MAXN], b[MAXN];
int dp[203][203][203];
queue<node> q;
priority_queue<node> q2;
void bfs(int la, int lb) {
q.push({0, 0, 0});
while (!q.empty()) {
node now = q.front();
q.pop();
if (now.i == la || now.j == lb) {
q2.push(now);
continue;
}
int l = 0, r = 0;
if (a[now.i] == '(') l++;
if (b[now.j] == '(') r++;
if (now.dis <= la && now.ans.length() + 1 + now.dis + 1 <
dp[now.i + l][now.j + r][now.dis + 1]) {
dp[now.i + l][now.j + r][now.dis + 1] =
now.ans.length() + 1 + now.dis + 1;
q.push({now.i + l, now.j + r, now.dis + 1, now.ans + '('});
}
l = 0, r = 0;
if (!now.dis) continue;
if (a[now.i] == ')') l++;
if (b[now.j] == ')') r++;
if (now.ans.length() + now.dis < dp[now.i + l][now.j + r][now.dis - 1]) {
dp[now.i + l][now.j + r][now.dis - 1] = now.ans.length() + now.dis;
q.push({now.i + l, now.j + r, now.dis - 1, now.ans + ')'});
}
}
while (!q2.empty()) {
node now = q2.top();
q2.pop();
if (now.i == la && now.j == lb) {
while (now.dis) {
now.ans += ')';
now.dis--;
}
cout << now.ans << "\n";
return;
}
if (now.i == la) {
if (b[now.j] == '(') {
now.dis++;
now.j++;
now.ans += '(';
q2.push(now);
} else if (b[now.j] == ')' && now.dis) {
now.dis--;
now.j++;
now.ans += ')';
q2.push(now);
} else {
now.j++;
now.ans += "()";
q2.push(now);
}
} else {
if (a[now.i] == '(') {
now.dis++;
now.i++;
now.ans += '(';
q2.push(now);
} else if (a[now.i] == ')' && now.dis) {
now.dis--;
now.i++;
now.ans += ')';
q2.push(now);
} else {
now.i++;
now.ans += "()";
q2.push(now);
}
}
}
}
signed main() {
scanf("%s%s", a, b);
int la = strlen(a), lb = strlen(b);
if (la < lb) swap(a, b), swap(la, lb);
memset(dp, 0x7f, sizeof(dp));
bfs(la, lb);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9 + 69;
int dp[205][205][405];
string s, t, ans;
int f(int is, int it, int cnt) {
if (cnt < 0 || cnt >= s.size() + t.size()) return inf;
if (is == s.size() && it == t.size() && cnt >= 0) return cnt;
if (is > s.size() || it > t.size()) return inf;
if (dp[is][it][cnt] != -1) return dp[is][it][cnt];
int res = inf;
res = min(res, 1 + f(is, it, cnt + 1));
if (is < s.size() && it < t.size() && s[is] == t[it])
res = min(res, 1 + f(is + 1, it + 1, cnt + (s[is] == '(' ? 1 : -1)));
else {
if (is < s.size())
res = min(res, 1 + f(is + 1, it, cnt + (s[is] == '(' ? 1 : -1)));
if (it < t.size())
res = min(res, 1 + f(is, it + 1, cnt + (t[it] == '(' ? 1 : -1)));
}
return dp[is][it][cnt] = res;
}
void b(int is, int it, int cnt) {
if (is == s.size() && it == t.size() && cnt >= 0) {
for (int i = 0; i < cnt; i++) ans.push_back(')');
return;
}
int res = f(is, it, cnt);
if (res == 1 + f(is, it, cnt + 1)) {
ans.push_back('(');
b(is, it, cnt + 1);
return;
}
if (is < s.size() && it < t.size() && s[is] == t[it]) {
if (res == 1 + f(is + 1, it + 1, cnt + (s[is] == '(' ? 1 : -1))) {
ans.push_back(s[is]);
b(is + 1, it + 1, cnt + (s[is] == '(' ? 1 : -1));
return;
}
} else {
if (is < s.size() &&
res == 1 + f(is + 1, it, cnt + (s[is] == '(' ? 1 : -1))) {
ans.push_back(s[is]);
b(is + 1, it, cnt + (s[is] == '(' ? 1 : -1));
return;
}
if (it < t.size() &&
res == 1 + f(is, it + 1, cnt + (t[it] == '(' ? 1 : -1))) {
ans.push_back(t[it]);
b(is, it + 1, cnt + (t[it] == '(' ? 1 : -1));
return;
}
}
return;
}
int main() {
memset(dp, -1, sizeof(dp));
cin >> s >> t;
f(0, 0, 0);
b(0, 0, 0);
cout << ans << endl;
return 0;
}
| ### Prompt
Your challenge is to write a Cpp solution to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9 + 69;
int dp[205][205][405];
string s, t, ans;
int f(int is, int it, int cnt) {
if (cnt < 0 || cnt >= s.size() + t.size()) return inf;
if (is == s.size() && it == t.size() && cnt >= 0) return cnt;
if (is > s.size() || it > t.size()) return inf;
if (dp[is][it][cnt] != -1) return dp[is][it][cnt];
int res = inf;
res = min(res, 1 + f(is, it, cnt + 1));
if (is < s.size() && it < t.size() && s[is] == t[it])
res = min(res, 1 + f(is + 1, it + 1, cnt + (s[is] == '(' ? 1 : -1)));
else {
if (is < s.size())
res = min(res, 1 + f(is + 1, it, cnt + (s[is] == '(' ? 1 : -1)));
if (it < t.size())
res = min(res, 1 + f(is, it + 1, cnt + (t[it] == '(' ? 1 : -1)));
}
return dp[is][it][cnt] = res;
}
void b(int is, int it, int cnt) {
if (is == s.size() && it == t.size() && cnt >= 0) {
for (int i = 0; i < cnt; i++) ans.push_back(')');
return;
}
int res = f(is, it, cnt);
if (res == 1 + f(is, it, cnt + 1)) {
ans.push_back('(');
b(is, it, cnt + 1);
return;
}
if (is < s.size() && it < t.size() && s[is] == t[it]) {
if (res == 1 + f(is + 1, it + 1, cnt + (s[is] == '(' ? 1 : -1))) {
ans.push_back(s[is]);
b(is + 1, it + 1, cnt + (s[is] == '(' ? 1 : -1));
return;
}
} else {
if (is < s.size() &&
res == 1 + f(is + 1, it, cnt + (s[is] == '(' ? 1 : -1))) {
ans.push_back(s[is]);
b(is + 1, it, cnt + (s[is] == '(' ? 1 : -1));
return;
}
if (it < t.size() &&
res == 1 + f(is, it + 1, cnt + (t[it] == '(' ? 1 : -1))) {
ans.push_back(t[it]);
b(is, it + 1, cnt + (t[it] == '(' ? 1 : -1));
return;
}
}
return;
}
int main() {
memset(dp, -1, sizeof(dp));
cin >> s >> t;
f(0, 0, 0);
b(0, 0, 0);
cout << ans << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int dp[201][201][201];
string s, t;
string Res = "";
int solve(int i, int j, int op) {
if (i == s.size() && j == t.size()) return op;
if (dp[i][j][op] != -1) return dp[i][j][op];
int x1 = 1e9, x2 = 1e9;
if (op == 0)
x1 = solve(i + (i + 1 <= s.size() && s[i] == '('),
j + (j + 1 <= t.size() && t[j] == '('), op + 1) +
1;
else if (i == s.size()) {
if (t[j] == '(')
x1 = solve(i, j + 1, op + 1) + 1;
else
x1 = solve(i, j + 1, op - 1) + 1;
} else if (j == t.size()) {
if (s[i] == '(')
x1 = solve(i + 1, j, op + 1) + 1;
else
x1 = solve(i + 1, j, op - 1) + 1;
} else {
if (s[i] == '(' && t[j] == '(')
x1 = solve(i + 1, j + 1, op + 1) + 1;
else if (s[i] == ')' && t[j] == ')')
x1 = solve(i + 1, j + 1, op - 1) + 1;
else {
x1 = solve(i + (s[i] == '('), j + (t[j] == '('), op + 1) + 1;
x2 = solve(i + (s[i] == ')'), j + (t[j] == ')'), op - 1) + 1;
}
}
return dp[i][j][op] = min(x1, x2);
}
void get_path(int i, int j, int op) {
if (i == s.size() && j == t.size()) return;
int optimal = dp[i][j][op];
int x1 = 1e9, x2 = 1e9;
if (op == 0) {
get_path(i + (i + 1 <= s.size() && s[i] == '('),
j + (j + 1 <= t.size() && t[j] == '('), op + 1);
Res += '(';
} else if (i == s.size()) {
if (t[j] == '(') {
get_path(i, j + 1, op + 1);
Res += '(';
} else {
get_path(i, j + 1, op - 1);
Res += ')';
}
} else if (j == t.size()) {
if (s[i] == '(') {
get_path(i + 1, j, op + 1);
Res += '(';
} else {
get_path(i + 1, j, op - 1);
Res += ')';
}
} else {
if (s[i] == '(' && t[j] == '(') {
get_path(i + 1, j + 1, op + 1);
Res += '(';
} else if (s[i] == ')' && t[j] == ')') {
get_path(i + 1, j + 1, op - 1);
Res += ')';
} else {
x1 = solve(i + (s[i] == '('), j + (t[j] == '('), op + 1) + 1;
x2 = solve(i + (s[i] == ')'), j + (t[j] == ')'), op - 1) + 1;
if (optimal == x1) {
get_path(i + (s[i] == '('), j + (t[j] == '('), op + 1);
Res += '(';
} else {
get_path(i + (s[i] == ')'), j + (t[j] == ')'), op - 1);
Res += ')';
}
}
}
}
int main() {
cin >> s >> t;
memset(dp, -1, sizeof dp);
solve(0, 0, 0);
get_path(0, 0, 0);
reverse(Res.begin(), Res.end());
int op = 0, cl = 0;
for (int i = 0; i < Res.size(); i++) {
if (Res[i] == '(')
op++;
else
cl++;
}
for (int i = 0; i < op - cl; i++) Res += ')';
cout << Res;
return 0;
}
| ### Prompt
Develop a solution in Cpp to the problem described below:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int dp[201][201][201];
string s, t;
string Res = "";
int solve(int i, int j, int op) {
if (i == s.size() && j == t.size()) return op;
if (dp[i][j][op] != -1) return dp[i][j][op];
int x1 = 1e9, x2 = 1e9;
if (op == 0)
x1 = solve(i + (i + 1 <= s.size() && s[i] == '('),
j + (j + 1 <= t.size() && t[j] == '('), op + 1) +
1;
else if (i == s.size()) {
if (t[j] == '(')
x1 = solve(i, j + 1, op + 1) + 1;
else
x1 = solve(i, j + 1, op - 1) + 1;
} else if (j == t.size()) {
if (s[i] == '(')
x1 = solve(i + 1, j, op + 1) + 1;
else
x1 = solve(i + 1, j, op - 1) + 1;
} else {
if (s[i] == '(' && t[j] == '(')
x1 = solve(i + 1, j + 1, op + 1) + 1;
else if (s[i] == ')' && t[j] == ')')
x1 = solve(i + 1, j + 1, op - 1) + 1;
else {
x1 = solve(i + (s[i] == '('), j + (t[j] == '('), op + 1) + 1;
x2 = solve(i + (s[i] == ')'), j + (t[j] == ')'), op - 1) + 1;
}
}
return dp[i][j][op] = min(x1, x2);
}
void get_path(int i, int j, int op) {
if (i == s.size() && j == t.size()) return;
int optimal = dp[i][j][op];
int x1 = 1e9, x2 = 1e9;
if (op == 0) {
get_path(i + (i + 1 <= s.size() && s[i] == '('),
j + (j + 1 <= t.size() && t[j] == '('), op + 1);
Res += '(';
} else if (i == s.size()) {
if (t[j] == '(') {
get_path(i, j + 1, op + 1);
Res += '(';
} else {
get_path(i, j + 1, op - 1);
Res += ')';
}
} else if (j == t.size()) {
if (s[i] == '(') {
get_path(i + 1, j, op + 1);
Res += '(';
} else {
get_path(i + 1, j, op - 1);
Res += ')';
}
} else {
if (s[i] == '(' && t[j] == '(') {
get_path(i + 1, j + 1, op + 1);
Res += '(';
} else if (s[i] == ')' && t[j] == ')') {
get_path(i + 1, j + 1, op - 1);
Res += ')';
} else {
x1 = solve(i + (s[i] == '('), j + (t[j] == '('), op + 1) + 1;
x2 = solve(i + (s[i] == ')'), j + (t[j] == ')'), op - 1) + 1;
if (optimal == x1) {
get_path(i + (s[i] == '('), j + (t[j] == '('), op + 1);
Res += '(';
} else {
get_path(i + (s[i] == ')'), j + (t[j] == ')'), op - 1);
Res += ')';
}
}
}
}
int main() {
cin >> s >> t;
memset(dp, -1, sizeof dp);
solve(0, 0, 0);
get_path(0, 0, 0);
reverse(Res.begin(), Res.end());
int op = 0, cl = 0;
for (int i = 0; i < Res.size(); i++) {
if (Res[i] == '(')
op++;
else
cl++;
}
for (int i = 0; i < op - cl; i++) Res += ')';
cout << Res;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
string s, t;
cin >> s >> t;
int s1 = s.size(), t1 = t.size();
const int INF = 1000000000;
int DP[s1 + 1][t1 + 1][s1 + t1];
int P[s1 + 1][t1 + 1][s1 + t1][3];
for (int i = 0; i <= s1; i++)
for (int j = 0; j <= t1; j++)
for (int k = 0; k < s1 + t1; k++) DP[i][j][k] = INF;
DP[0][0][0] = 0;
for (int i = 0; i <= s1; i++)
for (int j = 0; j <= t1; j++) {
for (int k = 0; k < t1 + s1 - 1; k++) {
int ti = i + (i < s1 && s[i] == '('), tj = j + (j < t1 && t[j] == '(');
if (DP[ti][tj][k + 1] > DP[i][j][k] + 1) {
DP[ti][tj][k + 1] = DP[i][j][k] + 1;
P[ti][tj][k + 1][0] = i;
P[ti][tj][k + 1][1] = j;
P[ti][tj][k + 1][2] = k;
}
}
for (int k = t1 + s1 - 1; k >= 1; k--) {
int ti = i + (i < s1 && s[i] == ')'), tj = j + (j < t1 && t[j] == ')');
if (DP[ti][tj][k - 1] > DP[i][j][k] + 1) {
DP[ti][tj][k - 1] = DP[i][j][k] + 1;
P[ti][tj][k - 1][0] = i;
P[ti][tj][k - 1][1] = j;
P[ti][tj][k - 1][2] = k;
}
}
}
string r = "";
int a = s1, b = t1, c = 0;
while (a || b || c) {
int nA = P[a][b][c][0], nB = P[a][b][c][1], nC = P[a][b][c][2];
r = (nC < c ? "(" : ")") + r;
a = nA;
b = nB;
c = nC;
}
cout << r << endl;
}
| ### Prompt
Construct a Cpp code solution to the problem outlined:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
string s, t;
cin >> s >> t;
int s1 = s.size(), t1 = t.size();
const int INF = 1000000000;
int DP[s1 + 1][t1 + 1][s1 + t1];
int P[s1 + 1][t1 + 1][s1 + t1][3];
for (int i = 0; i <= s1; i++)
for (int j = 0; j <= t1; j++)
for (int k = 0; k < s1 + t1; k++) DP[i][j][k] = INF;
DP[0][0][0] = 0;
for (int i = 0; i <= s1; i++)
for (int j = 0; j <= t1; j++) {
for (int k = 0; k < t1 + s1 - 1; k++) {
int ti = i + (i < s1 && s[i] == '('), tj = j + (j < t1 && t[j] == '(');
if (DP[ti][tj][k + 1] > DP[i][j][k] + 1) {
DP[ti][tj][k + 1] = DP[i][j][k] + 1;
P[ti][tj][k + 1][0] = i;
P[ti][tj][k + 1][1] = j;
P[ti][tj][k + 1][2] = k;
}
}
for (int k = t1 + s1 - 1; k >= 1; k--) {
int ti = i + (i < s1 && s[i] == ')'), tj = j + (j < t1 && t[j] == ')');
if (DP[ti][tj][k - 1] > DP[i][j][k] + 1) {
DP[ti][tj][k - 1] = DP[i][j][k] + 1;
P[ti][tj][k - 1][0] = i;
P[ti][tj][k - 1][1] = j;
P[ti][tj][k - 1][2] = k;
}
}
}
string r = "";
int a = s1, b = t1, c = 0;
while (a || b || c) {
int nA = P[a][b][c][0], nB = P[a][b][c][1], nC = P[a][b][c][2];
r = (nC < c ? "(" : ")") + r;
a = nA;
b = nB;
c = nC;
}
cout << r << endl;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long memo[212][212][312];
string a, b;
long long pd(long long ap, long long bp, long long first) {
if (ap == a.size() && bp == b.size() && first == 0) return 0;
long long &resp = memo[ap][bp][first];
if (resp == -1) {
resp = LLONG_MAX;
long long aa = 0, bb = 0;
if (ap < a.size() && a[ap] == '(') aa++;
if (bp < b.size() && b[bp] == '(') bb++;
if (first == 0 || aa != 0 || bb != 0)
resp = 1 + pd(ap + aa, bp + bb, first + 1);
if (first != 0) {
long long aa = 0, bb = 0;
if (ap < a.size() && a[ap] == ')') aa++;
if (bp < b.size() && b[bp] == ')') bb++;
resp = min(resp, 1 + pd(ap + aa, bp + bb, first - 1));
}
}
return resp;
}
int32_t main() {
ios::sync_with_stdio(false);
cin.tie(NULL);
cin >> a >> b;
for (long long i = 0; i < 212; i++) {
for (long long j = 0; j < 212; j++) {
for (long long k = 0; k < 312; k++) {
memo[i][j][k] = -1;
}
}
}
pd(0, 0, 0);
long long ap = 0, bp = 0, first = 0;
while (!(ap == a.size() && bp == b.size() && first == 0)) {
char agora;
long long aa = 0, bb = 0;
if (ap < a.size() && a[ap] == '(') aa++;
if (bp < b.size() && b[bp] == '(') bb++;
if (first == 0 || aa != 0 || bb != 0) {
if (pd(ap, bp, first) == 1 + pd(ap + aa, bp + bb, first + 1)) {
cout << '(';
ap = ap + aa;
bp = bp + bb;
first = first + 1;
continue;
}
}
if (first != 0) {
long long aa = 0, bb = 0;
if (ap < a.size() && a[ap] == ')') aa++;
if (bp < b.size() && b[bp] == ')') bb++;
if (pd(ap, bp, first) == 1 + pd(ap + aa, bp + bb, first - 1)) {
cout << ')';
ap = ap + aa;
bp = bp + bb;
first = first - 1;
continue;
}
}
}
cout << endl;
return 0;
}
| ### Prompt
In Cpp, your task is to solve the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long memo[212][212][312];
string a, b;
long long pd(long long ap, long long bp, long long first) {
if (ap == a.size() && bp == b.size() && first == 0) return 0;
long long &resp = memo[ap][bp][first];
if (resp == -1) {
resp = LLONG_MAX;
long long aa = 0, bb = 0;
if (ap < a.size() && a[ap] == '(') aa++;
if (bp < b.size() && b[bp] == '(') bb++;
if (first == 0 || aa != 0 || bb != 0)
resp = 1 + pd(ap + aa, bp + bb, first + 1);
if (first != 0) {
long long aa = 0, bb = 0;
if (ap < a.size() && a[ap] == ')') aa++;
if (bp < b.size() && b[bp] == ')') bb++;
resp = min(resp, 1 + pd(ap + aa, bp + bb, first - 1));
}
}
return resp;
}
int32_t main() {
ios::sync_with_stdio(false);
cin.tie(NULL);
cin >> a >> b;
for (long long i = 0; i < 212; i++) {
for (long long j = 0; j < 212; j++) {
for (long long k = 0; k < 312; k++) {
memo[i][j][k] = -1;
}
}
}
pd(0, 0, 0);
long long ap = 0, bp = 0, first = 0;
while (!(ap == a.size() && bp == b.size() && first == 0)) {
char agora;
long long aa = 0, bb = 0;
if (ap < a.size() && a[ap] == '(') aa++;
if (bp < b.size() && b[bp] == '(') bb++;
if (first == 0 || aa != 0 || bb != 0) {
