output
stringlengths 52
181k
| instruction
stringlengths 296
182k
|
|---|---|
#include <bits/stdc++.h>
using namespace std;
string a;
int t, n;
int main() {
cin >> t;
while (t--) {
cin >> a;
cin >> n;
int l = a.size(), f = 1;
string b(l, '1');
for (int i = 0; i < l; i++) {
if (a[i] == '0') {
if (i - n >= 0) {
b[i - n] = '0';
}
if (i + n < l) {
b[i + n] = '0';
}
}
}
for (int i = 0; i < l; i++) {
if (a[i] == '1') {
if ((i - n >= 0 && b[i - n] == '1') || (i + n < l && b[i + n] == '1')) {
continue;
} else {
f = 0;
break;
}
}
}
if (f == 1) {
cout << b << endl;
} else {
cout << "-1" << endl;
}
}
return 0;
}
| ### Prompt
Create a solution in Cpp for the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
string a;
int t, n;
int main() {
cin >> t;
while (t--) {
cin >> a;
cin >> n;
int l = a.size(), f = 1;
string b(l, '1');
for (int i = 0; i < l; i++) {
if (a[i] == '0') {
if (i - n >= 0) {
b[i - n] = '0';
}
if (i + n < l) {
b[i + n] = '0';
}
}
}
for (int i = 0; i < l; i++) {
if (a[i] == '1') {
if ((i - n >= 0 && b[i - n] == '1') || (i + n < l && b[i + n] == '1')) {
continue;
} else {
f = 0;
break;
}
}
}
if (f == 1) {
cout << b << endl;
} else {
cout << "-1" << endl;
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
inline int Isdigit(char c) {
if (c < '0' || c > '9') return 0;
return 1;
}
template <class T>
T read() {
register T x = 0, flag = 1;
register char ch;
while (!Isdigit(ch = getchar()))
if (ch == '-') flag = -1;
while (Isdigit(ch)) x = x * 10 + (ch & 15), ch = getchar();
return x * flag;
}
template <class T>
inline void write(T x) {
if (x < 0) putchar('-'), x = -x;
if (x >= 10) write(x / 10);
putchar(x % 10 + '0');
}
template <class T>
inline bool Chkmax(T& x, const T& y) {
return x < y ? x = y, true : false;
}
template <class T>
inline bool Chkmin(T& x, const T& y) {
return x > y ? x = y, true : false;
}
const int maxn = 202020;
int T, x;
string s;
int tag[maxn];
int Id(int x) {
if (x < 0 || x >= s.size()) return 0;
if (tag[x] == 0) return 0;
return 1;
}
int main() {
T = read<int>();
while (T--) {
cin >> s;
x = read<int>();
for (int i = 0; i < s.size(); i++) tag[i] = -1;
for (int i = 0; i < s.size(); i++) {
if (s[i] == '0') {
if (i - x >= 0) tag[i - x] = 0;
if (i + x <= s.size() - 1) tag[i + x] = 0;
}
}
int ok = 0;
for (int i = 0; i < s.size(); i++) {
if (s[i] == '1') {
if ((Id(i - x) | Id(x + i)) == 0) ok = 1;
}
}
if (ok == 1)
puts("-1");
else {
for (int i = 0; i < s.size(); i++) {
if (tag[i] == -1)
printf("%d", 1);
else
printf("%d", 0);
}
puts("");
}
}
}
| ### Prompt
Your challenge is to write a cpp solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
inline int Isdigit(char c) {
if (c < '0' || c > '9') return 0;
return 1;
}
template <class T>
T read() {
register T x = 0, flag = 1;
register char ch;
while (!Isdigit(ch = getchar()))
if (ch == '-') flag = -1;
while (Isdigit(ch)) x = x * 10 + (ch & 15), ch = getchar();
return x * flag;
}
template <class T>
inline void write(T x) {
if (x < 0) putchar('-'), x = -x;
if (x >= 10) write(x / 10);
putchar(x % 10 + '0');
}
template <class T>
inline bool Chkmax(T& x, const T& y) {
return x < y ? x = y, true : false;
}
template <class T>
inline bool Chkmin(T& x, const T& y) {
return x > y ? x = y, true : false;
}
const int maxn = 202020;
int T, x;
string s;
int tag[maxn];
int Id(int x) {
if (x < 0 || x >= s.size()) return 0;
if (tag[x] == 0) return 0;
return 1;
}
int main() {
T = read<int>();
while (T--) {
cin >> s;
x = read<int>();
for (int i = 0; i < s.size(); i++) tag[i] = -1;
for (int i = 0; i < s.size(); i++) {
if (s[i] == '0') {
if (i - x >= 0) tag[i - x] = 0;
if (i + x <= s.size() - 1) tag[i + x] = 0;
}
}
int ok = 0;
for (int i = 0; i < s.size(); i++) {
if (s[i] == '1') {
if ((Id(i - x) | Id(x + i)) == 0) ok = 1;
}
}
if (ok == 1)
puts("-1");
else {
for (int i = 0; i < s.size(); i++) {
if (tag[i] == -1)
printf("%d", 1);
else
printf("%d", 0);
}
puts("");
}
}
}
``` |
#include <bits/stdc++.h>
const long long int mod = 1e9 + 7;
using namespace std;
long long int power(long long int x, long long int y, long long int p) {
long long int res = 1;
x = x % p;
while (y > 0) {
if (y & 1) res = (res * x) % p;
y = y >> 1;
x = (x * x) % p;
}
return res;
}
long long int imod(long long int a, long long int m) {
return power(a, m - 2, m);
}
void myfun() {
long long int n, x, ans = 1, i, cn;
string s;
cin >> s;
cin >> x;
n = s.length();
long long int arr[n];
for (i = 0; i < n; i++) arr[i] = 1;
for (i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) arr[i - x] = 0;
if (i + x < n) arr[i + x] = 0;
}
}
for (i = 0; i < n; i++) {
cn = 0;
if (s[i] == '1') {
if (((i - x) >= 0) && ((i + x) < n)) {
if (arr[i - x] != 1 && arr[i + x] != 1) ans = 0;
} else if ((i - x) >= 0) {
if (arr[i - x] == 0) ans = false;
} else if ((i + x) < n) {
if (arr[i + x] == 0) ans = false;
} else
ans = false;
}
}
if (ans) {
for (i = 0; i < n; i++) cout << arr[i];
cout << "\n";
} else
cout << "-1\n";
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
;
long long int t = 1;
cin >> t;
while (t--) myfun();
return 0;
}
| ### Prompt
Your task is to create a CPP solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
const long long int mod = 1e9 + 7;
using namespace std;
long long int power(long long int x, long long int y, long long int p) {
long long int res = 1;
x = x % p;
while (y > 0) {
if (y & 1) res = (res * x) % p;
y = y >> 1;
x = (x * x) % p;
}
return res;
}
long long int imod(long long int a, long long int m) {
return power(a, m - 2, m);
}
void myfun() {
long long int n, x, ans = 1, i, cn;
string s;
cin >> s;
cin >> x;
n = s.length();
long long int arr[n];
for (i = 0; i < n; i++) arr[i] = 1;
for (i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) arr[i - x] = 0;
if (i + x < n) arr[i + x] = 0;
}
}
for (i = 0; i < n; i++) {
cn = 0;
if (s[i] == '1') {
if (((i - x) >= 0) && ((i + x) < n)) {
if (arr[i - x] != 1 && arr[i + x] != 1) ans = 0;
} else if ((i - x) >= 0) {
if (arr[i - x] == 0) ans = false;
} else if ((i + x) < n) {
if (arr[i + x] == 0) ans = false;
} else
ans = false;
}
}
if (ans) {
for (i = 0; i < n; i++) cout << arr[i];
cout << "\n";
} else
cout << "-1\n";
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
;
long long int t = 1;
cin >> t;
while (t--) myfun();
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
cerr << ' ' << *it << " = " << a;
err(++it, args...);
}
template <typename T, typename U>
inline void min_self(T& x, U y) {
if (y < x) x = y;
}
template <typename T, typename U>
inline void max_self(T& x, U y) {
if (x < y) x = y;
}
template <class T1, class T2>
ostream& operator<<(ostream& out, pair<T1, T2> pair) {
return out << "(" << pair.first << ", " << pair.second << ")";
}
template <class T>
ostream& operator<<(ostream& out, vector<T> vec) {
out << "(";
for (auto& v : vec) out << v << ", ";
return out << ")";
}
template <class T>
ostream& operator<<(ostream& out, set<T> vec) {
out << "(";
for (auto& v : vec) out << v << ", ";
return out << ")";
}
template <class L, class R>
ostream& operator<<(ostream& out, map<L, R> vec) {
out << "(";
for (auto& v : vec) out << "[" << v.first << ", " << v.second << "]";
return out << ")";
}
template <class A, class B>
istream& operator>>(istream& in, pair<A, B>& a) {
return in >> a.first >> a.second;
}
template <class A>
istream& operator>>(istream& in, vector<A>& a) {
for (A& i : a) in >> i;
return in;
}
long long XX[4] = {-1, 0, 1, 0};
long long YY[4] = {0, -1, 0, 1};
void solve() {
string s;
cin >> s;
int n, x;
cin >> x;
n = s.length();
vector<int> a(n, -1);
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i >= x) a[i - x] = 0;
if (i + x < n) a[i + x] = 0;
}
}
for (int i = 0; i < n; i++) {
if (s[i] == '1') {
int cnt = 0;
if (i >= x && a[i - x] == -1) {
cnt++;
}
if (i + x < n && a[i + x] == -1) {
cnt++;
}
if (cnt == 0) {
cout << -1 << '\n';
return;
}
}
}
for (int i : a) {
if (i == -1)
cout << '1';
else
cout << '0';
}
cout << '\n';
return;
}
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
long long t = 1;
cin >> t;
while (t--) {
solve();
}
return 0;
}
| ### Prompt
Develop a solution in CPP to the problem described below:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
cerr << ' ' << *it << " = " << a;
err(++it, args...);
}
template <typename T, typename U>
inline void min_self(T& x, U y) {
if (y < x) x = y;
}
template <typename T, typename U>
inline void max_self(T& x, U y) {
if (x < y) x = y;
}
template <class T1, class T2>
ostream& operator<<(ostream& out, pair<T1, T2> pair) {
return out << "(" << pair.first << ", " << pair.second << ")";
}
template <class T>
ostream& operator<<(ostream& out, vector<T> vec) {
out << "(";
for (auto& v : vec) out << v << ", ";
return out << ")";
}
template <class T>
ostream& operator<<(ostream& out, set<T> vec) {
out << "(";
for (auto& v : vec) out << v << ", ";
return out << ")";
}
template <class L, class R>
ostream& operator<<(ostream& out, map<L, R> vec) {
out << "(";
for (auto& v : vec) out << "[" << v.first << ", " << v.second << "]";
return out << ")";
}
template <class A, class B>
istream& operator>>(istream& in, pair<A, B>& a) {
return in >> a.first >> a.second;
}
template <class A>
istream& operator>>(istream& in, vector<A>& a) {
for (A& i : a) in >> i;
return in;
}
long long XX[4] = {-1, 0, 1, 0};
long long YY[4] = {0, -1, 0, 1};
void solve() {
string s;
cin >> s;
int n, x;
cin >> x;
n = s.length();
vector<int> a(n, -1);
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i >= x) a[i - x] = 0;
if (i + x < n) a[i + x] = 0;
}
}
for (int i = 0; i < n; i++) {
if (s[i] == '1') {
int cnt = 0;
if (i >= x && a[i - x] == -1) {
cnt++;
}
if (i + x < n && a[i + x] == -1) {
cnt++;
}
if (cnt == 0) {
cout << -1 << '\n';
return;
}
}
}
for (int i : a) {
if (i == -1)
cout << '1';
else
cout << '0';
}
cout << '\n';
return;
}
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
long long t = 1;
cin >> t;
while (t--) {
solve();
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 1000;
int str[maxn];
string w;
int x;
int main() {
int T;
cin >> T;
while (T--) {
cin >> w;
cin >> x;
int len = w.length();
for (int i = 0; i < len; i++) {
str[i] = -1;
}
for (int i = 0; i < len; i++) {
if (w[i] == '0') {
if (i + x < len) {
str[i + x] = 0;
}
if (i - x >= 0) {
str[i - x] = 0;
}
}
}
int flag = 1;
for (int i = 0; i < len; i++) {
if (w[i] == '1') {
if (i + x < len && i - x >= 0) {
if (str[i + x] == 0 && str[i - x] == 0) {
flag = 0;
break;
}
if (str[i + x] != 0) {
str[i + x] = 1;
}
if (str[i - x] != 0) {
str[i - x] = 1;
}
} else if (i + x >= len && i - x >= 0) {
if (str[i - x] == 0) {
flag = 0;
break;
}
if (str[i - x] != 0) {
str[i - x] = 1;
}
} else if (i + x < len && i - x < 0) {
if (str[i + x] == 0) {
flag = 0;
break;
}
if (str[i + x] != 0) {
str[i + x] = 1;
}
} else if (i + x >= len && i - x < 0) {
flag = 0;
}
}
}
for (int i = 0; i < len; i++) {
if (str[i] == -1) {
str[i] = 0;
}
}
if (flag) {
for (int i = 0; i < len; i++) {
cout << str[i];
}
printf("\n");
} else {
printf("-1\n");
}
}
return 0;
}
| ### Prompt
Please formulate a CPP solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 1000;
int str[maxn];
string w;
int x;
int main() {
int T;
cin >> T;
while (T--) {
cin >> w;
cin >> x;
int len = w.length();
for (int i = 0; i < len; i++) {
str[i] = -1;
}
for (int i = 0; i < len; i++) {
if (w[i] == '0') {
if (i + x < len) {
str[i + x] = 0;
}
if (i - x >= 0) {
str[i - x] = 0;
}
}
}
int flag = 1;
for (int i = 0; i < len; i++) {
if (w[i] == '1') {
if (i + x < len && i - x >= 0) {
if (str[i + x] == 0 && str[i - x] == 0) {
flag = 0;
break;
}
if (str[i + x] != 0) {
str[i + x] = 1;
}
if (str[i - x] != 0) {
str[i - x] = 1;
}
} else if (i + x >= len && i - x >= 0) {
if (str[i - x] == 0) {
flag = 0;
break;
}
if (str[i - x] != 0) {
str[i - x] = 1;
}
} else if (i + x < len && i - x < 0) {
if (str[i + x] == 0) {
flag = 0;
break;
}
if (str[i + x] != 0) {
str[i + x] = 1;
}
} else if (i + x >= len && i - x < 0) {
flag = 0;
}
}
}
for (int i = 0; i < len; i++) {
if (str[i] == -1) {
str[i] = 0;
}
}
if (flag) {
for (int i = 0; i < len; i++) {
cout << str[i];
}
printf("\n");
} else {
printf("-1\n");
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
#pragma GCC optimize(3, "Ofast", "inline")
using namespace std;
const int maxn = 5e5 + 10;
const int maxm = 5e5 + 10;
const long long INF = 1e18 + 10;
const long long mod = 1e9 + 7;
const double pi = acos(-1.0);
const double eps = 1e-8;
int dir[4][2] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};
int t;
char s[maxn];
int x;
void solve() {
int f = 1;
char ans[maxn];
int len = strlen(s);
for (int i = 0; i < len; i++) ans[i] = '1';
for (int i = 0; i < len; i++) {
if (s[i] == '0') {
if (i >= x) ans[i - x] = '0';
if (i + x < len) ans[i + x] = '0';
}
}
for (int i = 0; i < len; i++) {
int f1 = 0;
int f2 = 0;
if (s[i] == '1') {
if (i >= x) f1 = (ans[i - x] == '1');
if (i + x < len) f2 = (ans[i + x] == '1');
}
if (s[i] == '0') f1 = 1;
f = (f1 || f2);
if (!f) break;
}
if (!f)
puts("-1");
else {
for (int i = 0; i < len; i++) cout << ans[i];
cout << endl;
}
}
int main() {
cin >> t;
while (t--) {
cin >> s >> x;
solve();
}
return 0;
}
| ### Prompt
Construct a CPP code solution to the problem outlined:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
#pragma GCC optimize(3, "Ofast", "inline")
using namespace std;
const int maxn = 5e5 + 10;
const int maxm = 5e5 + 10;
const long long INF = 1e18 + 10;
const long long mod = 1e9 + 7;
const double pi = acos(-1.0);
const double eps = 1e-8;
int dir[4][2] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};
int t;
char s[maxn];
int x;
void solve() {
int f = 1;
char ans[maxn];
int len = strlen(s);
for (int i = 0; i < len; i++) ans[i] = '1';
for (int i = 0; i < len; i++) {
if (s[i] == '0') {
if (i >= x) ans[i - x] = '0';
if (i + x < len) ans[i + x] = '0';
}
}
for (int i = 0; i < len; i++) {
int f1 = 0;
int f2 = 0;
if (s[i] == '1') {
if (i >= x) f1 = (ans[i - x] == '1');
if (i + x < len) f2 = (ans[i + x] == '1');
}
if (s[i] == '0') f1 = 1;
f = (f1 || f2);
if (!f) break;
}
if (!f)
puts("-1");
else {
for (int i = 0; i < len; i++) cout << ans[i];
cout << endl;
}
}
int main() {
cin >> t;
while (t--) {
cin >> s >> x;
solve();
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
void solve() {
string str;
cin >> str;
long long x, n = str.size();
cin >> x;
string ans = "";
for (int i = 0; i < n; i++) ans += "1";
vector<bool> s(n, false);
for (int i = 0; i < n; i++) {
if (str[i] == '0') {
if (i - x >= 0) ans[i - x] = '0';
if (i + x < n) ans[i + x] = '0';
}
}
for (int i = 0; i < n; i++) {
if (str[i] == '1') {
int count = 0;
if (i - x >= 0 && ans[i - x] == '1') count++;
if (i + x < n && ans[i + x] == '1') count++;
if (count == 0) {
cout << -1 << endl;
return;
}
}
}
cout << ans << endl;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;
cin >> t;
for (int i = 1; i <= t; i++) {
solve();
}
}
| ### Prompt
Please formulate a Cpp solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void solve() {
string str;
cin >> str;
long long x, n = str.size();
cin >> x;
string ans = "";
for (int i = 0; i < n; i++) ans += "1";
vector<bool> s(n, false);
for (int i = 0; i < n; i++) {
if (str[i] == '0') {
if (i - x >= 0) ans[i - x] = '0';
if (i + x < n) ans[i + x] = '0';
}
}
for (int i = 0; i < n; i++) {
if (str[i] == '1') {
int count = 0;
if (i - x >= 0 && ans[i - x] == '1') count++;
if (i + x < n && ans[i + x] == '1') count++;
if (count == 0) {
cout << -1 << endl;
return;
}
}
}
cout << ans << endl;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;
cin >> t;
for (int i = 1; i <= t; i++) {
solve();
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int t;
cin >> t;
while (t--) {
string s;
cin >> s;
int x;
cin >> x;
int n = s.size();
string a(n, '1');
for (int i = 0; i < n; ++i) {
if (s[i] == '1') continue;
if (i - x >= 0) a[i - x] = '0';
if (i + x < n) a[i + x] = '0';
}
string w;
w = a;
for (int i = 0; i < n; ++i) {
if (i - x >= 0 && a[i - x] == '1') {
w[i] = '1';
continue;
}
if (i + x < n && a[i + x] == '1') {
w[i] = '1';
continue;
}
w[i] = '0';
}
if (w != s) {
cout << -1 << '\n';
continue;
}
cout << a << '\n';
}
return 0;
}
| ### Prompt
Your task is to create a CPP solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int t;
cin >> t;
while (t--) {
string s;
cin >> s;
int x;
cin >> x;
int n = s.size();
string a(n, '1');
for (int i = 0; i < n; ++i) {
if (s[i] == '1') continue;
if (i - x >= 0) a[i - x] = '0';
if (i + x < n) a[i + x] = '0';
}
string w;
w = a;
for (int i = 0; i < n; ++i) {
if (i - x >= 0 && a[i - x] == '1') {
w[i] = '1';
continue;
}
if (i + x < n && a[i + x] == '1') {
w[i] = '1';
continue;
}
w[i] = '0';
}
if (w != s) {
cout << -1 << '\n';
continue;
}
cout << a << '\n';
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
std::ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
;
long long t;
cin >> t;
while (t--) {
long long n;
string a;
string ans = "";
long long x;
cin >> a >> x;
n = a.size();
for (long long i = 0; i < n; i++) ans += '1';
bool ok = false;
for (long long i = 0; i < n; i++) {
if (a[i] == '1') {
;
} else {
if (i < x && i + x >= n)
;
if (i >= x) {
ans[i - x] = '0';
}
if (i + x < n) {
ans[i + x] = '0';
}
}
}
long long mx = 1e14;
for (long long i = 0; i < n; i++) {
long long p = 0, q = 0;
if (a[i] == '1') {
if (i < x && i + 1 >= n) {
p = 0;
break;
}
if (i >= x) {
if (ans[i - x] == '1') p++;
}
if (i + x < n) {
if (ans[i + x] == '1') p++;
}
q = 1e14;
} else if (a[i] == '0') {
if (i < x && i + x >= n) q = 1e14;
if (i >= x) {
if (ans[i - x] == '0') q++;
}
if (i + x < n) {
if (ans[i + x] == '0') q++;
}
p = 1e14;
}
mx = min({mx, p, q});
}
if (mx == 0)
cout << -1 << endl;
else
cout << ans << endl;
}
}
| ### Prompt
Please provide a cpp coded solution to the problem described below:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
std::ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
;
long long t;
cin >> t;
while (t--) {
long long n;
string a;
string ans = "";
long long x;
cin >> a >> x;
n = a.size();
for (long long i = 0; i < n; i++) ans += '1';
bool ok = false;
for (long long i = 0; i < n; i++) {
if (a[i] == '1') {
;
} else {
if (i < x && i + x >= n)
;
if (i >= x) {
ans[i - x] = '0';
}
if (i + x < n) {
ans[i + x] = '0';
}
}
}
long long mx = 1e14;
for (long long i = 0; i < n; i++) {
long long p = 0, q = 0;
if (a[i] == '1') {
if (i < x && i + 1 >= n) {
p = 0;
break;
}
if (i >= x) {
if (ans[i - x] == '1') p++;
}
if (i + x < n) {
if (ans[i + x] == '1') p++;
}
q = 1e14;
} else if (a[i] == '0') {
if (i < x && i + x >= n) q = 1e14;
if (i >= x) {
if (ans[i - x] == '0') q++;
}
if (i + x < n) {
if (ans[i + x] == '0') q++;
}
p = 1e14;
}
mx = min({mx, p, q});
}
if (mx == 0)
cout << -1 << endl;
else
cout << ans << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long nmod = 998244353;
const double PI = 3.14159265358979323846;
void solve() {
long long n;
string s;
long long x;
cin >> s >> x;
n = s.size();
s = 'G' + s;
string ans = string(n + 1, 'G');
for (long long j = 1; j <= n; j++) {
if (s[j] == '0') {
if (j - x >= 1) ans[j - x] = '0';
if (j + x <= n) ans[j + x] = '0';
} else {
if (j - x >= 1 && ans[j - x] == 'G') ans[j - x] = '1';
if (j + x <= n && ans[j + x] == 'G') ans[j + x] = '1';
}
}
for (long long i = 1; i <= n; i++) {
if (s[i] == '0') {
if (i - x >= 1 && ans[i - x] == '1') {
cout << -1;
return;
} else if (i + x <= n && ans[i + x] == '1') {
cout << -1;
return;
}
} else {
if ((i - x >= 1 && ans[i - x] == '1') ||
(i + x <= n && ans[i + x] == '1')) {
continue;
} else {
cout << -1;
return;
}
}
}
string res = ans.substr(1, n);
for (long long i = 0; i < n; i++)
if (res[i] == 'G') res[i] = '0';
cout << res;
}
int32_t main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
;
long long t;
cin >> t;
while (t--) {
solve();
cout << endl;
}
return 0;
}
| ### Prompt
Please formulate a Cpp solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long nmod = 998244353;
const double PI = 3.14159265358979323846;
void solve() {
long long n;
string s;
long long x;
cin >> s >> x;
n = s.size();
s = 'G' + s;
string ans = string(n + 1, 'G');
for (long long j = 1; j <= n; j++) {
if (s[j] == '0') {
if (j - x >= 1) ans[j - x] = '0';
if (j + x <= n) ans[j + x] = '0';
} else {
if (j - x >= 1 && ans[j - x] == 'G') ans[j - x] = '1';
if (j + x <= n && ans[j + x] == 'G') ans[j + x] = '1';
}
}
for (long long i = 1; i <= n; i++) {
if (s[i] == '0') {
if (i - x >= 1 && ans[i - x] == '1') {
cout << -1;
return;
} else if (i + x <= n && ans[i + x] == '1') {
cout << -1;
return;
}
} else {
if ((i - x >= 1 && ans[i - x] == '1') ||
(i + x <= n && ans[i + x] == '1')) {
continue;
} else {
cout << -1;
return;
}
}
}
string res = ans.substr(1, n);
for (long long i = 0; i < n; i++)
if (res[i] == 'G') res[i] = '0';
cout << res;
}
int32_t main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
;
long long t;
cin >> t;
while (t--) {
solve();
cout << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 5;
int a[MAXN], b[MAXN];
int main() {
int T;
scanf("%d", &T);
while (T--) {
getchar();
int c, x, n = 0;
while ((c = getchar()) != '\n') a[++n] = c - '0';
scanf("%d", &x);
int ok = 1;
for (int i = 1; i <= n; i++) b[i] = 1;
for (int i = 1; i <= n; i++) {
if (!a[i]) {
if (i > x) b[i - x] = 0;
if (i + x <= n) b[i + x] = 0;
}
}
for (int i = 1; i <= n; i++) {
if (a[i]) {
if (!((i > x && b[i - x]) || (i + x <= n && b[i + x]))) {
ok = 0;
break;
}
}
}
if (ok) {
for (int i = 1; i <= n; i++) {
printf("%d", b[i]);
}
printf("\n");
} else
printf("-1\n");
}
return 0;
}
| ### Prompt
Generate a Cpp solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 5;
int a[MAXN], b[MAXN];
int main() {
int T;
scanf("%d", &T);
while (T--) {
getchar();
int c, x, n = 0;
while ((c = getchar()) != '\n') a[++n] = c - '0';
scanf("%d", &x);
int ok = 1;
for (int i = 1; i <= n; i++) b[i] = 1;
for (int i = 1; i <= n; i++) {
if (!a[i]) {
if (i > x) b[i - x] = 0;
if (i + x <= n) b[i + x] = 0;
}
}
for (int i = 1; i <= n; i++) {
if (a[i]) {
if (!((i > x && b[i - x]) || (i + x <= n && b[i + x]))) {
ok = 0;
break;
}
}
}
if (ok) {
for (int i = 1; i <= n; i++) {
printf("%d", b[i]);
}
printf("\n");
} else
printf("-1\n");
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int T, x, n;
string s;
void solve() {
cin >> s >> x;
n = s.length();
bool a[n], ans = 0;
for (int i = 0; i < n; i++) a[i] = 1;
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) a[i - x] = 0;
if (i + x < n) a[i + x] = 0;
}
}
for (int i = 0; i < n; i++) {
if (s[i] == '1') {
if (!((i - x >= 0 && a[i - x]) || (i + x < n && a[i + x]))) {
ans = 1;
break;
}
}
}
if (ans)
cout << -1;
else
for (int i = 0; i < n; i++) cout << a[i];
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cin >> T;
while (T--) {
solve();
cout << '\n';
}
}
| ### Prompt
Your task is to create a cpp solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int T, x, n;
string s;
void solve() {
cin >> s >> x;
n = s.length();
bool a[n], ans = 0;
for (int i = 0; i < n; i++) a[i] = 1;
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) a[i - x] = 0;
if (i + x < n) a[i + x] = 0;
}
}
for (int i = 0; i < n; i++) {
if (s[i] == '1') {
if (!((i - x >= 0 && a[i - x]) || (i + x < n && a[i + x]))) {
ans = 1;
break;
}
}
}
if (ans)
cout << -1;
else
for (int i = 0; i < n; i++) cout << a[i];
