output stringlengths 52 181k | instruction stringlengths 296 182k |
|---|---|
#include <bits/stdc++.h>
using namespace std;
std::mt19937 rng(
(int)std::chrono::steady_clock::now().time_since_epoch().count());
const double EPS = 1e-9;
const double PI = acos(-1);
const int MOD = 1000000007;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int n, a;
cout << 'a' << endl;
c... | ### Prompt
Create a solution in Cpp for the following problem:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a single spell s,... |
#include <bits/stdc++.h>
using namespace std;
double PI = 3.141592653589793238462643383279;
const double EPS = 1e-9;
const long long MOD = 1000000007;
const int inf = 1 << 30;
const long long linf = 1LL << 60;
int main() {
cin.tie(0);
ios::sync_with_stdio(false);
cout << "a" << endl;
int n;
cin >> n;
if (n ... | ### Prompt
Please formulate a Cpp solution to the following problem:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a single sp... |
#include <bits/stdc++.h>
using namespace std;
int f(string s) {
cout << s << endl;
int w;
cin >> w;
if (w == 0) exit(0);
return w;
}
int main() {
const int N = 300;
int st = f(string(N, 'a'));
int n = 2 * N - (st + f(string(N, 'b')));
string t(n, 'a');
int A = N - st, B = n - A;
st = B;
for (int... | ### Prompt
Generate a CPP solution to the following problem:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a single spell s, w... |
#include <bits/stdc++.h>
using namespace std;
using LL = long long;
int main() {
ios::sync_with_stdio(false);
int n;
cout << "a" << endl;
cin >> n;
if (n == 0) return 0;
string s;
for (int i = 0; i < n; ++i) s.push_back('b');
cout << s << endl;
int base;
cin >> base;
if (base == 0) return 0;
s =... | ### Prompt
Please formulate a cpp solution to the following problem:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a single sp... |
#include <bits/stdc++.h>
using namespace std;
char s[300];
int main() {
for (int i = 0; i < 300; i++) {
cout << 'a';
}
cout << endl;
int a;
cin >> a;
a = 300 - a;
int cnt = 0;
for (int i = 0; i < 300; i++) {
cout << 'b';
}
cout << endl;
int b;
cin >> b;
b = 300 - b;
for (int i = 0; i... | ### Prompt
Generate a cpp solution to the following problem:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a single spell s, w... |
#include <bits/stdc++.h>
using namespace std;
char rev(char c) { return (c == 'a' ? 'b' : 'a'); }
int query(const string s) {
cout << s << "\n";
fflush(stdout);
int ret;
cin >> ret;
if (!ret) exit(0);
return ret;
}
int main() {
int n = query("a") + 1;
string s = "a";
for (int i = 1; i < n; ++i) s += "... | ### Prompt
Your challenge is to write a cpp solution to the following problem:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a... |
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int a, b, n;
string x;
int main() {
for (int i = 0; i < 300; i++) x.push_back('a');
cout << x << '\n';
cout.flush();
cin >> a;
if (a == 0) return 0;
x = "";
for (int i = 0; i < 300; i++) x.push_back('b');
cout << x << '\n';
cout.flus... | ### Prompt
Create a solution in Cpp for the following problem:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a single spell s,... |
#include <bits/stdc++.h>
using namespace std;
signed main() {
long long v1, v2;
cout << "a" << '\n';
fflush(stdout);
cin >> v1;
string ans = "";
long long len;
len = v1 + 1;
if (len > 300) {
ans = "";
for (long long i = 0; i < v1; i++) {
ans += 'b';
}
cout << ans << '\n';
fflus... | ### Prompt
Please formulate a Cpp solution to the following problem:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a single sp... |
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
const long long inf = 1ll << 50;
const long long maxn = 100005;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout << "a" << endl;
long long x;
cin >> x;
if (x == 300) {
cout << string(300, 'b') << endl;
... | ### Prompt
Please create a solution in CPP to the following problem:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a single sp... |
#include <bits/stdc++.h>
using namespace std;
template <typename _T>
inline void _DBG(const char *s, _T x) {
cerr << s << " = " << x << "\n";
}
template <typename _T, typename... args>
void _DBG(const char *s, _T x, args... a) {
while (*s != ',') cerr << *s++;
cerr << " = " << x << ',';
_DBG(s + 1, a...);
}
int... | ### Prompt
Please formulate a CPP solution to the following problem:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a single sp... |
#include <bits/stdc++.h>
using namespace std;
int len = 0;
int t[505];
int solve() {
string s = "";
for (int i = 0; i < len; i++) {
if (t[i] == 0) {
s += 'a';
} else {
s += 'b';
}
}
cout << s << endl;
cout.flush();
int v;
cin >> v;
return v;
}
bool flag[505];
int vk[505];
int mai... | ### Prompt
Your challenge is to write a cpp solution to the following problem:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a... |
#include <bits/stdc++.h>
using namespace std;
using namespace std::chrono;
const long long int mod = 1000000007ll;
const long long int smod = 998244353ll;
const long long int inf = 10000000000000007ll;
const double eps = 1e-10;
struct less_key {
bool operator()(const std::pair<long long int, long long int> &left,
... | ### Prompt
In CPP, your task is to solve the following problem:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a single spell s... |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cout << "a\n";
cout.flush();
int len;
cin >> len;
if (len == 0) return 0;