if (pd(ap, bp, first) == 1 + pd(ap + aa, bp + bb, first + 1)) {
cout << '(';
ap = ap + aa;
bp = bp + bb;
first = first + 1;
continue;
}
}
if (first != 0) {
long long aa = 0, bb = 0;
if (ap < a.size() && a[ap] == ')') aa++;
if (bp < b.size() && b[bp] == ')') bb++;
if (pd(ap, bp, first) == 1 + pd(ap + aa, bp + bb, first - 1)) {
cout << ')';
ap = ap + aa;
bp = bp + bb;
first = first - 1;
continue;
}
}
}
cout << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
int pwr(int a, int b) {
int ans = 1;
while (b) {
if (b & 1) ans = (ans * 1LL * a) % mod;
a = (a * 1LL * a) % mod;
b >>= 1;
}
return ans;
}
const int N = 205;
const int M = 405;
int dp[N][N][M];
pair<pair<int, int>, int> par[N][N][M];
int main() {
ios_base::sync_with_stdio(0);
cin.tie(NULL);
cout.tie(NULL);
string s, t;
cin >> s >> t;
int n = (int)(s).size(), m = (int)(t).size();
s = '0' + s;
t = '0' + t;
s += "#";
t += "#";
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= m; j++) {
for (int k = 0; k <= M - 1; k++) {
dp[i][j][k] = 1e9;
}
}
}
dp[0][0][0] = 0;
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= m; j++) {
for (int k = 0; k <= M - 1; k++) {
if (dp[i][j][k] == 1e9) continue;
if (k + 1 < M) {
if (dp[i + (s[i + 1] == '(')][j + (t[j + 1] == '(')][k + 1] >
1 + dp[i][j][k]) {
dp[i + (s[i + 1] == '(')][j + (t[j + 1] == '(')][k + 1] =
1 + dp[i][j][k];
par[i + (s[i + 1] == '(')][j + (t[j + 1] == '(')][k + 1] = {{i, j},
k};
}
}
if (k > 0) {
if (dp[i + (s[i + 1] == ')')][j + (t[j + 1] == ')')][k - 1] >
1 + dp[i][j][k]) {
dp[i + (s[i + 1] == ')')][j + (t[j + 1] == ')')][k - 1] =
1 + dp[i][j][k];
par[i + (s[i + 1] == ')')][j + (t[j + 1] == ')')][k - 1] = {{i, j},
k};
}
}
}
}
}
string ans = "";
int kk = 0;
for (int k = 1; k <= M - 1; k++) {
if (k + dp[n][m][k] < kk + dp[n][m][kk]) kk = k;
}
for (int i = 1; i <= kk; i++) {
ans += ")";
}
while (n || m || kk) {
auto x = par[n][m][kk];
if (x.second < kk)
ans += "(";
else
ans += ")";
n = x.first.first, m = x.first.second, kk = x.second;
}
reverse(ans.begin(), ans.end());
cout << ans;
return 0;
}
| ### Prompt
Construct a CPP code solution to the problem outlined:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
int pwr(int a, int b) {
int ans = 1;
while (b) {
if (b & 1) ans = (ans * 1LL * a) % mod;
a = (a * 1LL * a) % mod;
b >>= 1;
}
return ans;
}
const int N = 205;
const int M = 405;
int dp[N][N][M];
pair<pair<int, int>, int> par[N][N][M];
int main() {
ios_base::sync_with_stdio(0);
cin.tie(NULL);
cout.tie(NULL);
string s, t;
cin >> s >> t;
int n = (int)(s).size(), m = (int)(t).size();
s = '0' + s;
t = '0' + t;
s += "#";
t += "#";
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= m; j++) {
for (int k = 0; k <= M - 1; k++) {
dp[i][j][k] = 1e9;
}
}
}
dp[0][0][0] = 0;
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= m; j++) {
for (int k = 0; k <= M - 1; k++) {
if (dp[i][j][k] == 1e9) continue;
if (k + 1 < M) {
if (dp[i + (s[i + 1] == '(')][j + (t[j + 1] == '(')][k + 1] >
1 + dp[i][j][k]) {
dp[i + (s[i + 1] == '(')][j + (t[j + 1] == '(')][k + 1] =
1 + dp[i][j][k];
par[i + (s[i + 1] == '(')][j + (t[j + 1] == '(')][k + 1] = {{i, j},
k};
}
}
if (k > 0) {
if (dp[i + (s[i + 1] == ')')][j + (t[j + 1] == ')')][k - 1] >
1 + dp[i][j][k]) {
dp[i + (s[i + 1] == ')')][j + (t[j + 1] == ')')][k - 1] =
1 + dp[i][j][k];
par[i + (s[i + 1] == ')')][j + (t[j + 1] == ')')][k - 1] = {{i, j},
k};
}
}
}
}
}
string ans = "";
int kk = 0;
for (int k = 1; k <= M - 1; k++) {
if (k + dp[n][m][k] < kk + dp[n][m][kk]) kk = k;
}
for (int i = 1; i <= kk; i++) {
ans += ")";
}
while (n || m || kk) {
auto x = par[n][m][kk];
if (x.second < kk)
ans += "(";
else
ans += ")";
n = x.first.first, m = x.first.second, kk = x.second;
}
reverse(ans.begin(), ans.end());
cout << ans;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long power_mod(long long num, long long g) {
if (g == 0) return 1;
if (g % 2 == 1)
return (num * power_mod((num * num) % 1000000007, g / 2)) % 1000000007;
return power_mod((num * num) % 1000000007, g / 2);
}
long long power(long long num, long long g) {
if (g == 0) return 1;
if (g % 2 == 1) return (num * power((num * num), g / 2));
return power((num * num), g / 2);
}
long long a[205][205][205];
long long l, m;
string s, t;
long long solve(long long x, long long y, long long z) {
if (a[x][y][z] > -1) return a[x][y][z];
if (x == l && y == m) return a[x][y][z] = z;
if (x == l) {
if (t[y] == '(') return a[x][y][z] = solve(x, y + 1, z + 1) + 1;
if (z > 0) return a[x][y][z] = solve(x, y + 1, z - 1) + 1;
return a[x][y][z] = solve(x, y + 1, z) + 2;
}
if (y == m) {
if (s[x] == '(') return a[x][y][z] = solve(x + 1, y, z + 1) + 1;
if (z > 0) return a[x][y][z] = solve(x + 1, y, z - 1) + 1;
return a[x][y][z] = solve(x + 1, y, z) + 2;
}
if (s[x] == t[y]) {
if (s[x] == '(') return a[x][y][z] = solve(x + 1, y + 1, z + 1) + 1;
if (z > 0) return a[x][y][z] = solve(x + 1, y + 1, z - 1) + 1;
return a[x][y][z] = solve(x + 1, y + 1, z) + 2;
}
if (z > 0) {
if (s[x] == '(') {
return a[x][y][z] =
min(solve(x + 1, y, z + 1) + 1, solve(x, y + 1, z - 1) + 1);
} else {
return a[x][y][z] =
min(solve(x + 1, y, z - 1) + 1, solve(x, y + 1, z + 1) + 1);
}
}
if (s[x] == '(') return a[x][y][z] = solve(x + 1, y, z + 1) + 1;
return a[x][y][z] = solve(x, y + 1, z + 1) + 1;
}
void solve1(long long x, long long y, long long z) {
if (x == l && y == m && z == 0) return;
if (x == l && y == m) {
cout << ")";
solve1(x, y, z - 1);
return;
}
if (x == l) {
if (t[y] == '(') {
cout << "(";
solve1(x, y + 1, z + 1);
return;
}
if (z > 0) {
cout << ")";
solve1(x, y + 1, z - 1);
return;
}
cout << "()";
solve1(x, y + 1, z);
return;
}
if (y == m) {
if (s[x] == '(') {
cout << "(";
solve1(x + 1, y, z + 1);
return;
}
if (z > 0) {
cout << ")";
solve1(x + 1, y, z - 1);
return;
}
cout << "()";
solve1(x + 1, y, z);
return;
}
if (s[x] == t[y]) {
if (t[y] == '(') {
cout << "(";
solve1(x + 1, y + 1, z + 1);
return;
}
if (z > 0) {
cout << ")";
solve1(x + 1, y + 1, z - 1);
return;
}
cout << "()";
solve1(x + 1, y + 1, z);
return;
}
if (z > 0) {
if (s[x] == '(') {
if (a[x + 1][y][z + 1] < a[x][y + 1][z - 1]) {
cout << "(";
solve1(x + 1, y, z + 1);
} else {
cout << ")";
solve1(x, y + 1, z - 1);
}
return;
} else {
if (a[x + 1][y][z - 1] < a[x][y + 1][z + 1]) {
cout << ")";
solve1(x + 1, y, z - 1);
} else {
cout << "(";
solve1(x, y + 1, z + 1);
}
return;
}
}
cout << "(";
if (s[x] == '(') {
solve1(x + 1, y, z + 1);
} else
solve1(x, y + 1, z + 1);
}
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
cin >> s >> t;
l = s.size();
m = t.size();
for (long long i = 0; i < l + 1; i++)
for (long long j = 0; j < m + 1; j++)
for (long long k = 0; k < max(l, m) + 1; k++) a[i][j][k] = -1;
solve(0, 0, 0);
solve1(0, 0, 0);
return 0;
}
| ### Prompt
Generate a cpp solution to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long power_mod(long long num, long long g) {
if (g == 0) return 1;
if (g % 2 == 1)
return (num * power_mod((num * num) % 1000000007, g / 2)) % 1000000007;
return power_mod((num * num) % 1000000007, g / 2);
}
long long power(long long num, long long g) {
if (g == 0) return 1;
if (g % 2 == 1) return (num * power((num * num), g / 2));
return power((num * num), g / 2);
}
long long a[205][205][205];
long long l, m;
string s, t;
long long solve(long long x, long long y, long long z) {
if (a[x][y][z] > -1) return a[x][y][z];
if (x == l && y == m) return a[x][y][z] = z;
if (x == l) {
if (t[y] == '(') return a[x][y][z] = solve(x, y + 1, z + 1) + 1;
if (z > 0) return a[x][y][z] = solve(x, y + 1, z - 1) + 1;
return a[x][y][z] = solve(x, y + 1, z) + 2;
}
if (y == m) {
if (s[x] == '(') return a[x][y][z] = solve(x + 1, y, z + 1) + 1;
if (z > 0) return a[x][y][z] = solve(x + 1, y, z - 1) + 1;
return a[x][y][z] = solve(x + 1, y, z) + 2;
}
if (s[x] == t[y]) {
if (s[x] == '(') return a[x][y][z] = solve(x + 1, y + 1, z + 1) + 1;
if (z > 0) return a[x][y][z] = solve(x + 1, y + 1, z - 1) + 1;
return a[x][y][z] = solve(x + 1, y + 1, z) + 2;
}
if (z > 0) {
if (s[x] == '(') {
return a[x][y][z] =
min(solve(x + 1, y, z + 1) + 1, solve(x, y + 1, z - 1) + 1);
} else {
return a[x][y][z] =
min(solve(x + 1, y, z - 1) + 1, solve(x, y + 1, z + 1) + 1);
}
}
if (s[x] == '(') return a[x][y][z] = solve(x + 1, y, z + 1) + 1;
return a[x][y][z] = solve(x, y + 1, z + 1) + 1;
}
void solve1(long long x, long long y, long long z) {
if (x == l && y == m && z == 0) return;
if (x == l && y == m) {
cout << ")";
solve1(x, y, z - 1);
return;
}
if (x == l) {
if (t[y] == '(') {
cout << "(";
solve1(x, y + 1, z + 1);
return;
}
if (z > 0) {
cout << ")";
solve1(x, y + 1, z - 1);
return;
}
cout << "()";
solve1(x, y + 1, z);
return;
}
if (y == m) {
if (s[x] == '(') {
cout << "(";
solve1(x + 1, y, z + 1);
return;
}
if (z > 0) {
cout << ")";
solve1(x + 1, y, z - 1);
return;
}
cout << "()";
solve1(x + 1, y, z);
return;
}
if (s[x] == t[y]) {
if (t[y] == '(') {
cout << "(";
solve1(x + 1, y + 1, z + 1);
return;
}
if (z > 0) {
cout << ")";
solve1(x + 1, y + 1, z - 1);
return;
}
cout << "()";
solve1(x + 1, y + 1, z);
return;
}
if (z > 0) {
if (s[x] == '(') {
if (a[x + 1][y][z + 1] < a[x][y + 1][z - 1]) {
cout << "(";
solve1(x + 1, y, z + 1);
} else {
cout << ")";
solve1(x, y + 1, z - 1);
}
return;
} else {
if (a[x + 1][y][z - 1] < a[x][y + 1][z + 1]) {
cout << ")";
solve1(x + 1, y, z - 1);
} else {
cout << "(";
solve1(x, y + 1, z + 1);
}
return;
}
}
cout << "(";
if (s[x] == '(') {
solve1(x + 1, y, z + 1);
} else
solve1(x, y + 1, z + 1);
}
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
cin >> s >> t;
l = s.size();
m = t.size();
for (long long i = 0; i < l + 1; i++)
for (long long j = 0; j < m + 1; j++)
for (long long k = 0; k < max(l, m) + 1; k++) a[i][j][k] = -1;
solve(0, 0, 0);
solve1(0, 0, 0);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 205;
const int INF = (int)1e5 + 5;
int dp[N][N][N];
string s, t;
int n, m;
int solve(int i, int j, int open) {
if (i == n && j == m) {
return open;
}
int &ans = dp[i][j][open];
if (ans != -1) return ans;
ans = INF;
if (i == n && j != m) {
int cost = 0;
for (int k = j; k < m; k++) {
if (t[k] == '(')
open++, cost++;
else {
if (open > 0)
open--, cost++;
else
cost += 2;
}
}
return ans = cost + open;
}
if (j == m && i != n) {
int cost = 0;
for (int k = i; k < n; k++) {
if (s[k] == '(')
open++, cost++;
else {
if (open > 0)
open--, cost++;
else
cost += 2;
}
}
return ans = cost + open;
}
ans = INF;
if (s[i] == t[j]) {
if (s[i] == '(')
ans = min(ans, 1 + solve(i + 1, j + 1, open + 1));
else {
if (open > 0)
ans = min(ans, 1 + solve(i + 1, j + 1, open - 1));
else
ans = min(ans, 2 + solve(i + 1, j + 1, 0));
}
} else {
if (s[i] == '(') {
ans = min(ans, 1 + solve(i + 1, j, open + 1));
if (open > 0) ans = min(ans, 1 + solve(i, j + 1, open - 1));
} else {
if (open > 0) ans = min(ans, 1 + solve(i + 1, j, open - 1));
ans = min(ans, 1 + solve(i, j + 1, open + 1));
}
}
return ans;
}
string res = "";
void build(int i, int j, int open) {
if (i == n && j == m) {
for (int k = 0; k < open; k++) res += ')';
return;
}
if (i == n && j != m) {
for (int k = j; k < m; k++) {
if (t[k] == '(') {
open++;
res += '(';
} else {
if (open > 0) {
res += ')';
open--;
} else
res += "()";
}
}
for (int k = 0; k < open; k++) res += ')';
return;
}
if (j == m && i != n) {
for (int k = i; k < n; k++) {
if (s[k] == '(') {
open++;
res += '(';
} else {
if (open > 0) {
res += ')';
open--;
} else
res += "()";
}
}
for (int k = 0; k < open; k++) res += ')';
return;
}
if (s[i] == t[j]) {
if (s[i] == '(') {
res += '(';
build(i + 1, j + 1, open + 1);
return;
} else {
if (open > 0) {
res += ')';
build(i + 1, j + 1, open - 1);
return;
} else {
res += "()";
build(i + 1, j + 1, 0);
return;
}
}
} else {
if (s[i] == '(') {
int ans1 = INF, ans2 = INF;
ans1 = 1 + solve(i + 1, j, open + 1);
if (open > 0) ans2 = 1 + solve(i, j + 1, open - 1);
if (ans1 < ans2) {
res += '(';
build(i + 1, j, open + 1);
return;
} else {
res += ')';
build(i, j + 1, open - 1);
return;
}
} else {
int ans1 = INF, ans2 = INF;
if (open > 0) ans1 = 1 + solve(i + 1, j, open - 1);
ans2 = 1 + solve(i, j + 1, open + 1);
if (ans1 < ans2) {
res += ')';
build(i + 1, j, open - 1);
return;
} else {
res += '(';
build(i, j + 1, open + 1);
return;
}
}
}
}
int32_t main() {
ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
cin >> s >> t;
n = s.size(), m = t.size();
memset(dp, -1, sizeof dp);
int ans = solve(0, 0, 0);
build(0, 0, 0);
cout << res << '\n';
return 0;
}
| ### Prompt
Generate a CPP solution to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 205;
const int INF = (int)1e5 + 5;
int dp[N][N][N];
string s, t;
int n, m;
int solve(int i, int j, int open) {
if (i == n && j == m) {
return open;
}
int &ans = dp[i][j][open];
if (ans != -1) return ans;
ans = INF;
if (i == n && j != m) {
int cost = 0;
for (int k = j; k < m; k++) {
if (t[k] == '(')
open++, cost++;
else {
if (open > 0)
open--, cost++;
else
cost += 2;
}
}
return ans = cost + open;
}
if (j == m && i != n) {
int cost = 0;
for (int k = i; k < n; k++) {
if (s[k] == '(')
open++, cost++;
else {
if (open > 0)
open--, cost++;
else
cost += 2;
}
}
return ans = cost + open;
}
ans = INF;
if (s[i] == t[j]) {
if (s[i] == '(')
ans = min(ans, 1 + solve(i + 1, j + 1, open + 1));
else {
if (open > 0)
ans = min(ans, 1 + solve(i + 1, j + 1, open - 1));
else
ans = min(ans, 2 + solve(i + 1, j + 1, 0));
}
} else {
if (s[i] == '(') {
ans = min(ans, 1 + solve(i + 1, j, open + 1));
if (open > 0) ans = min(ans, 1 + solve(i, j + 1, open - 1));
} else {
if (open > 0) ans = min(ans, 1 + solve(i + 1, j, open - 1));
ans = min(ans, 1 + solve(i, j + 1, open + 1));
}
}
return ans;
}
string res = "";
void build(int i, int j, int open) {
if (i == n && j == m) {
for (int k = 0; k < open; k++) res += ')';
return;
}
if (i == n && j != m) {
for (int k = j; k < m; k++) {
if (t[k] == '(') {
open++;
res += '(';
} else {
if (open > 0) {
res += ')';
open--;
} else
res += "()";
}
}
for (int k = 0; k < open; k++) res += ')';
return;
}
if (j == m && i != n) {
for (int k = i; k < n; k++) {
if (s[k] == '(') {
open++;
res += '(';
} else {
if (open > 0) {
res += ')';
open--;
} else
res += "()";
}
}
for (int k = 0; k < open; k++) res += ')';
return;
}
if (s[i] == t[j]) {
if (s[i] == '(') {
res += '(';
build(i + 1, j + 1, open + 1);
return;
} else {
if (open > 0) {
res += ')';
build(i + 1, j + 1, open - 1);
return;
} else {
res += "()";
build(i + 1, j + 1, 0);
return;
}
}
} else {
if (s[i] == '(') {
int ans1 = INF, ans2 = INF;
ans1 = 1 + solve(i + 1, j, open + 1);
if (open > 0) ans2 = 1 + solve(i, j + 1, open - 1);
if (ans1 < ans2) {
res += '(';
build(i + 1, j, open + 1);
return;
} else {
res += ')';
build(i, j + 1, open - 1);
return;
}
} else {
int ans1 = INF, ans2 = INF;
if (open > 0) ans1 = 1 + solve(i + 1, j, open - 1);
ans2 = 1 + solve(i, j + 1, open + 1);
if (ans1 < ans2) {