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cin >> T;
while (T--) {
solve();
cout << '\n';
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long mod_power(long long n, long long x, long long mod) {
long long res = 1;
n = n % mod;
if (n == 0) {
return 0;
}
while (x > 0) {
if (x & 1) {
res = (res * n) % mod;
}
x = x >> 1;
n = (n * n) % mod;
}
return res;
}
long long power(long long n, long long x) {
long long res = 1;
while (x > 0) {
if (x & 1) {
res = res * n;
}
x = x >> 1;
n = n * n;
}
return res;
}
long long countDigit(long long n) {
if (n == 0) return 0;
return 1 + countDigit(n / 10);
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long t;
cin >> t;
while (t--) {
string s;
cin >> s;
long long x;
cin >> x;
string w = "";
long long n = s.size();
for (long long i = 0; i < n; i++) {
w += '1';
}
for (long long i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) {
w[i - x] = '0';
}
if (i + x < n) {
w[i + x] = '0';
}
}
}
bool okay = true;
for (long long i = 0; i < n; i++) {
if (s[i] == '1') {
bool check1 = false, check2 = false;
if (i - x >= 0) {
if (w[i - x] == '1') {
check1 = true;
}
}
if (i + x < n) {
if (w[i + x] == '1') {
check2 = true;
}
}
if (check1 == true || check2 == true) {
} else {
okay = false;
break;
}
}
}
if (okay == true) {
cout << w << "\n";
} else {
cout << "-1\n";
}
}
return 0;
}
| ### Prompt
In cpp, your task is to solve the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long mod_power(long long n, long long x, long long mod) {
long long res = 1;
n = n % mod;
if (n == 0) {
return 0;
}
while (x > 0) {
if (x & 1) {
res = (res * n) % mod;
}
x = x >> 1;
n = (n * n) % mod;
}
return res;
}
long long power(long long n, long long x) {
long long res = 1;
while (x > 0) {
if (x & 1) {
res = res * n;
}
x = x >> 1;
n = n * n;
}
return res;
}
long long countDigit(long long n) {
if (n == 0) return 0;
return 1 + countDigit(n / 10);
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long t;
cin >> t;
while (t--) {
string s;
cin >> s;
long long x;
cin >> x;
string w = "";
long long n = s.size();
for (long long i = 0; i < n; i++) {
w += '1';
}
for (long long i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) {
w[i - x] = '0';
}
if (i + x < n) {
w[i + x] = '0';
}
}
}
bool okay = true;
for (long long i = 0; i < n; i++) {
if (s[i] == '1') {
bool check1 = false, check2 = false;
if (i - x >= 0) {
if (w[i - x] == '1') {
check1 = true;
}
}
if (i + x < n) {
if (w[i + x] == '1') {
check2 = true;
}
}
if (check1 == true || check2 == true) {
} else {
okay = false;
break;
}
}
}
if (okay == true) {
cout << w << "\n";
} else {
cout << "-1\n";
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9 + 1000;
const long long INF64 = 1e18;
const int N = 1001000;
const int MOD = 1e9 + 7;
const double EPS = 1E-9;
long long gcd(long long a, long long b) { return a == 0 ? b : gcd(b % a, a); }
int add(int first, int second, int m) {
first += second;
while (first >= m) first -= m;
while (first < 0) first += m;
return first;
}
int sub(int first, int second, int m) { return add(first, -second, m); }
int mul(long long first, long long second, int m) {
first %= m;
second %= m;
return (first * 1ll * second) % m;
}
int binpow(int first, int second, int m) {
int z = 1;
while (second) {
if (second & 1) z = mul(z, first, m);
first = mul(first, first, m);
second >>= 1;
}
return z;
}
int divide(int first, int second, int m) {
return mul(first, binpow(second, m - 2, m), m);
}
long long binpow(long long first, long long second) {
long long z = 1;
while (second) {
if (second & 1) z *= first;
first = first * first;
second >>= 1;
}
return z;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int t;
cin >> t;
while (t--) {
string s;
int first;
cin >> s >> first;
string res(s.size(), ' ');
int done = 0;
int cur_idx = 0;
int n = s.size();
for (int i = 0; i < n; ++i) {
if (s[i] == '0') {
if (i - first >= 0 && i + first < n) {
res[i - first] = res[i + first] = '0';
} else if (i - first < 0 && i + first < n) {
res[i + first] = '0';
} else if (i - first >= 0 && i + first >= n) {
res[i - first] = '0';
}
}
}
for (int i = 0; i < n; ++i) {
if (res[i] == ' ') {
res[i] = '1';
}
}
bool ok = true;
for (int i = 0; i < n; ++i) {
if (i - first >= 0 && res[i - first] == '1') {
ok &= (s[i] == '1');
} else if (i + first < n && res[i + first] == '1') {
ok &= (s[i] == '1');
} else {
ok &= (s[i] == '0');
}
}
cout << (ok ? res : "-1") << endl;
}
}
| ### Prompt
Construct a cpp code solution to the problem outlined:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9 + 1000;
const long long INF64 = 1e18;
const int N = 1001000;
const int MOD = 1e9 + 7;
const double EPS = 1E-9;
long long gcd(long long a, long long b) { return a == 0 ? b : gcd(b % a, a); }
int add(int first, int second, int m) {
first += second;
while (first >= m) first -= m;
while (first < 0) first += m;
return first;
}
int sub(int first, int second, int m) { return add(first, -second, m); }
int mul(long long first, long long second, int m) {
first %= m;
second %= m;
return (first * 1ll * second) % m;
}
int binpow(int first, int second, int m) {
int z = 1;
while (second) {
if (second & 1) z = mul(z, first, m);
first = mul(first, first, m);
second >>= 1;
}
return z;
}
int divide(int first, int second, int m) {
return mul(first, binpow(second, m - 2, m), m);
}
long long binpow(long long first, long long second) {
long long z = 1;
while (second) {
if (second & 1) z *= first;
first = first * first;
second >>= 1;
}
return z;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int t;
cin >> t;
while (t--) {
string s;
int first;
cin >> s >> first;
string res(s.size(), ' ');
int done = 0;
int cur_idx = 0;
int n = s.size();
for (int i = 0; i < n; ++i) {
if (s[i] == '0') {
if (i - first >= 0 && i + first < n) {
res[i - first] = res[i + first] = '0';
} else if (i - first < 0 && i + first < n) {
res[i + first] = '0';
} else if (i - first >= 0 && i + first >= n) {
res[i - first] = '0';
}
}
}
for (int i = 0; i < n; ++i) {
if (res[i] == ' ') {
res[i] = '1';
}
}
bool ok = true;
for (int i = 0; i < n; ++i) {
if (i - first >= 0 && res[i - first] == '1') {
ok &= (s[i] == '1');
} else if (i + first < n && res[i + first] == '1') {
ok &= (s[i] == '1');
} else {
ok &= (s[i] == '0');
}
}
cout << (ok ? res : "-1") << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
int a[100005] = {};
for (int i = 0; i < 100005; ++i) {
a[i] = 1;
}
string s;
cin >> s;
int x;
cin >> x;
int n = s.size();
for (int i = 0; i < s.size(); ++i) {
char c = s[i];
if (c == '0') {
if (i >= x) {
a[i - x] = 0;
}
if (i + x < n) {
a[i + x] = 0;
}
}
}
string res = "";
bool resb = true;
for (int i = 0; i < s.size(); ++i) {
char c = s[i];
if (c == '1') {
bool ok = false;
if (i >= x && a[i - x] == 1) {
ok = true;
}
if (i + x < n && a[i + x] == 1) {
ok = true;
}
if (!ok) {
resb = false;
break;
}
}
}
for (int i = 0; i < n; ++i) {
if (a[i] == 0) {
res += "0";
} else {
res += "1";
}
}
if (resb) {
cout << res << endl;
} else {
cout << -1 << endl;
}
}
return 0;
}
| ### Prompt
Your challenge is to write a CPP solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
int a[100005] = {};
for (int i = 0; i < 100005; ++i) {
a[i] = 1;
}
string s;
cin >> s;
int x;
cin >> x;
int n = s.size();
for (int i = 0; i < s.size(); ++i) {
char c = s[i];
if (c == '0') {
if (i >= x) {
a[i - x] = 0;
}
if (i + x < n) {
a[i + x] = 0;
}
}
}
string res = "";
bool resb = true;
for (int i = 0; i < s.size(); ++i) {
char c = s[i];
if (c == '1') {
bool ok = false;
if (i >= x && a[i - x] == 1) {
ok = true;
}
if (i + x < n && a[i + x] == 1) {
ok = true;
}
if (!ok) {
resb = false;
break;
}
}
}
for (int i = 0; i < n; ++i) {
if (a[i] == 0) {
res += "0";
} else {
res += "1";
}
}
if (resb) {
cout << res << endl;
} else {
cout << -1 << endl;
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long maxn = 1e6 + 50;
vector<long long> vc, vans;
signed main() {
ios_base::sync_with_stdio(0), cin.tie(0);
long long t;
cin >> t;
while (t--) {
vc.clear();
vans.clear();
string s, ans, cmp;
cin >> s;
long long x, n = s.size();
cin >> x;
for (long long i = 0; i < n; i++) {
if (s[i] == '0') vc.push_back(i);
ans.push_back('1');
cmp.push_back('1');
}
for (auto i : vc) {
if (i + x < n) ans[i + x] = '0';
if (i - x >= 0) ans[i - x] = '0';
}
for (long long i = 0; i < n; i++) {
if ((i + x >= n || ans[i + x] == '0') and
(i - x < 0 || ans[i - x] == '0')) {
cmp[i] = '0';
}
}
if (cmp != s)
cout << -1 << "\n";
else
cout << ans << "\n";
}
}
| ### Prompt
Create a solution in CPP for the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long maxn = 1e6 + 50;
vector<long long> vc, vans;
signed main() {
ios_base::sync_with_stdio(0), cin.tie(0);
long long t;
cin >> t;
while (t--) {
vc.clear();
vans.clear();
string s, ans, cmp;
cin >> s;
long long x, n = s.size();
cin >> x;
for (long long i = 0; i < n; i++) {
if (s[i] == '0') vc.push_back(i);
ans.push_back('1');
cmp.push_back('1');
}
for (auto i : vc) {
if (i + x < n) ans[i + x] = '0';
if (i - x >= 0) ans[i - x] = '0';
}
for (long long i = 0; i < n; i++) {
if ((i + x >= n || ans[i + x] == '0') and
(i - x < 0 || ans[i - x] == '0')) {
cmp[i] = '0';
}
}
if (cmp != s)
cout << -1 << "\n";
else
cout << ans << "\n";
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
cerr << '{';
__print(x.first);
cerr << ',';
__print(x.second);
cerr << '}';
}
template <typename T>
void __print(const T &x) {
int f = 0;
cerr << '{';
for (auto i : x) cerr << (f++ ? "," : ""), __print(i);
cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
__print(t);
if (sizeof...(v)) cerr << ", ";
_print(v...);
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t, x;
string s;
cin >> t;
while (t--) {
cin >> s >> x;
int n = s.size();
string res(n, '1');
for (int i = 0; i < s.size(); i++) {
if (s[i] == '0') {
if (i - x >= 0) res[i - x] = '0';
if (i + x < n) res[i + x] = '0';
}
}
bool ok = true;
for (int i = 0; i < n && ok; i++)
if (s[i] == '1' && ((i - x < 0 || res[i - x] != '1') &&
(i + x >= n || res[i + x] != '1')))
ok = false;
if (ok)
cout << res << "\n";
else
cout << "-1\n";
}
}
| ### Prompt
Create a solution in Cpp for the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
cerr << '{';
__print(x.first);
cerr << ',';
__print(x.second);
cerr << '}';
}
template <typename T>
void __print(const T &x) {
int f = 0;
cerr << '{';
for (auto i : x) cerr << (f++ ? "," : ""), __print(i);
cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
__print(t);
if (sizeof...(v)) cerr << ", ";
_print(v...);
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t, x;
string s;
cin >> t;
while (t--) {
cin >> s >> x;
int n = s.size();
string res(n, '1');
for (int i = 0; i < s.size(); i++) {
if (s[i] == '0') {
if (i - x >= 0) res[i - x] = '0';
if (i + x < n) res[i + x] = '0';
}
}
bool ok = true;
for (int i = 0; i < n && ok; i++)
if (s[i] == '1' && ((i - x < 0 || res[i - x] != '1') &&
(i + x >= n || res[i + x] != '1')))
ok = false;
if (ok)
cout << res << "\n";
else
cout << "-1\n";
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
std::cin.tie(NULL);
std::cout.tie(NULL);
;
int t;
cin >> t;
while (t--) {
string s;
cin >> s;
int n;
cin >> n;
bool check = 1;
string tmp(int((s).size()), '1');
for (int i = (0); i < (int((s).size())); i += (1))
if (s[i] == '0') {
if (i - n >= 0) tmp[i - n] = '0';
if (i + n < int((tmp).size())) tmp[i + n] = '0';
}
for (int i = (0); i < (int((tmp).size())); i += (1)) {
bool check1 = (i - n >= 0 && tmp[i - n] == '1');
bool check2 = (i + n < int((tmp).size()) && tmp[i + n] == '1');
if (check1 || check2) {
if (s[i] != '1') {
check = 0;
break;
}
} else if (s[i] != '0') {
check = 0;
break;
}
}
cout << (check ? tmp : "-1") << '\n';
}
return 0;
}
| ### Prompt
Your challenge is to write a cpp solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
std::cin.tie(NULL);
std::cout.tie(NULL);
;
int t;
cin >> t;
while (t--) {
string s;
cin >> s;
int n;
cin >> n;
bool check = 1;
string tmp(int((s).size()), '1');
for (int i = (0); i < (int((s).size())); i += (1))
if (s[i] == '0') {
if (i - n >= 0) tmp[i - n] = '0';
if (i + n < int((tmp).size())) tmp[i + n] = '0';
}
for (int i = (0); i < (int((tmp).size())); i += (1)) {
bool check1 = (i - n >= 0 && tmp[i - n] == '1');
bool check2 = (i + n < int((tmp).size()) && tmp[i + n] == '1');
if (check1 || check2) {
if (s[i] != '1') {
check = 0;
break;
}
} else if (s[i] != '0') {
check = 0;
break;
}
}
cout << (check ? tmp : "-1") << '\n';
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
void solve() {
string s, w, t;
long long n, x, k = 0, cnt = 0, mx = -1000000007, mn = 1000000007;
cin >> s >> x;
n = s.length();
for (int i = 0; i < n; i++) w.push_back('1');
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i >= x) w[i - x] = '0';
if (i + x < n) w[i + x] = '0';
}
}
bool works = 1;
for (int i = 0; i < n; i++) {
if (s[i] == '1') {
if (i >= x && w[i - x] == '1') continue;
if (i + x < n && w[i + x] == '1') continue;
works = 0;
}
}
if (works)
cout << w;
else
cout << -1;
cout << "\n";
}
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
long long n, tests = 1, prev = 0, i = 0;
cin >> tests;
while (tests--) {
solve();
}
cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
std::mt19937_64 rng(
std::chrono::steady_clock::now().time_since_epoch().count());
return 0;
}
| ### Prompt
Your task is to create a Cpp solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void solve() {
string s, w, t;
long long n, x, k = 0, cnt = 0, mx = -1000000007, mn = 1000000007;
cin >> s >> x;
n = s.length();
for (int i = 0; i < n; i++) w.push_back('1');
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i >= x) w[i - x] = '0';
if (i + x < n) w[i + x] = '0';
}
}
bool works = 1;
for (int i = 0; i < n; i++) {
if (s[i] == '1') {
if (i >= x && w[i - x] == '1') continue;
if (i + x < n && w[i + x] == '1') continue;
works = 0;
}
}
if (works)
cout << w;
else
cout << -1;
cout << "\n";
}
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
long long n, tests = 1, prev = 0, i = 0;
cin >> tests;
while (tests--) {
solve();
}
cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
std::mt19937_64 rng(
std::chrono::steady_clock::now().time_since_epoch().count());
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int t = 1;
cin >> t;
while (t--) {
long long n, x;
string w, s;
cin >> s >> x;
n = s.size();
w.resize(n, '-');
bool ans = 0;
for (long long i = 0; i < x; i++) {
if (s[i] == '1') {
if (i + x >= n)
ans = 1;
else {
w[i + x] = '1';
}
} else {
if (i + x >= n)
;
else
w[i + x] = '0';
}
if (ans) break;
}
for (long long i = x; i + x < n; i++) {
if (s[i] == '1') {
if (w[i - x] == '-' || w[i - x] == '1')
w[i - x] = '1';
else if (w[i + x] == '-' || w[i + x] == '1')
w[i + x] = '1';
else
ans = 1;
} else {
if (w[i - x] == '1' || w[i + x] == '1')
ans = 1;
else {
w[i - x] = w[i + x] = '0';
}
}
if (ans) break;
}
for (long long i = n - x; i < n; i++) {
if (s[i] == '1') {
if (i - x >= 0 && (w[i - x] == '-' || w[i - x] == '1'))
w[i - x] = '1';
else
ans = 1;
} else {
if (i - x >= 0 && (w[i - x] == '-' || w[i - x] == '0'))
w[i - x] = '0';
else if (i - x >= 0)
ans = 1;
}
if (ans) break;
}
for (int i = 0; i < n; i++) {
if (w[i] == '-') w[i] = '0';
}
if (ans)
cout << -1 << endl;
else
cout << w << endl;
}
return 0;
}
| ### Prompt
Develop a solution in CPP to the problem described below:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int t = 1;
cin >> t;
while (t--) {
long long n, x;
string w, s;
cin >> s >> x;
n = s.size();
w.resize(n, '-');
bool ans = 0;
for (long long i = 0; i < x; i++) {
if (s[i] == '1') {
if (i + x >= n)
ans = 1;
else {
w[i + x] = '1';
}
} else {
if (i + x >= n)
;
else
w[i + x] = '0';
}
if (ans) break;
}
for (long long i = x; i + x < n; i++) {
if (s[i] == '1') {
if (w[i - x] == '-' || w[i - x] == '1')
w[i - x] = '1';
else if (w[i + x] == '-' || w[i + x] == '1')
w[i + x] = '1';
else
ans = 1;
} else {
if (w[i - x] == '1' || w[i + x] == '1')
ans = 1;
else {
w[i - x] = w[i + x] = '0';
}
}
if (ans) break;
}
for (long long i = n - x; i < n; i++) {
if (s[i] == '1') {
if (i - x >= 0 && (w[i - x] == '-' || w[i - x] == '1'))
w[i - x] = '1';
else
ans = 1;
} else {
if (i - x >= 0 && (w[i - x] == '-' || w[i - x] == '0'))
w[i - x] = '0';
else if (i - x >= 0)
ans = 1;
}
if (ans) break;
}
for (int i = 0; i < n; i++) {
if (w[i] == '-') w[i] = '0';
}
if (ans)
cout << -1 << endl;
else
cout << w << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
char s[N], w[N];
int n, x;
int main() {
int T;
cin >> T;
while (T--) {
cin >> s + 1 >> x;
n = strlen(s + 1);
for (int i = 1; i <= n; i++) w[i] = '1';
for (int i = 1; i <= n; i++) {
if (s[i] == '0') {
if (i - x >= 1) w[i - x] = '0';
if (i + x <= n) w[i + x] = '0';
}
}
int flag = 1;
for (int i = 1; i <= n; i++) {
if (s[i] == '1') {
if (i - x >= 1 && w[i - x] == '1') continue;
if (i + x <= n && w[i + x] == '1') continue;
flag = 0;
}
}
if (flag)
for (int i = 1; i <= n; i++) cout << w[i];
else
cout << "-1";
cout << "\n";
}
return 0;
}
| ### Prompt
In CPP, your task is to solve the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
char s[N], w[N];
int n, x;
int main() {
int T;
cin >> T;
while (T--) {
cin >> s + 1 >> x;
n = strlen(s + 1);
for (int i = 1; i <= n; i++) w[i] = '1';
for (int i = 1; i <= n; i++) {
if (s[i] == '0') {
if (i - x >= 1) w[i - x] = '0';
if (i + x <= n) w[i + x] = '0';
}
}
int flag = 1;
for (int i = 1; i <= n; i++) {
if (s[i] == '1') {
if (i - x >= 1 && w[i - x] == '1') continue;
if (i + x <= n && w[i + x] == '1') continue;
flag = 0;
}
}
if (flag)
for (int i = 1; i <= n; i++) cout << w[i];
else
cout << "-1";
cout << "\n";
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
const int LIM = 100005;
long long power(long long x, long long n) {
long long res = 1;
while (n) {
if (n & 1) {
res = res * x % mod;
}
x = x * x % mod;
n >>= 1;
}
return (res % mod);
}
int32_t main() {
long long tt = 1, n, x, k, y, z, i, j;
cin >> tt;
while (tt--) {
string s, w = "", second = "";
cin >> s;
cin >> x;
n = s.length();
for (i = 0; i < n; i++) {
w += "1";
second += "0";
}
for (i = 0; i < n; i++) {
if (s[i] == '0') {
if (i + x < n) w[i + x] = '0';
if (i - x >= 0) w[i - x] = '0';
}
}
long long f = 0;
for (i = 0; i < n; i++) {
if (w[i] == '1') {
if (i + x < n) second[i + x] = '1';
if (i - x >= 0) second[i - x] = '1';
}
}
if (second == s)
cout << w << '\n';
else
cout << -1 << '\n';
}
}
| ### Prompt
Please create a solution in Cpp to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
const int LIM = 100005;
long long power(long long x, long long n) {
long long res = 1;
while (n) {
if (n & 1) {
res = res * x % mod;
}
x = x * x % mod;
n >>= 1;
}
return (res % mod);
}
int32_t main() {
long long tt = 1, n, x, k, y, z, i, j;
cin >> tt;
while (tt--) {
string s, w = "", second = "";
cin >> s;
cin >> x;
n = s.length();
for (i = 0; i < n; i++) {
w += "1";
second += "0";
}
for (i = 0; i < n; i++) {
if (s[i] == '0') {
if (i + x < n) w[i + x] = '0';
if (i - x >= 0) w[i - x] = '0';
}
}
long long f = 0;
for (i = 0; i < n; i++) {
if (w[i] == '1') {
if (i + x < n) second[i + x] = '1';
if (i - x >= 0) second[i - x] = '1';
}
}
if (second == s)
cout << w << '\n';
else
cout << -1 << '\n';
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void MAX(T &x) {
x = std::numeric_limits<T>::max();
}
template <typename T>
inline void MIN(T &x) {
x = std::numeric_limits<T>::min();
}
inline int cset(const int x) { return __builtin_popcount(x); }
inline int cset(const long long x) { return __builtin_popcountll(x); }
inline int cset(const unsigned long long x) { return __builtin_popcountll(x); }
using namespace std;
template <class c>
struct rge {
c b, e;
};
template <class c>
rge<c> range(c i, c j) {
return rge<c>{i, j};
}
template <class c>
auto dud(c *x) -> decltype(cout << *x, 0);
template <class c>
char dud(...);
struct debug {
~debug() { cout << '\n'; }
template <class c>
typename enable_if<sizeof dud<c>(0) != 1, debug &>::type operator<<(c i) {
cout << boolalpha << i;
return *this;
}
template <class c>
typename enable_if<sizeof dud<c>(0) == 1, debug &>::type operator<<(c i) {
return *this << range(begin(i), end(i));
}
template <class c, class b>
debug &operator<<(pair<b, c> d) {
return *this << "(" << d.first << ", " << d.second << ")";
}
template <class c>
debug &operator<<(rge<c> d) {
*this << "[";
for (auto it = d.b; it != d.e; ++it) *this << ", " + 2 * (it == d.b) << *it;
return *this << "]";
}
};
const long double PI = acos(-1);
const int M = (int)1e9 + 7;
signed main(void) {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int t;
cin >> t;
while (t--) {
string s;
cin >> s;
int n = s.size();
s = '*' + s;
int x;
cin >> x;
vector<int> ans(n + 1, -1);
for (int i = 1; i <= n; i++) {
if (s[i] == '0') {
if (i > x) ans[i - x] = 0;
if (i + x <= n) ans[i + x] = 0;
}
}
for (int i = 1; i <= n; i++) {
if (s[i] == '1') {
if (i > x && ans[i - x] == -1) ans[i - x] = 1;
if (i + x <= n && ans[i + x] == -1) ans[i + x] = 1;
}
}
for (int i = 1; i <= n; i++)
if (ans[i] == -1) ans[i] = 0;
string s1 = "";
for (int i = 1; i <= n; i++) {
if (i > x && ans[i - x] == 1)
s1 += '1';
else if (i + x <= n && ans[i + x] == 1)
s1 += '1';
else
s1 += '0';
}
if (s1 == s.substr(1)) {
for (int i = 1; i <= n; i++) cout << ans[i];
cout << '\n';
} else {
cout << "-1" << '\n';
}
}
}
| ### Prompt
Your challenge is to write a Cpp solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void MAX(T &x) {
x = std::numeric_limits<T>::max();
}
template <typename T>
inline void MIN(T &x) {
x = std::numeric_limits<T>::min();
}
inline int cset(const int x) { return __builtin_popcount(x); }
inline int cset(const long long x) { return __builtin_popcountll(x); }
inline int cset(const unsigned long long x) { return __builtin_popcountll(x); }
using namespace std;
template <class c>
struct rge {
c b, e;
};
template <class c>
rge<c> range(c i, c j) {
return rge<c>{i, j};
}
template <class c>
auto dud(c *x) -> decltype(cout << *x, 0);
template <class c>
char dud(...);
struct debug {
~debug() { cout << '\n'; }
template <class c>
typename enable_if<sizeof dud<c>(0) != 1, debug &>::type operator<<(c i) {
cout << boolalpha << i;
return *this;
}
template <class c>
typename enable_if<sizeof dud<c>(0) == 1, debug &>::type operator<<(c i) {
return *this << range(begin(i), end(i));
}
template <class c, class b>
debug &operator<<(pair<b, c> d) {
return *this << "(" << d.first << ", " << d.second << ")";
}
template <class c>
debug &operator<<(rge<c> d) {
*this << "[";
for (auto it = d.b; it != d.e; ++it) *this << ", " + 2 * (it == d.b) << *it;
return *this << "]";
}
};
const long double PI = acos(-1);
const int M = (int)1e9 + 7;
signed main(void) {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int t;
cin >> t;
while (t--) {
string s;
cin >> s;
int n = s.size();
s = '*' + s;
int x;
cin >> x;
vector<int> ans(n + 1, -1);
for (int i = 1; i <= n; i++) {
if (s[i] == '0') {
if (i > x) ans[i - x] = 0;
if (i + x <= n) ans[i + x] = 0;
}
}
for (int i = 1; i <= n; i++) {
if (s[i] == '1') {
if (i > x && ans[i - x] == -1) ans[i - x] = 1;
if (i + x <= n && ans[i + x] == -1) ans[i + x] = 1;
}
}
for (int i = 1; i <= n; i++)
if (ans[i] == -1) ans[i] = 0;
string s1 = "";
for (int i = 1; i <= n; i++) {
if (i > x && ans[i - x] == 1)
s1 += '1';