char b[len + 1];
for (int i = 0; i < len; i++) b[i] = 'b';
b[len] = '\0';
int fin_len;
cout << b << "\... | ### Prompt
Generate a CPP solution to the following problem:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a single spell s, w... |
#include <bits/stdc++.h>
using namespace std;
class DebugStream {
} LOG;
template <typename T>
DebugStream &operator<<(DebugStream &s, const T &) {
return s;
}
mt19937 rnd(time(NULL));
const int maxN = 1e5 + 9, maxV = 1e6 + 9, MOD = 1e9 + 7, SQ = 335, lg = 20,
bs = 29;
int query(string st) {
cout << st <<... | ### Prompt
Create a solution in cpp for the following problem:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a single spell s,... |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cout.tie(0);
cin.tie(0);
cout << "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
"aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
"aaaaaaaaaaaaaaaaaaaaaaaaaaaa... | ### Prompt
Your task is to create a CPP solution to the following problem:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a sin... |
#include <bits/stdc++.h>
using namespace std;
int N, A, B, C[305];
string s;
int query(string s) {
printf("%s\n", s.c_str());
fflush(stdout);
int a;
scanf("%d", &a);
if (a == 0 || a == -1) exit(0);
return a;
}
void cal() {
if (A == 0 || B == 0) {
char c = (B == 0) ? 'a' : 'b';
for (int i = 0; i < ... | ### Prompt
Please create a solution in Cpp to the following problem:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a single sp... |
#include <bits/stdc++.h>
using namespace ::std;
const long double PI = acos(-1);
const long long MOD = 1000000000 + 7;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long add(long long a, long long b, long long m = MOD) {
if (a >= m) a %= m;
if (b >= m) b %= m;
if (a < 0) a +=... | ### Prompt
In cpp, your task is to solve the following problem:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a single spell s... |
#include <bits/stdc++.h>
using namespace std;
char s[1000], ans[1000];
namespace judger {
char s[3000], buf[300];
int cnt;
int n;
void init(int _n) {
n = _n;
for (int i = 0; i < n; ++i) s[i] = rand() % 2 + 'a';
printf("spell: %s\n", s);
}
} // namespace judger
inline void submit(char s[]) {
printf("%s\n", s);
... | ### Prompt
Develop a solution in CPP to the problem described below:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a single sp... |
#include <bits/stdc++.h>
using namespace std;
const long long INF(0x3f3f3f3f3f3f3f3fll);
const long long inf(0x3f3f3f3f);
template <typename T>
void read(T &res) {
bool flag = false;
char ch;
while (!isdigit(ch = getchar())) (ch == '-') && (flag = true);
for (res = ch - 48; isdigit(ch = getchar());
res =... | ### Prompt
Please create a solution in Cpp to the following problem:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a single sp... |
#include <bits/stdc++.h>
using namespace std;
using ld = long double;
using ll = long long;
using vi = vector<int>;
using pii = pair<int, int>;
const int MX = 1e5;
int query(const string& t) {
cout << t << '\n';
cout.flush();
int d;
cin >> d;
if (d == 0) exit(0);
return d;
}
int main() {
ios_base::sync_wi... | ### Prompt
Construct a Cpp code solution to the problem outlined:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a single spell... |
#include <bits/stdc++.h>
using namespace std;
int f(string s) {
cout << s << endl;
int r;
cin >> r;
if (r == 0) exit(0);
return r;
}
void solve() {
int qa = f(string(300, 'a'));
int qb = f(string(300, 'b'));
int A = 300 - qa;
int B = 300 - qb;
int last = B;
int l = A + B;
string ans(l, 'a');
f... | ### Prompt
Your task is to create a cpp solution to the following problem:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a sin... |
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
vector<int> v;
int main() {
int tota, totb, n;
cout << string(300, 'a') << endl;
fflush(stdout);
cin >> tota;
if (!tota) {
return 0;
}
tota = 300 - tota;
cout << string(300, 'b') << endl;
fflush(stdout);
cin >> totb;
if ... | ### Prompt
Please provide a cpp coded solution to the problem described below:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a... |
#include <bits/stdc++.h>
using namespace std;
const long long INF = (long long)1e18;
const int mod = (int)1e9 + 7;
const int MAXN = (int)3e5 + 5;
long long n;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
string a, b;
for (int i = 0; i < 300; i++) {
a += 'a';
b +=... | ### Prompt
In cpp, your task is to solve the following problem:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a single spell s... |
#include <bits/stdc++.h>
using namespace std;
int main() {
string aaa = "";
string bbb = "";
for (int i = 0; i < 300; i++) {
aaa += 'a';
bbb += 'b';
}
int s, t;
cout << aaa << endl;
cout.flush();
cin >> s;
if (s == 0) return 0;
cout << bbb << endl;
cout.flush();
cin >> t;
if (t == 0) r... | ### Prompt
Please provide a cpp coded solution to the problem described below:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a... |
#include <bits/stdc++.h>
const long long N = 1e6;
using namespace std;
int query(string s) {
cout << s << endl;
fflush(stdout);
int n;
cin >> n;
return n;
}
string gena(int n) {
string s;
for (int i = 0; i < n; i++) s += 'a';
return s;
}
string genb(int n) {
string s;
for (int i = 0; i < n; i++) s +... | ### Prompt
Your challenge is to write a cpp solution to the following problem:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a... |
#include <bits/stdc++.h>