res += ')';
build(i + 1, j, open - 1);
return;
} else {
res += '(';
build(i, j + 1, open + 1);
return;
}
}
}
}
int32_t main() {
ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
cin >> s >> t;
n = s.size(), m = t.size();
memset(dp, -1, sizeof dp);
int ans = solve(0, 0, 0);
build(0, 0, 0);
cout << res << '\n';
return 0;
}
``` |
#include <bits/stdc++.h>
const long long inf = 0x3f3f3f3f3f3f3f3LL;
const long long mod = (long long)1e9 + 7;
using namespace std;
template <class T>
void smin(T& a, T val) {
if (a > val) a = val;
}
template <class T>
void smax(T& a, T val) {
if (a < val) a = val;
}
const long long N = 5 * (long long)1e5 + 10;
string s, t;
long long n, m;
long long dp[205][205][205];
vector<char> v;
long long solve(long long idx1, long long idx2, long long balance) {
if (balance > max(n, m) or balance < 0) return inf;
if (idx1 >= n and idx2 >= m and balance == 0) {
return 0;
}
long long& ans = dp[idx1][idx2][balance];
if (ans != -1) {
return ans;
}
ans = inf;
ans =
1 + solve(idx1 + (s[idx1] == '('), idx2 + (t[idx2] == '('), balance + 1);
smin(ans, 1 + solve(idx1 + (s[idx1] == ')'), idx2 + (t[idx2] == ')'),
balance - 1));
return ans;
}
string vals = "";
void trace(long long idx1, long long idx2, long long balance) {
if (idx1 >= n and idx2 >= m and balance == 0) {
cout << vals;
exit(0);
}
long long& ans = dp[idx1][idx2][balance];
if (ans == 1 + solve(idx1 + (s[idx1] == '('), idx2 + (t[idx2] == '('),
balance + 1)) {
vals += "(";
trace(idx1 + (s[idx1] == '('), idx2 + (t[idx2] == '('), balance + 1);
return;
}
vals += ")";
trace(idx1 + (s[idx1] == ')'), idx2 + (t[idx2] == ')'), balance - 1);
}
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
cin >> s >> t;
n = (long long)(s.size()), m = (long long)(t.size());
s = s + '#';
t = t + '#';
memset(dp, -1, sizeof dp);
solve(0, 0, 0);
trace(0, 0, 0);
return 0;
}
| ### Prompt
Your challenge is to write a Cpp solution to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
const long long inf = 0x3f3f3f3f3f3f3f3LL;
const long long mod = (long long)1e9 + 7;
using namespace std;
template <class T>
void smin(T& a, T val) {
if (a > val) a = val;
}
template <class T>
void smax(T& a, T val) {
if (a < val) a = val;
}
const long long N = 5 * (long long)1e5 + 10;
string s, t;
long long n, m;
long long dp[205][205][205];
vector<char> v;
long long solve(long long idx1, long long idx2, long long balance) {
if (balance > max(n, m) or balance < 0) return inf;
if (idx1 >= n and idx2 >= m and balance == 0) {
return 0;
}
long long& ans = dp[idx1][idx2][balance];
if (ans != -1) {
return ans;
}
ans = inf;
ans =
1 + solve(idx1 + (s[idx1] == '('), idx2 + (t[idx2] == '('), balance + 1);
smin(ans, 1 + solve(idx1 + (s[idx1] == ')'), idx2 + (t[idx2] == ')'),
balance - 1));
return ans;
}
string vals = "";
void trace(long long idx1, long long idx2, long long balance) {
if (idx1 >= n and idx2 >= m and balance == 0) {
cout << vals;
exit(0);
}
long long& ans = dp[idx1][idx2][balance];
if (ans == 1 + solve(idx1 + (s[idx1] == '('), idx2 + (t[idx2] == '('),
balance + 1)) {
vals += "(";
trace(idx1 + (s[idx1] == '('), idx2 + (t[idx2] == '('), balance + 1);
return;
}
vals += ")";
trace(idx1 + (s[idx1] == ')'), idx2 + (t[idx2] == ')'), balance - 1);
}
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
cin >> s >> t;
n = (long long)(s.size()), m = (long long)(t.size());
s = s + '#';
t = t + '#';
memset(dp, -1, sizeof dp);
solve(0, 0, 0);
trace(0, 0, 0);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 203;
const int inf = 1e9;
inline int read() {
int p = 0;
int f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-') f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9') {
p = p * 10 + ch - '0';
ch = getchar();
}
return p * f;
}
char a[N], b[N];
struct node {
int i, j, k;
node() {}
node(int _i, int _j, int _k) {
i = _i;
j = _j;
k = _k;
}
};
node g[N][N][N];
char ch[N][N][N];
int nw = 0;
bool f[N][N][N];
queue<node> q;
vector<char> ans;
int main() {
scanf("%s%s", a + 1, b + 1);
int alen = strlen(a + 1);
int blen = strlen(b + 1);
q.push({0, 0, 0});
memset(f, 0, sizeof(f));
f[0][0][0] = 1;
while (!q.empty()) {
int i = q.front().i;
int j = q.front().j;
int k = q.front().k;
q.pop();
if (k < alen + blen) {
int x = i + (a[i + 1] == '(');
int y = j + (b[j + 1] == '(');
int z = k + 1;
if (!f[x][y][z]) {
f[x][y][z] = 1;
q.push(node(x, y, z));
ch[x][y][z] = '(';
g[x][y][z] = node(i, j, k);
}
}
if (k) {
int x = i + (a[i + 1] == ')');
int y = j + (b[j + 1] == ')');
int z = k - 1;
if (!f[x][y][z]) {
f[x][y][z] = 1;
q.push(node(x, y, z));
ch[x][y][z] = ')';
g[x][y][z] = node(i, j, k);
}
}
if (i == alen && j == blen && k == 0) {
while (i || j || k) {
ans.push_back(ch[i][j][k]);
int x = g[i][j][k].i;
int y = g[i][j][k].j;
int z = g[i][j][k].k;
i = x;
j = y;
k = z;
}
}
}
for (int i = ans.size() - 1; i >= 0; i--) printf("%c", ans[i]);
printf("\n");
return 0;
}
| ### Prompt
Generate a Cpp solution to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 203;
const int inf = 1e9;
inline int read() {
int p = 0;
int f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-') f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9') {
p = p * 10 + ch - '0';
ch = getchar();
}
return p * f;
}
char a[N], b[N];
struct node {
int i, j, k;
node() {}
node(int _i, int _j, int _k) {
i = _i;
j = _j;
k = _k;
}
};
node g[N][N][N];
char ch[N][N][N];
int nw = 0;
bool f[N][N][N];
queue<node> q;
vector<char> ans;
int main() {
scanf("%s%s", a + 1, b + 1);
int alen = strlen(a + 1);
int blen = strlen(b + 1);
q.push({0, 0, 0});
memset(f, 0, sizeof(f));
f[0][0][0] = 1;
while (!q.empty()) {
int i = q.front().i;
int j = q.front().j;
int k = q.front().k;
q.pop();
if (k < alen + blen) {
int x = i + (a[i + 1] == '(');
int y = j + (b[j + 1] == '(');
int z = k + 1;
if (!f[x][y][z]) {
f[x][y][z] = 1;
q.push(node(x, y, z));
ch[x][y][z] = '(';
g[x][y][z] = node(i, j, k);
}
}
if (k) {
int x = i + (a[i + 1] == ')');
int y = j + (b[j + 1] == ')');
int z = k - 1;
if (!f[x][y][z]) {
f[x][y][z] = 1;
q.push(node(x, y, z));
ch[x][y][z] = ')';
g[x][y][z] = node(i, j, k);
}
}
if (i == alen && j == blen && k == 0) {
while (i || j || k) {
ans.push_back(ch[i][j][k]);
int x = g[i][j][k].i;
int y = g[i][j][k].j;
int z = g[i][j][k].k;
i = x;
j = y;
k = z;
}
}
}
for (int i = ans.size() - 1; i >= 0; i--) printf("%c", ans[i]);
printf("\n");
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
struct node {
int i, j, k;
};
ostream& operator<<(ostream& dout, node& x) {
dout << x.i << " " << x.j << " " << x.k << "\n";
return dout;
}
int main() {
string s, t;
cin >> s >> t;
int m = s.length(), n = t.length(), i, j, k;
char dp[m + 1][n + 1][max(m, n) + 1];
node dp1[m + 1][n + 1][max(m, n) + 1];
for (i = 0; i <= m; i++)
for (j = 0; j <= n; j++)
for (k = 0; k <= max(m, n); k++) dp[i][j][k] = 'n';
dp[0][0][0] = 's';
queue<node> q;
node nd0 = {0, 0, 0}, nd1;
q.push(nd0);
char ch1;
while (!q.empty()) {
nd0 = q.front();
i = nd0.i, j = nd0.j, k = nd0.k;
q.pop();
nd1 = nd0;
nd1.k++;
if (i + 1 <= m && s[i + 1 - 1] == '(') nd1.i++;
if (nd1.i <= m && nd1.j <= n && nd1.k >= 0 && nd1.k <= max(m, n) &&
dp[nd1.i][nd1.j][nd1.k] == 'n') {
dp[nd1.i][nd1.j][nd1.k] = '(';
dp1[nd1.i][nd1.j][nd1.k] = nd0;
q.push(nd1);
}
nd1 = nd0;
nd1.k++;
if (j + 1 <= n && t[j + 1 - 1] == '(') nd1.j++;
if (nd1.i <= m && nd1.j <= n && nd1.k >= 0 && nd1.k <= max(m, n) &&
dp[nd1.i][nd1.j][nd1.k] == 'n') {
dp[nd1.i][nd1.j][nd1.k] = '(';
dp1[nd1.i][nd1.j][nd1.k] = nd0;
q.push(nd1);
}
nd1 = nd0;
nd1.k++;
if (i + 1 <= m && s[i + 1 - 1] == '(' && j + 1 <= n && t[j + 1 - 1] == '(')
nd1.i++, nd1.j++;
if (nd1.i <= m && nd1.j <= n && nd1.k >= 0 && nd1.k <= max(m, n) &&
dp[nd1.i][nd1.j][nd1.k] == 'n') {
dp[nd1.i][nd1.j][nd1.k] = '(';
dp1[nd1.i][nd1.j][nd1.k] = nd0;
q.push(nd1);
}
nd1 = nd0;
nd1.k--;
if (i + 1 <= m && s[i + 1 - 1] == ')') nd1.i++;
if (nd1.i <= m && nd1.j <= n && nd1.k >= 0 && nd1.k <= max(m, n) &&
dp[nd1.i][nd1.j][nd1.k] == 'n') {
dp[nd1.i][nd1.j][nd1.k] = ')';
dp1[nd1.i][nd1.j][nd1.k] = nd0;
q.push(nd1);
}
nd1 = nd0;
nd1.k--;
if (j + 1 <= n && t[j + 1 - 1] == ')') nd1.j++;
if (nd1.i <= m && nd1.j <= n && nd1.k >= 0 && nd1.k <= max(m, n) &&
dp[nd1.i][nd1.j][nd1.k] == 'n') {
dp[nd1.i][nd1.j][nd1.k] = ')';
dp1[nd1.i][nd1.j][nd1.k] = nd0;
q.push(nd1);
}
nd1 = nd0;
nd1.k--;
if (i + 1 <= m && s[i + 1 - 1] == ')' && j + 1 <= n && t[j + 1 - 1] == ')')
nd1.i++, nd1.j++;
if (nd1.i <= m && nd1.j <= n && nd1.k >= 0 && nd1.k <= max(m, n) &&
dp[nd1.i][nd1.j][nd1.k] == 'n') {
dp[nd1.i][nd1.j][nd1.k] = ')';
dp1[nd1.i][nd1.j][nd1.k] = nd0;
q.push(nd1);
}
}
string res = "";
char ch;
i = m, j = n, k = 0;
while (i > 0 || j > 0 || k > 0) {
res += dp[i][j][k];
int ti = dp1[i][j][k].i;
int tj = dp1[i][j][k].j;
int tk = dp1[i][j][k].k;
i = ti, j = tj, k = tk;
}
reverse(res.begin(), res.end());
cout << res;
}
| ### Prompt
Please provide a cpp coded solution to the problem described below:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
struct node {
int i, j, k;
};
ostream& operator<<(ostream& dout, node& x) {
dout << x.i << " " << x.j << " " << x.k << "\n";
return dout;
}
int main() {
string s, t;
cin >> s >> t;
int m = s.length(), n = t.length(), i, j, k;
char dp[m + 1][n + 1][max(m, n) + 1];
node dp1[m + 1][n + 1][max(m, n) + 1];
for (i = 0; i <= m; i++)
for (j = 0; j <= n; j++)
for (k = 0; k <= max(m, n); k++) dp[i][j][k] = 'n';
dp[0][0][0] = 's';
queue<node> q;
node nd0 = {0, 0, 0}, nd1;
q.push(nd0);
char ch1;
while (!q.empty()) {
nd0 = q.front();
i = nd0.i, j = nd0.j, k = nd0.k;
q.pop();
nd1 = nd0;
nd1.k++;
if (i + 1 <= m && s[i + 1 - 1] == '(') nd1.i++;
if (nd1.i <= m && nd1.j <= n && nd1.k >= 0 && nd1.k <= max(m, n) &&
dp[nd1.i][nd1.j][nd1.k] == 'n') {
dp[nd1.i][nd1.j][nd1.k] = '(';
dp1[nd1.i][nd1.j][nd1.k] = nd0;
q.push(nd1);
}
nd1 = nd0;
nd1.k++;
if (j + 1 <= n && t[j + 1 - 1] == '(') nd1.j++;
if (nd1.i <= m && nd1.j <= n && nd1.k >= 0 && nd1.k <= max(m, n) &&
dp[nd1.i][nd1.j][nd1.k] == 'n') {
dp[nd1.i][nd1.j][nd1.k] = '(';
dp1[nd1.i][nd1.j][nd1.k] = nd0;
q.push(nd1);
}
nd1 = nd0;
nd1.k++;
if (i + 1 <= m && s[i + 1 - 1] == '(' && j + 1 <= n && t[j + 1 - 1] == '(')
nd1.i++, nd1.j++;
if (nd1.i <= m && nd1.j <= n && nd1.k >= 0 && nd1.k <= max(m, n) &&
dp[nd1.i][nd1.j][nd1.k] == 'n') {
dp[nd1.i][nd1.j][nd1.k] = '(';
dp1[nd1.i][nd1.j][nd1.k] = nd0;
q.push(nd1);
}
nd1 = nd0;
nd1.k--;
if (i + 1 <= m && s[i + 1 - 1] == ')') nd1.i++;
if (nd1.i <= m && nd1.j <= n && nd1.k >= 0 && nd1.k <= max(m, n) &&
dp[nd1.i][nd1.j][nd1.k] == 'n') {
dp[nd1.i][nd1.j][nd1.k] = ')';
dp1[nd1.i][nd1.j][nd1.k] = nd0;
q.push(nd1);
}
nd1 = nd0;
nd1.k--;
if (j + 1 <= n && t[j + 1 - 1] == ')') nd1.j++;
if (nd1.i <= m && nd1.j <= n && nd1.k >= 0 && nd1.k <= max(m, n) &&
dp[nd1.i][nd1.j][nd1.k] == 'n') {
dp[nd1.i][nd1.j][nd1.k] = ')';
dp1[nd1.i][nd1.j][nd1.k] = nd0;
q.push(nd1);
}
nd1 = nd0;
nd1.k--;
if (i + 1 <= m && s[i + 1 - 1] == ')' && j + 1 <= n && t[j + 1 - 1] == ')')
nd1.i++, nd1.j++;
if (nd1.i <= m && nd1.j <= n && nd1.k >= 0 && nd1.k <= max(m, n) &&
dp[nd1.i][nd1.j][nd1.k] == 'n') {
dp[nd1.i][nd1.j][nd1.k] = ')';
dp1[nd1.i][nd1.j][nd1.k] = nd0;
q.push(nd1);
}
}
string res = "";
char ch;
i = m, j = n, k = 0;
while (i > 0 || j > 0 || k > 0) {
res += dp[i][j][k];
int ti = dp1[i][j][k].i;
int tj = dp1[i][j][k].j;
int tk = dp1[i][j][k].k;
i = ti, j = tj, k = tk;
}
reverse(res.begin(), res.end());
cout << res;
}
``` |
#include <bits/stdc++.h>
using namespace std;
inline long long read() {
char c = getchar();
long long f = 1, x = 0;
while (c < '0' || c > '9') {
if (c == '-') f = -1;
c = getchar();
}
while (c >= '0' && c <= '9') x = x * 10 + (c - '0'), c = getchar();
return x * f;
}
void MOD(int &x) {
if (x >= 1000000007) x -= 1000000007;
}
int f[210][210][410], g[210][210][410];
char s[210], t[210];
int num;
int dfs(int i, int j, int k) {
int &w = f[i][j][k];
if (w != (1 << 30)) return w;
if (k) {
int c = dfs(i - (s[i - 1] == '('), j - (t[j - 1] == '('), k - 1) + 1;
if (c < w) {
w = c;
g[i][j][k] = '(';
}
}
if (k < num) {
int c = dfs(i - (s[i - 1] == ')'), j - (t[j - 1] == ')'), k + 1) + 1;
if (c < w) {
w = c;
g[i][j][k] = ')';
}
}
return w;
}
char ans[1000];
int main() {
scanf("%s%s", s, t);
int n = strlen(s), m = strlen(t);
num = n + m;
for (int i = 0; i <= n; i++)
for (int j = 0; j <= m; j++)
for (int k = 0; k <= n + m; k++) f[i][j][k] = (1 << 30);
f[0][0][0] = 0;
dfs(n, m, 0);
int i = n, j = m, k = 0;
int cnt = 0;
while (i || j || k) {
ans[++cnt] = g[i][j][k];
if (g[i][j][k] == '(') {
i -= (s[i - 1] == '(');
j -= (t[j - 1] == '(');
k -= 1;
} else {
i -= (s[i - 1] == ')');
j -= (t[j - 1] == ')');
k += 1;
}
}
reverse(ans + 1, ans + cnt + 1);
printf("%s\n", ans + 1);
}
| ### Prompt
Generate a CPP solution to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
inline long long read() {
char c = getchar();
long long f = 1, x = 0;
while (c < '0' || c > '9') {
if (c == '-') f = -1;
c = getchar();
}
while (c >= '0' && c <= '9') x = x * 10 + (c - '0'), c = getchar();
return x * f;
}
void MOD(int &x) {
if (x >= 1000000007) x -= 1000000007;
}
int f[210][210][410], g[210][210][410];
char s[210], t[210];
int num;
int dfs(int i, int j, int k) {
int &w = f[i][j][k];
if (w != (1 << 30)) return w;
if (k) {
int c = dfs(i - (s[i - 1] == '('), j - (t[j - 1] == '('), k - 1) + 1;
if (c < w) {
w = c;
g[i][j][k] = '(';
}
}
if (k < num) {
int c = dfs(i - (s[i - 1] == ')'), j - (t[j - 1] == ')'), k + 1) + 1;
if (c < w) {
w = c;
g[i][j][k] = ')';
}
}
return w;
}
char ans[1000];
int main() {
scanf("%s%s", s, t);
int n = strlen(s), m = strlen(t);
num = n + m;
for (int i = 0; i <= n; i++)
for (int j = 0; j <= m; j++)
for (int k = 0; k <= n + m; k++) f[i][j][k] = (1 << 30);
f[0][0][0] = 0;
dfs(n, m, 0);
int i = n, j = m, k = 0;
int cnt = 0;
while (i || j || k) {
ans[++cnt] = g[i][j][k];
if (g[i][j][k] == '(') {
i -= (s[i - 1] == '(');
j -= (t[j - 1] == '(');
k -= 1;
} else {
i -= (s[i - 1] == ')');
j -= (t[j - 1] == ')');
k += 1;
}
}
reverse(ans + 1, ans + cnt + 1);
printf("%s\n", ans + 1);
}
``` |
#include <bits/stdc++.h>
const int N = 1000006;
const int M = 1000 * 1000 * 1000 + 7;
using namespace std;
void debug(string var, int val) { cout << var << " : " << val << endl; }
int gcd(int f, int s) {
if (s == 0)
return f;
else
return gcd(s, f % s);
}
string s, s1;
int dp[201][201][500];
int calc(int ind, int ind1, int r) {
if (ind == s.length() && ind1 == s1.length()) return dp[ind][ind1][r] = r;
int& rem = dp[ind][ind1][r];
if (rem != -1) return rem;
rem = 1e9;
if (ind == s.length()) {
if (s1[ind1] == '(')
rem = min(rem, calc(ind, ind1 + 1, r + 1) + 1);
else {
if (r > 0)
rem = min(rem, calc(ind, ind1 + 1, r - 1) + 1);
else
rem = min(rem, calc(ind, ind1, r + 1) + 1);
}
} else if (ind1 == s1.length()) {
if (s[ind] == '(')
rem = min(rem, calc(ind + 1, ind1, r + 1) + 1);
