output stringlengths 52 181k | instruction stringlengths 296 182k |
|---|---|
#include <bits/stdc++.h>
using namespace std;
const int TAM = 1e5 + 10;
struct Nodo {
int id, s, a, b;
};
inline bool byS(const Nodo& x, const Nodo& y) { return x.s < y.s; }
inline bool byID(const Nodo& x, const Nodo& y) { return x.id < y.id; }
int n;
Nodo nodo[TAM];
int main() {
cin >> n;
for (int i = 0; i < n; ... | ### Prompt
Construct a cpp code solution to the problem outlined:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely, y... |
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 100;
pair<int, int> s[MAXN];
int a[MAXN], b[MAXN];
bool used[MAXN];
int main() {
int n;
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%d", &s[i].first);
s[i].second = i;
}
sort(s, s + n);
fill(a, a + n, -1);
fill(b, b + n... | ### Prompt
Create a solution in cpp for the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely, you ... |
#include <bits/stdc++.h>
using namespace std;
int n;
int data[100009], a[100009], b[100009], visa[100009], visb[100009], vis[100009],
indexes[100009];
vector<pair<int, int> > in, outing;
int main() {
scanf("%d", &n);
int q;
for (int i = 0; i < n; i++) {
scanf("%d", &q);
in.push_back(make_pair(q, i));
... | ### Prompt
Please provide a CPP coded solution to the problem described below:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two.... |
#include <bits/stdc++.h>
using namespace std;
const int limN = 1e5 + 5;
int N;
int nums[limN];
int ord[limN];
int A[limN], B[limN];
int main() {
scanf("%d", &N);
for (int i = 0; i < N; i++) {
scanf("%d", &nums[i]);
ord[i] = i;
}
sort(ord, ord + N,
[](const int a, const int b) { return nums[a] < n... | ### Prompt
Create a solution in cpp for the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely, you ... |
#include <bits/stdc++.h>
struct cancerut {
int a, b, init, poz;
};
bool cmpgay(cancerut a, cancerut b) { return a.init < b.init; }
bool cmpgay2(cancerut a, cancerut b) { return a.poz < b.poz; }
cancerut v[100005];
int main() {
int n;
scanf("%d", &n);
printf("YES\n");
for (int i = 0; i < n; i++) {
scanf("%... | ### Prompt
Your challenge is to write a CPP solution to the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two.... |
#include <bits/stdc++.h>
using namespace std;
const int mxN = 1e5;
int n, a[mxN], b[mxN];
pair<int, int> ps[mxN];
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cin >> n;
for (int i = 0; i < n; ++i) cin >> ps[i].first, ps[i].second = i;
sort(ps, ps + n);
for (int i = 0; i < (n + 2) / 3; ++i) {
a... | ### Prompt
Please provide a CPP coded solution to the problem described below:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two.... |
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
pair<int, int> arr[N];
set<int> a, b;
int aa[N], bb[N], used[N];
int n;
int main() {
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%d", &arr[i].first);
arr[i].second = i;
}
sort(arr, arr + n);
int cur = 0;
for (int i = 0; i... | ### Prompt
Please provide a Cpp coded solution to the problem described below:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two.... |
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
using namespace std;
const double PI = acos(-1.0);
int32_t main() {
ios::sync_with_stdio(0);
cin.tie(0);
long long n;
cin >> n;
pair<long long, long long> p[n];
for (long long i = 0; i < n; i++) {
cin >> p[i].first;
p[i].... | ### Prompt
Generate a CPP solution to the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely, you ne... |
#include <bits/stdc++.h>
using namespace std;
long long rdtsc() {
long long tmp;
asm("rdtsc" : "=A"(tmp));
return tmp;
}
const int maxn = (int)1e5;
pair<int, int> s[maxn];
int s0[maxn], ans[2][maxn];
int main() {
srand(rdtsc());
int n;
while (scanf("%d", &n) >= 1) {
for (int i = 0; i < n; i++)
sca... | ### Prompt
Develop a solution in CPP to the problem described below:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely... |
#include <bits/stdc++.h>
using namespace std;
int vet[100001], tmp[100001], n, m, resa[100001], resb[100001], ma, mb, p;
map<int, int> a, b, pos;
int main() {
scanf("%d", &n);
for (int i = 0; i < n; ++i) scanf("%d", &vet[i]), tmp[i] = vet[i];
sort(vet, vet + n);
for (int i = 0; i < n; ++i) pos[vet[i]] = i;
m ... | ### Prompt
Please create a solution in cpp to the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely... |
#include <bits/stdc++.h>
using namespace std;
const long maxn = 110000;
struct ss {
long n, x;
} s[maxn] = {}, a[maxn] = {};
long i, n, m;
bool comp(ss a, ss b) { return a.x < b.x; }
bool cmp(ss a, ss b) { return a.n < b.n; }
int main() {
ios::sync_with_stdio(false);
cin >> n;
for (i = 0; i < n; s[i].n = i++) c... | ### Prompt
Please create a solution in cpp to the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely... |
#include <bits/stdc++.h>
using namespace std;
struct NODE {
int val, mark;
NODE() {}
bool operator<(const NODE &p) const { return val < p.val; }
};
const int maxn = 100007;
int n;
NODE s[maxn];
int a[maxn], b[maxn];
int A[maxn], B[maxn];
bool BUG(int a[maxn]) {
map<int, int> Map;
for (int i = 1; i <= n; i++) ... | ### Prompt
Construct a cpp code solution to the problem outlined:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely, y... |
#include <bits/stdc++.h>
