output stringlengths 52 181k | instruction stringlengths 296 182k |
|---|---|
#include <bits/stdc++.h>
int main() {
int a, b, c, d;
scanf("%d%d%d%d", &a, &b, &c, &d);
if (c > d * 2 || d > c * 2 || d >= b)
printf("-1\n");
else
printf("%d\n%d\n%d\n", a * 2, b * 2, c > d ? c : d);
return 0;
}
| ### Prompt
Please create a solution in Cpp to the following problem:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smalles... |
#include <bits/stdc++.h>
using namespace std;
int main() {
int x1, x2, x3, xm;
cin >> x1 >> x2 >> x3 >> xm;
if (x2 * 2 <= xm * 2 || xm > x3 * 2 || x3 > xm * 2)
cout << -1 << endl;
else {
cout << x1 * 2 << endl;
cout << x2 * 2 << endl;
cout << min(xm, x3) * 2 << endl;
}
return 0;
}
| ### Prompt
Develop a solution in CPP to the problem described below:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smalles... |
#include <bits/stdc++.h>
using namespace std;
int main() {
int v1, v2, v3, vm;
scanf("%d %d %d %d", &v1, &v2, &v3, &vm);
int flag = 0;
for (int i3 = v3; i3 <= 2 * v3; i3++) {
if (vm <= i3 && 2 * vm >= i3) {
for (int i2 = v2; i2 <= 2 * v2; i2++) {
for (int i1 = v1; i1 <= 2 * v1; i1++) {
... | ### Prompt
Your challenge is to write a CPP solution to the following problem:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into t... |
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int MAXN = 100009;
int t, n;
int main() {
int a, b, c, d;
cin >> a >> b >> c >> d;
int x1 = -1;
for (int i = 1; i <= 10000; i++) {
if (i >= a && i >= d && a * 2 >= i && d * 2 < i) {
x1 = i;
break;
}
}
int x2 ... | ### Prompt
Construct a cpp code solution to the problem outlined:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smallest c... |
#include <bits/stdc++.h>
using namespace std;
int main() {
long long n, i, j, k, l, m;
ios_base::sync_with_stdio(0);
cin.tie(0);
long long a, c, b, d;
cin >> a >> b >> c >> d;
for (i = 1; i <= 200; i++) {
for (j = 1; j <= 200; j++) {
for (k = 1; k <= 200; k++) {
if (i >= a && i <= 2 * a &&... | ### Prompt
Construct a Cpp code solution to the problem outlined:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smallest c... |
#include <bits/stdc++.h>
using namespace std;
const int INF = 1000000007;
const double EPS = 1e-9;
int main() {
int v1, v2, v3, vm;
cin >> v1 >> v2 >> v3 >> vm;
int a1, a2, a3;
a3 = max(v3, vm);
if (min(v3, vm) * 2 < a3) {
cout << -1 << endl;
return 0;
}
a2 = max(vm * 2 + 1, max(a3 + 1, v2));
if... | ### Prompt
Your task is to create a cpp solution to the following problem:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the s... |
#include <bits/stdc++.h>
using namespace std;
int main() {
int v1, v2, v3, vm;
scanf("%d", &v1) == 0, scanf("%d", &v2) == 0, scanf("%d", &v3) == 0,
scanf("%d", &vm) == 0;
int rv1 = v1 * 2;
int rv2 = v2 * 2;
int rv3 = -1;
for (decltype(1) i = 1; i < 200; i++) {
if (i >= v3 && v3 * 2 >= i) {
i... | ### Prompt
Construct a Cpp code solution to the problem outlined:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smallest c... |
#include <bits/stdc++.h>
using namespace std;
int main() {
int a, b, c, d, i, k, h;
scanf("%d %d %d %d", &a, &b, &c, &d);
for (i = a; i <= 2 * a; i++)
for (k = b; k <= 2 * b; k++)
for (h = c; h <= 2 * c; h++)
if (h < k && k < i && d <= h && h <= 2 * d && 2 * d < k) {
printf("%d\n%d\n%d... | ### Prompt
Your task is to create a CPP solution to the following problem:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the s... |
#include <bits/stdc++.h>
using namespace std;
signed main(void) {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
long long fat, mot, son, mas;
cin >> fat >> mot >> son >> mas;
long long ans1, ans2, ans3;
long long flag = 0, flag2 = 0, flag3 = 0;
for (__typeof(2 * son + 1) i = (son) - ((son) ... | ### Prompt
Please create a solution in Cpp to the following problem:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smalles... |
#include <bits/stdc++.h>
using namespace std;
template <class T>
T sqr(T x) {
return x * x;
}
long double pi = 3.1415926535897932384626433832795;
long long mod = 1e9 + 7;
const int N = 3e5 + 10;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long bin_pow(long long n, long long p) {
l... | ### Prompt
Construct a CPP code solution to the problem outlined:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smallest c... |
#include <bits/stdc++.h>
using namespace std;
int fact[100005];
int invfact[100005];
bool prime[100005];
vector<long long> divisor[100005];
long long pw(long long a, long long b) {
long long res = 1;
while (b) {
if (b & 1ll) {
res = res * a % 1000000007;
}
b >>= 1;
a = a * a % 1000000007;
}
... | ### Prompt
Construct a Cpp code solution to the problem outlined:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smallest c... |
#include <bits/stdc++.h>
using namespace std;
int main() {
int a, b, c, d;
cin >> a >> b >> c >> d;
if (c > 2 * d || d > 2 * c || d >= b)
cout << -1 << endl;
else
cout << 2 * a << endl << 2 * b << endl << max(c, d);
return 0;
}
| ### Prompt
In CPP, your task is to solve the following problem:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smallest car... |
#include <bits/stdc++.h>
using namespace std;
int main() {
vector<int> v;
int n;
for (int i = 0; i < 3; i++) {
int t;
cin >> t;
v.push_back(t);
}
cin >> n;
for (int i = 0; i <= 200; i++) {
for (int j = i + 1; j <= 200; j++) {
for (int k = j + 1; k <= 200; k++) {
if (k < max(v[0... | ### Prompt
Please formulate a cpp solution to the following problem:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smalles... |
#include <bits/stdc++.h>
using namespace std;
template <typename T>
using V = vector<T>;
bool ok(int a, int b) { return a <= b && 2 * a >= b; }
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int v1, v2, v3, vm;
cin >> v1 >> v2 >> v3 >> vm;
for (int c1 = 1; c1 <= 400; ++c1) {
for (int c2 = c1 + 1; c2 <=... | ### Prompt
Please formulate a Cpp solution to the following problem:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smalles... |