using namespace std;
int q(string s) {
int x;
cout << s << endl;
cin >> x;
if (!x || x == -1)
exit(0);
else
return x;
}
int main() {
int l = q("a");
string s(l, 'b');
int best = q(s);
++l, s += 'b';
for (int i = 0; i < l; i++) {
s[i] = 'a';
int now = q(s);
... | ### Prompt
Your task is to create a Cpp solution to the following problem:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a sin... |
#include <bits/stdc++.h>
using namespace std;
int cnta, cntb, x, n, cntfind;
bool isb[310];
int main() {
for (int i = 1; i <= 300; i++) printf("a");
puts("");
fflush(stdout);
scanf("%d", &x);
if (x == 0) {
return 0;
}
cnta = 300 - x;
for (int i = 1; i <= 300; i++) printf("b");
puts("");
fflush(s... | ### Prompt
Please provide a cpp coded solution to the problem described below:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a... |
#include <bits/stdc++.h>
using namespace std;
int main() {
int na, p = 0, t1, t2;
string s(300, 'a');
cout << s << endl;
cin >> na;
if (na == 0) return 0;
na = 300 - na;
s.assign(na, 'a');
if (na == 0) s = "b";
cout << s << endl;
cin >> t1;
if (t1 == 0) return 0;
while (1) {
string s0 = s.su... | ### Prompt
Please formulate a cpp solution to the following problem:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a single sp... |
#include <bits/stdc++.h>
using namespace std;
long long power(long long x, long long y, long long md = 1000000007) {
long long res = 1;
x %= md;
while (y > 0) {
if (y & 1) res = (res * x) % md;
x = (x * x) % md;
y = y >> 1;
}
return res % md;
}
struct node {
vector<long long> v, w;
};
long long ... | ### Prompt
Generate a CPP solution to the following problem:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a single spell s, w... |
#include <bits/stdc++.h>
using namespace std;
signed main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
cout << "a" << endl;
int len;
cin >> len;
string s = "a", opt = "a";
for (int i = 0; i < len; i++) s += "a", opt += "a";
if (len == 300) {
for (int i = 0; i < 300; i++) c... | ### Prompt
Create a solution in cpp for the following problem:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a single spell s,... |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(NULL);
string a = "";
string b = "";
for (int i = 0; i < 300; i++) {
a += 'a';
b += 'b';
}
cout << a << endl;
int q1;
cin >> q1;
if (q1 == 0) return 0;
cout << b << endl;
int q2;
cin >> q2;... | ### Prompt
In CPP, your task is to solve the following problem:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a single spell s... |
#include <bits/stdc++.h>
using namespace std;
int main() {
int prev, cur;
string a = "", b = "";
for (int i = 0; i < 300; i++) a += 'a', b += 'b';
cout << a << endl;
cin >> prev;
if (prev == 0) return 0;
cout << b << endl;
cin >> cur;
if (cur == 0) return 0;
int l = 600 - prev - cur;
string s = ""... | ### Prompt
Your challenge is to write a Cpp solution to the following problem:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a... |
#include <bits/stdc++.h>
using namespace std;
int main() {
int ac, bc, i, pd, d, n;
map<char, char> m;
map<char, int> map;
map['a'] = m['b'] = 0;
m['a'] = 'b';
m['b'] = 'a';
string t(300, 'a');
cout << t << "\n";
fflush(stdout);
cin >> ac;
if (ac == 0) return 0;
ac = 300 - ac;
if (ac == 0) {
... | ### Prompt
Create a solution in CPP for the following problem:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a single spell s,... |
#include <bits/stdc++.h>
using namespace std;
void out(string s) {
cout << s << endl;
cout.flush();
}
int getdist(string i) {
out(i);
int res;
scanf("%d", &res);
if (res == 0) exit(0);
return res;
}
string gets(int pos, int length) {
string s = "";
for (int i = (0); i < (length); i++) {
if (i == p... | ### Prompt
Generate a cpp solution to the following problem:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a single spell s, w... |
#include <bits/stdc++.h>
using namespace std;
int n, ca = 0, cb = 0;
int main() {
for (int i = (1), _x = (300); i <= _x; i++) cout << "a";
cout << endl;
cin >> ca;
if (ca == 0) return 0;
ca = 300 - ca;
for (int i = (1), _x = (300); i <= _x; i++) cout << "b";
cout << endl;
cin >> cb;
if (cb == 0) retur... | ### Prompt
Generate a CPP solution to the following problem:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a single spell s, w... |
#include <bits/stdc++.h>
using namespace std;
template <typename T1, typename T2>
inline void minaj(T1 &x, T2 y) {
x = (x > y ? y : x);
}
template <typename T1, typename T2>
inline void maxaj(T1 &x, T2 y) {
x = (x < y ? y : x);
}
const int mod = 1e9 + 7;
int add(int x, int y) {
x += y;
if (x >= mod) return x - ... | ### Prompt
Please create a solution in Cpp to the following problem:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a single sp... |
#include <bits/stdc++.h>
using namespace std;
int ask(string s) {
for (int i = 0; i < s.length(); i++) printf("%c", s[i]);
printf("\n");
fflush(stdout);
int x;
scanf("%d", &x);
if (x == 0) exit(0);
return x;
}
int main() {
int a = ask("a");
int n = a + 1;
n = min(n, 300);
string s = "";
for (int... | ### Prompt
Create a solution in CPP for the following problem:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a single spell s,... |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
int ans1, ans2, tam;
cout << 'a' << endl;
cin >> ans1;
for (int i = 0; i < ans1; i++) cout << 'b';
cout << endl;
cin >> ans2;
if (ans2 == 0) {
for (int i = 0; i < ans1; i++) cout << 'b';
cout << endl;