using namespace std;
const int NMAX = 200000;
vector<pair<int, int> > v;
int s[NMAX + 2], a[NMAX + 2], b[NMAX + 2], a_ind = 1, b_ind = 1;
int N;
int main() {
cin >> N;
cout << "YES\n";
for (int i = 1; i <= N; ++i) cin >> s[i], v.push_back({s[i], i});
sort(v.begin(), v.end());
for (int... | ### Prompt
In cpp, your task is to solve the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely, you... |
#include <bits/stdc++.h>
using namespace std;
int n, a[100005], b[100005];
pair<int, int> s[100005];
int main() {
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%d", &s[i].first);
s[i].second = i;
}
sort(s, s + n);
int i;
for (i = 0; i < ((n + 2) / 3); i++) {
a[s[i].second] = i;
b[s[i... | ### Prompt
Your challenge is to write a cpp solution to the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two.... |
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1000 * 1000;
int a[MAXN], b[MAXN];
pair<int, int> s[MAXN];
int main() {
int n;
cin >> n;
for (int i = 0; i < n; i++) cin >> s[i].first;
for (int i = 0; i < n; i++) s[i].second = i;
sort(s, s + n);
cout << "YES" << endl;
int l1 = (n + 1) / 3, l... | ### Prompt
Construct a CPP code solution to the problem outlined:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely, y... |
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5;
pair<int, int> v[MAXN];
int a[MAXN], b[MAXN];
int main() {
int n, p, q;
cin >> n;
for (int i = 0; i < n; ++i) {
cin >> v[i].first;
v[i].second = i;
}
sort(v, v + n);
cout << "YES\n";
p = n / 3;
q = 2 * n / 3;
for (int i = 0; i... | ### Prompt
Please provide a CPP coded solution to the problem described below:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two.... |
#include <bits/stdc++.h>
using namespace std;
inline int ___INT() {
int ret;
scanf("%d", &ret);
return ret;
}
void solve(vector<int>& A) {
vector<int> B = A;
map<int, int> M;
sort(B.begin(), B.end());
int n = A.size();
int zeros = 1 + (n + 2) / 3;
vector<int> U, V;
for (int i = 0; i < zeros; ++i) {
... | ### Prompt
Generate a CPP solution to the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely, you ne... |
#include <bits/stdc++.h>
using namespace std;
const int N = 200 * 1000;
int a[N];
int b[N];
pair<int, int> s[N];
int ansA[N];
int ansB[N];
int n;
void solve() {
cin >> n;
for (int i = 0; i < n; ++i) {
cin >> s[i].first;
s[i].second = i;
}
sort(s, s + n);
int size = n / 3;
if (n % 3 != 0) size++;
f... | ### Prompt
Your task is to create a cpp solution to the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Pre... |
#include <bits/stdc++.h>
using namespace std;
int n, s[112345];
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.setf(ios::fixed);
cout.precision(20);
cin >> n;
for (int i = 0; i < n; ++i) cin >> s[i];
vector<int> tmp_s(s, s + n);
sort(tmp_s.begin(), tmp_s.end());
int sz = (tmp_s.size() + 2... | ### Prompt
Please create a solution in CPP to the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely... |
#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
pair<int, int> a[N];
int b[N], c[N];
map<int, bool> hb, hc;
int main() {
int n;
cin >> n;
for (int i = 0; i < n; ++i) {
cin >> a[i].first;
a[i].second = i;
}
sort(a, a + n);
int st = (n + 2) / 3, z = 0;
for (int i = st; i < n; ++i... | ### Prompt
Develop a solution in CPP to the problem described below:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely... |
#include <bits/stdc++.h>
using namespace std;
const int TAM = 1e5 + 10;
struct Nodo {
int id, s, a, b;
};
inline bool byS(const Nodo& x, const Nodo& y) { return x.s < y.s; }
inline bool byID(const Nodo& x, const Nodo& y) { return x.id < y.id; }
int n;
Nodo nodo[TAM];
int main() {
scanf("%d", &n);
for (int i = 0; ... | ### Prompt
Your task is to create a cpp solution to the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Pre... |
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007, dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};
const double eps = 1e-8;
void read(int &k) {
k = 0;
char x = getchar();
while (x < '0' || x > '9') x = getchar();
while (x >= '0' && x <= '9') {
k = k * 10 - 48 + x;
x = getchar();
}
}... | ### Prompt
Develop a solution in Cpp to the problem described below:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely... |
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 101010;
pair<int, int> arr[MAXN];
pair<int, int> a[MAXN], b[MAXN];
int N;
bool cmp(const pair<int, int> &a, const pair<int, int> &b) {
return a.second < b.second;
}
int main() {
ios_base::sync_with_stdio(0);
cin >> N;
for (int i = 0; i < N; i++) {
... | ### Prompt
In CPP, your task is to solve the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely, you... |
#include <bits/stdc++.h>
using namespace std;
vector<pair<int, int> > targ;
int n;
int sol[2][100000];
int sol2[2][100000];
int first;
int inv[100000];
int main() {
ios::sync_with_stdio(0);
cin >> n;
targ.resize(n);
for (int c = 0; c < n; c++) {
cin >> targ[c].first;
targ[c].second = c;
}
sort(targ.... | ### Prompt
In cpp, your task is to solve the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely, you... |
#include <bits/stdc++.h>
using namespace std;
int runGCJ(istream& in, ostream& out) {
int T;
in >> T;
for (int t = 1; t <= T; ++t) {
long long res = 0;
out << "Case #" << t << ": " << res << endl;