else if (i + x <= n && ans[i + x] == 1)
s1 += '1';
else
s1 += '0';
}
if (s1 == s.substr(1)) {
for (int i = 1; i <= n; i++) cout << ans[i];
cout << '\n';
} else {
cout << "-1" << '\n';
}
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int t;
cin >> t;
while (t--) {
string s;
cin >> s;
string tmp = "";
string tmp2 = "";
int n;
cin >> n;
for (int i = 0; i < s.size(); i++) {
tmp.push_back('1');
tmp2.push_back('1');
}
for (int i = 0; i < s.size(); i++) {
if (s[i] == '0') {
int flag = 2;
int chkpos = i + n;
if (chkpos >= 0 && chkpos < s.size()) {
tmp[chkpos] = '0';
}
chkpos = i - n;
if (chkpos >= 0 && chkpos < s.size()) {
tmp[chkpos] = '0';
}
}
}
for (int i = 0; i < s.size(); i++) {
int flag = 2;
int chkpos = i + n;
if (chkpos >= 0 && chkpos < s.size()) {
if (tmp[chkpos] == '1') flag--;
}
chkpos = i - n;
if (chkpos >= 0 && chkpos < s.size()) {
if (tmp[chkpos] == '1') flag--;
}
if (flag == 2) {
tmp2[i] = '0';
}
}
if (tmp2 == s)
cout << tmp << '\n';
else
cout << -1 << '\n';
}
}
| ### Prompt
Please provide a CPP coded solution to the problem described below:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int t;
cin >> t;
while (t--) {
string s;
cin >> s;
string tmp = "";
string tmp2 = "";
int n;
cin >> n;
for (int i = 0; i < s.size(); i++) {
tmp.push_back('1');
tmp2.push_back('1');
}
for (int i = 0; i < s.size(); i++) {
if (s[i] == '0') {
int flag = 2;
int chkpos = i + n;
if (chkpos >= 0 && chkpos < s.size()) {
tmp[chkpos] = '0';
}
chkpos = i - n;
if (chkpos >= 0 && chkpos < s.size()) {
tmp[chkpos] = '0';
}
}
}
for (int i = 0; i < s.size(); i++) {
int flag = 2;
int chkpos = i + n;
if (chkpos >= 0 && chkpos < s.size()) {
if (tmp[chkpos] == '1') flag--;
}
chkpos = i - n;
if (chkpos >= 0 && chkpos < s.size()) {
if (tmp[chkpos] == '1') flag--;
}
if (flag == 2) {
tmp2[i] = '0';
}
}
if (tmp2 == s)
cout << tmp << '\n';
else
cout << -1 << '\n';
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
string s;
cin >> s;
int x;
cin >> x;
string w;
for (int i = 0; i < s.length(); i++) {
w.push_back('1');
}
for (int i = 0; i < s.length(); i++) {
if (s[i] == '0') {
if (i - x >= 0) w[i - x] = '0';
if (i + x < s.length()) w[i + x] = '0';
}
}
string temp;
for (int i = 0; i < s.length(); i++) {
if (i - x >= 0 && w[i - x] == '1')
temp.push_back('1');
else if (i + x < s.length() && w[i + x] == '1')
temp.push_back('1');
else
temp.push_back('0');
}
if (temp != s)
cout << "-1" << endl;
else
cout << w << endl;
}
}
| ### Prompt
Create a solution in cpp for the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
string s;
cin >> s;
int x;
cin >> x;
string w;
for (int i = 0; i < s.length(); i++) {
w.push_back('1');
}
for (int i = 0; i < s.length(); i++) {
if (s[i] == '0') {
if (i - x >= 0) w[i - x] = '0';
if (i + x < s.length()) w[i + x] = '0';
}
}
string temp;
for (int i = 0; i < s.length(); i++) {
if (i - x >= 0 && w[i - x] == '1')
temp.push_back('1');
else if (i + x < s.length() && w[i + x] == '1')
temp.push_back('1');
else
temp.push_back('0');
}
if (temp != s)
cout << "-1" << endl;
else
cout << w << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int M = 1e9 + 7, N = 205001;
int mm(int x, int y) {
x %= M, y %= M;
return (x * y) % M;
}
int po(int x, int y) {
if (!y) return 1;
int a = po(x, y / 2) % M;
if (y % 2) return mm(a, mm(a, x));
return mm(a, a);
}
void solve() {
int n, k;
string s;
cin >> s >> k;
n = s.size();
string a(n, '1');
for (int i = 0; 0 < n ? i < n : i > n; 0 < n ? i += 1 : i -= 1) {
if (s[i] == '0' and i + k < n) a[i + k] = '0';
if (s[i] == '0' and i - k >= 0) a[i - k] = '0';
}
for (int i = 0; 0 < n ? i < n : i > n; 0 < n ? i += 1 : i -= 1) {
if (s[i] == '1')
if (!((i + k < n and a[i + k] == '1') or
(i - k >= 0 and a[i - k] == '1'))) {
cout << "-1\n";
return;
}
}
cout << a << "\n";
}
signed main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t = 1;
cin >> t;
while (t--) solve();
return 0;
}
| ### Prompt
In CPP, your task is to solve the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int M = 1e9 + 7, N = 205001;
int mm(int x, int y) {
x %= M, y %= M;
return (x * y) % M;
}
int po(int x, int y) {
if (!y) return 1;
int a = po(x, y / 2) % M;
if (y % 2) return mm(a, mm(a, x));
return mm(a, a);
}
void solve() {
int n, k;
string s;
cin >> s >> k;
n = s.size();
string a(n, '1');
for (int i = 0; 0 < n ? i < n : i > n; 0 < n ? i += 1 : i -= 1) {
if (s[i] == '0' and i + k < n) a[i + k] = '0';
if (s[i] == '0' and i - k >= 0) a[i - k] = '0';
}
for (int i = 0; 0 < n ? i < n : i > n; 0 < n ? i += 1 : i -= 1) {
if (s[i] == '1')
if (!((i + k < n and a[i + k] == '1') or
(i - k >= 0 and a[i - k] == '1'))) {
cout << "-1\n";
return;
}
}
cout << a << "\n";
}
signed main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t = 1;
cin >> t;
while (t--) solve();
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int32_t main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
long long int T;
cin >> T;
while (T--) {
string s;
cin >> s;
int k;
cin >> k;
int len = s.length();
char ans[len];
for (int i = 0; i < len; i++) ans[i] = '1';
for (int i = 0; i < len; i++) {
if (s[i] == '0') {
if (i + k < len) ans[i + k] = '0';
if (i - k >= 0) ans[i - k] = '0';
}
}
bool possible = true;
for (int i = 0; i < len; i++) {
if (s[i] == '1') {
if (i + k < len && ans[i + k] == '1') continue;
if (i - k >= 0 && ans[i - k] == '1') continue;
possible = false;
}
}
if (!possible)
cout << -1 << endl;
else {
for (int i = 0; i < len; i++) cout << ans[i];
cout << endl;
}
}
return 0;
}
| ### Prompt
Generate a cpp solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int32_t main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
long long int T;
cin >> T;
while (T--) {
string s;
cin >> s;
int k;
cin >> k;
int len = s.length();
char ans[len];
for (int i = 0; i < len; i++) ans[i] = '1';
for (int i = 0; i < len; i++) {
if (s[i] == '0') {
if (i + k < len) ans[i + k] = '0';
if (i - k >= 0) ans[i - k] = '0';
}
}
bool possible = true;
for (int i = 0; i < len; i++) {
if (s[i] == '1') {
if (i + k < len && ans[i + k] == '1') continue;
if (i - k >= 0 && ans[i - k] == '1') continue;
possible = false;
}
}
if (!possible)
cout << -1 << endl;
else {
for (int i = 0; i < len; i++) cout << ans[i];
cout << endl;
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
template <typename T>
void DEBUG(string label, T value) {
cerr << "[" << label << " = " << value << "]\n";
}
void solve() {
string s;
int x;
cin >> s >> x;
int n = (int)s.size();
string ans(n, ' ');
for (int i = 0; i < x; ++i) {
int e = i + x;
if (e < n) {
ans[e] = s[i];
}
}
for (int i = n - 1; i >= n - 1 - x + 1; --i) {
int e = i - x;
if (e >= 0) {
if (ans[e] != ' ' && ans[e] != s[i]) {
cout << -1 << '\n';
return;
}
ans[e] = s[i];
}
}
for (int i = 0; i < n; ++i) {
if (s[i] == '1') continue;
int one = i - x;
int two = i + x;
if (one < 0 && two >= n) {
continue;
}
if (one >= 0) {
if (ans[one] == '1') {
cout << -1 << '\n';
return;
} else {
ans[one] = '0';
}
}
if (two < n) {
if (ans[two] == '1') {
cout << -1 << '\n';
return;
} else {
ans[two] = '0';
}
}
}
for (int i = 0; i < n; ++i) {
if (s[i] == '0') {
continue;
}
int one = i - x;
int two = i + x;
if (one < 0 && two >= n) {
cout << -1 << '\n';
return;
}
if (one >= 0 && two < n) {
if (ans[one] == '0' && ans[two] == '0') {
cout << -1 << '\n';
return;
}
} else if (one >= 0) {
if (ans[one] == '0') {
cout << -1 << '\n';
return;
}
ans[one] = '1';
} else if (two <= n) {
if (ans[two] == '0') {
cout << -1 << '\n';
return;
}
ans[two] = '1';
}
}
for (auto i : ans) {
if (i == ' ') {
cout << '1';
} else {
cout << i;
}
}
cout << '\n';
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t;
cin >> t;
while (t--) {
solve();
}
}
| ### Prompt
Please create a solution in Cpp to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
template <typename T>
void DEBUG(string label, T value) {
cerr << "[" << label << " = " << value << "]\n";
}
void solve() {
string s;
int x;
cin >> s >> x;
int n = (int)s.size();
string ans(n, ' ');
for (int i = 0; i < x; ++i) {
int e = i + x;
if (e < n) {
ans[e] = s[i];
}
}
for (int i = n - 1; i >= n - 1 - x + 1; --i) {
int e = i - x;
if (e >= 0) {
if (ans[e] != ' ' && ans[e] != s[i]) {
cout << -1 << '\n';
return;
}
ans[e] = s[i];
}
}
for (int i = 0; i < n; ++i) {
if (s[i] == '1') continue;
int one = i - x;
int two = i + x;
if (one < 0 && two >= n) {
continue;
}
if (one >= 0) {
if (ans[one] == '1') {
cout << -1 << '\n';
return;
} else {
ans[one] = '0';
}
}
if (two < n) {
if (ans[two] == '1') {
cout << -1 << '\n';
return;
} else {
ans[two] = '0';
}
}
}
for (int i = 0; i < n; ++i) {
if (s[i] == '0') {
continue;
}
int one = i - x;
int two = i + x;
if (one < 0 && two >= n) {
cout << -1 << '\n';
return;
}
if (one >= 0 && two < n) {
if (ans[one] == '0' && ans[two] == '0') {
cout << -1 << '\n';
return;
}
} else if (one >= 0) {
if (ans[one] == '0') {
cout << -1 << '\n';
return;
}
ans[one] = '1';
} else if (two <= n) {
if (ans[two] == '0') {
cout << -1 << '\n';
return;
}
ans[two] = '1';
}
}
for (auto i : ans) {
if (i == ' ') {
cout << '1';
} else {
cout << i;
}
}
cout << '\n';
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t;
cin >> t;
while (t--) {
solve();
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int T = 1;
cin >> T;
while (T--) {
string s;
int x;
cin >> s >> x;
string ans;
int n = s.size();
for (int i = 0; i < n; i++) ans.push_back('1');
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) ans[i - x] = '0';
if (i + x < n) ans[i + x] = '0';
}
}
bool f = 1;
for (int i = 0; i < n; i++) {
if (s[i] == '1') {
bool t = 0;
if (i + x < n && ans[i + x] == '1') t = 1;
if (i - x >= 0 && ans[i - x] == '1') t = 1;
if (!t) f = 0;
}
}
if (f)
cout << ans << endl;
else
cout << "-1\n";
}
return 0;
}
| ### Prompt
Generate a CPP solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int T = 1;
cin >> T;
while (T--) {
string s;
int x;
cin >> s >> x;
string ans;
int n = s.size();
for (int i = 0; i < n; i++) ans.push_back('1');
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) ans[i - x] = '0';
if (i + x < n) ans[i + x] = '0';
}
}
bool f = 1;
for (int i = 0; i < n; i++) {
if (s[i] == '1') {
bool t = 0;
if (i + x < n && ans[i + x] == '1') t = 1;
if (i - x >= 0 && ans[i - x] == '1') t = 1;
if (!t) f = 0;
}
}
if (f)
cout << ans << endl;
else
cout << "-1\n";
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int i, t, x, vis[100005];
scanf("%d", &t);
while (t--) {
int l;
char s[100005], w[100005], ss[100005];
memset(vis, 0, sizeof(vis));
scanf("%s", s);
scanf("%d", &x);
l = strlen(s);
for (i = 0; i < l; i++) {
if (s[i] == '0') {
if (i >= x && vis[i - x] == 0) {
vis[i - x] = 1;
w[i - x] = '0';
}
if ((i + x) < l && vis[i + x] == 0) {
vis[i + x] = 1;
w[i + x] = '0';
}
}
}
for (i = 0; i < l; i++) {
if (s[i] == '1') {
if (i >= x && vis[i - x] == 0) {
vis[i - x] = 1;
w[i - x] = '1';
} else if (vis[i + x] == 0 && (i + x) < l) {
vis[i + x] = 1;
w[i + x] = '1';
}
}
}
for (i = 0; i < l; i++) {
if (vis[i] == 0) w[i] = '1';
}
w[l] = '\0';
for (i = 0; i < l; i++) {
ss[i] = '0';
if (i >= x) {
if (w[i - x] == '1') {
ss[i] = '1';
}
}
if ((i + x) < l) {
if (w[i + x] == '1') {
ss[i] = '1';
}
}
}
ss[i] = '\0';
if (strcmp(ss, s) == 0) {
for (i = 0; i < l; i++) printf("%c", w[i]);
} else
printf("-1");
printf("\n");
}
return 0;
}
| ### Prompt
Your challenge is to write a Cpp solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int i, t, x, vis[100005];
scanf("%d", &t);
while (t--) {
int l;
char s[100005], w[100005], ss[100005];
memset(vis, 0, sizeof(vis));
scanf("%s", s);
scanf("%d", &x);
l = strlen(s);
for (i = 0; i < l; i++) {
if (s[i] == '0') {
if (i >= x && vis[i - x] == 0) {
vis[i - x] = 1;
w[i - x] = '0';
}
if ((i + x) < l && vis[i + x] == 0) {
vis[i + x] = 1;
w[i + x] = '0';
}
}
}
for (i = 0; i < l; i++) {
if (s[i] == '1') {
if (i >= x && vis[i - x] == 0) {
vis[i - x] = 1;
w[i - x] = '1';
} else if (vis[i + x] == 0 && (i + x) < l) {
vis[i + x] = 1;
w[i + x] = '1';
}
}
}
for (i = 0; i < l; i++) {
if (vis[i] == 0) w[i] = '1';
}
w[l] = '\0';
for (i = 0; i < l; i++) {
ss[i] = '0';
if (i >= x) {
if (w[i - x] == '1') {
ss[i] = '1';
}
}
if ((i + x) < l) {
if (w[i + x] == '1') {
ss[i] = '1';
}
}
}
ss[i] = '\0';
if (strcmp(ss, s) == 0) {
for (i = 0; i < l; i++) printf("%c", w[i]);
} else
printf("-1");
printf("\n");
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
bool ok;
int n, x;
string s;
bool w[300000];
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int t;
cin >> t;
while (t--) {
cin >> s >> x;
n = s.size();
ok = true;
for (int i = 0; i < n; i++) w[i] = 1;
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i + x < n) w[i + x] = 0;
if (i - x >= 0) w[i - x] = 0;
}
}
for (int i = 0; i < n; i++) {
if (s[i] == '1') {
if (i + x < n && w[i + x]) continue;
if (i - x >= 0 && w[i - x]) continue;
ok = false;
break;
}
}
if (!ok)
cout << "-1\n";
else {
for (int i = 0; i < n; i++) cout << w[i];
cout << "\n";
}
}
}
| ### Prompt
Please provide a Cpp coded solution to the problem described below:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
bool ok;
int n, x;
string s;
bool w[300000];
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int t;
cin >> t;
while (t--) {
cin >> s >> x;
n = s.size();
ok = true;
for (int i = 0; i < n; i++) w[i] = 1;
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i + x < n) w[i + x] = 0;
if (i - x >= 0) w[i - x] = 0;
}
}
for (int i = 0; i < n; i++) {
if (s[i] == '1') {
if (i + x < n && w[i + x]) continue;
if (i - x >= 0 && w[i - x]) continue;
ok = false;
break;
}
}
if (!ok)
cout << "-1\n";
else {
for (int i = 0; i < n; i++) cout << w[i];
cout << "\n";
}
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long int t = 1, T;
long long int mod = 1e9 + 7;
void print(vector<long long int> a) {
for (long long int x : a) {
cout << x << ' ';
}
cout << '\n';
return;
}
void swap(long long int &n, long long int &m) {
n = n ^ m;
m = n ^ m;
n = n ^ m;
}
void solve() {
string s;
cin >> s;
long long int x;
cin >> x;
long long int n = s.size();
string ans(n, '2');
for (long long int i = 0; i < n; i++) {
if (s[i] == '1') {
if (i + x >= n && i - x < 0) {
cout << -1 << '\n';
return;
} else if (i + x >= n && i - x >= 0) {
ans[i - x] = '1';
} else if (i + x < n && i - x < 0) {
ans[i + x] = '1';
} else if (i + x < n && i - x >= 0) {
ans[i + x] = '3';
ans[i - x] = '3';
}
}
}
for (long long int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i + x < n && i - x >= 0) {
if (ans[i - x] == '1' || ans[i + x] == '1') {
cout << -1 << '\n';
return;
}
ans[i - x] = '0';
ans[i + x] = '0';
} else if (i + x >= n && i - x >= 0) {
if (ans[i - x] == '1') {
cout << -1 << '\n';
return;
}
ans[i - x] = '0';
} else if (i + x < n && i - x < 0) {
if (ans[i + x] == '1') {
cout << -1 << '\n';
return;
}
ans[i + x] = '0';
}
}
}
for (long long int i = 0; i < n; i++) {
if (s[i] == '1') {
if (i + x < n && i - x >= 0) {
if (ans[i + x] == '0' && ans[i - x] == '0') {
cout << -1 << '\n';
return;
}
if (ans[i + x] != '0') ans[i + x] = '1';
if (ans[i - x] != '0') ans[i - x] = '1';
}
}
}
for (long long int i = 0; i < n; i++) {
if (ans[i] == '2') ans[i] = '1';
}
cout << ans << '\n';
}
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
cin >> t;
T = t;
while (t--) {
solve();
}
return 0;
}
| ### Prompt
In Cpp, your task is to solve the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long int t = 1, T;
long long int mod = 1e9 + 7;
void print(vector<long long int> a) {
for (long long int x : a) {
cout << x << ' ';
}
cout << '\n';
return;
}
void swap(long long int &n, long long int &m) {
n = n ^ m;
m = n ^ m;
n = n ^ m;
}
void solve() {
string s;
cin >> s;
long long int x;
cin >> x;
long long int n = s.size();
string ans(n, '2');
for (long long int i = 0; i < n; i++) {
if (s[i] == '1') {
if (i + x >= n && i - x < 0) {
cout << -1 << '\n';
return;
} else if (i + x >= n && i - x >= 0) {
ans[i - x] = '1';
} else if (i + x < n && i - x < 0) {
ans[i + x] = '1';
} else if (i + x < n && i - x >= 0) {
ans[i + x] = '3';
ans[i - x] = '3';
}
}
}
for (long long int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i + x < n && i - x >= 0) {
if (ans[i - x] == '1' || ans[i + x] == '1') {
cout << -1 << '\n';
return;
}
ans[i - x] = '0';
ans[i + x] = '0';
} else if (i + x >= n && i - x >= 0) {
if (ans[i - x] == '1') {
cout << -1 << '\n';
return;
}
ans[i - x] = '0';
} else if (i + x < n && i - x < 0) {
if (ans[i + x] == '1') {
cout << -1 << '\n';
return;
}
ans[i + x] = '0';
}
}
}
for (long long int i = 0; i < n; i++) {
if (s[i] == '1') {
if (i + x < n && i - x >= 0) {
if (ans[i + x] == '0' && ans[i - x] == '0') {
cout << -1 << '\n';
return;
}
if (ans[i + x] != '0') ans[i + x] = '1';
if (ans[i - x] != '0') ans[i - x] = '1';
}
}
}
for (long long int i = 0; i < n; i++) {
if (ans[i] == '2') ans[i] = '1';
}
cout << ans << '\n';
}
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
cin >> t;
T = t;
while (t--) {
solve();
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
void parseArray(long long* A, long long n) {
for (long long K = 0; K < n; K++) {
cin >> A[K];
}
}
long long modInverse(long long a, long long b) {
return 1 < a ? b - modInverse(b % a, a) * b / a : 1;
}
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) { return (a * b) / gcd(a, b); }
long double dist(long double x, long double y, long double a, long double b) {
return sqrt((x - a) * (x - a) + (y - b) * (y - b));
}
void debug(long long* a, long long n) {
for (long long k = 0; k < n; k++) {
cout << a[k] << " ";
}
cout << "\n";
}
int main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
;
long long t;
cin >> t;
while (t--) {
string s;
cin >> s;
long long m;
cin >> m;
long long n = s.length();
char ans[n + 1];
memset(ans, 0, sizeof ans);
for (int k = 0; k < n; k++) {
if (s[k] == '1') {
if (k - m >= 0) ans[k - m] = '1';
if (k + m < n) ans[k + m] = '1';
}
}
for (int k = 0; k < n; k++) {
if (s[k] == '0') {
if (k + m < n) ans[k + m] = '0';
if (k - m >= 0) ans[k - m] = '0';
}
}
for (int k = 0; k < n; k++)
if (ans[k] == 0) ans[k] = '0';
bool flag = true;
for (int k = 0; k < n; k++) {
if (s[k] == '1') {
bool temp = false;
if (k + m < n && ans[k + m] == '1') temp = true;
if (k - m >= 0 && ans[k - m] == '1') temp = true;
flag = flag && temp;
} else {
int temp = 0;
int cnt = 0;
if (k + m < n) cnt++;
if (k - m >= 0) cnt++;
if (k + m < n && ans[k + m] == '0') temp++;
if (k - m >= 0 && ans[k - m] == '0') temp++;
if (temp != cnt) flag = false;
}
}
if (flag)
cout << ans << "\n";
else
cout << "-1\n";
}
return 0;
}
| ### Prompt
Generate a CPP solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void parseArray(long long* A, long long n) {
for (long long K = 0; K < n; K++) {
cin >> A[K];
}
}
long long modInverse(long long a, long long b) {
return 1 < a ? b - modInverse(b % a, a) * b / a : 1;
}
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) { return (a * b) / gcd(a, b); }
long double dist(long double x, long double y, long double a, long double b) {
return sqrt((x - a) * (x - a) + (y - b) * (y - b));
}
void debug(long long* a, long long n) {
for (long long k = 0; k < n; k++) {
cout << a[k] << " ";
}
cout << "\n";
}
int main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
;
long long t;
cin >> t;
while (t--) {
string s;
cin >> s;
long long m;
cin >> m;
long long n = s.length();
char ans[n + 1];
memset(ans, 0, sizeof ans);
for (int k = 0; k < n; k++) {
if (s[k] == '1') {
if (k - m >= 0) ans[k - m] = '1';
if (k + m < n) ans[k + m] = '1';
}
}
for (int k = 0; k < n; k++) {
if (s[k] == '0') {
if (k + m < n) ans[k + m] = '0';
if (k - m >= 0) ans[k - m] = '0';
}
}
for (int k = 0; k < n; k++)
if (ans[k] == 0) ans[k] = '0';
bool flag = true;
for (int k = 0; k < n; k++) {
if (s[k] == '1') {
bool temp = false;
if (k + m < n && ans[k + m] == '1') temp = true;
if (k - m >= 0 && ans[k - m] == '1') temp = true;
flag = flag && temp;
} else {
int temp = 0;
int cnt = 0;
if (k + m < n) cnt++;
if (k - m >= 0) cnt++;
if (k + m < n && ans[k + m] == '0') temp++;
if (k - m >= 0 && ans[k - m] == '0') temp++;
if (temp != cnt) flag = false;
}
}
if (flag)
cout << ans << "\n";
else
cout << "-1\n";
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
void solve() {
long long t;
cin >> t;
while (t--) {
string s;
cin >> s;
long long x;
cin >> x;
long long n = s.size();
string ans(n, '1');
for (long long i = 0; i < n; ++i) {
if (s[i] == '0') {
if ((i - x) >= 0) ans[i - x] = '0';
if ((i + x) < n) ans[i + x] = '0';
}
}
string res(n, '0');
for (long long i = 0; i < n; ++i) {
long long l = i - x;
long long r = i + x;
if (l >= 0) {
if (ans[l] == '1') res[i] = '1';
}
if (r < n) {
if (ans[r] == '1') res[i] = '1';
}
}
if (res == s)
cout << ans << "\n";
else
cout << -1 << "\n";
}
}
int main() {
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
}
solve();
return 0;
}
| ### Prompt
Please formulate a CPP solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void solve() {
long long t;
cin >> t;
while (t--) {
string s;
cin >> s;
long long x;
cin >> x;
long long n = s.size();
string ans(n, '1');
for (long long i = 0; i < n; ++i) {
if (s[i] == '0') {
if ((i - x) >= 0) ans[i - x] = '0';
if ((i + x) < n) ans[i + x] = '0';
}
}
string res(n, '0');
for (long long i = 0; i < n; ++i) {
long long l = i - x;
long long r = i + x;
if (l >= 0) {
if (ans[l] == '1') res[i] = '1';
}
if (r < n) {
if (ans[r] == '1') res[i] = '1';
}
}
if (res == s)
cout << ans << "\n";
else
cout << -1 << "\n";
}
}
int main() {
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
}
solve();
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
using ll = long long int;
using ld = long double;
using pll = pair<ll, ll>;
using pii = pair<int, int>;
using vll = vector<ll>;
using vi = vector<int>;
const ll mod = (ll)(1e9) + 7LL;
const ll M = 988244353LL;
void solve() {
string s;
cin >> s;
int x;
cin >> x;
int n = s.length();
vector<int> confirm(n, 0);
string ans = string(n, '0');
bool flag = true;
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i >= x) {
if (confirm[i - x] == 1 && ans[i - x] == '1') {
flag = false;
break;
} else {
ans[i - x] = '0';
confirm[i - x] = 1;
}
}
if (i + x < n) {
ans[i + x] = '0';
confirm[i + x] = 1;
}
} else {
if (i >= x) {
if (confirm[i - x] == 1 && ans[i - x] == '0') {
if (i + x < n) {
ans[i + x] = '1';
confirm[i + x] = 1;
} else {
flag = false;
break;
}
} else {
confirm[i - x] = 1;
ans[i - x] = '1';
}
} else if (i + x < n) {
confirm[i + x] = 1;
ans[i + x] = '1';
} else {
flag = false;
break;
}
}
}
if (flag) {
cout << ans << endl;
} else {
cout << -1 << endl;
}
}
int main() {
ll tc = 1;
cin >> tc;
while (tc--) {
solve();
}
return 0;
}
| ### Prompt
Create a solution in Cpp for the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long int;
using ld = long double;
using pll = pair<ll, ll>;
using pii = pair<int, int>;
using vll = vector<ll>;
using vi = vector<int>;
const ll mod = (ll)(1e9) + 7LL;
const ll M = 988244353LL;
void solve() {
string s;
cin >> s;
int x;
cin >> x;
int n = s.length();
vector<int> confirm(n, 0);
string ans = string(n, '0');
bool flag = true;
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i >= x) {
if (confirm[i - x] == 1 && ans[i - x] == '1') {
flag = false;
break;
} else {
ans[i - x] = '0';
confirm[i - x] = 1;
}
}
if (i + x < n) {
ans[i + x] = '0';