#include <bits/stdc++.h>
using namespace std;
int main() {
int a, b, c, d;
cin >> a >> b >> c >> d;
int s, m, l;
s = c;
l = a;
for (int i = c; i <= 2 * c; i++) {
for (int j = b; j <= 2 * b; j++) {
for (int k = a; k <= 2 * a; k++) {
if (d <= i && d <= j && d <= k) {
if (i != j && ... | ### Prompt
Please provide a CPP coded solution to the problem described below:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into t... |
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
int main() {
int v1, v2, v3, vm;
scanf("%d%d%d%d", &v1, &v2, &v3, &vm);
for (int i = v3; i <= 2 * v3; i++) {
if (vm <= i && 2 * vm >= i)
for (int j = max(i * 2 + 1, v2); j <= 2 * v2; j++) {
for (int k = max(j + 1, v1); k <=... | ### Prompt
Develop a solution in CPP to the problem described below:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smalles... |
#include <bits/stdc++.h>
using namespace std;
int main() {
int a, b, c, d;
while (~scanf("%d%d%d%d", &a, &b, &c, &d)) {
if (d <= 2 * c && 2 * d >= c && d < b) {
c = max(c, d);
if (2 * c >= b) {
b = 2 * c + 1;
}
a = max(a, b + 1);
printf("%d\n%d\n%d\n", a, b, c);
} else ... | ### Prompt
In Cpp, your task is to solve the following problem:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smallest car... |
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using namespace std;
const int N = (int)1e5 + 5;
const int INF = 0x3f3f3f3f;
int a[10];
bool ok(int a, int b) { return (2 * a >= b); }
int main() {
cin >> a[1] >> a[2] >> a[3] >> a[4];
for (int f = a[1]; f <= a[1] * 2; f++) {
... | ### Prompt
Your task is to create a Cpp solution to the following problem:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the s... |
#include <bits/stdc++.h>
using namespace std;
int v1, v2, v3, vm;
int main() {
scanf("%d%d%d%d", &v1, &v2, &v3, &vm);
int flag = 0;
for (int i = v3; i <= 2 * v3; i++) {
if (vm <= i && 2 * vm >= i) {
for (int j = v2; j <= 2 * v2; j++) {
if (vm <= j && 2 * vm < j) {
for (int k = v1; k <=... | ### Prompt
Please create a solution in Cpp to the following problem:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smalles... |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int a, b, c, m;
cin >> a >> b >> c >> m;
for (int i = 0, ThxDem = 300; i < ThxDem; ++i) {
for (int j = 0, ThxDem = i; j < ThxDem; ++j) {
for (int k = 0, ThxDem = j; k < ThxDem; ++k) {
... | ### Prompt
Your task is to create a cpp solution to the following problem:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the s... |
#include <bits/stdc++.h>
const long long MX = 100 * 1000 * 2 + 1000, inf = 0x7FFFFFFF,
mod = 1000 * 1000 * 1000 + 7;
using namespace std;
long long n, siz[5];
pair<long long, long long> inp[5];
int32_t main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
for (long long i = 0; i < 4; i++) {... | ### Prompt
Please provide a cpp coded solution to the problem described below:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into t... |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int a, b, c, d;
cin >> a >> b >> c >> d;
for (size_t i = 1; i <= 200; i++) {
if (a <= i && 2 * a >= i)
for (size_t j = 1; j < i; j++) {
if (b <= j && 2 * b >= j)
... | ### Prompt
Develop a solution in CPP to the problem described below:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smalles... |
#include <bits/stdc++.h>
using namespace std;
int main() {
int v1, v2, v3, vm;
cin >> v1 >> v2 >> v3 >> vm;
for (int a1 = 1; a1 <= 210; a1++)
for (int a2 = 1; a2 <= 210; a2++)
for (int a3 = 1; a3 <= 210; a3++) {
if (a1 <= 2 * v1 && 2 * v1 <= 2 * a1 && a2 <= 2 * v2 &&
2 * v2 <= 2 * a2... | ### Prompt
Develop a solution in cpp to the problem described below:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smalles... |
#include <bits/stdc++.h>
using namespace std;
int main() {
int v1, v2, v3, v4;
cin >> v1 >> v2 >> v3 >> v4;
if (v3 * 2 < v4 || v3 > 2 * v4 || v4 >= v2) {
cout << "-1" << endl;
return 0;
}
cout << v1 * 2 << endl;
cout << v2 * 2 << endl;
cout << max(v3, v4) << endl;
}
| ### Prompt
In CPP, your task is to solve the following problem:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smallest car... |
#include <bits/stdc++.h>
using namespace std;
int a, b, c, d, ans = 0;
int aa, bb, cc;
int main() {
cin >> a >> b >> c >> d;
cc = max(c, d);
bb = max(2 * d + 1, b);
bb = max(cc + 1, bb);
aa = max(a, bb + 1);
if (c * 2 < cc || b * 2 < bb || a * 2 < aa || d * 2 < cc)
cout << -1 << endl;
else
cout <<... | ### Prompt
Generate a Cpp solution to the following problem:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smallest car an... |
#include <bits/stdc++.h>
int32_t main() {
std::ios::sync_with_stdio(false);
int a, b, c, d;
std::cin >> a >> b >> c >> d;
for (int i = 0; i <= 200; ++i) {
for (int j = i + 1; j <= 200; ++j) {
for (int q = j + 1; q <= 200; ++q) {
if (i >= c && 2 * c >= i && j >= b && 2 * b >= j && q >= a &&
... | ### Prompt
Please create a solution in cpp to the following problem:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smalles... |
#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1.0);
const double eps = 1e-11;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const long long N = 100010;
int main() {
int a, b, c, d;
scanf("%d%d%d%d", &a, &b, &c, &d);
for (int i = a; i <= a * 2; i++)
for (int j = b; j <= b * 2; j++) {
... | ### Prompt
Please create a solution in cpp to the following problem:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smalles... |
#include <bits/stdc++.h>
using namespace std;
int main() {
int a, b, c, d;
cin >> a >> b >> c >> d;
if (d > c * 2 || c > d * 2) {
cout << -1 << endl;
return 0;
}
if (d * 2 >= b || max(c, d) >= b) {
b *= 2;
a *= 2;
}
if (d * 2 >= b || max(c, d) >= b) {
cout << -1 << endl;
return 0;
... | ### Prompt
Your task is to create a CPP solution to the following problem:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the s... |
#include <bits/stdc++.h>
using namespace std;
bool fit(long long a, long long b) { return a <= b; }