}
return 0;
}
bool cmpf(pair<int, int> a, pair<int, int> b) { return a.second < b.second; }
int runCF(ist... | ### Prompt
Your challenge is to write a cpp solution to the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two.... |
#include <bits/stdc++.h>
using namespace std;
int n;
const int max_n = 100005;
int s[max_n];
int a[max_n];
int b[max_n];
int idx[max_n];
bool cmp(const int &aa, const int &bb) { return s[aa] < s[bb]; }
int main() {
int n;
while (scanf("%d", &n) == 1) {
printf("YES\n");
for (int i = 0; i < n; i++) {
sc... | ### Prompt
Please formulate a Cpp solution to the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely... |
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:60777216")
using namespace std;
int n;
int s[111111];
int a[111111];
int r1[111111];
int r2[111111];
int main() {
cin >> n;
for (int i = (0); i < (n); i++) scanf("%d", s + i), a[i] = s[i];
sort(a, a + n);
int tn = n / 3;
if (tn * 3 != n) tn++;
for (in... | ### Prompt
Develop a solution in cpp to the problem described below:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely... |
#include <bits/stdc++.h>
using namespace std;
int A[1 << 20];
int B[1 << 20];
int main() {
int n;
scanf("%d", &n);
int i;
vector<pair<int, int> > v;
for (i = (0); i < (n); ++i) {
int a;
scanf("%d", &a);
v.push_back(pair<int, int>(a, i));
}
sort((v).begin(), (v).end());
int k = (n + 2) / 3;
... | ### Prompt
Please create a solution in CPP to the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely... |
#include <bits/stdc++.h>
using namespace std;
template <class A, class B>
ostream& operator<<(ostream& out, const pair<A, B>& a) {
return out << "(" << a.first << ", " << a.second << ")";
}
template <class A>
ostream& operator<<(ostream& out, const vector<A>& a) {
out << "[";
for (auto it = a.begin(); it != a.end... | ### Prompt
Create a solution in Cpp for the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely, you ... |
#include <bits/stdc++.h>
using namespace std;
struct Data {
int Value, Number;
};
bool operator<(Data A, Data B) { return A.Value < B.Value; }
int Get() {
char c;
while (c = getchar(), c < '0' || c > '9')
;
int X = 0;
while (c >= '0' && c <= '9') {
X = X * 10 + c - 48;
c = getchar();
}
return ... | ### Prompt
Generate a cpp solution to the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely, you ne... |
#include <bits/stdc++.h>
using namespace std;
bool cmp(pair<int, int> i, pair<int, int> j) { return i.second < j.second; }
const int MAXN = 100000;
pair<int, int> s[MAXN];
int a[MAXN];
int b[MAXN];
int main() {
int n;
cin >> n;
for (typeof(0) i = (0); i < (n); ++i) {
cin >> s[i].first;
s[i].second = i;
... | ### Prompt
In CPP, your task is to solve the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely, you... |
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const auto INF = 0x3f3f3f3f;
const int N = 1e5 + 5;
struct node {
int id, x;
} s[N];
int a[N], b[N];
bool cmp(const struct node &u, const struct node &v) { return u.x < v.x; }
int main() {
int n;
scanf("%d", &n);
int m = (n / 3) + 1;
for (... | ### Prompt
Develop a solution in cpp to the problem described below:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely... |
#include <bits/stdc++.h>
using namespace std;
const int MAX = 1e5 + 5;
int n, k;
struct st {
int s, a, b, p;
};
st v[MAX];
int main() {
cin >> n;
k = n / 3;
if (n % 3 == 2) {
++k;
}
for (int i = 1; i <= n; ++i) {
cin >> v[i].s;
v[i].p = i;
}
sort(v + 1, v + n + 1, [&](const st &a, const st &... | ### Prompt
Create a solution in Cpp for the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely, you ... |
#include <bits/stdc++.h>
using namespace std;
const int inf = (int)(1e9);
const int mod = inf + 7;
const double pi = acos(-1.0);
const double eps = 1e-9;
int n, a[100100], b[100100], ansa[100100], ansb[100100];
pair<int, int> second[100100];
int main() {
cin >> n;
for (int i = 0; i < n; i++) {
scanf("%d", &seco... | ### Prompt
In Cpp, your task is to solve the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely, you... |
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
int s[maxn], a[maxn], b[maxn];
int n;
int main() {
cin >> n;
vector<pair<int, int>> e(n);
for (int i = 1; i <= n; ++i) {
cin >> s[i];
e[i - 1] = {s[i], i};
}
sort(begin(e), end(e));
int idx = 0;
for (int i = 0; i < 3; ++i) {
... | ### Prompt
Your task is to create a Cpp solution to the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Pre... |
#include <bits/stdc++.h>
using std::min;
using std::sort;
int const N = 111111;
struct node {
int num, id;
bool operator<(const node& tmp) const { return num < tmp.num; }
} Data[N];
int n, a[N], b[N];
int main() {
while (~scanf("%d", &n)) {
for (int i = 0; i < n; i++) scanf("%d", &Data[i].num), Data[i].id = i... | ### Prompt
Your challenge is to write a cpp solution to the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two.... |
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int a[N], x[N], y[N];
pair<int, int> b[N];
int main() {
int n;
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%d", &a[i]);
b[i] = {a[i], i};
}
sort(b, b + n);
int m = n / 3 + !!(n % 3);
for (int i = 0, no; i < m; i++) {
... | ### Prompt
Please create a solution in CPP to the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely... |