else {
if (r > 0)
rem = min(rem, calc(ind + 1, ind1, r - 1) + 1);
else
rem = min(rem, calc(ind, ind1, r + 1) + 1);
}
} else {
if (s[ind] == s1[ind1]) {
if (s[ind] == '(')
rem = min(rem, calc(ind + 1, ind1 + 1, r + 1) + 1);
else {
if (r > 0)
rem = min(rem, min(rem, calc(ind + 1, ind1 + 1, r - 1) + 1));
else
rem = min(rem, min(rem, calc(ind, ind1, r + 1) + 1));
}
} else {
if (s[ind] == '(') {
rem = min(rem, calc(ind + 1, ind1, r + 1) + 1);
if (r > 0) rem = min(rem, calc(ind, ind1 + 1, r - 1) + 1);
} else {
rem = min(rem, calc(ind, ind1 + 1, r + 1) + 1);
if (r > 0) rem = min(rem, calc(ind + 1, ind1, r - 1) + 1);
}
}
}
return rem;
}
string ans = "";
int abuhijleh;
void buildPath(int ind, int ind1, int r) {
if (ind == s.length() && ind1 == s1.length()) {
for (int i = 0; i < abuhijleh; i++) ans += ")";
return;
}
abuhijleh--;
if (s.length() != ind && s1.length() != ind1 &&
dp[ind + 1][ind1 + 1][r + 1] == abuhijleh && s[ind] == '(' &&
s1[ind1] == '(') {
ans += "(";
buildPath(ind + 1, ind1 + 1, r + 1);
return;
}
if (s.length() != ind && dp[ind + 1][ind1][r + 1] == abuhijleh &&
s[ind] == '(') {
ans += "(";
buildPath(ind + 1, ind1, r + 1);
return;
}
if (s1.length() != ind1 && dp[ind][ind1 + 1][r + 1] == abuhijleh &&
s1[ind1] == '(') {
ans += "(";
buildPath(ind, ind1 + 1, r + 1);
return;
}
if (dp[ind][ind1][r + 1] == abuhijleh && r == 0) {
ans += "(";
buildPath(ind, ind1, r + 1);
return;
}
if (r > 0 && s.length() != ind && s1.length() != ind1 &&
dp[ind + 1][ind1 + 1][r - 1] == abuhijleh && s[ind] == ')' &&
s1[ind1] == ')') {
ans += ")";
buildPath(ind + 1, ind1 + 1, r - 1);
return;
}
if (r > 0 && s.length() != ind && dp[ind + 1][ind1][r - 1] == abuhijleh &&
s[ind] == ')') {
ans += ")";
buildPath(ind + 1, ind1, r - 1);
return;
}
if (r > 0 && s1.length() != ind1 && dp[ind][ind1 + 1][r - 1] == abuhijleh &&
s1[ind1] == ')') {
ans += ")";
buildPath(ind, ind1 + 1, r - 1);
return;
}
}
int main() {
cin >> s >> s1;
memset(dp, -1, sizeof dp);
abuhijleh = calc(0, 0, 0);
buildPath(0, 0, 0);
cout << ans << endl;
return 0;
}
| ### Prompt
Please formulate a cpp solution to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
const int N = 1000006;
const int M = 1000 * 1000 * 1000 + 7;
using namespace std;
void debug(string var, int val) { cout << var << " : " << val << endl; }
int gcd(int f, int s) {
if (s == 0)
return f;
else
return gcd(s, f % s);
}
string s, s1;
int dp[201][201][500];
int calc(int ind, int ind1, int r) {
if (ind == s.length() && ind1 == s1.length()) return dp[ind][ind1][r] = r;
int& rem = dp[ind][ind1][r];
if (rem != -1) return rem;
rem = 1e9;
if (ind == s.length()) {
if (s1[ind1] == '(')
rem = min(rem, calc(ind, ind1 + 1, r + 1) + 1);
else {
if (r > 0)
rem = min(rem, calc(ind, ind1 + 1, r - 1) + 1);
else
rem = min(rem, calc(ind, ind1, r + 1) + 1);
}
} else if (ind1 == s1.length()) {
if (s[ind] == '(')
rem = min(rem, calc(ind + 1, ind1, r + 1) + 1);
else {
if (r > 0)
rem = min(rem, calc(ind + 1, ind1, r - 1) + 1);
else
rem = min(rem, calc(ind, ind1, r + 1) + 1);
}
} else {
if (s[ind] == s1[ind1]) {
if (s[ind] == '(')
rem = min(rem, calc(ind + 1, ind1 + 1, r + 1) + 1);
else {
if (r > 0)
rem = min(rem, min(rem, calc(ind + 1, ind1 + 1, r - 1) + 1));
else
rem = min(rem, min(rem, calc(ind, ind1, r + 1) + 1));
}
} else {
if (s[ind] == '(') {
rem = min(rem, calc(ind + 1, ind1, r + 1) + 1);
if (r > 0) rem = min(rem, calc(ind, ind1 + 1, r - 1) + 1);
} else {
rem = min(rem, calc(ind, ind1 + 1, r + 1) + 1);
if (r > 0) rem = min(rem, calc(ind + 1, ind1, r - 1) + 1);
}
}
}
return rem;
}
string ans = "";
int abuhijleh;
void buildPath(int ind, int ind1, int r) {
if (ind == s.length() && ind1 == s1.length()) {
for (int i = 0; i < abuhijleh; i++) ans += ")";
return;
}
abuhijleh--;
if (s.length() != ind && s1.length() != ind1 &&
dp[ind + 1][ind1 + 1][r + 1] == abuhijleh && s[ind] == '(' &&
s1[ind1] == '(') {
ans += "(";
buildPath(ind + 1, ind1 + 1, r + 1);
return;
}
if (s.length() != ind && dp[ind + 1][ind1][r + 1] == abuhijleh &&
s[ind] == '(') {
ans += "(";
buildPath(ind + 1, ind1, r + 1);
return;
}
if (s1.length() != ind1 && dp[ind][ind1 + 1][r + 1] == abuhijleh &&
s1[ind1] == '(') {
ans += "(";
buildPath(ind, ind1 + 1, r + 1);
return;
}
if (dp[ind][ind1][r + 1] == abuhijleh && r == 0) {
ans += "(";
buildPath(ind, ind1, r + 1);
return;
}
if (r > 0 && s.length() != ind && s1.length() != ind1 &&
dp[ind + 1][ind1 + 1][r - 1] == abuhijleh && s[ind] == ')' &&
s1[ind1] == ')') {
ans += ")";
buildPath(ind + 1, ind1 + 1, r - 1);
return;
}
if (r > 0 && s.length() != ind && dp[ind + 1][ind1][r - 1] == abuhijleh &&
s[ind] == ')') {
ans += ")";
buildPath(ind + 1, ind1, r - 1);
return;
}
if (r > 0 && s1.length() != ind1 && dp[ind][ind1 + 1][r - 1] == abuhijleh &&
s1[ind1] == ')') {
ans += ")";
buildPath(ind, ind1 + 1, r - 1);
return;
}
}
int main() {
cin >> s >> s1;
memset(dp, -1, sizeof dp);
abuhijleh = calc(0, 0, 0);
buildPath(0, 0, 0);
cout << ans << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int maxN = 2e2 + 7;
string A, B, C = "()";
int a, b, c[] = {1, -1};
int dp[maxN][maxN][maxN];
int get(int i, int j, int k) {
if (k > 200 || k < 0) return 1e9;
if (i == a && j == b && !k) return 0;
int &ans = dp[i][j][k];
if (ans + 1) return ans;
ans = 1e9;
for (int x = 0; x < 2; x++) {
ans = min(ans, get(i + (C[x] == A[i]), j + (C[x] == B[j]), k + c[x]) + 1);
}
return ans;
}
void trace(int i, int j, int k) {
if (!k && i == a && j == b) return;
int res = get(i, j, k);
for (int x = 0; x < 2; x++) {
if (res == get(i + (C[x] == A[i]), j + (C[x] == B[j]), k + c[x]) + 1) {
cout << C[x];
trace(i + (C[x] == A[i]), j + (C[x] == B[j]), k + c[x]);
break;
}
}
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
if (fopen("test.inp", "r")) freopen("test.inp", "r", stdin);
memset(dp, -1, sizeof dp);
cin >> A >> B;
a = A.size(), b = B.size();
A = A + '#', B = B + '#';
trace(0, 0, 0);
}
| ### Prompt
Your challenge is to write a Cpp solution to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxN = 2e2 + 7;
string A, B, C = "()";
int a, b, c[] = {1, -1};
int dp[maxN][maxN][maxN];
int get(int i, int j, int k) {
if (k > 200 || k < 0) return 1e9;
if (i == a && j == b && !k) return 0;
int &ans = dp[i][j][k];
if (ans + 1) return ans;
ans = 1e9;
for (int x = 0; x < 2; x++) {
ans = min(ans, get(i + (C[x] == A[i]), j + (C[x] == B[j]), k + c[x]) + 1);
}
return ans;
}
void trace(int i, int j, int k) {
if (!k && i == a && j == b) return;
int res = get(i, j, k);
for (int x = 0; x < 2; x++) {
if (res == get(i + (C[x] == A[i]), j + (C[x] == B[j]), k + c[x]) + 1) {
cout << C[x];
trace(i + (C[x] == A[i]), j + (C[x] == B[j]), k + c[x]);
break;
}
}
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
if (fopen("test.inp", "r")) freopen("test.inp", "r", stdin);
memset(dp, -1, sizeof dp);
cin >> A >> B;
a = A.size(), b = B.size();
A = A + '#', B = B + '#';
trace(0, 0, 0);
}
``` |
#include <bits/stdc++.h>
struct State {
int16_t val;
int16_t prev_i, prev_j, prev_balance;
char best_ch;
State() {}
State(int32_t val_, int32_t prev_i_, int32_t prev_j_, int32_t prev_balance_,
char best_ch_) {
val = val_;
prev_i = prev_i_;
prev_j = prev_j_;
prev_balance = prev_balance_;
best_ch = best_ch_;
}
void relax(State& other, int32_t i, int32_t j, int32_t prev_balance_,
char ch) {
if (other.val == INT16_MAX) return;
if (val == INT16_MAX || other.val < val - 1) {
val = other.val + 1;
prev_i = i;
prev_j = j;
prev_balance = prev_balance_;
best_ch = ch;
}
}
};
int main() {
std::string str1, str2;
std::cin >> str1 >> str2;
int32_t n = str1.size(), m = str2.size();
State*** dp = new State**[n + 1];
for (int32_t i = 0; i <= n; i++) {
dp[i] = new State*[m + 1];
for (int32_t j = 0; j <= m; j++) dp[i][j] = new State[1000];
}
dp[0][0][0] = {0, -1, -1, -1, ' '};
for (int32_t i = 0; i <= n; i++) {
for (int32_t j = 0; j <= m; j++) {
for (int32_t k = 0; k < 1000; k++) {
if (i == 0 && j == 0 && k == 0) continue;
dp[i][j][k] = {INT16_MAX, -1, -1, -1, ' '};
if (k < 1000 - 1) {
if (i > 0 && str1[i - 1] == ')')
dp[i][j][k].relax(dp[i - 1][j][k + 1], i - 1, j, k + 1, ')');
if (j > 0 && str2[j - 1] == ')')
dp[i][j][k].relax(dp[i][j - 1][k + 1], i, j - 1, k + 1, ')');
if ((i > 0 && str1[i - 1] == ')') && (j > 0 && str2[j - 1] == ')'))
dp[i][j][k].relax(dp[i - 1][j - 1][k + 1], i - 1, j - 1, k + 1,
')');
}
if (k != 0) {
if (i > 0 && str1[i - 1] == '(')
dp[i][j][k].relax(dp[i - 1][j][k - 1], i - 1, j, k - 1, '(');
if (j > 0 && str2[j - 1] == '(')
dp[i][j][k].relax(dp[i][j - 1][k - 1], i, j - 1, k - 1, '(');
if ((i > 0 && str1[i - 1] == '(') && (j > 0 && str2[j - 1] == '('))
dp[i][j][k].relax(dp[i - 1][j - 1][k - 1], i - 1, j - 1, k - 1,
'(');
}
}
if ((i == n || str1[i] == '(') && (j == m || str2[j] == '(')) {
for (int32_t k = 1000 - 1; k >= 0; k--) {
dp[i][j][k].relax(dp[i][j][k + 1], i, j, k + 1, ')');
}
}
if ((i == n || str1[i] == ')') && (j == m || str2[j] == ')')) {
for (int32_t k = 1; k < 1000; k++) {
dp[i][j][k].relax(dp[i][j][k - 1], i, j, k - 1, '(');
}
}
}
}
int32_t cur_i = n, cur_j = m, cur_balance = 0;
std::string answer;
while (cur_i != 0 || cur_j != 0 || cur_balance != 0) {
answer.push_back(dp[cur_i][cur_j][cur_balance].best_ch);
int32_t cur_i_ = dp[cur_i][cur_j][cur_balance].prev_i;
int32_t cur_j_ = dp[cur_i][cur_j][cur_balance].prev_j;
int32_t cur_balance_ = dp[cur_i][cur_j][cur_balance].prev_balance;
cur_i = cur_i_;
cur_j = cur_j_;
cur_balance = cur_balance_;
}
std::reverse(answer.begin(), answer.end());
std::cout << answer;
return 0;
}
| ### Prompt
Your task is to create a cpp solution to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
struct State {
int16_t val;
int16_t prev_i, prev_j, prev_balance;
char best_ch;
State() {}
State(int32_t val_, int32_t prev_i_, int32_t prev_j_, int32_t prev_balance_,
char best_ch_) {
val = val_;
prev_i = prev_i_;
prev_j = prev_j_;
prev_balance = prev_balance_;
best_ch = best_ch_;
}
void relax(State& other, int32_t i, int32_t j, int32_t prev_balance_,
char ch) {
if (other.val == INT16_MAX) return;
if (val == INT16_MAX || other.val < val - 1) {
val = other.val + 1;
prev_i = i;
prev_j = j;
prev_balance = prev_balance_;
best_ch = ch;
}
}
};
int main() {
std::string str1, str2;
std::cin >> str1 >> str2;
int32_t n = str1.size(), m = str2.size();
State*** dp = new State**[n + 1];
for (int32_t i = 0; i <= n; i++) {
dp[i] = new State*[m + 1];
for (int32_t j = 0; j <= m; j++) dp[i][j] = new State[1000];
}
dp[0][0][0] = {0, -1, -1, -1, ' '};
for (int32_t i = 0; i <= n; i++) {
for (int32_t j = 0; j <= m; j++) {
for (int32_t k = 0; k < 1000; k++) {
if (i == 0 && j == 0 && k == 0) continue;
dp[i][j][k] = {INT16_MAX, -1, -1, -1, ' '};
if (k < 1000 - 1) {
if (i > 0 && str1[i - 1] == ')')
dp[i][j][k].relax(dp[i - 1][j][k + 1], i - 1, j, k + 1, ')');
if (j > 0 && str2[j - 1] == ')')
dp[i][j][k].relax(dp[i][j - 1][k + 1], i, j - 1, k + 1, ')');
if ((i > 0 && str1[i - 1] == ')') && (j > 0 && str2[j - 1] == ')'))
dp[i][j][k].relax(dp[i - 1][j - 1][k + 1], i - 1, j - 1, k + 1,
')');
}
if (k != 0) {
if (i > 0 && str1[i - 1] == '(')
dp[i][j][k].relax(dp[i - 1][j][k - 1], i - 1, j, k - 1, '(');
if (j > 0 && str2[j - 1] == '(')
dp[i][j][k].relax(dp[i][j - 1][k - 1], i, j - 1, k - 1, '(');
if ((i > 0 && str1[i - 1] == '(') && (j > 0 && str2[j - 1] == '('))
dp[i][j][k].relax(dp[i - 1][j - 1][k - 1], i - 1, j - 1, k - 1,
'(');
}
}
if ((i == n || str1[i] == '(') && (j == m || str2[j] == '(')) {
for (int32_t k = 1000 - 1; k >= 0; k--) {
dp[i][j][k].relax(dp[i][j][k + 1], i, j, k + 1, ')');
}
}
if ((i == n || str1[i] == ')') && (j == m || str2[j] == ')')) {
for (int32_t k = 1; k < 1000; k++) {
dp[i][j][k].relax(dp[i][j][k - 1], i, j, k - 1, '(');
}
}
}
}
int32_t cur_i = n, cur_j = m, cur_balance = 0;
std::string answer;
while (cur_i != 0 || cur_j != 0 || cur_balance != 0) {
answer.push_back(dp[cur_i][cur_j][cur_balance].best_ch);
int32_t cur_i_ = dp[cur_i][cur_j][cur_balance].prev_i;
int32_t cur_j_ = dp[cur_i][cur_j][cur_balance].prev_j;
int32_t cur_balance_ = dp[cur_i][cur_j][cur_balance].prev_balance;
cur_i = cur_i_;
cur_j = cur_j_;
cur_balance = cur_balance_;
}
std::reverse(answer.begin(), answer.end());
std::cout << answer;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 200;
int dp[N + 10][N + 10][2 * N + 10];
pair<pair<int, int>, int> par[N + 10][N + 10][2 * N + 10];
const int inf = 1e6;
int main() {
string s, t;
cin >> s >> t;
int szs = s.size(), szt = t.size();
for (int i = 0; i <= N; i++) {
for (int j = 0; j <= N; j++) {
for (int k = 0; k <= 2 * N; k++) {
dp[i][j][k] = inf;
}
}
}
dp[0][0][0] = 0;
for (int i = 0; i <= szs; i++) {
for (int j = 0; j <= szt; j++) {
if (i == szs && j == szt) {
for (int k = 2 * N; k > 0; k--) {
int c = dp[i][j][k - 1];
dp[i][j][k - 1] = min(dp[i][j][k - 1], dp[i][j][k] + 1);
if (dp[i][j][k - 1] < c) {
par[i][j][k - 1] = make_pair(make_pair(i, j), k);
}
}
} else if (i == szs) {
if (t[j] == '(') {
for (int k = 2 * N; k > 0; k--) {
int c = dp[i][j][k - 1];
dp[i][j][k - 1] = min(dp[i][j][k - 1], dp[i][j][k] + 1);
if (dp[i][j][k - 1] < c) {
par[i][j][k - 1] = make_pair(make_pair(i, j), k);
}
}
for (int k = 0; k < 2 * N; k++) {
int c = dp[i][j + 1][k + 1];
dp[i][j + 1][k + 1] = min(dp[i][j + 1][k + 1], dp[i][j][k] + 1);
if (dp[i][j + 1][k + 1] < c) {
par[i][j + 1][k + 1] = make_pair(make_pair(i, j), k);
}
}
} else {
for (int k = 0; k < 2 * N; k++) {
int c = dp[i][j][k + 1];
dp[i][j][k + 1] = min(dp[i][j][k + 1], dp[i][j][k] + 1);
if (c > dp[i][j][k + 1]) {
par[i][j][k + 1] = make_pair(make_pair(i, j), k);
}
}
for (int k = 2 * N; k > 0; k--) {
int c = dp[i][j + 1][k - 1];
dp[i][j + 1][k - 1] = min(dp[i][j + 1][k - 1], dp[i][j][k] + 1);
if (c > dp[i][j + 1][k - 1]) {
par[i][j + 1][k - 1] = make_pair(make_pair(i, j), k);
}
}
}
} else if (j == szt) {
if (s[i] == '(') {
for (int k = 2 * N; k > 0; k--) {
int c = dp[i][j][k - 1];
dp[i][j][k - 1] = min(dp[i][j][k - 1], dp[i][j][k] + 1);
if (c > dp[i][j][k - 1]) {
par[i][j][k - 1] = make_pair(make_pair(i, j), k);
}
}
for (int k = 0; k < 2 * N; k++) {
int c = dp[i + 1][j][k + 1];
dp[i + 1][j][k + 1] = min(dp[i + 1][j][k + 1], dp[i][j][k] + 1);
if (c > dp[i + 1][j][k + 1]) {
par[i + 1][j][k + 1] = make_pair(make_pair(i, j), k);
}
}
} else {
for (int k = 0; k < 2 * N; k++) {
int c = dp[i][j][k + 1];
dp[i][j][k + 1] = min(dp[i][j][k + 1], dp[i][j][k] + 1);
if (c > dp[i][j][k + 1]) {
par[i][j][k + 1] = make_pair(make_pair(i, j), k);
}
}
for (int k = 2 * N; k > 0; k--) {
int c = dp[i + 1][j][k - 1];
dp[i + 1][j][k - 1] = min(dp[i + 1][j][k - 1], dp[i][j][k] + 1);
if (c > dp[i + 1][j][k - 1]) {
par[i + 1][j][k - 1] = make_pair(make_pair(i, j), k);
}
}
}
} else {
if (s[i] == '(' && t[j] == ')') {
for (int k = 0; k < 2 * N; k++) {
int c = dp[i + 1][j][k + 1];
dp[i + 1][j][k + 1] = min(dp[i + 1][j][k + 1], dp[i][j][k] + 1);