confirm[i + x] = 1;
}
} else {
if (i >= x) {
if (confirm[i - x] == 1 && ans[i - x] == '0') {
if (i + x < n) {
ans[i + x] = '1';
confirm[i + x] = 1;
} else {
flag = false;
break;
}
} else {
confirm[i - x] = 1;
ans[i - x] = '1';
}
} else if (i + x < n) {
confirm[i + x] = 1;
ans[i + x] = '1';
} else {
flag = false;
break;
}
}
}
if (flag) {
cout << ans << endl;
} else {
cout << -1 << endl;
}
}
int main() {
ll tc = 1;
cin >> tc;
while (tc--) {
solve();
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int a, x;
scanf("%d", &a);
while (a--) {
string str;
cin >> str;
int wrong = 1, right, changdu = str.length();
cin >> x;
string str2(changdu, '1');
for (int i = 0; i < changdu; i++) {
if (str[i] == '0') {
if (i - x >= 0) str2[i - x] = '0';
if (i + x < changdu) str2[i + x] = '0';
}
}
for (int i = 0; i < changdu; i++) {
right = 0;
if ((i - x) >= 0 && str2[i - x] == '1') right = 1;
if ((i + x) < changdu && str2[i + x] == '1') right = 1;
if (str[i] == '1') {
if (right == 0) {
wrong = 0;
cout << -1;
break;
}
}
}
if (wrong) cout << str2;
cout << endl;
}
}
| ### Prompt
Create a solution in Cpp for the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int a, x;
scanf("%d", &a);
while (a--) {
string str;
cin >> str;
int wrong = 1, right, changdu = str.length();
cin >> x;
string str2(changdu, '1');
for (int i = 0; i < changdu; i++) {
if (str[i] == '0') {
if (i - x >= 0) str2[i - x] = '0';
if (i + x < changdu) str2[i + x] = '0';
}
}
for (int i = 0; i < changdu; i++) {
right = 0;
if ((i - x) >= 0 && str2[i - x] == '1') right = 1;
if ((i + x) < changdu && str2[i + x] == '1') right = 1;
if (str[i] == '1') {
if (right == 0) {
wrong = 0;
cout << -1;
break;
}
}
}
if (wrong) cout << str2;
cout << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long ml(long long a, long long b) {
return ((a % 1000000007) * (b % 1000000007)) % 1000000007;
}
int main() {
long long t;
cin >> t;
while (t--) {
string s;
long long x;
cin >> s >> x;
long long n = s.size();
int a[n];
memset(a, -1, sizeof(a));
for (long long i = 0; i < n; i = i + 1) {
if (s[i] == '1') {
if (i - x >= 0) {
}
}
if (s[i] == '0') {
if (i - x >= 0) {
a[i - x] = 0;
}
if (i + x < n) {
a[i + x] = 0;
}
}
}
bool k = true;
for (long long i = 0; i < n; i = i + 1) {
if (s[i] == '1') {
if (i - x >= 0 && a[i - x] != 0)
a[i - x] = 1;
else if (i + x < n && a[i + x] != 0)
a[i + x] = 1;
else {
k = false;
}
}
}
if (!k)
cout << -1 << endl;
else {
for (long long i = 0; i < n; i = i + 1)
if (a[i] == -1) a[i] = 1;
for (long long i = 0; i < n; i = i + 1) cout << a[i];
cout << endl;
}
}
}
| ### Prompt
Your challenge is to write a Cpp solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long ml(long long a, long long b) {
return ((a % 1000000007) * (b % 1000000007)) % 1000000007;
}
int main() {
long long t;
cin >> t;
while (t--) {
string s;
long long x;
cin >> s >> x;
long long n = s.size();
int a[n];
memset(a, -1, sizeof(a));
for (long long i = 0; i < n; i = i + 1) {
if (s[i] == '1') {
if (i - x >= 0) {
}
}
if (s[i] == '0') {
if (i - x >= 0) {
a[i - x] = 0;
}
if (i + x < n) {
a[i + x] = 0;
}
}
}
bool k = true;
for (long long i = 0; i < n; i = i + 1) {
if (s[i] == '1') {
if (i - x >= 0 && a[i - x] != 0)
a[i - x] = 1;
else if (i + x < n && a[i + x] != 0)
a[i + x] = 1;
else {
k = false;
}
}
}
if (!k)
cout << -1 << endl;
else {
for (long long i = 0; i < n; i = i + 1)
if (a[i] == -1) a[i] = 1;
for (long long i = 0; i < n; i = i + 1) cout << a[i];
cout << endl;
}
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
string constructstring(string s, int x) {
string str;
int n = s.size();
for (int i = 0; i < n; i++) {
if ((i - x >= 0 && s[i - x] == '1') || (i + x < n && s[i + x] == '1')) {
str.push_back('1');
} else {
str.push_back('0');
}
}
return str;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int t, x;
cin >> t;
while (t--) {
string s;
cin >> s;
cin >> x;
string str = s;
for (auto &c : str) c = '1';
int n = s.size();
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) {
str[i - x] = '0';
}
if (i + x < n) {
str[i + x] = '0';
}
}
}
if (s == constructstring(str, x)) {
cout << str << '\n';
} else {
cout << "-1\n";
}
}
return 0;
}
| ### Prompt
Generate a cpp solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
string constructstring(string s, int x) {
string str;
int n = s.size();
for (int i = 0; i < n; i++) {
if ((i - x >= 0 && s[i - x] == '1') || (i + x < n && s[i + x] == '1')) {
str.push_back('1');
} else {
str.push_back('0');
}
}
return str;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int t, x;
cin >> t;
while (t--) {
string s;
cin >> s;
cin >> x;
string str = s;
for (auto &c : str) c = '1';
int n = s.size();
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) {
str[i - x] = '0';
}
if (i + x < n) {
str[i + x] = '0';
}
}
}
if (s == constructstring(str, x)) {
cout << str << '\n';
} else {
cout << "-1\n";
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
const long long int MAXN = 1e5 + 5;
const long long int mod = 1e9 + 7;
long long int it = 0, ans = 0;
template <typename temp>
void sp(temp x) {
cout << x << " ";
}
template <typename temp>
void np(temp x) {
cout << x << "\n";
}
template <typename Arg1, typename... Args>
void sp(Arg1 arg1, Args... args) {
cout << arg1 << " ";
sp(args...);
}
template <typename Arg1, typename... Args>
void np(Arg1 arg1, Args... args) {
cout << arg1 << "\n";
np(args...);
}
template <typename temp>
void spv(temp T, long long int n) {
for (int i = 0; i < n; i++) sp(T[i]);
}
template <typename temp>
void npv(temp T, long long int n) {
for (int i = 0; i < n; i++) np(T[i]);
}
template <typename T>
T modpow(T base, T exp, T modulus) {
base %= modulus;
T result = 1;
while (exp > 0) {
if (exp & 1) result = (result * base) % modulus;
base = (base * base) % modulus;
exp >>= 1;
}
return result;
}
template <typename T>
T modinv(T base, T exp) {
return modpow(base, exp - 2, exp);
}
long long int spf[MAXN];
void sieve() {
spf[1] = 1;
for (auto i = 2; i < MAXN; i++) spf[i] = i;
for (auto i = 4; i < MAXN; i += 2) spf[i] = 2;
for (auto i = 3; i * i < MAXN; i++) {
if (spf[i] == i) {
for (auto j = i * i; j < MAXN; j += i)
if (spf[j] == j) spf[j] = i;
}
}
return;
}
vector<long long int> getFactorization(long long int x) {
vector<long long int> res;
while (x != 1) {
res.push_back(spf[x]);
x = x / spf[x];
}
return res;
}
void check() {
string s;
long long int k;
cin >> s >> k;
long long int n = s.size();
string w(n, '1');
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i + k < n) w[i + k] = '0';
if (i - k >= 0) w[i - k] = '0';
}
}
for (int i = 0; i < n; i++) {
bool ok = false;
ok = ok || (i + k < n && w[i + k] == '1');
ok = ok || (i - k >= 0 && w[i - k] == '1');
if (s[i] != ok + '0') {
np(-1);
return;
}
}
np(w);
return;
}
int32_t main() {
ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
;
long long int t = 1;
cin >> t;
while (t--) check();
return 0;
}
| ### Prompt
In CPP, your task is to solve the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
const long long int MAXN = 1e5 + 5;
const long long int mod = 1e9 + 7;
long long int it = 0, ans = 0;
template <typename temp>
void sp(temp x) {
cout << x << " ";
}
template <typename temp>
void np(temp x) {
cout << x << "\n";
}
template <typename Arg1, typename... Args>
void sp(Arg1 arg1, Args... args) {
cout << arg1 << " ";
sp(args...);
}
template <typename Arg1, typename... Args>
void np(Arg1 arg1, Args... args) {
cout << arg1 << "\n";
np(args...);
}
template <typename temp>
void spv(temp T, long long int n) {
for (int i = 0; i < n; i++) sp(T[i]);
}
template <typename temp>
void npv(temp T, long long int n) {
for (int i = 0; i < n; i++) np(T[i]);
}
template <typename T>
T modpow(T base, T exp, T modulus) {
base %= modulus;
T result = 1;
while (exp > 0) {
if (exp & 1) result = (result * base) % modulus;
base = (base * base) % modulus;
exp >>= 1;
}
return result;
}
template <typename T>
T modinv(T base, T exp) {
return modpow(base, exp - 2, exp);
}
long long int spf[MAXN];
void sieve() {
spf[1] = 1;
for (auto i = 2; i < MAXN; i++) spf[i] = i;
for (auto i = 4; i < MAXN; i += 2) spf[i] = 2;
for (auto i = 3; i * i < MAXN; i++) {
if (spf[i] == i) {
for (auto j = i * i; j < MAXN; j += i)
if (spf[j] == j) spf[j] = i;
}
}
return;
}
vector<long long int> getFactorization(long long int x) {
vector<long long int> res;
while (x != 1) {
res.push_back(spf[x]);
x = x / spf[x];
}
return res;
}
void check() {
string s;
long long int k;
cin >> s >> k;
long long int n = s.size();
string w(n, '1');
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i + k < n) w[i + k] = '0';
if (i - k >= 0) w[i - k] = '0';
}
}
for (int i = 0; i < n; i++) {
bool ok = false;
ok = ok || (i + k < n && w[i + k] == '1');
ok = ok || (i - k >= 0 && w[i - k] == '1');
if (s[i] != ok + '0') {
np(-1);
return;
}
}
np(w);
return;
}
int32_t main() {
ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
;
long long int t = 1;
cin >> t;
while (t--) check();
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int t;
cin >> t;
for (int q = 0; q < t; q++) {
string s;
cin >> s;
int x;
cin >> x;
string ans(s.length(), '*');
for (int i = 0; i < s.length(); i++) {
if (s[i] == '0') {
if (i - x >= 0) ans[i - x] = '0';
if (i + x < s.length()) ans[i + x] = '0';
}
if (i - x < 0 && i + x >= s.length()) {
ans[i] = '0';
}
}
bool f = true;
for (int i = 0; i < s.length(); i++) {
if (s[i] == '1') {
if (i - x >= 0 && i + x < s.length()) {
if (ans[i - x] == '0' && ans[i + x] == '0') {
f = false;
break;
} else if (ans[i - x] == '0') {
ans[i + x] = '1';
} else if (ans[i + x] == '0') {
ans[i - x] = '1';
} else {
ans[i - x] = '1';
}
} else if (i - x >= 0) {
if (ans[i - x] == '0') {
f = false;
break;
} else {
ans[i - x] = '1';
}
} else if (i + x < s.length()) {
if (ans[i + x] == '0') {
f = false;
break;
} else {
ans[i + x] = '1';
}
} else {
f = false;
break;
}
}
}
if (f) {
for (int i = 0; i < s.length(); i++) {
if (ans[i] == '*') {
ans[i] = '1';
}
}
cout << ans;
} else {
cout << -1;
}
cout << "\n";
}
return 0;
}
| ### Prompt
Your challenge is to write a cpp solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int t;
cin >> t;
for (int q = 0; q < t; q++) {
string s;
cin >> s;
int x;
cin >> x;
string ans(s.length(), '*');
for (int i = 0; i < s.length(); i++) {
if (s[i] == '0') {
if (i - x >= 0) ans[i - x] = '0';
if (i + x < s.length()) ans[i + x] = '0';
}
if (i - x < 0 && i + x >= s.length()) {
ans[i] = '0';
}
}
bool f = true;
for (int i = 0; i < s.length(); i++) {
if (s[i] == '1') {
if (i - x >= 0 && i + x < s.length()) {
if (ans[i - x] == '0' && ans[i + x] == '0') {
f = false;
break;
} else if (ans[i - x] == '0') {
ans[i + x] = '1';
} else if (ans[i + x] == '0') {
ans[i - x] = '1';
} else {
ans[i - x] = '1';
}
} else if (i - x >= 0) {
if (ans[i - x] == '0') {
f = false;
break;
} else {
ans[i - x] = '1';
}
} else if (i + x < s.length()) {
if (ans[i + x] == '0') {
f = false;
break;
} else {
ans[i + x] = '1';
}
} else {
f = false;
break;
}
}
}
if (f) {
for (int i = 0; i < s.length(); i++) {
if (ans[i] == '*') {
ans[i] = '1';
}
}
cout << ans;
} else {
cout << -1;
}
cout << "\n";
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int32_t main() {
ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
long long t;
cin >> t;
while (t--) {
string s;
long long x;
cin >> s;
cin >> x;
long long n = s.length();
long long arr[n];
for (long long i = 0; i < n; i++) {
arr[i] = 1;
}
for (long long i = 0; i < n; i++) {
if (s[i] == '0') {
if (i >= x) {
arr[i - x] = 0;
}
if ((i + x) < n) {
arr[i + x] = 0;
}
}
}
long long ar[n];
for (long long i = 0; i < n; i++) {
if ((i - x) >= 0 && arr[i - x] == 1) {
ar[i] = 1;
} else if ((i + x) < n && arr[i + x] == 1) {
ar[i] = 1;
} else {
ar[i] = 0;
}
}
long long k = 0;
for (long long i = 0; i < n; i++) {
if (s[i] != '0' && ar[i] == 0) {
cout << "-1" << ("\n");
k = 1;
break;
} else if (s[i] == '0' && ar[i] != 0) {
cout << "-1" << ("\n");
k = 1;
break;
}
}
if (k != 1) {
for (long long i = 0; i < n; i++) {
cout << arr[i];
}
cout << ("\n");
}
}
return 0;
}
| ### Prompt
Your task is to create a cpp solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int32_t main() {
ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
long long t;
cin >> t;
while (t--) {
string s;
long long x;
cin >> s;
cin >> x;
long long n = s.length();
long long arr[n];
for (long long i = 0; i < n; i++) {
arr[i] = 1;
}
for (long long i = 0; i < n; i++) {
if (s[i] == '0') {
if (i >= x) {
arr[i - x] = 0;
}
if ((i + x) < n) {
arr[i + x] = 0;
}
}
}
long long ar[n];
for (long long i = 0; i < n; i++) {
if ((i - x) >= 0 && arr[i - x] == 1) {
ar[i] = 1;
} else if ((i + x) < n && arr[i + x] == 1) {
ar[i] = 1;
} else {
ar[i] = 0;
}
}
long long k = 0;
for (long long i = 0; i < n; i++) {
if (s[i] != '0' && ar[i] == 0) {
cout << "-1" << ("\n");
k = 1;
break;
} else if (s[i] == '0' && ar[i] != 0) {
cout << "-1" << ("\n");
k = 1;
break;
}
}
if (k != 1) {
for (long long i = 0; i < n; i++) {
cout << arr[i];
}
cout << ("\n");
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int flag;
int main() {
int T;
cin >> T;
for (; T--;) {
int n, x;
string s;
cin >> s >> x;
n = s.size();
vector<int> pd(n, -1);
flag = 1;
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) pd[i - x] = 0;
if (i + x < n) pd[i + x] = 0;
}
}
for (int i = 0; i < n; i++)
if (pd[i] < 0) pd[i] = 1;
string ans;
for (int i = 0; i < n; i++) {
int w = 0;
if (i - x >= 0 && pd[i - x]) w = 1;
if (i + x < n && pd[i + x]) w = 1;
if (w != s[i] - '0') flag = 0;
ans += (pd[i] + '0');
}
if (flag)
cout << ans << endl;
else
cout << -1 << endl;
}
return 0;
}
| ### Prompt
Generate a cpp solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int flag;
int main() {
int T;
cin >> T;
for (; T--;) {
int n, x;
string s;
cin >> s >> x;
n = s.size();
vector<int> pd(n, -1);
flag = 1;
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) pd[i - x] = 0;
if (i + x < n) pd[i + x] = 0;
}
}
for (int i = 0; i < n; i++)
if (pd[i] < 0) pd[i] = 1;
string ans;
for (int i = 0; i < n; i++) {
int w = 0;
if (i - x >= 0 && pd[i - x]) w = 1;
if (i + x < n && pd[i + x]) w = 1;
if (w != s[i] - '0') flag = 0;
ans += (pd[i] + '0');
}
if (flag)
cout << ans << endl;
else
cout << -1 << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int t;
cin >> t;
while (t--) {
int x;
string s;
cin >> s >> x;
int n = s.size();
string ans = string(n, '1');
for (int i = 0; i < n; ++i) {
if (s[i] == '0') {
if (i - x >= 0) ans[i - x] = '0';
if (i + x < n) ans[i + x] = '0';
}
}
for (int i = 0; i < n; ++i) {
if (s[i] == '1') {
if ((i - x >= 0 && ans[i - x] == '1') ||
(i + x < n && ans[i + x] == '1')) {
} else {
ans = "-1";
break;
}
}
if (i - x >= 0 && ans[i - x] == '1' && s[i] == '0') {
ans = "-1";
break;
}
if (i + x < n && ans[i + x] == '1' && s[i] == '0') {
ans = "-1";
break;
}
}
cout << ans << '\n';
}
return 0;
}
| ### Prompt
In Cpp, your task is to solve the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int t;
cin >> t;
while (t--) {
int x;
string s;
cin >> s >> x;
int n = s.size();
string ans = string(n, '1');
for (int i = 0; i < n; ++i) {
if (s[i] == '0') {
if (i - x >= 0) ans[i - x] = '0';
if (i + x < n) ans[i + x] = '0';
}
}
for (int i = 0; i < n; ++i) {
if (s[i] == '1') {
if ((i - x >= 0 && ans[i - x] == '1') ||
(i + x < n && ans[i + x] == '1')) {
} else {
ans = "-1";
break;
}
}
if (i - x >= 0 && ans[i - x] == '1' && s[i] == '0') {
ans = "-1";
break;
}
if (i + x < n && ans[i + x] == '1' && s[i] == '0') {
ans = "-1";
break;
}
}
cout << ans << '\n';
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
long long t;
cin >> t;
while (t--) {
long long i, j, x;
string s;
cin >> s >> x;
long long n = s.length();
string ans = "";
for (i = 0; i < n; i++) {
ans += '1';
}
for (i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) ans[i - x] = '0';
if (i + x < n) ans[i + x] = '0';
}
}
string t = "";
for (i = 0; i < n; i++) {
t += '1';
}
for (i = 0; i < n; i++) {
long long f = 0;
if (i - x >= 0) {
if (ans[i - x] == '0') {
f++;
}
} else {
f++;
}
if (i + x < n) {
if (ans[i + x] == '0') {
f++;
}
} else {
f++;
}
if (f == 2) {
t[i] = '0';
}
}
if (t == s) {
cout << ans << endl;
} else {
cout << -1 << endl;
}
}
}
| ### Prompt
Please create a solution in Cpp to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
long long t;
cin >> t;
while (t--) {
long long i, j, x;
string s;
cin >> s >> x;
long long n = s.length();
string ans = "";
for (i = 0; i < n; i++) {
ans += '1';
}
for (i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) ans[i - x] = '0';
if (i + x < n) ans[i + x] = '0';
}
}
string t = "";
for (i = 0; i < n; i++) {
t += '1';
}
for (i = 0; i < n; i++) {
long long f = 0;
if (i - x >= 0) {
if (ans[i - x] == '0') {
f++;
}
} else {
f++;
}
if (i + x < n) {
if (ans[i + x] == '0') {
f++;
}
} else {
f++;
}
if (f == 2) {
t[i] = '0';
}
}
if (t == s) {
cout << ans << endl;
} else {
cout << -1 << endl;
}
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
int a[maxn];
int main() {
int T;
cin >> T;
while (T--) {
int f = 0;
int x;
string s;
cin >> s >> x;
for (int i = 0; i < s.size(); i++) {
a[i] = 1;
}
for (int i = 0; i < s.size(); i++) {
if (s[i] == '0') {
if (i + x < s.size()) {
a[i + x] = 0;
}
if (i - x >= 0) {
a[i - x] = 0;
}
}
}
for (int i = 0; i < s.size(); i++) {
int flag = 0;
if (s[i] == '1') {
if (i + x < s.size() && a[i + x]) {
flag = 1;
}
if (i - x >= 0 && a[i - x]) {
flag = 1;
}
if (!flag) {
f = 1;
break;
}
}
}
if (f) {
cout << "-1"
<< "\n";
continue;
}
for (int i = 0; i < s.size(); i++) {
cout << a[i];
}
cout << "\n";
}
return 0;
}
| ### Prompt
Generate a Cpp solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
int a[maxn];
int main() {
int T;
cin >> T;
while (T--) {
int f = 0;
int x;
string s;
cin >> s >> x;
for (int i = 0; i < s.size(); i++) {
a[i] = 1;
}
for (int i = 0; i < s.size(); i++) {
if (s[i] == '0') {
if (i + x < s.size()) {
a[i + x] = 0;
}
if (i - x >= 0) {
a[i - x] = 0;
}
}
}
for (int i = 0; i < s.size(); i++) {
int flag = 0;
if (s[i] == '1') {
if (i + x < s.size() && a[i + x]) {
flag = 1;
}
if (i - x >= 0 && a[i - x]) {
flag = 1;
}
if (!flag) {
f = 1;
break;
}
}
}
if (f) {
cout << "-1"
<< "\n";
continue;
}
for (int i = 0; i < s.size(); i++) {
cout << a[i];
}
cout << "\n";
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
void solve() {
string s;
cin >> s;
long long int n = s.size();
long long int x;
cin >> x;
long long int a[n + 1];
for (long long int i = 0; i < n; i++) {
a[i] = 1;
}
for (long long int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) a[i - x] = 0;
if (i + x < n) a[i + x] = 0;
}
}
for (long long int i = 0; i < n; i++) {
if (s[i] == '1') {
bool ok = false;
if (i + x < n) {
if (a[i + x] == 1) ok = true;
}
if (i - x >= 0) {
if (a[i - x] == 1) ok = true;
}
if (ok == false) {
cout << "-1\n";
return;
}
}
}
for (long long int i = 0; i < n; i++) {
cout << a[i];
}
cout << endl;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(NULL);
long long int t = 1;
cin >> t;
long long int ctr = 0;
while (t--) {
ctr++;
solve();
}
}
| ### Prompt
Generate a CPP solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void solve() {
string s;
cin >> s;
long long int n = s.size();
long long int x;
cin >> x;
long long int a[n + 1];
for (long long int i = 0; i < n; i++) {
a[i] = 1;
}
for (long long int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) a[i - x] = 0;
if (i + x < n) a[i + x] = 0;
}
}
for (long long int i = 0; i < n; i++) {
if (s[i] == '1') {
bool ok = false;
if (i + x < n) {
if (a[i + x] == 1) ok = true;
}
if (i - x >= 0) {
if (a[i - x] == 1) ok = true;
}
if (ok == false) {
cout << "-1\n";
return;
}
}
}
for (long long int i = 0; i < n; i++) {
cout << a[i];
}
cout << endl;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(NULL);
long long int t = 1;
cin >> t;
long long int ctr = 0;
while (t--) {
ctr++;
solve();
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
void solve() {
string s;
cin >> s;
long long x;
cin >> x;
long long n = s.size();
string c = s;
for (long long i = 0; i < n; i++) {
c[i] = '1';
}
for (long long i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) {
c[i - x] = '0';
}
if (i + x < n) {
c[i + x] = '0';
}
}
}
string total = s;
for (long long i = 0; i < n; i++) {
long long tot = 0;
if (i - x >= 0) {
tot += 1 * (c[i - x] == '1');
}
if (i + x < n) {
tot += 1 * (c[i + x] == '1');
}
if (tot > 0) {
total[i] = '1';
} else {
total[i] = '0';
}
}
bool can = false;
if (total == s) {
can = true;
}
if (can) {
cout << c << endl;
} else {
cout << -1 << endl;
}
}
int main() {
ifstream fin("text.in");
ofstream fout("text.out");
long long t = 1;
cin >> t;
while (t--) solve();
return 0;
}
| ### Prompt
Create a solution in CPP for the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void solve() {
string s;
cin >> s;
long long x;
cin >> x;
long long n = s.size();
string c = s;
for (long long i = 0; i < n; i++) {
c[i] = '1';
}
for (long long i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) {
c[i - x] = '0';
}
if (i + x < n) {
c[i + x] = '0';
}
}
}
string total = s;
for (long long i = 0; i < n; i++) {
long long tot = 0;
if (i - x >= 0) {
tot += 1 * (c[i - x] == '1');
}
if (i + x < n) {
tot += 1 * (c[i + x] == '1');
}
if (tot > 0) {
total[i] = '1';
} else {
total[i] = '0';
}
}
bool can = false;
if (total == s) {
can = true;
}
if (can) {
cout << c << endl;
} else {
cout << -1 << endl;
}
}
int main() {
ifstream fin("text.in");
ofstream fout("text.out");
long long t = 1;
cin >> t;
while (t--) solve();
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
string s;
cin >> s;
int x;
cin >> x;
int n = s.length();
char a[n + 5], w[n + 5];
for (int i = 0; i < n; i++) a[i + 1] = s[i];
for (int i = 1; i <= n; i++) w[i] = '1';
for (int i = 1; i <= n; i++) {
bool flag = false;
if (i > x) {
if (a[i] == '1') {
flag = true;
}
}
if (i + x <= n) {
if (a[i] == '1') {
flag = true;
}
}
if (flag == false) {
if (i > x) w[i - x] = '0';
if (i + x <= n) w[i + x] = '0';
}
}
for (int i = 1; i <= n; i++) {
bool flag = false;
if (i > x) {
if (w[i - x] == '1') {
a[i] = '1';
flag = true;
}
}
if (i + x <= n) {
if (w[i + x] == '1') {
a[i] = '1';
flag = true;
}
}
if (flag == false) a[i] = '0';
}
int c = 0;
for (int i = 1; i <= n; i++) {
if (s[i - 1] == a[i]) c++;
}
if (c == n) {
for (int i = 1; i <= n; i++) cout << w[i];
} else
cout << -1;
cout << endl;
}
}
| ### Prompt
Your challenge is to write a cpp solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
string s;
cin >> s;
int x;
cin >> x;
int n = s.length();
char a[n + 5], w[n + 5];
for (int i = 0; i < n; i++) a[i + 1] = s[i];
for (int i = 1; i <= n; i++) w[i] = '1';
for (int i = 1; i <= n; i++) {
bool flag = false;
if (i > x) {
if (a[i] == '1') {
flag = true;
}
}
if (i + x <= n) {
if (a[i] == '1') {
flag = true;
}
}
if (flag == false) {
if (i > x) w[i - x] = '0';
if (i + x <= n) w[i + x] = '0';
}
}
for (int i = 1; i <= n; i++) {
bool flag = false;
if (i > x) {
if (w[i - x] == '1') {
a[i] = '1';
flag = true;
}
}
if (i + x <= n) {
if (w[i + x] == '1') {
a[i] = '1';
flag = true;
}
}
if (flag == false) a[i] = '0';
}
int c = 0;
for (int i = 1; i <= n; i++) {
if (s[i - 1] == a[i]) c++;
}
if (c == n) {
for (int i = 1; i <= n; i++) cout << w[i];
} else
cout << -1;
cout << endl;
}
}
``` |
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long q;