#include <bits/stdc++.h>
using namespace std;
pair<int, int> s[100001];
int a[100001];
int b[100001];
int as[100001];
int bs[100001];
int main() {
int n, i;
scanf("%d", &n);
for (i = 1; i <= n; i++) scanf("%d", &s[i].first), s[i].second = i;
sort(s + 1, s + n + 1);
for (i = 1; i <= n / 3; i++) a[i] = i - 1, b... | ### Prompt
Please formulate a CPP solution to the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely... |
#include <bits/stdc++.h>
using namespace std;
int a[100005], b[100005], s[100005];
struct Node {
int index, n;
};
Node node[100005];
bool cmp(const Node &a, const Node &b) { return a.n < b.n; }
int main() {
int n;
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
scanf("%d", &s[i]);
node[i].index = i, nod... | ### Prompt
Please create a solution in CPP to the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely... |
#include <bits/stdc++.h>
using namespace std;
const int INF = 1E9 + 7;
template <class C>
void mini(C& a4, C b4) {
a4 = min(a4, b4);
}
template <class C>
void maxi(C& a4, C b4) {
a4 = max(a4, b4);
}
vector<pair<int, int> > v;
int n, a[100007], b[100007], c, p1, p2, p3;
int main() {
ios_base::sync_with_stdio(false... | ### Prompt
In Cpp, your task is to solve the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely, you... |
#include <bits/stdc++.h>
using namespace std;
const int maxN = 1e5 + 100;
int s[maxN], cs[maxN], b[maxN], c[maxN];
int n;
bool cmp(int x, int y) { return s[x] < s[y]; }
bool cmp_b(int x, int y) { return b[x] < b[y]; }
bool cmp_c(int x, int y) { return c[x] < c[y]; }
int main() {
cin >> n;
for (int i = 0; i < n; i++... | ### Prompt
Please provide a cpp coded solution to the problem described below:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two.... |
#include <bits/stdc++.h>
using namespace std;
const double EPS = 1e-9;
const double PI = acos(-1);
const int MAXN = (int)1e6 + 7;
const int INF = (int)1e9 + 7;
const int dx[] = {-1, 0, 1, 0};
const int dy[] = {0, -1, 0, 1};
int n;
int A[MAXN], B[MAXN], a[MAXN], b[MAXN];
pair<int, int> second[MAXN];
int main() {
ios_b... | ### Prompt
In cpp, your task is to solve the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely, you... |
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 10;
int N, arr[MAXN], tid[MAXN], a[MAXN], b[MAXN];
int main() {
scanf("%d", &N);
for (int i = (0), iend = (N); i < iend; ++i) scanf("%d", &arr[i]);
iota(tid, tid + N, 0);
sort(tid, tid + N,
[](const int& x, const int& y) { return arr[x]... | ### Prompt
Please provide a CPP coded solution to the problem described below:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two.... |
#include <bits/stdc++.h>
using namespace std;
int n;
int data[100009], a[100009], b[100009], visa[100009], visb[100009], vis[100009],
indexes[100009];
vector<pair<int, int> > in, outing;
int main() {
scanf("%d", &n);
int q;
for (int i = 0; i < n; i++) {
scanf("%d", &q);
in.push_back(make_pair(q, i));
... | ### Prompt
Please formulate a cpp solution to the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely... |
#include <bits/stdc++.h>
using namespace std;
const int MAXn = 100 * 1000 + 1000;
int n, a[MAXn], b[MAXn], a_ans[MAXn], b_ans[MAXn];
pair<int, int> s[MAXn];
int saghf(int x, int y) {
if (x % y == 0) return (int)x / y;
return ((int)x / y + 1);
}
int main() {
ios::sync_with_stdio(false);
cin >> n;
for (int i = ... | ### Prompt
Create a solution in cpp for the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely, you ... |
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
int s[N], a[N], b[N], n, m;
pair<int, int> X[N];
int main(void) {
scanf("%d", &n);
for (int i = 0; i < n; ++i) scanf("%d", &s[i]), X[i] = make_pair(s[i], i);
sort(X, X + n);
int tot = 0;
int TTT = (n / 3) + !!(n % 3);
if (TTT * 2 > n) TTT-... | ### Prompt
Your task is to create a cpp solution to the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Pre... |
#include <bits/stdc++.h>
using namespace std;
vector<pair<int, int> > v;
int n;
int a[1 << 20];
int b[1 << 20];
int q;
int main() {
ios_base::sync_with_stdio(0);
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> q;
v.push_back(make_pair(q, i));
}
sort(v.begin(), v.end());
for (int i = 0; i < n / 3; i+... | ### Prompt
Please formulate a CPP solution to the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely... |
#include <bits/stdc++.h>
using namespace std;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
cerr << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
const char* comma = strchr(names + 1, ',');
cerr.write(nam... | ### Prompt
In CPP, your task is to solve the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely, you... |
#include <bits/stdc++.h>
using namespace std;
int main() {
string s;
cin >> s;
int i = 0;
while (i <= s.size() - 1) {
if (s[i] == '1') {
i++;
if (s[i] == '4') {
i++;
if (s[i] == '4') i++;
}
} else
break;
}
if (i == s.size())
cout << "YES";
else
cout ... | ### Prompt
Please provide a cpp coded solution to the problem described below:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. De... |
#include <bits/stdc++.h>
using namespace std;
bool solve(string s, int p) {
if (s[p] == '\0') {
return true;
}
for (int i = p; i < (int)s.length(); i++) {
string t = s.substr(p, i - p + 1);