bool like(long long a, long long b) { return 2 * a >= b; }
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long v1, v2, v3, v4;
cin >> v1 >> v2 >> v3 >> v4;
for (long long a = 1; a <= 201; a+... | ### Prompt
Please provide a CPP coded solution to the problem described below:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into t... |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
int a, b, c, d;
cin >> a >> b >> c >> d;
for (int i = 1; i <= 200; i++) {
for (int j = 1; j <= 200; j++) {
for (int z = 1; z <= 200; z++) {
if (i >= a && j >= b && z >= c && z >= d && i >... | ### Prompt
Your challenge is to write a cpp solution to the following problem:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into t... |
#include <bits/stdc++.h>
using namespace std;
int main() {
int a, b, c, d;
scanf("%d %d %d %d", &a, &b, &c, &d);
a *= 2;
b *= 2;
if (2 * d >= b or max(c, d) >= b or 2 * min(c, d) < max(c, d))
return printf("-1\n"), 0;
for (int i = max(c, d); i <= 2 * min(c, d); ++i) {
if (2 * c >= i and i >= c and 2... | ### Prompt
Please provide a Cpp coded solution to the problem described below:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into t... |
#include <bits/stdc++.h>
int main() {
int max, a, b, c, d;
scanf("%d %d %d %d", &a, &b, &c, &d);
if (d >= b || 2 * d < c || 2 * c < d) {
printf("-1");
} else {
max = c > d ? c : d;
printf("%d\n%d\n%d", 2 * a, 2 * b, max);
}
return 0;
}
| ### Prompt
Construct a cpp code solution to the problem outlined:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smallest c... |
#include <bits/stdc++.h>
using namespace std;
void car(int a, int b, int c, int m) {
int s, p, d;
s = c;
if (m > 2 * c) {
cout << -1;
} else if (2 * m < c) {
cout << -1;
} else if (m > c) {
s = m;
p = b;
if (b > 2 * m) {
p = b;
if (2 * a > p) {
d = 2 * a;
cout <... | ### Prompt
Create a solution in CPP for the following problem:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smallest car ... |
#include <bits/stdc++.h>
using namespace std;
int main() {
int v1, v2, v3, vm;
cin >> v1 >> v2 >> v3 >> vm;
for (auto i = 200; i >= 0; i--) {
for (auto j = 200; j >= 0; j--) {
for (auto k = 200; k >= 0; k--) {
if (i > j && j > k && i >= v1 && 2 * v1 >= i &&
!(i >= vm && 2 * vm >= i) ... | ### Prompt
Your task is to create a cpp solution to the following problem:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the s... |
#include <bits/stdc++.h>
using namespace std;
void out() {
cout << -1;
exit(0);
}
int main() {
int v1, v2, v3, v4;
cin >> v1 >> v2 >> v3 >> v4;
for (int s1 = 1; s1 < 220; s1++) {
for (int s2 = 1; s2 < s1; s2++) {
for (int s3 = 1; s3 < s2; s3++) {
if (v1 <= s1 and v1 * 2 >= s1)
if (... | ### Prompt
In Cpp, your task is to solve the following problem:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smallest car... |
#include <bits/stdc++.h>
using namespace std;
bool comp(long long x, long long y) {
if (x >= y) return true;
return false;
}
bool compare(const pair<long long, long long> &i,
const pair<long long, long long> &j) {
return i.first <= j.first;
}
bool compare1(const pair<long long, long long> &i,
... | ### Prompt
In Cpp, your task is to solve the following problem:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smallest car... |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base ::sync_with_stdio(0);
cin.tie(0);
int a, b, c, d;
cin >> a >> b >> c >> d;
for (int i = c; i <= 10000; ++i) {
if (d <= i and 2 * d >= i and c <= i and 2 * c >= i and 2 * d < 2 * a and
2 * d < 2 * b) {
cout << 2 * a << '\n' <... | ### Prompt
Please formulate a Cpp solution to the following problem:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smalles... |
#include <bits/stdc++.h>
using namespace std;
class Global {
public:
int n, i, k, ot = -1;
vector<int> v, w, r, o;
char ch;
void pref(vector<int>& v) {
int x;
for (i = x = 0; i < v.size(); ++i) {
x += v[i];
v[i] = x;
}
}
void start() {
cin >> n >> k;
v.resize(n);
for (i ... | ### Prompt
Please provide a Cpp coded solution to the problem described below:
Arkady decided to buy roses for his girlfriend.
A flower shop has white, orange and red roses, and the total amount of them is n. Arkady thinks that red roses are not good together with white roses, so he won't buy a bouquet containing bo... |
#include <bits/stdc++.h>
using namespace std;
template <class T>
bool umin(T& a, T b) {
if (a > b) {
a = b;
return 1;
}
return 0;
}
template <class T>
bool umax(T& a, T b) {
if (a < b) {
a = b;
return 1;
}
return 0;
}
char s[200009];
int arr[200009];
vector<int> adj[3], par[3];
int main() {
... | ### Prompt
Create a solution in Cpp for the following problem:
Arkady decided to buy roses for his girlfriend.
A flower shop has white, orange and red roses, and the total amount of them is n. Arkady thinks that red roses are not good together with white roses, so he won't buy a bouquet containing both red and white... |
#include <bits/stdc++.h>
using namespace std;
int dbg = 0;
const int N = 2e5 + 1;
const int lg = 20;
const long long mod = 0xfffff;
void solve() {
int n, k;
cin >> n >> k;
vector<int> b(n, 0);
for (int i = 0; i < n; i++) cin >> b[i];
string s;
cin >> s;
if (k == 1) {
cout << -1 << "\n";
return;
... | ### Prompt
Develop a solution in Cpp to the problem described below:
Arkady decided to buy roses for his girlfriend.
A flower shop has white, orange and red roses, and the total amount of them is n. Arkady thinks that red roses are not good together with white roses, so he won't buy a bouquet containing both red and... |
#include <bits/stdc++.h>
using namespace std;
int main() {
string s;
long long n, b;
cin >> n >> b;
vector<pair<long long, char> > v(n);
for (long long i = 0; i < n; i++) {
cin >> v[i].first;
}
cin >> s;
for (long long i = 0; i < n; i++) {
v[i].second = s[i];
}
bool ORANGE = false;
bool RE... | ### Prompt
Please formulate a cpp solution to the following problem:
Arkady decided to buy roses for his girlfriend.