} els... | ### Prompt
In Cpp, your task is to solve the following problem:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a single spell s... |
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
using ll = long long;
const char E = '\n';
const int N = 305;
const ll mod = 1e9 + 7;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int a, b, n;
for (int i = 1; i <= 300; i++) {
cout << 'a';
}
cout << endl;
... | ### Prompt
Construct a CPP code solution to the problem outlined:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a single spell... |
#include <bits/stdc++.h>
char s[1100];
int n, x, minx;
int main() {
printf("a\n");
fflush(stdout);
scanf("%d", &x);
if (x == 0) return 0;
if (x == 300) {
for (int i = 1; i <= 300; i++) putchar('b');
putchar('\n');
fflush(stdout);
return 0;
}
n = x + 1;
for (int i = 1; i <= n; i++) putcha... | ### Prompt
Develop a solution in Cpp to the problem described below:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a single sp... |
#include <bits/stdc++.h>
using namespace std;
template <typename T>
static ostream& operator<<(ostream& stream, const vector<T>& t);
template <typename T>
static istream& read(T, T, istream& = cin);
int test(string s) {
assert(s.back() != ' ');
cout << s << endl;
int ans;
cin >> ans;
if (ans == 0) exit(0);
... | ### Prompt
Create a solution in Cpp for the following problem:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a single spell s,... |
#include <bits/stdc++.h>
using namespace std;
template <class T1, class T2>
ostream& operator<<(ostream& out, pair<T1, T2> p) {
out << '(' << p.first << ',' << p.second << ')';
return out;
}
template <class T1, class T2>
istream& operator>>(istream& in, pair<T1, T2>& p) {
in >> p.first >> p.second;
return in;
}... | ### Prompt
Create a solution in Cpp for the following problem:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a single spell s,... |
#include <bits/stdc++.h>
using namespace std;
const int dx[] = {-1, 0, 1, 0};
const int dy[] = {0, -1, 0, 1};
inline int read() {
int ans = 0;
bool f = 1;
char ch = getchar();
while (!isdigit(ch)) f ^= ch == '-', ch = getchar();
while (isdigit(ch)) ans = ((ans << 2) + ans << 1) + (ch ^ 48), ch = getchar();
... | ### Prompt
Please provide a cpp coded solution to the problem described below:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a... |
#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
const long long INF = mod * mod;
const long double eps = 1e-6;
const long double pi = acos(-1.0);
int query(string s) {
cout << s << endl;
int x;
cin >> x;
return x;
}
void solve() {
int z = query("a");
if (z == 0) return;
if (... | ### Prompt
Please provide a cpp coded solution to the problem described below:
This is an interactive problem.
After completing the last level of the enchanted temple, you received a powerful artifact of the 255th level. Do not rush to celebrate, because this artifact has a powerful rune that can be destroyed with a... |
#include <bits/stdc++.h>
using namespace std;
long long int* getLPS(string& p) {
long long int len = p.size();
long long int* LPS = new long long int[len];
long long int i = 1, j = 0;
LPS[0] = 0;
while (i < len) {
if (p[i] == p[j]) {
LPS[i] = j + 1;
i++;
j++;
} else {
if (j != ... | ### Prompt
Your challenge is to write a cpp solution to the following problem:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting ... |
#include <bits/stdc++.h>
using std::bitset;
using std::cerr;
using std::cin;
using std::cout;
using std::deque;
using std::endl;
using std::fill;
using std::fixed;
using std::ios_base;
using std::iterator;
using std::map;
using std::max_element;
using std::min_element;
using std::multimap;
using std::multiset;
using st... | ### Prompt
Generate a cpp solution to the following problem:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of lowercase Engli... |
#include <bits/stdc++.h>
using namespace std;
constexpr int NMAX = 1000010;
char strng[NMAX], interp[NMAX * 2];
char restr1[NMAX], restr2[NMAX];
int palSpan[NMAX * 2];
void solve(char *strng, char *result) {
int n = strlen(strng);
for (int i = 0; i < 2 * n; i += 2) {
interp[i] = strng[i / 2];
if (i + 1 < 2 ... | ### Prompt
Your challenge is to write a CPP solution to the following problem:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting ... |
#include <bits/stdc++.h>
using namespace std;
vector<int> manacher(string s) {
string T = "#";
for (char c : s) T += c, T += '#';
vector<int> P((T).size());
int c = 0, R = 0;
for (auto(i) = (1); (i) < ((T).size() - 1); ++(i)) {
if (R > i) P[i] = min(R - i, P[2 * c - i]);
for (int r = i + 1 + P[i], l =... | ### Prompt
Please provide a cpp coded solution to the problem described below:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting ... |
#include <bits/stdc++.h>
using namespace std;
long long pw(long long b, long long r, long long md) {
long long ans = 1;
while (r) {
if (r & 1) ans = (ans * b) % md;
b = (b * b) % md;
r >>= 1;
}
return ans;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long t;
c... | ### Prompt
Please provide a Cpp coded solution to the problem described below:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting ... |
#include <bits/stdc++.h>
using namespace std;
const int infi = 1e9;
const long long int infl = 1e18;
const int MOD = 1e9 + 7;
long long int log_2(long long int n) {
return (n > 1ll) ? 1ll + log_2(n / 2) : 0ll;
}
inline void add(long long int &a, long long int b) {
a += b;
if (a >= MOD) a -= MOD;