if (t == "1" || t == "14" || t == "144") {
if (solve(s, i + 1)) {
return true;
}
} else ... | ### Prompt
Please provide a Cpp coded solution to the problem described below:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. De... |
#include <bits/stdc++.h>
using namespace std;
stack<int> st;
int n;
int main() {
cin >> n;
while (n) {
st.push(n % 10);
n /= 10;
}
bool ok = true;
bool need1 = true, need4 = false;
int c4 = 0;
while (!st.empty()) {
int m = st.top();
st.pop();
if (m == 1 && need1) {
need4 = true;
... | ### Prompt
Construct a Cpp code solution to the problem outlined:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if it... |
#include <bits/stdc++.h>
using namespace std;
bool fun(long long n) {
if (n == 1 || n == 14 || n == 144) return 1;
if (n % 10 == 1 && fun(n / 10)) return 1;
if (n % 100 == 14 && fun(n / 100)) return 1;
if (n % 1000 == 144 && fun(n / 1000)) return 1;
return 0;
}
int main() {
ios_base::sync_with_stdio(false);... | ### Prompt
Please provide a CPP coded solution to the problem described below:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. De... |
#include <bits/stdc++.h>
using namespace std;
int main() {
string n;
cin >> n;
int tam = n.size();
for (int i = 0; i < tam; i++) {
if (n[i] == '1' || (n[i] == '4' && (n[i - 1] == '1' || n[i - 2] == '1'))) {
} else {
cout << "NO";
return 0;
}
}
cout << "YES";
}
| ### Prompt
Your challenge is to write a cpp solution to the following problem:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. De... |
#include <bits/stdc++.h>
using namespace std;
const long long inf = 1e18;
long long add(long long x, long long y) {
long long res = x + y;
return (res >= 1000000007 ? res - 1000000007 : res);
}
long long mul(long long x, long long y) {
long long res = x * y;
return (res >= 1000000007 ? res % 1000000007 : res);
... | ### Prompt
Please provide a CPP coded solution to the problem described below:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. De... |
#include <bits/stdc++.h>
using namespace std;
const int bfsz = 1 << 16;
char bf[bfsz + 5];
int rsz = 0;
int ptr = 0;
char gc() {
if (rsz <= 0) {
ptr = 0;
rsz = fread(bf, 1, bfsz, stdin);
if (rsz <= 0) return EOF;
}
--rsz;
return bf[ptr++];
}
void ga(char &c) {
c = EOF;
while (!isalpha(c)) c = gc... | ### Prompt
Construct a Cpp code solution to the problem outlined:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if it... |
#include <bits/stdc++.h>
using namespace std;
int main() {
unsigned long long int n, i, f = 0, f1 = 0, d, v = 10, c = 0;
cin >> n;
vector<int> digits;
while (n > 0) {
d = (n % v);
n = n / v;
digits.push_back(d);
c++;
}
reverse(digits.begin(), digits.end());
for (i = 0; i < c;) {
if (di... | ### Prompt
Your task is to create a cpp solution to the following problem:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determ... |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
string x;
cin >> x;
int flag = 0;
for (int i = 0; i < x.length(); i++) {
if (x[i] != '1' && x[i] != '4') {
cout << "NO";
return 0;
}
if (i + 2 < x.length() && x[i] == '1' && x[... | ### Prompt
Create a solution in CPP for the following problem:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if it is... |
#include <bits/stdc++.h>
using namespace std;
long long choose(long long n, long long k);
bool isprime(long long n);
void setprime(bool isprime[], long long n);
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
string s;
cin >> s;
long long count = 0;
for (long long i = s.size() - 1; i >... | ### Prompt
Create a solution in Cpp for the following problem:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if it is... |
#include <bits/stdc++.h>
using namespace std;
string str;
int main() {
ios_base::sync_with_stdio(false);
cin.tie();
cin >> str;
for (int i = 0; i <= str.size() - 1; i++) {
if (str[i] == '1' || (str[i] == '4' && i > 0 && str[i - 1] == '1') ||
(str[i] == '4' && i > 1 && str[i - 1] == '4' && str[i - 2]... | ### Prompt
Please formulate a Cpp solution to the following problem:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if... |
#include <bits/stdc++.h>
using namespace std;
int main() {
long long i;
string s;
cin >> s;
for (i = 0; i < s.length(); i++) {
if (s[i] != '1' && s[i] != '4') {
cout << "NO";
break;
} else if (s[i] == '1')
continue;
else {
if (i == 0) {
cout << "NO";
break;
... | ### Prompt
Construct a CPP code solution to the problem outlined:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if it... |
#include <bits/stdc++.h>
using namespace std;
int main() {
string a;
cin >> a;
int n = a.size(), i, s = 0;
for (i = 0; i <= n - 1; i++) {
if ((a[i] == '1') && (a[i + 1] == '4') && (a[i + 2] == '4'))
i = i + 2;
else if ((a[i] == '1') && (a[i + 1] == '4'))
i = i + 1;
else if (a[i] == '1' &... | ### Prompt
Your challenge is to write a cpp solution to the following problem:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. De... |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
string s;
cin >> s;
int c = 0;
int cnt = 0;
for (int i = 0; i < s.size(); i++) {
if (s[i] != '1' && s[i] != '4') {
c++;
} else if (s[i] == '4' && s[i + 1] == '4' && s[i + 2] == '4')... | ### Prompt
Construct a CPP code solution to the problem outlined:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if it... |