A flower shop has white, orange and red roses, and the total amount of them is n. Arkady thinks that red roses are not good together with white roses, so he won't buy a bouquet containing both red and... |
#include <bits/stdc++.h>
const long long LINF = 1e18;
const int INF = 1e9;
const int M = 1e9 + 7;
const double EPS = 1e-9;
using namespace std;
struct Rose {
int beauty;
char col;
} a[202000];
bool cmp(Rose a, Rose b) {
if (a.beauty <= b.beauty) return 0;
return 1;
}
int main(int argc, const char* argv[]) {
s... | ### Prompt
In cpp, your task is to solve the following problem:
Arkady decided to buy roses for his girlfriend.
A flower shop has white, orange and red roses, and the total amount of them is n. Arkady thinks that red roses are not good together with white roses, so he won't buy a bouquet containing both red and whit... |
#include <bits/stdc++.h>
using namespace std;
long long n, k;
vector<long long> r, w, o;
vector<long long> v;
string col;
void panic() {
cout << -1 << endl;
exit(0);
}
long long solve_ro() {
long long anw = 0;
if (o.size() == 0 || r.size() == 0) {
return -1;
}
anw += o[0] + r[0];
int o_ptr = 1, r_ptr ... | ### Prompt
Please formulate a CPP solution to the following problem:
Arkady decided to buy roses for his girlfriend.
A flower shop has white, orange and red roses, and the total amount of them is n. Arkady thinks that red roses are not good together with white roses, so he won't buy a bouquet containing both red and... |
#include <bits/stdc++.h>
using namespace std;
const int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
const int dxx[8] = {-1, -1, 0, 1, 1, 1, 0, -1},
dyy[8] = {0, 1, 1, 1, 0, -1, -1, -1};
const long long mod = 1000000007;
const int base = 311;
const int N = 200005;
int n, k, b[N], a[N];
vector<int> vec[3];
vo... | ### Prompt
Develop a solution in CPP to the problem described below:
Arkady decided to buy roses for his girlfriend.
A flower shop has white, orange and red roses, and the total amount of them is n. Arkady thinks that red roses are not good together with white roses, so he won't buy a bouquet containing both red and... |
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, k;
cin >> n >> k;
vector<int> arr(n);
for (int i = 0; i < n; i++) cin >> arr[i];
vector<char> str(n);
int k_orange = 0;
int k_red = 0;
int k_white = 0;
for (int i = 0; i < n; i++) {
cin >> str[i];
if (str[i] == 'O') {
k_or... | ### Prompt
In Cpp, your task is to solve the following problem:
Arkady decided to buy roses for his girlfriend.
A flower shop has white, orange and red roses, and the total amount of them is n. Arkady thinks that red roses are not good together with white roses, so he won't buy a bouquet containing both red and whit... |
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 36;
const int INF = 1e9 + 7;
long long a[N];
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
long long n, k;
cin >> n >> k;
for (int i(0); i < n; i++) {
cin >> a[i];
}
string s;
cin >> s;
vector<int> o, r, ... | ### Prompt
Your challenge is to write a cpp solution to the following problem:
Arkady decided to buy roses for his girlfriend.
A flower shop has white, orange and red roses, and the total amount of them is n. Arkady thinks that red roses are not good together with white roses, so he won't buy a bouquet containing bo... |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int n, k;
cin >> n >> k;
int a[n];
for (int i = 0; i < n; i++) cin >> a[i];
string s;
cin >> s;
if (k == 1) {
cout << -1;
return 0;
}
unordered_map<char, vector<int... | ### Prompt
Please provide a CPP coded solution to the problem described below:
Arkady decided to buy roses for his girlfriend.
A flower shop has white, orange and red roses, and the total amount of them is n. Arkady thinks that red roses are not good together with white roses, so he won't buy a bouquet containing bo... |
#include <bits/stdc++.h>
using namespace std;
int n, k;
int Beauty[200002];
string s;
int Rose[200002], SpRose[200002], szRose;
int RedRose[200002], SpRedRose[200002], szRedRose;
int WhiteRose[200002], SpWhiteRose[200002], szWhiteRose;
int sol = -1;
void Welcome_to_Rose() {
ios_base::sync_with_stdio(false);
cin.tie... | ### Prompt
Your challenge is to write a cpp solution to the following problem:
Arkady decided to buy roses for his girlfriend.
A flower shop has white, orange and red roses, and the total amount of them is n. Arkady thinks that red roses are not good together with white roses, so he won't buy a bouquet containing bo... |
#include <bits/stdc++.h>
using namespace std;
int c[200005], B[200005], O[200005], R[200005];
char s[200005];
int n, k, z[200005];
long long solve(int *x, int *y) {
if (!x[0] || !y[0]) return 0;
long long ans = x[1] + y[1];
z[0] = 0;
for (int i = 2; i <= x[0]; i++) z[++z[0]] = x[i];
for (int i = 2; i <= y[0];... | ### Prompt
Please formulate a Cpp solution to the following problem:
Arkady decided to buy roses for his girlfriend.
A flower shop has white, orange and red roses, and the total amount of them is n. Arkady thinks that red roses are not good together with white roses, so he won't buy a bouquet containing both red and... |
#include <bits/stdc++.h>
using namespace std;
long long n, k, a[200005], W[200005], O[200005], R[2000005], ans;
long long r = 0;
long long w = 0;
long long o = 0;
string s;
bool kt(long long a, long long b) { return (a > b); }
int32_t main() {
cin >> n >> k;
for (long long i = 1; i <= n; i++) {
cin >> a[i];
}... | ### Prompt
Construct a CPP code solution to the problem outlined:
Arkady decided to buy roses for his girlfriend.
A flower shop has white, orange and red roses, and the total amount of them is n. Arkady thinks that red roses are not good together with white roses, so he won't buy a bouquet containing both red and wh... |
#include <bits/stdc++.h>
using namespace std;
class Solution {
int n, k;
vector<int> A[3];
public:
void run() {
cin >> n >> k;
if (k < 2) return cout << -1 << '\n', void();
{
vector<int> B;
B.resize(n);
for (int i = 0; i < n; ++i) cin >> B[i];
string s;
cin >> s;
... | ### Prompt
Create a solution in Cpp for the following problem:
Arkady decided to buy roses for his girlfriend.