}
inline void sub... | ### Prompt
Develop a solution in CPP to the problem described below:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of lowerca... |
#include <bits/stdc++.h>
using namespace std;
const long long maxn = 2e6 + 5;
long long T, n, d[maxn];
char s[maxn], t[maxn];
vector<long long> out;
int main() {
scanf("%lld", &T);
while (T--) {
scanf("%s", s + 1);
n = strlen(s + 1);
long long L = 1, R = n;
out.clear();
while (R - L + 1 > 1 && s... | ### Prompt
Your task is to create a cpp solution to the following problem:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of l... |
#include <bits/stdc++.h>
using namespace std;
vector<long long> manacher1(string s) {
long long n = s.size();
vector<long long> d1(n);
for (long long i = 0, l = 0, r = -1; i < n; i++) {
long long k = (i > r) ? 1 : min(d1[l + r - i], r - i + 1);
while (0 <= i - k && i + k < n && s[i - k] == s[i + k]) {
... | ### Prompt
Please create a solution in cpp to the following problem:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of lowerca... |
#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e6 + 5;
const int inf = 2e9;
char ma[maxn << 1];
int mp[maxn << 1];
inline void manacher(string &s, int len) {
int p = 0, r = 0, mid = 0;
ma[p++] = '$', ma[p++] = '#';
for (int i = 0; i < len; i++) ma[p++] = s[i], ma[p++] = '#';
for (int i = 0; i <... | ### Prompt
Construct a cpp code solution to the problem outlined:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of lowercase ... |
#include <bits/stdc++.h>
using namespace std;
long long lpp(const string &s) {
string kmprev = s;
std::reverse(kmprev.begin(), kmprev.end());
string kmp = s + "#" + kmprev;
vector<long long> lps(kmp.size(), 0);
for (long long i = 1; i < (long long)lps.size(); ++i) {
long long prev_idx = lps[i - 1];
wh... | ### Prompt
Develop a solution in cpp to the problem described below:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of lowerca... |
#include <bits/stdc++.h>
using namespace std;
string solve_palindrome(const string& s) {
string a = s;
reverse(a.begin(), a.end());
a = s + "#" + a;
int c, pref[a.size() + 1];
memset(pref, 0, sizeof(pref));
c = 0;
for (int i = 1; i < a.size(); i++) {
while (c != 0 && a[c] != a[i]) c = pref[c - 1];
... | ### Prompt
Develop a solution in Cpp to the problem described below:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of lowerca... |
#include <bits/stdc++.h>
using namespace std;
bool pal(string s) {
long long j = s.size() - 1;
long long sw = 0;
for (long long i = 0; i < s.size() / 2; i++) {
if (s[i] != s[j]) {
sw = 1;
break;
}
j--;
}
if (sw)
return false;
else
return true;
}
string man(string x) {
strin... | ### Prompt
Generate a cpp solution to the following problem:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of lowercase Engli... |
#include <bits/stdc++.h>
using namespace std;
const long long p = 53;
const long long mod = 2000000011;
map<long long, set<long long> > ind;
signed main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
long long q;
cin >> q;
for (long long i = 0; i < q; ++i) {
string s;
cin >> s;
long long l = 0, r =... | ### Prompt
Create a solution in Cpp for the following problem:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of lowercase Eng... |
#include <bits/stdc++.h>
using namespace std;
int check(vector<int> a) {
int ans = 1;
for (int i = 0; i < a.size(); i++) {
if (i + 1 == a[i]) ans = max(ans, a[i] - 1);
}
return ans;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(NULL);
cout.tie(NULL);
int t;
cin >> t;
while (t--) {
str... | ### Prompt
Generate a cpp solution to the following problem:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of lowercase Engli... |
#include <bits/stdc++.h>
using namespace std;
vector<int> manacher(string s) {
string T = "#";
for (char c : s) T += c, T += '#';
vector<int> P((T).size());
int c = 0, r = 0;
for (auto(i) = (1); (i) < ((T).size() + 1); ++(i)) {
if (r > i) {
P[i] = min(r - i, P[2 * c - i]);
}
while (i + 1 + P... | ### Prompt
Please provide a cpp coded solution to the problem described below:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting ... |
#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) { cer... | ### Prompt
Create a solution in cpp for the following problem:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of lowercase Eng... |
#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
const long long max_n = 1e5 + 5;
const long long max_c = 1e4 + 5;
const long long inf = 1e9 + 7;
string solve(string &s) {
long long n = s.size();
vector<bool> is_good(n + 1, 0);
is_good[0] = 1;
for (long long i = 0; i < n; ++i) {
... | ### Prompt
Please create a solution in cpp to the following problem:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of lowerca... |
#include <bits/stdc++.h>
using namespace std;
template <class T>
ostream& operator<<(ostream& cout, deque<T> V) {
cout << "[ ";
for (auto v : V) cout << v << ' ';
return cout << ']';
}
template <class T>
ostream& operator<<(ostream& cout, vector<T> V) {
cout << "[ ";
for (auto v : V) cout << v << ' ';
retur... | ### Prompt
In cpp, your task is to solve the following problem:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of lowercase En... |
#include <bits/stdc++.h>
using namespace std;
long long int* getLPS(string& p) {
long long int len = p.size();
long long int* LPS = new long long int[len];
long long int i = 1, j = 0;
LPS[0] = 0;
while (i < len) {
if (p[i] == p[j]) {
LPS[i] = j + 1;