if (c > dp[i + 1][j][k + 1]) {
par[i + 1][j][k + 1] = make_pair(make_pair(i, j), k);
}
}
for (int k = 2 * N; k > 0; k--) {
int c = dp[i][j + 1][k - 1];
dp[i][j + 1][k - 1] = min(dp[i][j + 1][k - 1], dp[i][j][k] + 1);
if (c > dp[i][j + 1][k - 1]) {
par[i][j + 1][k - 1] = make_pair(make_pair(i, j), k);
}
}
} else if (s[i] == '(' && t[j] == '(') {
for (int k = 2 * N; k > 0; k--) {
int c = dp[i][j][k - 1];
dp[i][j][k - 1] = min(dp[i][j][k - 1], dp[i][j][k] + 1);
if (c > dp[i][j][k - 1]) {
par[i][j][k - 1] = make_pair(make_pair(i, j), k);
}
}
for (int k = 0; k < 2 * N; k++) {
int c = dp[i + 1][j + 1][k + 1];
dp[i + 1][j + 1][k + 1] =
min(dp[i + 1][j + 1][k + 1], dp[i][j][k] + 1);
if (c > dp[i + 1][j + 1][k + 1]) {
par[i + 1][j + 1][k + 1] = make_pair(make_pair(i, j), k);
}
}
} else if (s[i] == ')' && t[j] == ')') {
for (int k = 0; k < 2 * N; k++) {
int c = dp[i][j][k + 1];
dp[i][j][k + 1] = min(dp[i][j][k + 1], dp[i][j][k] + 1);
if (c > dp[i][j][k + 1]) {
par[i][j][k + 1] = make_pair(make_pair(i, j), k);
}
}
for (int k = 2 * N; k > 0; k--) {
int c = dp[i + 1][j + 1][k - 1];
dp[i + 1][j + 1][k - 1] =
min(dp[i + 1][j + 1][k - 1], dp[i][j][k] + 1);
if (c > dp[i + 1][j + 1][k - 1]) {
par[i + 1][j + 1][k - 1] = make_pair(make_pair(i, j), k);
}
}
} else {
for (int k = 0; k < 2 * N; k++) {
int c = dp[i][j + 1][k + 1];
dp[i][j + 1][k + 1] = min(dp[i][j + 1][k + 1], dp[i][j][k] + 1);
if (c > dp[i][j + 1][k + 1]) {
par[i][j + 1][k + 1] = make_pair(make_pair(i, j), k);
}
}
for (int k = 2 * N; k > 0; k--) {
int c = dp[i + 1][j][k - 1];
dp[i + 1][j][k - 1] = min(dp[i + 1][j][k - 1], dp[i][j][k] + 1);
if (c > dp[i + 1][j][k - 1]) {
par[i + 1][j][k - 1] = make_pair(make_pair(i, j), k);
}
}
}
}
}
}
vector<int> seq;
seq.push_back(0);
int bal = 0;
while (szs > 0 || szt > 0 || bal > 0) {
pair<pair<int, int>, int> A = par[szs][szt][bal];
szs = A.first.first;
szt = A.first.second;
bal = A.second;
seq.push_back(bal);
}
reverse(seq.begin(), seq.end());
string ans = "";
for (int sz = seq.size(), i = 1; i < sz; i++) {
if (seq[i] > seq[i - 1]) {
ans += "(";
} else {
ans += ")";
}
}
cout << ans << "\n";
}
| ### Prompt
Please formulate a cpp solution to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 200;
int dp[N + 10][N + 10][2 * N + 10];
pair<pair<int, int>, int> par[N + 10][N + 10][2 * N + 10];
const int inf = 1e6;
int main() {
string s, t;
cin >> s >> t;
int szs = s.size(), szt = t.size();
for (int i = 0; i <= N; i++) {
for (int j = 0; j <= N; j++) {
for (int k = 0; k <= 2 * N; k++) {
dp[i][j][k] = inf;
}
}
}
dp[0][0][0] = 0;
for (int i = 0; i <= szs; i++) {
for (int j = 0; j <= szt; j++) {
if (i == szs && j == szt) {
for (int k = 2 * N; k > 0; k--) {
int c = dp[i][j][k - 1];
dp[i][j][k - 1] = min(dp[i][j][k - 1], dp[i][j][k] + 1);
if (dp[i][j][k - 1] < c) {
par[i][j][k - 1] = make_pair(make_pair(i, j), k);
}
}
} else if (i == szs) {
if (t[j] == '(') {
for (int k = 2 * N; k > 0; k--) {
int c = dp[i][j][k - 1];
dp[i][j][k - 1] = min(dp[i][j][k - 1], dp[i][j][k] + 1);
if (dp[i][j][k - 1] < c) {
par[i][j][k - 1] = make_pair(make_pair(i, j), k);
}
}
for (int k = 0; k < 2 * N; k++) {
int c = dp[i][j + 1][k + 1];
dp[i][j + 1][k + 1] = min(dp[i][j + 1][k + 1], dp[i][j][k] + 1);
if (dp[i][j + 1][k + 1] < c) {
par[i][j + 1][k + 1] = make_pair(make_pair(i, j), k);
}
}
} else {
for (int k = 0; k < 2 * N; k++) {
int c = dp[i][j][k + 1];
dp[i][j][k + 1] = min(dp[i][j][k + 1], dp[i][j][k] + 1);
if (c > dp[i][j][k + 1]) {
par[i][j][k + 1] = make_pair(make_pair(i, j), k);
}
}
for (int k = 2 * N; k > 0; k--) {
int c = dp[i][j + 1][k - 1];
dp[i][j + 1][k - 1] = min(dp[i][j + 1][k - 1], dp[i][j][k] + 1);
if (c > dp[i][j + 1][k - 1]) {
par[i][j + 1][k - 1] = make_pair(make_pair(i, j), k);
}
}
}
} else if (j == szt) {
if (s[i] == '(') {
for (int k = 2 * N; k > 0; k--) {
int c = dp[i][j][k - 1];
dp[i][j][k - 1] = min(dp[i][j][k - 1], dp[i][j][k] + 1);
if (c > dp[i][j][k - 1]) {
par[i][j][k - 1] = make_pair(make_pair(i, j), k);
}
}
for (int k = 0; k < 2 * N; k++) {
int c = dp[i + 1][j][k + 1];
dp[i + 1][j][k + 1] = min(dp[i + 1][j][k + 1], dp[i][j][k] + 1);
if (c > dp[i + 1][j][k + 1]) {
par[i + 1][j][k + 1] = make_pair(make_pair(i, j), k);
}
}
} else {
for (int k = 0; k < 2 * N; k++) {
int c = dp[i][j][k + 1];
dp[i][j][k + 1] = min(dp[i][j][k + 1], dp[i][j][k] + 1);
if (c > dp[i][j][k + 1]) {
par[i][j][k + 1] = make_pair(make_pair(i, j), k);
}
}
for (int k = 2 * N; k > 0; k--) {
int c = dp[i + 1][j][k - 1];
dp[i + 1][j][k - 1] = min(dp[i + 1][j][k - 1], dp[i][j][k] + 1);
if (c > dp[i + 1][j][k - 1]) {
par[i + 1][j][k - 1] = make_pair(make_pair(i, j), k);
}
}
}
} else {
if (s[i] == '(' && t[j] == ')') {
for (int k = 0; k < 2 * N; k++) {
int c = dp[i + 1][j][k + 1];
dp[i + 1][j][k + 1] = min(dp[i + 1][j][k + 1], dp[i][j][k] + 1);
if (c > dp[i + 1][j][k + 1]) {
par[i + 1][j][k + 1] = make_pair(make_pair(i, j), k);
}
}
for (int k = 2 * N; k > 0; k--) {
int c = dp[i][j + 1][k - 1];
dp[i][j + 1][k - 1] = min(dp[i][j + 1][k - 1], dp[i][j][k] + 1);
if (c > dp[i][j + 1][k - 1]) {
par[i][j + 1][k - 1] = make_pair(make_pair(i, j), k);
}
}
} else if (s[i] == '(' && t[j] == '(') {
for (int k = 2 * N; k > 0; k--) {
int c = dp[i][j][k - 1];
dp[i][j][k - 1] = min(dp[i][j][k - 1], dp[i][j][k] + 1);
if (c > dp[i][j][k - 1]) {
par[i][j][k - 1] = make_pair(make_pair(i, j), k);
}
}
for (int k = 0; k < 2 * N; k++) {
int c = dp[i + 1][j + 1][k + 1];
dp[i + 1][j + 1][k + 1] =
min(dp[i + 1][j + 1][k + 1], dp[i][j][k] + 1);
if (c > dp[i + 1][j + 1][k + 1]) {
par[i + 1][j + 1][k + 1] = make_pair(make_pair(i, j), k);
}
}
} else if (s[i] == ')' && t[j] == ')') {
for (int k = 0; k < 2 * N; k++) {
int c = dp[i][j][k + 1];
dp[i][j][k + 1] = min(dp[i][j][k + 1], dp[i][j][k] + 1);
if (c > dp[i][j][k + 1]) {
par[i][j][k + 1] = make_pair(make_pair(i, j), k);
}
}
for (int k = 2 * N; k > 0; k--) {
int c = dp[i + 1][j + 1][k - 1];
dp[i + 1][j + 1][k - 1] =
min(dp[i + 1][j + 1][k - 1], dp[i][j][k] + 1);
if (c > dp[i + 1][j + 1][k - 1]) {
par[i + 1][j + 1][k - 1] = make_pair(make_pair(i, j), k);
}
}
} else {
for (int k = 0; k < 2 * N; k++) {
int c = dp[i][j + 1][k + 1];
dp[i][j + 1][k + 1] = min(dp[i][j + 1][k + 1], dp[i][j][k] + 1);
if (c > dp[i][j + 1][k + 1]) {
par[i][j + 1][k + 1] = make_pair(make_pair(i, j), k);
}
}
for (int k = 2 * N; k > 0; k--) {
int c = dp[i + 1][j][k - 1];
dp[i + 1][j][k - 1] = min(dp[i + 1][j][k - 1], dp[i][j][k] + 1);
if (c > dp[i + 1][j][k - 1]) {
par[i + 1][j][k - 1] = make_pair(make_pair(i, j), k);
}
}
}
}
}
}
vector<int> seq;
seq.push_back(0);
int bal = 0;
while (szs > 0 || szt > 0 || bal > 0) {
pair<pair<int, int>, int> A = par[szs][szt][bal];
szs = A.first.first;
szt = A.first.second;
bal = A.second;
seq.push_back(bal);
}
reverse(seq.begin(), seq.end());
string ans = "";
for (int sz = seq.size(), i = 1; i < sz; i++) {
if (seq[i] > seq[i - 1]) {
ans += "(";
} else {
ans += ")";
}
}
cout << ans << "\n";
}
``` |
#include <bits/stdc++.h>
using namespace std;
int dp[405][205][205][2];
string a;
string b;
string ans;
string fans;
int recur(int par, int la, int lb) {
if (par > 400) {
return 1000000000;
}
if (par == 0 && la == 0 && lb == 0) {
return 0;
}
if (dp[par][la][lb][0] >= 0) return dp[par][la][lb][0];
int tpar = par + 1;
int tla = la, tlb = lb;
if (la > 0 && a[la - 1] == ')') tla--;
if (lb > 0 && b[lb - 1] == ')') tlb--;
int val = recur(tpar, tla, tlb) + 1;
if (val > 0 && (dp[par][la][lb][0] == -1 || val < dp[par][la][lb][0])) {
dp[par][la][lb][0] = val;
dp[par][la][lb][1] = ')';
}
if (par > 0) {
tpar = par - 1;
tla = la, tlb = lb;
if (la > 0 && a[la - 1] == '(') tla--;
if (lb > 0 && b[lb - 1] == '(') tlb--;
int val = recur(tpar, tla, tlb) + 1;
if (val > 0 && (dp[par][la][lb][0] == -1 || val < dp[par][la][lb][0])) {
dp[par][la][lb][0] = val;
dp[par][la][lb][1] = '(';
}
}
return dp[par][la][lb][0];
}
void recons(int par, int la, int lb) {
if (par == 0 && la == 0 && lb == 0) return;
char c = dp[par][la][lb][1];
if (c == ')') {
if (la > 0 && a[la - 1] == ')') la--;
if (lb > 0 && b[lb - 1] == ')') lb--;
ans.push_back(')');
recons(par + 1, la, lb);
} else {
if (la > 0 && a[la - 1] == '(') la--;
if (lb > 0 && b[lb - 1] == '(') lb--;
ans.push_back('(');
recons(par - 1, la, lb);
}
}
int main() {
for (int i = 0; i <= 401; i++)
for (int j = 0; j <= 201; j++)
for (int k = 0; k <= 201; k++) {
dp[i][j][k][0] = dp[i][j][k][1] = -1;
}
cin >> a;
cin >> b;
recur(0, a.size(), b.size());
recons(0, a.size(), b.size());
reverse(ans.begin(), ans.end());
cout << ans << endl;
return 0;
}
| ### Prompt
Develop a solution in Cpp to the problem described below:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int dp[405][205][205][2];
string a;
string b;
string ans;
string fans;
int recur(int par, int la, int lb) {
if (par > 400) {
return 1000000000;
}
if (par == 0 && la == 0 && lb == 0) {
return 0;
}
if (dp[par][la][lb][0] >= 0) return dp[par][la][lb][0];
int tpar = par + 1;
int tla = la, tlb = lb;
if (la > 0 && a[la - 1] == ')') tla--;
if (lb > 0 && b[lb - 1] == ')') tlb--;
int val = recur(tpar, tla, tlb) + 1;
if (val > 0 && (dp[par][la][lb][0] == -1 || val < dp[par][la][lb][0])) {
dp[par][la][lb][0] = val;
dp[par][la][lb][1] = ')';
}
if (par > 0) {
tpar = par - 1;
tla = la, tlb = lb;
if (la > 0 && a[la - 1] == '(') tla--;
if (lb > 0 && b[lb - 1] == '(') tlb--;
int val = recur(tpar, tla, tlb) + 1;
if (val > 0 && (dp[par][la][lb][0] == -1 || val < dp[par][la][lb][0])) {
dp[par][la][lb][0] = val;
dp[par][la][lb][1] = '(';
}
}
return dp[par][la][lb][0];
}
void recons(int par, int la, int lb) {
if (par == 0 && la == 0 && lb == 0) return;
char c = dp[par][la][lb][1];
if (c == ')') {
if (la > 0 && a[la - 1] == ')') la--;
if (lb > 0 && b[lb - 1] == ')') lb--;
ans.push_back(')');
recons(par + 1, la, lb);
} else {
if (la > 0 && a[la - 1] == '(') la--;
if (lb > 0 && b[lb - 1] == '(') lb--;
ans.push_back('(');
recons(par - 1, la, lb);
}
}
int main() {
for (int i = 0; i <= 401; i++)
for (int j = 0; j <= 201; j++)
for (int k = 0; k <= 201; k++) {
dp[i][j][k][0] = dp[i][j][k][1] = -1;
}
cin >> a;
cin >> b;
recur(0, a.size(), b.size());
recons(0, a.size(), b.size());
reverse(ans.begin(), ans.end());
cout << ans << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 205;
int n, m, dp[MAX_N][MAX_N][MAX_N];
char a[MAX_N], b[MAX_N];
int f(int i, int j, int k) {
if (k > 200) return 123456;
if (i == n && j == m) return k;
if (dp[i][j][k] != -1) return dp[i][j][k];
int ans = f(i + (a[i] == '('), j + (b[j] == '('), k + 1);
if (k) ans = min(ans, f(i + (a[i] == ')'), j + (b[j] == ')'), k - 1));
return dp[i][j][k] = 1 + ans;
}
int main() {
scanf("%s%s", a, b);
n = strlen(a), m = strlen(b);
for (int i = 0; i <= n; i++)
for (int j = 0; j <= m; j++)
for (int k = 0; k <= 200; k++) dp[i][j][k] = -1;
int i = 0, j = 0, k = 0;
while (i < n || j < m)
if (f(i, j, k) == 1 + f(i + (a[i] == '('), j + (b[j] == '('), k + 1)) {
putchar('(');
i += a[i] == '(';
j += b[j] == '(';
k++;
} else {
putchar(')');
i += a[i] == ')';
j += b[j] == ')';
k--;
}
while (k--) putchar(')');
putchar('\n');
return 0;
}
| ### Prompt
Please provide a cpp coded solution to the problem described below:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 205;
int n, m, dp[MAX_N][MAX_N][MAX_N];
char a[MAX_N], b[MAX_N];
int f(int i, int j, int k) {
if (k > 200) return 123456;
if (i == n && j == m) return k;
if (dp[i][j][k] != -1) return dp[i][j][k];
int ans = f(i + (a[i] == '('), j + (b[j] == '('), k + 1);
if (k) ans = min(ans, f(i + (a[i] == ')'), j + (b[j] == ')'), k - 1));
return dp[i][j][k] = 1 + ans;
}
int main() {
scanf("%s%s", a, b);
n = strlen(a), m = strlen(b);
for (int i = 0; i <= n; i++)
for (int j = 0; j <= m; j++)
for (int k = 0; k <= 200; k++) dp[i][j][k] = -1;
int i = 0, j = 0, k = 0;
while (i < n || j < m)
if (f(i, j, k) == 1 + f(i + (a[i] == '('), j + (b[j] == '('), k + 1)) {
putchar('(');
i += a[i] == '(';
j += b[j] == '(';
k++;
} else {
putchar(')');
i += a[i] == ')';
j += b[j] == ')';
k--;
}
while (k--) putchar(')');
putchar('\n');
return 0;
}
``` |
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
const int N = 210;
const int inf = 1e9 + 7;
int dp[N][N][N];
vector<int> pr[N][N][N];
string t = "";
void rec(int i, int j, int b) {
if (i == 0 && j == 0 && b == 0) return;
if (pr[i][j][b][2] > b)
t += ')';
else
t += '(';
rec(pr[i][j][b][0], pr[i][j][b][1], pr[i][j][b][2]);
}
int32_t main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
string s;
cin >> s;
string s1;
cin >> s1;
int n = s.size();
int m = s1.size();
for (int i = 0; i <= n; i++)
for (int j = 0; j <= m; j++)
for (int b = 0; b < N; b++) dp[i][j][b] = inf;
dp[0][0][0] = 0;
for (int i = 0; i <= n; i++)
for (int j = 0; j <= m; j++)
for (int b = 0; b <= max(n, m); b++) {
int ni = i + (s[i] == '(');
int nj = j + (s1[j] == '(');
if (dp[ni][nj][b + 1] > dp[i][j][b] + 1) {
dp[ni][nj][b + 1] = min(dp[ni][nj][b + 1], dp[i][j][b] + 1);
pr[ni][nj][b + 1] = {i, j, b};
}
if (b) {
ni = i + (s[i] == ')');
nj = j + (s1[j] == ')');
if (dp[ni][nj][b - 1] > dp[i][j][b] + 1) {
dp[ni][nj][b - 1] = min(dp[ni][nj][b - 1], dp[i][j][b] + 1);
pr[ni][nj][b - 1] = {i, j, b};
}
}
}
int ans = 0;
for (int b = 0; b < N; b++)
if (dp[n][m][ans] + ans > dp[n][m][b] + b) ans = b;
rec(n, m, ans);
reverse(t.begin(), t.end());
while (ans--) t += ')';
cout << t;
return 0;
}
| ### Prompt
Please create a solution in CPP to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
const int N = 210;
const int inf = 1e9 + 7;
int dp[N][N][N];
vector<int> pr[N][N][N];
string t = "";
void rec(int i, int j, int b) {
if (i == 0 && j == 0 && b == 0) return;
if (pr[i][j][b][2] > b)
t += ')';
else
t += '(';
rec(pr[i][j][b][0], pr[i][j][b][1], pr[i][j][b][2]);
}
int32_t main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
string s;
cin >> s;
string s1;
cin >> s1;
int n = s.size();
int m = s1.size();
for (int i = 0; i <= n; i++)
for (int j = 0; j <= m; j++)
for (int b = 0; b < N; b++) dp[i][j][b] = inf;
dp[0][0][0] = 0;
for (int i = 0; i <= n; i++)
for (int j = 0; j <= m; j++)
for (int b = 0; b <= max(n, m); b++) {
int ni = i + (s[i] == '(');
int nj = j + (s1[j] == '(');
if (dp[ni][nj][b + 1] > dp[i][j][b] + 1) {
dp[ni][nj][b + 1] = min(dp[ni][nj][b + 1], dp[i][j][b] + 1);
pr[ni][nj][b + 1] = {i, j, b};
}
if (b) {
ni = i + (s[i] == ')');
nj = j + (s1[j] == ')');
if (dp[ni][nj][b - 1] > dp[i][j][b] + 1) {
dp[ni][nj][b - 1] = min(dp[ni][nj][b - 1], dp[i][j][b] + 1);
pr[ni][nj][b - 1] = {i, j, b};
}
}
}
int ans = 0;
for (int b = 0; b < N; b++)
if (dp[n][m][ans] + ans > dp[n][m][b] + b) ans = b;
rec(n, m, ans);
reverse(t.begin(), t.end());
while (ans--) t += ')';
cout << t;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
string s1, s2;
cin >> s1 >> s2;
int n = s1.size();
int m = s2.size();