cin >> q;
for (long long i = 0; i < q; i++) {
string s, w;
w = "";
long long n, x;
cin >> s >> x;
bool ans = 1;
n = s.length();
w.append(s.length(), '1');
for (long long j = 0; j < n; j++)
if (s[j] == '1') {
bool t1 = 0;
bool t2 = 0;
if ((j - x) >= 0) {
if (w[j - x] == '0') t1 = 1;
}
if ((j + x) < n) {
if (w[j + x] == '0') t2 = 1;
}
if (t1 && t2) {
ans = 0;
break;
}
} else {
if ((j - x) >= 0) w[j - x] = '0';
if ((j + x) < n) w[j + x] = '0';
}
bool t = 1;
for (long long j = 0; j < n; j++) {
if (w[j] == '1') {
if (((j - x) < 0 || s[j - x] == '1') &&
((j + x) >= n || s[j + x] == '1')) {
} else {
t = 0;
break;
}
}
if (s[j] == '0') {
if (((j - x) < 0 || w[j - x] == '0') &&
((j + x) >= n || w[j + x] == '0')) {
} else {
t = 0;
break;
}
} else {
if (((j - x) >= 0 && w[j - x] == '1') ||
((j + x) < n && w[j + x] == '1')) {
} else {
t = 0;
break;
}
}
}
if (t && ans)
cout << w << '\n';
else
cout << -1 << '\n';
}
}
| ### Prompt
Construct a Cpp code solution to the problem outlined:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long q;
cin >> q;
for (long long i = 0; i < q; i++) {
string s, w;
w = "";
long long n, x;
cin >> s >> x;
bool ans = 1;
n = s.length();
w.append(s.length(), '1');
for (long long j = 0; j < n; j++)
if (s[j] == '1') {
bool t1 = 0;
bool t2 = 0;
if ((j - x) >= 0) {
if (w[j - x] == '0') t1 = 1;
}
if ((j + x) < n) {
if (w[j + x] == '0') t2 = 1;
}
if (t1 && t2) {
ans = 0;
break;
}
} else {
if ((j - x) >= 0) w[j - x] = '0';
if ((j + x) < n) w[j + x] = '0';
}
bool t = 1;
for (long long j = 0; j < n; j++) {
if (w[j] == '1') {
if (((j - x) < 0 || s[j - x] == '1') &&
((j + x) >= n || s[j + x] == '1')) {
} else {
t = 0;
break;
}
}
if (s[j] == '0') {
if (((j - x) < 0 || w[j - x] == '0') &&
((j + x) >= n || w[j + x] == '0')) {
} else {
t = 0;
break;
}
} else {
if (((j - x) >= 0 && w[j - x] == '1') ||
((j + x) < n && w[j + x] == '1')) {
} else {
t = 0;
break;
}
}
}
if (t && ans)
cout << w << '\n';
else
cout << -1 << '\n';
}
}
``` |
#include <bits/stdc++.h>
static const auto fast = []() {
std::ios_base::sync_with_stdio(false);
std::cin.tie(0);
std::cout.tie(0);
return 0;
}();
using namespace std;
int32_t main() {
long long T;
cin >> T;
while (T--) {
string s;
cin >> s;
long long k;
cin >> k;
long long n = s.size();
string ans(s.size(), '1');
for (long long i = 0; i < s.size(); i++) {
if (s[i] == '0') {
if (i + k < n) ans[i + k] = '0';
if (i - k >= 0) ans[i - k] = '0';
}
}
bool flag = 0;
for (long long i = 0; i < s.size(); i++) {
if (s[i] == '1') {
long long z = 0;
if (i + k >= n || ans[i + k] == '0') {
z++;
}
if (i - k < 0 || ans[i - k] == '0') {
z++;
}
if (z == 2) {
flag = 1;
break;
}
}
}
if (flag) {
cout << "-1" << endl;
} else {
cout << ans << endl;
}
}
}
| ### Prompt
Your task is to create a Cpp solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
static const auto fast = []() {
std::ios_base::sync_with_stdio(false);
std::cin.tie(0);
std::cout.tie(0);
return 0;
}();
using namespace std;
int32_t main() {
long long T;
cin >> T;
while (T--) {
string s;
cin >> s;
long long k;
cin >> k;
long long n = s.size();
string ans(s.size(), '1');
for (long long i = 0; i < s.size(); i++) {
if (s[i] == '0') {
if (i + k < n) ans[i + k] = '0';
if (i - k >= 0) ans[i - k] = '0';
}
}
bool flag = 0;
for (long long i = 0; i < s.size(); i++) {
if (s[i] == '1') {
long long z = 0;
if (i + k >= n || ans[i + k] == '0') {
z++;
}
if (i - k < 0 || ans[i - k] == '0') {
z++;
}
if (z == 2) {
flag = 1;
break;
}
}
}
if (flag) {
cout << "-1" << endl;
} else {
cout << ans << endl;
}
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long f, a, b, x, c, q, d, w, a1, b1, c1, a2, g, b2, c2, n, m, i, j, k, l,
o, p, t, y, r, q1, q2, mx, mx2, mn = 1000000000000, mn2 = 1000000000000;
string s, s1, s2, s3;
bool used[100005];
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> t;
for (y = 0; y < t; y++) {
cin >> s >> x;
q = 0;
s2 = "";
for (i = 0; i < s.size(); i++) {
used[i] = 0;
s2 = s2 + '0';
}
for (i = 0; i < s.size(); i++) {
if (s[i] == '0') {
if (i - x >= 0) {
if (s2[i - x] == '1') q = 1;
s2[i - x] = '0';
}
if (i + x < s.size()) {
s2[i + x] = '0';
used[i + x] = 1;
}
}
if (s[i] == '1') {
if (i - x >= 0) {
if (used[i - x] == 1)
;
else {
s2[i - x] = '1';
continue;
}
}
if (i + x < s.size()) {
s2[i + x] = '1';
continue;
}
q = 1;
}
}
if (q == 1)
cout << -1;
else
cout << s2;
cout << "\n";
}
}
| ### Prompt
Please create a solution in cpp to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long f, a, b, x, c, q, d, w, a1, b1, c1, a2, g, b2, c2, n, m, i, j, k, l,
o, p, t, y, r, q1, q2, mx, mx2, mn = 1000000000000, mn2 = 1000000000000;
string s, s1, s2, s3;
bool used[100005];
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> t;
for (y = 0; y < t; y++) {
cin >> s >> x;
q = 0;
s2 = "";
for (i = 0; i < s.size(); i++) {
used[i] = 0;
s2 = s2 + '0';
}
for (i = 0; i < s.size(); i++) {
if (s[i] == '0') {
if (i - x >= 0) {
if (s2[i - x] == '1') q = 1;
s2[i - x] = '0';
}
if (i + x < s.size()) {
s2[i + x] = '0';
used[i + x] = 1;
}
}
if (s[i] == '1') {
if (i - x >= 0) {
if (used[i - x] == 1)
;
else {
s2[i - x] = '1';
continue;
}
}
if (i + x < s.size()) {
s2[i + x] = '1';
continue;
}
q = 1;
}
}
if (q == 1)
cout << -1;
else
cout << s2;
cout << "\n";
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
void solve() {
string s;
int x;
cin >> s >> x;
int len = s.length();
int arr[len];
for (int i = 0; i < len; i++) {
arr[i] = -1;
}
for (int i = 0; i < len; i++) {
if (s[i] == '0') {
int bef = i - x;
int aft = i + x;
if ((bef >= 0)) {
arr[bef] = 0;
}
if (aft < len) {
arr[aft] = 0;
}
}
}
for (int i = 0; i < len; i++) {
if (s[i] == '1') {
bool first = false;
bool second = false;
int bef = i - x;
int aft = i + x;
if ((bef >= 0) && ((arr[bef] == -1) || (arr[bef] == 1))) {
arr[bef] = 1;
first = true;
}
if ((aft < len) && ((arr[aft] == -1) || (arr[aft] == 1))) {
arr[aft] = 1;
second = true;
}
if ((first == false) && (second == false)) {
cout << -1 << endl;
return;
}
}
}
for (int i = 0; i < len; i++) {
if (arr[i] == -1) {
arr[i] = 1;
}
cout << arr[i];
}
cout << endl;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int t;
cin >> t;
while (t--) {
solve();
}
return 0;
}
| ### Prompt
Your challenge is to write a CPP solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void solve() {
string s;
int x;
cin >> s >> x;
int len = s.length();
int arr[len];
for (int i = 0; i < len; i++) {
arr[i] = -1;
}
for (int i = 0; i < len; i++) {
if (s[i] == '0') {
int bef = i - x;
int aft = i + x;
if ((bef >= 0)) {
arr[bef] = 0;
}
if (aft < len) {
arr[aft] = 0;
}
}
}
for (int i = 0; i < len; i++) {
if (s[i] == '1') {
bool first = false;
bool second = false;
int bef = i - x;
int aft = i + x;
if ((bef >= 0) && ((arr[bef] == -1) || (arr[bef] == 1))) {
arr[bef] = 1;
first = true;
}
if ((aft < len) && ((arr[aft] == -1) || (arr[aft] == 1))) {
arr[aft] = 1;
second = true;
}
if ((first == false) && (second == false)) {
cout << -1 << endl;
return;
}
}
}
for (int i = 0; i < len; i++) {
if (arr[i] == -1) {
arr[i] = 1;
}
cout << arr[i];
}
cout << endl;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int t;
cin >> t;
while (t--) {
solve();
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
char ch[N];
int s[N], w[N];
int T, n, x;
int main() {
scanf("%d", &T);
for (; T; T--) {
scanf("%s", ch + 1);
n = strlen(ch + 1);
scanf("%d", &x);
for (int i = 1; i <= n; i++) {
w[i] = 1;
s[i] = ch[i] - '0';
}
for (int i = 1; i <= n; i++)
if (!s[i]) {
if (i + x <= n) w[i + x] = 0;
if (i - x >= 1) w[i - x] = 0;
}
bool bz = 1;
for (int i = 1; i <= n; i++)
if (s[i])
if (((i + x <= n && !w[i + x]) || i + x > n) &&
((i - x >= 1 && !w[i - x]) || i - x < 1)) {
bz = 0;
break;
}
if (!bz)
puts("-1");
else {
for (int i = 1; i <= n; i++) printf("%d", w[i]);
puts("");
}
}
}
| ### Prompt
Your task is to create a Cpp solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
char ch[N];
int s[N], w[N];
int T, n, x;
int main() {
scanf("%d", &T);
for (; T; T--) {
scanf("%s", ch + 1);
n = strlen(ch + 1);
scanf("%d", &x);
for (int i = 1; i <= n; i++) {
w[i] = 1;
s[i] = ch[i] - '0';
}
for (int i = 1; i <= n; i++)
if (!s[i]) {
if (i + x <= n) w[i + x] = 0;
if (i - x >= 1) w[i - x] = 0;
}
bool bz = 1;
for (int i = 1; i <= n; i++)
if (s[i])
if (((i + x <= n && !w[i + x]) || i + x > n) &&
((i - x >= 1 && !w[i - x]) || i - x < 1)) {
bz = 0;
break;
}
if (!bz)
puts("-1");
else {
for (int i = 1; i <= n; i++) printf("%d", w[i]);
puts("");
}
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
char s[100005], w[100005];
bool vis[100005];
int main() {
int ts;
scanf("%d", &ts);
while (ts--) {
int n, x;
scanf("%s %d", s, &x);
n = strlen(s);
for (int i = 0; i < n; i++) vis[i] = 0;
bool f = 1;
for (int i = 0; i < n && f; i++) {
if (s[i] == '1') {
if ((i - x) >= 0 || (i + x) < n) {
if ((i - x) >= 0 && (!vis[i - x] || w[i - x] == '1'))
vis[i - x] = 1, w[i - x] = '1';
else if ((i + x) < n && (!vis[i + x] || w[i + x] == '1'))
vis[i + x] = 1, w[i + x] = '1';
else
f = 0;
} else
f = 0;
} else {
if ((i - x) >= 0 && vis[i - x] && w[i - x] == '1')
f = 0;
else if ((i + x) < n && vis[i + x] && w[i + x] == '1')
f = 0;
else {
if ((i - x) >= 0) vis[i - x] = 1, w[i - x] = '0';
if ((i + x) < n) vis[i + x] = 1, w[i + x] = '0';
}
}
}
for (int i = 0; i < n; i++)
if (!vis[i]) w[i] = '0';
if (f) {
for (int i = 0; i < n; i++) printf("%c", w[i]);
puts("");
} else
puts("-1");
}
return 0;
}
| ### Prompt
Develop a solution in Cpp to the problem described below:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
char s[100005], w[100005];
bool vis[100005];
int main() {
int ts;
scanf("%d", &ts);
while (ts--) {
int n, x;
scanf("%s %d", s, &x);
n = strlen(s);
for (int i = 0; i < n; i++) vis[i] = 0;
bool f = 1;
for (int i = 0; i < n && f; i++) {
if (s[i] == '1') {
if ((i - x) >= 0 || (i + x) < n) {
if ((i - x) >= 0 && (!vis[i - x] || w[i - x] == '1'))
vis[i - x] = 1, w[i - x] = '1';
else if ((i + x) < n && (!vis[i + x] || w[i + x] == '1'))
vis[i + x] = 1, w[i + x] = '1';
else
f = 0;
} else
f = 0;
} else {
if ((i - x) >= 0 && vis[i - x] && w[i - x] == '1')
f = 0;
else if ((i + x) < n && vis[i + x] && w[i + x] == '1')
f = 0;
else {
if ((i - x) >= 0) vis[i - x] = 1, w[i - x] = '0';
if ((i + x) < n) vis[i + x] = 1, w[i + x] = '0';
}
}
}
for (int i = 0; i < n; i++)
if (!vis[i]) w[i] = '0';
if (f) {
for (int i = 0; i < n; i++) printf("%c", w[i]);
puts("");
} else
puts("-1");
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
;
signed long long t;
cin >> t;
while (t--) {
string s;
cin >> s;
signed long long x;
cin >> x;
int n = s.length();
vector<bool> vis(n, false);
string w(n, '1');
bool ans = true;
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) {
if (vis[i - x] == true && w[i - x] == '1') {
ans = false;
break;
} else {
vis[i - x] = true;
w[i - x] = '0';
}
}
if (i + x < n) {
if (vis[i + x] == true && w[i + x] == '1') {
ans = false;
break;
} else {
vis[i + x] = true;
w[i + x] = '0';
}
}
} else {
if (i - x >= 0 && w[i - x] == '1')
continue;
else if (i + x < n && w[i + x] == '1') {
vis[i + x] = true;
continue;
} else {
ans = false;
break;
}
}
}
if (ans == false) {
cout << -1 << endl;
} else
cout << w << endl;
}
return 0;
}
| ### Prompt
In Cpp, your task is to solve the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
;
signed long long t;
cin >> t;
while (t--) {
string s;
cin >> s;
signed long long x;
cin >> x;
int n = s.length();
vector<bool> vis(n, false);
string w(n, '1');
bool ans = true;
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) {
if (vis[i - x] == true && w[i - x] == '1') {
ans = false;
break;
} else {
vis[i - x] = true;
w[i - x] = '0';
}
}
if (i + x < n) {
if (vis[i + x] == true && w[i + x] == '1') {
ans = false;
break;
} else {
vis[i + x] = true;
w[i + x] = '0';
}
}
} else {
if (i - x >= 0 && w[i - x] == '1')
continue;
else if (i + x < n && w[i + x] == '1') {
vis[i + x] = true;
continue;
} else {
ans = false;
break;
}
}
}
if (ans == false) {
cout << -1 << endl;
} else
cout << w << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
char s[N];
int x;
char ans[N];
int main() {
int t;
scanf("%d", &t);
while (t--) {
scanf("%s", (s + 1));
scanf("%d", &x);
int len = strlen((s + 1));
for (int i = 1; i <= len; i++) {
ans[i] = '1';
}
for (int i = 1; i <= len; i++) {
if (s[i] == '0') {
if (i - x >= 1) {
ans[i - x] = '0';
}
if (i + x <= len) {
ans[i + x] = '0';
}
}
}
int flag = 1;
for (int i = 1; i <= len; i++) {
if (s[i] == '1') {
int tmp = 0;
if (i - x >= 0 && ans[i - x] == '1') tmp = 1;
if (i + x <= len && ans[i + x] == '1') tmp = 1;
if (!tmp) {
flag = 0;
break;
}
}
}
if (flag) {
for (int i = 1; i <= len; i++) {
printf("%c", ans[i]);
}
} else {
printf("-1");
}
printf("\n");
}
return 0;
}
| ### Prompt
Construct a cpp code solution to the problem outlined:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
char s[N];
int x;
char ans[N];
int main() {
int t;
scanf("%d", &t);
while (t--) {
scanf("%s", (s + 1));
scanf("%d", &x);
int len = strlen((s + 1));
for (int i = 1; i <= len; i++) {
ans[i] = '1';
}
for (int i = 1; i <= len; i++) {
if (s[i] == '0') {
if (i - x >= 1) {
ans[i - x] = '0';
}
if (i + x <= len) {
ans[i + x] = '0';
}
}
}
int flag = 1;
for (int i = 1; i <= len; i++) {
if (s[i] == '1') {
int tmp = 0;
if (i - x >= 0 && ans[i - x] == '1') tmp = 1;
if (i + x <= len && ans[i + x] == '1') tmp = 1;
if (!tmp) {
flag = 0;
break;
}
}
}
if (flag) {
for (int i = 1; i <= len; i++) {
printf("%c", ans[i]);
}
} else {
printf("-1");
}
printf("\n");
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
void solve() {
string s;
cin >> s;
long long n;
cin >> n;
long long m = s.size();
char c[m];
for (long long i = 0; i < m; i++) {
c[i] = '1';
}
for (long long i = 0; i < m; i++)
if (s[i] == '0') {
if (i - n >= 0) c[i - n] = '0';
if (i + n < m) c[i + n] = '0';
}
long long flag = 0;
for (long long i = 0; i < m; i++) {
if (s[i] == '1') {
if ((i - n >= 0 && c[i - n] == '1') || (i + n < m && c[i + n] == '1')) {
} else {
flag = 1;
break;
}
}
}
if (flag == 1)
cout << "-1";
else
for (long long i = 0; i < m; i++) cout << c[i];
}
signed main() {
long long n;
cin >> n;
while (n--) {
solve();
cout << endl;
}
}
| ### Prompt
Develop a solution in cpp to the problem described below:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void solve() {
string s;
cin >> s;
long long n;
cin >> n;
long long m = s.size();
char c[m];
for (long long i = 0; i < m; i++) {
c[i] = '1';
}
for (long long i = 0; i < m; i++)
if (s[i] == '0') {
if (i - n >= 0) c[i - n] = '0';
if (i + n < m) c[i + n] = '0';
}
long long flag = 0;
for (long long i = 0; i < m; i++) {
if (s[i] == '1') {
if ((i - n >= 0 && c[i - n] == '1') || (i + n < m && c[i + n] == '1')) {
} else {
flag = 1;
break;
}
}
}
if (flag == 1)
cout << "-1";
else
for (long long i = 0; i < m; i++) cout << c[i];
}
signed main() {
long long n;
cin >> n;
while (n--) {
solve();
cout << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long MAXN = 2e5 + 69;
long long t;
string s;
long long n;
long long x;
long long ans[MAXN];
signed main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> t;
while (t--) {
cin >> s >> x;
n = s.size();
s = ' ' + s;
for (long long i = 1; i <= n; i++) ans[i] = 1;
for (long long i = 1; i <= n; i++) {
if (s[i] == '0') {
if (i > x) ans[i - x] = 0;
if (i + x <= n) ans[i + x] = 0;
}
}
for (long long i = 1; i <= n; i++) {
long long ok = 1;
if (s[i] == '1') {
ok = 0;
if (i > x && ans[i - x] == 1) ok = 1;
if (i + x <= n && ans[i + x] == 1) ok = 1;
}
if (ok == 0) {
cout << -1 << '\n';
goto s;
}
}
for (long long i = 1; i <= n; i++) cout << ans[i];
cout << '\n';
s:;
}
}
| ### Prompt
Create a solution in cpp for the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long MAXN = 2e5 + 69;
long long t;
string s;
long long n;
long long x;
long long ans[MAXN];
signed main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> t;
while (t--) {
cin >> s >> x;
n = s.size();
s = ' ' + s;
for (long long i = 1; i <= n; i++) ans[i] = 1;
for (long long i = 1; i <= n; i++) {
if (s[i] == '0') {
if (i > x) ans[i - x] = 0;
if (i + x <= n) ans[i + x] = 0;
}
}
for (long long i = 1; i <= n; i++) {
long long ok = 1;
if (s[i] == '1') {
ok = 0;
if (i > x && ans[i - x] == 1) ok = 1;
if (i + x <= n && ans[i + x] == 1) ok = 1;
}
if (ok == 0) {
cout << -1 << '\n';
goto s;
}
}
for (long long i = 1; i <= n; i++) cout << ans[i];
cout << '\n';
s:;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main(int argc, char **argv) {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
long long int t;
cin >> t;
while (t > 0) {
string s;
cin >> s;
long long int x;
cin >> x;
string w(s.size(), '1');
for (long long int i = 0; i < s.size(); i++) {
if (s[i] == '0') {
if (i - x >= 0) w[i - x] = '0';
if (i + x <= s.size() - 1) w[i + x] = '0';
}
}
long long int flag = 1;
for (long long int i = 0; i < s.size(); i++) {
if (s[i] == '1') {
if (!((i - x >= 0 && w[i - x] == '1') ||
(i + x <= s.size() - 1 && w[i + x] == '1'))) {
flag = 0;
break;
}
}
}
if (flag)
cout << w << "\n";
else
cout << -1 << "\n";
t--;
}
return 0;
}
| ### Prompt
Please create a solution in CPP to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main(int argc, char **argv) {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
long long int t;
cin >> t;
while (t > 0) {
string s;
cin >> s;
long long int x;
cin >> x;
string w(s.size(), '1');
for (long long int i = 0; i < s.size(); i++) {
if (s[i] == '0') {
if (i - x >= 0) w[i - x] = '0';
if (i + x <= s.size() - 1) w[i + x] = '0';
}
}
long long int flag = 1;
for (long long int i = 0; i < s.size(); i++) {
if (s[i] == '1') {
if (!((i - x >= 0 && w[i - x] == '1') ||
(i + x <= s.size() - 1 && w[i + x] == '1'))) {
flag = 0;
break;
}
}
}
if (flag)
cout << w << "\n";
else
cout << -1 << "\n";
t--;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t = 1;
cin >> t;
while (t--) {
string s;
cin >> s;
int x;
cin >> x;
string w;
int len = s.length();
for (int i = 0; i < len; i++) w += "1";
for (int i = 0; i < len; i++) {
if (s[i] == '0') {
if (i - x >= 0) w[i - x] = '0';
if (i + x < len) w[i + x] = '0';
}
}
for (int i = 0; i < len; i++) {
if (s[i] == '1') {
if (i - x >= 0 && w[i - x] == '1')
continue;
else if (i + x < len && w[i + x] == '1')
continue;
else {
w = "-1";
break;
}
}
}
cout << w << "\n";
}
cerr << "time taken : " << (float)clock() / CLOCKS_PER_SEC << " secs" << endl;
return 0;
}
| ### Prompt
In Cpp, your task is to solve the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t = 1;
cin >> t;
while (t--) {
string s;
cin >> s;
int x;
cin >> x;
string w;
int len = s.length();
for (int i = 0; i < len; i++) w += "1";
for (int i = 0; i < len; i++) {
if (s[i] == '0') {
if (i - x >= 0) w[i - x] = '0';
if (i + x < len) w[i + x] = '0';
}
}
for (int i = 0; i < len; i++) {
if (s[i] == '1') {
if (i - x >= 0 && w[i - x] == '1')
continue;
else if (i + x < len && w[i + x] == '1')
continue;
else {
w = "-1";
break;
}
}
}
cout << w << "\n";
}
cerr << "time taken : " << (float)clock() / CLOCKS_PER_SEC << " secs" << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int ans[100005];
void solve() {
string s;
cin >> s;
int n = s.length();
s = "#" + s;
int x;
cin >> x;
memset(ans, -1, sizeof ans);
for (int i = 1; i <= n; ++i) {
if (s[i] == '0') {
if (i - x > 0) ans[i - x] = 0;
if (i + x <= n) ans[i + x] = 0;
}
}
for (int i = 1; i <= n; ++i)
if (ans[i] == -1) ans[i] = 1;
for (int i = 1; i <= n; ++i) {
bool ok = 1;
if (s[i] == '0') {
if (i - x > 0 && ans[i - x] != 0) ok = 0;
if (i + x <= n && ans[i + x] != 0) ok = 0;
} else {
ok = 0;
if (i - x > 0 && ans[i - x] == 1) ok = 1;
if (i + x <= n && ans[i + x] == 1) ok = 1;
}
if (!ok) {
cout << -1 << endl;
return;
}
}
for (int i = 1; i <= n; ++i) cout << ans[i];
cout << endl;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int t = 1;
cin >> t;
while (t--) solve();
}
| ### Prompt
Generate a cpp solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int ans[100005];
void solve() {
string s;
cin >> s;
int n = s.length();
s = "#" + s;
int x;
cin >> x;
memset(ans, -1, sizeof ans);
for (int i = 1; i <= n; ++i) {
if (s[i] == '0') {
if (i - x > 0) ans[i - x] = 0;
if (i + x <= n) ans[i + x] = 0;
}
}
for (int i = 1; i <= n; ++i)
if (ans[i] == -1) ans[i] = 1;
for (int i = 1; i <= n; ++i) {
bool ok = 1;
if (s[i] == '0') {
if (i - x > 0 && ans[i - x] != 0) ok = 0;
if (i + x <= n && ans[i + x] != 0) ok = 0;
} else {
ok = 0;
if (i - x > 0 && ans[i - x] == 1) ok = 1;
if (i + x <= n && ans[i + x] == 1) ok = 1;
}
if (!ok) {
cout << -1 << endl;
return;
}
}
for (int i = 1; i <= n; ++i) cout << ans[i];
cout << endl;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int t = 1;
cin >> t;
while (t--) solve();
}
``` |
#include <bits/stdc++.h>
using namespace std;
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
const int INF = 1e9 + 23;
const long long INFL = 2e18 + 23;
inline void solve() {
int n, x;
string s;
cin >> s >> x;
n = s.size();
vector<int> ans(n, -1);
bool ok = true;
for (int i = 0; i < n; i++) {
if (s[i] == '1') continue;
int l = i - x;
int r = i + x;
if (l >= 0) ans[l] = 0;
if (r < n) ans[r] = 0;
}
for (int i = 0; i < n; i++) {
if (s[i] == '0') continue;
int l = i - x;
int r = i + x;
bool done = false;
if (l >= 0 and (ans[l] == -1 or ans[l] == 1)) ans[l] = 1, done = true;
if (r < n and (ans[r] == -1 or ans[r] == 1)) ans[r] = 1, done = true;
if (!done) {
cout << "-1\n";
return;
}
}
for (int i = 0; i < n; i++) {
if (ans[i] == -1) ans[i] = 1;
cout << ans[i];
}
cout << "\n";
}
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int tc;
cin >> tc;
for (int nc = 1; nc <= tc; nc++) {
solve();
}
return 0;
}
| ### Prompt
In cpp, your task is to solve the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
const int INF = 1e9 + 23;
const long long INFL = 2e18 + 23;
inline void solve() {
int n, x;
string s;
cin >> s >> x;
n = s.size();
vector<int> ans(n, -1);
bool ok = true;
for (int i = 0; i < n; i++) {
if (s[i] == '1') continue;
int l = i - x;
int r = i + x;
if (l >= 0) ans[l] = 0;
if (r < n) ans[r] = 0;
}
for (int i = 0; i < n; i++) {
if (s[i] == '0') continue;
int l = i - x;
int r = i + x;
bool done = false;
if (l >= 0 and (ans[l] == -1 or ans[l] == 1)) ans[l] = 1, done = true;
if (r < n and (ans[r] == -1 or ans[r] == 1)) ans[r] = 1, done = true;
if (!done) {
cout << "-1\n";
return;
}
}
for (int i = 0; i < n; i++) {
if (ans[i] == -1) ans[i] = 1;
cout << ans[i];
}
cout << "\n";
}
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int tc;
cin >> tc;
for (int nc = 1; nc <= tc; nc++) {
solve();
}
return 0;
}
``` |
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
string func(string s, long long x) {
string res;
for (long long i = 0; i < s.size(); i++) {
if (i - x >= 0 && s[i - x] == '1' || i + x < s.size() && s[i + x] == '1')
res += '1';
else
res += '0';
}
return res;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
long long t;
cin >> t;
while (t--) {
string s;
cin >> s;
long long x, n = s.size();
cin >> x;
string ans(n, '1');
for (long long i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) ans[i - x] = '0';
if (i + x < n) ans[i + x] = '0';
}
}
if (func(ans, x) == s)
cout << ans;
else
cout << -1;
cout << "\n";
}
return 0;
}
| ### Prompt
Your task is to create a Cpp solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