i++;
j++;
} else {
if (j != ... | ### Prompt
Construct a cpp code solution to the problem outlined:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of lowercase ... |
#include <bits/stdc++.h>
using namespace std;
int f[2000010];
int prefix(string s) {
s += '#' + string(s.rbegin(), s.rend());
int k = 0;
for (int i = 1; i < s.size(); i++) {
while (k && s[i] != s[k]) k = f[k];
if (s[i] == s[k]) k++;
f[i + 1] = k;
}
return k;
}
int main() {
ios_base::sync_with_st... | ### Prompt
Generate a Cpp solution to the following problem:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of lowercase Engli... |
#include <bits/stdc++.h>
using namespace std;
int T, ans, mid, r, az;
int b[2000005];
string s;
string ycl(string t, int x) {
string res = "@#";
for (int i = x; i < (int)t.length() - x; i++) res += t[i], res += '#';
return res;
}
int main() {
scanf("%d", &T);
while (T--) {
string t = "";
cin >> s;
... | ### Prompt
Generate a Cpp solution to the following problem:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of lowercase Engli... |
#include <bits/stdc++.h>
using namespace std;
template <class T>
void printvec(const vector<T>& v) {
for (auto x : v) {
cout << x << " ";
}
cout << endl;
}
template <class T>
void printtree(const vector<vector<T> >& tree) {
for (long long i = 0; i < tree.size(); ++i) {
cout << i + 1 << ": ";
printve... | ### Prompt
Develop a solution in CPP to the problem described below:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of lowerca... |
#include <bits/stdc++.h>
#pragma GCC optimize(2)
using namespace std;
const long long mod = 1e9 + 7;
const long long maxn = 2e6 + 10;
long long p[2] = {1000000007, 998244353};
long long seed[2] = {500000004, 500000004};
int tt, n, k;
string s;
char t[maxn * 2];
int pos = 0, maxright = 0, maxlen = 0, r[maxn * 2], mxlen;... | ### Prompt
Please provide a Cpp coded solution to the problem described below:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting ... |
#include <bits/stdc++.h>
using namespace std;
int n, m, cas, p[1000005 << 1];
char rS[1000005], S[1000005], T[1000005 << 1];
bool f[1000005];
inline void Build(char S[], char T[]) {
m = strlen(S + 1);
for (int i = 1; i <= m; i++) rS[i] = S[m - i + 1];
T[0] = '$';
T[1] = '#';
for (int i = 1; i <= m; i++) T[i <... | ### Prompt
Generate a Cpp solution to the following problem:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of lowercase Engli... |
#include <bits/stdc++.h>
using namespace std;
long long int mod = 1000000007;
int zalgo(string g) {
int l = 0, r = 0;
string res = g;
reverse(res.begin(), res.end());
string t = g + "#" + res;
int l2 = t.length();
vector<int> z(t.size(), 0);
for (int i = 1; i < l2; i++) {
if (i > r) {
l = i;
... | ### Prompt
Your task is to create a CPP solution to the following problem:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of l... |
#include <bits/stdc++.h>
using namespace std;
void pankajgaur_98() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
long long lps[2000005];
long long get_lps(string& s) {
long long l = s.length();
lps[0] = 0;
long long i = 1, j = 0;
while (i < l) {
if (s.at(i) == s.at(j)) {
lps[i] = j ... | ### Prompt
Generate a Cpp solution to the following problem:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of lowercase Engli... |
#include <bits/stdc++.h>
using namespace std;
using db = double;
using ll = long long;
using uint = unsigned int;
using pii = pair<int, int>;
template <class T>
void cmax(T& x, const T& y) {
if (x < y) x = y;
}
template <class T>
void cmin(T& x, const T& y) {
if (x > y) x = y;
}
template <class T>
void sort(vector<... | ### Prompt
In Cpp, your task is to solve the following problem:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of lowercase En... |
#include <bits/stdc++.h>
using namespace std;
int n, len[2000006];
char A[2000006], s[2000006];
void manacher(int n, char *s) {
int pos = 0, mx = 0;
for (int i = n; i; i--) s[i << 1] = s[i], s[(i << 1) - 1] = '#';
s[0] = '$', s[n = n << 1 | 1] = '#', s[n + 1] = '@';
for (int i = 1; i <= n; i++) {
len[i] = m... | ### Prompt
Please provide a Cpp coded solution to the problem described below:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting ... |
#include <bits/stdc++.h>
long long int M = 1000000007;
using namespace std;
long long int pi[2000005];
string kmp(string s) {
string a = s;
reverse(a.begin(), a.end());
a = s + "#" + a;
long long int n = a.size(), j = 0;
for (long long int i = 1; i < n; i++) {
j = pi[i - 1];
while (j > 0 && a[i] != a[... | ### Prompt
Please create a solution in cpp to the following problem:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of lowerca... |
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1000005;
int d1[MAXN];
int d2[MAXN];
void manacher(string& s) {
int l = 0, r = -1, n = s.size();
for (int i = 0, colchin = n; i < colchin; ++i) {
int k = i > r ? 1 : min(d1[l + r - i], r - i);
while (i + k < n && i - k >= 0 && s[i + k] == s[i - ... | ### Prompt
Your task is to create a CPP solution to the following problem:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of l... |
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
template <typename T, typename TT>
inline ostream &operator<<(ostream &os, const pair<T, TT> &t) {
return os << t.first << " " << t.second;
}
template <typename T>
inline ostream &operator<<(ostream &os, const vector<T> &t) {
for (auto i : ... | ### Prompt