#include <bits/stdc++.h>
using namespace std;
int main() {
string s;
int flag = 1;
cin >> s;
for (int i = 0; i < s.length(); i++) {
if (s[i] == '1') {
if (i == s.length() - 1) break;
if (s[i + 1] == '4') {
if (i == s.length() - 2) break;
if (s[i + 2] == '4') {
i += 2;
... | ### Prompt
Construct a CPP code solution to the problem outlined:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if it... |
#include <bits/stdc++.h>
int main() {
int N, A, B, C;
char Error = 0;
scanf("%d", &N);
while (N) {
A = N % 10;
B = N % 100;
C = N % 1000;
if (A == 1) {
N /= 10;
} else if (B == 14) {
N /= 100;
} else if (C == 144) {
N /= 1000;
} else {
Error = 1;
break;
... | ### Prompt
Please provide a Cpp coded solution to the problem described below:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. De... |
#include <bits/stdc++.h>
using namespace std;
char n[1000000];
int main() {
bool num[1000000];
cin >> n;
for (int i = 0; i + 2 < strlen(n); i++) {
if (n[i] == '1' && n[i + 1] == '4' && n[i + 2] == '4') {
num[i] = true;
num[i + 1] = true;
num[i + 2] = true;
}
}
for (int i = 0; i + 1 <... | ### Prompt
In CPP, your task is to solve the following problem:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if it i... |
#include <bits/stdc++.h>
using namespace std;
int main() {
char str[100];
cin >> str;
int i, j, len, flag = 0, n = 0;
len = strlen(str);
for (i = 0;;) {
if (i > len - 1) {
break;
} else {
if (str[i] == '1' && str[i + 1] == '4' && str[i + 2] == '4') {
i += 3;
continue;
... | ### Prompt
Generate a cpp solution to the following problem:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if it is a... |
#include <bits/stdc++.h>
using namespace std;
bool sortinrev(const pair<int, int> &a, const pair<int, int> &b) {
return (a.first > b.first);
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int n, t, i, j;
cin >> n;
int d = 0;
int tru = 1;
while (n > 0) {
d = n % 10;
if (d == 1)
... | ### Prompt
Develop a solution in cpp to the problem described below:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if... |
#include <bits/stdc++.h>
using namespace std;
int main() {
char str[15];
int cnt = 0;
cin >> str;
for (int i = 0; i <= strlen(str) - 1;) {
if (str[i] == '1' && str[i + 1] != '4')
i++;
else if (str[i] == '1' && str[i + 1] == '4' && str[i + 2] != '4')
i += 2;
else if (str[i] == '1' && str[... | ### Prompt
Create a solution in CPP for the following problem:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if it is... |
#include <bits/stdc++.h>
int main() {
int i, len, count4 = 0, one = 0;
char str[20];
scanf("%s", str);
len = strlen(str);
for (i = 0; i < len; i++) {
if (str[i] == '1')
one = 1;
else if (str[i] == '4') {
while (str[i] == '4') {
count4++;
i++;
}
i--;
} else {... | ### Prompt
Please formulate a Cpp solution to the following problem:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if... |
#include <bits/stdc++.h>
using namespace std;
int main() {
string s;
cin >> s;
for (int i = 0; i < s.length(); i++) {
if (s[i] == '1' && s[i + 1] == '4' && s[i + 2] == '4')
i += 2;
else if (s[i] == '1' && s[i + 1] == '4')
i++;
else if (s[i] == '1')
true;
else {
cout << "NO"... | ### Prompt
Please formulate a Cpp solution to the following problem:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if... |
#include <bits/stdc++.h>
int64_t TN = 1;
using namespace std;
void solution() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int c = 0;
string s;
cin >> s;
for (int i = 0; i < (s).size(); i++) {
if (s[i] == '1') {
c++;
}
}
if (c == (s).size()) {
cout << "YES" << endl;
return;
}
c ... | ### Prompt
Develop a solution in CPP to the problem described below:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if... |
#include <bits/stdc++.h>
using namespace std;
bool is_magical_num(string num) {
int f = 0;
int l = num.size();
for (int i = 0; i < l; i++) {
if (num[0] != '1') {
return false;
}
if (num[i] == '1') {
f = 0;
} else if (num[i] == '4') {
f++;
if (f >= 3) return false;
} els... | ### Prompt
Generate a cpp solution to the following problem:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if it is a... |
#include <bits/stdc++.h>
using namespace std;
int main() {
long int n;
cin >> n;
int flag = 0;
while (flag == 0 && n != 0) {
if ((n % 10) == 1)
n = n / 10;
else if ((n % 10) == 4) {
if (((n / 10) % 10) == 1)
n = n / 100;
else if (((n / 10) % 10) == 4) {
if (((n / 100) %... | ### Prompt
Please create a solution in CPP to the following problem:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if... |
#include <bits/stdc++.h>
using namespace std;
bool func(string n) {
for (int i = 0; i < (int)n.size(); i++) {
if (n[i] != '1' && n[i] != '4') {
return false;
}
}
if (n[0] == '4') {
return false;
}
if (n.find("444") != n.npos) {
return false;
}
return true;
}
int main() {
string n;
... | ### Prompt
Construct a Cpp code solution to the problem outlined:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if it... |
#include <bits/stdc++.h>
using namespace std;
int main() {
char n;
bool ok = true;
int cnt = 2;
while (cin >> n) {
if (n == '1')
cnt = 0;
else if (n == '4') {
if (cnt == 2) ok = false;
cnt++;
} else
ok = false;
}
puts(ok ? "YES" : "NO");
return 0;
}
| ### Prompt
In CPP, your task is to solve the following problem:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if it i... |