A flower shop has white, orange and red roses, and the total amount of them is n. Arkady thinks that red roses are not good together with white roses, so he won't buy a bouquet containing both red and white... |
#include <bits/stdc++.h>
using namespace std;
long long getTop(long long prefix[], long long n, long long size) {
if (n == 0 || size == 0) return 0;
return prefix[n - 1];
}
const long long maxn = 200005;
long long redPrefix[maxn], orangePrefix[maxn], whitePrefix[maxn];
vector<long long> red, orange, white;
long lon... | ### Prompt
Please create a solution in Cpp to the following problem:
Arkady decided to buy roses for his girlfriend.
A flower shop has white, orange and red roses, and the total amount of them is n. Arkady thinks that red roses are not good together with white roses, so he won't buy a bouquet containing both red and... |
#include <bits/stdc++.h>
using namespace std;
const long long inf = 1e18;
const int N = 2e5 + 1;
auto rnd = bind(uniform_int_distribution<int>(1, 10000), mt19937(time(0)));
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int n, k;
cin >> n >> k;
vector<int> a(n);
vector<char> c(... | ### Prompt
Develop a solution in CPP to the problem described below:
Arkady decided to buy roses for his girlfriend.
A flower shop has white, orange and red roses, and the total amount of them is n. Arkady thinks that red roses are not good together with white roses, so he won't buy a bouquet containing both red and... |
#include <bits/stdc++.h>
using namespace std;
string str;
vector<long long> red, orange, white;
long long i, k;
long long f(vector<long long> red, vector<long long> orange) {
vector<pair<long long, char>> ro;
long long ro_sum = 0, wo_sum = 0;
bool seen_red = false, seen_orange = false;
if (red.size() + orange.s... | ### Prompt
In cpp, your task is to solve the following problem:
Arkady decided to buy roses for his girlfriend.
A flower shop has white, orange and red roses, and the total amount of them is n. Arkady thinks that red roses are not good together with white roses, so he won't buy a bouquet containing both red and whit... |
#include <bits/stdc++.h>
using namespace std;
long long k, n;
long long b[200011];
vector<long long> a[3];
string s;
int main() {
cin >> n >> k;
for (long long i = (long long)(0); i < (long long)(n); i++) {
cin >> b[i];
}
cin >> s;
if (k == 1) {
cout << -1 << endl;
return 0;
}
for (long long i... | ### Prompt
Create a solution in CPP for the following problem:
Arkady decided to buy roses for his girlfriend.
A flower shop has white, orange and red roses, and the total amount of them is n. Arkady thinks that red roses are not good together with white roses, so he won't buy a bouquet containing both red and white... |
#include <bits/stdc++.h>
using namespace std;
struct output {};
struct input {};
template <typename T>
output operator,(output l, const T& r) {
cout << r;
return l;
}
template <typename T>
input operator,(input l, T& r) {
cin >> r;
return l;
}
input I;
output P;
char sp = ' ', nl = '\n';
string spp = ", ", sdp ... | ### Prompt
Construct a CPP code solution to the problem outlined:
Arkady decided to buy roses for his girlfriend.
A flower shop has white, orange and red roses, and the total amount of them is n. Arkady thinks that red roses are not good together with white roses, so he won't buy a bouquet containing both red and wh... |
#include <bits/stdc++.h>
using namespace std;
const int N = (int)3e5;
int main(void) {
int n, k;
char buf;
long long a[N];
vector<long long> red, white, orange;
scanf("%d%d", &n, &k);
for (int i = 0; i < n; i++) scanf("%I64d", a + i);
scanf("%c", &buf);
for (int i = 0; i < n; i++) {
scanf("%c", &buf... | ### Prompt
In cpp, your task is to solve the following problem:
Arkady decided to buy roses for his girlfriend.
A flower shop has white, orange and red roses, and the total amount of them is n. Arkady thinks that red roses are not good together with white roses, so he won't buy a bouquet containing both red and whit... |
#include <bits/stdc++.h>
using namespace std;
int s[200010];
pair<int, int> q1[200010], q2[200010];
int num1 = 0, num2 = 0;
char st[200010];
int main() {
int n, k;
scanf("%d%d", &n, &k);
for (int i = 1; i <= n; ++i) scanf("%d", &s[i]);
scanf("%s", st + 1);
for (int i = 1; i <= n; ++i)
if (st[i] == 'R')
... | ### Prompt
Construct a CPP code solution to the problem outlined:
Arkady decided to buy roses for his girlfriend.
A flower shop has white, orange and red roses, and the total amount of them is n. Arkady thinks that red roses are not good together with white roses, so he won't buy a bouquet containing both red and wh... |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
int n, k;
cin >> n >> k;
vector<int> arr(n);
for (int i = 0; i < n; ++i) cin >> arr[i];
string s;
cin >> s;
vector<int> red, orange, white;
for (int i = 0; i < n; ++i) {
if (s[i] == 'W') whit... | ### Prompt
Please create a solution in Cpp to the following problem:
Arkady decided to buy roses for his girlfriend.
A flower shop has white, orange and red roses, and the total amount of them is n. Arkady thinks that red roses are not good together with white roses, so he won't buy a bouquet containing both red and... |
#include <bits/stdc++.h>
using namespace std;
const int MAX = 2e5 + 5;
const int INF = (1 << 31) - 1;
vector<int> values[3];
string s;
int b[MAX];
int n, k;
int roll(int id) {
int total = values[id].back() + values[2].back();
int ptr_i = (int)values[id].size() - 2;
int ptr_2 = (int)values[2].size() - 2;
int lef... | ### Prompt
Please formulate a CPP solution to the following problem:
Arkady decided to buy roses for his girlfriend.
A flower shop has white, orange and red roses, and the total amount of them is n. Arkady thinks that red roses are not good together with white roses, so he won't buy a bouquet containing both red and... |
#include <bits/stdc++.h>
using namespace std;
const int N = (int)(2e5 + 5);
int n, k;
int a[N];
string second;
vector<int> R, O, W;
int solve(const vector<int> &a, const vector<int> &b) {
int sum = -1;
if (a.size() > 0 && b.size() > 0 && a.size() + b.size() >= k) {
sum = a[0] + b[0];
int i = 1;
int j = ... | ### Prompt
Please provide a CPP coded solution to the problem described below:
Arkady decided to buy roses for his girlfriend.