Your task is to create a Cpp solution to the following problem:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of l... |
#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) { return !b ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
string get_lps(string s) {
string rs = s;
reverse(rs.begin(), rs.end());
s = s + '#' + rs;
int n = s.length();
vector... | ### Prompt
Your challenge is to write a CPP solution to the following problem:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting ... |
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pli = pair<ll, int>;
using ld = long double;
const int MN = 1e6 + 3, LOG = 18, MOD = 998244353;
char s[MN];
int main() {
int t;
scanf("%d", &t);
while (t-... | ### Prompt
Create a solution in Cpp for the following problem:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of lowercase Eng... |
#include <bits/stdc++.h>
using namespace std;
inline long long gcd(long long a, long long b) {
while (b != 0) {
long long c = a % b;
a = b;
b = c;
}
return a < 0 ? -a : a;
}
inline long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
inline long long lowbit(long long x) { return x &... | ### Prompt
Please create a solution in CPP to the following problem:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of lowerca... |
#include <bits/stdc++.h>
using namespace std;
mt19937 rng(1213);
const int maxn = 1e6 + 10, MOD = 1e9 + 7;
string s;
long long prefix[maxn], suffix[maxn], power[maxn];
long long exp(long long int base, long long exponent) {
long long res = 1;
while (exponent) {
if (exponent & 1) res = (base * res) % MOD;
ex... | ### Prompt
Your task is to create a Cpp solution to the following problem:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of l... |
#include <bits/stdc++.h>
using namespace std;
const int M = (int)(2e6 + 239);
int pref[M], c;
string solve_pal(const string& s) {
string a = s;
reverse(a.begin(), a.end());
a = s + "#" + a;
c = 0;
for (int i = 1; i < (int)a.size(); i++) {
while (c != 0 && a[c] != a[i]) c = pref[c - 1];
if (a[c] == a[i... | ### Prompt
Please formulate a cpp solution to the following problem:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of lowerca... |
#include <bits/stdc++.h>
using namespace std;
const int LETTER = 26;
static int C = 13331;
template <int M, class B>
struct A {
int x;
B b;
A(int x = 0) : x(x), b(x) {}
A(int x, B b) : x(x), b(b) {}
A operator+(A o) {
int y = x + o.x;
return {y - (y >= M) * M, b + o.b};
}
A operator-(A o) {
in... | ### Prompt
Develop a solution in cpp to the problem described below:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of lowerca... |
#include <bits/stdc++.h>
using namespace std;
inline void Boost() {
ios::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
}
const int NMax = 1e6 + 50;
string convert(const string &s) {
string ret = "@";
for (int i = 0; i < (int)s.size(); ++i) {
ret += "#" + s.substr(i, 1);
}
ret += "#$";
retur... | ### Prompt
In CPP, your task is to solve the following problem:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of lowercase En... |
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using pr = pair<int, int>;
using prll = pair<long, long>;
using V = vector<int>;
using Vll = vector<long long>;
template <typename T, typename TT>
pr operator+(pair<T, TT> a, pair<T, TT> b) {
return {a.first + b.first, a.secon... | ### Prompt
Please provide a cpp coded solution to the problem described below:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting ... |
#include <bits/stdc++.h>
using namespace std;
inline long long gcd(long long a, long long b) {
while (b != 0) {
long long c = a % b;
a = b;
b = c;
}
return a < 0 ? -a : a;
}
inline long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
inline long long lowbit(long long x) { return x &... | ### Prompt
Create a solution in Cpp for the following problem:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of lowercase Eng... |
#include <bits/stdc++.h>
using namespace std;
bool is(long long l, long long r, string& s) {
for (long long i = l; i < l + (r - l) / 2; i++) {
if (s[i] != s[r + l - i - 1]) {
return 0;
}
}
return 1;
}
vector<long long> v = {31, 73, 239, 43},
u = {1000000007, 998244353, 1000000009};... | ### Prompt
Please create a solution in cpp to the following problem:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of lowerca... |
#include <bits/stdc++.h>
using namespace std;
const long long p1 = 1000000007, p2 = 1000000009;
int main() {
int t;
scanf("%d", &t);
while (t--) {
char s[1000010];
scanf("%s", s);
int len = strlen(s);
int cont = 0;
while ((s[cont] == s[len - cont - 1]) && (cont < len - cont - 1)) cont++;
i... | ### Prompt
Your challenge is to write a CPP solution to the following problem:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting ... |
#include <bits/stdc++.h>
using namespace std;
string pal(string s) {
string t = s;
reverse(t.begin(), t.end());
s = s + "#" + t;
long long int l = s.length();
long long int a[l + 1], c = 0;
a[0] = 0;
for (int i = 1; i < l; i++) {
while (c != 0 && s[c] != s[i]) c = a[c - 1];
if (s[c] == s[i]) c++;
... | ### Prompt
Please create a solution in cpp to the following problem:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of lowerca... |
#include <bits/stdc++.h>
using namespace std;
long long int gcd(long long int a, long long int b) {
if (a == 0) return b;
return gcd(b % a, a);
}
int bin(long int a[], long int low, long int high, long d, long day[]) {
if (low > high) return -1;
long mid = (low + high) / 2;