string func(string s, long long x) {
string res;
for (long long i = 0; i < s.size(); i++) {
if (i - x >= 0 && s[i - x] == '1' || i + x < s.size() && s[i + x] == '1')
res += '1';
else
res += '0';
}
return res;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
long long t;
cin >> t;
while (t--) {
string s;
cin >> s;
long long x, n = s.size();
cin >> x;
string ans(n, '1');
for (long long i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) ans[i - x] = '0';
if (i + x < n) ans[i + x] = '0';
}
}
if (func(ans, x) == s)
cout << ans;
else
cout << -1;
cout << "\n";
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long q, n, x, w[100005];
char s[100005];
signed main() {
cin >> q;
while (q--) {
scanf("%s%lld", s + 1, &x);
n = strlen(s + 1);
for (long long i = 1; i <= n; w[i] = -1, i++)
;
for (long long i = 1; i <= n; i++)
if (s[i] == '0') {
if (i - x >= 1) w[i - x] = 0;
if (i + x <= n) w[i + x] = 0;
}
bool flag = true;
for (long long i = 1; i <= n; i++)
if (s[i] == '1') {
bool f = false;
if (i - x >= 1 && (w[i - x] == 1 || w[i - x] == -1))
w[i - x] = 1, f = true;
else if (i + x <= n && (w[i + x] == 1 || w[i + x] == -1))
w[i + x] = 1, f = true;
if (!f) {
flag = false;
break;
}
}
if (!flag) {
puts("-1");
continue;
}
for (long long i = 1; i <= n; i++)
if (w[i] == -1)
cout << 1;
else
cout << w[i];
puts("");
}
return 0;
}
| ### Prompt
Construct a Cpp code solution to the problem outlined:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long q, n, x, w[100005];
char s[100005];
signed main() {
cin >> q;
while (q--) {
scanf("%s%lld", s + 1, &x);
n = strlen(s + 1);
for (long long i = 1; i <= n; w[i] = -1, i++)
;
for (long long i = 1; i <= n; i++)
if (s[i] == '0') {
if (i - x >= 1) w[i - x] = 0;
if (i + x <= n) w[i + x] = 0;
}
bool flag = true;
for (long long i = 1; i <= n; i++)
if (s[i] == '1') {
bool f = false;
if (i - x >= 1 && (w[i - x] == 1 || w[i - x] == -1))
w[i - x] = 1, f = true;
else if (i + x <= n && (w[i + x] == 1 || w[i + x] == -1))
w[i + x] = 1, f = true;
if (!f) {
flag = false;
break;
}
}
if (!flag) {
puts("-1");
continue;
}
for (long long i = 1; i <= n; i++)
if (w[i] == -1)
cout << 1;
else
cout << w[i];
puts("");
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
void solve() {
string s;
long long x;
cin >> s >> x;
long long n = s.size();
string w(s.size(), '1');
for (long long i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) w[i - x] = '0';
if (i + x < n) w[i + x] = '0';
}
}
string ns = w;
for (long long i = 0; i < n; i++) {
if ((i - x >= 0 && w[i - x] == '1') || (i + x < n && w[i + x] == '1'))
ns[i] = '1';
else
ns[i] = '0';
}
if (ns == s) {
cout << w << "\n";
} else {
cout << -1 << "\n";
}
}
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
long long t = 1;
cin >> t;
for (long long i = 1; i <= t; i++) {
solve();
}
return 0;
}
| ### Prompt
Construct a CPP code solution to the problem outlined:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void solve() {
string s;
long long x;
cin >> s >> x;
long long n = s.size();
string w(s.size(), '1');
for (long long i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) w[i - x] = '0';
if (i + x < n) w[i + x] = '0';
}
}
string ns = w;
for (long long i = 0; i < n; i++) {
if ((i - x >= 0 && w[i - x] == '1') || (i + x < n && w[i + x] == '1'))
ns[i] = '1';
else
ns[i] = '0';
}
if (ns == s) {
cout << w << "\n";
} else {
cout << -1 << "\n";
}
}
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
long long t = 1;
cin >> t;
for (long long i = 1; i <= t; i++) {
solve();
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int MOD = (int)1e9 + 7;
const int inf = (int)1e9;
const long long linf = (long long)1e18 + 1;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
;
int t;
cin >> t;
while (t--) {
string s;
cin >> s;
int x;
cin >> x;
string t(s.length(), '1');
int n = s.length();
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i + x < n) t[i + x] = '0';
if (i - x >= 0) t[i - x] = '0';
}
}
bool ok = true;
for (int i = 0; i < n; i++) {
if (s[i] == '1') {
bool f1 = 1, f2 = 1;
if (((i + x) >= n) || (t[i + x] == '0')) f1 = 0;
if (((i - x) < 0) || (t[i - x] == '0')) f2 = 0;
if (!f1 && !f2) ok = false;
}
}
if (!ok)
cout << -1 << endl;
else
cout << t << endl;
}
return 0;
}
| ### Prompt
Please create a solution in cpp to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int MOD = (int)1e9 + 7;
const int inf = (int)1e9;
const long long linf = (long long)1e18 + 1;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
;
int t;
cin >> t;
while (t--) {
string s;
cin >> s;
int x;
cin >> x;
string t(s.length(), '1');
int n = s.length();
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i + x < n) t[i + x] = '0';
if (i - x >= 0) t[i - x] = '0';
}
}
bool ok = true;
for (int i = 0; i < n; i++) {
if (s[i] == '1') {
bool f1 = 1, f2 = 1;
if (((i + x) >= n) || (t[i + x] == '0')) f1 = 0;
if (((i - x) < 0) || (t[i - x] == '0')) f2 = 0;
if (!f1 && !f2) ok = false;
}
}
if (!ok)
cout << -1 << endl;
else
cout << t << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int t, x;
string s, ans;
signed main() {
cin >> t;
while (t--) {
cin >> s >> x;
ans = s;
for (int i = 0; i < s.size(); i++) ans[i] = '1';
for (int i = 0; i < s.size(); i++) {
if (s[i] == '0') {
if (i - x >= 0) ans[i - x] = '0';
if (i + x < s.size()) ans[i + x] = '0';
}
}
bool tf = 1;
for (int i = 0; i < s.size(); i++) {
if (s[i] == '1') {
if ((i - x >= 0 && ans[i - x] == '1') ||
(i + x < s.size() && ans[i + x] == '1'))
continue;
else
tf = 0;
}
}
if (tf)
cout << ans << "\n";
else
cout << -1 << "\n";
}
}
| ### Prompt
Please provide a cpp coded solution to the problem described below:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int t, x;
string s, ans;
signed main() {
cin >> t;
while (t--) {
cin >> s >> x;
ans = s;
for (int i = 0; i < s.size(); i++) ans[i] = '1';
for (int i = 0; i < s.size(); i++) {
if (s[i] == '0') {
if (i - x >= 0) ans[i - x] = '0';
if (i + x < s.size()) ans[i + x] = '0';
}
}
bool tf = 1;
for (int i = 0; i < s.size(); i++) {
if (s[i] == '1') {
if ((i - x >= 0 && ans[i - x] == '1') ||
(i + x < s.size() && ans[i + x] == '1'))
continue;
else
tf = 0;
}
}
if (tf)
cout << ans << "\n";
else
cout << -1 << "\n";
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cout.precision(15);
long long int t;
cin >> t;
while (t--) {
string s;
cin >> s;
long long int x;
cin >> x;
long long int n = (long long int)s.length();
vector<int> v(n, 2);
for (long long int i = x; i < n - x; i++) {
if (s[i - x] == '1' && s[i + x] == '1')
v[i] = 1;
else
v[i] = 0;
}
for (long long int i = 0; i < x; i++) {
if ((i + x) < n && s[i + x] == '0')
v[i] = 0;
else
v[i] = 1;
}
for (long long int i = n - x; i < n; i++) {
if ((i - x) >= 0 && s[i - x] == '0')
v[i] = 0;
else
v[i] = 1;
}
string ans;
for (long long int i = 0; i < n; i++) {
if (i - x >= 0 && v[i - x] == 1)
ans += "1";
else if (i + x < n && v[i + x] == 1)
ans += "1";
else
ans += "0";
}
if (ans != s)
cout << "-1\n";
else {
string a;
for (long long int i = 0; i < n; i++) a += ('0' + v[i]);
cout << a << "\n";
}
}
}
| ### Prompt
Your challenge is to write a Cpp solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cout.precision(15);
long long int t;
cin >> t;
while (t--) {
string s;
cin >> s;
long long int x;
cin >> x;
long long int n = (long long int)s.length();
vector<int> v(n, 2);
for (long long int i = x; i < n - x; i++) {
if (s[i - x] == '1' && s[i + x] == '1')
v[i] = 1;
else
v[i] = 0;
}
for (long long int i = 0; i < x; i++) {
if ((i + x) < n && s[i + x] == '0')
v[i] = 0;
else
v[i] = 1;
}
for (long long int i = n - x; i < n; i++) {
if ((i - x) >= 0 && s[i - x] == '0')
v[i] = 0;
else
v[i] = 1;
}
string ans;
for (long long int i = 0; i < n; i++) {
if (i - x >= 0 && v[i - x] == 1)
ans += "1";
else if (i + x < n && v[i + x] == 1)
ans += "1";
else
ans += "0";
}
if (ans != s)
cout << "-1\n";
else {
string a;
for (long long int i = 0; i < n; i++) a += ('0' + v[i]);
cout << a << "\n";
}
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9 + 7;
const int MOD = 1e9 + 7;
double PI = 4 * atan(1);
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
;
int t;
cin >> t;
while (t--) {
string s;
cin >> s;
int l = s.size(), x, c = 0;
cin >> x;
string w = string(l, '1');
for (int i = 0; i < l; i++) {
if (s[i] == '1')
c++;
else {
c = 0;
if (i - x >= 0) w[i - x] = '0';
if (i + x < l) w[i + x] = '0';
}
}
string p = string(l, '0');
for (int i = 0; i < l; i++) {
if (((i + x) < l && w[i + x] == '1') || ((i - x) >= 0 && w[i - x] == '1'))
p[i] = '1';
}
if (p == s)
cout << w << '\n';
else
cout << -1 << '\n';
}
return 0;
}
| ### Prompt
Develop a solution in cpp to the problem described below:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9 + 7;
const int MOD = 1e9 + 7;
double PI = 4 * atan(1);
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
;
int t;
cin >> t;
while (t--) {
string s;
cin >> s;
int l = s.size(), x, c = 0;
cin >> x;
string w = string(l, '1');
for (int i = 0; i < l; i++) {
if (s[i] == '1')
c++;
else {
c = 0;
if (i - x >= 0) w[i - x] = '0';
if (i + x < l) w[i + x] = '0';
}
}
string p = string(l, '0');
for (int i = 0; i < l; i++) {
if (((i + x) < l && w[i + x] == '1') || ((i - x) >= 0 && w[i - x] == '1'))
p[i] = '1';
}
if (p == s)
cout << w << '\n';
else
cout << -1 << '\n';
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int t;
cin >> t;
while (t--) {
string s;
int x;
cin >> s >> x;
string w;
for (int i = 0; i < s.size(); i++) {
if ((i - x < 0 || s[i - x] == '1') &&
(i + x >= s.size() || s[i + x] == '1')) {
w += '1';
} else {
w += '0';
}
}
bool possible = true;
assert(w.size() == s.size());
for (int i = 0; i < s.size(); i++) {
if (i - x >= 0 && w[i - x] == '1' && s[i] != '1') {
possible = false;
}
if (i + x < w.size() && w[i + x] == '1' && s[i] != '1') {
possible = false;
}
if ((i + x >= w.size() || w[i + x] == '0') &&
(i - x < 0 || w[i - x] == '0') && s[i] != '0') {
possible = false;
}
}
if (possible) {
cout << w << '\n';
} else {
cout << "-1\n";
}
}
}
| ### Prompt
Construct a cpp code solution to the problem outlined:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int t;
cin >> t;
while (t--) {
string s;
int x;
cin >> s >> x;
string w;
for (int i = 0; i < s.size(); i++) {
if ((i - x < 0 || s[i - x] == '1') &&
(i + x >= s.size() || s[i + x] == '1')) {
w += '1';
} else {
w += '0';
}
}
bool possible = true;
assert(w.size() == s.size());
for (int i = 0; i < s.size(); i++) {
if (i - x >= 0 && w[i - x] == '1' && s[i] != '1') {
possible = false;
}
if (i + x < w.size() && w[i + x] == '1' && s[i] != '1') {
possible = false;
}
if ((i + x >= w.size() || w[i + x] == '0') &&
(i - x < 0 || w[i - x] == '0') && s[i] != '0') {
possible = false;
}
}
if (possible) {
cout << w << '\n';
} else {
cout << "-1\n";
}
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
void solve() {
string s;
cin >> s;
int x;
cin >> x;
int n = s.size();
string p;
map<int, int> mp;
for (int i = 0; i < n; i++) p.push_back('%');
int ans = 0;
int flag = 0, f1 = 0, f2 = 0;
for (int i = 0; i < n; i++) {
flag = 0;
f1 = 0;
f2 = 0;
if (s[i] == '0') {
if (i - x >= 0) {
flag = 1;
if (!mp[i - x]) {
p[i - x] = '0';
mp[i - x] = 1;
} else {
if (p[i - x] == '1') {
ans = -1;
break;
}
}
}
if (i + x < n) {
p[i + x] = '0';
mp[i + x] = 1;
}
} else {
if (i - x >= 0) {
flag = 1;
if (!mp[i - x]) {
p[i - x] = '1';
mp[i - x] = 1;
} else if (p[i - x] == '1')
p[i - x] = '1';
else {
f1 = 1;
}
}
if (flag == 0 || f1 == 1) {
if (i + x < n) {
mp[i + x] = 1;
p[i + x] = '1';
} else {
f2 = 1;
}
}
if (f2 == 1) {
ans = -1;
break;
}
}
}
for (int i = 0; i < n; i++) {
if (p[i] == '%') p[i] = '1';
}
if (ans == -1)
cout << "-1" << endl;
else
cout << p << endl;
}
int main() {
int t;
cin >> t;
while (t--) solve();
}
| ### Prompt
In cpp, your task is to solve the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void solve() {
string s;
cin >> s;
int x;
cin >> x;
int n = s.size();
string p;
map<int, int> mp;
for (int i = 0; i < n; i++) p.push_back('%');
int ans = 0;
int flag = 0, f1 = 0, f2 = 0;
for (int i = 0; i < n; i++) {
flag = 0;
f1 = 0;
f2 = 0;
if (s[i] == '0') {
if (i - x >= 0) {
flag = 1;
if (!mp[i - x]) {
p[i - x] = '0';
mp[i - x] = 1;
} else {
if (p[i - x] == '1') {
ans = -1;
break;
}
}
}
if (i + x < n) {
p[i + x] = '0';
mp[i + x] = 1;
}
} else {
if (i - x >= 0) {
flag = 1;
if (!mp[i - x]) {
p[i - x] = '1';
mp[i - x] = 1;
} else if (p[i - x] == '1')
p[i - x] = '1';
else {
f1 = 1;
}
}
if (flag == 0 || f1 == 1) {
if (i + x < n) {
mp[i + x] = 1;
p[i + x] = '1';
} else {
f2 = 1;
}
}
if (f2 == 1) {
ans = -1;
break;
}
}
}
for (int i = 0; i < n; i++) {
if (p[i] == '%') p[i] = '1';
}
if (ans == -1)
cout << "-1" << endl;
else
cout << p << endl;
}
int main() {
int t;
cin >> t;
while (t--) solve();
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int gcd(int a, int b) {
if (b == 0) return a;
a %= b;
return gcd(b, a);
}
bool compare(long long i, long long j) { return i > j; }
bool mysort(pair<pair<long long, long long>, long long> a,
pair<pair<long long, long long>, long long> b) {
if (a.first.first != b.first.first) {
return a.first.first > b.first.first;
}
return a.first.second > b.first.second;
}
struct HASH {
std::size_t operator()(const pair<int, int> &x) const {
return hash<long long>()((long long)x.first ^ (long long)x.second << 32);
}
};
int main() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
long long t;
cin >> t;
while (t--) {
string str;
cin >> str;
long long x;
cin >> x;
vector<int> ob(str.size(), 0);
vector<char> ans(str.size(), '#');
long long n = str.size();
int flag = 0;
for (int i = 0; i < str.size(); i++) {
if (str[i] == '0') {
if (i + x < n) ans[i + x] = '0';
if (i - x >= 0) ans[i - x] = '0';
}
}
for (int i = 0; i < n; i++) {
if (ans[i] == '#') {
ans[i] = '1';
}
}
for (int i = 0; i < n; i++) {
if (str[i] == '0') {
if (i + x < n && ans[i + x] != '0') {
flag = 1;
break;
}
if (i - x >= 0 && ans[i - x] != '0') {
flag = 1;
break;
}
} else {
int count = 0;
if (i + x < n && ans[i + x] == '1') {
count++;
}
if (i - x >= 0 && ans[i - x] == '1') {
count++;
}
if (count >= 1) {
continue;
} else {
flag = 1;
break;
}
}
}
if (flag) {
cout << -1 << endl;
;
} else {
string a = "";
for (int i = 0; i < n; i++) {
a += ans[i];
}
cout << a << endl;
;
}
}
}
| ### Prompt
Generate a Cpp solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int gcd(int a, int b) {
if (b == 0) return a;
a %= b;
return gcd(b, a);
}
bool compare(long long i, long long j) { return i > j; }
bool mysort(pair<pair<long long, long long>, long long> a,
pair<pair<long long, long long>, long long> b) {
if (a.first.first != b.first.first) {
return a.first.first > b.first.first;
}
return a.first.second > b.first.second;
}
struct HASH {
std::size_t operator()(const pair<int, int> &x) const {
return hash<long long>()((long long)x.first ^ (long long)x.second << 32);
}
};
int main() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
long long t;
cin >> t;
while (t--) {
string str;
cin >> str;
long long x;
cin >> x;
vector<int> ob(str.size(), 0);
vector<char> ans(str.size(), '#');
long long n = str.size();
int flag = 0;
for (int i = 0; i < str.size(); i++) {
if (str[i] == '0') {
if (i + x < n) ans[i + x] = '0';
if (i - x >= 0) ans[i - x] = '0';
}
}
for (int i = 0; i < n; i++) {
if (ans[i] == '#') {
ans[i] = '1';
}
}
for (int i = 0; i < n; i++) {
if (str[i] == '0') {
if (i + x < n && ans[i + x] != '0') {
flag = 1;
break;
}
if (i - x >= 0 && ans[i - x] != '0') {
flag = 1;
break;
}
} else {
int count = 0;
if (i + x < n && ans[i + x] == '1') {
count++;
}
if (i - x >= 0 && ans[i - x] == '1') {
count++;
}
if (count >= 1) {
continue;
} else {
flag = 1;
break;
}
}
}
if (flag) {
cout << -1 << endl;
;
} else {
string a = "";
for (int i = 0; i < n; i++) {
a += ans[i];
}
cout << a << endl;
;
}
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int z;
cin >> z;
for (int i = 0; i < z; i++) {
string s;
cin >> s;
int x;
cin >> x;
int l = s.size();
string s1(l, '1');
string s2(l, '1');
for (int j = 1; j <= l; j++) {
if (s[j - 1] == '0') {
if (j - x > 0) {
s1[j - x - 1] = '0';
}
if (j + x <= l) {
s1[j + x - 1] = '0';
}
}
}
for (int k = 1; k <= l; k++) {
if (k - x > 0 && s1[k - x - 1] == '1') {
continue;
} else if (k + x <= l && s1[k + x - 1] == '1') {
continue;
} else {
s2[k - 1] = '0';
}
}
int nan = 1;
for (int p = 0; p < l; p++) {
if (s2[p] != s[p]) {
nan = 0;
break;
}
}
if (nan == 1) {
cout << s1 << endl;
} else {
cout << -1 << endl;
}
}
return 0;
}
| ### Prompt
Your challenge is to write a Cpp solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int z;
cin >> z;
for (int i = 0; i < z; i++) {
string s;
cin >> s;
int x;
cin >> x;
int l = s.size();
string s1(l, '1');
string s2(l, '1');
for (int j = 1; j <= l; j++) {
if (s[j - 1] == '0') {
if (j - x > 0) {
s1[j - x - 1] = '0';
}
if (j + x <= l) {
s1[j + x - 1] = '0';
}
}
}
for (int k = 1; k <= l; k++) {
if (k - x > 0 && s1[k - x - 1] == '1') {
continue;
} else if (k + x <= l && s1[k + x - 1] == '1') {
continue;
} else {
s2[k - 1] = '0';
}
}
int nan = 1;
for (int p = 0; p < l; p++) {
if (s2[p] != s[p]) {
nan = 0;
break;
}
}
if (nan == 1) {
cout << s1 << endl;
} else {
cout << -1 << endl;
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
string solve(string s, int x) {
string w(s.size(), '2');
for (int i = 0; i < x && i + x < s.size(); ++i) {
w[i + x] = s[i];
}
for (int i = (int)s.size() - 1; i >= (int)s.size() - x && i - x >= 0; --i) {
w[i - x] = s[i];
}
for (int i = 0; i < s.size(); ++i) {
if (s[i] == '0') {
if (i - x < 0 && i + x >= s.size()) continue;
if (i - x >= 0 && w[i - x] == '1') return "-1";
if (i + x < s.size() && w[i + x] == '1') return "-1";
if (i - x >= 0 && w[i - x] == '2') w[i - x] = '0';
if (i + x < s.size() && w[i + x] == '2') w[i + x] = '0';
} else {
if (i - x >= 0 && w[i - x] == '1') continue;
if (i + x < s.size() && w[i + x] == '1') continue;
if (i - x >= 0 && w[i - x] == '2') {
w[i - x] = '1';
continue;
}
if (i + x < s.size() && w[i + x] == '2') {
w[i + x] = '1';
continue;
}
return "-1";
}
}
for (int i = 0; i < w.size(); ++i) {
if (w[i] == '2') {
w[i] = '1';
}
}
return w;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int T;
cin >> T;
while (T--) {
string s;
int x;
cin >> s >> x;
cout << solve(s, x) << '\n';
}
return 0;
}
| ### Prompt
Construct a Cpp code solution to the problem outlined:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
string solve(string s, int x) {
string w(s.size(), '2');
for (int i = 0; i < x && i + x < s.size(); ++i) {
w[i + x] = s[i];
}
for (int i = (int)s.size() - 1; i >= (int)s.size() - x && i - x >= 0; --i) {
w[i - x] = s[i];
}
for (int i = 0; i < s.size(); ++i) {
if (s[i] == '0') {
if (i - x < 0 && i + x >= s.size()) continue;
if (i - x >= 0 && w[i - x] == '1') return "-1";
if (i + x < s.size() && w[i + x] == '1') return "-1";
if (i - x >= 0 && w[i - x] == '2') w[i - x] = '0';
if (i + x < s.size() && w[i + x] == '2') w[i + x] = '0';
} else {
if (i - x >= 0 && w[i - x] == '1') continue;
if (i + x < s.size() && w[i + x] == '1') continue;
if (i - x >= 0 && w[i - x] == '2') {
w[i - x] = '1';
continue;
}
if (i + x < s.size() && w[i + x] == '2') {
w[i + x] = '1';
continue;
}
return "-1";
}
}
for (int i = 0; i < w.size(); ++i) {
if (w[i] == '2') {
w[i] = '1';
}
}
return w;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int T;
cin >> T;
while (T--) {
string s;
int x;
cin >> s >> x;
cout << solve(s, x) << '\n';
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
int x;
int n;
string s;
cin >> t;
while (t--) {
cin >> s;
cin >> x;
n = s.length();
vector<char> ans(n, '1');
vector<int> depended(n, -1);
int flag;
for (int i = 0; i < n; i++) {
if (s[i] == '1') {
flag = 1;
if (i >= x) {
if ((depended[i - x] == -1 || depended[i - x] == 1)) {
flag = 0;
depended[i - x] = 1;
} else {
flag = 1;
}
} else if (flag == 1) {
if (i + x < n) {
if ((depended[i + x] == -1 || depended[i + x] == 1)) {
flag = 0;
depended[i + x] = 1;
}
}
}
} else if (s[i] == '0') {
int flag1 = 0, flag2 = 0;
if (i - x >= 0) {
if (depended[i - x] == -1 || depended[i - x] == 0) {
depended[i - x] = 0;
flag1 = 0;
ans[i - x] = '0';
}
}
if (i + x < n) {
if (depended[i + x] == -1 || depended[i + x] == 0) {
depended[i + x] = 0;
flag2 = 0;
ans[i + x] = '0';
}
}
}
}
string temp;
for (int i = 0; i < n; i++) {
int flag = 1;
if (i + x < n) {
if (ans[i + x] == '1') {
temp = temp + '1';
flag = 0;
}
}
if (flag == 1) {
if (i - x >= 0) {
if (ans[i - x] == '1') {
temp = temp + '1';
flag = 0;
}
}
}
if (flag == 1) {
temp = temp + '0';
}
}
if (temp == s) {
for (int i = 0; i < n; i++) {
cout << ans[i];
}
cout << endl;
} else {
cout << -1 << endl;
}
}
}
| ### Prompt
Please provide a Cpp coded solution to the problem described below:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
int x;
int n;
string s;
cin >> t;
while (t--) {
cin >> s;
cin >> x;
n = s.length();
vector<char> ans(n, '1');
vector<int> depended(n, -1);
int flag;
for (int i = 0; i < n; i++) {
if (s[i] == '1') {
flag = 1;
if (i >= x) {
if ((depended[i - x] == -1 || depended[i - x] == 1)) {
flag = 0;
depended[i - x] = 1;
} else {
flag = 1;
}
} else if (flag == 1) {
if (i + x < n) {
if ((depended[i + x] == -1 || depended[i + x] == 1)) {
flag = 0;
depended[i + x] = 1;
}
}
}
} else if (s[i] == '0') {
int flag1 = 0, flag2 = 0;
if (i - x >= 0) {
if (depended[i - x] == -1 || depended[i - x] == 0) {
depended[i - x] = 0;
flag1 = 0;
ans[i - x] = '0';
}
}
if (i + x < n) {
if (depended[i + x] == -1 || depended[i + x] == 0) {
depended[i + x] = 0;
flag2 = 0;
ans[i + x] = '0';
}
}
}
}
string temp;
for (int i = 0; i < n; i++) {
int flag = 1;
if (i + x < n) {
if (ans[i + x] == '1') {
temp = temp + '1';
flag = 0;
}
}
if (flag == 1) {
if (i - x >= 0) {
if (ans[i - x] == '1') {
temp = temp + '1';
flag = 0;
}
}
}
if (flag == 1) {
temp = temp + '0';
}
}
if (temp == s) {
for (int i = 0; i < n; i++) {
cout << ans[i];
}
cout << endl;
} else {
cout << -1 << endl;
}
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long int t;
cin >> t;
while (t--) {
string second;
cin >> second;
long long int x;
cin >> x;
long long int n = second.size();
string ans;
for (long long int i = 0; i < n; i++) ans += '.';
for (long long int i = 0; i < n; i++) {
if (second[i] == '0') {
if (i - x >= 0) ans[i - x] = '0';
if (i + x < n) ans[i + x] = '0';
}
}
long long int temp = 0;
for (long long int i = 0; i < n; i++) {
if (second[i] == '1') {
if (i - x >= 0 && i + x < n && ans[i - x] == '0' && ans[i + x] == '0') {
temp = 1;
break;
}
if (i - x >= 0 && ans[i - x] == '0' && i + x >= n) {
temp = 1;
break;
}
if (i - x < 0 && i + x < n && ans[i + x] == '0') {
temp = 1;
break;
};
if (i - x < 0 && i + x >= n) {
temp = 1;
break;
}
}
}
if (temp == 1) {
cout << "-1\n";
} else {
for (long long int i = 0; i < n; i++) {
if (ans[i] == '.') ans[i] = '1';
}
cout << ans << "\n";
}
}
}
| ### Prompt
Please provide a cpp coded solution to the problem described below:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long int t;
cin >> t;
while (t--) {
string second;
cin >> second;
long long int x;
cin >> x;
long long int n = second.size();
string ans;
for (long long int i = 0; i < n; i++) ans += '.';
for (long long int i = 0; i < n; i++) {
if (second[i] == '0') {