#include <bits/stdc++.h>
using namespace std;
int main() {
string s;
cin >> s;
if (s[0] != '1') {
cout << "NO";
return 0;
}
for (int i = 1; i < s.size(); i++) {
if (s[i] == '4' && s[i - 1] == '4' && s[i + 1] == '4') {
cout << "NO";
return 0;
}
if (s[i] != '4' && s[i] != '1') {
... | ### Prompt
Your task is to create a CPP solution to the following problem:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determ... |
#include <bits/stdc++.h>
using namespace std;
char str[20];
int len;
int DFS(int sta) {
if (sta == len) return 1;
if (str[sta] != '1') return 0;
if (DFS(sta + 1)) return 1;
if (sta + 1 >= len || str[sta + 1] != '4') return 0;
if (DFS(sta + 2)) return 1;
if (sta + 2 >= len || str[sta + 2] != '4') return 0;
... | ### Prompt
Develop a solution in Cpp to the problem described below:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if... |
#include <bits/stdc++.h>
using namespace std;
int main() {
int x;
cin >> x;
int temp;
int x1 = 0;
int x2 = 0;
bool flag = true;
while (x != 0) {
temp = x % 10;
x = x / 10;
if ((temp == 4) && (x1 == 0)) {
x1 = 1;
x2 = 1;
} else if ((temp == 1)) {
x1 = 0;
x2 = 0;
... | ### Prompt
Your challenge is to write a Cpp solution to the following problem:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. De... |
#include <bits/stdc++.h>
using namespace std;
int main() {
string s;
cin >> s;
for (int i = 0; i < s.length(); ++i) {
if (s[i] != '1' && s[i] != '4') {
cout << "NO";
return 0;
}
if (i < s.length() - 2) {
if (s[i] == '4' && s[i + 1] == '4' && s[i + 2] == '4') {
cout << "NO";
... | ### Prompt
Construct a cpp code solution to the problem outlined:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if it... |
#include <bits/stdc++.h>
using namespace std;
bool dp[100];
bool isMagicNumber(const string& s) {
memset(dp, false, sizeof(dp));
int n = s.length();
dp[0] = true;
for (int i = 1; i <= n; i++) {
dp[i] = (dp[i - 1] && (s[i - 1] == '1')) ||
(i >= 2 && dp[i - 2] && s[i - 2] == '1' && s[i - 1] == '4'... | ### Prompt
Generate a CPP solution to the following problem:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if it is a... |
#include <bits/stdc++.h>
using namespace std;
bool f(string s) {
bool ans = false;
if (s.size() == 0) {
return false;
}
if (s == "1" or s == "14" or s == "144") {
return true;
}
if (s.size() >= 1 && s[s.size() - 1] == '1')
ans = (ans or f(s.substr(0, s.size() - 1)));
if (s.size() >= 2 && s[s.s... | ### Prompt
Please provide a cpp coded solution to the problem described below:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. De... |
#include <bits/stdc++.h>
using namespace std;
bool recurse(string s, int i, int n) {
if (i == n) {
cout << "YES\n";
return true;
}
if (s[i] == '1') {
bool f1 = recurse(s, i + 1, n);
if (f1) return true;
if (s[i + 1] == '4') {
bool f2 = recurse(s, i + 2, n);
if (f2) return true;
... | ### Prompt
Your challenge is to write a CPP solution to the following problem:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. De... |
#include <bits/stdc++.h>
using namespace std;
int fx[] = {+0, +0, +1, -1, -1, +1, -1, +1};
int fy[] = {-1, +1, +0, +0, +1, +1, -1, -1};
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
long long i, diff, one = 0, four = 0, remain, four_div;
string str;
cin >> str;
for (i = 0; i < str.size(); i++) {
... | ### Prompt
Please provide a CPP coded solution to the problem described below:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. De... |
#include <bits/stdc++.h>
using namespace std;
bool verif(bool flag[], string cad) {
for (int i = 0; i < cad.size(); i++)
if (!flag[i]) return false;
return true;
}
int main() {
string cad;
getline(cin, cad);
for (int i = 0; i < cad.size(); i++) {
if (cad[i] == '1' || cad[i] == '4') {
;
} els... | ### Prompt
Create a solution in Cpp for the following problem:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if it is... |
#include <bits/stdc++.h>
int main() {
char s[1000001];
int i, j, k, l, f = 1;
scanf("%s", s);
for (i = 0; i < strlen(s); i++) {
if (s[0] != '1') {
f = 0;
break;
} else if (s[i] != '1' && s[i] != '4') {
f = 0;
break;
} else if (s[i] == '4' && s[i + 1] == '4' && s[i + 2] == '4'... | ### Prompt
In CPP, your task is to solve the following problem:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if it i... |
#include <bits/stdc++.h>
using namespace std;
int main() {
int i, n, found = 1;
;
string s;
cin >> s;
n = s.size();
for (i = 0; i < n;) {
if (s[i] == '2' || s[i] == '3' || s[i] == '5' || s[i] == '6' ||
s[i] == '7' || s[i] == '8' || s[i] == '9') {
found = 0;
break;
}
if (s[i] ... | ### Prompt
Please provide a CPP coded solution to the problem described below:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. De... |
#include <bits/stdc++.h>
using namespace std;
int main() {
string a;
int b = 1;
cin >> a;
for (int i = 0; i < a.size(); i++) {
if (a[0] != '1') {
b = 0;
} else if (a[i] != '1' && a[i] != '4') {
b = 0;
} else if (a.find("444") != a.npos) {
b = 0;
}
}
if (b == 0) {
cout <... | ### Prompt
Construct a cpp code solution to the problem outlined:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if it... |
#include <bits/stdc++.h>
using namespace std;
int main() {
int i, n, k = 0, res = 1, n2;
string s;
cin >> s;