A flower shop has white, orange and red roses, and the total amount of them is n. Arkady thinks that red roses are not good together with white roses, so he won't buy a bouquet containing bo... |
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
const int mod = 1e9 + 7, mod1 = 1e9 + 17, bas = 29;
const int nmax = 1e5 + 10;
int st[nmax][2], h[nmax][2];
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int n, k;
cin >> n >> k;
pair<int, int> a[n];
for (int i ... | ### Prompt
Create a solution in CPP for the following problem:
Arkady decided to buy roses for his girlfriend.
A flower shop has white, orange and red roses, and the total amount of them is n. Arkady thinks that red roses are not good together with white roses, so he won't buy a bouquet containing both red and white... |
#include <bits/stdc++.h>
int b[200010];
int b1[200010];
int b2[200010];
int b3[200010];
using namespace std;
int main() {
int n, k;
scanf("%d%d", &n, &k);
for (int i = 0; i < n; i++) {
scanf("%d", &b[i]);
}
char* c;
c = (char*)malloc(n + 1);
scanf("%s", c);
int n1 = 0, n2 = 0, n3 = 0;
for (int i =... | ### Prompt
Your task is to create a Cpp solution to the following problem:
Arkady decided to buy roses for his girlfriend.
A flower shop has white, orange and red roses, and the total amount of them is n. Arkady thinks that red roses are not good together with white roses, so he won't buy a bouquet containing both r... |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(0), cin.tie(0);
int N, K;
cin >> N >> K;
if (K == 1) {
cout << -1 << '\n';
exit(0);
}
vector<int> B(N);
for (int i = 0; i < N; i++) {
cin >> B[i];
}
string C;
cin >> C;
vector<int> reds;
vector<int> white... | ### Prompt
Please provide a CPP coded solution to the problem described below:
Arkady decided to buy roses for his girlfriend.
A flower shop has white, orange and red roses, and the total amount of them is n. Arkady thinks that red roses are not good together with white roses, so he won't buy a bouquet containing bo... |
#include <bits/stdc++.h>
using namespace std;
int a[200500];
vector<int> r;
vector<int> o;
vector<int> w;
vector<pair<int, char> > ro;
vector<pair<int, char> > wo;
int main() {
int n;
cin >> n;
int k;
cin >> k;
if (k == 1) {
cout << -1;
return 0;
}
for (int i = 0; i < n; i++) {
cin >> a[i];
... | ### Prompt
In cpp, your task is to solve the following problem:
Arkady decided to buy roses for his girlfriend.
A flower shop has white, orange and red roses, and the total amount of them is n. Arkady thinks that red roses are not good together with white roses, so he won't buy a bouquet containing both red and whit... |
#include <bits/stdc++.h>
using namespace std;
vector<int> r, o, w;
int n, k, b[200001], ans = -1;
char s[200001];
bool cmp(int x, int y) { return x > y; }
int rua(vector<int> x, vector<int> y) {
int _1 = 1, _2 = 1, res = x[0] + y[0];
for (int i = 3; i <= k; i++) {
if (_1 >= x.size()) {
res += y[_2++];
... | ### Prompt
Construct a Cpp code solution to the problem outlined:
Arkady decided to buy roses for his girlfriend.
A flower shop has white, orange and red roses, and the total amount of them is n. Arkady thinks that red roses are not good together with white roses, so he won't buy a bouquet containing both red and wh... |
#include <bits/stdc++.h>
using namespace std;
long long solve(vector<long long> a, vector<long long> b, long long k) {
if (a.size() + b.size() < k) return -1;
long long ans = -1;
long long suma = 0;
vector<long long> preb;
preb.push_back(0);
for (int i = 0; i < b.size(); i++) {
long long x = preb[i] + b... | ### Prompt
Generate a Cpp solution to the following problem:
Arkady decided to buy roses for his girlfriend.
A flower shop has white, orange and red roses, and the total amount of them is n. Arkady thinks that red roses are not good together with white roses, so he won't buy a bouquet containing both red and white r... |
#include <bits/stdc++.h>
const long long LINF = 1e18;
const int INF = 1e9;
const int M = 1e9 + 7;
const double EPS = 1e-9;
using namespace std;
int c[200200];
vector<int> orange;
vector<int> red;
vector<int> white;
string s;
int n, k;
long long ansR() {
if (!red.size() || !orange.size() || (orange.size() + red.size()... | ### Prompt
Generate a Cpp solution to the following problem:
Arkady decided to buy roses for his girlfriend.
A flower shop has white, orange and red roses, and the total amount of them is n. Arkady thinks that red roses are not good together with white roses, so he won't buy a bouquet containing both red and white r... |
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
const long long mod = 1e9 + 7, mod1 = 1e9 + 17, bas = 29;
const long long nmax = 1e5 + 10;
long long st[nmax][2], h[nmax][2];
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long n, k;
cin >> n >> k;
pair<long ... | ### Prompt
Please create a solution in CPP to the following problem:
Arkady decided to buy roses for his girlfriend.
A flower shop has white, orange and red roses, and the total amount of them is n. Arkady thinks that red roses are not good together with white roses, so he won't buy a bouquet containing both red and... |
#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e18 + 1;
const int MAX = 200001;
const long long MOD = 1000000007;
struct trio {
int first;
int second;
int third;
};
struct long_trio {
long long first;
long long second;
long long third;
};
double dist_point_line(int x1, int y1, int x2, ... | ### Prompt
Your task is to create a Cpp solution to the following problem:
Arkady decided to buy roses for his girlfriend.
A flower shop has white, orange and red roses, and the total amount of them is n. Arkady thinks that red roses are not good together with white roses, so he won't buy a bouquet containing both r... |
#include <bits/stdc++.h>
using namespace std;
struct box {
int v, n;
};
int a, b = 0, c = 0, d, e, q[200007];
box q1[200007], q2[200007];
int mw = 0, mr = 0, mo = 0, nw = 0, nr = 0, no = 0, bw = 0, br = 0, bo = 0;
string s;
bool cmp(box x1, box x2) { return x1.v > x2.v; }
int main() {
cin >> a >> d;
if (d == 1) {... | ### Prompt
Generate a cpp solution to the following problem:
Arkady decided to buy roses for his girlfriend.
A flower shop has white, orange and red roses, and the total amount of them is n. Arkady thinks that red roses are not good together with white roses, so he won't buy a bouquet containing both red and white r... |
#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1000000007;
long long POW(long long a, long long b, long long MMM = MOD) {
long long ret = 1;
for (; b; b >>= 1, a = (a * a) % MMM)
if (b & 1) ret = (ret * a) % MMM;
return ret;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a ... | ### Prompt
Please create a solution in cpp to the following problem:
Arkady decided to buy roses for his girlfriend.