pair<pair<int, int>, pair<int, char> > parent[n + 1][m + 1][205];
int dp[n + 5][m + 5][205];
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= m; j++) {
for (int bal = 0; bal <= 200; bal++) {
dp[i][j][bal] = 600;
}
}
}
dp[0][0][0] = 0;
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= m; j++) {
for (int bal = 0; bal <= 200; bal++) {
if (dp[i][j][bal] == 600) continue;
int newI = i + (i < n && s1[i] == '(');
int newJ = j + (j < m && s2[j] == '(');
if (bal + 1 < 205 && dp[newI][newJ][bal + 1] > dp[i][j][bal] + 1) {
dp[newI][newJ][bal + 1] = dp[i][j][bal] + 1;
parent[newI][newJ][bal + 1] =
make_pair(make_pair(i, j), make_pair(bal, '('));
}
newI = i + (i < n && s1[i] == ')');
newJ = j + (j < m && s2[j] == ')');
if (bal > 0 && dp[newI][newJ][bal - 1] > dp[i][j][bal] + 1) {
dp[newI][newJ][bal - 1] = dp[i][j][bal] + 1;
parent[newI][newJ][bal - 1] =
make_pair(make_pair(i, j), make_pair(bal, ')'));
}
}
}
}
int cBal = 0;
for (int bal = 0; bal <= 200; bal++) {
if (dp[n][m][bal] + bal < dp[n][m][cBal] + cBal) {
cBal = bal;
}
}
int ci = n, cj = m;
string res = string(cBal, ')');
while (ci > 0 || cj > 0 || cBal != 0) {
int nci = parent[ci][cj][cBal].first.first;
int ncj = parent[ci][cj][cBal].first.second;
int ncBal = parent[ci][cj][cBal].second.first;
res += parent[ci][cj][cBal].second.second;
ci = nci;
cj = ncj;
cBal = ncBal;
}
reverse(res.begin(), res.end());
cout << res << endl;
return 0;
}
| ### Prompt
Please create a solution in Cpp to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
string s1, s2;
cin >> s1 >> s2;
int n = s1.size();
int m = s2.size();
pair<pair<int, int>, pair<int, char> > parent[n + 1][m + 1][205];
int dp[n + 5][m + 5][205];
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= m; j++) {
for (int bal = 0; bal <= 200; bal++) {
dp[i][j][bal] = 600;
}
}
}
dp[0][0][0] = 0;
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= m; j++) {
for (int bal = 0; bal <= 200; bal++) {
if (dp[i][j][bal] == 600) continue;
int newI = i + (i < n && s1[i] == '(');
int newJ = j + (j < m && s2[j] == '(');
if (bal + 1 < 205 && dp[newI][newJ][bal + 1] > dp[i][j][bal] + 1) {
dp[newI][newJ][bal + 1] = dp[i][j][bal] + 1;
parent[newI][newJ][bal + 1] =
make_pair(make_pair(i, j), make_pair(bal, '('));
}
newI = i + (i < n && s1[i] == ')');
newJ = j + (j < m && s2[j] == ')');
if (bal > 0 && dp[newI][newJ][bal - 1] > dp[i][j][bal] + 1) {
dp[newI][newJ][bal - 1] = dp[i][j][bal] + 1;
parent[newI][newJ][bal - 1] =
make_pair(make_pair(i, j), make_pair(bal, ')'));
}
}
}
}
int cBal = 0;
for (int bal = 0; bal <= 200; bal++) {
if (dp[n][m][bal] + bal < dp[n][m][cBal] + cBal) {
cBal = bal;
}
}
int ci = n, cj = m;
string res = string(cBal, ')');
while (ci > 0 || cj > 0 || cBal != 0) {
int nci = parent[ci][cj][cBal].first.first;
int ncj = parent[ci][cj][cBal].first.second;
int ncBal = parent[ci][cj][cBal].second.first;
res += parent[ci][cj][cBal].second.second;
ci = nci;
cj = ncj;
cBal = ncBal;
}
reverse(res.begin(), res.end());
cout << res << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
string s, t;
long long n, m, mx, dp[205][205][825];
long long solve(int i, int j, int op) {
if (i == n && j == m) return op;
if (op > mx) return INT_MAX;
long long &ret = dp[i][j][op];
if (~ret) return ret;
long long ans = INT_MAX;
long long x = (i < n && s[i] == '(' ? i + 1 : i);
long long y = (j < m && t[j] == '(' ? j + 1 : j);
ans = min(ans, solve(x, y, op + 1) + 1);
if (op) {
x = (i < n && s[i] == ')' ? i + 1 : i);
y = (j < m && t[j] == ')' ? j + 1 : j);
ans = min(ans, solve(x, y, op - 1) + 1);
}
return ret = ans;
}
void path(int i, int j, int op) {
if (i == n && j == m) {
while (op--) cout << ')';
return;
}
if (op > mx) return;
long long x = (i < n && s[i] == '(' ? i + 1 : i);
long long y = (j < m && t[j] == '(' ? j + 1 : j);
long long ans1 = solve(x, y, op + 1) + 1, ans2 = INT_MAX;
if (op) {
x = (i < n && s[i] == ')' ? i + 1 : i);
y = (j < m && t[j] == ')' ? j + 1 : j);
ans2 = solve(x, y, op - 1) + 1;
}
if (ans1 < ans2) {
cout << '(';
x = (i < n && s[i] == '(' ? i + 1 : i);
y = (j < m && t[j] == '(' ? j + 1 : j);
path(x, y, op + 1);
} else {
cout << ')';
x = (i < n && s[i] == ')' ? i + 1 : i);
y = (j < m && t[j] == ')' ? j + 1 : j);
path(x, y, op - 1);
}
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> s >> t;
memset(dp, -1, sizeof dp);
n = s.size(), m = t.size();
mx = (n + m) * 2;
solve(0, 0, 0);
path(0, 0, 0);
return 0;
}
| ### Prompt
Develop a solution in cpp to the problem described below:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
string s, t;
long long n, m, mx, dp[205][205][825];
long long solve(int i, int j, int op) {
if (i == n && j == m) return op;
if (op > mx) return INT_MAX;
long long &ret = dp[i][j][op];
if (~ret) return ret;
long long ans = INT_MAX;
long long x = (i < n && s[i] == '(' ? i + 1 : i);
long long y = (j < m && t[j] == '(' ? j + 1 : j);
ans = min(ans, solve(x, y, op + 1) + 1);
if (op) {
x = (i < n && s[i] == ')' ? i + 1 : i);
y = (j < m && t[j] == ')' ? j + 1 : j);
ans = min(ans, solve(x, y, op - 1) + 1);
}
return ret = ans;
}
void path(int i, int j, int op) {
if (i == n && j == m) {
while (op--) cout << ')';
return;
}
if (op > mx) return;
long long x = (i < n && s[i] == '(' ? i + 1 : i);
long long y = (j < m && t[j] == '(' ? j + 1 : j);
long long ans1 = solve(x, y, op + 1) + 1, ans2 = INT_MAX;
if (op) {
x = (i < n && s[i] == ')' ? i + 1 : i);
y = (j < m && t[j] == ')' ? j + 1 : j);
ans2 = solve(x, y, op - 1) + 1;
}
if (ans1 < ans2) {
cout << '(';
x = (i < n && s[i] == '(' ? i + 1 : i);
y = (j < m && t[j] == '(' ? j + 1 : j);
path(x, y, op + 1);
} else {
cout << ')';
x = (i < n && s[i] == ')' ? i + 1 : i);
y = (j < m && t[j] == ')' ? j + 1 : j);
path(x, y, op - 1);
}
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> s >> t;
memset(dp, -1, sizeof dp);
n = s.size(), m = t.size();
mx = (n + m) * 2;
solve(0, 0, 0);
path(0, 0, 0);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
string s, t;
int dp[222][222][222];
int skad[222][222][222];
int n, m;
int reku(int a, int b, int bal) {
if (bal > 210 || bal < 0) return 1e9;
if (a == 0 && b == 0) return bal;
if (a > n && b > m) return 1e9;
if (dp[a][b][bal] != 1e9) return dp[a][b][bal];
if (a == 0 || b == 0) {
int lewo = 1e9;
int prawo = 1e9;
if (a == 0) {
int xd;
xd = b;
if (t[xd - 1] == ')') {
xd--;
lewo = reku(0, xd, bal + 1);
} else
lewo = reku(0, b, bal + 1);
xd = b;
if (t[xd - 1] == '(') {
xd--;
prawo = reku(0, xd, bal - 1);
} else
prawo = reku(0, b, bal - 1);
dp[a][b][bal] = min(dp[a][b][bal], 1 + min(lewo, prawo));
} else if (b == 0) {
int xd = a;
if (s[xd - 1] == ')') {
xd--;
lewo = reku(xd, 0, bal + 1);
} else
lewo = reku(a, 0, bal + 1);
xd = a;
if (s[xd - 1] == '(') {
xd--;
prawo = reku(xd, 0, bal - 1);
} else
prawo = reku(a, 0, bal - 1);
dp[a][b][bal] = min(dp[a][b][bal], 1 + min(lewo, prawo));
}
return dp[a][b][bal];
}
int res = 1e9;
int pot1 = 1;
if (true) {
int xd1, xd2;
xd1 = a;
xd2 = b;
if (xd1 > 0) {
if (s[xd1 - 1] == ')') xd1--;
}
if (xd2 > 0) {
if (t[xd2 - 1] == ')') xd2--;
}
pot1 += reku(xd1, xd2, bal + 1);
res = min(res, pot1);
}
if (bal > 0) {
int pot2 = 1;
int xd1, xd2;
xd1 = a;
xd2 = b;
if (xd1 > 0) {
if (s[xd1 - 1] == '(') xd1--;
}
if (xd2 > 0) {
if (t[xd2 - 1] == '(') xd2--;
}
pot2 += reku(xd1, xd2, bal - 1);
res = min(res, pot2);
}
dp[a][b][bal] = res;
return res;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cin >> s >> t;
n = s.length();
m = t.length();
for (int i = 0; i < 222; i++) {
for (int j = 0; j < 222; j++) {
for (int k = 0; k < 222; k++) {
dp[i][j][k] = 1e9;
}
}
}
string ans;
int wyn = reku(n, m, 0);
int bal = 0;
for (int i = wyn; i >= 1; i--) {
if (bal == 0) {
ans += ")";
bal++;
if (s[n - 1] == ')') n--;
if (t[m - 1] == ')') m--;
continue;
}
int xd1 = n;
int xd2 = m;
if (s[n - 1] == ')') xd1--;
if (t[m - 1] == ')') xd2--;
if (dp[xd1][xd2][bal + 1] == i - 1) {
n = xd1;
m = xd2;
bal++;
ans += ")";
continue;
} else {
xd1 = n;
xd2 = m;
if (s[n - 1] == '(') xd1--;
if (t[m - 1] == '(') xd2--;
n = xd1;
m = xd2;
bal--;
ans += "(";
continue;
}
}
for (int i = wyn - 1; i >= 0; i--) cout << ans[i];
return 0;
}
| ### Prompt
Your challenge is to write a Cpp solution to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
string s, t;
int dp[222][222][222];
int skad[222][222][222];
int n, m;
int reku(int a, int b, int bal) {
if (bal > 210 || bal < 0) return 1e9;
if (a == 0 && b == 0) return bal;
if (a > n && b > m) return 1e9;
if (dp[a][b][bal] != 1e9) return dp[a][b][bal];
if (a == 0 || b == 0) {
int lewo = 1e9;
int prawo = 1e9;
if (a == 0) {
int xd;
xd = b;
if (t[xd - 1] == ')') {
xd--;
lewo = reku(0, xd, bal + 1);
} else
lewo = reku(0, b, bal + 1);
xd = b;
if (t[xd - 1] == '(') {
xd--;
prawo = reku(0, xd, bal - 1);
} else
prawo = reku(0, b, bal - 1);
dp[a][b][bal] = min(dp[a][b][bal], 1 + min(lewo, prawo));
} else if (b == 0) {
int xd = a;
if (s[xd - 1] == ')') {
xd--;
lewo = reku(xd, 0, bal + 1);
} else
lewo = reku(a, 0, bal + 1);
xd = a;
if (s[xd - 1] == '(') {
xd--;
prawo = reku(xd, 0, bal - 1);
} else
prawo = reku(a, 0, bal - 1);
dp[a][b][bal] = min(dp[a][b][bal], 1 + min(lewo, prawo));
}
return dp[a][b][bal];
}
int res = 1e9;
int pot1 = 1;
if (true) {
int xd1, xd2;
xd1 = a;
xd2 = b;
if (xd1 > 0) {
if (s[xd1 - 1] == ')') xd1--;
}
if (xd2 > 0) {
if (t[xd2 - 1] == ')') xd2--;
}
pot1 += reku(xd1, xd2, bal + 1);
res = min(res, pot1);
}
if (bal > 0) {
int pot2 = 1;
int xd1, xd2;
xd1 = a;
xd2 = b;
if (xd1 > 0) {
if (s[xd1 - 1] == '(') xd1--;
}
if (xd2 > 0) {
if (t[xd2 - 1] == '(') xd2--;
}
pot2 += reku(xd1, xd2, bal - 1);
res = min(res, pot2);
}
dp[a][b][bal] = res;
return res;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cin >> s >> t;
n = s.length();
m = t.length();
for (int i = 0; i < 222; i++) {
for (int j = 0; j < 222; j++) {
for (int k = 0; k < 222; k++) {
dp[i][j][k] = 1e9;
}
}
}
string ans;
int wyn = reku(n, m, 0);
int bal = 0;
for (int i = wyn; i >= 1; i--) {
if (bal == 0) {
ans += ")";
bal++;
if (s[n - 1] == ')') n--;
if (t[m - 1] == ')') m--;
continue;
}
int xd1 = n;
int xd2 = m;
if (s[n - 1] == ')') xd1--;
if (t[m - 1] == ')') xd2--;
if (dp[xd1][xd2][bal + 1] == i - 1) {
n = xd1;
m = xd2;
bal++;
ans += ")";
continue;
} else {
xd1 = n;
xd2 = m;
if (s[n - 1] == '(') xd1--;
if (t[m - 1] == '(') xd2--;
n = xd1;
m = xd2;
bal--;
ans += "(";
continue;
}
}
for (int i = wyn - 1; i >= 0; i--) cout << ans[i];
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 210;
const int inf = 0x3f3f3f3f;
char s[N], t[N];
int n, m;
int dp[N][N][N * 2];
struct state {
int i, j, k;
state() {}
state(int _i, int _j, int _k) : i(_i), j(_j), k(_k) {}
};
state pre[N][N][N * 2];
string ans;
inline void upd(int i, int j, int k, int _i, int _j, int _k) {
if (_i < 0 || _j < 0 || _i > n || _j > m || _k > n + m || _k < 0) return;
if (dp[i][j][k] > dp[_i][_j][_k] + 1) {
dp[i][j][k] = dp[_i][_j][_k] + 1;
pre[i][j][k] = state(_i, _j, _k);
}
}
int main() {
scanf("%s", s);
scanf("%s", t);
n = strlen(s), m = strlen(t);
memset(dp, inf, sizeof(dp));
dp[0][0][0] = 0;
for (int i = 0; i <= n; ++i) {
for (int j = 0; j <= m; ++j) {
for (int k = 0; k <= n + m; ++k) {
upd(i, j, k, i - (i && s[i - 1] == '('), j - (j && t[j - 1] == '('),
k - 1);
upd(i, j, k, i - (i && s[i - 1] == ')'), j - (j && t[j - 1] == ')'),
k + 1);
}
}
}
int p1 = n, p2 = m, pk = 0;
for (int k = 0; k <= n + m; ++k) {
if (dp[n][m][pk] + pk > dp[n][m][k] + k) {
pk = k;
}
}
for (int i = 0; i < pk; ++i) ans += ")";
while (p1 || p2 || pk) {
state now = pre[p1][p2][pk];
if (pk == now.k)
ans += ")(";
else if (pk < now.k)
ans += ")";
else
ans += "(";
p1 = now.i;
p2 = now.j;
pk = now.k;
}
reverse(ans.begin(), ans.end());
cout << ans << endl;
return 0;
}
| ### Prompt
In CPP, your task is to solve the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 210;
const int inf = 0x3f3f3f3f;
char s[N], t[N];
int n, m;
int dp[N][N][N * 2];
struct state {
int i, j, k;
state() {}
state(int _i, int _j, int _k) : i(_i), j(_j), k(_k) {}
};
state pre[N][N][N * 2];
string ans;
inline void upd(int i, int j, int k, int _i, int _j, int _k) {
if (_i < 0 || _j < 0 || _i > n || _j > m || _k > n + m || _k < 0) return;
if (dp[i][j][k] > dp[_i][_j][_k] + 1) {
dp[i][j][k] = dp[_i][_j][_k] + 1;
pre[i][j][k] = state(_i, _j, _k);
}
}
int main() {
scanf("%s", s);
scanf("%s", t);
n = strlen(s), m = strlen(t);
memset(dp, inf, sizeof(dp));
dp[0][0][0] = 0;
for (int i = 0; i <= n; ++i) {
for (int j = 0; j <= m; ++j) {
for (int k = 0; k <= n + m; ++k) {
upd(i, j, k, i - (i && s[i - 1] == '('), j - (j && t[j - 1] == '('),
k - 1);
upd(i, j, k, i - (i && s[i - 1] == ')'), j - (j && t[j - 1] == ')'),
k + 1);
}
}
}
int p1 = n, p2 = m, pk = 0;
for (int k = 0; k <= n + m; ++k) {
if (dp[n][m][pk] + pk > dp[n][m][k] + k) {
pk = k;
}
}
for (int i = 0; i < pk; ++i) ans += ")";
while (p1 || p2 || pk) {
state now = pre[p1][p2][pk];
if (pk == now.k)
ans += ")(";
else if (pk < now.k)
ans += ")";
else
ans += "(";
p1 = now.i;
p2 = now.j;
pk = now.k;
}
reverse(ans.begin(), ans.end());
cout << ans << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e9 + 7;
const int N = 2e2 + 10;
int dp[N][N][N];
tuple<int, int, int> pre[N][N][N];
char s1[N];
char s2[N];
int main() {
memset((dp), (0x3f), sizeof(dp));
scanf("%s", s1 + 1);
scanf("%s", s2 + 1);
int n = strlen(s1 + 1);
int m = strlen(s2 + 1);
dp[0][0][0] = 0;
int minv = INF;
queue<tuple<int, int, int>> q;
q.emplace(0, 0, 0);
while (!q.empty()) {
auto [o, x, y] = q.front();
q.pop();
string str = "()";
for (auto& c : str) {
int oo = o, xx = x, yy = y;
if (c == '(')
oo++;
else
oo--;
if (oo < 0) continue;
if (oo == N) continue;
if (s1[xx + 1] == c) xx++;
if (s2[yy + 1] == c) yy++;
if (dp[oo][xx][yy] > dp[o][x][y] + 1) {
dp[oo][xx][yy] = dp[o][x][y] + 1;
pre[oo][xx][yy] = {o, x, y};
q.emplace(oo, xx, yy);
}
}
}
int o = 0, x = n, y = m;
string ans;
while (o > 0 || x > 0 || y > 0) {
auto [oo, xx, yy] = pre[o][x][y];
if (o > oo)
ans.push_back('(');
else
ans.push_back(')');
tie(o, x, y) = pre[o][x][y];
}
reverse(ans.begin(), ans.end());
puts(ans.c_str());
return 0;
}
| ### Prompt
Generate a Cpp solution to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e9 + 7;
const int N = 2e2 + 10;
int dp[N][N][N];
tuple<int, int, int> pre[N][N][N];
char s1[N];
char s2[N];
int main() {
memset((dp), (0x3f), sizeof(dp));
scanf("%s", s1 + 1);
scanf("%s", s2 + 1);
int n = strlen(s1 + 1);
int m = strlen(s2 + 1);
dp[0][0][0] = 0;
int minv = INF;
queue<tuple<int, int, int>> q;
q.emplace(0, 0, 0);
while (!q.empty()) {
auto [o, x, y] = q.front();
q.pop();
string str = "()";
for (auto& c : str) {
int oo = o, xx = x, yy = y;
if (c == '(')
oo++;
else
oo--;
if (oo < 0) continue;
if (oo == N) continue;
if (s1[xx + 1] == c) xx++;
if (s2[yy + 1] == c) yy++;
if (dp[oo][xx][yy] > dp[o][x][y] + 1) {
dp[oo][xx][yy] = dp[o][x][y] + 1;
pre[oo][xx][yy] = {o, x, y};