if (i - x >= 0) ans[i - x] = '0';
if (i + x < n) ans[i + x] = '0';
}
}
long long int temp = 0;
for (long long int i = 0; i < n; i++) {
if (second[i] == '1') {
if (i - x >= 0 && i + x < n && ans[i - x] == '0' && ans[i + x] == '0') {
temp = 1;
break;
}
if (i - x >= 0 && ans[i - x] == '0' && i + x >= n) {
temp = 1;
break;
}
if (i - x < 0 && i + x < n && ans[i + x] == '0') {
temp = 1;
break;
};
if (i - x < 0 && i + x >= n) {
temp = 1;
break;
}
}
}
if (temp == 1) {
cout << "-1\n";
} else {
for (long long int i = 0; i < n; i++) {
if (ans[i] == '.') ans[i] = '1';
}
cout << ans << "\n";
}
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int t;
cin >> t;
while (t--) {
string s;
cin >> s;
int x;
cin >> x;
string w = string(s.length(), '2');
bool ok = 1;
for (int i = 0; i < s.length(); i++) {
if (i - x < 0 && i + x >= (int)s.length() && s[i] == '1') {
ok = 0;
break;
}
}
for (int i = 0; i < s.length(); i++) {
if (i - x >= 0 && i + x >= (int)s.length()) w[i - x] = s[i];
}
for (int i = 0; i < s.length(); i++) {
if (i - x < 0 && i + x < (int)s.length()) {
if ((s[i] - '0') + (w[i + x] - '0') == 1) {
ok = 0;
break;
}
w[i + x] = s[i];
}
}
for (int i = 0; i < s.length(); i++) {
if (i - x >= 0 && i + x < (int)s.length()) {
if (s[i] == '0') {
if (w[i + x] == '1' || w[i - x] == '1') {
ok = 0;
break;
}
w[i + x] = w[i - x] = '0';
} else {
if (w[i - x] != '0')
w[i - x] = '1';
else if (w[i + x] != '0')
w[i + x] = '1';
else {
ok = 0;
break;
}
}
}
}
if (!ok)
cout << "-1\n";
else {
for (int i = 0; i < w.length(); i++) {
w[i] = min(w[i], '1');
}
cout << w << "\n";
}
}
}
| ### Prompt
Please create a solution in CPP to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int t;
cin >> t;
while (t--) {
string s;
cin >> s;
int x;
cin >> x;
string w = string(s.length(), '2');
bool ok = 1;
for (int i = 0; i < s.length(); i++) {
if (i - x < 0 && i + x >= (int)s.length() && s[i] == '1') {
ok = 0;
break;
}
}
for (int i = 0; i < s.length(); i++) {
if (i - x >= 0 && i + x >= (int)s.length()) w[i - x] = s[i];
}
for (int i = 0; i < s.length(); i++) {
if (i - x < 0 && i + x < (int)s.length()) {
if ((s[i] - '0') + (w[i + x] - '0') == 1) {
ok = 0;
break;
}
w[i + x] = s[i];
}
}
for (int i = 0; i < s.length(); i++) {
if (i - x >= 0 && i + x < (int)s.length()) {
if (s[i] == '0') {
if (w[i + x] == '1' || w[i - x] == '1') {
ok = 0;
break;
}
w[i + x] = w[i - x] = '0';
} else {
if (w[i - x] != '0')
w[i - x] = '1';
else if (w[i + x] != '0')
w[i + x] = '1';
else {
ok = 0;
break;
}
}
}
}
if (!ok)
cout << "-1\n";
else {
for (int i = 0; i < w.length(); i++) {
w[i] = min(w[i], '1');
}
cout << w << "\n";
}
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
int n;
for (int o = 0; o < t; o++) {
string s;
cin >> s;
int x;
cin >> x;
int n = s.size();
char w[n];
for (int i = 0; i < n; i++) {
w[i] = '1';
}
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0 && i - x <= n - 1) {
w[i - x] = '0';
}
if (i + x >= 0 && i + x <= n - 1) {
w[i + x] = '0';
}
}
}
int q = 1, m = 2, p = 2, b = 0;
for (int i = 0; i < n; i++) {
m = 2;
p = 2;
if (i - x >= 0 && i - x <= n - 1) {
if (w[i - x] == '1') {
m = 1;
} else {
m = 0;
}
}
if (i + x >= 0 && i + x <= n - 1) {
if (w[i + x] == '1') {
p = 1;
} else {
p = 0;
}
}
if (s[i] == '0') {
if (p == 1 || m == 1) {
cout << -1 << endl;
b = 1;
break;
}
}
if (s[i] == '1') {
if (p != 1 && m != 1) {
cout << -1 << endl;
b = 1;
break;
}
}
}
if (b == 0) {
for (int i = 0; i < n; i++) {
cout << w[i];
}
cout << endl;
}
}
return 0;
}
| ### Prompt
Develop a solution in Cpp to the problem described below:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
int n;
for (int o = 0; o < t; o++) {
string s;
cin >> s;
int x;
cin >> x;
int n = s.size();
char w[n];
for (int i = 0; i < n; i++) {
w[i] = '1';
}
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0 && i - x <= n - 1) {
w[i - x] = '0';
}
if (i + x >= 0 && i + x <= n - 1) {
w[i + x] = '0';
}
}
}
int q = 1, m = 2, p = 2, b = 0;
for (int i = 0; i < n; i++) {
m = 2;
p = 2;
if (i - x >= 0 && i - x <= n - 1) {
if (w[i - x] == '1') {
m = 1;
} else {
m = 0;
}
}
if (i + x >= 0 && i + x <= n - 1) {
if (w[i + x] == '1') {
p = 1;
} else {
p = 0;
}
}
if (s[i] == '0') {
if (p == 1 || m == 1) {
cout << -1 << endl;
b = 1;
break;
}
}
if (s[i] == '1') {
if (p != 1 && m != 1) {
cout << -1 << endl;
b = 1;
break;
}
}
}
if (b == 0) {
for (int i = 0; i < n; i++) {
cout << w[i];
}
cout << endl;
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
void IMPLEMENT_CODE() {
char s[200005];
cin >> s;
long long int l = strlen(s);
long long int x;
cin >> x;
char ans[l + 1];
char o = '1';
char z = '0';
for (long long int i = 0; i < l; ++i) {
ans[i] = z;
}
ans[l] = '\0';
for (long long int i = 0; i < l; i++) {
if (s[i] == '1') {
if (i + x < l) {
if (i + 2 * x < l && s[i + 2 * x] == '1')
ans[i + x] = o;
else if (i + 2 * x >= l)
ans[i + x] = o;
}
if (i - x >= 0) {
if (i - 2 * x >= 0 && s[i - 2 * x] == '1')
ans[i - x] = o;
else if (i - 2 * x < 0)
ans[i - x] = o;
}
}
}
char p[l + 1];
p[l] = '\0';
for (long long int i = 0; i < l; ++i) {
p[i] = z;
}
for (long long int i = 0; i < l; i++) {
if (i + x < l && ans[i + x] == '1') p[i] = o;
if (i - x >= 0 && ans[i - x] == '1') p[i] = o;
}
if (strcmp(p, s) == 0)
cout << ans << endl;
else {
int q = 1;
for (long long int i = 0; i < l; i++) {
if (s[i] == o && p[i] != o) {
q = 0;
break;
}
}
if (q == 1)
cout << ans << endl;
else
cout << -1 << endl;
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(0);
;
long long int testCases;
cin >> testCases;
while (testCases--) {
IMPLEMENT_CODE();
}
return 0;
}
| ### Prompt
Develop a solution in Cpp to the problem described below:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void IMPLEMENT_CODE() {
char s[200005];
cin >> s;
long long int l = strlen(s);
long long int x;
cin >> x;
char ans[l + 1];
char o = '1';
char z = '0';
for (long long int i = 0; i < l; ++i) {
ans[i] = z;
}
ans[l] = '\0';
for (long long int i = 0; i < l; i++) {
if (s[i] == '1') {
if (i + x < l) {
if (i + 2 * x < l && s[i + 2 * x] == '1')
ans[i + x] = o;
else if (i + 2 * x >= l)
ans[i + x] = o;
}
if (i - x >= 0) {
if (i - 2 * x >= 0 && s[i - 2 * x] == '1')
ans[i - x] = o;
else if (i - 2 * x < 0)
ans[i - x] = o;
}
}
}
char p[l + 1];
p[l] = '\0';
for (long long int i = 0; i < l; ++i) {
p[i] = z;
}
for (long long int i = 0; i < l; i++) {
if (i + x < l && ans[i + x] == '1') p[i] = o;
if (i - x >= 0 && ans[i - x] == '1') p[i] = o;
}
if (strcmp(p, s) == 0)
cout << ans << endl;
else {
int q = 1;
for (long long int i = 0; i < l; i++) {
if (s[i] == o && p[i] != o) {
q = 0;
break;
}
}
if (q == 1)
cout << ans << endl;
else
cout << -1 << endl;
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(0);
;
long long int testCases;
cin >> testCases;
while (testCases--) {
IMPLEMENT_CODE();
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
string str;
cin >> str;
int n = str.length();
string ans(n, '#');
int x;
cin >> x;
bool res = 1;
for (int i = 0; res && i < n; i++) {
if (str[i] == '1') {
if (i - x >= 0 && (ans[i - x] == '#' || ans[i - x] == '1')) {
if (ans[i - x] == '#') ans[i - x] = '1';
} else if (i + x < n && (ans[i + x] == '#' || ans[i + x] == '1')) {
if (ans[i + x] == '#') ans[i + x] = '1';
} else
res = 0;
} else {
if (i - x < 0 || (ans[i - x] == '#' || ans[i - x] == '0')) {
if (i - x >= 0 && ans[i - x] == '#') ans[i - x] = '0';
} else
res = 0;
if (i + x >= n || (ans[i + x] == '#' || ans[i + x] == '0')) {
if (i + x < n && ans[i + x] == '#') ans[i + x] = '0';
} else
res = 0;
}
}
if (!res)
cout << -1 << "\n";
else {
for (int i = 0; i < n; i++)
if (ans[i] == '#') ans[i] = '1';
cout << ans << "\n";
}
}
return 0;
}
| ### Prompt
In Cpp, your task is to solve the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
string str;
cin >> str;
int n = str.length();
string ans(n, '#');
int x;
cin >> x;
bool res = 1;
for (int i = 0; res && i < n; i++) {
if (str[i] == '1') {
if (i - x >= 0 && (ans[i - x] == '#' || ans[i - x] == '1')) {
if (ans[i - x] == '#') ans[i - x] = '1';
} else if (i + x < n && (ans[i + x] == '#' || ans[i + x] == '1')) {
if (ans[i + x] == '#') ans[i + x] = '1';
} else
res = 0;
} else {
if (i - x < 0 || (ans[i - x] == '#' || ans[i - x] == '0')) {
if (i - x >= 0 && ans[i - x] == '#') ans[i - x] = '0';
} else
res = 0;
if (i + x >= n || (ans[i + x] == '#' || ans[i + x] == '0')) {
if (i + x < n && ans[i + x] == '#') ans[i + x] = '0';
} else
res = 0;
}
}
if (!res)
cout << -1 << "\n";
else {
for (int i = 0; i < n; i++)
if (ans[i] == '#') ans[i] = '1';
cout << ans << "\n";
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t, x;
cin >> t;
for (int i = 0; i < t; i++) {
string s;
cin >> s >> x;
bool possible = true;
for (int j = 0; j < s.size(); j++) {
if (s[j] == '1') {
possible = false;
if (j > x - 1) {
if (j < 2 * x or s[j - 2 * x] == '1') possible = true;
}
if (j < s.size() - x) {
if (j > (int)s.size() - 2 * x - 1 or s[j + 2 * x] == '1')
possible = true;
}
if (!possible) {
break;
}
}
}
if (!possible) {
cout << -1 << "\n";
} else {
for (int j = 0; j < s.size(); j++) {
if ((j > x - 1 and s[j - x] == '0') or
(j < s.size() - x and s[j + x] == '0')) {
cout << 0;
} else {
cout << 1;
}
}
cout << "\n";
}
}
}
| ### Prompt
Create a solution in cpp for the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t, x;
cin >> t;
for (int i = 0; i < t; i++) {
string s;
cin >> s >> x;
bool possible = true;
for (int j = 0; j < s.size(); j++) {
if (s[j] == '1') {
possible = false;
if (j > x - 1) {
if (j < 2 * x or s[j - 2 * x] == '1') possible = true;
}
if (j < s.size() - x) {
if (j > (int)s.size() - 2 * x - 1 or s[j + 2 * x] == '1')
possible = true;
}
if (!possible) {
break;
}
}
}
if (!possible) {
cout << -1 << "\n";
} else {
for (int j = 0; j < s.size(); j++) {
if ((j > x - 1 and s[j - x] == '0') or
(j < s.size() - x and s[j + x] == '0')) {
cout << 0;
} else {
cout << 1;
}
}
cout << "\n";
}
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
int x;
string a;
cin >> a >> x;
string w(a.size(), '1');
for (int i = 0; i < a.size(); i++) {
if (a[i] == '0') {
if (i - x >= 0) w[i - x] = '0';
if (i + x < a.size()) w[i + x] = '0';
}
}
bool ff = 0;
for (int i = 0; i < a.size(); i++) {
bool f = 0;
ff = 0;
f = f or (i - x >= 0 and w[i - x] == '1');
f = f or (i + x < a.size() and w[i + x] == '1');
if (a[i] != f + '0') {
ff = 1;
cout << "-1\n";
break;
}
if (ff) break;
}
if (ff)
continue;
else
cout << w << endl;
}
return 0;
}
| ### Prompt
Please formulate a Cpp solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
int x;
string a;
cin >> a >> x;
string w(a.size(), '1');
for (int i = 0; i < a.size(); i++) {
if (a[i] == '0') {
if (i - x >= 0) w[i - x] = '0';
if (i + x < a.size()) w[i + x] = '0';
}
}
bool ff = 0;
for (int i = 0; i < a.size(); i++) {
bool f = 0;
ff = 0;
f = f or (i - x >= 0 and w[i - x] == '1');
f = f or (i + x < a.size() and w[i + x] == '1');
if (a[i] != f + '0') {
ff = 1;
cout << "-1\n";
break;
}
if (ff) break;
}
if (ff)
continue;
else
cout << w << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long t;
cin >> t;
while (t--) {
string s;
long long x;
cin >> s;
long long n = s.size();
cin >> x;
bool chk = true;
string w;
for (long long i = 0; i < n; i++) {
w += '2';
if (i >= x and s[i - x] == '0')
w[i] = '0';
else if (i + x < n and s[i + x] == '0')
w[i] = '0';
else
w[i] = '1';
}
for (long long i = 0; i < n; i++) {
if (s[i] == '1') {
long long one = 0;
if (i >= x and w[i - x] == '1')
one++;
else if (i + x < n and w[i + x] == '1')
one++;
if (one == 0) {
chk = false;
}
}
}
if (chk)
cout << w << endl;
else
cout << -1 << endl;
}
return 0;
}
| ### Prompt
Create a solution in CPP for the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long t;
cin >> t;
while (t--) {
string s;
long long x;
cin >> s;
long long n = s.size();
cin >> x;
bool chk = true;
string w;
for (long long i = 0; i < n; i++) {
w += '2';
if (i >= x and s[i - x] == '0')
w[i] = '0';
else if (i + x < n and s[i + x] == '0')
w[i] = '0';
else
w[i] = '1';
}
for (long long i = 0; i < n; i++) {
if (s[i] == '1') {
long long one = 0;
if (i >= x and w[i - x] == '1')
one++;
else if (i + x < n and w[i + x] == '1')
one++;
if (one == 0) {
chk = false;
}
}
}
if (chk)
cout << w << endl;
else
cout << -1 << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
int x, i, j, p = 0, l;
string a;
cin >> a;
cin >> x;
l = a.size();
string w(l, '1');
for (i = 0; i < l; i++) {
if (a[i] == '0') {
if (i - x >= 0) w[i - x] = '0';
if (i + x < l) w[i + x] = '0';
}
}
for (i = 0; i < l; i++) {
if (a[i] == '1') {
if ((i - x >= 0 && w[i - x] == '1') || (i + x < l && w[i + x] == '1'))
continue;
else {
p = 1;
break;
}
}
}
if (p == 1)
cout << "-1" << endl;
else
cout << w << endl;
}
}
| ### Prompt
Construct a cpp code solution to the problem outlined:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
int x, i, j, p = 0, l;
string a;
cin >> a;
cin >> x;
l = a.size();
string w(l, '1');
for (i = 0; i < l; i++) {
if (a[i] == '0') {
if (i - x >= 0) w[i - x] = '0';
if (i + x < l) w[i + x] = '0';
}
}
for (i = 0; i < l; i++) {
if (a[i] == '1') {
if ((i - x >= 0 && w[i - x] == '1') || (i + x < l && w[i + x] == '1'))
continue;
else {
p = 1;
break;
}
}
}
if (p == 1)
cout << "-1" << endl;
else
cout << w << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int N = 200005;
const long long MOD = 1000000007;
const int INF = 0x3f3f3f3f;
int a[N];
string s, t;
void go() {
int n, x;
cin >> s >> x;
n = s.size();
t.clear();
for (int i = 1; i <= n; ++i) {
t.push_back('1');
}
for (int i = 0; i < n; ++i) {
if (s[i] == '0') {
if (i - x >= 0) t[i - x] = '0';
if (i + x < n) t[i + x] = '0';
}
}
bool ok = 1;
for (int i = 0; i < n; ++i) {
if (i - x >= 0 && t[i - x] == '1') {
if (s[i] == '0') {
ok = 0;
break;
}
} else if (i + x < n && t[i + x] == '1') {
if (s[i] == '0') {
ok = 0;
break;
}
} else {
if (s[i] == '1') {
ok = 0;
break;
}
}
}
if (ok)
cout << t << '\n';
else
cout << -1 << '\n';
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t;
cin >> t;
while (t--) {
go();
}
}
| ### Prompt
Create a solution in Cpp for the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int N = 200005;
const long long MOD = 1000000007;
const int INF = 0x3f3f3f3f;
int a[N];
string s, t;
void go() {
int n, x;
cin >> s >> x;
n = s.size();
t.clear();
for (int i = 1; i <= n; ++i) {
t.push_back('1');
}
for (int i = 0; i < n; ++i) {
if (s[i] == '0') {
if (i - x >= 0) t[i - x] = '0';
if (i + x < n) t[i + x] = '0';
}
}
bool ok = 1;
for (int i = 0; i < n; ++i) {
if (i - x >= 0 && t[i - x] == '1') {
if (s[i] == '0') {
ok = 0;
break;
}
} else if (i + x < n && t[i + x] == '1') {
if (s[i] == '0') {
ok = 0;
break;
}
} else {
if (s[i] == '1') {
ok = 0;
break;
}
}
}
if (ok)
cout << t << '\n';
else
cout << -1 << '\n';
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t;
cin >> t;
while (t--) {
go();
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int t;
cin >> t;
while (t--) {
string s;
cin >> s;
string ans((int)s.size(), '1');
int x;
cin >> x;
int n = s.size();
for (int i = 0; i < (int)s.size(); ++i) {
if (s[i] & 1) continue;
if (i >= x) {
ans[i - x] = '0';
}
if (i + x < n) {
ans[i + x] = '0';
}
}
string ch((int)ans.size(), '1');
for (int i = 0; i < (int)ch.size(); ++i) {
bool ok1 = 0, ok2 = 0;
if (i >= x && ans[i - x] & 1) ok1 = 1;
if (i + x < n && ans[i + x] & 1) ok2 = 1;
if ((!ok1) && (!ok2)) ch[i] = '0';
}
cout << (ch == s ? ans : "-1") << "\n";
}
}
| ### Prompt
Develop a solution in CPP to the problem described below:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int t;
cin >> t;
while (t--) {
string s;
cin >> s;
string ans((int)s.size(), '1');
int x;
cin >> x;
int n = s.size();
for (int i = 0; i < (int)s.size(); ++i) {
if (s[i] & 1) continue;
if (i >= x) {
ans[i - x] = '0';
}
if (i + x < n) {
ans[i + x] = '0';
}
}
string ch((int)ans.size(), '1');
for (int i = 0; i < (int)ch.size(); ++i) {
bool ok1 = 0, ok2 = 0;
if (i >= x && ans[i - x] & 1) ok1 = 1;
if (i + x < n && ans[i + x] & 1) ok2 = 1;
if ((!ok1) && (!ok2)) ch[i] = '0';
}
cout << (ch == s ? ans : "-1") << "\n";
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int t;
cin >> t;
while (t--) {
string s;
cin >> s;
unsigned long long x;
cin >> x;
unsigned long long n = s.length();
char w[n];
for (unsigned long long i = 0; i < n; ++i) w[i] = '-';
for (unsigned long long i = 0; i < n; ++i) {
if (s[i] == '0') {
if ((i + x) < n) w[i + x] = '0';
if (i >= x) w[i - x] = '0';
}
}
bool flag = true;
unsigned long long k = 0;
for (unsigned long long i = 0; i < n; ++i) {
k = 0;
if (s[i] == '1') {
if ((i + x) < n) {
if (w[i + x] == '-') k++;
}
if (i >= x) {
if (w[i - x] == '-') k++;
}
if (k == 0) {
flag = false;
break;
}
}
}
if (!flag)
cout << -1 << endl;
else {
for (unsigned long long i = 0; i < n; ++i) {
if (w[i] == '-') w[i] = '1';
}
for (unsigned long long i = 0; i < n; ++i) cout << w[i];
cout << endl;
}
}
}
| ### Prompt
Create a solution in cpp for the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int t;
cin >> t;
while (t--) {
string s;
cin >> s;
unsigned long long x;
cin >> x;
unsigned long long n = s.length();
char w[n];
for (unsigned long long i = 0; i < n; ++i) w[i] = '-';
for (unsigned long long i = 0; i < n; ++i) {
if (s[i] == '0') {
if ((i + x) < n) w[i + x] = '0';
if (i >= x) w[i - x] = '0';
}
}
bool flag = true;
unsigned long long k = 0;
for (unsigned long long i = 0; i < n; ++i) {
k = 0;
if (s[i] == '1') {
if ((i + x) < n) {
if (w[i + x] == '-') k++;
}
if (i >= x) {
if (w[i - x] == '-') k++;
}
if (k == 0) {
flag = false;
break;
}
}
}
if (!flag)
cout << -1 << endl;
else {
for (unsigned long long i = 0; i < n; ++i) {
if (w[i] == '-') w[i] = '1';
}
for (unsigned long long i = 0; i < n; ++i) cout << w[i];
cout << endl;
}
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long a[100005];
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
long long i, j, k, x, y, z, m, n, t, b, c, l, r;
cin >> t;
while (t--) {
string s, ss;
cin >> s;
ss = s;
cin >> n;
long long sz = s.size();
string ans(sz, '1');
for (i = 0; i < sz; i++) {
if (s[i] == '0') {
if (i - n >= 0) ans[i - n] = '0';
if (i + n < sz) ans[i + n] = '0';
}
}
string tmp(sz, '.');
for (i = 0; i < sz; i++) {
if (i - n >= 0 && ans[i - n] == '1')
tmp[i] = '1';
else if (i + n < sz && ans[i + n] == '1')
tmp[i] = '1';
else
tmp[i] = '0';
}
if (tmp == ss)
cout << ans << '\n';
else
cout << -1 << '\n';
}
}
| ### Prompt
Please provide a cpp coded solution to the problem described below:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long a[100005];
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
long long i, j, k, x, y, z, m, n, t, b, c, l, r;
cin >> t;
while (t--) {
string s, ss;
cin >> s;
ss = s;
cin >> n;
long long sz = s.size();
string ans(sz, '1');
for (i = 0; i < sz; i++) {
if (s[i] == '0') {
if (i - n >= 0) ans[i - n] = '0';
if (i + n < sz) ans[i + n] = '0';
}
}
string tmp(sz, '.');
for (i = 0; i < sz; i++) {
if (i - n >= 0 && ans[i - n] == '1')
tmp[i] = '1';
else if (i + n < sz && ans[i + n] == '1')
tmp[i] = '1';
else
tmp[i] = '0';
}
if (tmp == ss)
cout << ans << '\n';
else
cout << -1 << '\n';
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using vll = vector<ll>;
using vvll = vector<vll>;
using vc = vector<char>;
using vvc = vector<vc>;
using pll = pair<ll, ll>;
ll mod = 1000000007;
ll INF = 1e15;
void yorn(bool a) {
if (a)
cout << "Yes" << endl;
else
cout << "No" << endl;
}
ll gcd(long long a, long long b) {
if (b == 0) return a;
return gcd(b, a % b);
}
ll lcm(long long a, long long b) { return a / gcd(a, b) * b; }
void fix_cout() { cout << fixed << setprecision(20); }
void chmax(ll &a, ll b) {
if (a < b) a = b;
}
void chmin(ll &a, ll b) {
if (a > b) a = b;
}
void solve() {
string s;
cin >> s;
ll x;
cin >> x;
bool ok = true;
ll n = s.size();
for (int i = 0; i < n; i++) {
if (s[i] == '1' && i - x < 0 && i + x >= n) {
cout << -1 << endl;
return;
}
}
string ans(n, 'o');
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) ans[i - x] = '0';
if (i + x < n) ans[i + x] = '0';
}
}
for (int i = 0; i < n; i++) {
if (s[i] == '1') {
if (i - x < 0) {
if (ans[i + x] == '0') {
cout << -1 << endl;
return;
}
ans[i + x] = '1';
} else if (i + x >= n) {
if (ans[i - x] == '0') {
cout << -1 << endl;
return;
}
ans[i - x] = '1';
} else {
if (ans[i - x] == '0' && ans[i + x] == '0') {
cout << -1 << endl;
return;
}
if (ans[i - x] == 'o') ans[i - x] = '1';
if (ans[i + x] == 'o') ans[i + x] = '1';
}
}
}
for (int i = 0; i < n; i++)
if (ans[i] == 'o') ans[i] = '0';
cout << ans << endl;
}
int main() {
ll t;
cin >> t;
while (t--) {
solve();
}
}
| ### Prompt
Please formulate a Cpp solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using vll = vector<ll>;
using vvll = vector<vll>;
using vc = vector<char>;
using vvc = vector<vc>;
using pll = pair<ll, ll>;
ll mod = 1000000007;
ll INF = 1e15;
void yorn(bool a) {
if (a)
cout << "Yes" << endl;
else
cout << "No" << endl;
}
ll gcd(long long a, long long b) {
if (b == 0) return a;
return gcd(b, a % b);
}
ll lcm(long long a, long long b) { return a / gcd(a, b) * b; }
void fix_cout() { cout << fixed << setprecision(20); }
void chmax(ll &a, ll b) {
if (a < b) a = b;
}
void chmin(ll &a, ll b) {
if (a > b) a = b;
}
void solve() {
string s;
cin >> s;
ll x;
cin >> x;
bool ok = true;
ll n = s.size();
for (int i = 0; i < n; i++) {
if (s[i] == '1' && i - x < 0 && i + x >= n) {
cout << -1 << endl;
return;
}
}
string ans(n, 'o');
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) ans[i - x] = '0';
if (i + x < n) ans[i + x] = '0';
}
}
for (int i = 0; i < n; i++) {
if (s[i] == '1') {
if (i - x < 0) {
if (ans[i + x] == '0') {
cout << -1 << endl;
return;
}
ans[i + x] = '1';
} else if (i + x >= n) {
if (ans[i - x] == '0') {
cout << -1 << endl;
return;
}
ans[i - x] = '1';
} else {
if (ans[i - x] == '0' && ans[i + x] == '0') {
cout << -1 << endl;
return;
}
if (ans[i - x] == 'o') ans[i - x] = '1';
if (ans[i + x] == 'o') ans[i + x] = '1';
}
}
}
for (int i = 0; i < n; i++)
if (ans[i] == 'o') ans[i] = '0';
cout << ans << endl;
}
int main() {
ll t;
cin >> t;
while (t--) {
solve();
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
void solve() {
string s;
int first;
cin >> s >> first;
string w = string(s.length(), 'w');
for (int i = 0; i < s.length(); i++) {
if (s[i] == '0') {
if (i - first >= 0) w[i - first] = '0';
if (i + first < s.length()) w[i + first] = '0';
}
}
for (int i = 0; i < s.length(); i++) {
if (w[i] == 'w') w[i] = '1';
}
string newS = "";
for (int i = 0; i < s.length(); i++) {
bool one = false;
if (i - first >= 0 && w[i - first] == '1') one = true;
if (i + first < s.length() && w[i + first] == '1') one = true;
if (one) {
newS += "1";
} else
newS += "0";
}
if (s == newS) {
cout << w << "\n";
} else {
cout << "-1\n";
}
}
int main() {
ios::sync_with_stdio(false);
cin.tie(NULL);
int t = 1;
cin >> t;
while (t--) {
solve();
}
return 0;
}
| ### Prompt
Construct a CPP code solution to the problem outlined:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void solve() {
string s;
int first;
cin >> s >> first;
string w = string(s.length(), 'w');
for (int i = 0; i < s.length(); i++) {
if (s[i] == '0') {
if (i - first >= 0) w[i - first] = '0';
if (i + first < s.length()) w[i + first] = '0';
}
}
for (int i = 0; i < s.length(); i++) {
if (w[i] == 'w') w[i] = '1';
}
string newS = "";