if (a[mid] > (a[low] + d) && (a[mid... | ### Prompt
Your challenge is to write a CPP solution to the following problem:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting ... |
#include <bits/stdc++.h>
using namespace std;
string s;
int p[2000005];
int Manachers(string s) {
if (s == "") return 0;
string ss = "";
ss += '$';
int cnt = 0;
for (int i = 0; i < s.length(); i++) {
ss += s[i];
ss += '#';
}
int len = ss.length();
ss[len - 1] = '%';
p[0] = 0;
int mx = 0, id ... | ### Prompt
Construct a CPP code solution to the problem outlined:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of lowercase ... |
#include <bits/stdc++.h>
using namespace std;
int longest_palindrome_prefix(const string &s) {
string kmprev = s;
std::reverse(kmprev.begin(), kmprev.end());
string kmp = s + "#" + kmprev;
vector<int> lps(kmp.size(), 0);
for (int i = 1; i < (int)lps.size(); ++i) {
int prev_idx = lps[i - 1];
while (pre... | ### Prompt
Develop a solution in cpp to the problem described below:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of lowerca... |
#include <bits/stdc++.h>
using namespace std;
long long spf[1000005];
vector<long long> primes, prev[1000005], next[1000005];
void sieve() {
for (long long i = 2; i < 1000005; ++i) {
if (spf[i] == 0) {
spf[i] = i;
for (long long j = 2 * i; j <= 1000005; j += i) {
if (spf[j] == 0) spf[j] = i;
... | ### Prompt
Please create a solution in Cpp to the following problem:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of lowerca... |
#include <bits/stdc++.h>
int next[2000009] = {-1};
std::string find_palindrome(std::string &s) {
std::string temp_s = s;
std::reverse(temp_s.begin(), temp_s.end());
temp_s += ('#' + s);
int i = 0, j = -1;
int len = temp_s.size();
while (i < len) {
if (j != -1 && temp_s[i] != temp_s[j]) {
j = next[... | ### Prompt
Generate a cpp solution to the following problem:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of lowercase Engli... |
#include <bits/stdc++.h>
#pragma GCC diagnostic ignored "-Wunused-variable"
#pragma GCC diagnostic ignored "-Wunused-parameter"
#pragma GCC diagnostic ignored "-Wunused-but-set-variable"
using namespace std;
char ch[1000006];
struct Manachar {
int n;
vector<int> P, Common, indices;
Manachar(int _n) {
n = _n;
... | ### Prompt
Your challenge is to write a CPP solution to the following problem:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting ... |
#include <bits/stdc++.h>
using namespace std;
long long int power(long long int x, long long int y) {
if (y == 0) return 1;
long long int temp = power(x, y / 2);
temp = (temp * temp) % 1000000007;
if (y % 2) {
return (temp * x) % 1000000007;
}
return temp;
}
long long int modinv(long long int x) { retur... | ### Prompt
Generate a cpp solution to the following problem:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of lowercase Engli... |
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
long long H[N], H2[N], x = 2463564, p = 1e9 + 9, pot[N];
string s;
bool pal(int a, int b) {
return ((((H[b + 1] - H[a] * pot[b - a + 1]) % p + p) % p) ==
(((H2[a] - H2[b + 1] * pot[b - a + 1]) % p + p) % p));
}
int main() {
ios_base::syn... | ### Prompt
Construct a CPP code solution to the problem outlined:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of lowercase ... |
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
template <typename t>
void fff(const char *x, t &&val1) {
cout << x << " : " << val1 << endl;
}
template <typename t1, typename... t2>
void fff(const c... | ### Prompt
Please create a solution in Cpp to the following problem:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of lowerca... |
#include <bits/stdc++.h>
using namespace std;
template <typename t>
bool ckmin(t& a, const t& b) {
return a > b ? a = b, true : false;
}
template <typename t>
bool ckmax(t& a, const t& b) {
return a < b ? a = b, true : false;
}
vector<int> P;
void manacher(string& s) {
string T = "#";
for (auto((i)) = (0); ((i)... | ### Prompt
Create a solution in Cpp for the following problem:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of lowercase Eng... |
#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353, x = 8641 + 97532;
long long int hashh[26];
long long modpow(long long int a, int b) {
long long int ans = 1;
for (; b; b >>= 1, a = a * a % mod) {
if (b & 1) ans = ans * a % mod;
}
return ans;
}
int f(string s) {
long long int now1 =... | ### Prompt
Generate a cpp solution to the following problem:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of lowercase Engli... |
#include <bits/stdc++.h>
using namespace std;
const int m = 2e6 + 29;
int pref[m];
string solve_palindrome(string a) {
string t = a + '#';
reverse(a.begin(), a.end());
t = t + a;
int c = 0, i, l = t.size();
for (i = 1; i < l; i++) {
while (c != 0 and t[c] != t[i]) {
c = pref[c - 1];
}
if (t[... | ### Prompt
Please provide a cpp coded solution to the problem described below:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting ... |
#include <bits/stdc++.h>
using namespace std;
const int M = (int)(2e6 + 239);
int pref[M];
string palindrome(const string &s) {
string a = s;
reverse(a.begin(), a.end());
a = s + "#" + a;
int c = 0;
for (int i = 1; i < (int)a.size(); i++) {
while (c != 0 && a[c] != a[i]) c = pref[c - 1];
if (a[i] == a... | ### Prompt
Generate a cpp solution to the following problem:
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of lowercase Engli... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.