q.emplace(oo, xx, yy);
}
}
}
int o = 0, x = n, y = m;
string ans;
while (o > 0 || x > 0 || y > 0) {
auto [oo, xx, yy] = pre[o][x][y];
if (o > oo)
ans.push_back('(');
else
ans.push_back(')');
tie(o, x, y) = pre[o][x][y];
}
reverse(ans.begin(), ans.end());
puts(ans.c_str());
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 205;
long long n1, n2, dp[N][N][N];
string s1, s2, s = "";
long long go(long long i, long long j, long long k) {
if (k < 0 || k >= N) return 1e10;
if (i == n1 && j == n2 && k == 0) return 0;
long long &ans = dp[i][j][k];
if (ans != -1) return ans;
ans = 1e10;
long long x, y;
x = i;
y = j;
if (i < n1 && s1[i] == '(') x++;
if (j < n2 && s2[j] == '(') y++;
ans = min(ans, 1 + go(x, y, k + 1));
x = i;
y = j;
if (i < n1 && s1[i] == ')') x++;
if (j < n2 && s2[j] == ')') y++;
ans = min(ans, 1 + go(x, y, k - 1));
return ans;
}
void build(long long i, long long j, long long k) {
long long x1, x2, y1, y2;
x1 = x2 = i;
y1 = y2 = j;
if (s1[i] == '(') x1++;
if (s2[j] == '(') y1++;
if (s1[i] == ')') x2++;
if (s2[j] == ')') y2++;
if (i == n1 && j == n2 && k == 0) return;
if (go(i, j, k) == (1 + go(x1, y1, k + 1))) {
s += "(";
build(x1, y1, k + 1);
} else {
s += ")";
build(x2, y2, k - 1);
}
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
long long n, i, j, x, y, t;
cin >> s1 >> s2;
n1 = s1.size();
n2 = s2.size();
memset(dp, -1, sizeof(dp));
go(0, 0, 0);
x = y = 0;
if (s1[0] == '(') x++;
if (s2[0] == '(') y++;
build(0, 0, 0);
cout << s;
}
| ### Prompt
In CPP, your task is to solve the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 205;
long long n1, n2, dp[N][N][N];
string s1, s2, s = "";
long long go(long long i, long long j, long long k) {
if (k < 0 || k >= N) return 1e10;
if (i == n1 && j == n2 && k == 0) return 0;
long long &ans = dp[i][j][k];
if (ans != -1) return ans;
ans = 1e10;
long long x, y;
x = i;
y = j;
if (i < n1 && s1[i] == '(') x++;
if (j < n2 && s2[j] == '(') y++;
ans = min(ans, 1 + go(x, y, k + 1));
x = i;
y = j;
if (i < n1 && s1[i] == ')') x++;
if (j < n2 && s2[j] == ')') y++;
ans = min(ans, 1 + go(x, y, k - 1));
return ans;
}
void build(long long i, long long j, long long k) {
long long x1, x2, y1, y2;
x1 = x2 = i;
y1 = y2 = j;
if (s1[i] == '(') x1++;
if (s2[j] == '(') y1++;
if (s1[i] == ')') x2++;
if (s2[j] == ')') y2++;
if (i == n1 && j == n2 && k == 0) return;
if (go(i, j, k) == (1 + go(x1, y1, k + 1))) {
s += "(";
build(x1, y1, k + 1);
} else {
s += ")";
build(x2, y2, k - 1);
}
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
long long n, i, j, x, y, t;
cin >> s1 >> s2;
n1 = s1.size();
n2 = s2.size();
memset(dp, -1, sizeof(dp));
go(0, 0, 0);
x = y = 0;
if (s1[0] == '(') x++;
if (s2[0] == '(') y++;
build(0, 0, 0);
cout << s;
}
``` |
#include <bits/stdc++.h>
bool local = false;
using namespace std;
template <class T>
string toString(T x) {
ostringstream sout;
sout << x;
return sout.str();
}
template <class T>
void print(T v) {
cout << ((int)(v).size()) << "\n";
for (auto x : v) cout << x << ' ';
cout << "\n";
};
template <class T>
void print1(T v) {
cout << ((int)(v).size()) << "\n";
for (auto x : v) cout << x + 1 << ' ';
cout << "\n";
};
template <class T1, class T2>
ostream &operator<<(ostream &o, pair<T1, T2> x) {
return o << x.first << ' ' << x.second;
}
template <class T1, class T2>
istream &operator>>(istream &o, pair<T1, T2> &x) {
return o >> x.first >> x.second;
}
template <class T1, class T2>
pair<T1, T2> operator+(pair<T1, T2> a, pair<T1, T2> b) {
a.first += b.first;
a.second += b.second;
return a;
}
template <class T1, class T2>
pair<T1, T2> operator-(pair<T1, T2> a, pair<T1, T2> b) {
a.first -= b.first;
a.second -= b.second;
return a;
}
template <class T1, class T2>
void operator+=(pair<T1, T2> &a, pair<T1, T2> b) {
a.first += b.first;
a.second += b.second;
}
template <class T1, class T2>
void operator-=(pair<T1, T2> &a, pair<T1, T2> b) {
a.first -= b.first;
a.second -= b.second;
}
int nint() {
int x;
cin >> x;
return x;
}
double getTime() { return clock() / (double)CLOCKS_PER_SEC; };
mt19937 rnd(0);
int rand(int n) { return rnd() % n; }
int rand(int l, int r) { return rnd() % (r - l + 1) + l; }
const int mod = 1000000000 + 7;
void initIO() {
if (local) {
freopen("input.txt", "r", stdin);
srand((int)time(0));
rnd.seed((int)time(0));
} else {
{
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
};
}
}
void solve();
int main() {
initIO();
int tc = 1;
for (int tt = 0; tt < (tc); tt++) {
solve();
}
if (local)
cout << "\n"
<< "time = " << getTime() << "\n";
}
int n;
int m;
int table[210][210][420];
int ne[210][210][420];
string s1, s2;
int getans(int p1, int p2, int b) {
if (b < 0 || b > n + m) return mod;
if (p1 == n && p2 == m && b == 0) return 0;
int &ans = table[p1][p2][b];
if (ans != -1) return ans;
ans = getans(min(n, p1 + (s1[p1] == '(')), min(m, p2 + (s2[p2] == '(')),
b + 1) +
1;
ne[p1][p2][b] = 1;
if (b) {
int t = getans(min(n, p1 + (s1[p1] == ')')), min(m, p2 + (s2[p2] == ')')),
b - 1) +
1;
if (t < ans) {
ans = t;
ne[p1][p2][b] = 2;
}
}
return ans;
}
void solve() {
cin >> s1 >> s2;
n = ((int)(s1).size());
m = ((int)(s2).size());
s1 += " ";
s2 += " ";
memset(table, -1, sizeof(table));
getans(0, 0, 0);
int p1 = 0, p2 = 0, b = 0;
while (p1 < n || p2 < m || b) {
int o = ne[p1][p2][b];
if (o == 1) {
cout << "(";
if (p1 < n && s1[p1] == '(') p1++;
if (p2 < m && s2[p2] == '(') p2++;
b++;
} else {
cout << ")";
if (p1 < n && s1[p1] == ')') p1++;
if (p2 < m && s2[p2] == ')') p2++;
b--;
}
}
}
| ### Prompt
Generate a cpp solution to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
bool local = false;
using namespace std;
template <class T>
string toString(T x) {
ostringstream sout;
sout << x;
return sout.str();
}
template <class T>
void print(T v) {
cout << ((int)(v).size()) << "\n";
for (auto x : v) cout << x << ' ';
cout << "\n";
};
template <class T>
void print1(T v) {
cout << ((int)(v).size()) << "\n";
for (auto x : v) cout << x + 1 << ' ';
cout << "\n";
};
template <class T1, class T2>
ostream &operator<<(ostream &o, pair<T1, T2> x) {
return o << x.first << ' ' << x.second;
}
template <class T1, class T2>
istream &operator>>(istream &o, pair<T1, T2> &x) {
return o >> x.first >> x.second;
}
template <class T1, class T2>
pair<T1, T2> operator+(pair<T1, T2> a, pair<T1, T2> b) {
a.first += b.first;
a.second += b.second;
return a;
}
template <class T1, class T2>
pair<T1, T2> operator-(pair<T1, T2> a, pair<T1, T2> b) {
a.first -= b.first;
a.second -= b.second;
return a;
}
template <class T1, class T2>
void operator+=(pair<T1, T2> &a, pair<T1, T2> b) {
a.first += b.first;
a.second += b.second;
}
template <class T1, class T2>
void operator-=(pair<T1, T2> &a, pair<T1, T2> b) {
a.first -= b.first;
a.second -= b.second;
}
int nint() {
int x;
cin >> x;
return x;
}
double getTime() { return clock() / (double)CLOCKS_PER_SEC; };
mt19937 rnd(0);
int rand(int n) { return rnd() % n; }
int rand(int l, int r) { return rnd() % (r - l + 1) + l; }
const int mod = 1000000000 + 7;
void initIO() {
if (local) {
freopen("input.txt", "r", stdin);
srand((int)time(0));
rnd.seed((int)time(0));
} else {
{
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
};
}
}
void solve();
int main() {
initIO();
int tc = 1;
for (int tt = 0; tt < (tc); tt++) {
solve();
}
if (local)
cout << "\n"
<< "time = " << getTime() << "\n";
}
int n;
int m;
int table[210][210][420];
int ne[210][210][420];
string s1, s2;
int getans(int p1, int p2, int b) {
if (b < 0 || b > n + m) return mod;
if (p1 == n && p2 == m && b == 0) return 0;
int &ans = table[p1][p2][b];
if (ans != -1) return ans;
ans = getans(min(n, p1 + (s1[p1] == '(')), min(m, p2 + (s2[p2] == '(')),
b + 1) +
1;
ne[p1][p2][b] = 1;
if (b) {
int t = getans(min(n, p1 + (s1[p1] == ')')), min(m, p2 + (s2[p2] == ')')),
b - 1) +
1;
if (t < ans) {
ans = t;
ne[p1][p2][b] = 2;
}
}
return ans;
}
void solve() {
cin >> s1 >> s2;
n = ((int)(s1).size());
m = ((int)(s2).size());
s1 += " ";
s2 += " ";
memset(table, -1, sizeof(table));
getans(0, 0, 0);
int p1 = 0, p2 = 0, b = 0;
while (p1 < n || p2 < m || b) {
int o = ne[p1][p2][b];
if (o == 1) {
cout << "(";
if (p1 < n && s1[p1] == '(') p1++;
if (p2 < m && s2[p2] == '(') p2++;
b++;
} else {
cout << ")";
if (p1 < n && s1[p1] == ')') p1++;
if (p2 < m && s2[p2] == ')') p2++;
b--;
}
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
string s, t;
cin >> s >> t;
int s_size = s.size();
int t_size = t.size();
vector<vector<vector<int>>> dist(
1 + max(s_size, t_size),
vector<vector<int>>(1 + s_size, vector<int>(1 + t_size, 1000 * 1000)));
vector<vector<vector<array<int, 3>>>> prev(
1 + max(s_size, t_size),
vector<vector<array<int, 3>>>(
1 + s_size, vector<array<int, 3>>(1 + t_size, {0, 0, 0})));
dist[0][0][0] = 0;
queue<array<int, 3>> q;
q.push({0, 0, 0});
while (!q.empty()) {
auto curr = q.front();
q.pop();
if (curr[0] + 1 < dist.size()) {
int next1 = curr[1] + (curr[1] < s_size && s[curr[1]] == '(');
int next2 = curr[2] + (curr[2] < t_size && t[curr[2]] == '(');
if (dist[curr[0] + 1][next1][next2] == 1000 * 1000) {
dist[curr[0] + 1][next1][next2] = dist[curr[0]][curr[1]][curr[2]] + 1;
prev[curr[0] + 1][next1][next2] = curr;
q.push({curr[0] + 1, next1, next2});
}
}
if (curr[0] - 1 >= 0) {
int next1 = curr[1] + (curr[1] < s_size && s[curr[1]] == ')');
int next2 = curr[2] + (curr[2] < t_size && t[curr[2]] == ')');
if (dist[curr[0] - 1][next1][next2] == 1000 * 1000) {
dist[curr[0] - 1][next1][next2] = dist[curr[0]][curr[1]][curr[2]] + 1;
prev[curr[0] - 1][next1][next2] = curr;
q.push({curr[0] - 1, next1, next2});
}
}
}
array<int, 3> curr = {0, s_size, t_size};
string res = "";
while (curr != array<int, 3>{0, 0, 0}) {
if (prev[curr[0]][curr[1]][curr[2]][0] + 1 == curr[0]) {
res += '(';
} else {
res += ')';
}
curr = prev[curr[0]][curr[1]][curr[2]];
}
reverse(res.begin(), res.end());
cout << res << "\n";
}
| ### Prompt
Your challenge is to write a Cpp solution to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
string s, t;
cin >> s >> t;
int s_size = s.size();
int t_size = t.size();
vector<vector<vector<int>>> dist(
1 + max(s_size, t_size),
vector<vector<int>>(1 + s_size, vector<int>(1 + t_size, 1000 * 1000)));
vector<vector<vector<array<int, 3>>>> prev(
1 + max(s_size, t_size),
vector<vector<array<int, 3>>>(
1 + s_size, vector<array<int, 3>>(1 + t_size, {0, 0, 0})));
dist[0][0][0] = 0;
queue<array<int, 3>> q;
q.push({0, 0, 0});
while (!q.empty()) {
auto curr = q.front();
q.pop();
if (curr[0] + 1 < dist.size()) {
int next1 = curr[1] + (curr[1] < s_size && s[curr[1]] == '(');
int next2 = curr[2] + (curr[2] < t_size && t[curr[2]] == '(');
if (dist[curr[0] + 1][next1][next2] == 1000 * 1000) {
dist[curr[0] + 1][next1][next2] = dist[curr[0]][curr[1]][curr[2]] + 1;
prev[curr[0] + 1][next1][next2] = curr;
q.push({curr[0] + 1, next1, next2});
}
}
if (curr[0] - 1 >= 0) {
int next1 = curr[1] + (curr[1] < s_size && s[curr[1]] == ')');
int next2 = curr[2] + (curr[2] < t_size && t[curr[2]] == ')');
if (dist[curr[0] - 1][next1][next2] == 1000 * 1000) {
dist[curr[0] - 1][next1][next2] = dist[curr[0]][curr[1]][curr[2]] + 1;
prev[curr[0] - 1][next1][next2] = curr;
q.push({curr[0] - 1, next1, next2});
}
}
}
array<int, 3> curr = {0, s_size, t_size};
string res = "";
while (curr != array<int, 3>{0, 0, 0}) {
if (prev[curr[0]][curr[1]][curr[2]][0] + 1 == curr[0]) {
res += '(';
} else {
res += ')';
}
curr = prev[curr[0]][curr[1]][curr[2]];
}
reverse(res.begin(), res.end());
cout << res << "\n";
}
``` |
#include <bits/stdc++.h>
using namespace std;
int dp[205][205][205];
int from[205][205][205];
string S, T;
string ans;
void get(int i, int j, int b) {
if (i == 0 && j == 0 && b == 0) return;
int fi = from[i][j][b] / (205 * 205);
int fj = from[i][j][b] % (205 * 205) / 205;
int fb = from[i][j][b] % 205;
get(fi, fj, fb);
if (fb < b)
ans.push_back('(');
else
ans.push_back(')');
}
int main() {
cin >> S >> T;
memset(dp, 0x3f, sizeof(dp));
dp[0][0][0] = 0;
for (int i = 0; i <= S.size(); i++) {
for (int j = 0; j <= T.size(); j++) {
for (int b = 0; b < 205; b++) {
if (dp[i][j][b] + 1 < dp[i][j][b + 1])
dp[i][j][b + 1] = dp[i][j][b] + 1,
from[i][j][b + 1] = (i * 205 * 205 + j * 205 + b);
for (char c : "()") {
int diff = (c == '(') ? 1 : -1;
if (b + diff < 205 && b + diff >= 0 && i + (S[i] == c) <= S.size() &&
j + (T[j] == c) <= T.size()) {
if (dp[i][j][b] + 1 <
dp[i + (S[i] == c)][j + (T[j] == c)][b + diff])
dp[i + (S[i] == c)][j + (T[j] == c)]
[b + diff] = dp[i][j][b] + 1,
from[i + (S[i] == c)][j + (T[j] == c)][b + diff] =
(i * 205 * 205 + j * 205 + b);
}
}
}
}
}
int mini = 0;
for (int i = 1; i < 205; i++)
if (dp[S.size()][T.size()][i] + i < dp[S.size()][T.size()][mini] + mini)
mini = i;
get(S.size(), T.size(), mini);
for (int i = 0; i < mini; i++) ans.push_back(')');
cout << ans << endl;
return 0;
}
| ### Prompt
Your challenge is to write a cpp solution to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int dp[205][205][205];
int from[205][205][205];
string S, T;
string ans;
void get(int i, int j, int b) {
if (i == 0 && j == 0 && b == 0) return;
int fi = from[i][j][b] / (205 * 205);
int fj = from[i][j][b] % (205 * 205) / 205;
int fb = from[i][j][b] % 205;
get(fi, fj, fb);
if (fb < b)
ans.push_back('(');
else
ans.push_back(')');
}
int main() {
cin >> S >> T;
memset(dp, 0x3f, sizeof(dp));
dp[0][0][0] = 0;
for (int i = 0; i <= S.size(); i++) {
for (int j = 0; j <= T.size(); j++) {
for (int b = 0; b < 205; b++) {
if (dp[i][j][b] + 1 < dp[i][j][b + 1])
dp[i][j][b + 1] = dp[i][j][b] + 1,
from[i][j][b + 1] = (i * 205 * 205 + j * 205 + b);
for (char c : "()") {
int diff = (c == '(') ? 1 : -1;
if (b + diff < 205 && b + diff >= 0 && i + (S[i] == c) <= S.size() &&
j + (T[j] == c) <= T.size()) {
if (dp[i][j][b] + 1 <
dp[i + (S[i] == c)][j + (T[j] == c)][b + diff])
dp[i + (S[i] == c)][j + (T[j] == c)]
[b + diff] = dp[i][j][b] + 1,
from[i + (S[i] == c)][j + (T[j] == c)][b + diff] =
(i * 205 * 205 + j * 205 + b);
}
}
}
}
}
int mini = 0;
for (int i = 1; i < 205; i++)
if (dp[S.size()][T.size()][i] + i < dp[S.size()][T.size()][mini] + mini)
mini = i;
get(S.size(), T.size(), mini);
for (int i = 0; i < mini; i++) ans.push_back(')');
cout << ans << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long modulo(long long base, long long exponent, long long modulus);
long long choose(int n, int k);
int inverse(int a, int m);
void build();
void fileio();
int ncr(int n, int r);
const int nax = 1e6 + 10;
const int LG = log2(nax) + 1;
vector<int> fact(nax);
int dp[405][405][405];
int n, m;
string a, b;
int dpsolve(int open, int x, int y) {
if (open < 0 || open >= (n + m)) return 100000;
if (x >= n && y >= m) {
if (open == 0) return 0;
}
int &ans = dp[open][x][y];
if (ans != -1) return ans;
ans = 100000;
{
ans = min(ans, dpsolve(open + 1, x + (x < n && a[x] == '('),