A flower shop has white, orange and red roses, and the total amount of them is n. Arkady thinks that red roses are not good together with white roses, so he won't buy a bouquet containing both red and... |
#include <bits/stdc++.h>
using namespace std;
long long a[200005];
long long r[200005];
long long o[200005];
long long w[200005];
long long n, k;
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> n >> k;
for (long long i = 1; i <= n; i++) {
cin >> a[i];
}
long long cnt... | ### Prompt
Please provide a Cpp coded solution to the problem described below:
Arkady decided to buy roses for his girlfriend.
A flower shop has white, orange and red roses, and the total amount of them is n. Arkady thinks that red roses are not good together with white roses, so he won't buy a bouquet containing bo... |
#include <bits/stdc++.h>
using namespace std;
string s;
vector<long long> a;
vector<long long> r;
vector<long long> o;
vector<long long> w;
long long n, k;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
long long i, j;
cin >> n >> k;
for (i = 0; i < n; i++) {
long long q;
cin >> q;
a.push_... | ### Prompt
Please provide a Cpp coded solution to the problem described below:
Arkady decided to buy roses for his girlfriend.
A flower shop has white, orange and red roses, and the total amount of them is n. Arkady thinks that red roses are not good together with white roses, so he won't buy a bouquet containing bo... |
#include <bits/stdc++.h>
using namespace std;
const long long inf = 1e18;
int n, k;
vector<int> mas;
string c;
vector<int> ms[3];
vector<long long> pref[3];
int hsh(char a) {
if (a == 'W') return 0;
if (a == 'O') return 1;
if (a == 'R') return 2;
return 4;
}
void calc(int v) {
pref[v].resize(ms[v].size());
... | ### Prompt
Your task is to create a Cpp solution to the following problem:
Arkady decided to buy roses for his girlfriend.
A flower shop has white, orange and red roses, and the total amount of them is n. Arkady thinks that red roses are not good together with white roses, so he won't buy a bouquet containing both r... |
#include <bits/stdc++.h>
using namespace std;
vector<bool> visited;
vector<vector<int>> edges;
long long ans;
int n;
int magic = 550;
int maxk = 2;
vector<vector<int>> depth, children;
void solve(int u) {
int d = children[u].size();
vector<vector<int>> values(d + 2);
for (int i = 0; i < d; ++i)
for (int j = 1... | ### Prompt
Develop a solution in CPP to the problem described below:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k o... |
#include <bits/stdc++.h>
using namespace std;
const int LIMIT = 20;
int n, dep[300001];
vector<int> adj[300001];
int dp[300001][LIMIT], dp2[300001][LIMIT];
void dfs(int x, int pre) {
vector<int> child;
for (auto y : adj[x]) {
if (y != pre) {
dfs(y, x);
child.push_back(y);
dep[x] = max(dep[x], ... | ### Prompt
In Cpp, your task is to solve the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of its... |
#include <bits/stdc++.h>
using namespace std;
int n;
vector<int> vec[300005];
int niche[300005];
int nije[300005][21], dp[300005][21];
long long ans;
void dfs(int s, int p = 0) {
niche[s] = 1;
for (int v : vec[s]) {
if (v == p) continue;
dfs(v, s);
niche[s] = max(niche[s], niche[v] + 1);
}
nije[s][1... | ### Prompt
Your task is to create a Cpp solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at lea... |
#include <bits/stdc++.h>
const int MAXN = 300010;
int head[MAXN], nxt[MAXN << 1], to[MAXN << 1], tot;
void adde(int b, int e) {
nxt[++tot] = head[b];
to[head[b] = tot] = e;
nxt[++tot] = head[e];
to[head[e] = tot] = b;
}
std::vector<int> dp[MAXN];
int deg[MAXN];
std::priority_queue<int, std::vector<int>, std::gr... | ### Prompt
Create a solution in cpp for the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of its ... |
#include <bits/stdc++.h>
using namespace std;
vector<int> g[300005];
int n;
int f[300005][32];
int w[300005];
int a[300005];
int m;
long long r;
void dfs(int x, int p) {
w[x] = 1;
for (int y : g[x])
if (y != p) {
dfs(y, x);
w[x] = max(w[x], 1 + w[y]);
}
f[x][1] = n;
for (int d = 2, ThxDem = ... | ### Prompt
Please create a solution in cpp to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k o... |
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3fffffff;
const int SINF = 0x7fffffff;
const long long LINF = 0x3fffffffffffffff;
const long long SLINF = 0x7fffffffffffffff;
const int MAXN = 300007;
const int MAXL = 21;
int n;
long long ans;
vector<int> ch[MAXN];
vector<int> dp[MAXN];
vector<int> tmp[MA... | ### Prompt
Please formulate a CPP solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k o... |
#include <bits/stdc++.h>
using namespace std;
inline void setmin(int &x, int y) {
if (y < x) x = y;
}
inline void setmax(int &x, int y) {
if (y > x) x = y;
}
inline void setmin(long long &x, long long y) {
if (y < x) x = y;
}
inline void setmax(long long &x, long long y) {
if (y > x) x = y;
}
const int N = 3000... | ### Prompt
Please provide a cpp coded solution to the problem described below:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at... |
#include <bits/stdc++.h>
using namespace std;
char nc() {
static char buf[100000], *p1 = buf, *p2 = buf;
return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
? EOF
: *p1++;
}
void Read(int& x) {
char c = nc();
for (; c < '0' || c > '9'; c = nc())
;
for (x... | ### Prompt
In CPP, your task is to solve the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of its... |
#include <bits/stdc++.h>
using namespace std;
const int MOD = 95542721;
const int INF = 2e9;
const long long INF64 = 4e18;
const long double EPS = 1e-16;
const long long MD = 1551513443;
const long long T = 2543;
const int N = 300010;
const int M = 100;
const int DEBUG = 0;
const int MAGIC = 320;
int n;
long long ans;
... | ### Prompt
Construct a CPP code solution to the problem outlined:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of i... |
#include <bits/stdc++.h>
using std::make_pair;
using std::max;
using std::pair;
using std::vector;
const int w = 20, N = 300005;
int to[N << 1], edge, last[N], Next[N << 1], dp[N][w], b[N], n, x, y, f[N],
g[N];
long long ans, sum;
vector<pair<int, int> > v[N];
void add(int x, int y) {
to[++edge] = y;
Next[edge]... | ### Prompt
Generate a cpp solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of its ch... |