for (int i = 0; i < s.length(); i++) {
bool one = false;
if (i - first >= 0 && w[i - first] == '1') one = true;
if (i + first < s.length() && w[i + first] == '1') one = true;
if (one) {
newS += "1";
} else
newS += "0";
}
if (s == newS) {
cout << w << "\n";
} else {
cout << "-1\n";
}
}
int main() {
ios::sync_with_stdio(false);
cin.tie(NULL);
int t = 1;
cin >> t;
while (t--) {
solve();
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
long long x, i;
string s, str;
cin >> s;
cin >> x;
long long n = s.length();
for (i = 0; i < n; i++) {
str += '1';
}
bool temp = 0;
for (i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) str[i - x] = '0';
if (i + x < n) str[i + x] = '0';
}
}
for (i = 0; i < n; i++) {
if (s[i] == '1') {
if ((i - x >= 0 && str[i - x] == '1') ||
(i + x < n && str[i + x] == '1')) {
continue;
} else {
temp = 1;
break;
}
}
}
if (temp)
cout << "-1" << endl;
else
cout << str << endl;
}
}
| ### Prompt
Develop a solution in Cpp to the problem described below:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
long long x, i;
string s, str;
cin >> s;
cin >> x;
long long n = s.length();
for (i = 0; i < n; i++) {
str += '1';
}
bool temp = 0;
for (i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) str[i - x] = '0';
if (i + x < n) str[i + x] = '0';
}
}
for (i = 0; i < n; i++) {
if (s[i] == '1') {
if ((i - x >= 0 && str[i - x] == '1') ||
(i + x < n && str[i + x] == '1')) {
continue;
} else {
temp = 1;
break;
}
}
}
if (temp)
cout << "-1" << endl;
else
cout << str << endl;
}
}
``` |
#include <bits/stdc++.h>
template <class T>
void rd(T &x) {
x = 0;
T f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-') f = -1;
ch = getchar();
}
while (ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
x *= f;
}
using namespace std;
const int N = 1e5 + 10;
int T;
char s[N];
int x, n;
int ans[N];
int main() {
rd(T);
while (T--) {
scanf("%s", s + 1);
n = strlen(s + 1);
rd(x);
for (int i = 1; i <= n; i++) ans[i] = 1;
for (int i = 1; i <= n; i++) {
if (s[i] - '0') continue;
if (i > x) ans[i - x] = 0;
if (i + x <= n) ans[i + x] = 0;
}
bool flg = 0;
for (int i = 1; i <= n; i++) {
if (s[i] - '1') continue;
bool tmp = 0;
if (i > x && ans[i - x]) tmp = 1;
if (i + x <= n && ans[i + x]) tmp = 1;
if (!tmp) flg = 1;
}
if (flg)
printf("-1\n");
else {
for (int i = 1; i <= n; i++) printf("%d", ans[i]);
puts("");
}
}
return 0;
}
| ### Prompt
Please provide a CPP coded solution to the problem described below:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
template <class T>
void rd(T &x) {
x = 0;
T f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-') f = -1;
ch = getchar();
}
while (ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
x *= f;
}
using namespace std;
const int N = 1e5 + 10;
int T;
char s[N];
int x, n;
int ans[N];
int main() {
rd(T);
while (T--) {
scanf("%s", s + 1);
n = strlen(s + 1);
rd(x);
for (int i = 1; i <= n; i++) ans[i] = 1;
for (int i = 1; i <= n; i++) {
if (s[i] - '0') continue;
if (i > x) ans[i - x] = 0;
if (i + x <= n) ans[i + x] = 0;
}
bool flg = 0;
for (int i = 1; i <= n; i++) {
if (s[i] - '1') continue;
bool tmp = 0;
if (i > x && ans[i - x]) tmp = 1;
if (i + x <= n && ans[i + x]) tmp = 1;
if (!tmp) flg = 1;
}
if (flg)
printf("-1\n");
else {
for (int i = 1; i <= n; i++) printf("%d", ans[i]);
puts("");
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e7;
const int N = 2e6 + 13, M = N;
int n, m;
void solve() {
string s;
long long x;
cin >> s >> x;
int n = s.size(), i;
string w(n, '0');
bool ok = true;
for (i = 0; i < n; i++) {
if (s[i] == '1') {
ok = false;
if (i + x < n) {
if (i + x + x >= n || s[i + x + x] == '1') {
w[i + x] = '1';
ok = 1;
}
}
if (i - x >= 0) {
if (i - x - x < 0 || s[i - x - x] == '1') {
w[i - x] = '1';
ok = 1;
}
}
if (!ok) break;
}
}
string ans = "-1";
if (ok) {
ans = w;
}
cout << ans << "\n";
}
int main() {
ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int t = 1;
cin >> t;
while (t--) {
solve();
}
return 0;
}
| ### Prompt
In Cpp, your task is to solve the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e7;
const int N = 2e6 + 13, M = N;
int n, m;
void solve() {
string s;
long long x;
cin >> s >> x;
int n = s.size(), i;
string w(n, '0');
bool ok = true;
for (i = 0; i < n; i++) {
if (s[i] == '1') {
ok = false;
if (i + x < n) {
if (i + x + x >= n || s[i + x + x] == '1') {
w[i + x] = '1';
ok = 1;
}
}
if (i - x >= 0) {
if (i - x - x < 0 || s[i - x - x] == '1') {
w[i - x] = '1';
ok = 1;
}
}
if (!ok) break;
}
}
string ans = "-1";
if (ok) {
ans = w;
}
cout << ans << "\n";
}
int main() {
ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int t = 1;
cin >> t;
while (t--) {
solve();
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
void print1() { cout << "1"; }
void print0() { cout << "0"; }
int main() {
int t;
cin >> t;
while (t--) {
string s;
cin >> s;
long long int x;
cin >> x;
vector<char> vec(s.length(), '2');
for (int i = 0; i < s.length(); i++) {
if (s[i] == '0') {
if (i - x >= 0) vec[i - x] = '0';
if (i + x < s.length()) vec[i + x] = '0';
}
}
long long int a = 1;
int b = 0, c = 0, d = b + c;
for (int i = 0; i < s.length(); i++) {
if (s[i] == '1') {
if ((i - x >= 0 && vec[i - x] == '0') || i - x < 0) {
if (i + x < s.length() && vec[i + x] == '0' || i + x >= s.length()) {
a = 0;
break;
}
}
}
}
if (a == 0) {
cout << -1 << endl;
continue;
}
for (int i = 0; i < s.length(); i++) {
if (vec[i] == '2')
print1();
else
print0();
}
cout << endl;
}
return 0;
}
| ### Prompt
Create a solution in cpp for the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void print1() { cout << "1"; }
void print0() { cout << "0"; }
int main() {
int t;
cin >> t;
while (t--) {
string s;
cin >> s;
long long int x;
cin >> x;
vector<char> vec(s.length(), '2');
for (int i = 0; i < s.length(); i++) {
if (s[i] == '0') {
if (i - x >= 0) vec[i - x] = '0';
if (i + x < s.length()) vec[i + x] = '0';
}
}
long long int a = 1;
int b = 0, c = 0, d = b + c;
for (int i = 0; i < s.length(); i++) {
if (s[i] == '1') {
if ((i - x >= 0 && vec[i - x] == '0') || i - x < 0) {
if (i + x < s.length() && vec[i + x] == '0' || i + x >= s.length()) {
a = 0;
break;
}
}
}
}
if (a == 0) {
cout << -1 << endl;
continue;
}
for (int i = 0; i < s.length(); i++) {
if (vec[i] == '2')
print1();
else
print0();
}
cout << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int32_t main() {
long long int t;
cin >> t;
while (t--) {
string s;
cin >> s;
long long int k;
cin >> k;
long long int n = s.size();
string ans(n, '1');
for (long long int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - k >= 0) ans[i - k] = '0';
if (i + k < n) ans[i + k] = '0';
}
}
long long int flag = 0;
for (long long int i = 0; i < n; i++) {
bool a = false;
if (s[i] == '1') {
if (i - k >= 0 and ans[i - k] == '1') a = true;
if (i + k < n and ans[i + k] == '1') a = true;
if (!a) {
flag = 1;
break;
}
}
}
if (flag == 1)
cout << -1 << endl;
else
cout << ans << endl;
}
}
| ### Prompt
Your task is to create a CPP solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int32_t main() {
long long int t;
cin >> t;
while (t--) {
string s;
cin >> s;
long long int k;
cin >> k;
long long int n = s.size();
string ans(n, '1');
for (long long int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - k >= 0) ans[i - k] = '0';
if (i + k < n) ans[i + k] = '0';
}
}
long long int flag = 0;
for (long long int i = 0; i < n; i++) {
bool a = false;
if (s[i] == '1') {
if (i - k >= 0 and ans[i - k] == '1') a = true;
if (i + k < n and ans[i + k] == '1') a = true;
if (!a) {
flag = 1;
break;
}
}
}
if (flag == 1)
cout << -1 << endl;
else
cout << ans << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
inline long long read() {
long long nm = 0;
bool fh = true;
char cw = getchar();
for (; !isdigit(cw); cw = getchar()) fh ^= (cw == '-');
for (; isdigit(cw); cw = getchar()) nm = nm * 10 + (cw - '0');
return fh ? nm : -nm;
}
char ch[200200], nw[200200];
int n, x, w[200200];
inline void solve() {
scanf("%s", ch + 1), x = read(), n = strlen(ch + 1);
for (int i = 1; i <= n; i++) w[i] = 1;
for (int i = 1; i <= n; i++)
if (ch[i] == '0') {
if (i > x) w[i - x] = 0;
if (i + x <= n) w[i + x] = 0;
}
for (int i = 1; i <= n; i++) {
nw[i] = '0';
if (i > x && w[i - x] == 1) nw[i] = '1';
if (i + x <= n && w[i + x] == 1) nw[i] = '1';
}
for (int i = 1; i <= n; i++)
if (nw[i] ^ ch[i]) {
puts("-1");
return;
}
for (int i = 1; i <= n; i++) printf("%d", w[i]);
puts("");
}
int main() {
for (int Cas = read(); Cas; --Cas) solve();
return 0;
}
| ### Prompt
Generate a Cpp solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
inline long long read() {
long long nm = 0;
bool fh = true;
char cw = getchar();
for (; !isdigit(cw); cw = getchar()) fh ^= (cw == '-');
for (; isdigit(cw); cw = getchar()) nm = nm * 10 + (cw - '0');
return fh ? nm : -nm;
}
char ch[200200], nw[200200];
int n, x, w[200200];
inline void solve() {
scanf("%s", ch + 1), x = read(), n = strlen(ch + 1);
for (int i = 1; i <= n; i++) w[i] = 1;
for (int i = 1; i <= n; i++)
if (ch[i] == '0') {
if (i > x) w[i - x] = 0;
if (i + x <= n) w[i + x] = 0;
}
for (int i = 1; i <= n; i++) {
nw[i] = '0';
if (i > x && w[i - x] == 1) nw[i] = '1';
if (i + x <= n && w[i + x] == 1) nw[i] = '1';
}
for (int i = 1; i <= n; i++)
if (nw[i] ^ ch[i]) {
puts("-1");
return;
}
for (int i = 1; i <= n; i++) printf("%d", w[i]);
puts("");
}
int main() {
for (int Cas = read(); Cas; --Cas) solve();
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
string find(string s, int x) {
int i;
string st;
for (i = 0; i < s.length(); i++) {
if (i - x >= 0 && s[i - x] == '1')
st.push_back('1');
else if (i + x < s.length() && s[i + x] == '1')
st.push_back('1');
else
st.push_back('0');
}
return st;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t, i, x, k;
string str;
cin >> t;
while (t--) {
cin >> str;
cin >> x;
string str1;
for (i = 0; i < str.length(); i++) str1.push_back('1');
for (i = 0; i < str.length(); i++) {
if (str[i] == '0') {
if (i - x >= 0) str1[i - x] = '0';
if (i + x < str.length()) str1[i + x] = '0';
}
}
for (i = 0; i < str.length(); i++) {
if (str[i] == '0') {
if (i - x >= 0 && str1[i - x] == '1') {
break;
}
if (i + x < str1.length() && str1[i + x] == '1') break;
} else {
if (i - x >= 0 && str1[i - x] == '1') continue;
if (i + x < str1.length() && str1[i + x] == '1')
continue;
else
break;
}
}
if (i != str.length())
cout << -1 << '\n';
else
cout << str1 << '\n';
}
}
| ### Prompt
Develop a solution in cpp to the problem described below:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
string find(string s, int x) {
int i;
string st;
for (i = 0; i < s.length(); i++) {
if (i - x >= 0 && s[i - x] == '1')
st.push_back('1');
else if (i + x < s.length() && s[i + x] == '1')
st.push_back('1');
else
st.push_back('0');
}
return st;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t, i, x, k;
string str;
cin >> t;
while (t--) {
cin >> str;
cin >> x;
string str1;
for (i = 0; i < str.length(); i++) str1.push_back('1');
for (i = 0; i < str.length(); i++) {
if (str[i] == '0') {
if (i - x >= 0) str1[i - x] = '0';
if (i + x < str.length()) str1[i + x] = '0';
}
}
for (i = 0; i < str.length(); i++) {
if (str[i] == '0') {
if (i - x >= 0 && str1[i - x] == '1') {
break;
}
if (i + x < str1.length() && str1[i + x] == '1') break;
} else {
if (i - x >= 0 && str1[i - x] == '1') continue;
if (i + x < str1.length() && str1[i + x] == '1')
continue;
else
break;
}
}
if (i != str.length())
cout << -1 << '\n';
else
cout << str1 << '\n';
}
}
``` |
#include <bits/stdc++.h>
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
int main() {
std::ios_base::sync_with_stdio(false);
std::cin.tie(nullptr);
std::cout.tie(nullptr);
long long int t;
std::cin >> t;
while (t--) {
bool is_succes = true;
std::string s;
long long int x;
std::cin >> s >> x;
std::string w(s.size(), '1');
for (long long int i = 0; i < s.size(); ++i) {
if (s[i] == '0') {
if (i - x >= 0) w[i - x] = '0';
if (i + x <= s.size()) w[i + x] = '0';
}
}
for (long long int i = 0; i < s.size(); ++i) {
if (s[i] == '1') {
if ((i + x <= w.size() && w[i + x] == '1') ||
(i - x >= 0 && w[i - x] == '1'))
continue;
else {
std::cout << "-1\n";
is_succes = false;
break;
}
}
if (s[i] == '0') {
if ((i + x <= w.size() && w[i + x] == '1') ||
(i - x >= 0 && w[i - x] == '1')) {
std::cout << "-1\n";
is_succes = false;
break;
} else {
continue;
}
}
}
if (is_succes) std::cout << w << "\n";
}
return 0;
}
| ### Prompt
Create a solution in Cpp for the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
int main() {
std::ios_base::sync_with_stdio(false);
std::cin.tie(nullptr);
std::cout.tie(nullptr);
long long int t;
std::cin >> t;
while (t--) {
bool is_succes = true;
std::string s;
long long int x;
std::cin >> s >> x;
std::string w(s.size(), '1');
for (long long int i = 0; i < s.size(); ++i) {
if (s[i] == '0') {
if (i - x >= 0) w[i - x] = '0';
if (i + x <= s.size()) w[i + x] = '0';
}
}
for (long long int i = 0; i < s.size(); ++i) {
if (s[i] == '1') {
if ((i + x <= w.size() && w[i + x] == '1') ||
(i - x >= 0 && w[i - x] == '1'))
continue;
else {
std::cout << "-1\n";
is_succes = false;
break;
}
}
if (s[i] == '0') {
if ((i + x <= w.size() && w[i + x] == '1') ||
(i - x >= 0 && w[i - x] == '1')) {
std::cout << "-1\n";
is_succes = false;
break;
} else {
continue;
}
}
}
if (is_succes) std::cout << w << "\n";
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
long long t;
cin >> t;
while (t--) {
string s;
long long x;
cin >> s >> x;
string sol = "";
long long n = s.size();
for (int i = 0; i < n; i++) sol += "2";
bool possible = true;
for (int i = 0; possible and i < n; i++) {
if (i >= x and i + x < n) {
if (s[i] == '1') {
if (sol[i - x] == '0' and sol[i + x] == '0') possible = false;
if (sol[i - x] == '0') {
sol[i + x] = '1';
} else {
sol[i - x] = '1';
sol[i + x] = '3';
if (i + 2 * x >= n) sol[i + x] = '1';
}
} else {
if (sol[i - x] == '1' or sol[i + x] == '1') possible = false;
sol[i + x] = '0';
sol[i - x] = '0';
}
} else if (i >= x) {
if (s[i] == '1') {
if (sol[i - x] == '0') possible = false;
sol[i - x] = '1';
} else {
if (sol[i - x] == '1') possible = false;
sol[i - x] = '0';
}
} else if (i + x < n) {
if (s[i] == '1') {
if (sol[i + x] == '0') possible = false;
sol[i + x] = '1';
} else {
if (sol[i + x] == '1') possible = false;
sol[i + x] = '0';
}
} else {
if (sol[i] == '1' or s[i] == '1') possible = false;
sol[i] = '0';
}
}
if (possible)
cout << sol << '\n';
else
cout << "-1\n";
}
return 0;
}
| ### Prompt
Please create a solution in cpp to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
long long t;
cin >> t;
while (t--) {
string s;
long long x;
cin >> s >> x;
string sol = "";
long long n = s.size();
for (int i = 0; i < n; i++) sol += "2";
bool possible = true;
for (int i = 0; possible and i < n; i++) {
if (i >= x and i + x < n) {
if (s[i] == '1') {
if (sol[i - x] == '0' and sol[i + x] == '0') possible = false;
if (sol[i - x] == '0') {
sol[i + x] = '1';
} else {
sol[i - x] = '1';
sol[i + x] = '3';
if (i + 2 * x >= n) sol[i + x] = '1';
}
} else {
if (sol[i - x] == '1' or sol[i + x] == '1') possible = false;
sol[i + x] = '0';
sol[i - x] = '0';
}
} else if (i >= x) {
if (s[i] == '1') {
if (sol[i - x] == '0') possible = false;
sol[i - x] = '1';
} else {
if (sol[i - x] == '1') possible = false;
sol[i - x] = '0';
}
} else if (i + x < n) {
if (s[i] == '1') {
if (sol[i + x] == '0') possible = false;
sol[i + x] = '1';
} else {
if (sol[i + x] == '1') possible = false;
sol[i + x] = '0';
}
} else {
if (sol[i] == '1' or s[i] == '1') possible = false;
sol[i] = '0';
}
}
if (possible)
cout << sol << '\n';
else
cout << "-1\n";
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long int ll_max(long long int a, long long int b, long long int c) {
return max(a, max(b, c));
}
int int_max(int a, int b, int c) { return max(a, max(b, c)); }
long long int ll_min(long long int a, long long int b, long long int c) {
return min(a, min(b, c));
}
int int_min(int a, int b, int c) { return min(a, min(b, c)); }
long long int max(int a, long long int b) { return max((long long int)a, b); }
long long int min(int a, long long int b) { return min((long long int)a, b); }
long long int min(long long int a, int b) { return min(a, (long long int)b); }
long long int max(long long int a, int b) { return max(a, (long long int)b); }
long long int maximum(long long int a[], long long int n) {
long long int max1 = LLONG_MIN;
for (long long int i = 0; i < n; i++) {
max1 = max(max1, a[i]);
}
return max1;
}
long long int minimum(long long int a[], long long int n) {
long long int min1 = LLONG_MAX;
for (long long int i = 0; i < n; i++) {
min1 = min(min1, a[i]);
}
return min1;
}
long long int dx[] = {0, 0, 1, -1};
long long int dy[] = {1, -1, 0, 0};
long long int power(long long int x, long long int y, long long int m) {
if (y == 0) return 1;
long long int a = power(x, y / 2, m);
if (y % 2) {
return (a * ((a * x) % m)) % m;
} else {
return (a * a) % m;
}
}
long long int mod_inverse(long long int x, long long int m) {
return power(x, m - 2, m);
}
long long int fact(long long int n, long long int m) {
if (n <= 1) return 1;
return (fact(n - 1, m) * n) % m;
}
long long int ncr(long long int n, long long int r, long long int m) {
if (r > n) return 0;
long long int n1 = 1, d1 = 1, d2 = 1;
n1 = fact(n, m);
d1 = fact(r, m);
d2 = fact(n - r, m);
long long int ans = mod_inverse((d1 * d2) % m, m);
ans = (ans * n1) % m;
return ans;
}
int gcd(int a, int b) {
if (a == 0 || b == 0) return max(a, b);
if (a < b) return gcd(b, a);
if (a % b == 0) return b;
return gcd(b, a % b);
}
int ispal(string s) {
int len = s.size();
int flag = 1;
for (int i = 0; i < len; ++i) {
if (s[i] != s[len - i - 1]) {
flag = 0;
break;
}
}
return flag;
}
long long int sroot(long long int n, long long int low = 1,
long long int high = 1e9 + 1) {
if (low == high) return low;
if (low == high - 1) {
if (high * high <= n)
return high;
else
return low;
}
long long int mid = (low + high) / 2;
long long int a = mid * mid;
if (a > n)
return sroot(n, low, mid - 1);
else
return sroot(n, mid, high);
}
long long int croot(long long int n, long long int low = 1,
long long int high = 1e6 + 1) {
if (low == high) return low;
if (low == high - 1) {
if (high * high * high <= n)
return high;
else
return low;
}
long long int mid = (low + high) / 2;
long long int a = mid * mid * mid;
if (a > n)
return croot(n, low, mid - 1);
else
return croot(n, mid, high);
}
bool sortbysec(const pair<long long int, long long int> &a,
const pair<long long int, long long int> &b) {
return (a.second < b.second);
}
class solver {
public:
void solve() {
string s;
cin >> s;
long long int x;
cin >> x;
long long int n = s.length();
long long int dp1[n + 5], dp2[n + 5];
string w = "";
for (long long int i = 0; i < n; i++) w += "1";
long long int f = 0;
for (long long int i = 0; i < n; i++) {
if (s[i] == '0') {
long long int g = 0;
if (i - x >= 0) {
w[i - x] = '0';
}
if (i + x < n) {
w[i + x] = '0';
}
}
}
if (f == 0) {
for (long long int i = 0; i < n; i++) {
long long int g = 0;
if (s[i] == '1') {
if (i - x >= 0) {
if (w[i - x] == '1') g = 1;
}
if (i + x < n) {
if (w[i + x] == '1') g = 1;
}
if (g == 0) f = -1;
}
}
}
if (f == -1)
cout << "-1\n";
else
cout << w << "\n";
}
};
int main() {
std::ios::sync_with_stdio(false);
std::cin.tie(NULL);
std::cout.tie(NULL);
int test = 1, c = 1;
cin >> test;
while (test--) {
solver o;
o.solve();
}
}
| ### Prompt
Please create a solution in cpp to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long int ll_max(long long int a, long long int b, long long int c) {
return max(a, max(b, c));
}
int int_max(int a, int b, int c) { return max(a, max(b, c)); }
long long int ll_min(long long int a, long long int b, long long int c) {
return min(a, min(b, c));
}
int int_min(int a, int b, int c) { return min(a, min(b, c)); }
long long int max(int a, long long int b) { return max((long long int)a, b); }
long long int min(int a, long long int b) { return min((long long int)a, b); }
long long int min(long long int a, int b) { return min(a, (long long int)b); }
long long int max(long long int a, int b) { return max(a, (long long int)b); }
long long int maximum(long long int a[], long long int n) {
long long int max1 = LLONG_MIN;
for (long long int i = 0; i < n; i++) {
max1 = max(max1, a[i]);
}
return max1;
}
long long int minimum(long long int a[], long long int n) {
long long int min1 = LLONG_MAX;
for (long long int i = 0; i < n; i++) {
min1 = min(min1, a[i]);
}
return min1;
}
long long int dx[] = {0, 0, 1, -1};
long long int dy[] = {1, -1, 0, 0};
long long int power(long long int x, long long int y, long long int m) {
if (y == 0) return 1;
long long int a = power(x, y / 2, m);
if (y % 2) {
return (a * ((a * x) % m)) % m;
} else {
return (a * a) % m;
}
}
long long int mod_inverse(long long int x, long long int m) {
return power(x, m - 2, m);
}
long long int fact(long long int n, long long int m) {
if (n <= 1) return 1;
return (fact(n - 1, m) * n) % m;
}
long long int ncr(long long int n, long long int r, long long int m) {
if (r > n) return 0;
long long int n1 = 1, d1 = 1, d2 = 1;
n1 = fact(n, m);
d1 = fact(r, m);
d2 = fact(n - r, m);
long long int ans = mod_inverse((d1 * d2) % m, m);
ans = (ans * n1) % m;
return ans;
}
int gcd(int a, int b) {
if (a == 0 || b == 0) return max(a, b);
if (a < b) return gcd(b, a);
if (a % b == 0) return b;
return gcd(b, a % b);
}
int ispal(string s) {
int len = s.size();
int flag = 1;
for (int i = 0; i < len; ++i) {
if (s[i] != s[len - i - 1]) {
flag = 0;
break;
}
}
return flag;
}
long long int sroot(long long int n, long long int low = 1,
long long int high = 1e9 + 1) {
if (low == high) return low;
if (low == high - 1) {
if (high * high <= n)
return high;
else
return low;
}
long long int mid = (low + high) / 2;
long long int a = mid * mid;
if (a > n)
return sroot(n, low, mid - 1);
else
return sroot(n, mid, high);
}
long long int croot(long long int n, long long int low = 1,
long long int high = 1e6 + 1) {
if (low == high) return low;
if (low == high - 1) {
if (high * high * high <= n)
return high;
else
return low;
}
long long int mid = (low + high) / 2;
long long int a = mid * mid * mid;
if (a > n)
return croot(n, low, mid - 1);
else
return croot(n, mid, high);
}
bool sortbysec(const pair<long long int, long long int> &a,
const pair<long long int, long long int> &b) {
return (a.second < b.second);
}
class solver {
public:
void solve() {
string s;
cin >> s;
long long int x;
cin >> x;
long long int n = s.length();
long long int dp1[n + 5], dp2[n + 5];
string w = "";
for (long long int i = 0; i < n; i++) w += "1";
long long int f = 0;
for (long long int i = 0; i < n; i++) {
if (s[i] == '0') {
long long int g = 0;
if (i - x >= 0) {
w[i - x] = '0';
}
if (i + x < n) {
w[i + x] = '0';
}
}
}
if (f == 0) {
for (long long int i = 0; i < n; i++) {
long long int g = 0;
if (s[i] == '1') {
if (i - x >= 0) {
if (w[i - x] == '1') g = 1;
}
if (i + x < n) {
if (w[i + x] == '1') g = 1;
}
if (g == 0) f = -1;
}
}
}
if (f == -1)
cout << "-1\n";
else
cout << w << "\n";
}
};
int main() {
std::ios::sync_with_stdio(false);
std::cin.tie(NULL);
std::cout.tie(NULL);
int test = 1, c = 1;
cin >> test;
while (test--) {
solver o;
o.solve();
}
}
``` |
Subsets and Splits