if (s[0] != '1') res = 0;
for (i = 0; i < s.length(); i++)
if (s[i] != '1' && s[i] != '4') res = 0;
for (i = 0; i < s.length(); i++) {
if (s[i] == '4') {
k = 0;
while (s[i] == '4'... | ### Prompt
In Cpp, your task is to solve the following problem:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if it i... |
#include <bits/stdc++.h>
using namespace std;
int main() {
long long int n;
cin >> n;
int flag = 0;
while (n > 0) {
if (n % 1000 == 144) {
n = n / 1000;
} else if (n % 100 == 14) {
n = n / 100;
} else if (n % 10 == 1) {
n = n / 10;
} else {
flag = -1;
cout << "NO" <... | ### Prompt
Construct a cpp code solution to the problem outlined:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if it... |
#include <bits/stdc++.h>
using namespace std;
int main() {
string x;
int i = 0;
cin >> x;
int l = x.size() - 1;
while (i <= l) {
if (x[i] == '1' && x[i + 1] == '4' && x[i + 2] == '4')
i += 3;
else if (x[i] == '1' && x[i + 1] == '4')
i += 2;
else if (x[i] == '1')
i++;
else {
... | ### Prompt
Please provide a Cpp coded solution to the problem described below:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. De... |
#include <bits/stdc++.h>
using namespace std;
int INF = 1000000007;
vector<int> a;
void writeln(int a) { printf("%d\n", a); }
void writeln(int a, int b) { printf("%d %d\n", a, b); }
void writeln(int a, int b, int c) { printf("%d %d %d\n", a, b, c); }
void writeln(int a, int b, int c, int d) {
printf("%d %d %d %d\n", ... | ### Prompt
Please create a solution in Cpp to the following problem:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if... |
#include <bits/stdc++.h>
using namespace std;
int main() {
string s;
cin >> s;
for (int i = (int)s.length() - 1; i >= 0;) {
if (s[i] == '1')
i--;
else if (s[i] == '4') {
if (s[i - 1] == '1')
i -= 2;
else if (s[i - 1] == '4' && s[i - 2] == '1') {
i -= 3;
} else {
... | ### Prompt
Your task is to create a CPP solution to the following problem:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determ... |
#include <bits/stdc++.h>
using namespace std;
int main() {
string s;
cin >> s;
int n = (int)s.length();
int i = 0;
while (i < n) {
if ((i <= n - 3) && s[i] == '1' && s[i + 1] == '4' && s[i + 2] == '4') {
i += 3;
} else if (i <= n - 2 && s[i] == '1' && s[i + 1] == '4') {
i += 2;
} else ... | ### Prompt
Your challenge is to write a cpp solution to the following problem:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. De... |
#include <bits/stdc++.h>
using namespace std;
int main() {
string s;
while (cin >> s) {
int i = 0;
int len = s.length();
while (i < len) {
if (s[i] == '1') {
if (s[i + 1] == '4') {
if (s[i + 2] == '4') {
i += 3;
} else
i += 2;
} else
... | ### Prompt
Your task is to create a cpp solution to the following problem:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determ... |
#include <bits/stdc++.h>
using namespace std;
int main() {
string s;
cin >> s;
int flag = 0;
for (long long int i = 0; i < s.length(); i++) {
if (s[0] != '1') {
cout << "NO";
flag = 0;
break;
}
if (s[i] != '1' && s[i] != '4') {
cout << "NO";
flag = 0;
break;
}... | ### Prompt
In cpp, your task is to solve the following problem:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if it i... |
#include <bits/stdc++.h>
using namespace std;
int main() {
long long int n;
cin >> n;
int c = 0;
int f = 1;
while (n != 0) {
int d = n % 10;
if (d == 4) {
c++;
if (c >= 3) {
f = 0;
break;
}
} else if (d == 1) {
c = 0;
} else {
f = 0;
break;
... | ### Prompt
Develop a solution in Cpp to the problem described below:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if... |
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
signed main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
string s;
cin >> s;
bool valid = true;
long long int i = 0;
for (; i < (long long int)s.length() - 2; ++i) {
if (s[i] == '1' && s[i + 1] == '4' && s[i + 2] ==... | ### Prompt
Construct a Cpp code solution to the problem outlined:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if it... |
#include <bits/stdc++.h>
using namespace std;
int main() {
string s;
cin >> s;
bool magic = true;
int i = 0;
while (i < s.length()) {
if (s[i] == '1' && s[i + 1] == '4' && s[i + 2] == '4') {
i += 3;
} else if (s[i] == '1' && s[i + 1] == '4') {
i += 2;
} else if (s[i] == '1') {
i ... | ### Prompt
Generate a Cpp solution to the following problem:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if it is a... |
#include <bits/stdc++.h>
int main() {
long long a;
int i = 0, n = 1;
scanf("%lld", &a);
int b[11] = {0};
while (a != 0) {
b[i] = a % 10;
a /= 10;
i++;
}
i--;
int j;
j = i;
int t = 0;
if (b[i] != 1)
n = 0;
else {
for (i--; i >= 0; i--) {
if (b[i] == 1) {
t = 0;
... | ### Prompt
Construct a CPP code solution to the problem outlined:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if it... |
#include <bits/stdc++.h>
using namespace std;
int main() {
long long int n;
scanf("%lld", &n);
long long int c = 0;
bool flag = true;
while (n > 9) {
long long int r = n % 10;
if (r == 4)
c++;
else
c = 0;
if (c == 3) {
flag = false;
break;
}
if (r != 1 & r != 4)... | ### Prompt
Your challenge is to write a cpp solution to the following problem:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. De... |
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