#include <bits/stdc++.h>
using namespace std;
const int N = 300005;
vector<int> e[N];
int d[N], q[N], n;
int f[N][20];
long long sum;
void dfs(int x, int fa) {
for (auto i : e[x])
if (i != fa) {
dfs(i, x);
d[x] = max(d[x], d[i]);
}
d[x]++;
sum += d[x];
f[x][1] = n;
for (int j = (int)(2); j... | ### Prompt
In cpp, your task is to solve the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of its... |
#include <bits/stdc++.h>
using namespace std;
vector<vector<pair<int, int> > > vec(323456);
vector<vector<int> > adj(312345);
int lev[312345], maxdepth[312345];
long long bigg = 0;
int jaja[312345][19];
vector<int> wow;
int n;
int dfs(int cur, int par) {
int i;
lev[cur] = 1;
maxdepth[cur] = 0;
for (i = 0; i < a... | ### Prompt
Generate a cpp solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of its ch... |
#include <bits/stdc++.h>
using namespace std;
long long ans;
int dp[300000 + 5][20], n;
vector<int> e[300000 + 5];
inline void upd(int &x, int y) {
if (x < y) x = y;
}
int dfs(int now, int fa) {
int mxdep = 0;
for (auto v : e[now])
if (v != fa) upd(mxdep, dfs(v, now));
mxdep++;
ans += mxdep;
dp[now][1] ... | ### Prompt
Your task is to create a CPP solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at lea... |
#include <bits/stdc++.h>
using namespace std;
const long long N = 300005;
long long n, dp[N][20][2], ans, h[N];
vector<vector<long long> > gr;
priority_queue<long long> pq[N][20];
void dfs(long long u, long long par) {
dp[u][1][0] = dp[u][1][1] = n;
for (long long i = 0; i < gr[u].size(); ++i) {
long long v = g... | ### Prompt
Please formulate a CPP solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k o... |
#include <bits/stdc++.h>
using namespace std;
bool debug = 0;
int n, m, k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
string direc = "RDLU";
long long ln, lk, lm;
void etp(bool f = 0) {
puts(f ? "YES" : "NO");
exit(0);
}
void addmod(int &x, int y, int mod = 1000000007) {
assert(y >= 0);
x += y;
if (x >... | ### Prompt
Please formulate a cpp solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k o... |
#include <bits/stdc++.h>
using namespace std;
struct point {
int id, d;
};
vector<point> p[300010];
inline int read() {
int x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-') f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();... | ### Prompt
Please formulate a Cpp solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k o... |
#include <bits/stdc++.h>
using namespace std;
int n, x, y, dep[300010], v[300010 * 2], nxt[300010 * 2], h[300010], ec,
f[300010][30], a[300010], g[300010], fa[300010];
long long r;
struct no {
int x, y;
};
vector<no> vc[300010];
void add(int x, int y) {
v[++ec] = y;
nxt[ec] = h[x];
h[x] = ec;
}
int cp(int x... | ### Prompt
Please provide a cpp coded solution to the problem described below:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at... |
#include <bits/stdc++.h>
using namespace std;
char nc() {
static char buf[100000], *p1 = buf, *p2 = buf;
return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
? EOF
: *p1++;
}
void Read(int& x) {
char c = nc();
for (; c < '0' || c > '9'; c = nc())
;
for (x... | ### Prompt
Please provide a CPP coded solution to the problem described below:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at... |
#include <bits/stdc++.h>
std::vector<int> edges[300000];
int parent[300000];
int dp[300000];
int64_t sum = 0;
int subtree[300000];
void dfs_parent(int node, int par) {
parent[node] = par;
dp[node] = 1;
subtree[node] = 1;
for (int child : edges[node]) {
if (child == par) continue;
dfs_parent(child, node)... | ### Prompt
Your task is to create a CPP solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at lea... |
#include <bits/stdc++.h>
using std::vector;
const int N = 300005;
int n, f[N][30], g[N][30], tmp[N];
long long ans;
vector<int> e[N];
int dfs(int u, int fa) {
int sz = 0, mx = 0;
for (int v : e[u])
if (v != fa) mx = std::max(mx, dfs(v, u)), ++sz;
++mx;
f[u][1] = n;
ans += n - 1 + mx;
for (int i = 2; i <... | ### Prompt
Generate a CPP solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of its ch... |
#include <bits/stdc++.h>
using namespace std;
inline char gc() {
static char buf[100000], *p1 = buf, *p2 = buf;
return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
? EOF
: *p1++;
}
inline int read() {
int x = 0;
char ch = getchar();
bool positive = 1;
for ... | ### Prompt
Please create a solution in cpp to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k o... |
#include <bits/stdc++.h>
using namespace std;
const int maxn = 3e5;
int dp[maxn + 5][25];
vector<int> g[maxn + 5];
int a[maxn + 5], mx[maxn + 5];
int len;
int n;
long long ans = 0;
void dfs(int k, int fa) {
for (auto u : g[k]) {
if (u == fa) continue;
dfs(u, k);
}
dp[k][1] = n;
for (int dep = 2; dep <= ... | ### Prompt
Generate a CPP solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of its ch... |
#include <bits/stdc++.h>
using namespace std;
const int B = 550;
const int N = B * B;
const int C = 24;
const int D = 4;
int dp[N];
int max_depth[N];
int n;
vector<int> edges[N];
void dfs(int v, int par, int k) {
vector<int> pool;
int ma = 1;
for (int i = (int)(0); i < (int)(edges[v].size()); i++) {
int w = e... | ### Prompt
In Cpp, your task is to solve the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of its... |
#include <bits/stdc++.h>
using namespace std;
int n;
long long int Ans = 0;
int Par[300005];
int Reach[300005];
int dp[300005][25];
vector<int> adj[300005];
vector<pair<int, int> > Update[300005];
long long int dfs1(int u, int p) {
long long int mx = 0;
Par[u] = p;
for (int i = 0; i < adj[u].size(); i++) {
in... | ### Prompt
Develop a solution in cpp to the problem described below:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k o... |
#include <bits/stdc++.h>
template <class T>
std::ostream &operator<<(std::ostream &s, const std::vector<T> &v) {
for (const T &i : v) {
s << i << " ";
}
return s;
}
struct Vertex {
std::vector<Vertex *> chld;
std::vector<uint32_t> heapDepth;
uint64_t ans;
Vertex *par;
std::vector<uint32_t> *dp;
Ve... | ### Prompt
Generate a CPP solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of its ch... |
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