y + (y < m && b[y] == '(')) +
1);
}
{
ans = min(ans, dpsolve(open - 1, x + (x < n && a[x] == ')'),
y + (y < m && b[y] == ')')) +
1);
}
return ans;
}
string f;
void backtrack(int open, int x, int y) {
if (open < 0 || open >= (n + m)) return;
int &ans = dp[open][x][y];
if (ans == dpsolve(open + 1, x + (x < n && a[x] == '('),
y + (y < m && b[y] == '(')) +
1) {
f += '(';
backtrack(open + 1, x + (x < n && a[x] == '('), y + (y < m && b[y] == '('));
return;
}
if (ans == dpsolve(open - 1, x + (x < n && a[x] == ')'),
y + (y < m && b[y] == ')')) +
1) {
f += ')';
backtrack(open - 1, x + (x < n && a[x] == ')'), y + (y < m && b[y] == ')'));
return;
}
}
int32_t main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> a >> b;
n = a.size();
m = b.size();
memset(dp, -1, sizeof(dp));
dpsolve(0, 0, 0);
backtrack(0, 0, 0);
cout << f << '\n';
return 0;
}
int ncr(int n, int r) {
if (r > n || r < 0 || n < 0) return 0;
int ans = fact[n];
int temp = (fact[n - r] * fact[r]) % 1000000007;
ans = (ans * inverse(temp, 1000000007)) % 1000000007;
return ans;
}
void fileio() {
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
}
void build() {
fact[0] = 1;
for (int i = 1; i < nax; i++) fact[i] = (fact[i - 1] * i) % 1000000007;
}
long long modulo(long long base, long long exponent, long long modulus) {
if (modulus == 1) return 0;
long long result = 1;
base = base % modulus;
while (exponent > 0) {
if (exponent % 2 == 1) {
result = (result * base) % modulus;
}
exponent = exponent >> 1;
base = (base * base) % modulus;
}
return result;
}
long long choose(int n, int k) {
if (k == 0) return 1;
return (n * choose(n - 1, k - 1)) / k;
}
void EE(int a, int b, int &co1, int &co2) {
if (a % b == 0) {
co1 = 0;
co2 = 1;
return;
}
EE(b, a % b, co1, co2);
int temp = co1;
co1 = co2;
co2 = temp - co2 * (a / b);
}
int inverse(int a, int m) {
int x, y;
EE(a, m, x, y);
if (x < 0) x += m;
return x;
}
| ### Prompt
Create a solution in cpp for the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long modulo(long long base, long long exponent, long long modulus);
long long choose(int n, int k);
int inverse(int a, int m);
void build();
void fileio();
int ncr(int n, int r);
const int nax = 1e6 + 10;
const int LG = log2(nax) + 1;
vector<int> fact(nax);
int dp[405][405][405];
int n, m;
string a, b;
int dpsolve(int open, int x, int y) {
if (open < 0 || open >= (n + m)) return 100000;
if (x >= n && y >= m) {
if (open == 0) return 0;
}
int &ans = dp[open][x][y];
if (ans != -1) return ans;
ans = 100000;
{
ans = min(ans, dpsolve(open + 1, x + (x < n && a[x] == '('),
y + (y < m && b[y] == '(')) +
1);
}
{
ans = min(ans, dpsolve(open - 1, x + (x < n && a[x] == ')'),
y + (y < m && b[y] == ')')) +
1);
}
return ans;
}
string f;
void backtrack(int open, int x, int y) {
if (open < 0 || open >= (n + m)) return;
int &ans = dp[open][x][y];
if (ans == dpsolve(open + 1, x + (x < n && a[x] == '('),
y + (y < m && b[y] == '(')) +
1) {
f += '(';
backtrack(open + 1, x + (x < n && a[x] == '('), y + (y < m && b[y] == '('));
return;
}
if (ans == dpsolve(open - 1, x + (x < n && a[x] == ')'),
y + (y < m && b[y] == ')')) +
1) {
f += ')';
backtrack(open - 1, x + (x < n && a[x] == ')'), y + (y < m && b[y] == ')'));
return;
}
}
int32_t main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> a >> b;
n = a.size();
m = b.size();
memset(dp, -1, sizeof(dp));
dpsolve(0, 0, 0);
backtrack(0, 0, 0);
cout << f << '\n';
return 0;
}
int ncr(int n, int r) {
if (r > n || r < 0 || n < 0) return 0;
int ans = fact[n];
int temp = (fact[n - r] * fact[r]) % 1000000007;
ans = (ans * inverse(temp, 1000000007)) % 1000000007;
return ans;
}
void fileio() {
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
}
void build() {
fact[0] = 1;
for (int i = 1; i < nax; i++) fact[i] = (fact[i - 1] * i) % 1000000007;
}
long long modulo(long long base, long long exponent, long long modulus) {
if (modulus == 1) return 0;
long long result = 1;
base = base % modulus;
while (exponent > 0) {
if (exponent % 2 == 1) {
result = (result * base) % modulus;
}
exponent = exponent >> 1;
base = (base * base) % modulus;
}
return result;
}
long long choose(int n, int k) {
if (k == 0) return 1;
return (n * choose(n - 1, k - 1)) / k;
}
void EE(int a, int b, int &co1, int &co2) {
if (a % b == 0) {
co1 = 0;
co2 = 1;
return;
}
EE(b, a % b, co1, co2);
int temp = co1;
co1 = co2;
co2 = temp - co2 * (a / b);
}
int inverse(int a, int m) {
int x, y;
EE(a, m, x, y);
if (x < 0) x += m;
return x;
}
``` |
#include <bits/stdc++.h>
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-O3")
#pragma GCC optimize("Ofast")
using namespace std;
const int oo = 2e9;
const long long OO = 4e18;
const long double pi = arg(complex<long double>(-1, 0));
const long double pi2 = pi + pi;
const int md = 0x3b800001;
const int MD = 1e9 + 7;
inline long long time() {
return chrono ::system_clock().now().time_since_epoch().count();
}
mt19937 rnd(time());
mt19937_64 RND(time());
template <typename t>
inline void umin(t &a, t b) {
a = min(a, b);
}
template <typename t>
inline void umax(t &a, t b) {
a = max(a, b);
}
const int N = 228;
string s, t;
int n, m, f[N][N][N + N];
char pr[N][N][N + N];
void solve() {
memset(f, 127, sizeof f);
cin >> s >> t;
n = int(s.size()), m = int(t.size());
f[0][0][0] = 0;
pr[0][0][0] = '#';
for (int i = 0; i < n + 1; ++i)
for (int j = 0; j < m + 1; ++j) {
for (int b = 0; b < n + m + 1; ++b)
if (b) {
int pi = i, pj = j, pb = b - 1;
if (i && s[i - 1] == '(') --pi;
if (j && t[j - 1] == '(') --pj;
if (f[i][j][b] > f[pi][pj][pb] + 1) {
f[i][j][b] = f[pi][pj][pb] + 1;
pr[i][j][b] = '(';
}
}
for (int b = n + m - 1; ~b; --b) {
int pi = i, pj = j, pb = b + 1;
if (i && s[i - 1] == ')') --pi;
if (j && t[j - 1] == ')') --pj;
if (f[i][j][b] > f[pi][pj][pb] + 1) {
f[i][j][b] = f[pi][pj][pb] + 1;
pr[i][j][b] = ')';
}
}
}
if (f[n][m][0] > 700) return void(cout << "wat\n");
int i = n, j = m, p = 0;
string ans = "";
for (; i || j || p;) {
char c = pr[i][j][p];
if (c == '(') {
--p;
if (i && s[i - 1] == '(') --i;
if (j && t[j - 1] == '(') --j;
} else {
++p;
if (i && s[i - 1] == ')') --i;
if (j && t[j - 1] == ')') --j;
}
if (c != '#') ans += c;
}
reverse(ans.begin(), ans.end());
cout << ans << endl;
}
int main() {
ios_base ::sync_with_stdio(0);
solve();
return 0;
}
| ### Prompt
Your task is to create a cpp solution to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-O3")
#pragma GCC optimize("Ofast")
using namespace std;
const int oo = 2e9;
const long long OO = 4e18;
const long double pi = arg(complex<long double>(-1, 0));
const long double pi2 = pi + pi;
const int md = 0x3b800001;
const int MD = 1e9 + 7;
inline long long time() {
return chrono ::system_clock().now().time_since_epoch().count();
}
mt19937 rnd(time());
mt19937_64 RND(time());
template <typename t>
inline void umin(t &a, t b) {
a = min(a, b);
}
template <typename t>
inline void umax(t &a, t b) {
a = max(a, b);
}
const int N = 228;
string s, t;
int n, m, f[N][N][N + N];
char pr[N][N][N + N];
void solve() {
memset(f, 127, sizeof f);
cin >> s >> t;
n = int(s.size()), m = int(t.size());
f[0][0][0] = 0;
pr[0][0][0] = '#';
for (int i = 0; i < n + 1; ++i)
for (int j = 0; j < m + 1; ++j) {
for (int b = 0; b < n + m + 1; ++b)
if (b) {
int pi = i, pj = j, pb = b - 1;
if (i && s[i - 1] == '(') --pi;
if (j && t[j - 1] == '(') --pj;
if (f[i][j][b] > f[pi][pj][pb] + 1) {
f[i][j][b] = f[pi][pj][pb] + 1;
pr[i][j][b] = '(';
}
}
for (int b = n + m - 1; ~b; --b) {
int pi = i, pj = j, pb = b + 1;
if (i && s[i - 1] == ')') --pi;
if (j && t[j - 1] == ')') --pj;
if (f[i][j][b] > f[pi][pj][pb] + 1) {
f[i][j][b] = f[pi][pj][pb] + 1;
pr[i][j][b] = ')';
}
}
}
if (f[n][m][0] > 700) return void(cout << "wat\n");
int i = n, j = m, p = 0;
string ans = "";
for (; i || j || p;) {
char c = pr[i][j][p];
if (c == '(') {
--p;
if (i && s[i - 1] == '(') --i;
if (j && t[j - 1] == '(') --j;
} else {
++p;
if (i && s[i - 1] == ')') --i;
if (j && t[j - 1] == ')') --j;
}
if (c != '#') ans += c;
}
reverse(ans.begin(), ans.end());
cout << ans << endl;
}
int main() {
ios_base ::sync_with_stdio(0);
solve();
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
const char* comma = strchr(names + 1, ',');
cerr.write(names, comma - names) << " : " << arg1 << " | ";
__f(comma + 1, args...);
}
int d[205][205][405], c[205][205][405], i, n, m, ans;
string a, b;
int dp(int x, int y, int z) {
if (z < 0 || z > 400) return 1e8;
x = min(x, n);
y = min(y, m);
if (x == n && y == m) return z;
if (d[x][y][z] != -1) return d[x][y][z];
int one = 1 + dp(x + (a[x] == '('), y + (b[y] == '('), z + 1);
int two = 1 + dp(x + (a[x] == ')'), y + (b[y] == ')'), z - 1);
d[x][y][z] = one;
if (two < one) {
d[x][y][z] = two;
c[x][y][z] = 1;
}
return d[x][y][z];
}
void path(int x, int y, int z) {
x = min(x, n);
y = min(y, m);
if (x == n && y == m) {
for (i = 0; i < z; i++) cout << ')';
return;
}
if (c[x][y][z]) {
cout << ')';
path(x + (a[x] == ')'), y + (b[y] == ')'), z - 1);
} else {
cout << '(';
path(x + (a[x] == '('), y + (b[y] == '('), z + 1);
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
cin >> a >> b;
n = a.size();
m = b.size();
a += "a";
b += "b";
memset(d, -1, sizeof(d));
ans = dp(0, 0, 0);
path(0, 0, 0);
return 0;
}
| ### Prompt
Please formulate a Cpp solution to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
const char* comma = strchr(names + 1, ',');
cerr.write(names, comma - names) << " : " << arg1 << " | ";
__f(comma + 1, args...);
}
int d[205][205][405], c[205][205][405], i, n, m, ans;
string a, b;
int dp(int x, int y, int z) {
if (z < 0 || z > 400) return 1e8;
x = min(x, n);
y = min(y, m);
if (x == n && y == m) return z;
if (d[x][y][z] != -1) return d[x][y][z];
int one = 1 + dp(x + (a[x] == '('), y + (b[y] == '('), z + 1);
int two = 1 + dp(x + (a[x] == ')'), y + (b[y] == ')'), z - 1);
d[x][y][z] = one;
if (two < one) {
d[x][y][z] = two;
c[x][y][z] = 1;
}
return d[x][y][z];
}
void path(int x, int y, int z) {
x = min(x, n);
y = min(y, m);
if (x == n && y == m) {
for (i = 0; i < z; i++) cout << ')';
return;
}
if (c[x][y][z]) {
cout << ')';
path(x + (a[x] == ')'), y + (b[y] == ')'), z - 1);
} else {
cout << '(';
path(x + (a[x] == '('), y + (b[y] == '('), z + 1);
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
cin >> a >> b;
n = a.size();
m = b.size();
a += "a";
b += "b";
memset(d, -1, sizeof(d));
ans = dp(0, 0, 0);
path(0, 0, 0);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 210;
string a, b;
int n, m, dp[MAXN][MAXN][MAXN];
int solve(int i, int j, int k, bool print) {
if (k < 0 || k >= MAXN) return 1e9;
int& DP = dp[i][j][k];
if (DP != -1 && !print) return DP;
if (i == n && j == m && k == 0) return DP = 0;
int open;
if (i == n && j == m)
open = 1e9;
else
open = solve(i + (a[i] == '('), j + (b[j] == '('), k + 1, false);
int close = solve(i + (a[i] == ')'), j + (b[j] == ')'), k - 1, false);
DP = min(open, close) + 1;
if (!print) return DP;
if (open < close) {
cout << '(';
solve(i + (a[i] == '('), j + (b[j] == '('), k + 1, true);
} else {
cout << ')';
solve(i + (a[i] == ')'), j + (b[j] == ')'), k - 1, true);
}
return DP;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> a >> b;
n = a.size();
m = b.size();
a += '*';
b += '*';
for (int i = 0; i <= n; i++)
for (int j = 0; j <= m; j++)
for (int k = 0; k < MAXN; k++) dp[i][j][k] = -1;
solve(0, 0, 0, true);
return 0;
}
| ### Prompt
Your challenge is to write a CPP solution to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 210;
string a, b;
int n, m, dp[MAXN][MAXN][MAXN];
int solve(int i, int j, int k, bool print) {
if (k < 0 || k >= MAXN) return 1e9;
int& DP = dp[i][j][k];
if (DP != -1 && !print) return DP;
if (i == n && j == m && k == 0) return DP = 0;
int open;
if (i == n && j == m)
open = 1e9;
else
open = solve(i + (a[i] == '('), j + (b[j] == '('), k + 1, false);
int close = solve(i + (a[i] == ')'), j + (b[j] == ')'), k - 1, false);
DP = min(open, close) + 1;
if (!print) return DP;
if (open < close) {
cout << '(';
solve(i + (a[i] == '('), j + (b[j] == '('), k + 1, true);
} else {
cout << ')';
solve(i + (a[i] == ')'), j + (b[j] == ')'), k - 1, true);
}
return DP;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> a >> b;
n = a.size();
m = b.size();
a += '*';
b += '*';
for (int i = 0; i <= n; i++)
for (int j = 0; j <= m; j++)
for (int k = 0; k < MAXN; k++) dp[i][j][k] = -1;
solve(0, 0, 0, true);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 205, M = 4 * N, INF = 0x3f3f3f3f;
int dp[N][N][M];
char ch[N][N][M];
string s, t;
int n, m;
int rec(int i, int j, int op) {
if (i == n and j == m) {
if (op == 0) return 0;
dp[i][j][op] = 1 + rec(i, j, op - 1);
ch[i][j][op] = ')';
return dp[i][j][op];
}
if (dp[i][j][op] < INF) return dp[i][j][op];
int ii = i;
if (i < n && s[i] == '(') ++ii;
int jj = j;
if (j < m && t[j] == '(') ++jj;
if (op + 1 < M) {
int temp = 1 + rec(ii, jj, op + 1);
if (temp < dp[i][j][op]) {
dp[i][j][op] = temp;
ch[i][j][op] = '(';
}
}
ii = i, jj = j;
if (i < n && s[i] == ')') ++ii;
if (j < m && t[j] == ')') ++jj;
if (op > 0) {
int temp = 1 + rec(ii, jj, op - 1);
if (temp < dp[i][j][op]) {
dp[i][j][op] = temp;
ch[i][j][op] = ')';
}
}
return dp[i][j][op];
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> s >> t;
n = s.size(), m = t.size();
memset(dp, INF, sizeof dp);
int len = rec(0, 0, 0);
string res;
for (int i = 0, j = 0, k = 0, op = 0; k < len; ++k) {
char c = ch[i][j][op];
res += c;
if (i < n && s[i] == c) ++i;
if (j < m && t[j] == c) ++j;
if (c == '(')
++op;
else
--op;
}
cout << res;
return 0;
}
| ### Prompt
Your challenge is to write a CPP solution to the following problem:
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 205, M = 4 * N, INF = 0x3f3f3f3f;
int dp[N][N][M];
char ch[N][N][M];
string s, t;
int n, m;
int rec(int i, int j, int op) {
if (i == n and j == m) {
if (op == 0) return 0;
dp[i][j][op] = 1 + rec(i, j, op - 1);
ch[i][j][op] = ')';
return dp[i][j][op];
}
if (dp[i][j][op] < INF) return dp[i][j][op];
int ii = i;
if (i < n && s[i] == '(') ++ii;
int jj = j;
if (j < m && t[j] == '(') ++jj;
if (op + 1 < M) {
int temp = 1 + rec(ii, jj, op + 1);
if (temp < dp[i][j][op]) {
dp[i][j][op] = temp;
ch[i][j][op] = '(';
}
}
ii = i, jj = j;
if (i < n && s[i] == ')') ++ii;
if (j < m && t[j] == ')') ++jj;
if (op > 0) {
int temp = 1 + rec(ii, jj, op - 1);
if (temp < dp[i][j][op]) {
dp[i][j][op] = temp;
ch[i][j][op] = ')';
}
}
return dp[i][j][op];
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> s >> t;
n = s.size(), m = t.size();
memset(dp, INF, sizeof dp);
int len = rec(0, 0, 0);
string res;
for (int i = 0, j = 0, k = 0, op = 0; k < len; ++k) {
char c = ch[i][j][op];
res += c;
if (i < n && s[i] == c) ++i;
if (j < m && t[j] == c) ++j;
if (c == '(')
++op;
else
--op;
}
cout << res;
return 0;
}
``` |
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